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[ "### ashmelev's blog\n\nBy ashmelev, history, 2 years ago, translation,", null, "991A - If at first you don't succeed...\n\nEditorial\n\nSolution1\n\nSolution2\n\n991B - Getting an A\n\nEditorial\n\nSolution\n\n991C - Candies\n\nEditorial\n\nSolution\n\n991D - Bishwock\n\nEditorial\n\nSolution1\n\nSolution2\n\n991E - Bus Number\n\nEditorial\n\nSolution\n\n991F - Concise and clear\n\nEditorial\n\nSolution", null, "Tutorial of Codeforces Round #491 (Div. 2)", null, "491,\nBy ashmelev, history, 2 years ago, translation,", null, "Hello!\n\nCodeforces Round #491 (Div.2) will start this Saturday, June 23, 18:35 (UTC+3). This round will be rated for the participants with rating lower than 2100 and other contestants can join it out of competition.\n\nThis round problems have a significant intersection with NNSU Programming Contest 2018. Please do not participate in the round if you participated in the NNSU contest or tried to upsolve the problems.\n\nDuring the round you have to help student Vasya to manage the difficulties caused by the end of the academic year. There will be 6 problems and 2 hours to solve them. If you solve all the problems in 25 minutes you will be able to watch the second half of South Korea — Mexico at the FIFA World Cup.\n\nThe scoring is unusual a bit: 500-1000-1250-1500-2000-2750\n\nGreat thanks to Mikhail MikeMirzayanov Mirzayanov for the well-known platforms; Nikolay KAN Kalinin — for the help with problems and the round coordination; Mikhail mike_live Krivonosov, Alexey Livace Ilyukhov, Nikita neckbosov Bosov, Andrew GreenGrape Rayskiy and Alexey Aleks5d Upirvitskiy — for the round testing; Arseniy arsor Sorokin — for the statements translation. And good luck to all contestants!\n\nUPD: The round is over, thank you for the participation!\n\nUPD: Congratulations to the winners!\n\nDiv. 1:\n\nDiv. 2:\n\n1. adamczh218 — solved all the problems, well done!\n2. Daniar\n3. Help.me\n4. shoemakerjo\n5. Toki_Time-Tinker\n\nUPD: The editorial is published", null, "Announcement of Codeforces Round #491 (Div. 2)", null, "491,\nBy ashmelev, 8 years ago, translation,", null, "#### Problem 175F - Gnomes of Might and Magic\n\nConstruct the graph with vertices corresponding to castles and edges to roads. Note, that a degree of each graph vertex does not exceed 4, so the amount of edges does not exceed E <= 2 * N.\n\nIn order to solve the problem let's find out how we can handle each query with time O(_log_(_N_)). Consider the query to find amount of gnomes destroyed by the Mission of Death. Assume that the edge weight is the amount of gnomes on the appropriate road. We must find the shortest path between two vertices and among them the path with the smallest amount of edges. So, the way is described by two numbers — G and R — amount of gnomes and edges (for now we do not consider lexicographical minimality). One edge between vertex u and v is described by (_C_(_u_,_v_), 1), where C(_u_,_v_) — amount of gnomes on the edge (_u_,_v_). Consider two vertices s and t. We want to find the path between them. The shortest path between them can be one of the following :\n\n• the path along the Evil Shortcut, if both vertices are on the same Shortcut\n• the path s -> g1 -> g2 -> t, where g1 и g2 — vertices on the Good Path and g1 and s are on the same Shortcut, g2 and t are on the same Shortcut (some of these 4 vertices can be identical).\n\nSo, it is necessary to determine quickly the shortest path between vertices e1 and e2 along the Evil Shortcut and between vertices g1 and g2 along the Good Path. In order to find distance between vertices along the Evil Shortcut construct for each Evil Shortcut the segment tree etree[i], which can be used to update and get the sum on the segment. For each vertex let's store the number of Evil Shortcut it belongs to and it index number on that Shortcut (it is necessary to consider cases with vertices of the Good Path, because they belong to two Shortcuts). So, the distance between two vertices of one Shortcut is the sum of edges weights of the approproate segment tree with boundaries equal to indexes of considered vertices. Similarly, construct the segment tree gtree for Good Path, by cutting it in the first vertex (for paths where it was inner vertex we will do two queries). For each pair of adjacent vertices of the Good Path we will store the minimum of two distances: distance along the edge of the Good Path or the distance along the Evil Shortcut. So for each two vertices of the Good Path the corresponding query to the segment tree returns the length (a pair of numbers) of the shortest path between two arbitrary vertices. Such a data structure allows to find the shortest path between two arbitrary vertices of the Good Path or one Evil Shortcut in time O(_log_(_N_)). It is necessary to update appropriate element of the tree with value (_0_, 1) for each pair of adjacent vertices of the Good Path initially.\n\nNow consider the query for addition of one gnome to the edge. In order to handle this query quickly it is necessary to store amount of gnomes gsum[i] along each road of the Good Path and total number of gnomes esum[i] along each Evil Shortcut. If the gnome stands on the edge i of the Good Path, then increase gsum[i] by 1. Similarly if the gnome stands on the edge i of the Evil Shortcut, then increase esum[i] by 1 and update the appropriate value in the segment tree etree[i]. It is necessary to put new shortest path between corresponding adjacent vertices into the segment tree gtree after any of these actions because it may have changed. Note, that after removal of gnomes from the edge we can do opposite operations with the same asymptotics. So, the query for updating amount of gnomes on the edge can be handled in time O(_log_(_N_)).\n\nNow let's figure out how we can find the shortest path between two arbitrary vertices s and t. Construct the vector of the Good Path vertices sg those we can reach from s along the edges of the Evil Shortcut which it belongs to (there will be one or two vertices in the sg ). Similarly, the vector tg contains the vertices of the Good Path those we can reach from t along the edges of the Evil Shortcut. Consider the union of vectors tsg = sg + tg. For each vertex u of the uninon let's find the vertex v1 which is the first one among vertices tsg on the path from u if we move along the edges of Good Path in the order of indexes increasing (i.e. in the order the vertices are given in the input data). Similarly, v2 is the first vertex on the opposite direction. Add to the adjacenty list of the vertex u vertices v1 and v2 with distances to them. Also, find edges (distances along the Evil Shortcuts) from s to the vertices of vector sg, and from vertices of the tg to the vertex t. If vertices s and t are on the same Evil Shortcut (and none of them is on the Good Path), then add one more direct edge from s to t, corresponding to the path along the Evil Shortcut. So we have constructed the graph which consists of not more than 6 vertices and no more than 13 edges. Start the Deijkstra algorithm on that graph to find shortest path from s to t. Note, that paths of the initial graph corresponding to edges of the new graph do not intersect in inner vertices (except, probably, a direct edge from s to t). In order to find lexicographically minimal path it is necessary to know the first inner vertex on the path corresponding to one of the edges. So we can, for example, add one more option to each path of the new graph — the vector of indexes of first vertices of the initial graph when moving along edges of a new graph. In this case 3 options (amount of gnomes, amount of edges and the vector of indexes) uniquely determines the optimal path of the Death Mission.\n\nNow it is necessary to print the answer and find out how to handle gnome destruction on the path. Restore vertices on the optimal path. The edge between each pair of adjacent vertices is the path along the Evil Shortcut or the path between two vertices of the Good Path. In order to find gnomes on the Evil Shortcut consider the multiset eset[i] containing indexes of edges with gnomes of that Shortcut for each Evil Shortcut. Then after adding a gnome to the edge it is necessary to add the gnome index to the multiset of the Shortcut. Now all destructions of gnomes from the Evil Shortcut i, between edges l and r, can be handled by searching of iterator of the first gnome, using eset[i]._lower_bound_(_l_) and iteration, saving obtained indexes of gnomes into the list, until the value will not exceed r. After that it is necessary to remove all gnomes in the list using the algorithm described above (similarly to addition). In order to handle paths between two vertices of the Good Path, consider set gset with indexes of vertices of the Good Path that there is no path without gnomes between two adjacent vertices corresponding to indexes. After each set updating (adding/removal of the gnome) it is necessary to add the check: if the minimum distance (corresponding to the value in the segment tree gtree) is 0 (by the amount of gnomes), then it is necessary to remove appropriate index in the gset, in the opposite case it is necessary to add the index. After invocation of similar procedure (invoke gset._lower_bound_ and iterate required amount of times) we will get the list of vertices indexes going through each one we will have to destroy at least one gnome to the path to the next vertex of the Good Path (obvious, that pairs of vertices between those exists a path without gnomes should not be considered because the Death Mission will go along it without destruction of gnomes). Consider index i from this list. If for the correspondent vertex the optimal path to the next vertex of the Good Path is on the Evil Shortcut, i.e. esum[i] < gsum[i], then it is necessary to destroy all gnomes from that Shortcut using the algorithm described above. In the opposite case it is necessary to set gsum[i] = 0 and update corresponding value in the segment tree of the Shortcut gtree.\n\nSo one gnome can be removed from the path of Death Mission in O(_log_(_N_)). The amount of gnome removals does not exceed amount of added gnomes (i.e. it is not more than Q), so the complexity of one Death Mission handling is also O(_log_(_N_)). The total complexity is O(_log_(_N_) * Q).", null, "Tutorial of Codeforces Round #115", null, "By ashmelev, 8 years ago, translation,", null, "#### Problem 175A - Robot Bicorn Attack\n\nGo over all possible partitions of the given string into 3 substrings (for example, go over a pair of indexes — the ends of the first and the second substrings). If all three substrings of the partition satisfy constraints (there are no leading zeroes and the corresponding number does not exceed 1000000), then update the current answer with the sum of obtained numbers (if it is necessary). The total amount of partitions is O(N^2), where N is the length of the string. The check of one partition takes O(N).\n\n#### Problem 175B - Plane of Tanks: Pro\n\nThe solution is to simulate actions described in the statement. Find the best result for each player and count total amount of players N. Then find a number of players C for each player , those results are not better than best result of the considering player (this can be done by going over all players). Then it is necessary to determine players category using numbers C and N. In order to avoid rounding error use the following approach:\n\n• if C * 100 >= N * 99, then the player belongs to category \"pro\"\n• if C * 100 >= N * 90, then the player is \"hardcore\"\n• if C * 100 >= N * 80, then the player is \"average\"\n• if C * 100 >= N * 50, then the player is \"random\"\n\nIn other case the player is \"noob\".\n\n#### Problem 175C - Geometry Horse\n\nObvious, that figures should be destroyed in the cost increasing order. Sort figures type in ascending order of their costs. Consider two pointers – position i in the array P (current factor) and position j in the array of figures type. Consider the current answer and the number of figures G, those need to be destroyed to move to the next value of factor. If the number of figures F of the current type does not exceed G, then add F * (i + 1) * Сj to the answer and reduce G by F and increase pointer j by 1. In other case add G * (i + 1) * Cj to the answer, reduce F by G, increase i by 1 and set G = Pi – P(i-1). Continue iteration until all figure types are considered.\n\n#### Problem 175D - Plane of Tanks: Duel\n\nFirst consider case when at least one probability of not-piercing is equal to 100%. If Vasya does not pierce the enemy tank with probability 100%, then the answer is 0. If the enemy tank cannot pierce Vasya's tank with probability 100% then the answer is 1. Then consider that the probability of shot does not pierce tank does not exceed 99%. It can be checked that the probability that tank will stay alive after D = 5000 shots is less than 10^-6 (for any damage value, probability of not-piercing and amount of hit points). For each tank calculate the following table dp[i][j] — probability that tank will have j hit points after i shots to him. dp[hp] = 1, where hp – is the initial amount of hit points. In order to calculate the line dp[i+1] it is necessary to go over all possible damages for each value of j in the line dp[i], which shot (i+1)-th damages (considering cases when the shot does not pierce the tank armour) and update appropriate values of the line (i+1):\n\ndp[i + 1][max(0, j – x)] += dp[i][j] * (1 – p) / (r – l + 1)\n\n\nwhere p – is the probability of not-piercing, x – possible shot damage. Let's dpv – calculcated table for Vasya's tank and dpe is the table for enemy's tank. Now it is necessary to find probability that enemy's tank will be destroyed after i shots of Vasya's tank: pk[i] = dpe[i] – dpe[i-1]. Vasya wins the following way: Vasya fired (K — 1) shots and do not destroy enemy tank and is still alive also. After that Vasya fires K-th shot and destroy the enemy. Go over K and calculate the probability of Vasya's victory with the K-th shot. In order to do this find how many shots T can fire the enemy before Vasya makes K-th shot (here is the only place in the solution where we must use the gun recharge speed):\n\nT = ((K – 1) * dtv + dte - 1) / dte\n\n\nwhere dtv is the time required to recharge the gun, dte is the time of enemy gun recharge. Then the probability of victory is (1 – dpv[T]) * pk[K]. The answer for the problem is the sum all these probabilities for each K from 1 to D. The algorithmic complexity of the algorithm is O(D * hp * (r – l)).\n\n#### Задача 175E - Power Defence\n\nIf we reflect the position of towers alogn the line OX then the interval where towers will affect the Villain will not change. So, we can consider that towers can be built in the points (_x_, 1), not more than two towers in the same point. If there exists point X that there are towers to the left and to the right from X but in the point X there is no tower, then abscissas of all towers \"to the right\" can be reduced by 1 and the answer will not become worse. The same way, we can prove that there are no adjacent points with exactly one tower in each one. Now it is easy to check that in order to construct the optimal solution it is enough to check 13 successive points.\n\nGo over positions of freezing towers. In the worst case there are approximately 200000 cases to put freezing towers into 13 points. Consider the case when we fixed positions of several freezing towers. Let's calculate how much damage can hit fire-tower or electric-tower in the point for each empty points (points where we can put two towers are splitted into two) and save numbers into the array d (_d_[k].f and d[k].e – damage by fire and electricity in the point k correspondingly). Sort the array d in the order of decreasing of the value d[k].f. Then optimal position of the rest towers can be found using dynamic programming. Designate dp[i][j] – the maximum possible damage, which can be hitted if we have i fire-towers and j electric-towers. Designate p — the amount of towers have been set already: p = cf – i + ce – j. If i = 0 then we used first p values of array d and the answer is the sum of j maximum elemets of d[k].e, starting from index p. Otherwise we put one fire-tower or one electric tower in the position p. It is necessary to put the tower into position p because in the opposite case the d[p].f will decrease. Then:\n\ndp[i][j] += max(dp[i - 1][j] + d[p].f, d[i][j – 1] + d[p].e)\n\n\nThe answer is the value dp[cf][ce], which is calculated in O(cf * ce * log(ce))\n\nComment1: exhaustive search can be reduced in 2 times because any two symmetric towers arrangements are equivalent and have the same answer. However, this optimization is not required with a given constraints.\n\nComment2: the formula of Villain speed decrease 1 / (K + 1) allows to calculate the tower damage for a case when all freezing towers are fixed easily. Freezing towers can be taken into account separately.", null, "Tutorial of Codeforces Round #115", null, "By ashmelev, 9 years ago, translation,", null, "Problem D.\n\nFirst, let’s divide graph to connected components (provinces). Next, we consider only new graph on these components – for each province we assign a vertex of the graph. Let the total number of provinces is n. Initially the graph is empty since there are no roads between different provinces. Also for each province we have a limit of the number of tunnels that can be constructed from this province: ki = min (k, ci) where ci – the number of cities which was originally contained in the province (component) i. The resulting provinces graph should be connected after building tunnels and roads. When k = 1 we have to build at least n - 2 roads, otherwise the graph will have at least 3 components and at least 2 tunnels should be constructed from one of them which is prohibited.\n\nFurther, we assume that k > = 2. Let’s calculate the largest number of tunnels we can build. Let s – the sum of all numbers ki. Obviously we cannot build more than s / 2 tunnels since each tunnel connects exactly two provinces. The following statement is true: we can construct s / 2 (rounded to the lower side) of tunnels or to make a graph connected by constructing n - 1 tunnels (if s / 2 > = n - 1). Let’s consider vertices which have ki > 1. We may connect these vertices into a chain using tunnels. After that let’s start to connecting vertices with ki = 1to the chain (using tunnels too) while it is possible. Suppose we had less than s / 2 tunnels built and we are unable to build one more tunnel. It means that we have exactly one vertex j with degree no more than kj - 2. Thus kj > 1 and this vertex is included into the chain and all the vertices with ki = 1 are attached to this chain too (otherwise we could build another tunnel), so the graph is connected.\n\nIf after building the tunnels we have a connected graph then the answer is 0. Otherwise the graph consists of n - s / 2 components, that is we need to build at least n - s / 2 - 1 roads. In fact such a number of roads will be enough. Let’s draw each of n - s / 2 - 1 roads the following way. First, choose 2 different connected components in the current graph. Because we have built tunnels (and possibly roads) only between different components each of the chosen components is a tree. So these components have vertices with degree not greater than 1. Now, let’s choose one such vertex in each of the selected components and connect the components through these vertices (i.e. the vertices are merged into one keeping the edges from them). Thus we have a new vertex (province) with no more than two tunnels constructed from it, so we did not violate the terms since k >= 2. Thus we can get a connected graph by building additional n - s / 2 - 1 roads which is the answer for the problem.\n\nProblem E.\n\nWhen x = 2 then the answer is 0. Further we assume that x > 2.\n\nIn order to uniquely identify the desired number of items t we must choose a set of numbers ai so that for every y from 2 to x the representation in modes ai is unique, i.e. sets of numbers b (y, i) = y / ai (rounded up) are pairwise distinct among all y. Note that for each i function b(y, i) is monotone by y. Hence if for some i and numbers y and z (y < z) holds b(y, i) = b(z, i) then b(y, i) = b(y + 1, i) too. So to select the set of numbers ai it is sufficient to guarantee that for each y from 2 to x - 1 exists a number j such that b(y, j) < b(y + 1, j). It is easy to see that b(y, j) < b(y + 1, j) if and only if y is divisible by aj. Thus, it is necessary that for each y from 2 to x - 1 exists number ai so that y is a multiple of ai. If some number ai is equal to 1 Vasya can just see the list in this mode to find the desired number and the answer for the problem is 1. Otherwise it is necessary and enough to take all the primes pi < x in our set ai to find the number. Indeed if we will not use some prime number pi then we will be unable to distinguish numbers pi and (pi + 1) (since pi is not divisible by some of the selected numbers). On the contrary if we will use all primes less than x then any number from 2 to x - 1 will be divisible by at least one of them. Thus we need to check whether there are all prime numbers less than x among ai. Since the number of primes from 1 to x is about O(x / ln (x)) for large x all prime numbers less than x cannot be in the set of numbers ai. For example the following statement is true: if x > 20 * n then the answer is -1. This means that we can use the Sieve of Eratosthenes to find all primes less than x for x <= 20 * n and to check whether there is at least one number from them which does not occur in ai. If such number exists then the answer for the problem is -1 otherwise the answer is the number of primes less than x.\n\nProblem F.\n\nIf for some velocity v1 we were able to go from point A to point B and receive no more than k hits, then for any velocity v2 > = v1 we also will be able to go from A to B. So we can use  the binary search algorithm to find the answer.\n\nSuppose we have fixed speed of the tank v. Now we have to count how many enemy tanks will be able to shoot at our tank during the ride. Let’s consider enemy tank i located at the point P on the plane. It may aim at our tank in two ways: turn the turret at point B or rotate the turret at point A and then start turning it from point A to point B. In the first case we may just compare the time required for the tank  to move from A to B with the time required the enemy to aim the turret to point B. If the enemy tank will be able to take aim to B before we can reach this point then the enemy can make a shot. Next consider the second possible enemy strategy.  Let’s draw perpendicular PQ to the line AB. So we have divided the segment AB into 2 parts: AQ and QB (if Q does not lie on the segment AB then one of the parts will be empty and the other is a segment of AB. In this case let Q denote the end of segment AB closest to the base of the perpendicular). Let’s consider the first part of the segment - AQ (before the base of the perpendicular). It is easy to check that while the angular velocity of the turret is a constant, the linear velocity of the enemy sight along the segment AQ is monotonely decreasing. Given the fact that the speed of our tank along AB is constant we find that the difference between the coordinates of the enemy’s sight and the tank at the AQ interval is a convex function of time (second derivative is negative). Also this fact can be verified by finding the second derivative of this function explicitly. Thus we can use the ternary search algorithm for finding the minimum of this function on a time interval corresponding to time when our tank rides at segment AQ. When the minimum value of this function is negative the enemy is able to take aim at our tank and perform a shoot. Otherwise, the tank will ride ahead the enemy sight on the whole interval AQ. (Using similar statements we can find for example the minimum value of the time difference between reaching a point D of the interval AQ by the enemy sight and by our tank). It is possible to avoid the ternary search by finding a moment when the speed of the sight is equal to the speed of our tank and check who is closer to point B at this moment. But in this case we are to carefully handle the cases where one velocity is always greater than the other on the whole interval.\n\nNow let’s consider the second part of the segment - QB (after the base of the perpendicular). If the enemy is unable to shoot in our tank at the first part of the segment (AQ) then at the time of sighting the enemy on point Q our tank will be located closer to point B than the sight. Similarly the first part of segment AB, we can prove that the linear speed of sight along QB is\nmonotonely increasing. So if at some point C of segment QB the sight of the enemy tank has caught our tank then speed of the sight should be higher than speed of our tank at that moment (otherwise the enemy would not be able to catch the tank). Due to the monotonicity of the sight velocity on the remaining segment CB the sight will be faster than the tank and the sight will reach point B before our tank. Accordingly, if the enemy's sight has reached point B after our tank then the tank was ahead the sight on the whole interval QB too. Thus, to determine whether the enemy can shoot it is sufficient to check only point B.\n\nPerforming these calculations for each of the n enemies we get the number of hits on our tank and comparing this value with the number k we go to the desired branch of the binary search.", null, "Tutorial of Codeforces Beta Round #66", null, "By ashmelev, 9 years ago, translation,", null, "Problem A.\n\nLet’s call the monster dimension sizes as x1, x2, x3.\n\n1. O(min(k, x1 + x2 + x3)) solution\n\nWe can make at most (x1 – 1) + (x2 – 1) + (x3 – 1) cuttings, so we may assume that k <= x1 + x2 + x3 – 3. For each of the three dimensions we will store an integer ai – number of cuttings performed through the corresponding dimension. Let’s perform the following actions k times: consider all numbers ai which we may increase (ai <  xi - 1) , call these dimensions “uncompleted”. Select the minimum number aj among all uncompleted ai and perform a cut through the corresponding dimension (aj will be increased by 1 as the result). Now let’s consider the resulting set after k actions: {a1, a2, a3}. Using the described algorithm we grant that the maximum element of this set is as little as possible and the minimum element is as big as possible. Because the sum a1 + a2 + a3 = k is fixed, we get the maximum product (a1 + 1) * (a2 + 1) * (a3 + 1) which is the answer for the problem.\n\n2. O(1) solution\n\nInstead of simulation all the k actions we may quickly determine values of numbers ai after using the algorithm described above.\n\nLet x – the smallest value among (x– 1). When x * 3 >= k on each of the algorithm iterations all three dimensions will be uncompleted. It means that during the first (k / 3) * 3 steps each of numbers ai will be increased by (k / 3). Then 0, 1 or 2 numbers ai will be increased by 1 subject to value of k % 3. So, we have found values ai.\n\nOtherwise (x * 3 < k) during the first x * 3 steps each of ai will be increased by x. After that we will have at most two uncompleted dimensions which can be processed a similar way (we should choose the minimum value x among the remaining uncompleted dimensions).\n\nProblem B.\n\nWe may assume that we have exactly n awarded places but some of them give 0 points.\n\nLet’s sort places by amount of points (bi) and racers by amount of currently gained points (ai). First let’s find the best place Vasya can reach. In the best case he will get b0 points (where b0 is the greatest number among bi). So, let the total Vasya’s point is v. Now we should to distribute other prize points to racers so that the number of racers better than Vasya will be minimal. For doing that we are to give maximum number of prizes so that corresponding racers will have less than v points. Note that if we can give k prizes keeping this property, than we can give k “cheapest” prizes too. The following statement is also true: if we can get a prize with t points to racer i and to racer jai > aj, then it is better to give this prize to racer i. Formally: if there exists a way to give k prizes where this prize will get racer j, than there exists a way to give k prizes where this prize will get racer i. It can be proven the following way. Consider a way to give k prizes where racer j got prize with t points, racer i – s points or didn’t get prize at all. In the first case we can swap prizes for racers i and j: ai > aj and ai + s < v (since racer i have got the prize), so aj + s < v, and ai + t < v i.e. this change is acceptable. In the second case we can just give the prize t to racer i instead of racer j. In the both cases we get a way to give k prizes where racer i receive prize t.\n\nNow including this statement we may give prizes to racers using the following greedy algorithm. Let’s begin to give prizes starting from the cheapest one and check the racers starting from the best one (of course excluding Vasya and the best prize). For each prize i we will go through the racers until applicable racer (j) found: bi + aj < v. If no such racers remain we are unable to give more prizes without violating the rule (racers should have less than v points). In this case we should stop the algorithm and the answer is n – k where k is the number of prizes we have given. If we have found such racer j we can give him prize bi and go to the next prize. Complexity of this step is O(n).\n\nSimilarly we can find the worst place for Vasya. For doing that we should give him the cheapest prize (note it may have positive points though). After that we should distribute the prizes iterating over prizes from the largest to cheapest and check racers from the worst to the best one trying to make sum of racer’s points more than v.\n\nTotal complexity of the described algorithm is O(n * log(n)) because we have to sort prizes and racers.\n\nProblem C.\n\nIt is easy to see that if we put some symbol c at position p of the string s it will not affect symbols at positions (p+2) and greater. So we have a standard DP problem. State of the dynamic is described by three parameters: p – the number of already processed symbols (or the index of currently processed symbol of the string), c – the previous symbol, t – the number of allowed symbol changes. To calculate the answer for a state we should choose the best value among all symbols for current position (when t > 0) or just go to the index (p + 1) with current symbol s[p]. Thus we get the followings formulas:\n\nd[n][*][*] = 0\n\nd[p][c][t] = d[p + 1][s[p]][t] + bonus[c][s[p]] when t = 0\n\nd[p][c][t] = max(d[p + 1][c’][t – (c’ <> s[p])] + bonus[c][c’])\n\nwhere n  is the length of string s.\n\nComputation complexity of the algorithm is O(n * k * h^2), where h is the alphabet size (h = 26 for current problem).", null, "Tutorial of Codeforces Beta Round #66", null, "By ashmelev, 9 years ago, translation,", null, "Good afternoon!\n\nAuthors of the today's contest are Evgeny Lazarev (Nizhny Novgorod STU) and me (Alexey Shmelev, Nizhny Novgorod SU)\n\nToday you will help a boy Vasya find himself in the world of computer games.\n\nPlease note, that the round will be held in unusual format - 6 problems with costs 500, 1000, 1000, 1500, 1500 and 2000 points.\n\n### The round duration is increased to 2 hours and 30 minutes.\n\nWe say thanks to Marya Belova for statements translations, Artem Rakhov and Alexander Kouprin for help in contest preparation, writing alternative solutions and preparing of intricate tests.\n\nGood luck and successful submission!\n\nUPD: We apologize for inaccuracy in problem E and incorrect answers to several contestant clarifications.", null, "Announcement of Codeforces Beta Round #66", null, "" ]

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[ "# How do you evaluate the integral int8x+3 dx?\n\nSep 2, 2014\n\nWhen taking integrals, you will normally solve them one term at a time. You will do the inverse of the power rule so the answer would be:\n\n$F \\left(x\\right) = 4 {x}^{2} + 3 x + C$\n\nIntegrals are the inverse of derivatives so you follow the rules in reverse. The $8 x$ can be written as $8 {x}^{1}$. To take the derivative of this you would multiply the coefficient by one then subtract one from the exponent, so if:\n\n$f \\left(x\\right) = {x}^{n}$ then $f ' \\left(x\\right) = n {x}^{n - 1}$\n\nTo reverse the power rule, you will first add one to the exponent then divide the whole term by the new term:\n\n$F \\left(x\\right) = \\frac{{x}^{n + 1}}{n + 1}$\n\nBoth terms in this problem can be solved with the power rule.\n\nDue to this being a indefinite integral, not having any bounds, you will have to put $+ C$ do to the possibility of a constant being dropped when a derivative was taken. In other words:\n\n$f \\left(x\\right) = 6 {x}^{3} + 5$ and $g \\left(x\\right) = 6 {x}^{3} + 25$\n\nwould have the same derivative because the constant becomes zero and the additive identity property states that anything added to zero is unchanged." ]

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[ "# TOM LEE-GOUGH\n\naccountancy services\n\nDarlington based, general accountancy practice. Get in contact to find out how we can help you.\n\n## Python Basics\n\nIn some future posts, I will make worked examples of API interactions and some automations. My preferred programming language is Python. I cannot remember why I chose to learn Python, but it was pretty strightforward to pick up. I initially learned about coding when recording macros within Excel, then trying to improve them by writing the VBA (Visual Basic). In this post, I will cover some of the basic concepts that I will use in later posts.\n\nI won't cover installing Python or running Python on your system. There are lots of tutorials and guides out there and that's part of the fun of learning a new programming language. For reference, I mostly use Debian (operating system) and Python 3.\n\n## General\n\nPython uses whitespace to separate out code. Broadly, one piece of logic per line is best. It makes it easier to understand.\n\nPython also uses indents to group code together:\n\n``````def some_function(i):\ndoes something to i\nreturns result\n``````\n``````def another_function(a):\nthis function is different to above\nreturn result\n``````\n\n## Variables\n\nAny one that has taken GCSE maths would have come across variables. It's not a popular subject, but algebra is all variables. A variable is a container. In algebra it is a number, in programming it can be anything. For example, the formula Y = X2 + 4 both Y and X are variables. To plot a graph of Y, you would plug values into X and solve for Y.\n\nHere's how you can assign a variable in Python:\n\n``````# assign a string to a variable called \"name\"\nname = 'guitar'\n# assign a number to a variable\nstrings = 6\n# assign a list to a variable\ntypes = ['electric', 'acoustic']\n``````\n\n## Arrays\n\nAn array is a collection of data. Python has several types. I mainly use lists and dicts.\n\n### Lists\n\nLists are pretty simple. A list is a series of items that are separated by commas within square brackets. They can contain any data types, and can be mixed.\n\n``````# list of strings\n['hello', 'this', 'is', 'a', 'string']\n# list of numbers\n[1, 2, 3, 4.5, 6.7 ]\n# list of variables\n[first, second, third]\n# mix list\n['string', string, '1', 1]\n``````\n\n### Dicts / JSON\n\nA dict (or dictionary) is a more comprevensive data item. The syntax is reasonably easy to understand, and features heavily when dealing with API. Javascript has a similar array type called JSON (JavaScript Object Notation). Many API will return a JSON object for you to use.\n\nAgain, dicts can hold all different types of data. However, the most interesting part of a dict is that each data item has a key. The key refers to the data stored. A dict is how one might intuitively think of a database record. Of course, you can have lists of dicts.\n\nNot all dicts in a list have to have the same keys, but it is useful if they do.\n\n``````books = [\n{\n'title': '1984',\n'author': 'George Orwell'\n},\n{\n'title': 'Brave New World',\n'author': 'Aldous Huxley'\n}\n]\n``````\n\n## Functions\n\nA function is an action that you regularly perform within a script. This saves you time in your code, as you only have to write the thing once. There are lots of built in functions in Python, you will definitely write your own.\n\nUsing the formula from above, we could write it as a function:\n\n``````def solve_y(x):\ny = (x * x) + 4\nreturn y\n``````\n\nFunctions start def then the name of the function followed by brackets and a colon. In the example we can pass a number to the function and the function performs the action. The \"return\" part means that the function will give the result of the function back. This can be assigned to a variable or something else.\n\nHere are some results for our function:\n\n``````>>> solve_y(4)\n20\n>>> solve_y(2)\n8\n``````\n\nOur solve_y function can be writted using the in-built power function. It raises any number to the power of another number:\n\n``````def solve_y(x):\ny = pow(x, 2) + 4\nreturn y\n``````\n\nIn our particular case, it doesn't really make it a lot simpler. However, it a bit clearer from a BODMAS point of view.\n\n### For Loop\n\nFor loops are used lots. A for loop iterates over an item, assigning each of results to a variable. It's kind of hard for non-programmers to get your head around, however, it very powerful. Here are some examples:\n\n``````>>> for i in range(5):\n... i\n...\n0\n1\n2\n3\n4\n``````\n\nrange(x) is a function that returns numbers incremented by 1. The range function starts at 0 by default and will count the number of items. Our for loop will assign the result of the range function to the variable i. This is then displayed on the screen." ]

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[ "# Inverse matrix (order n) Calculator\n\n## Calculates the inverse matrix of a square matrix of order n.\n\n (enter a data after click each cell in matrix) matrix A {aij} 6digit10digit14digit18digit22digit26digit30digit34digit38digit42digit46digit50digit Inverse matrix A-1\nInverse matrix (order n)\n [1-1] /1 Disp-Num5103050100200", null, "", null, "2014/03/02 18:02   Under 20 years old / High-school/ University/ Grad student / A little /\nPurpose of use\nfor a university homework, to test/see if a n-order diagonal matrix has a pattern of inversion", null, "", null, "Thank you for your questionnaire.\n\nSending completion", null, "To improve this 'Inverse matrix (order n) Calculator', please fill in questionnaire.\nAge\n\nOccupation\n\nUseful?\n\nPurpose of use?", null, "" ]

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[ "Feet To Meters\n\n# 60800 ft to m60800 Foot to Meters\n\nft\n=\nm\n\n## How to convert 60800 foot to meters?\n\n 60800 ft * 0.3048 m = 18531.84 m 1 ft\nA common question is How many foot in 60800 meter? And the answer is 199475.065617 ft in 60800 m. Likewise the question how many meter in 60800 foot has the answer of 18531.84 m in 60800 ft.\n\n## How much are 60800 feet in meters?\n\n60800 feet equal 18531.84 meters (60800ft = 18531.84m). Converting 60800 ft to m is easy. Simply use our calculator above, or apply the formula to change the length 60800 ft to m.\n\n## Convert 60800 ft to common lengths\n\nUnitLengths\nNanometer1.853184e+13 nm\nMicrometer18531840000.0 µm\nMillimeter18531840.0 mm\nCentimeter1853184.0 cm\nInch729600.0 in\nFoot60800.0 ft\nYard20266.6666666 yd\nMeter18531.84 m\nKilometer18.53184 km\nMile11.5151515152 mi\nNautical mile10.0063930885 nmi\n\n## What is 60800 feet in m?\n\nTo convert 60800 ft to m multiply the length in feet by 0.3048. The 60800 ft in m formula is [m] = 60800 * 0.3048. Thus, for 60800 feet in meter we get 18531.84 m.\n\n## 60800 Foot Conversion Table", null, "## Alternative spelling\n\n60800 ft to Meters, 60800 ft in Meters, 60800 ft in Meter, 60800 Foot in m, 60800 ft to m, 60800 ft in m, 60800 Feet to m, 60800 Feet in m, 60800 Foot to Meters, 60800 Foot in Meters, 60800 Feet to Meters, 60800 Feet in Meters," ]

[ null, "https://feet-to-meters.appspot.com/image/60800.png", null ]

[ "# probability of two events occurring in same time\n\nThis is very practical question, I am not fully aware in mathematical sense what my variables are.\n\nApologies if question should be presented as mathematical, but if I were able to do that, I think I would be able to answer it.\n\nConsider I have two computer networks, let's call them A and O networks. I will only need O network, when A network fails. Let's also assume they fail independently.\n\nI want probability for both networks being down same time in any given hour.\n\nLet's say that A network fails twice a year, for 3h each time. For simplicity let's assume O network has similar, but independent failure mode.\n\nIs that 2*3h failure probability equivalent for 1*6h? That is, is my probability of failure 6/(365*24)?\n\nAnd if so, is the probability for A and O network to fail at same time (6/(365*24))**2? That is 1 failure every 243 years?\n\nWhat about if I want probability of no failure inside specific window? In 3h window is the probability then: 1/(1-((1-((6.0/(365*24))**2))**3))/24/365 - that is, once every 81 years?\n\nAm I even looking at the problem correctly? If not, how can I analyse the availability requirements for A and O networks?\n\n• Very roughly speaking, the chance a network is down at a given hour is 6/8760 or 1/1460, so the chance both are down is roughly that squared or 1/2131600, or 1 in 2,131,600 hours, which is about once every 243 years. I would be suspicious of answers that are too far away from this estimate. – user1566 Sep 20 '16 at 16:55\n• Problem is, I know I don't understand the problem. And now I have 3 distinct solutions, yours is same as mine, which in my case lends less credibility as I don't really trust myself. – ytti Sep 22 '16 at 10:02\n• I wrote simulation - p.ip.fi/HWVk - unsure if it is fair. I made failure count reset every year, and choose failure from normal distribution, and choose failure length at start of failure from normal distribution. With the parameters in the code now, I'm seeing sub-hour outages maybe every 250 years or so (the parameters are more aggressive than in my question) => p.ip.fi/n94f – ytti Sep 23 '16 at 12:35\n\nFor a rough estimate: There are 8760 hours in a year, and thus 2920 3-hour blocks. Assuming that the failures always occur in these blocks (and excluding cases where A might be dead from 1 to 4 PM and O dead from 2 to 5, for example), there's a 2/2920 = 0.07% probability of O at any given block of time. Since you claim that the failures are independent, this means that given that A is down, there is a 0.07% chance that O is down as well. Combining these, there is a $(2/2920)^2$ probability that both will be down at any given block; in the course of a year with 2920 block there are thus an average of 4/2920 = 0.0014 blocks where both A and B are down, and you really only expect it to occur once every thousand years or so.\n\nFor a more exact answer: The exact statistical model of your process is a Poisson process with dead time. There is a given rate at which failure occur, but once they occur there is a period of time (3 hours) where nothing else happens. There are some analytic ways of modeling this (see here for example), but the rough estimate above is going to be close, and would likely be easier to write a small Monte Carlo simulation in Python than it would be to apply the analytic formalism (though for such small rates that could take a while).\n\n• \"There are 8760 hours in a year, and thus 2920 3-hour blocks\" - that is not true. There is 8760 - 2 such blocks. You ignore the fact that blocks can cross. – Tim Sep 20 '16 at 12:30\n• @Tim I purposely ignored crossing blocks and stated so in the answer. And you're still assuming that 1 hour is an indivisible unit. If you include the fact that blocks can cross, you must count an infinite number of blocks: 1:00PM to 3:00PM, 1:01PM to 3:01PM, 1:15PM to 3:15PM, 1:15PM + 7 microseconds to 3:15PM + 7 microseconds, etc. Considering overlaps in blocks requires a difficult Poisson process + deadtime analysis; hence the approximation that I stated for my \"rough estimate.\" – jwimberley Sep 20 '16 at 12:58\n\nThe answer to your question depeneds on the distribution of failure times. You say that $A$ fails twice a year. Is it exactly twice a year, or on average? Is it more likely to fail at some specific part of the year?\n\nI will give a simple example for illustration.\n\nLet us assume that $T_A$ is the failure time of $A$, and that $T_A$ is distributed uniformly over $[0,1]$ (this interval will represent the year). This means, that the probability of $T_A$ being in some specific time interval is the proportion of this time interval out of the entire interval. For example, the probability of $T_A\\in(1/2,3/4)$ is 1/4.\n\nLet $T_O$ be defined similarly with regards to O, and suppose that $T_A$ and $T_O$ are independent. We assume further that the down time is (deterministiclly) a proportion of $d$ out of the entire time interval. To calculate the probability that both of them are down at the same time, you need to use $T_A$ and $T_O$'s mutual density, which (since they are independent) is uniform over the unit square $B=[0,1]\\times [0,1]\\in \\mathbb{R}^2$.\n\nNow, the probability that you want, is the total area of the set $\\{(x,y)\\in B:\\ |x-y|<d\\}$, divided by the total area of $B$ (which is 1).\n\nThere is a lot of information here that you might not be familiar with, and is taught on probability introductory course.\n\nAnyway, hope this helps\n\n• On average. I don't actually have real data of historic failures, but I think I would know how to model probability of failure in future if I had historic failure rates. Thanks, I will indeed need time and research to digest your answer. – ytti Sep 20 '16 at 13:33\n• I think that a 'Poisson point process' might be a good way to model the consecutive failure times. To read about this process, I recommend chapter 2 in \"Daley, Vere-Jones: an introduction to the theory of point processes\" (again, requires basic background in probability) – Snoop Catt Sep 20 '16 at 14:02\n\nSummary: @jwimberly is correct\n\nLet's take the problem very literally and assume that both networks MUST fail EXACTLY 2 times per year. Let's also briefly assume that the networks fail at the start of a given hour.\n\nThere are 8760 ways to choose \"A\" network's first failure hour and 8757 ways to choose \"A\" network's second failure (since it can't overlap the first failure). Since we can choose any pair of failure times in either order, we divide by two. Thus, there are 8760*8757/2 ways to choose the two failure times for network \"A\".\n\nThere also 8760*8757/2 ways to choose the two failure times for network \"O\". Of these ways, how many don't overlap the two \"A\" failures? There's 8754 start hours for the first failure (avoiding the 6 failure hours taken up by \"A\"), and 8751 start hours for the second failure (avoiding the 6 \"A\" failure hours and the 3 \"O\" first failure hours). Again, we can choose these in either order, for a total of 8754*8751/2\n\nDividing these, we get (8754*8751/2)/(8760*8757/2) = 4255903/4261740 as the chance of non-failure in a given year, and thus 1-4255903/4261740 = 5837/4261740 chance of failure, which is about 0.14%, which agrees with @jwimberley's computation. This is roughly 1/730, so you would expect a failure once every 730.125 years or so.\n\nAbove, we assumed the failures started exactly on the hour. Now, let's suppose the failures start any second of the year. There are 8760*3600 seconds in a year, and thus that many possible start times for one of \"A\"'s failures. Since 3 hours is 10800 seconds, there are 8760*3600-10800 seconds for \"A\"'s other failure. Again by symmetry, there are (8760*3600)*(8760*3600-10800)/2 total pairs of start seconds for \"A\"'s failure.\n\nThe same is true for \"O\"'s failure, so let's compute the number of ways \"O\" can fail without overlapping \"A\"'s failure. This is (8760*3600-10800*2) ways for one failure (excluding 6 hours) and (8760*3600-10800*3) for the other (excluding \"A\"'s six hours and \"O\"'s failure). Again by symmetry, there are (8760*3600-10800*2)*(8760*3600-10800*3)/2 total pairs of seconds in which \"O\"'s failures can start without overlapping \"A\"'s.\n\nTaking the ratio ((8760*3600-10800*2)*(8760*3600-10800*3)/2)/((8760*3600)*(8760*3600-10800)/2) we have 4255903/4261740 times when that doesn't happen, exactly the same as we got before. Of course, we could do the same computation for microseconds, but the above demonstrates we'd end up getting the same answer each time.\n\nUltimately, we're doing what @jwimberley did, but using discrete/combinatoric methods.\n\nMy comment of once every 243 years was wrong for the following reason: if you choose a random point in the year, the chances that both networks are down is indeed 1 in 2,131,600. However, there's no guarantee the double outage would last the entire hour. In fact, my guess is almost exactly 3 times too small, suggesting that, if there is a double outage at a given point in time, it will resolve itself within 1/3 hour or 20 minutes (however, I haven't done the math on this last part)\n\nMinutiae: in the Gregorian calendar, there are 8765.82 hours in a year, excluding leap seconds." ]

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[ "MUTATION AND MIGRATION\n\nWe have learned how selection can change the frequencies of alleles and genotypes in populations. Selection typically eliminates variation from within populations. (The general exception to this claim is with the class selection models we have called \"balancing\" selection where alleles are maintained in the population by overdominance, habitat-specific selection, or frequency dependent selection). If selection removes variation, soon there will be no more variation for selection to act on, and evolution will grind to a halt, right? This would be true if it were not for the reality of Mutation which will restore genetic variation eliminated by selection. Thus, mutations are the fundamental raw material of evolution.\n\nWe will be gin by considering what mutation will do as an evolutionary force acting by itself. Simply, mutation will change allele frequencies, and hence, genotype frequencies. Lets consider a \"fight\" between forward and backward mutation. Forward mutation changes the A allele to the a allele at a rate (u); backward mutation changes a to A at a rate (v). We can express the frequency (p) of the A allele in the next generation (pt+1) in terms of these opposing forward and reverse mutations, much like forward and reverse chemical equations: (pt+1) = pt(1-u) + qt(v). The first part on the right is accounts for alleles not mutated (1-u), and the second part accounts for the increase in p due to mutation from a to A (the frequency of a times the mutation rate to A). We can also describe the change in allele frequency between generations (Dp) as: Dp = (pt+1) - (pt). This is useful because it lets us calculate a theoretical equilibrium frequency which is defined as the point at which there is no more change in allele frequencies, i.e. when Dp = 0 which is when (pt+1) = (pt); from above: pt(1-u) + (1-p)t(v) = pt [remember, q=(1-p)]. Now solve for p and convince yourself that the equilibrium frequency = p = v/(u+v). Similarly the equilibrium frequency of q = u/(u+v).\n\nMUTATION AND SELECTION BALANCE\n\nIn the real world we will generally not find specific evolutionary forces acting alone; there will always be some other force that might counteract a specific force of interest. Our ability to detect these opposing evolutionary forces depends, of course, on the relative strengths of the two (or more) forces. However, it is instructive to examine the conditions where evolutionary forces oppose one another to give us a feel for the complexity of evolutionary processes. Here we will consider a simple case where mutation introduces a deleterious allele into the population and selection tries to eliminate it.\n\nAs above we define the mutation rate (u) as the mutation rate to the \"a\" allele. This will tend to increase the frequency of a (i.e., q will increase). In fact, q increases at a rate of u(1-q); remember, (1-q) = p, or the frequency of the A allele. This mutation pressure will increase the number of alleles which selection can act against.\n\nTo select against the a allele, we first will assume complete dominance, i.e., that the deleterious effects of the a allele are only observed in the aa homozygote. Under these conditions, the frequency of \"a\" (q) decreases by selection at a rate of -sq2(1-q), where s is the selection coefficient. We won't derive this for you, but note that the amount of change generated by this selection is a function of the frequency of the aa homozygote (q2) and the frequency of the A allele (1-q). In other words, the amount of change is proportional to the amount of genetic variation in the population, as we showed last lecture.\n\nIf we put these terms for mutation and selection together, the amount of change in the a allele is :\n\nDq = u(1-q) - sq2(1-q)\n\nNow, if the \"fight\" between selection and mutation is a \"draw\", we have a condition where there is no change in the frequency of the a allele since mutation is increasing q just as fast as selection is reducing it. Under these conditions, Dq = 0, and we say that the equilibrium allele frequency, q-hat, has been reached (in formal notation q-hat is q with a circumflex over it). We simply rearrange the above formula so that is becomes : u(1-q) = sq2(1-q). We solve this for q to give the equilibrium allele frequency , q-hat: q = sqrt(u/s) (sqrt stands for square root).\n\nMost mutation rates are fairly small numbers (about 10-6), so this equation suggests that deleterious alleles will be maintained in mutation selection balance at fairly low frequencies. However, this statement is exactly what this mode is intended to illustrate: we cannot say what that frequency is until we know BOTH mutation rate and selection coefficient. However, we will refer back to mutation selection balance several times during the course, so it is essential that you have a feel for dynamics of this evolutionary interaction.\n\nMIGRATION\n\nIn population genetics, the term \"migration\" is really meant to describe Gene flow, defined as the movement of alleles from one area (deme, population, region) to another. Gene flow assumes some form of dispersal or migration (wind pollination, seed dispersal, birds flying, etc.) but dispersal is not gene flow (genes must be transferred, not just their carriers)\n\nWe can describe gene flow (migration) in a manner similar to mutation. Consider two populations, x and y with frequencies of the A allele of px and py . Now consider that some individuals from population y migrate into population x. The proportion of these y individuals that become parents in population x in the next generation = m. After the migration event, population x can be considered to consist of migrant individuals (proportion m) and non-migrant individuals (proportion [1-m]). Thus the frequency of the A allele in population x in the next generation (px t+1) is just the frequency in the non-migrant portion (= px [1-m]) plus the frequency in the migrant portion (py m). Thus: px t+1 = px t [1-m] + py m.\n\nThe change in allele frequency due to gene flow is Dp = (px t+1) - px t which is just;\n\n[px t [1-m] + py m] - px t Multiplying through and canceling terms leaves us with:\n\nDp = -m(px t - py t). This makes intuitive sense: the change in p depends on the migration rate and the difference in p between the two populations. If we considered a grid or array of populations and focus on one of those populations as the recipient population with all other populations contributing equally to it, then py would be replaced by the average p for all the other populations. Many scenarios are possible.\n\nGENE FLOW AND SELECTION\n\nTo address the combined effects of gene flow and selection, we will invoke a \"fight\" similar to what we described for mutation selection balance above. Consider that some weak allele is wafting over to the other side of the tracks, so to speak, where they do not survive (e.g., fish swimming into New York harbor). There is an evolutionary pressure changing allele frequencies in one direction ( into the harbor), and an opposing evolutionary force eliminating those alleles (sewage killing off genetically intolerant fish). Depending on the relative strengths of these two opposing forces, an equilibrium condition can arise.\n\nLets consider the movement of the \"a\" allele, and assume that it is completely recessive in its phenotype of death-by-sewage. The change in allele frequency from the migration into the harbor can be defined as above: Dq = -m(qx t - qy t). (Note that we have changed p to q since we are considering the a allele; x and y refer to the two populations). The change in allele frequency due to selection against this allele is -sq2(1-q) (note that this is the same expression we used in the mutation selection balance above). Putting these two pieces together, we can write the expression for the change in allele frequency that is due to BOTH gene flow and selection: Dq = -m(qx t - qy t) - sqx2(1-qx). When the \"fight\" between gene flow and selection is a \"draw\", the system will be in equilibrium and there will be no change in q,\n\nand -m(qx t - qy t) = sq2(1-q).\n\nSo, back to the fish. Lets say that aa homozygotes drop dead the minute they enter the East river (but that AA and Aa fish are fine). Outside New York Harbor, the frequencies of the two alleles are equal (p = q= 0.5). It is discovered that in the East river, q = 0.1 over many generations, and the Mayor wants to know what proportion of the fish in the East river come from the outside each generation. This information gives us all we need: it's in equilibrium, qx = 0.1 (East River), qy = 0.5 (outside), and s = 1.0 since the aa's are dead. Plug in the values and you get m = 0.023. This says that 2.3% of the fish in the East river must come from outside each generation to maintain the allele frequency at q = 0.1" ]

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[ "Survey\n\n* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project\n\nDocument related concepts\nTranscript\n```IB Biology Statistical Analysis Practice Test Answer Sheet\n1.\na. Mean: The mean is the sum of the data points divided by the number of data\npoints.\nb. Standard deviation: A measure of the amount of variability (or spread) among the\nnumbers in a data set. One standard deviation from the mean in either direction\non the horizontal axis accounts for ~ 68% of the data points.\nc. Range: The largest value in the data set minus the smallest value in the data set.\nd. Variability: A general description of the difference between any two measurements\n2.\na. Groups\n1\nand\n4\nb. Groups\n3\nand\n5\nc. The mean of group 5 is not likely to be significantly different to the mean of\ngroup 3.\nfrequency (y)\n3.\nB\nC\nA\ncontinuous variable (x)\n4.\n± sd (standard deviation from the mean represents 68 % of all the data points.\nIn data with a high standard deviation, data are clustered closer to / further from the mean.\nIn data with a low standard deviation, data are clustered closer to / further from the mean.\nOverlapping standard deviations suggest two datasets are / are not significantly different.\n1\nIB Biology Statistical Analysis Practice Test Answer Sheet\n6.\nMean =\nStdev =\np=\nGroup A\n24\n25\n26\n23\n25\n25\n26\n27\n23\n23\nGroup B\n24\n29\n25\n23\n29\n32\n34\n31\n32\n29\n24.7\n1.4\n0.004\n28.8\n3.7\n(T-Test)\nGroup A\n2\nA - Abar (A - Abar)\n-0.7\n0.49\n0.3\n0.09\n1.3\n1.69\n-1.7\n2.89\n0.3\n0.09\n0.3\n0.09\n1.3\n1.69\n2.3\n5.29\n-1.7\n2.89\n-1.7\n2.89\nΣ=\n18.1\nGroup B\n2\nB - Bbar (B - Bbar)\n-4.8\n23.04\n0.2\n0.04\n-3.8\n14.44\n-5.8\n33.64\n0.2\n0.04\n3.2\n10.24\n5.2\n27.04\n2.2\n4.84\n3.2\n10.24\n0.2\n0.04\nΣ=\n123.6\n7. If P is less than 0.05, then the difference between the two means is significant.\nAssuming a null hypothesis has been stated that there is no significant difference\nbetween the means of the two groups, then the calculated P-value of 0.004 results\nin a rejection of this null hypothesis.\nWhen two means are subjected to a statistical test where p = 0.05, it indicates that the\nnull hypothesis will have a 95% chance of being true. However, the means of the two\ngroups in our example beat the odds and show less than a 5% chance that they would\nbe within 2 standard deviations of each other. Thus, these two means are statistically\ndifferent from each other.\n8. T-tests enable scientists to express probabilities that any two events will occur by\nchance alone. These probabilities help scientist’s better frame debates about topics\nsuch as global warming, evolution, etc. in objective and meaningful ways.\n9.\na.\nb.\nc.\nd.\ne.\nf.\nFalse.\nTrue\nFalse\nFalse\nFalse\nTrue\n10.\nThe presence of chlorophyll as a function of temperature (leaves changing color).\nRate of osmosis as a function of membrane surface area.\n2\n```" ]

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[ "While programming, it is sometimes necessary to find the direction a 角色 should point given two 速度 (x and y). This article presents the 程式 that will provide the correct direction as well as an explanation of how it works.\n\n## The Script\n\n```如果 <(y-vel) = > 那麼\n如果 <(x-vel) < > 那麼\n面朝 (-90 v) 度\n\n面朝 (90 v) 度\nend\n\n如果 <(y-vel) > > 那麼\n面朝 ([atan v] 數值 ((x-vel) / (y-vel))) 度\n\n面朝 ((180) + ([atan v] of ((x-vel) / (y-vel)))) 度\nend\nend\n```\n\n## How it Works\n\nThe bottom half calculates the direction working on the assumption that y is not zero (which would yield an error when x is divided by y). The mathematical function \"atan\" which stands for \"Arc Tangent\" and is the inverse of tangent, a trigonometric function used to determine the ratio of the two legs of a right triangle.\n\nThe script essentially treats the x and y axes as the two legs of a right triangle, using \"atan\" to derive the angle. However, as the angle has to be between -90 and 90, the \"y > 0\" conditional (or 菱形积木) must be added to point the sprite in the proper direction.\n\nIt is notable that typical geometry uses atan(y/x) (rather than x/y) to determine the angle; the flip made in the script is to match with Scratch's unusual degree measurement system (0 is typically right).\n\nThe top half of the script is a simple comparison. If y velocity is zero, the only possible directions — ignoring the chance that the x velocity could be zero — are right and left (90 and -90 degrees). Should both the x and y velocities be zero, the sprite will face right (90 degrees).\n\n## 用法示例\n\n• Finding out the direction the mouse is moving in\n• Predicting the position of the character in a platformer in a few seconds, maybe as a power up tool.\n• Finding the direction an arrow should point in given an x and y velocity\n\n## 参见\n\nCookies help us deliver our services. By using our services, you agree to our use of cookies." ]

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[ "Math and Arithmetic\nStatistics\nAlgebra\n\n# How many different ten digit number can be formed using the digits zero until nine excluding numbers starting with zero?\n\nIf you exclude numbers starting with zero then the first digit\n\nmust be between 1 and 9 (i.e. 9 combinations). The remaining 9\n\ndigits can be any value between 0 and 9 (i.e. 10\n\ncombinations).\n\nSo you can have 9x109 = 9,000,000,000 combinations.\n\nCopyright © 2020 Multiply Media, LLC. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply." ]

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[ "## Algebra 2 (1st Edition)\n\n$x=8,y=-3$\nIf we add twice the first equation to five times the second equation we get: $19x=152\\\\x=8$ Now plugging this into the first equation: $2(8)+5y=1\\\\16+5y=1\\\\5y=-15\\\\y=-3$" ]

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[ "X\nX\n\n# Calculate the Greatest Common Factor or GCF of 72 and 2915\n\nThe instructions to find the GCF of 72 and 2915 are the next:\n\n## 1. Decompose all numbers into prime factors\n\n 72 2 36 2 18 2 9 3 3 3 1\n 2915 5 583 11 53 53 1\n\n## 2. Write all numbers as the product of its prime factors\n\n Prime factors of 72 = 23 . 32 Prime factors of 2915 = 5 . 11 . 53\n\n## 3. Choose the common prime factors with the lowest exponent\n\nCommon prime factors: None\n\nCommon prime factors with the lowest exponent: None\n\n## 4. Calculate the Greatest Common Factor or GCF\n\nRemember, to find the GCF of several numbers you must multiply the common prime factors with the lowest exponent.\n\nSince there are not common prime factors the GCF is 1\n\nAlso calculates the:" ]

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[ "# zbMATH — the first resource for mathematics\n\nCompactification of the stacks of shtukas and geometric invariant theory. (Compactification des champs de chtoucas et théorie géométrique des invariants.) (French) Zbl 1144.11047\nAstérisque 313. Paris: Société Mathématique de France (ISBN 978-2-85629-243-3/pbk). 124 p. (2007).\nThis book contains the results announced previously by the author in [C. R., Math., Acad. Sci. Paris 340, No. 2, 147–150 (2005; Zbl 1082.11036)]. A crucial step in the proof of the Langlands correspondence for $$\\text{GL}_r$$ over function fields is the construction and study of a compactification of the stack of rank $$r$$ shtukas (by Drinfeld for $$r=2$$ and Lafforgue for arbitrary $$r$$).\nIn this text the author uses Geometric Invariant Theory (à la I. V. Dolgachev and Y. Hu [Publ. Math., Inst. Hautes Étud. Sci. 87, 5–56 (1998; Zbl 1001.14018)] and M. Thaddeus [J. Am. Math. Soc. 9, No. 3, 691–723 (1996; Zbl 0874.14042)]) to construct compactifications of the stack of rank $$r$$ shtukas. This novel approach does not only recover Lafforgue’s original compactification, but also produces several new ones.\nIt is hoped that this method will be more suitable to attack the problem of compactifying the stacks of $$G$$-shtukas for arbitrary reductive groups $$G$$.\n\n##### MSC:\n 11G09 Drinfel’d modules; higher-dimensional motives, etc. 11R39 Langlands-Weil conjectures, nonabelian class field theory 14D20 Algebraic moduli problems, moduli of vector bundles 11-02 Research exposition (monographs, survey articles) pertaining to number theory" ]

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[ "# 弧度\n\n${\\frac {\\pi }{180}}\\times deg=rad$", null, "$deg=rad\\times {\\frac {180}{\\pi }}$", null, "$\\lim _{h\\to 0}{\\frac {\\sin h}{h}}=1$", null, "$\\mathbf {0}$", null, "${\\tfrac {1}{12}}$", null, "${\\tfrac {1}{8}}$", null, "${\\tfrac {1}{6}}$", null, "${\\tfrac {1}{4}}$", null, "${\\tfrac {1}{2}}$", null, "${\\tfrac {3}{4}}$", null, "$\\mathbf {1}$", null, "## 參考\n\n• ISO 31-1\n• ISO/IEC 80000-3" ]

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[ "# What is 1/42 as a percentage?\n\n1 / 42 = 2.38%\n\n## Fraction to Percent Conversion Summary\n\nWe encourage you to check out our introduction to percentage page for a little recap of what percentage is. You can also learn about fractions in our fractions section of the website. Sometimes, you may want to express a fraction in the form of a percentage, or vice-versa. This page will cover the former case. Luckily for us, this problem only requires a bit of multiplication and division. We recommend that you use a calculator, but solving these problems by hand or in your head is possibly too! Here's how we discovered that 1 / 42 = 2.38% :\n\n• Step 1: Divide 1 by 42 to get the number as a decimal. 1 / 42 = 0.02\n• Step 2: Multiply 0.02 by 100. 0.02 times 100 = 2.38. That's all there is to it!\nNote that you can reverse steps 1 and 2 and still come to the same solution. If you multiply 1 by 100 and then divide the result by 42, you will still come to 2.38!\n\n## When are fractions useful?\n\nFractions are commonly used in everyday life. If you are splitting a bill or trying to score a test, you will often describe the problem using fractions. Sometimes, you may want to express the fraction as a percentage.\n\n## Convert 1 / 42 into a percentage or decimal\n\n### Fraction Conversion Table\n\nPercentage Fraction Decimal\n2.38% 1 / 42 0.02\nRemember: Converting a fraction to a percentage or decimal are all equivalent numbers. They are all used to represent a relationship between numbers. We can choose how we represent numbers by which style matches the appropriate situation.\n\n## Find the Denominator\n\nA percentage is a number out of 100, so we need to make our denominator 100!\n\nIf the original denominator is 42, we need to solve for how we can make the denominator 100.\n\nHint: percentages all have a denominator of 100!\n\nTo convert this fraction, we would divide 100 by 42, which gives us 2.38.\n\n## Find the Numerator\n\nNow, we multiply 2.38 by 1, our original numerator, which is equal to 2.38\n\n## Convert your Fraction to a Percent\n\n2.38 / 100 = 2.38%\n\nRemember, a percentage is any number out of 100. If we can balance 1 / 42 with a new denominator of 100, we can find the percentage of that fraction!" ]

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[ "Math Concepts\nTricks and Importance of Vedic Maths\n140.9k views\n\n 1 Introduction 2 Benefits of Vedic Maths 3 Addition Tricks 4 Subtraction Tricks 5 Squaring Tricks 6 Square Root Tricks 7 Multiplication Tricks 8 Conclusion 9 FAQs 10 External References\n\n\"Once the mind of a student develops an understanding of mental mathematics, he/she begins to think systematically and more creatively. In this article, we discussed some of the most basic Vedic maths tricks for beginners under different categories, with relevant examples and explanations.\"\n\nVedic Maths is a collection of techniques/sutras to solve mathematical problem sets in a fast and easy way. These tricks introduce wonderful applications of Arithmetical computation, theory of numbers, mathematical and algebraic operations, higher-level mathematics, calculus, and coordinate geometry, etc.", null, "It is very important to make children learn some of the Vedic maths tricks and concepts at an early stage to build a strong foundation for the child. It is one of the most refined and efficient mathematical systems possible.\n\nVedic maths was discovered in the mid-1900s and has certain specific principles to perform various calculations in mathematics. But the question that arises is that is mathematics only about performing calculations?\n\nHowever fascinating it might be to calculate faster using Vedic mathematics tricks, it fails to make a student understand the concepts, applications, and real-life scenarios of those particular problems.\n\n## What are the benefits of Vedic Maths?\n\nThese are the benefits of Vedic Maths:\n\n1. It helps a person to solve mathematical problems many times faster\n2. It helps in making intelligent decisions to both simple and complex problems\n3. It reduces the burden of memorizing difficult concepts\n4. It increases the concentration of a child and his determination to learn and develop his/her skills\n5. It helps in reducing silly mistakes which are often created by kids", null, "If you ever want to read it again as many times as you want, here is a downloadable PDF to explore more.\n\n## Vedic maths tricks for Addition\n\nThe addition is one of the most basic operations of Vedic mathematics. It states that,\n\n1. Find out the number which is closest to the 10s multiple because it is easier to add those numbers.\n\n7, 8, 9 close to 10\n\n21, 22, 23 close to 20\n\n67, 68, 69, are close to 70\n\n97, 98, 99, are close to 100 ....... and so on.\n\n1. Add the numbers which are the multiples of 10s\n2. Add/Subtract the deficiency of numbers.\n\nLet’s understand with the help of an example.\n\nSuppose, we have to add 27 and 98.\n\nSo, Vedic maths tells us to add 30 and 100 which is 130 and then subtract (3+2) i.e. the deficiency from 130. So the result will be 125.\n\nSimilarly, if we have to add 66 and 576.\n\nSo, Vedic maths tells us to add 70 and 580 which is 650 and then subtract (4+4) i.e. the deficiency from 650. So the result will be 642.\n\nThere is yet another trick to perform addition using Vedic maths which states to add hundreds with hundreds, tens with tens and ones with ones, and so on.\n\nFor example, Suppose we have the following question, 220 + 364 + 44 + 18 = ?\n\nVedic maths tells us to break the numbers as per their place values. So, we will break the addition into:\n\n200 + 300 = 500\n\n20 + 60 +40 +10 = 130\n\n4 + 4 +8 = 16\n\nRepeat the process:\n\n500 + 100 = 600\n\n30 + 10 = 40\n\nAnd at the units place we have 6.\n\nNow perform, 600 + 40 + 6 = 646\n\n## Vedic maths tricks for Subtraction\n\nFor subtraction using Vedic maths, we follow the rules given below,\n\n1. If the subtrahend is less than minuend, we subtract the numbers directly.\n2. If any digit in minuend is less than the corresponding digit in subtrahend, we use the concept of complements.\n\nLet us have a look at a few examples to understand these techniques.\n\n1) When subtrahend is less than minuend: If we have to subtract 47 from 98, then, we can directly subtract the digits in subtrahend from the corresponding digits in minuend.\n98 - 47 = 51\n\n2) When any digit in minuend is less than the corresponding digit in subtrahend: 896 - 239\n\nIn this case for the places where the digit in subtrahend is greater we use the complement symbol while subtracting as shown below,\n896 - 239 = 66$$\\bar3$$\nThe complement of 3 = $$\\bar3$$ = 10 - 3 = 7\nWhile replacing the value of $$\\bar3$$, we subtract 1 from the digit in the next place. Here, 1 will be subtracted from 6.\n\nTherefore, 896 - 239 = 657.\n\n## Vedic maths tricks for squares\n\nFor the numbers ending with 5:\n\nStep 1: Perform 5 × 5 = 25\n\nStep 2: Add 1 to the previous number and the result with the previous number.\n\nFor example, in the case of the square of 85, Add 1 to 8 = 9 and multiply 9 with 8 = 72\n\nStep 3: The result from step two will become the initial numbers of the final answer and the result of step one is the ending two numbers of the final answer. So the final answer will be 7225.\n\nSimilarly,\n\nWhat is the square of 195?\n\nStep 1: 5 × 5 = 25\n\nStep 2: 19 × 20 = 380\n\nStep 3: Combining the two results, which will give us 38025 which is the final answer.\n\n### Steps for Square Roots Math tricks\n\nFor performing square roots, we will have to keep some facts in mind:-\n\n• Squares of numbers from 1 to 9 are 1, 4, 9, 16, 25, 36, 49, 64, 81.\n\n• Square of a number cannot end with 2, 3, 7, and 8.\n\n• We can say that numbers ending with 2, 3, 7, and 8 cannot have a perfect square root.\n\n• The square root of a number ending with 1 (1, 81) ends with either 1 or 9\n\n• The square root of a number ending with 4 (4, 64) ends with either 2 or 8\n\n• The square root of a number ending with 9 (9, 49) ends with either 3 or 7\n\n• The square root of a number ending with 6 (16, 36) ends with either 4 or 6\n\n• If the number is of ‘n’ digits then the square root will be ‘n/2’ OR ‘(n+1)/2’ digits.\n\nLet us understand an example of finding a square root of 1764.\n\n1. The number ends with 4. Since it’s a perfect square, square root will end with 2 or 8.\n\n2. We need to find 2 perfect squares (In Multiples of 10) between which 1764 exists.\nThe numbers are 1600 (40) and 2500 (50).\n\n3. Find to whom 1764 is closer. It is closer to 40. Therefore the square root is nearer to 40. Now from Step 2, possibilities are 42 or 48 out of which 42 is closer to 40.\n\n4. Hence the square root = 42\n\n## Vedic maths tricks for Multiplication\n\nThere are a number of techniques to perform various types of multiplication calculations using Vedic maths tricks. Some of the most useful and the easiest ones are mentioned below:-\n\nThe trick of 11:\n\nStep 1:- Divide the number into two parts\n\nStep 2:- Add the two parts which will form the middle number\n\nLet us understand this with the help of an example which will clear the doubts.\n\nLet us say that we have to multiply 32 with 11.\n\nStep 1: Divide 32 into 3 and 2\n\nStep 2: The middle will be 3 + 2 = 5\n\nSo our answer to 32 × 11 would be 352.\n\nSimilarly, 75 × 11 = 7, 7 + 5, 5. Because 7+5 = 12, we will carry 1 to the previous digit and our final answer would be 825.\n\nMultiplying numbers which are close to powers of 10:\n\nIt would be easy to directly jump into the example and understand this concept through it. Suppose, we have 2 questions:\n\n1. Multiply 99 × 97\n\n2. Multiply 103 × 105\n\nOur approach should be something as shown in the picture.\n\nStep 1: Find by how much the number is more or less than the power of 100 and write it on the right side of the vertical line.", null, "Step 2: Either cross subtract or cross add. This will form the first part of the final answer.", null, "Step 3: Multiply the right side of the vertical line and this shall form the right side of the result.", null, "Similarly, suppose we have to find 996 × 997. Our approach will be as shown in the picture below.", null, "Multiplying 2 digit numbers:\n\nThis would again be simple if followed by a step approach through what is displayed in the picture. Suppose, we have to multiply 31 × 12. We have to follow the vertical and crosswise sutra. Our approach would be:-\n\nStep 1: Multiplying vertically the units place.", null, "Step 2: Multiplying in a cross pattern and adding the results. For example, 3 × 2 = 6 and 1 × 1 = 1 and 6 + 1 = 7.", null, "Step 3: Multiplying the tens digit vertically.", null, "Similarly, if we have to find 12 × 34, our approach will be as shown below:\n\nFirstly, multiply 2 × 4 = 8\n\nThen, (3 × 2) + (4 × 1) = 10. The zero remains and 1 is carried to the left.\n\nFinally, (3 × 1) = 3. Add to it the carried number so that it becomes, (3 + 1) = 4.\n\nOur final answer would be 408.", null, "## Conclusion\n\n1. Vedic Maths just showcases a process to do things faster. It does not teach a child the underlying philosophy or the background of the problem set given. Calculating faster is of no use if we fail to understand the meaning or the learning behind the problem set.\n2. These tricks can do wonders only if used properly after imbibing a proper learning experience. Practice makes a man perfect but Learning makes a man capable. Therefore, make Vedic Maths a habit only after understanding its nuances.\n3. Cuemath comes up with regular articles on varied topics, providing positive thinking and direction on different issues faced by children and parents. Comment in the box below any topic, doubts, or other feedback. We will surely research and reach out to you soon." ]

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[ " 基于需求学习的多阶段动态折扣和库存优化研究——以服装销售为例 Multi-Stage Dynamic Discount and Inventory Optimization Model Based on Demand Learning: Taking Garment Sales as an Example\n\nE-Commerce Letters\nVol. 08  No. 01 ( 2019 ), Article ID: 28453 , 8 pages\n10.12677/ECL.2019.81004\n\nMulti-Stage Dynamic Discount and Inventory Optimization Model Based on Demand Learning: Taking Garment Sales as an Example\n\nJiali Han\n\nCollege of Business Administration, Donghua University, Shanghai", null, "Received: Dec. 26th, 2018; accepted: Jan. 8th, 2019; published: Jan. 15th, 2019", null, "ABSTRACT\n\nAiming at the uncertainties of demand model, based on the known prior demand model, data assimilation technology is applied to continuously learn the parameters of demand model, and the state-valued demand parameters of the model are optimized by using the data of each observation time. A multi-stage discount and inventory model based on demand learning is established to formulate a reasonable discount strategy to meet the business sales objectives. Empirical research shows that the proposed model can effectively reduce the inventory level of merchants by making reasonable discounts under demand uncertainty, and provide scientific basis for discount promotion. Through sensitivity analysis of initial inventory and initial price, the optimal initial inventory and initial price can be found. It provides strong support for the determination of order quantity and initial pricing, and improves the control level of dynamic system.\n\nKeywords:Garment, Demand Learning, Parameter Prediction, Dynamic Discount", null, "", null, "", null, "Copyright © 2019 by author(s) and Hans Publishers Inc.\n\nThis work is licensed under the Creative Commons Attribution International License (CC BY).\n\nhttp://creativecommons.org/licenses/by/4.0/", null, "", null, "1. 引言\n\n2. 模型建立\n\n2.1. 算法介绍\n\n1) 预测。令系统状态向量 $\\theta ~N\\left({\\theta }_{0},{\\sigma }^{2}{P}_{0}\\right)$ ，那么状态预测为 ${\\theta }_{n}=A{\\theta }_{n-1}+\\omega$\n\n2) 计算卡尔曼增益 ${k}_{n}$ 。根据第n阶段的观测数据得到测量矩阵 ${H}_{n}$ 和测量噪声V，其中R是V的协方差，那么 ${k}_{n}={P}_{n}{{H}^{\\prime }}_{n}/\\left({H}_{n}{P}_{n}{{H}^{\\prime }}_{n}+R\\right)$\n\n3) 计算最优估计值。根据卡尔曼增益 ${k}_{n}$ 得到现在阶段n的最优化参数估算值： ${{\\theta }^{\\prime }}_{n}={\\theta }_{n}+{k}_{n}\\left({z}_{n}-{H}_{n}{\\theta }_{n}\\right)$ ，其中 ${z}_{n}={H}_{n}{\\theta }_{n}+V$ ，为第n阶段的系统测量值；\n\n4) 更新。为了要卡尔曼滤波器不断的运行下去直到系统过程结束，还要更新第n阶段下 ${\\theta }_{n}$ 的协方差矩阵 ${{P}^{\\prime }}_{n}$ ，计算公式为 ${{P}^{\\prime }}_{n}=\\left(1-{k}_{n}{H}_{n}\\right){P}_{n}$\n\n2.2. 基于参数更新的动态折扣和库存优化模型\n\n2.2.1. 动态折扣和库存优化模型建立\n\n${I}_{n+1}={I}_{n}-{Q}_{n}={I}_{0}-{\\sum }_{1}^{n}{Q}_{n}$ (1)\n\n${Q}_{n}={\\int }_{0}^{{t}_{n}}{q}_{n}\\left(t\\right)\\text{d}t={q}_{n}{t}_{n}={\\text{e}}^{\\alpha +\\beta {\\gamma }_{n}+ϵ}{t}_{n}=M{\\text{e}}^{\\alpha +\\beta {\\gamma }_{n}}{t}_{n}$ (2)\n\n${H}_{n}={\\int }_{0}^{{t}_{n}}{I}_{n}\\left(t\\right)h\\text{d}t={I}_{n}{t}_{n}h-\\frac{{q}_{n}{t}_{n}\\left(1+{t}_{n}\\right)}{2}h$ (3)\n\n${F}_{n}={Q}_{n}{p}_{0}{\\gamma }_{n}=M{\\text{e}}^{\\alpha +\\beta {\\gamma }_{n}}{t}_{n}{p}_{0}{\\gamma }_{n}$ (4)\n\n${R}_{n}={F}_{n}-{H}_{n}-c{Q}_{n}=M{\\text{e}}^{\\alpha +\\beta {\\gamma }_{n}}{t}_{n}\\left({p}_{0}{\\gamma }_{n}-c\\right)-{I}_{n}{t}_{n}h+\\frac{M{\\text{e}}^{\\alpha +\\beta {\\gamma }_{n}}{t}_{n}\\left(1+{t}_{n}\\right)}{2}h$ (5)\n\n$\\begin{array}{l}\\mathrm{max}{\\sum }_{n=1}^{n=m}{R}_{n}\\\\ \\text{s}\\text{.t}:{I}_{m+1}=0\\end{array}$ (6)\n\n2.2.2. 最优折扣求解\n\n${f}_{n}\\left({I}_{n}\\right)=\\mathrm{max}\\left\\{{R}_{n}+{f}_{n+1}\\left({I}_{n+1}\\right)\\right\\}$ (7)\n\n${f}_{m+1}\\left({I}_{m=1}\\right)=0,\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}n=m,m-1,\\cdots ,1$ (8)\n\n1) 2阶段模型求解\n\n$m=2$ 时，模型目标为： $\\text{max}\\left(E\\left({R}_{1}\\right)+E\\left({R}_{2}\\right)\\right)$ ，先验信息 $\\theta ={\\left(\\alpha ,\\beta \\right)}^{\\prime }~N\\left({\\theta }_{0},{\\sigma }^{2}{P}_{0}\\right)$ ，可以通过之前的历史数据得到，根据卡尔曼滤波重新更新得到 ${\\theta }_{1}$ ，以此类推，首先考虑第2阶段，进行 ${f}_{2}\\left({I}_{2}\\right)$ 的极值求解：\n\n${f}_{2}\\left({I}_{2}\\right)={R}_{2}=M{\\text{e}}^{\\alpha +\\beta {\\gamma }_{2}}{t}_{2}\\left({p}_{0}{\\gamma }_{2}-c\\right)-{I}_{2}{t}_{2}h+\\frac{M{\\text{e}}^{\\alpha +\\beta {\\gamma }_{2}}{t}_{2}\\left(1+{t}_{2}\\right)}{2}h$ (9)\n\n${\\gamma }_{2}^{\\ast }=\\frac{\\mathrm{ln}\\left(\\frac{{I}_{2}}{M{t}_{2}}\\right)-{\\alpha }_{1}}{{\\beta }_{1}}$ (10)\n\n$E\\left({R}_{2}^{\\ast }\\right)={f}_{2}\\left({I}_{2}\\right)={I}_{2}\\left({p}_{0}\\frac{\\mathrm{ln}\\left(\\frac{{I}_{2}}{M{t}_{2}}\\right)-{\\alpha }_{1}}{{\\beta }_{1}}-c\\right)-{I}_{2}{t}_{2}h+\\frac{{I}_{2}\\left(1+{t}_{2}\\right)}{2}h$ (11)", null, "(12)\n\n${\\gamma }_{1}^{\\ast }=\\frac{\\mathrm{ln}\\left(\\frac{{I}_{1}}{M\\left({t}_{2}{\\text{e}}^{\\frac{{\\beta }_{1}}{{\\beta }_{0}}+\\frac{2+{t}_{1}-{t}_{2}}{2{p}_{0}}h{\\beta }_{1}-{t}_{2}M}+{t}_{1}\\right)}\\right)-{\\alpha }_{0}}{{\\beta }_{0}}$ (13)\n\n2) 多阶段模型求解\n\n$m>2$ 时，参数更新下的动态规划求解相当困难。但是基于上述动态规划求解过程可以采取一个近似定价策略：要找到每个阶段的最优折扣 ${\\gamma }_{n}^{\\ast }$ ，就是要下式最大化：\n\n$\\begin{array}{l}E\\left({R}_{n}\\right)=M{\\text{e}}^{{\\alpha }_{n-1}+{\\beta }_{n-1}{\\gamma }_{n}}{t}_{n}\\left({p}_{0}{\\gamma }_{n}-c\\right)-{I}_{n}{t}_{n}h\\\\ \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}+\\frac{M{\\text{e}}^{{\\alpha }_{n-1}+{\\beta }_{n-1}{\\gamma }_{n}}{t}_{n}\\left(1+{t}_{n}\\right)}{2}h+\\frac{1}{2}\\frac{M{\\text{e}}^{{\\alpha }_{n}-1}}{{\\beta }_{n}{}^{3}}{\\sigma }_{{\\beta }_{n}}^{2}{\\sigma }_{{\\alpha }_{n}}^{2}\\end{array}$\n\n${\\gamma }_{n}^{\\ast }=\\frac{\\mathrm{ln}\\left(\\frac{{I}_{n}}{M\\left({t}_{n+1}{\\text{e}}^{\\frac{{\\beta }_{n}}{{\\beta }_{n-1}}+\\frac{2+{t}_{n}-{t}_{n+1}}{2{p}_{0}}h{\\beta }_{n}-{t}_{n+1}M}+{t}_{n}\\right)}\\right)-{\\alpha }_{n-1}}{{\\beta }_{n-1}}$ (14)\n\n3. 实证分析\n\n3.1. 数据源\n\n3.2. 预测误差分析\n\n$\\text{Error}=\\left({{\\theta }^{\\prime }}_{n}-{\\theta }_{n}\\right)/{\\theta }_{n}$ (15)\n\n3.3. 最优折扣策略计算\n\n$M=E\\left({\\text{e}}^{{ϵ}_{t}}\\right)={\\text{e}}^{{\\sigma }^{2}/2}=1.046$ ，参数更新频率为3周，即 ${t}_{n}=21$ 天，根据表1的各阶段参数的滤波估计值，计算最优折扣策略，期望销量和期望收益额(元)，结果如表2所示。", null, "Figure 1. Comparison of prediction errors under different assimilation frequencies", null, "Table 2. Optimal discount strategy and inventory change in each stage\n\n4. 参数灵敏度分析", null, "Table 3. Sensitivity analysis of initial inventory\n\n5. 结论\n\nMulti-Stage Dynamic Discount and Inventory Optimization Model Based on Demand Learning: Taking Garment Sales as an Example[J]. 电子商务评论, 2019, 08(01): 22-29. https://doi.org/10.12677/ECL.2019.81004\n\n1. 1. Golabi, K. (1985) Optimal Inventory Policies When Ordering Prices Are Random. Operations Research, 33, 575-588. https://doi.org/10.1287/opre.33.3.575\n\n2. 2. Bitran, G.R. and Mondschein, S.V. (1997) Periodic Pricing of Seasonal Products in Retailing. Management Science, 43, 64-79. https://doi.org/10.1287/mnsc.43.1.64\n\n3. 3. Besbes, O. and Zeevi, A. (2009) Dynamic Pricing without Knowing the Demand Function: Risk Bounds and Near-Optimal Algorithms. Operations Research, 57, 1407-1420. https://doi.org/10.1287/opre.1080.0640\n\n4. 4. Araman, V.F. and Caldentey, R. (2009) Dynamic Pricing for Perishable Products with Demand Learning. Operations Research, 57, 1169-1188. https://doi.org/10.1287/opre.1090.0725\n\n5. 5. Lin, K.Y. (2006) Dynamic Pricing with Real-Time Demand Learning. European Journal of Operational Research, 174, 522-538. https://doi.org/10.1016/j.ejor.2005.01.041\n\n6. 6. Huang, C.L. and Xin, L.I. (2006) Experiments of Soil Moisture Data Assimilation System Based on Ensemble Kalman Filter. Plateau Meteorology, 112, 888-900.\n\n7. 7. Carvalho, A.X. and Puterman, M.L. (2003) Dynamic Pricing and Reinforcement Learning. International Joint Conference on Neural Networks, 4, 2916-2921. https://doi.org/10.1109/IJCNN.2003.1224034" ]

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[ "## Evaluating Pi …\n\n… to 15 decimal places in one easy step (or not too difficult step).\n\nIn my previous post one of the examples given of the use of the Gaussian Quadrature function was to evaluate the value of Pi.  Let’s look at that a bit more closely.\n\nWikipedia has a nice article on the history of Pi, but like most such articles the methods described are either historical methods which are easy to understand but converge slowly, or recent methods which converge very quickly, but which are presented with little or no information about how they were derived.  For example the well known 20th Century Indian mathematician Srinivasa Ramanujan somehow derived the relationship:", null, "$\\frac{1}{\\pi} = \\frac{2 \\sqrt 2}{9801} \\sum_{k=0}^\\infty \\frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}\\!$\n\nand in 1987 the Chudnovsky brothers  derived:", null, "$\\frac{426880 \\sqrt{10005}}{\\pi} = \\sum_{k=0}^\\infty \\frac{(6k)! (13591409 + 545140134k)}{(3k)!(k!)^3 (-640320)^{3k}}\\!$\n\nwhich delivers 14 digits per term, but no indication is given of how these relationships were derived.\n\nThe early historical methods of calculating Pi, starting with Archimedes in the 3rd century BC, were based on the successive subdivision of polygons.", null, "Around AD 265, the Wei Kingdom mathematician Liu Hui provided a simple and rigorous iterative algorithmto calculate π to any degree of accuracy. He himself carried through the calculation to a 3072-gon and obtained an approximate value for π of 3.1416.", null, "A simple way to improve the accuracy of the polygon approach is to find an approximation of the area between the circle and the polygon, and add this to the known area of the polygon.  Starting with a hexagon of unit side, this may be divided into 6 equilateral triangles and six circular segments with 60 degree included angle:\n\nThe area of the triangle is 0.5 x 3^0.5 / 2 (0.5 x base x height)\n\nThe equation of the segment is (1 – x^2)^0.5 – 3^0.5 / 2\n\nIf this equation is integrated between x = -0.5 and x = 0.5, and added to the area of the equilateral triangle, this gives 1/6 the area of the circle of unit radius.  If the integration is evaluated using Gaussian Quadrature, with 8 integration points, the resulting area provides an estimate  of Pi with an error of 2.3E-11; not too bad:\n\nA further 4 significant figures can be obtained by simply changing the limits of integration to X = 0 to 0.5 (and adjusting the triangle area to suit, and multiplying the resulting area by 12):\n\nThe resulting estimate of Pi has an error of 7.6E-15, and matches the built in Excel Pi function to the 14th significant figure.\n\nThis entry was posted in Excel, Maths, Newton, UDFs, VBA and tagged , , , , . Bookmark the permalink.\n\n### 4 Responses to Evaluating Pi …\n\n1.", null, "paviavio says:\n\nThis would be very good piece for Pi Day.\n\nLike\n\n2.", null, "dougaj4 says:\n\n“This would be very good piece for Pi Day.”\n\nI didn’t think of that.\n\nPity there aren’t 31 days in April.\n\nI’ll just have to wait for 22/July 🙂\n\nLike\n\n3.", null, "خرید کریو says:\n\nthank you\n\nLike\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed." ]

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[ "# What is the P value in psychology?\n\nContents\n\nA p-value, or probability value, is a number describing how likely it is that your data would have occurred by random chance (i.e. that the null hypothesis is true). The level of statistical significance is often expressed as a p-value between 0 and 1.\n\n## How do you find p-value in psychology?\n\nIf your test statistic is positive, first find the probability that Z is greater than your test statistic (look up your test statistic on the Z-table, find its corresponding probability, and subtract it from one). Then double this result to get the p-value.\n\n## What is meant by p 0.05 Psychology?\n\nStatistical tests allow psychologists to work out the probability that their results could have occurred by chance, and in general psychologists use a probability level of 0.05. This means that there is a 5% probability that the results occurred by chance.\n\n## What is the true meaning of p-value?\n\nThe P value means the probability, for a given statistical model that, when the null hypothesis is true, the statistical summary would be equal to or more extreme than the actual observed results . … The smaller the P value, the greater statistical incompatibility of the data with the null hypothesis.\n\n## What does p .01 mean in psychology?\n\nIt’s given as a value between 0 and 1, and labelled p. … 01 means a 1% chance of getting the results if the null hypothesis were true; p = . 5 means 50% chance, p = . 99 means 99%, and so on. In psychology we usually look for p values lower than .\n\n## What does p-value of 0.5 mean?\n\nMathematical probabilities like p-values range from 0 (no chance) to 1 (absolute certainty). So 0.5 means a 50 per cent chance and 0.05 means a 5 per cent chance. In most sciences, results yielding a p-value of . … If the p-value is under . 01, results are considered statistically significant and if it’s below .\n\n## What does the p-value need to be to be significant?\n\nIf the p-value is 0.05 or lower, the result is trumpeted as significant, but if it is higher than 0.05, the result is non-significant and tends to be passed over in silence.\n\n## Why do psychologists use a 5% significance level?\n\nPsychologists normally use a 5% significant level as this is an acceptable level (within Social Sciences) to claim that results of an experiment are significant.\n\n## How do you interpret P values?\n\nThe smaller the p-value, the stronger the evidence that you should reject the null hypothesis.\n\n1. A p-value less than 0.05 (typically ≤ 0.05) is statistically significant. …\n2. A p-value higher than 0.05 (> 0.05) is not statistically significant and indicates strong evidence for the null hypothesis.\n\n## Why is a 5% level of significance used in psychology?\n\n1 mark for a limited or incomplete definition of a Type II error. 1 mark for a reason for why the 5% level of significance is used in psychological research. The 5% level is used as it strikes a balance between the risk of making the Type I and II errors (or similar).\n\nTHIS IS INTERESTING:  Best answer: How is psychology and anthropology related?\n\n## What is p-value example?\n\nP Value Definition\n\nA p value is used in hypothesis testing to help you support or reject the null hypothesis. The p value is the evidence against a null hypothesis. … For example, a p value of 0.0254 is 2.54%. This means there is a 2.54% chance your results could be random (i.e. happened by chance).\n\n## Is p-value of 0.001 significant?\n\nMost authors refer to statistically significant as P < 0.05 and statistically highly significant as P < 0.001 (less than one in a thousand chance of being wrong). … The significance level (alpha) is the probability of type I error.\n\n## How do you interpret p-value in regression?\n\nHow Do I Interpret the P-Values in Linear Regression Analysis? The p-value for each term tests the null hypothesis that the coefficient is equal to zero (no effect). A low p-value (< 0.05) indicates that you can reject the null hypothesis.\n\n## What does p 05 mean?\n\nP > 0.05 is the probability that the null hypothesis is true. 1 minus the P value is the probability that the alternative hypothesis is true. A statistically significant test result (P ≤ 0.05) means that the test hypothesis is false or should be rejected. A P value greater than 0.05 means that no effect was observed.\n\n## Is p-value of 0.1 Significant?\n\nThe smaller the p-value, the stronger the evidence for rejecting the H. This leads to the guidelines of p < 0.001 indicating very strong evidence against H, p < 0.01 strong evidence, p < 0.05 moderate evidence, p < 0.1 weak evidence or a trend, and p ≥ 0.1 indicating insufficient evidence.\n\n## What does p 01 tell us?\n\np=. 01, or the . 01 significance level, means that the likelihood that a mean (proportion) observed in a sample doesn’t really exist in the population (that is, the population mean is really 0) is 1%. This is sometimes expressed inversely, as the 99% confidence level.\n\nTHIS IS INTERESTING:  In what ways do you express your emotions?" ]

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[ "Volume 3, Issue 4\nAlternating Direction Finite Volume Element Methods for Three-Dimensional Parabolic Equations\n\nNumer. Math. Theor. Meth. Appl., 3 (2010), pp. 499-522.\n\nPublished online: 2010-03\n\nPreview Full PDF 356 3701\nExport citation\n\nCited by\n\n• Abstract\n\nThis paper presents alternating direction finite volume element methods for three-dimensional parabolic partial differential equations and gives four computational schemes, one is analogous to Douglas finite difference scheme with second-order splitting error, the other two schemes have third-order splitting error, and the last one is an extended LOD scheme. The $L^2$ norm and $H^1$ semi-norm error estimates are obtained for the first scheme and the second one, respectively. Finally, two numerical examples are provided to illustrate the efficiency and accuracy of the methods.\n\n• Keywords\n\nThree-dimensional parabolic equation, alternating direction method, finite volume element method, error estimate.\n\n65M08, 65M12, 65M15\n\n• BibTex\n• RIS\n• TXT\n@Article{NMTMA-3-499, author = {}, title = {Alternating Direction Finite Volume Element Methods for Three-Dimensional Parabolic Equations}, journal = {Numerical Mathematics: Theory, Methods and Applications}, year = {2010}, volume = {3}, number = {4}, pages = {499--522}, abstract = {\n\nThis paper presents alternating direction finite volume element methods for three-dimensional parabolic partial differential equations and gives four computational schemes, one is analogous to Douglas finite difference scheme with second-order splitting error, the other two schemes have third-order splitting error, and the last one is an extended LOD scheme. The $L^2$ norm and $H^1$ semi-norm error estimates are obtained for the first scheme and the second one, respectively. Finally, two numerical examples are provided to illustrate the efficiency and accuracy of the methods.\n\n}, issn = {2079-7338}, doi = {https://doi.org/10.4208/nmtma.2010.m99027}, url = {http://global-sci.org/intro/article_detail/nmtma/6011.html} }\nTY - JOUR T1 - Alternating Direction Finite Volume Element Methods for Three-Dimensional Parabolic Equations JO - Numerical Mathematics: Theory, Methods and Applications VL - 4 SP - 499 EP - 522 PY - 2010 DA - 2010/03 SN - 3 DO - http://doi.org/10.4208/nmtma.2010.m99027 UR - https://global-sci.org/intro/article_detail/nmtma/6011.html KW - Three-dimensional parabolic equation, alternating direction method, finite volume element method, error estimate. AB -\n\nThis paper presents alternating direction finite volume element methods for three-dimensional parabolic partial differential equations and gives four computational schemes, one is analogous to Douglas finite difference scheme with second-order splitting error, the other two schemes have third-order splitting error, and the last one is an extended LOD scheme. The $L^2$ norm and $H^1$ semi-norm error estimates are obtained for the first scheme and the second one, respectively. Finally, two numerical examples are provided to illustrate the efficiency and accuracy of the methods.\n\nTongke Wang. (2020). Alternating Direction Finite Volume Element Methods for Three-Dimensional Parabolic Equations. Numerical Mathematics: Theory, Methods and Applications. 3 (4). 499-522. doi:10.4208/nmtma.2010.m99027\nCopy to clipboard\nThe citation has been copied to your clipboard" ]

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[ "", null, "# For a circle with a diameter of 6 meters, what is the measurement of a central angle (in degrees) subtended by an arc with o length of 7 3 π meters?", null, "140\n\nStep-by-step explanation:\n\nIf it is an arc of 7pi/3, and the circumference is 6*pi or 6pi, then the arc takes up 7/18 of the circle, and since a circle has 360 degrees, the answr is 7/18 * 360 or 140.", null, "..............................", null, "", null, "", null, "Step-by-step explanation:\n\nWe know that,", null, "where,\n\nθ = central angle in radian.\n\nGiven,\n\ndiameter = 6 m, so radius = 3 m.", null, "Putting the values,", null, "", null, "", null, "", null, "Answer : The measurement of a central angle is,", null, "Step-by-step explanation :\n\nFormula used for angle subtended by an arc is:", null, "where,\n\ns = arc length =", null, "r = radius =", null, "Now put all the given values in the above formula, we get:", null, "", null, "", null, "Thus, the measurement of a central angle is,", null, "", null, "Step-by-step explanation:", null, "The measurement of a central angle (in degrees) subtended by an arc is", null, ".\n\nStep-by-step explanation:\n\nas", null, "meters\n\nso", null, "meters\n\nand", null, "meters\n\nUsing the formula", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Simplify", null, "", null, "", null, "Therefore, the measurement of a central angle (in degrees) subtended by an arc is", null, ".", null, "The measurement of the angle subtended by an arc with a length of 5/2 pi meters is 149.542°\n\nStep-by-step explanation:\n\nHere, the diameter of the circle = 6 m\n\nSo, radius  = D / 2  = 6 / 2  = 3 m\n\nAlso, the length of the arc = (", null, ") meters\n\nPutting π  = 3.14, we get\n\nThe length S of the arc =", null, "or, S  = 7.85 m\n\nLet us assume the arc subtends angle Ф at the center of the circle.\n\n⇒  S  = r Ф\n\nor, Ф =", null, "Now, 1 Radian  = 57.2958 Degrees\n\n⇒ 2. 62 Radian  = 2.61  x  ( 57.2958 Degrees)  =  149.542 °\n\nor, Ф    =  149.542°\n\nHence, the measurement of the angle subtended by an arc with a length of 5/2 pi meters is 149.542°", null, "Arc length = (5/2) * PI meters = 7.853981634 meters\ncircle circumference = 2 * PI * radius = 2 *PI * 3\n18.8495559215\narc length = (7.85381634 / 18.8495559215) * 360 =\n0.4166578975 *360 =\n150 degrees", null, "Arc length is given by:\nC=θ/360πd\nthus plugging in the values we get:\n5/2π=θ/360π*6\ndividing through by 6π we get:\n5/12=θ/360\nmultiplying both sides by 360\nθ=150°", null, "Arc length=5/2 pi\ndiameter=6m\nCircumference=pi x d=pi x 6m= 6pi\n\nCircumference/arc length=6pi/(5/2pie)=360degree/central angle\n\n6pie x central angle=360degree x 5/2pie\n\nCentral angle=360degree x 5/2pie/6pie\n\nCentral angle=360degree x 5/12\n\nCentral angle=150degree", null, "", null, "### Other questions on the subject: Mathematics", null, "Mathematics, 21.06.2019 14:10, hardwick744\nHow many real and imaginary solutions does the equation x^2-3x=-2x-7 have?", null, "Mathematics, 21.06.2019 16:00, vjacksongonzalez\nYou eat 8 strawberries and your friend eats 12 strawberries from a bowl. there are 20 strawberries left. which equation and solution give the original number of strawberries?", null, "Mathematics, 21.06.2019 19:00, NeonPlaySword\nIn he given figure if triangle abe similar to triangle acd show that triangle ade similar to triangle abc", null, "Mathematics, 21.06.2019 20:30, michaelmorin02\nR(1,3) s(3,1) t(5,2) is the traingle of which one scalene, isosceles, or equilititeral\nFor a circle with a diameter of 6 meters, what is the measurement of a central angle (in degrees) su...\n\n### Questions in other subjects:", null, "", null, "Mathematics, 29.09.2019 13:10", null, "", null, "", null, "", null, "", null, "English, 29.09.2019 13:10", null, "", null, "", null, "Questions on the website: 13578005" ]

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[ "", null, "Wednesday, March 24, 2021 3:53:40 PM\n\n# Difference Between Active And Reactive Power Pdf\n\nFile Name: difference between active and reactive power .zip\nSize: 2964Kb\nPublished: 24.03.2021", null, "", null, "The main difference between active and reactive power is that Active Power is actual or real power which is used in the circuit while Reactive power bounce back and forth between load and source which is theoretically useless. The following power triangle shows the relation between Active, Reactive and Apparent Power.\n\nThe required power supply is called the apparent power and is a complex value that can be expressed in a Pythagorean triangle relationship as indicated in the figure below. Apparent power is measured in volt-amperes VA - the AC system voltage multiplied with flowing current.\n\nActive and Reactive Power Introduction to Active and Reactive Power An understanding of the concepts of active and reactive power flow are critical to an understanding of power system dynamics. This chapter first reviews and summarizes the theory related to active and reactive power. Simple equations are then developed for the flow of both active. No reactive power is present; consequently, the angle between active power and complex power will be zero. So in this case, apparent power i.\n\n## Difference Between Active and Reactive Power – Watts vs VA\n\nIt is measured in kilowatt kW or MW. It is the actual outcomes of the electrical system which runs the electric circuits or load. Definition: The power which flows back and forth that means it moves in both the directions in the circuit or reacts upon itself, is called Reactive Power. It has been seen that power is consumed only in resistance. A pure inductor and a pure capacitor do not consume any power since in a half cycle whatever power is received from the source by these components, the same power is returned to the source.\n\nThis power which returns and flows in both the direction in the circuit, is called Reactive power. This reactive power does not perform any useful work in the circuit. In a purely resistive circuit, the current is in phase with the applied voltage, whereas in a purely inductive and capacitive circuit the current is 90 degrees out of phase, i. Hence, from all the above discussion, it is concluded that the current in phase with the voltage produces true or active power , whereas, the current 90 degrees out of phase with the voltage contributes to reactive power in the circuit.\n\nThe current I is divided into two components:. Therefore, the following expression shown below gives the active, reactive and apparent power respectively.\n\nThe current component, which is in phase with the circuit voltage and contributes to the active or true power of the circuit, is called an active component or watt-full component or in-phase component of the current. The current component, which is in quadrature or 90 degrees out of phase to the circuit voltage and contributes to the reactive power of the circuit, is called a reactive component of the current.\n\nThe poles of the rotating magnetic field of stator and rotor interlock each other. Thus, the rotor rotates at the speed of the rotating magnetic field of the stator or we can say the motor run at constant speed. It is very helpfull.", null, "## Definition of Active Power, Reactive Power and Apparent Power / Design of PV Farms\n\nIt is measured in kilowatt kW or MW. It is the actual outcomes of the electrical system which runs the electric circuits or load. Definition: The power which flows back and forth that means it moves in both the directions in the circuit or reacts upon itself, is called Reactive Power. It has been seen that power is consumed only in resistance. A pure inductor and a pure capacitor do not consume any power since in a half cycle whatever power is received from the source by these components, the same power is returned to the source. This power which returns and flows in both the direction in the circuit, is called Reactive power.\n\nThe most significant difference between the active and reactive power is that the active power is the actual power which is dissipated in the circuit. The other differences between the active and reactive power are explained below in the comparison chart. The right-angled triangle shown below shows the relation between the active, reactive and apparent power. Measures the power factor of the circuit. The power which is dissipated or do the useful work in the circuit is known as the active power. It is measured in watts or megawatts.\n\nInstantaneous power in an electric circuit is the rate of flow of energy past a given point of the circuit. In alternating current circuits, energy storage elements such as inductors and capacitors may result in periodic reversals of the direction of energy flow. The portion of power that, averaged over a complete cycle of the AC waveform , results in net transfer of energy in one direction is known as active power more commonly called real power to avoid ambiguity especially in discussions of loads with non-sinusoidal currents. The portion of power due to stored energy, which returns to the source in each cycle, is known as instantaneous reactive power , and its amplitude is the absolute value of reactive power. In a simple alternating current AC circuit consisting of a source and a linear load, both the current and voltage are sinusoidal.", null, "icel3.org › difference-between-active-and-reactive-power.\n\n## True, Reactive, and Apparent Power\n\nMany practical circuits contain a combination of resistive, inductive and capacitive elements. These elements cause the phase shift between the parameters of electrical supply such as voltage and current. Due to the behavior of voltage and currents, especially when subjected to these components, power quantity comes in different forms. In AC circuits, voltage and current amplitudes will change continuously over a time. Since the power is the voltage times the current, it will be maximized when currents and voltages are lined up with each other.\n\nThe main difference between active and reactive power is that Active Power is actual or real power which is used in the circuit while Reactive power bounce back and forth between load and source which is theoretically useless. The following power triangle shows the relation between Active, Reactive and Apparent Power. These all powers only induced in AC circuits when current is leading or lagging behind the voltage i. The average value of active power can be calculated by the following formulas.\n\nPower is the combination of voltage and current in electrical circuits. The abbreviations AC and DC are often used in electrical power systems as alternating current and direct current respectively. Both are different types of current used for the transmission of electrical energy. Active power and reactive power are the two most common terms used to describe the energy flow in electrical power systems.", null, "Если Цифровой крепости суждено стать любимой игрушкой АНБ, Стратмор хотел убедиться, что взломать ее невозможно.\n\n#### What is Reactive Power?\n\nОна никогда раньше не слышала выстрелов, разве что по телевизору, но не сомневалась в том, что это был за звук. Сьюзан словно пронзило током. В панике она сразу же представила себе самое худшее. Ей вспомнились мечты коммандера: черный ход в Цифровую крепость и величайший переворот в разведке, который он должен был вызвать. Она подумала о вирусе в главном банке данных, о его распавшемся браке, вспомнила этот странный кивок головы, которым он ее проводил, и, покачнувшись, ухватилась за перила. Коммандер.\n\nСьюзан стояла прямо и неподвижно, как статуя. Глаза ее были полны слез. - Сьюзан. По ее щеке скатилась слеза. - Что с тобой? - в голосе Стратмора слышалась мольба.\n\nСтратмор придвинулся ближе, держа беретту в вытянутой руке прямо перед. - Как ты узнал про черный ход. - Я же сказал. Я прочитал все, что вы доверили компьютеру. - Это невозможно.\n\nСтратмор задумался. - С какой стати он должен на него смотреть? - спросил. Сьюзан взглянула ему в. - Вы хотите отправить его домой.", null, "Знать ничего не знаю. - Не знаю, о ком вы говорите, - поправил его Беккер, подзывая проходившую мимо официантку." ]

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[ "This is “Market Supply and Market Demand”, section 7.1 from the book Theory and Applications of Microeconomics (v. 1.0). For details on it (including licensing), click here.\n\nHas this book helped you? Consider passing it on:\nCreative Commons supports free culture from music to education. Their licenses helped make this book available to you.\nDonorsChoose.org helps people like you help teachers fund their classroom projects, from art supplies to books to calculators.\n\n## 7.1 Market Supply and Market Demand\n\n### Learning Objectives\n\n1. How is the market demand curve derived?\n2. What is the slope of the market demand curve?\n3. How is the market supply curve derived?\n4. What is the slope of the market supply curve?\n5. What is the equilibrium of a perfectly competitive market?\n\nWe begin the chapter with the individual demand curve—sometimes also called the household demand curve—that is based on an individual’s choice among different goods. (In this chapter, we use the terms individual and household interchangeably.) We show how to build the market demand curve from these individual demand curves. Then we do the same thing for supply, showing how to build a market supply curve from the supply curves of individual firms. Finally, we put them together to obtain the market equilibrium.\n\n## Market Demand\n\nFigure 7.1 \"The Demand Curve of an Individual Household\" is an example of a household’s demand for chocolate bars each month. Taking the price of a chocolate bar as given, as well as its income and all other prices, the household decides how many chocolate bars to buy. Its choice is represented as a point on the household’s demand curve. For example, at \\$5, the household wishes to consume five chocolate bars each month. The remainder of the household income—which is its total income minus the \\$25 it spends on chocolate—is spent on other goods and services. If the price decreases to \\$3, the household buys eight bars every month. In other words, the quantity demanded by the household increases. Equally, if the price of a chocolate bar increases, the quantity demanded decreases. This is the law of demand in operation.\n\nOne way to summarize this behavior is to say that the household compares its marginal valuation from one more chocolate bar to price. The marginal valuation is a measure of how much the household would like one more chocolate bar. The household will keep buying chocolate bars up to the point where\n\nmarginal valuation = price.\n\nToolkit: Section 17.1 \"Individual Demand\"\n\nYou can review the foundations of individual demand and the idea of marginal valuation in the toolkit.\n\nFigure 7.1 The Demand Curve of an Individual Household", null, "The household demand curve shows the quantity of chocolate bars demanded by an individual household at each price. It has a negative slope: higher prices lead people to consume fewer chocolate bars.\n\nTable 7.1 Individual and Market Demand\n\nPrice (\\$) Household 1 Demand Household 2 Demand Market Demand\n1 17 10 27\n3 8 3 11\n5 5 2 7\n7 4 1.5 5.5\n\nIn most markets, many households purchase the good or the service traded. We need to add together all the demand curves of the individual households to obtain the market demand curve. To see how this works, look at Table 7.1 \"Individual and Market Demand\" and Figure 7.2 \"Market Demand\". Suppose that there are two households. Part (a) of Figure 7.2 \"Market Demand\" shows their individual demand curves. Household 1 has the demand curve from Figure 7.1 \"The Demand Curve of an Individual Household\". Household 2 demands fewer chocolate bars at every price. For example, at \\$5, household 2 buys 2 bars per month; at \\$3, it buys 3 bars per month. To get the market demand, we simply add together the demands of the two households at each price. For example, when the price is \\$5, the market demand is 7 chocolate bars (5 demanded by household 1 and 2 demanded by household 2). When the price is \\$3, the market demand is 11 chocolate bars (8 demanded by household 1 and 3 demanded by household 2). When we carry out the same calculation at every price, we get the market demand curve shown in part (b) of Figure 7.2 \"Market Demand\".\n\nToolkit: Section 17.9 \"Supply and Demand\"\n\nYou can review the market demand curve in the toolkit.\n\nFigure 7.2 Market Demand", null, "Market demand is obtained by adding together the individual demands of all the households in the economy.\n\nBecause the individual demand curves are downward sloping, the market demand curve is also downward sloping: the law of demand carries across to the market demand curve. As the price decreases, each household chooses to buy more of the product. Thus the quantity demanded increases as the price decreases. Although we used two households in this example, the same idea applies if there are 200 households or 20,000 households. In principle, we could add together the quantities demanded at each price and arrive at a market demand curve.\n\nThere is a second reason why demand curves slope down when we combine individual demand curves into a market demand curve. Think about the situation where each household has a unit demand curve: that is, each individual buys at most one unit of the product. As the price decreases, the number of individuals electing to buy increases, so the market demand curve slopes down.See Chapter 3 \"Everyday Decisions\" and Chapter 5 \"eBay and craigslist\" for discussions of unit demand. In general, both mechanisms come into play.\n\n• As price decreases, some households decide to enter the market; that is, these households buy some positive quantity other than zero.\n• As price decreases, households increase the quantity that they wish to purchase.\n\nWhen the price decreases, there are more buyers, and each buyer buys more.\n\n## Market Supply\n\nIn a competitive marketA market that satisfies two conditions: (1) there are many buyers and sellers, and (2) the goods the sellers produce are perfect substitutes., a single firm is only one of the many sellers producing and selling exactly the same product. The demand curve facing a firm exhibits perfectly elastic demand, which means that it sets its price equal to the price prevailing in the market, and it chooses its output such that this price equals its marginal costThe extra cost of producing an additional unit of output, which is equal to the change in cost divided by the change in quantity. of production.At the end of Chapter 6 \"Where Do Prices Come From?\", we derive the supply curve of a firm in a competitive market. If it were to try to set a higher price, it could not sell any output at all. If it were to set a lower price, it would be throwing away profits. Thus, for a competitive firm, the quantity produced satisfies this condition:\n\nprice = marginal cost.\n\nToolkit: Section 17.2 \"Elasticity\"\n\nFor more information on elasticity, see the toolkit.\n\nWe typically expect that marginal cost will increase as a firm produces more output. Marginal cost is the cost of producing one extra unit of output. The cost of producing an additional unit of output generally increases as firms produce a larger and larger quantity. In part, this is because firms start to hit constraints in their capacities to produce more product. For example, a factory might be able to produce more output only by running extra shifts at night, which require paying higher wages.\n\nIf marginal cost is increasing, then we know the following:\n\n• Given a price, there is only one level of output such that price equals marginal cost.\n• As the price increases, a firm will produce more.\n\nIndeed, the supply curve of an individual firm is the same as its marginal cost curve.\n\nFigure 7.3 \"The Supply Curve of an Individual Firm\" illustrates the supply curve for a firm. A firm supplies seven chocolate bars at \\$3 and eight chocolate bars at \\$5. From this we can deduce that the marginal cost of producing the seventh chocolate bar is \\$3. Similarly, the marginal cost of producing the eighth chocolate bar is \\$5.\n\nFigure 7.3 The Supply Curve of an Individual Firm", null, "A firm’s supply curve, which is the same as its marginal cost curve, shows the quantity of chocolate bars it is willing to supply at each price.\n\nJust as the market demand curve tells us the total amount demanded at each price, the market supply curve tells us the total amount supplied at each price. It is obtained analogously to the market demand curve: at each price we add together the quantity supplied by each firm to obtain the total quantity supplied at that price. If we perform this calculation for every price, then we get the market supply curve. Figure 7.4 \"Market Supply\" shows an example with two firms. At \\$3, firm 1 produces 7 bars, and firm 2 produces 3 bars. Thus the total supply at this price is 10 chocolate bars. At \\$5, firm 1 produces 8 bars, and firm 2 produces 5 bars. Thus the total supply at this price is 13 chocolate bars.\n\nThe market supply curve is increasing in price. As price increases, each firm in the market finds it profitable to increase output to ensure that price equals marginal cost. Moreover, as price increases, firms who choose not to produce and sell a product may be induced to enter into the market.A similar idea is in Chapter 5 \"eBay and craigslist\", where we show how to add together unit supply curves to obtain a market supply curve.\n\nFigure 7.4 Market Supply", null, "Market supply is obtained by adding together the individual supplies of all the firms in the economy.\n\nIn general, both mechanisms come into play. The market supply curve slopes up for two reasons:\n\n1. As the price increases, more firms decide to enter the market—that is, these firms produce some positive quantity other than zero.\n2. As the price increases, firms increase the quantity that they wish to produce.\n\nWhen the price increases, there are more firms in the market, and each firm produces more.\n\n## Market Equilibrium\n\nIn a perfectly competitive market, we combine the market demand and supply curves to obtain the supply-and-demand framework shown in Figure 7.5 \"Market Equilibrium\". The point where the curves cross is the market equilibrium.The definition of equilibrium is also presented in Chapter 5 \"eBay and craigslist\". At this point, there is a perfect match between the amount that buyers want to buy and the amount that sellers want to sell. The term equilibrium refers to the balancing of the forces of supply and demand in the market. At the equilibrium price, the suppliers of a good can sell as much as they wish, and demanders of a good can buy as much of the good as they wish. There are no disappointed buyers or sellers.\n\nToolkit: Section 17.9 \"Supply and Demand\"\n\nYou can review the definition and meaning of equilibrium in the supply-and-demand framework in the toolkit.\n\nFigure 7.5 Market Equilibrium", null, "In a competitive market, the equilibrium price and the equilibrium quantity are determined by the intersection of the supply and demand curves.\n\nBecause the demand curve has a negative slope and the supply curve has a positive slope, supply and demand will cross once. Both the equilibrium price and the equilibrium quantity will be positive. (More precisely, this is true as long as the vertical intercept of the demand curve is larger than the vertical intercept of the supply curve. If this is not the case, then the most that any buyer is willing to pay is less than the least any seller is willing to accept and there is no trade in the market.)\n\nTable 7.2 Market Equilibrium: An Example\n\nPrice (\\$) Market Supply Market Demand\n1 5 95\n5 25 75\n10 50 50\n20 100 0\n\nTable 7.2 \"Market Equilibrium: An Example\" shows an example of market equilibrium with market supply and market demand at four different prices. The equilibrium occurs at \\$10 and a quantity of 50 units. The table is based on the following equations:\n\nmarket demand = 100 − 5 × price\n\nand\n\nmarket supply = 5 × price.\n\nEquations such as these and diagrams such as Figure 7.5 \"Market Equilibrium\" are useful to economists who want to understand how the market works. Keep in mind, though, that firms and households in the market do not need any of this information. This is one of the beauties of the market. An individual firm or household needs to know only the price that is prevailing in the market.\n\n## Reaching the Market Equilibrium\n\nEconomists typically believe that a perfectly competitive market is likely to reach equilibrium for several reasons.\n\n• If the prevailing price is different from the equilibrium price, then there will be an imbalance between demand and supply, which gives buyers and sellers an incentive to behave differently. For example, if the prevailing price is less than the equilibrium price, demand will exceed supply. Disappointed buyers might start bidding the price up, or sellers might realize they could charge a higher price. The opposite is true if the prevailing price is too high: suppliers might be tempted to try decreasing prices, and buyers might look for better deals. These are informal stories because the supply and demand curves are based on the idea that firms and consumers take prices as given. Still, the idea that there will be pressure on prices away from equilibrium is a plausible one.\n• There is strong support for market predictions in the evidence from experimental markets.In Chapter 5 \"eBay and craigslist\", we explain that a double oral auction, in which buyers and sellers meet individually and bargain over prices, typically yields results very close to the market outcome in Figure 7.5 \"Market Equilibrium\".\n• The supply-and-demand framework generally provides reliable predictions about the movement of prices.\n\n### Key Takeaways\n\n• The market demand curve is obtained by adding together the demand curves of the individual households in an economy.\n• As the price increases, household demand decreases, so market demand is downward sloping.\n• The market supply curve is obtained by adding together the individual supply curves of all firms in an economy.\n• As the price increases, the quantity supplied by every firm increases, so market supply is upward sloping.\n• A perfectly competitive market is in equilibrium at the price where demand equals supply.\n\n### Checking Your Understanding\n\n1. In Table 7.2 \"Market Equilibrium: An Example\", market supply was equal to 5 × price. Suppose instead that market supply = 15 × price. Would the equilibrium price still be \\$10? If not, construct a new column in the table and find the new equilibrium price.\n2. Explain why supply and demand cross only once. Do they always cross at a positive price?" ]

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[ "To get the best deal on Tutoring, call 1-855-666-7440 (Toll Free)", null, "Top\n\n# Solving Equations with Variables on Both Sides\n\nAn equation is an expression with \"equal\" sign containing variables and constants joined together with the help of arithmetic operators. The variables in an equation can occur either on one side or on both the sides of \"equal\" sign.\nFor example: 3x - 4 = 5x and 7(y - 2) = 1 + 2y etc.\nFor solving this type of equations, we need to take the variables on one side.\n\nThere are certain rules for solving equations with variables on both the sides which are as follows:\n\n• We can add or subtract a number or a variable on both the sides.\n• We can multiply or divide all the terms on both the sides by a number or a variable.\n• While changing the side of a variable or a constant, its sign is reversed. That means, positive quantity becomes negative and negative quantity become positive.\n• When we change the side, multiplication operation becomes division and division operation becomes multiplication.\n\nLet us look at few examples:\n\nExample 1:\n\nSolve 5x - 8 = 3 + 6x\n\nSolution:\n\n5x - 8 = 3 + 6x\n5x - 6x = 3 + 8\n- x = 11\nx = - 11\n\nExample 2:\n\nSolve 3(t - 1) = 2t + 5\n\nSolution:\n\n3(t - 1) = 2t + 5\n3t - 3 = 2t + 5\n3t - 2t = 5 + 3\nt = 8\n\nExample 3:\n\n$\\frac{x+1}{2}=\\frac{x-2}{3}$\n\nSolution:\n\n$\\frac{x+1}{2}=\\frac{x-2}{3}$\n\nBy cross multiplication, we get\n3(x + 1) = 2 (x - 2)\n3x + 3 = 2x - 4\n3x - 2x = - 4 - 3\nx = - 7\n\n Related Calculators Solving Equations with Variables on both Sides Calculator Solve for a Variable Calculator Variable Equation Solver Linear Equations in Two Variables Calculator\n\n*AP and SAT are registered trademarks of the College Board." ]

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[ "# Hund's rule for comparing term symbol energies in excited state\n\nThis question was asked in an exam:\n\nThe lowest energy electronic state for excited state carbon atom is,\na. $$^1D_2$$\nb. $$^3D_1$$\nc. $$^3D_3$$\nd. $$^3D_2$$\n\nAlthough we have been taught Hund's rule for determining the relative energies for various term symbols in the ground state, I haven't read anything about the rules for excited state anywhere. I couldn't not find it in the book I use i.e. Atkins, Physical Chemistry.\n\nI am really confused about how to solve this.\n\n• summarising (a) lowest multiplicity lowest energy (b) with a given multiplicity largest L lowest energy (c) Atoms with less than 1/2 filled shell lowest J lowest energy. Does not the excited state just means an extra electron 'upstairs' such as $s^2p^2\\to s^1p^3$ – porphyrin May 3 at 8:55\n• But on subwhat shell are we going to apply the less than half or more than half filled subshell rule? – Kashish May 3 at 14:00\n• Hund's rules strictly assume LS coupling and application to excited states is not reliable. Of course you can try to apply them in a similar manner as for the ground state, but treat the outcome with caution – Paul Jun 19 at 8:19" ]

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[ "# Workshop Classification and set theory\n\nMarch 21 - 24, 2016\n\nProgram Committee: Aaron Tikuisis, Asger Törnquist\n\nWorkshop schedule\n\n### Monday March 21\n\n9:30-10:30 Damien Gaboriau\n\nTitle: Direct products and approximations in measured orbit equivalence\n\nAbstract: Hyperfiniteness (which plays a central role in orbit equivalence), consists in approximability by finite subrelations.\n\nIn a joint work with Robin Tucker-Drob we initiate a general study of the notion of \\textit{approximation} for countable standard probability measure preserving  equivalence \\ relations. We shall concentrate on non-approximability for actions of non-amenable direct products.\n\n11:00-12:00 Narutaka Ozawa\n\nTitle: C*-norms on B(\\ell_2) \\otimes B(\\ell_2) and related topics\n\nAbstract: M. Junge and G. Pisier [GAFA 1995] have shown that there are at least two distinct C*-norms on B(\\ell_2) \\otimes B(\\ell_2), which disproved E. Kirchberg's conjecture that the minimal tensor norm on tensor products of C*-algebras can be characterized by the injectivity property. Actually how many C*-norms exist on B(\\ell_2) \\otimes B(\\ell_2) remains an open problem since then.\n\nThis talk is based on the joint work with G. Pisier [arXiv:1404.7088] which shows that there are at least c distinct C*-norms, where c denotes the cardinal of the continuum. It is plausible that there are exactly 2^c distinct C*-norms, and probably logicians can help us.\n\n14:00-15:00 Todor Tsankov\n\nTitle: On metrizable universal minimal flows\n\nAbstract: To every topological group, one can associate a unique universal minimal flow (UMF): a flow that maps onto every minimal flow of the group. For some groups (for example, the locally compact ones), this flow is not metrizable and does not admit a concrete description. However, for many \"large\" Polish groups, the UMF is metrizable, can be computed, and carries interesting combinatorial information. The talk will concentrate on some new results that give a characterization of metrizable UMFs of Polish groups. It is based on two papers, one joint with I. Ben Yaacov and J. Melleray, and the other with J. Melleray and L. Nguyen Van Thé.\n\n15:15-15:45 Asger Törnquist\n\nTitle: The descriptive view of classification in operator algebras\n\nAbstract: I will give an overview of the achievements of the program to study the complexity of classification problems in operator algebras from the descriptive set-theoretic point of view. I will also discuss some of the remaining open problems.\n\n16:15-16:45 George Elliott\n\nTitle: A (one-eyed) bird's eye view of classification theory\n\nAbstract: A brief overview is given of two opposite extremes of functorial classification theory---the general abstract functor that covers all separable C*-algebras, and the pared-down, concretized, version of this that covers the suitably well-behaved unital simple case.\n\n17:00-17:30 Eusebio Gardella\n\nAbstract: It is a classical result due to Glimm that UHF (uniformly hyperfinite) C*-algebras are classified, by K-theory. In particular, they are classified by countable structures. A UHF operator algebra is, roughly speaking, an operator algebra obtained as an infinite tensor product of matrix algebras, where these matrix algebras are given a norm making them into an operator algebra (but not necessarily a C*-algebra). Such algebras have been studied by N. C. Phillips, who, among other things, exhibited uncountably many non-isomorphic UHF operator algebras. In this talk, which is based on joint work with Martino Lupini, we will show that UHF operator algebras are not classifiable by countable structures (even if they are assumed to have a very special and tractable form).\n\nOur proof relies on Borel complexity theory, and in particular Hjorth's theory of turbulence. Our methods also allow us to treat the case of (non-spatial) UHF $L^p$- operator algebras, for $1\\leq p<\\infty$. These are defined by letting the matrix algebras act on an $L^p$-space, and giving them the corresponding operator norm.\n\n### Tuesday March 22\n\n9:30-10:30 David Kyed\n\nTitle: Topologizing Lie algebra cohomology\n\nAbstract: I will explain how Lie algebra cohomology can be topologized in a way such that classical results, such as the van Est isomorphism, extend to the augmented context. Along the way I will define (most of) the objects involved, and the talk does not require prior knowledge about Lie algebra cohomology.\n\n11:00-12:00 Mikael Rørdam\n\nTitle: Just infinite C*-algebras\n\nAbstract: There is a well-established notion of just infinite groups, i.e., infinite groups for which all proper quotients are finite. The residually finite just infinite groups are particularly interesting. They are either branch groups (e.g., Grigorchuk's group of intermediate growth) or hereditarily just infinite groups (eg. Z, the infinite dihedral group, and SL_n(Z)). It is natural to consider the analogous notion for C*-algebras, whereby a C*-algebra is just infinite if it is infinite dimensional and all its proper quotients are finite dimensional. The study of these C*-algebras was motivated by a question of Grigorchuk if the group C*-algebra associated with his group might have this property. We give a classification of just infinite C*-algebras in terms of their primitive ideal space. We will discuss examples and properties of the residually finite dimensional just infinite C*-algebras; and we will also discuss the question of Grigorchuk.\n\nThis is joint work with R. Grigorchuk and M. Musat.\n\n14:00-15:00 David Kerr\n\nTitle: Actions of amenable groups on the Cantor set\n\nAbstract:  This talk will be an invitation to study the decriptive set theory of the space of actions of a countable amenable group on the Cantor set. The motivation comes from the effort to classify the crossed products of such actions, and specifically to determine when these crossed products are Z-stable, an issue which I will discuss in detail.\n\n15:15-15:45 David Schrittesser\n\nTitle: Absoluteness\n\nAbstract: Suppose brilliant mathematician Klaupaucius, despite all his efforts, finds he is unable to come up with a proof for his favorite conjecture and starts to suspect it might be logically independent of the axioms. So he consults his friend Trurl the logician, who, after asking a number of questions, declares that the statement in question is absolute (due to its simple logical structure) and hence cannot be shown to be independent with the standard methods.\n\nIn my talk I will give a short introduction to the notion of \"absoluteness\" alluded to in this little story.\n\n16:15-16:45 Hannes Thiel\n\nTitle: Cuntz semigroups of ultraproducts\n\nAbstract: The Cuntz semigroup is an invariant for C*-algebras that is constructed analogously to K_0-theory by using positive elements in place of projections. The Cuntz semigroup of a C*-algebra is an element in a category Cu of abstract Cuntz semigroups, introduced by Coward-Elliott-Ivanescu. We show that the category Cu admits products and ultraproducts and that the functor from C*-algebras to Cu preserves these. In other words, the Cuntz semigroup of an (ultra)product of C*-algebras is the (ultra)product of the Cuntz semigroups of the C*-algebras.\n\nThis is joint work with Ramon Antoine and Francesc Perera.\n\n17:00-17:30 Jiangchao Wu\n\nTitle: Furstenberg's x2, x3 problem and unitary representations of a metabelian group\n\nAbstract: In 1967, Furstenberg raised the question whether the (normalized) Lebesgue measure is the only non-atomic probabilistic measure on the unit circle that is ergodic under the x2 and x3 endomorphisms. Since then this problem has attracted much interest, but despite many impressive results, it stays open in general. I will introduce this problem and propose an operator algebraic approach that involves the group C*-algebra of a semidirect product of Z[1/6] by Z^2.\n\nThis is joint work with Huichi Huang.\n\n### Wednesday March 23\n\n9:30-10:30 Julien Melleray\n\nTitle: Sets of invariant measures of minimal homeomorphisms of a Cantor space\n\nAbstract: Given a compact set K of probability measures on a Cantor space X, one might ask when there exists a minimal homeomorphism g of X such that K is the set of all g-invariant measures. I will present an answer to that question, discuss an open problem related to that answer, as well as some questions about full groups of minimal homeomorphisms. This is joint work with Tomas Ibarlucia.\n\n10:45-11:45 Anush Tserunyan\n\nTitle: Finite index pairs of equivalence relations and treeability\n\nAbstract: I will give an overview of an open question of Jackson--Kechris--Louveau from the late 90s as to whether finite index extensions of countable treeable equivalence relations are themselves treeable. We will consider this question in both Borel and measurable settings and discuss relevant results and possible approaches in each.\n\n12:00-12:30 Alessandro  Vignati\n\nTitle: Set theory and automorphisms of C*-algebras\n\nAbstract: After the key results on the structure of the automorphisms group of the Calkin algebra, Farah and Coskey conjectured that under the Continuum Hypothesis there are wild automorphisms of corona algebras, while under the Proper Forcing Axiom the situation is conjectured to be quite rigid. McKenney and I recently verified the conjecture is presence of forcing axioms for a large class of algebras, although very much remains open. We present the current situation and then focus on some of the difficulties stopping from expanding our results, with, possibly exploring connections to perturbation theory.\n\n### Thursday March 24\n\n09:30-10:30 Chris Phillips\n\nTitle: Simple nuclear C*-algebras not equivariantly isomorphic to their opposites\n\nAbstract: We exhibit examples of simple separable nuclear C*-algebras, along with actions of the circle group and outer actions of the integers, which are not equivariantly isomorphic to their opposite algebras. In fact, the fixed point subalgebras are not isomorphic to their opposites. The C*-algebras we exhibit are well behaved from the perspective of structure and classification of nuclear C*-algebras: they are unital C*-algebras in the UCT class, with finite nuclear dimension. One is an AH-algebra with unique tracial state and absorbs the CAR algebra tensorially. The other is a Kirchberg algebra.\n\nThis is joint work with Marius Dadarlat and Ilan Hirshberg.\n\n11:00-12:00 Hiroshi Ando\n\nTitle: Descriptive analysis of self-adjoint operators and the Weyl-von Neumann equivalence relation\n\nAbstract: In a previous work, we studied the space $SA(H)$ of all (possibly unbounded) self-adjoint operators on a separable Hilbert space $H$ and in particular showed that the relation of unitary equivalence modulo compacts (Weyl-von Neumann equivalence) is unclassifiable by countable structures. Moreover, for a closed set $F$ of the real line, there may or may not be self-adjoint operators $A,B$ for which the essential spectrum coincide with $F$ but $A,B$ are not Weyl-von Neumann equivalent.\n\nIn this talk, we characterize which $F$ satisfies the conclusion of Weyl-von Neumann theorem. We also discuss the Borel complexity of Schatten class perturbations of unbounded self-adjoint operators, and see that they are all Borel reducible to the universal essentially K_{\\ sigma} equivalence relation $\\ell^{\\infty}$.\n\nThis is joint work with Yasumichi Matsuzawa (Shinshu University).\n\n14:00-15:00 Thomas Sinclair\n\nTitle: Model theory of C*-algebras as operator systems\n\nAbstract: I will discuss various aspects of the model-theoretic structure of the class of C*-algebras within the class of operator systems. This is based on joint work with Isaac Goldbring." ]

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[ "## The stationary thermistor problem with a current limiting device.(English)Zbl 0745.35042\n\n(Author’s summary.) The electrical heating of a conductor connected with a current limiting device of total resistance $$R$$ is studied under mixed boundary conditions. A theorem of existence is given using a transformation which permits the reduction of the nonlinear system governing the problem to the classical mixed boundary value problem for the Laplace equation.\n\n### MSC:\n\n 35Q60 PDEs in connection with optics and electromagnetic theory 78A30 Electro- and magnetostatics 35J05 Laplace operator, Helmholtz equation (reduced wave equation), Poisson equation 74F15 Electromagnetic effects in solid mechanics\n\n### Keywords:\n\nexistence; mixed boundary value problem; Laplace equation\nFull Text:\n\n### References:\n\n DOI: 10.1002/andp.19003060211 · JFM 31.0808.01 Cimatti, Quart. Appl. Math. XLVII pp 117– (1989) · Zbl 0694.35137 DOI: 10.1002/andp.19003060107 · JFM 31.0807.02 Protter, Maximum principle in partial differential equations (1967)\nThis reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching." ]

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[ "# Commented consensus/merkle.cpp Merkle root on Bitcoin Core Source Code\n\nIn this article we take a look at one of the most important components of block validation and overall blockchain integrity: the Merkle Root derivation process.\n\nWhile it's one of the most critical components in all of Bitcoin Core, its concept and implementation are very simple and straightforward.\n\nThe `consensus/merkle.h` header consists solely of 3 declarations, so let's jump straight into `consensus/merkle.cpp` and see how these functions are implemented.\n\n## `consensus/merkle.cpp`\n\nAt the top of merkle.cpp we find an interesting comment that's worth the read. I've taken a screenshot to preserve the formatting and the original tree drawings.\n\nFirst function we encounter is the Merkle Root computation subroutine.\n\n`uint256 ComputeMerkleRoot(std::vector<uint256> hashes, bool* mutated)`\n\nIt takes a `hashes` vector containing the input hashes and returns a `uint256` digest for the merkle root. The purpose of the `mutated` argument is explained in the vulnerability screenshot posted above: if this tree is ambiguous and can generate a collision then set `*mutated` to true to let the caller know.\n\nI'll comment the `ComputeMerkleRoot` from within the code itself. As always, my comments begin with >, original comments begin with // or /*\n\n``````> Note that a null mutated pointer will ignore mutated check.\nuint256 ComputeMerkleRoot(std::vector<uint256> hashes, bool* mutated) {\n> Assume no mutation. Tree collisions are the exception not the rule.\nbool mutation = false;\n> When we reach the end of computation, hashes should've been reduced\n> to a single hash containing the Merkle Root.\nwhile (hashes.size() > 1) {\n> Check that the mutated pointer is not null\nif (mutated) {\n> For every hash still in hashes, if this hash collides with the next\n> then set mutation to true. This loop repeats with every iteration of\n> the while loop so it checks the tree as it is compiled to the root.\nfor (size_t pos = 0; pos + 1 < hashes.size(); pos += 2) {\nif (hashes[pos] == hashes[pos + 1]) mutation = true;\n}\n}\n> If the size of hashes is odd then push the current hash into the back.\n> In the Merkle root algorithm when a leaf is dangling by itself then\n> it is hashed against itself. Thus the last leaf is pushed back here to\n> generate a tree with an even number of leaves.\nif (hashes.size() & 1) {\nhashes.push_back(hashes.back());\n}\n> Call SHA256D64 from crypto/sha256.cpp passing it a pointer for output,\n> a pointer for input and the size of the input.\nSHA256D64(hashes.begin(), hashes.begin(), hashes.size() / 2);\n> Once SHA256D64 is done we have digested half the tree so we reduce the size\n> by 2.\nhashes.resize(hashes.size() / 2);\n}\n> Copy the value of local variable mutation to *mutated parameter.\nif (mutated) *mutated = mutation;\n> If nothing was processed, return an empty uint256 blob\nif (hashes.size() == 0) return uint256();\n> Otherwise return the single value contained in hashes.\nreturn hashes;\n}``````\n\nNow we have a utility function that allows us to pass in a block to compute its Merkle root.\n\n``````uint256 BlockMerkleRoot(const CBlock& block, bool* mutated)\n{\nstd::vector<uint256> leaves;\nleaves.resize(block.vtx.size());\nfor (size_t s = 0; s < block.vtx.size(); s++) {\nleaves[s] = block.vtx[s]->GetHash();\n}\nreturn ComputeMerkleRoot(std::move(leaves), mutated);\n}``````\n\nIt declares a `leaves` `std::vector`, resizes `leaves` to the size of the block's number of transactions and copies each TX into `leaves`. Then it calls `ComputeMerkleRoot` to compute the Merkle root on the just assembled vector.\n\nNext up, another utility function to compute the Merkle root of the transaction witness hashes.\n\n``````uint256 BlockWitnessMerkleRoot(const CBlock& block, bool* mutated)\n{\nstd::vector<uint256> leaves;\nleaves.resize(block.vtx.size());\nleaves.SetNull(); // The witness hash of the coinbase is 0.\nfor (size_t s = 1; s < block.vtx.size(); s++) {\nleaves[s] = block.vtx[s]->GetWitnessHash();\n}\nreturn ComputeMerkleRoot(std::move(leaves), mutated);\n}``````\n\nIt works exactly the same way as `BlockMerkleRoot` except it calls `GetWitnessHash` instead of `GetHash` on each transaction.\n\nReturn to commented Bitcoin Core sources index." ]

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[ "", null, "", null, "# RS Aggarwal Solutions for Class 10 Maths Chapter 14 Height and Distance\n\nHeight is the measurement of an object in the vertical direction, and distance is the measurement of an object from a particular point in the horizontal direction. So if we know the height of the object and the linear distance, we can easily find out the angle by the trigonometric formula. It is given by tan = Height / distance. Height and Distance: One of the main application of trigonometry is to find the distance between two or more than two places or to find the height of the object or the angle subtended by any object at a given point without actually measuring the distance or heights or angles. Trigonometry is useful to astronomers, navigators, architects and surveyors etc. The angle of elevation of the sun that is better known as altitude angle can also be found by the same method. In this chapter from RS Aggarwal students can get to learn some problems on heights and distances. Problems should not involve more than two right triangles. Angles of elevation/depression should be only 35 degree, 45 degree and 60 degree." ]

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[ "Power is an electrical amount that is calculated in watts and is the velocity at which power is either being absorbed or fashioned by a circuit. We recognize that light bulbs and heaters attract energy and that the superior their price in watts the more force they will drink. Likewise, sequence and creator manufacture energy, and the superior their electrical rating the more influence they can carry to the consignment. The unit of electrical influence is the watt with its representation being a large letter “P” representing unvarying DC power or a miniature letter “p” indicating a time-varying AC supremacy.", null, "Electrical power is associated with energy which is the ability to do a job. It can also be definite as the pace at which force is relocate. If one joule of employment is either absorbed or carry at an even rate of one second, then the equivalent power will be correspondent to one watt so power, P can be definite as 1Joule/sec = 1Watt. Then we container says that one watt is equal to one joule per the following and electrical control should be defined as the speed of doing employment or the convey to Houston Electricity Rates .\n\nRegularly, we can classify power as being watts apiece second or joules. So if the power is measured in kilowatts (thousands of watts) and the time is measure in hours, then the unit of electrical energy is the kilowatt-hour, (kWh) and 1 kWh is the amount of electricity used by a device rated at 1000 watts in one hour.\n\nThen there are three probable formulas for manipulative electrical power in a track. If the designed power is optimistic, (+P) then the course or constituent attracts the power. But if the designed power is unhelpful, (-P) the course or component transport power in other terms it is a foundation of energy.\n\nWhat is the power rating\n\nElectrical machinery is given a ‘’power mark’’ in watts that designate the greatest rate at which the module covers the electrical energy into another figure of energy such as temperature, light, or activity. Electrical motors and other electrical systems have an effectiveness rating definite as the proportion of power transformed into occupation to the total power obsessive by the device. Efficiency is uttered as a decimal fraction but is normally defined as a fraction value such as 85% well-organized. So we container define the good organization as being identical to power output separated by power participation x 100%.\n\nThe good organization of an electrical mechanism or motor will forever be less than one (100%) due to electrical and automatic losses. If an electrical device has a competence rating of 85% then simply 85% of the effort power is misshapen into mechanical employment the other 15% is lost in temperature or other losses.\n\nA domestic electrical machine such as cleanses machines, driers, fridges, and freezers also have power good organization ratings that designate their energy procedure and cost. These ratings are agreed as ‘’A’’ for resourceful and ‘’G’’ for less proficient. So remember, the more vigor efficient is the tool, the less energy it will devour and the more money we will keep as well as creature helpful to the feeling." ]

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[ "# Thick Line segment\n\nHow is this 'wy' equation for thick line segment derived?", null, "", null, "The main part of it is simply Pythagoras's Theorem. The square root gives the length of the central line segment (which is the hypotenuse of a triangle formed by the change in x and change in y). The ratio between the hypotenuse and the change in x is the same as the ratio between the line width and the line width in y (they are similar triangles). Dividing by two is because $w_y$ is the half-width, not the full width. The -1 doesn't appear to make sense in a continuous context, so I assume it's to make sure the number of pixels is rounded down as part of the division. This prevents the line from jumping in thickness when it crosses a pixel boundary." ]

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[ "", null, "Statistics - Maple Programming Help\n\nHome : Support : Online Help : Graphics : Statistics : Statistics/BoxPlot\n\nStatistics\n\n BoxPlot\n create box plots from data\n\n Calling Sequence BoxPlot(X, options, plotoptions) BoxPlot['interactive'](X)\n\nParameters\n\n X - data options - (optional) equation(s) of the form option=value where option is one of datasetlabels, color, deciles, distance, mean, notched, offset, orientation, outliers, or width; specify options for generating the box plot plotoptions - options to be passed to the plots[display] command\n\nDescription\n\n • The BoxPlot command generates a box plot for the specified data.\n • The first parameter X is either a single data sample - given as e.g. a Vector - or a list of data samples. Note that the individual samples may be of variable size.\n • If the ['interactive'] option is used, then a dialog box appears that allows for customized creation of the plot.\n\nOptions\n\n The options argument can contain one or more of the options shown below. All unrecognized options will be passed to the plots[display] command. See plot/options for details.\n • datasetlabels=default or list\n Data set labels for the individual boxes. The labels appear along the axes.  By default, the labels are set to 1, 2, 3, etc.\n • color=name, list, or range\n This option specifies colors for the individual data sets. When a list of colors is given, each of the boxes is colored with the corresponding color in the list. If a range of colors is given, the colors are generated by selecting an appropriate number of equally spaced points in the corresponding hue range.\n • deciles=truefalse\n If this option is set to true then the deciles are included in the plot. Deciles are represented by points with symbol type set to box. The default value is true.\n • offset=realcons\n Initial offset along the x-axis. The default value is 0.\n Note: By default, the view wraps tightly around all visible plot objects and the horizontal axis is marked by data set labels, not regular coordinates, so this option will have no (visual) effect. It is meant for the case where this plot is combined with other plot elements.\n • distance=nonnegative\n This option controls the distance between the boxes. The default value is 0.25.\n • width=realcons\n This option controls the width of the boxes. The default value is 0.75.\n The following plot illustrates how the options offset, distance, and width are interpreted.", null, "– Note the lengths of the arrows labeled \"offset\", \"width\", and \"distance\" correspond to values for the offset, width, and distance options respectively.\n • mean=true or false\n If this option is set to true then the mean is included in the plot. The default value is true.\n • notched=true or false\n Draws notches on box-and-whisker plots. The default is false.\n • orientation=horizontal or vertical\n Indicate the orientation of the box plots. The default is vertical.\n • outliers=true or false\n Outliers are points that are farther than 3/2 times the interquartile range away from the upper and lower quartiles. If this option is set to false, then the outlying points are not included in the plot. In this case, the whiskers are extended past outliers to the maximum and minimum points.\n\nNotes\n\n • Note that the labels for the data sets are placed on the axes, and should not be confused for coordinates.\n\nExamples\n\n > $\\mathrm{with}\\left(\\mathrm{Statistics}\\right):$\n > $A≔\\left[\\mathrm{seq}\\left(\\mathrm{Sample}\\left(\\mathrm{Normal}\\left(\\mathrm{ln}\\left(i\\right),3\\right),100\\right),i=1..20\\right)\\right]:$\n > $\\mathrm{BoxPlot}\\left(A,\\mathrm{title}=\"Box Plot\",\\mathrm{deciles}=\\mathrm{false}\\right)$", null, "The commands to create the plot from the Plotting Guide are\n\n > $A≔\\left[\\mathrm{seq}\\left(\\mathrm{Sample}\\left(\\mathrm{Normal}\\left(\\mathrm{ln}\\left(i\\right),3\\right),10\\right),i=1..3\\right)\\right]:$\n > $\\mathrm{BoxPlot}\\left(A,\\mathrm{color}=\\left[\"Niagara Green\",\"Niagara Red\",\"Niagara Blue\"\\right],\\mathrm{deciles}=\\mathrm{false}\\right)$", null, "The BoxPlot command also accepts a Matrix. The columns are understood as individual data samples.\n\n > $M≔{\\mathrm{Matrix}\\left(A,\\mathrm{scan}=\\mathrm{columns}\\right)}^{\\mathrm{%T}}$\n ${M}{≔}\\left[\\begin{array}{ccc}{-6.15026459336058}& {3.92371013101162}& {4.18656682418225}\\\\ {-4.75323429938184}& {-2.37726783252569}& {1.80091744993437}\\\\ {0.437678454613428}& {3.85224337559343}& {-2.14973018520115}\\\\ {2.30262290447149}& {3.37829880614974}& {3.56948759515931}\\\\ {3.98256142065323}& {7.66312008588142}& {-3.07100377513460}\\\\ {-1.16195228843286}& {-1.17658650278167}& {-2.68832699450844}\\\\ {-0.801278731828643}& {1.40148133239068}& {3.78537967747607}\\\\ {-0.896338423031310}& {4.69554717110909}& {1.06825911132501}\\\\ {5.48507960338911}& {1.74466378733016}& {3.37507641016678}\\\\ {1.06042255049519}& {0.0600928337961485}& {3.55227242615577}\\end{array}\\right]$ (1)\n\nPlot options such as title are passed to the plots:-display command:\n\n > $\\mathrm{BoxPlot}\\left(M,\\mathrm{color}=\"Niagara Green\",\\mathrm{deciles}=\\mathrm{false},\\mathrm{title}=\"Box Plots\"\\right)$", null, "Compatibility\n\n • The Statistics[BoxPlot] command was updated in Maple 2017.\n • The offset option was updated in Maple 2017." ]

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[ "", null, "7 digit unique random number Oracle Community Random Numbers in Java. The easiest way to initialize a random number generator is to use the parameterless constructor, for example. Random generator = new\n\n## Generating random/unique strings numbers? BMC Communities\n\nGenerate 6-Digit Random Number (Java in General forum at. How to generate a random alpha-numeric string? for session identifiers should come from a random number generator for example. A predictable random number, Sometimes random numbers to be picked need to be unique. Use Java to generate unique random numbers. Generating Unique Random to Make a Random Number Generator..\n\nHow to generate Unique ID using Java Programming? including sample main() to generate 100 To use a Random number generator a \"seed\" value has to be given to I am talking about a level that I just finished 1 codeacademy free trial in Python and one in Java.. unique random numbers. generate a random number; Try\n\nRandom Numbers in Java. The easiest way to initialize a random number generator is to use the parameterless constructor, for example. Random generator = new This page allows you to generate random sets of integers using true This generator guarantees the numbers in each set will be unique within each set,\n\nSometimes random numbers to be picked need to be unique. Use Java to generate unique random numbers. The nextBoolean() method is used to get the next pseudorandom, uniformly distributed boolean value from this random number generator's sequence. The method call\n\nHow to Generate an unique ID in Java. By mkyong September 7, Example. UUID is the fastest I am trying to generate a unique 10 digit alpha numeric number, unique Number « Number « Java Data Type Q&A. Home; For example if i aget 5 number. Unique Random Number Generator :\n\nQuick Excel tip to generate unique random numbers in excel using the RAND formula. Use this formula tip when you want to generate sequential random numbers 25/07/2016 · Generate unique random numbers in Java Sagar S. ( generate a random number with a range ) Java Tutorials Arrays and Random Number Generator\n\nFor example if you want to generate five random integers the random number generator of choice is import java.util.Random; /** Generate random integers in a 26/08/2002 · I would like to generate 7 digit unique random number. I have it something like this : did you try this ?? java.util.Random generator = new java.util.Random();\n\nWhen you develop some applications in Java or on Android platform, you can sometimes need to generate a random alpha-numeric String to generate a session id, a unique Basically I have an array full of objects customer, one attribute a customer must have is a unique account number. //make me a random number generator called 'generator'\n\nThere are many ways to generate random numbers in java.Let’s see each with the help of example. Using Random class You can use java Random number generator in java. Sometimes random numbers to be picked need to be unique. Use Java to generate unique random numbers.\n\nThis page allows you to generate random sets of integers using true This generator guarantees the numbers in each set will be unique within each set, 26/08/2002 · I would like to generate 7 digit unique random number. I have it something like this : Skip did you try this ?? java.util.Random generator = new java.util.Random\n\n6/07/2017 · Visit and download code from here https://emergingwarriors.com/blog/201... Sometimes random numbers to be picked need to be unique. Use Java to generate unique random numbers. Generating Unique Random to Make a Random Number Generator.\n\n### Generate 6-Digit Random Number (Java in General forum at", null, "Generate random numbers in Java Learn Java Online. JavaScript random() Method JavaScript Math Object. Example. Return a random number between 0 (inclusive) and 1 (exclusive): Math.random();, Learn 6 ways how to implement random number(integer or double) generator in Java Random Number Generator in Java using for example, that I provided for a java.\n\nHow to generate a Random alpha-numeric String ? – All for. 4/08/2011 · Is there any mechanism by which Orchestrator can simply generate a random number or unique strings? * use xlst to call native java math lib (see example,, How do I create a unique ID in Java? already generates ids using \"cryptographically strong\" random number generator. Are there any examples where the.\n\n### 12 Digit unique random number generation in Java Stack", null, "Generating Unique Random Numbers in JAVA coderanch.com. There are many ways to generate random numbers in java.Let’s see each with the help of example. Using Random class You can use java Random number generator in java. https://en.wikipedia.org/wiki/List_of_random_number_generators How do I create a unique ID in Java? already generates ids using \"cryptographically strong\" random number generator. Are there any examples where the.", null, "10/05/2009 · Java Programming Tutorial - 26 - Random Number Generator Random class ( generate a random number with a range ) Java Game Development This page allows you to generate random sets of integers using true This generator guarantees the numbers in each set will be unique within each set,\n\nBasically I have an array full of objects customer, one attribute a customer must have is a unique account number. //make me a random number generator called 'generator' 28/12/2016 · Java - Generate Random Number In this Video i show how to generate random number whenever you start Java Tutorial 8 - Random Number Generator\n\nThe identifiers generated by UUID are actually universally unique identifiers. Example. import java.util //generate a random number String randomNum = Integer JavaScript random() Method JavaScript Math Object. Example. Return a random number between 0 (inclusive) and 1 (exclusive): Math.random();\n\nHi Friends, I need to generate a Exactly 6 digit random number.I have a method which generates 6 Digit Random#.But its generating repetative numbers How to generate a random alpha-numeric string? for session identifiers should come from a random number generator for example. A predictable random number\n\nLearn 6 ways how to implement random number(integer or double) generator in Java Random Number Generator in Java using for example, that I provided for a java 6/07/2017 · Visit and download code from here https://emergingwarriors.com/blog/201...\n\n4/08/2011 · Is there any mechanism by which Orchestrator can simply generate a random number or unique strings? * use xlst to call native java math lib (see example, Here's the source code for a complete Java Random class example. work with the Random class in Java. This example shows how to random number generator's\n\nJavaScript random() Method JavaScript Math Object. Example. Return a random number between 0 (inclusive) and 1 (exclusive): Math.random(); Random Numbers in Java. The easiest way to initialize a random number generator is to use the parameterless constructor, for example. Random generator = new\n\nBasically I have an array full of objects customer, one attribute a customer must have is a unique account number. //make me a random number generator called 'generator' There are many ways to generate random numbers in java.Let’s see each with the help of example. Using Random class You can use java Random number generator in java.\n\nHow to generate a random alpha-numeric string? for session identifiers should come from a random number generator for example. A predictable random number 26/08/2002 · I would like to generate 7 digit unique random number. I have it something like this : did you try this ?? java.util.Random generator = new java.util.Random();\n\nBasically I have an array full of objects customer, one attribute a customer must have is a unique account number. //make me a random number generator called 'generator' How to generate Unique ID using Java Programming? including sample main() to generate 100 To use a Random number generator a \"seed\" value has to be given to", null, "Sometimes random numbers to be picked need to be unique. Use Java to generate unique random numbers. Generating Unique Random to Make a Random Number Generator. Hi, I am new to Java i have a assignment and that is i need to generate 1,00,00,000 (one crore ) random numbers and which should be seed based random\n\n## How do I create a unique ID in Java? Stack Overflow", null, "How to generate Unique ID using Java Programming?. Hi, I am new to Java i have a assignment and that is i need to generate 1,00,00,000 (one crore ) random numbers and which should be seed based random, 6/09/2014 · Generate random number sequence without Generate unique random number sequence between 1 Aggregation vs Composition through example in Java;.\n\n### java.util.Random.nextBoolean() Method Example\n\nJava Generate Random Number - YouTube. java.util.Random.nextLong() Method Example method is used to return the next pseudorandom, uniformly distributed long value from this random number generator's, I am talking about a level that I just finished 1 codeacademy free trial in Python and one in Java.. unique random numbers. generate a random number; Try.\n\nThere are number of ways you could generate Unique Keys/IDs in Java. strong random number generator using instance a look at Example: Hi, I am new to Java i have a assignment and that is i need to generate 1,00,00,000 (one crore ) random numbers and which should be seed based random\n\nRandom Numbers in Java. The easiest way to initialize a random number generator is to use the parameterless constructor, for example. Random generator = new 12 Digit unique random number generation in Java. Generate 4 digit unique random number in Java using predefined What is an example of a proof by minimal\n\nGenerate unique random numbers in an I simply refer to it as the non-random random number generator. just a commonplace example of how you can use Streams to 25/07/2016 · Generate unique random numbers in Java Sagar S. ( generate a random number with a range ) Java Tutorials Arrays and Random Number Generator\n\nSometimes random numbers to be picked need to be unique. Use Java to generate unique random numbers. Here's the source code for a complete Java Random class example. work with the Random class in Java. This example shows how to random number generator's\n\nLearn 6 ways how to implement random number(integer or double) generator in Java Random Number Generator in Java using for example, that I provided for a java 26/08/2002 · I would like to generate 7 digit unique random number. I have it something like this : Skip did you try this ?? java.util.Random generator = new java.util.Random\n\n6/07/2017 · Visit and download code from here https://emergingwarriors.com/blog/201... How to generate a random alpha-numeric string? for session identifiers should come from a random number generator for example. A predictable random number\n\nSometimes random numbers to be picked need to be unique. Use Java to generate unique random numbers. 12 Digit unique random number generation in Java. Generate 4 digit unique random number in Java using predefined What is an example of a proof by minimal\n\n12 Digit unique random number generation in Java. Generate 4 digit unique random number in Java using predefined What is an example of a proof by minimal This page allows you to generate random sets of integers using true This generator guarantees the numbers in each set will be unique within each set,\n\njava.util.Random.nextFloat() Method Example uniformly distributed float value between 0.0 and 1.0 from this random number generator's sequence. The ideal PRNG for this problem is one which would generate a unique, random x is unique as long as \\(2x < p \\). For example, of unique random numbers is\n\nSometimes random numbers to be picked need to be unique. Use Java to generate unique random numbers. Generating Unique Random to Make a Random Number Generator. 26/08/2002 · I would like to generate 7 digit unique random number. I have it something like this : Skip did you try this ?? java.util.Random generator = new java.util.Random\n\n10/05/2009 · Java Programming Tutorial - 26 - Random Number Generator Random class ( generate a random number with a range ) Java Game Development This class provides a cryptographically strong random number generator random number generators, for example SecureRandom section in the Java\n\n### Creating a unique random number... (Beginning Java forum", null, "How do I create a unique ID in Java? Stack Overflow. There are many ways to generate random numbers in java.Let’s see each with the help of example. Using Random class You can use java Random number generator in java., Java – Generate random integers in a range. package com.mkyong.example.test; import java.util.Random; java8 random number. About the Author..", null, "### geekRai Generate random number sequence without duplicates", null, "RANDOM.ORG Integer Set Generator. The ideal PRNG for this problem is one which would generate a unique, random x is unique as long as \\(2x < p \\). For example, of unique random numbers is https://en.wikipedia.org/wiki/UUID There are number of ways you could generate Unique Keys/IDs in Java. strong random number generator using instance a look at Example:.", null, "• geekRai Generate random number sequence without duplicates\n• How to Generate an unique ID in Java – Mkyong.com\n• How to Generate a Sequence of Unique Random Integers\n\n• 10/05/2009 · Java Programming Tutorial - 26 - Random Number Generator Random class ( generate a random number with a range ) Java Game Development Quick Excel tip to generate unique random numbers in excel using the RAND formula. Use this formula tip when you want to generate sequential random numbers\n\nJava – Generate random integers in a range. package com.mkyong.example.test; import java.util.Random; java8 random number. About the Author. I am talking about a level that I just finished 1 codeacademy free trial in Python and one in Java.. unique random numbers. generate a random number; Try\n\nGenerate unique random numbers in an I simply refer to it as the non-random random number generator. just a commonplace example of how you can use Streams to 9/02/2006 · Does anyone know how to generate unique random numbers in Java? I need to generate a list of 10 random number below 50, and with no repeats. I feel...\n\nJava Math.random Examples Let's learn how to generate some random numbers in Java. Random numbers are since this is a Pseudo Random Number Generator JavaScript random() Method JavaScript Math Object. Example. Return a random number between 0 (inclusive) and 1 (exclusive): Math.random();\n\n6/09/2014 · Generate random number sequence without Generate unique random number sequence between 1 Aggregation vs Composition through example in Java; There are many ways to generate random numbers in java.Let’s see each with the help of example. Using Random class You can use java Random number generator in java.\n\nHow to Generate an unique ID in Java. By mkyong September 7, Example. UUID is the fastest I am trying to generate a unique 10 digit alpha numeric number, For example if you want to generate five random integers the random number generator of choice is import java.util.Random; /** Generate random integers in a\n\nQuick Excel tip to generate unique random numbers in excel using the RAND formula. Use this formula tip when you want to generate sequential random numbers How to generate a random alpha-numeric string? for session identifiers should come from a random number generator for example. A predictable random number\n\n10/05/2009 · Java Programming Tutorial - 26 - Random Number Generator Random class ( generate a random number with a range ) Java Game Development 17/05/2013 · Here is a Java code example of using both Random class You can even use ThreadLocalRandom from Java 1.7, which is a Random number generator isolated\n\n12 Digit unique random number generation in Java. Generate 4 digit unique random number in Java using predefined What is an example of a proof by minimal 9/02/2006 · Does anyone know how to generate unique random numbers in Java? I need to generate a list of 10 random number below 50, and with no repeats. I feel...\n\nHow to generate Unique ID using Java Programming? including sample main() to generate 100 To use a Random number generator a \"seed\" value has to be given to 9/02/2006 · Does anyone know how to generate unique random numbers in Java? I need to generate a list of 10 random number below 50, and with no repeats. I feel...\n\nSometimes random numbers to be picked need to be unique. Use Java to generate unique random numbers. This class provides a cryptographically strong random number generator random number generators, for example SecureRandom section in the Java" ]

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[ "# Go Math Worksheets 2 Similar Files Place Value Patterns Grade 4th", null, "go math worksheets 2 similar files place value patterns grade 4th.\n\nplace value patterns 4th grade worksheets number 5,place value patterns worksheets 4th grade number the best image collection,place value number patterns worksheets for grade 4th,place value patterns worksheets 4th grade number,place value patterns worksheets grade math 4th number,place value patterns 4th grade worksheets number go math 1 and vocabulary builder period,place value patterns 4th grade worksheets number,place value number patterns worksheets to 1 relationships 4th grade,place value patterns worksheets 4th grade number of a,place value number patterns worksheets 4th grade 5 worksheet round 6 digit numbers to the nearest." ]

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[ "## An Exponentially Weighted Covariance Matrix in R\n\nExponential weighting schemes can help navigate the trade-off between responsiveness and stability of the inherently noisy estimates we make from market data. We previously saw examples of calculating the exponentially weighted moving average of a vector, and estimating the correlation between SPY and TLT using an exponential weighting scheme [link]. In this article, we’ll implement …" ]

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[ "", null, "", null, "# If 1+5=12, 2+10=24, 3+15=36 then 5+25=??\n\n33,089 views", null, "posted Oct 3, 2014\n\nAns = 60\n\noutput = 2 * (first_number + second_number)\n\n``````2 * (1 + 5) = 12\n2 * (2 + 10) = 24\n2 * (3 + 15) = 36\n2 * (5 + 25) = 60\n``````", null, "answer Oct 13, 2014\nwhy to multiple 2\n\n7*1+5*1=12\n7*2+5*2=24\n7*3+5*3=36\n7*5+5*5=60", null, "answer Jun 12, 2015 by anonymous\n+1 vote\n\n`60`\n\n12*5 = 60", null, "answer Oct 3, 2014\n+1 vote", null, "answer Nov 25, 2015 by anonymous\n+1 vote\n\n1x2=2 5x2=10 so 2+10=12 1+5=12 2x2=4 10x2=20 so 4+20=24 multiply\nall using 2 u will get the answer same which is written as answer so 5x2=10 25x2=50 and 10+50 is 60 5+25=60", null, "answer Nov 28, 2015 by anonymous\n+1 vote\n\n(1 + 5) x 2 = 12\n(2 + 10) x 2 = 24\n(15 + 3) x 2 = 36\n(5 + 25) x 2 = 60", null, "answer Jun 6, 2016 by anonymous\n+1 vote\n\n1×12=12\n2×12=24\n3×12=36\n5x12=60\n\n60", null, "answer Sep 23, 2016 by\n\nIf 1+5=12, 2+10=24, 3+15=36 then 5+25=??\n\n(1+5)*2=6*2=12\n\n(2+10)*2=12*2=24\n\n(3+15)*2=18*2=36\n\n(5+25)*2=30*2=60", null, "answer Nov 25, 2016\n\nSimilar Puzzles\n+1 vote\n\nOn behalf of Sture Sjöstedt" ]

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[ "Great Pyramid Geometry\n\n# Great Pyramid Geometry\n\nG\n400 points\n\nThe Great Pyramid outside Cairo, Egypt has a square base measuring 756 feet on a side and a height of 480 feet. A] What is the volume of the great pyramid in cubic yards? B] The stones used to build the Great Pyramid were limestone blocks with an average volume of 1.5 cubic yards. How many of these blocks were needed to construct the Great Pyramid?\n\nGreat Pyramid\nGigi\n\n403.4k points\n\nThe volume of a square pyramid is given by the formula,\n\n$$V=\\frac{l^2 h}{3}$$\n\nconsider there are 3 ft in 1 yd so $l=756/3=252$ yds and $h=480/3=160$ yds. Plugging these values into the formula above will give your answer in cubic yds.\n\nTo find the amount of bricks used to build the pyramid simply divide the total volume found in part (a) by the volume of 1 block.\n\nZ\n0 points\n\n#### Oh Snap! This Answer is Locked\n\nFilename: BSA 310 All DQs.zip\n\nFilesize: < 2 MB\n\nPrint Length: 1 Pages/Slides\n\nWords: NA\n\nSurround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\\beta^2-1=0$$\n\nUse LaTeX to type formulas and markdown to format text. See example." ]

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[ "# 18 Grams Per Cubic Centimeter To Pounds Per Cubic Inch\n\nThis on the web oneway conversion tool converts density units from grams per cubic centimeter gcm3 into pounds per cubic inch lbin3 instantly online gram per cubic centimeter gcm3 06 pounds per cubic inch lbin3 ow many pounds per cubic inch lbin3 are in 1 gram per cubic centimeter 1 gcm3 how much of density from grams per cubic centimeter to pounds per.\n\n• ### Convert Grams Per Cubic Centimeter To Pounds Per\n\nThis on the web oneway conversion tool converts density units from grams per cubic centimeter gcm3 into pounds per cubic inch lbin3 instantly online gram per cubic centimeter gcm3 06 pounds per cubic inch lbin3 ow many pounds per cubic inch lbin3 are in 1 gram per cubic centimeter 1 gcm3 how much of density from grams per cubic centimeter to pounds per.\n\n→ Chat Online\n• ### 18GramsCubic Centimeter To PoundsCubic Inch\n\nConvert 18gramscubic centimeter to poundscubic inch gcm3 to lbsin3 with our conversion calculator and conversion tableso convert 18gcm3 to lbsin3 use direct conversion formula below8gcm3 0518518518519 lbsin3ou also can convert 18 gramscubic centimeter to other density popular units.\n\n→ Chat Online\n• ### Convert Grams Per Cubic Centimeter To Pounds Per\n\nHow to convert grams per cubic centimeter to pounds per cubic foot gcm to lbft lbft 6279606 gcmow many pounds per cubic foot in a gram per cubic centimeter if gcm 1 then lbft 6279606 1 6279606 lbftow many pounds per cubic foot in 42 grams per cubic centimeter.\n\n→ Chat Online\n• ### Grams Per Cubic Inch To Kilograms Per Liter Gin3 To\n\nConvert grams per cubic inch to kilograms per liter gin3 in kglrams per cubic inch and kilograms per liter both are the units of densityee the charts and tables conversion here.\n\n→ Chat Online\n• ### 1 GramsCubic Centimeter To PoundsCubic Foot 1\n\nConvert 1 gramscubic centimeter to poundscubic foot gcm3 to lbsft3 with our conversion calculator and conversion tableso convert 1 gcm3 to lbsft3 use direct conversion formula below gcm3 02421972534332 lbsft3ou also can convert 1 gramscubic centimeter to other.\n\n→ Chat Online\n• ### Cubic Inches To Cubic Centimeter Conversion In To\n\nCubic centimeter value will be converted automatically as you typehe decimals value is the number of digits to be calculated or rounded of the result of cubic inches to cubic centimeter conversionou can also check the cubic inches to cubic centimeter conversion chart below, or go back to cubic inches to cubic centimeter converter to top.\n\n→ Chat Online\n• ### Convert Pounds Per Square Inch To Kilograms Per\n\nConvert pounds per square inch to kilograms per square centimeter psi to kgfcm with the pressure conversion calculator, and learn the calculation formulaonvert pounds per square inch to kilograms per square centimeter psi to kgfcm with the pressure conversion calculator, and learn the calculation formula18 psi 155 kgf.\n\n→ Chat Online\n• ### What Is The Density Of Water In Grams Per Cubic Inch\n\nSilver has a density of 10 grams per cubic centimeter there are 167 cubic centimeters in a cubic inchranslates into 171 grams for a cubic inch of silver, which is about 6 ounces.\n\n→ Chat Online\n• ### Convert Grams Per Cubic Centimeter To Pounds Per\n\nConvert density unitsasily convert grams per cubic centimeter to pounds per cubic feet, convert gcm 3 to lbft 3 any other converters available for free.\n\n→ Chat Online\n• ### Lbin Pound Per Cubic Inchonversion Chart\n\n2019921this is a conversion chart for pound per cubic inch british and uo switch the unit simply find the one you want on the page and click itou can also go to the universal conversion page enter the value you want to convert pound per cubic inchhen click the convert me button.\n\n→ Chat Online\n• ### Cubic Centimeters To Grams Water Conversion\n\nWeight of 1 cubic centimeter cc, cm 3 of pure water at temperature 4 c 1 gram g 01 kilogram kg cubic centimeter cc, cm 3 1 ml milliliter 038140227 us fluid ounces flz 11000 l liter, the official si unit of volume many industries, the use of cubic centimetres cc was replaced by the millilitre mlubic centimeter is still widely used in the.\n\n→ Chat Online\n• ### Density Conversion Calculator Calculator Soup\n\nCalculator useonvert among mass density values along with mass concentration values mass divided by volumeunces and pounds are in the avoirdupois system, the standard everyday system in the united states where 1 ounce 116 poundow to convert units of densityonversions are performed by using a conversion factor.\n\n→ Chat Online\n• ### Conversion Of Gcm3 To Ounce Metriccubic Inch\n\nFree online density conversiononvert gcm3 to ounce metriccubic inch gramcubic centimeter to ozin3 with much by calculateplus.\n\n→ Chat Online\n• ### Cubic Centimeters Pounds Conversion NinjaUnits\n\nDefinition of cubic centimeters of water provided by wikipedia one cubic centimetre corresponds to a volume of 1 1,000,000 of a cubic metre, or 1 1,000 of a litre, or one millilitre 1 cm 3 1 mlhe mass of one cubic centimetre of water at 3 c the temperature at which it attains its maximum density is closely equal to one gram.\n\n→ Chat Online\n• ### Gcm3 To Lbsin3 Converter, Chart EndMemo\n\nConcentration solution unit conversion between gramcubic centimeter and poundcubic inch, poundcubic inch to gramcubic centimeter conversion in batch, gcm3 lbsin3 conversion chart lbsin3per 1 lbsin3 2767047 per lbsin3ppm 1 lbsin3 27679904291 ppm lbsin3ppb 1 lbsin3 27679904702 ppb.\n\n→ Chat Online\n• ### Conversion Of Density Units Rho Kg Cubic Meter Unit\n\n2017212grams per cubic centimeter grams per cup grams per liter grams per cubic meter kilograms per cubic meter liters per gram pounds per cubic foot pounds per cubic inch slugs per cubic foot reset conversions of other density unitsany people still use gcm 3 the dependence of the pressure on the water density is low per 1 bar 100000.\n\n→ Chat Online\n• ### Cubic Inches And Cubic Centimeters Converter In\n\nDisclaimerhilst every effort has been made in building this cubic inches and cubic centimeters converter, we are not to be held liable for any special, incidental, indirect or consequential damages or monetary losses of any kind arising out of or in connection with the use of the converter tools and information derived from the web site.\n\n→ Chat Online\n• ### Mercury Pound To Cubic Inches Of Mercury Converter\n\nConvert how many cubic inches cu in in3 of mercury are in 1 pound lb ne lb pound of mercury mass equals two point zero five cubic inches cu in in3 in volume of mercuryhis mercury calculator can be used to change a conversion factor from 1 pound lb equals 2 cubic inches cu in in3 exactlyonvert mercury measuring units.\n\n→ Chat Online" ]

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[ "# first_derivative#\n\nmetpy.calc.first_derivative(f, axis=None, x=None, delta=None)#\n\nCalculate the first derivative of a grid of values.\n\nWorks for both regularly-spaced data and grids with varying spacing.\n\nEither x or delta must be specified, or f must be given as an `xarray.DataArray` with attached coordinate and projection information. If f is an `xarray.DataArray`, and x or delta are given, f will be converted to a `pint.Quantity` and the derivative returned as a `pint.Quantity`, otherwise, if neither x nor delta are given, the attached coordinate information belonging to axis will be used and the derivative will be returned as an `xarray.DataArray`.\n\nThis uses 3 points to calculate the derivative, using forward or backward at the edges of the grid as appropriate, and centered elsewhere. The irregular spacing is handled explicitly, using the formulation as specified by [Bowen2005].\n\nParameters:\n• f (array-like) – Array of values of which to calculate the derivative\n\n• axis (int or str, optional) – The array axis along which to take the derivative. If f is ndarray-like, must be an integer. If f is a DataArray, can be a string (referring to either the coordinate dimension name or the axis type) or integer (referring to axis number), unless using implicit conversion to `pint.Quantity`, in which case it must be an integer. Defaults to 0. For reference, the current standard axis types are ‘time’, ‘vertical’, ‘y’, and ‘x’.\n\n• x (array-like, optional) – The coordinate values corresponding to the grid points in f\n\n• delta (array-like, optional) – Spacing between the grid points in f. Should be one item less than the size of f along axis.\n\nReturns:\n\narray-like – The first derivative calculated along the selected axis\n\nChanged in version 1.0: Changed signature from `(f, **kwargs)`" ]

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[ "# Point Slope Form Definition\n\nPoint slope equation of a line what is slope intercept form point slope form of a linear equation point slope form definition algebra the point slope equation of a line ppt 5 4 point slope form.", null, "Linear Functions And Equations Point Slope Form", null, "Point Slope Form Definition Algebra The Latest Trend In", null, "What S Point Slope Form Of A Linear Equation Virtual Nerd", null, "Things That Make You Love And Point Slope Form Of A Line", null, "Point Slope Equation Of A Line", null, "Learn Point Slope Form Tutorial Definition Example Formula", null, "Point Slope Form Simply Explained W 17 Examples", null, "Point Slope Form Simply Explained W 17 Examples", null, "Algebra 1 5 4 Point Slope Form Problem Writing An Equation In", null, "Point Slope Equation Of A Line", null, "Linear Functions And Equations Point Slope Form", null, "What Is Slope Intercept Form Definition Equation Examples", null, "What S Point Slope Form Of A Linear Equation Virtual Nerd Can Help", null, "Learn Two Point Form Tutorial Definition Example Formula", null, "Copy Of Point Slope Form Lessons Tes Teach", null, "Point Slope Form Simply Explained W 17 Examples", null, "Slope Wikipedia", null, "Math Concepts Explained Understanding Point Slope Form", null, "Slope Formula", null, "Linear Functions And Equations Point Slope Form", null, "Solving Linear Equations With One Variable Ppt Video Online", null, "", null, "Ppt 5 4 Point Slope Form Presentation Free", null, "Write An Equation Of A Line Using Point Slope Form Sutori", null, "Point Slope Form Simply Explained W 17 Examples\n\nLinear functions and equations point slope form slope formula point slope equation of a line solving linear equations with one variable ppt video online what is slope intercept form definition equation examples linear functions and equations point slope form." ]

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[ "# as.name & is.name Functions in R (2 Examples)\n\nThis tutorial explains how to apply the as.name and is.name functions in the R programming language.\n\nThe article will consist of two examples for the application of the as.name and is.name commands. More precisely, the tutorial contains the following content:\n\nLet’s get started…\n\n## Definitions & Basic R Syntaxes of as.name and is.name Functions\n\nDefinitions: You can find the definitions of the as.name and is.name functions below.\n\n• The as.name R function converts a character string to a name class object.\n• The is.name R function tests whether a data object has the class name.\n\nBasic R Syntaxes: Please find the basic R programming syntaxes of the as.name and is.name functions below.\n\n```as.name(x) # Basic R syntax of as.name function is.name(x) # Basic R syntax of is.name function```\n\nNote that the as.name and is.name functions are synonyms to the as.symbol and is.symbol functions. You can exchange these function names if you want.\n\nHowever, I’ll show in the following two examples how to use the as.name and is.name commands and functions in the R programming language.\n\n## Example Data\n\nThe following data will be used as basement for this R programming tutorial:\n\n```x <- \"my_name\" # Create character string object x # Print character string object # \"my_name\"```\n\nHave a look at the previous output of the RStudio console. It shows that our example data is a simple character string stored in the data object x.\n\nWe can use the class function to check the data type of our data object:\n\n```class(x) # Check class # \"character\"```\n\nOur example data has the class character.\n\n## Example 1: Convert Character String to name Class\n\nIn Example 1, I’ll show how to change the character class to the name class using the as.name function. Have a look at the following R code:\n\n`x_name <- as.name(x) # Apply as.name function`\n\nThe previous R syntax switched our data type from character to name. We can check the data type of our new data object using the class function:\n\n```class(x_name) # Check class # \"name\"```\n\nThe new data object has the name data mode.\n\n## Example 2: Check if Data Object has Class name\n\nWe can use the is.name function to return a logical value indicating whether a data object has the class name. Let’s apply the is.name function to our original data object x:\n\n```is.name(x) # Test character string # FALSE```\n\nAs the RStudio console output shows: Our original data object is not a name object.\n\nNow, let’s apply the is.name function to our new data object:\n\n```is.name(x_name) # Test name object # TRUE```\n\nThe is.name function returns the logical value TRUE, i.e. our new data object has the class name.\n\n## Video & Further Resources\n\nDo you need further info on the R programming syntax of this article? Then I can recommend to watch the following video of my YouTube channel. In the video, I’m explaining the R programming codes of this tutorial in a live session in R:\n\nFurthermore, you may have a look at the other articles of https://www.statisticsglobe.com/.\n\nYou learned in this article how to use the as.name and is.name functions in the R programming language. Don’t hesitate to let me know in the comments, if you have additional questions.\n\nSubscribe to my free statistics newsletter" ]

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[ "DAY 12\n0\nModern Web\n\n# 資料庫結構\n\n``````from werkzeug.security import generate_password_hash, check_password_hash\n\nclass Users(db.Model):\n__tablename__ = \"users\"\nid = db.Column(db.Integer, primary_key=True)\nemail = db.Column(db.String, unique=True, nullable=False)\nintroduction = db.Column(db.String)\nregister_time = db.Column(\ndb.DateTime, default=datetime.datetime.now, nullable=False\n)\nposts = db.relationship(\"Posts\")\n\nself.email = email\nself.introduction = introduction\n\n``````\n\n• `id` 是這個 table 的主鍵 (primary key)，如同他後面的參數 `primary_key=True` 所示。在他前面有一個 `db.Integer`，他表示了這個 column 的資料庫型別。有了前面這兩個參數，autoincrement (自動把 id 流水號) 就會自動存在。\n• `username` 是用來存使用者名稱的，他使用的資料庫型別是 `db.String`，此外，他還加上 `unique``nullable` 兩個參數，分別代表是否唯一及可否為空，那在此處當然是要唯一並不可為空，畢竟是使用者名稱，不這樣做也有點奇怪。\n• `password` 是存密碼用的，當然，我們存的不是明文，而是 hash 過的結果，這個部分會在等等看到。他使用的一樣是 `db.String`，而他也不能為空，但可以重複 (不設定就沒有限制)。\n• `email` 是電子郵件信箱，基本上我們會在註冊的時候發一封信給註冊者，然後就再也不會用到了，這就是 Flask-Mail 在這系列唯一的戲份。沒什麼好懷疑地，他是字串，然後不能重複，也不能為空。\n• `introduction` 是存使用者的自我介紹用的，畢竟我們的主題是部落格系統，讓作者可以自我介紹應該十分合理。這邊就不做限制，可以為空也可以跟別人一樣，雖然在正常情況下要跟別人一樣還蠻不容易的。\n• `is_admin` 是用來說明此使用者是不是管理員的，所以他使用的是 `db.Boolean`，也就是 `True` 或是 `False`，他不能為空，然後當然可以重複，最後他有一個 `default=False`，這代表沒有設定的時候他就會自動當他是 `False`\n• `register_time` 是用來存註冊時間。他使用的是 `db.Datetime` 這個資料庫型別，這個型別是用來對付 python 的 `datetime.datetime`，所以我們後面的 `default``datetime.datetime.now`。如果要用 `datetime.date` 的話，他搭配的資料型別是 `db.Date`。這裡要特別注意他傳入的是 `datetime.datetime.now` 這個函式，而非 `datetime.datetime.now()`\n• `posts``comments` 用到了 `db.relationship`，他分別是和在下兩個 table 裡面會看到的 `posts.id``comments.id` 有關聯，簡單來說我們可以用他找到這個使用者底下全部的文章和留言。\n\n``````class Posts(db.Model):\n__tablename__ = \"posts\"\nid = db.Column(db.Integer, primary_key=True)\nauthor_id = db.Column(db.Integer, db.ForeignKey(\"users.id\"), nullable=False)\ntitle = db.Column(db.String, nullable=False, unique=True)\ndescription = db.Column(db.String)\ncontent = db.Column(db.String, nullable=False)\ntime = db.Column(\ndb.DateTime, default=datetime.datetime.now, nullable=False\n)\n\ndef __init__(self, author_id, title, description, content):\nself.author_id = author_id\nself.title = title\nself.description = description\nself.content = content\n\nid = db.Column(db.Integer, primary_key=True)\nauthor_id = db.Column(db.Integer, db.ForeignKey(\"users.id\"), nullable=False)\npost_id = db.Column(db.Integer, db.ForeignKey(\"posts.id\"), nullable=False)\ncontent = db.Column(db.String, nullable=False)\ntime = db.Column(\ndb.DateTime, default=datetime.datetime.now, nullable=False\n)\n\ndef __init__(self, author_id, post_id, content):\nself.author_id = author_id\nself.post_id = post_id\nself.content = content\n``````\n\n• `author_id` 是這篇文章的作者的使用者 ID (`users.id`)，他是把 `users.id` 當外鍵 (foreign key)，然後讓 `users.posts` 可以存取到 posts 這個 table 的資料。而他當然不能為空 (如果使用者遭到刪除，文章和留言也會跟著刪除)。\n• `title` 是這篇文章的標題，不能為空，也不能重複。\n• `description` 是這篇文章的小標，有沒有都沒差。\n• `content` 是內文，當然不能為空。\n• `comments` 又是一個 relationship，他和下面的 `comments` 有關連，所以可想而知，`comments` 裡面一定也有一個跟這個 table 有關的外鍵。\n\n• `author_id` 跟上面 `posts` 裡面的一樣，就是用 `users.id` 作為外鍵讓 `users.comments` 可以存取這個 table。當然也不能為空。\n• `post_id` 就是用到剛剛 `posts` 那個 table 的 relationship，讓 `posts.comments` 可以存取到其下的留言。" ]

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[ "# VO2 Max and Aerobic Power\n\nOxygen is one of the vital elements of life because it acts as a fuel for aerobic respiration, which is the energy source in all organisms (the other fuel being glucose). Without energy from respiration, organisms simply die. As an organism (in this case me, a human) does work, it needs more energy. Thus it will need more fuel and particularly more oxygen since glucose can be stored in the body. The oxygen intake increases as the rate of work done increases, up to a limit known as your VO2 max.\nVO2 max is the maximum volume of oxygen uptake (hence the V in VO2) anyone can use. It is measured in millilitres per minute per kilogram of body mass (mlO2 min-1 kg-1). People who are more fit have higher VO2 max values and can exercise more intensely than those who are not very fit.\nFactors Affecting VO2 Max\n\nThe physical limitations that restrict the rate at which energy can be released aerobically are dependent upon:\n> The chemical ability of the muscular cellular tissue system to use oxygen in breaking down fuels and,\n> The combined ability of cardiovascular and pulmonary systems to transport the oxygen to the muscular tissue system.1\nThe Aim\nVO2 max can be measured in a variety of ways1. The aim of this experiment is to find out the subjects VO2 max and then covert it to the total aerobic power output.\nThe Method\nVO2 max can be measured fairly accurately by doing a “shuttle run” style test (known as “The Multistage Fitness Test” A.K.A. “The Beep Test”). Basically, someone has to run a 20 meter track at the starting speed of 8.5 kmh-1 for one minute. Once the person finishes the 20 meter track, they must run back at the same speed, and thus we get an oscillating pattern from one end of the track to the other until the minute is over.\nThe speed is increased by 1 kmh-1 every minute (so after one minute of running at 8.5 kmh-1, the person must run the second minute at 9.5 kmh-1). The same pattern is repeated only this time, because the person (the subject) is running at a higher speed during the same amount of time (one minute), they are going to cover a larger distance and therefore more of the 20 meter laps (in theory anyway).\nThis fact only works in theory because will most of the speeds, the subject can not run a set (integer) number of laps in exactly one minute. It turns out that if the subject runs at 8.5 kmh-1 for one minute, they will cover 7.08 laps. This is impractical (how can you tell that the subject ran 7.08 laps!) and so the number of laps must be rounded up or rounded down to an integer number of laps. Once a set number of integer laps are set, we work out the time taken to run the integer number of laps. (Refer to Columns and ).\nThe subject continues running until he or she can no longer keep up the pace. The speed that the subject sustains (i.e. the speed before the speed that the subject stops on) is known as the Maximum Aerobic Speed (MAS) and is measured in kmh-1. Once we have the MAS, we can work out the VO2 max in the following formula:\nVO2 max = 31 + 3.2 x (MAS – Subject’s Age [years]) + 0.15 x MAS x Age\nThe unit for VO2 max is: mlO2 min-1 kg-1.\n(Note: The above formula is a “conversion formula developed by researchers, to give an accurate measure of VO2 max”. See Activity Sheet 26)\nAfter calculating the VO2 max, we can convert it to maximum aerobic power output. Because the subject will be working with a high energy output, running requires a lot of energy; the only way to keep going is by aerobic respiration. Anaerobic respiration doesn’t provide the high amounts of energy that are needed in such exercises, especially for longer periods of time e.g. ten minutes.\nFor every litre of oxygen consumed, the subject’s muscles use 20kJ of energy. The total amount of oxygen consumed in a minute is the VO2 max multiplied by the body mass of the subject. This gives us the total oxygen intake of the subject in ml per minute (mlO2 min-1), since VO2 max is millilitres of oxygen per kilogram of body mass per minute. Once we have the total oxygen intake in mlO2 min-1, we multiply it by 20 (if 1 litre gives 20,000 J, then 1 millilitre will give 20 J) to get the total amount of energy used (i.e. power) in Joules per minute, (J min-1).\nPower (J min-1) = VO2 max (mlO2 min-1 kg-1) x Body Mass (kg) x 20 J mlO2-1\nPower (W, Js-1) = Power (J min-1) � 60 s min-1\nConventionally, the power output is measured in Watts per kilogram of body mass, W kg-1 (See Table 3 on Pg 37 of the text book, Salters Horners Advanced Physics), therefore we would need to divide the total power by the subject’s mass. However the aim of the experiment is to find out the total aerobic power output. (At least that is what the Activity Sheet 26 says, under the last bullet point in the Analysis section) This means that there is no need to divide the total power output by the subject’s mass. We just leave the total power output in Watts.\n(For Prepared Formulas and An Interactive VO2 max calculator, See File Formula Input Form.xls)\nInterestingly, the test given on the website1 for calculating VO2 max (look for The Multi-Stage Fitness Test) differs in some ways from the one suggested in Activity Sheet 26. One of the differences is the increase of the speed is\n0.5 kmh-1 every minute, not 1 kmh-1. Also, if the subject doesn’t complete a whole minute at the speed of 15.5 kmh-1, for example if the subject managed to complete 3 out of the 13 laps, then the subject would have a different (lower) VO2 max than if 10 laps were completed.\nLater on, I will discuss this issue and other differences in more detail (Under the Evaluation).\nThe Table (Table 1)\nIf you refer to Table 1 (File Table 1.xls) you will see all of the information needed and all of the calculations have been done beforehand. I compiled this table using Microsoft Excel(r). Below is a brief explanation of the table.\nColumn \nThe speed of the subject, in kmh-1.\nColumn \nThe speed of the subject given in ms-1. To convert speed from kmh-1 to ms-1, we multiply by 1,000 (converting km to m) and divide by 3600 (�[60 x 60] is converting hours to seconds). Simplified, converting kmh-1 to ms-1 we multiply by 10/36. Therefore:\nColumn = Column x 10/36\nColumn \nThe time taken to complete one lap can be worked out by the below formula\nVelocity (ms-1)= Distance (m) � Time (s), if we re-arrange the formula to make time the subject we get:\nTime (s) = Distance (m) � Velocity (ms-1).\nThe distance is of one lap is 20 m and the velocity has been calculated in Column . Column is just:\nColumn = 20m � Column .\nColumn \nThis is the number of laps made in one minute (60 seconds). If I know that it takes 8.47 s to run one lap, I can calculate the total number of laps made in 60 seconds be dividing 60 seconds by the time taken to run one lap. So:\n? of Laps in 60 seconds = 60 seconds � Time Taken to Run One Lap\nColumn = 60 s � Column \nColumn \nThis is just Column rounded up or down to give us an integer number of laps. It means I don’t have to deal with 7.08 laps and suchlike.\nColumn \nThis is the time taken to run the integer number of laps. We calculate this by multiplying the time taken to run one lap (Column ) by the integer number of laps (Column ).\nTime to Run Integer ? of Laps = Time to Run One Lap x Integer ? of Laps\nColumn = Column x Column \nColumn \nSince we know that each lap is 20 meters and we know how many integer laps the subject will run, we can find out the total distance covered during a specific speed by multiplying 20 meters by the integer number of laps.\nTotal Distance Covered During a Speed = 20 m x Integer ? of Laps\nColumn = 20 m x Column \nColumn \nColumn is the cumulative distance ran. Here, the distances are added up (accumulated) so that we know the total distance ran through the whole activity. The cumulative distance is the total of the previous distances (previous speeds) plus the distance of the current speed.\nColumn \nColumn is the cumulative time taken for the activity. It is in seconds and works in pretty much way as Column , i.e. the times of the previous runs are added to the current run to give the total cumulative time.\nColumn \nThis is the cumulative time presented in a more familiar and user friendly format, the minute : second style. This is just here to give a sense of how long 840 seconds are.\nThe Tape (or The Slideshow)\nIn order to make the subject run at the times listed on the Table, I will need to prepare a tape or some sort of timing device.\nAfter attempting to make a tape and failing miserably, I decided to use Microsoft PowerPoint(r) instead (making the tape proved to be a long winded, boring and fruitless exercise), because, with PowerPoint, I can set time intervals between slide transitions and add sounds on every slide transition, making it a visual as well as an aural aid and I can have a lot more fun making it! (i.e. I can have lots of “interesting” and slightly odd sounds on the slide show)\nI also realised that it would be more helpful (to the subject) if I had a sound in the middle of each lap and to have a marker on the middle of the lap (10 meters). In case the subject is going too slowly and doesn’t reach the middle marker when the middle bleep sounds, they can speed up to reach the end of the lap in time.\nHowever there is a slight disadvantage with using PowerPoint because the transition periods can only be set to 0.1 of a second (1 d.p.) and the lap times are given to 0.01 of a second (2 d.p.) and some of the half laps are to 0.001 of a second (3 d.p.). Therefore I have had to alter the timing of the transitions slightly so that there isn’t a cumulative error. For example, during the first speed (8.5 kmh-1), it takes 4.235 seconds to complete half a lap, but I can only have 4.2 and 4.3 as time intervals in PowerPoint, therefore I had to find a pattern that consisted of 4.2 and 4.3 time intervals to fit the 4.235 time interval as well as possible.\nThis technique took quite some time, however, using Excel helped greatly. With Excel, I could input different patterns (using 4.2 and 4.3 seconds) and view the sum automatically. If it wasn’t right (i.e. if the total time wasn’t near the time in Column in Table 1), I simply changed the pattern until I got the closest time.\nAlso I decided that there wasn’t a need to go beyond 13.5 kmh-1 because when we did a warm up to the test (a kind of preliminary), none of the subjects managed to run over a thousand meters. In order for any of the subjects to complete the 13.5 kmh-1 speed, they would need to run at least 1100 meters, therefore there was no point in extending the presentation beyond that speed.\n(See Files Timing.xls for the pattern generating, and Timing Presentation.ppt, listen out for a treat during the last few slides)\nThe Safety (Issues)\nIt is always important to consider safety in any situation, and it is especially important in this type of activity where there is a fairly high risk of an accident and or an injury occurring. Below is a set of guidelines that the subject and others present during the activity should follow.\nThe subject should begin with a five to ten minute warm-up period, before the test is started. It should consist of stretches and short runs where the subject should rapidly accelerate and then decelerate. This helps the subject to run better in the test and also helps avoiding any muscle cramps during the test.\nIn the (relatively) unlikely event of the subject falling, or hurting him/herself in any other way, the subject should stop running immediately. Also if the subject feels any pain or dizziness, they should stop. The subject should not continue with the test, even if they seem or appear to have recovered.\nThe Results\nThe results are in the Result Sheet (unsurprisingly). The results were gathered from the experiment, which was conducted with ten subjects, including myself. They are in order of VO2 max.\n(The Results Table can be found in Results.xls)\nThe Evaluation\nAfter completing the experiment, I worked out the aerobic power of the subjects very easily, with the help of Excel. I personally found the experiment enjoyable and it my got the heart pumping! (A somewhat rarity in physics, consider electricity… ok, maybe pacemakers, and the way they electrically shock people who have had heart attacks with two funny looking handle things, not much else though).\nHowever, there were many problems I encountered while conducting the experiment. To begin with, we couldn’t find a 20 meter track anywhere in our school, considering the fact that there had to be a socket close by since my timing mechanism uses a computer (there are three sports halls in the school, but they were all busy). Therefore, I had to settle for a smaller 10 meter track. The fact that I had midpoint bleeps in the timing mechanism meant that each bleep (midpoint and full-length bleeps) was a signal for the subjects to reach the end of the 10 meter track. This meant the timing was not affected, however the experiment could have been affected greatly. (See the Miscellaneous Calculations section)\nSecondly, in the Activity Sheet, it says, “For every litre of oxygen consumed, 20kJ of energy are transferred to the subject’s muscles”. However it fails to mention whether or not some of that energy is lost as heat and other ways of energy loss (e.g. fiction from the ground, the energy needed to stop at the end of every lap and even the energy needed to move the muscles themselves, i.e. contracting and relaxing of muscles). There are no suggestions or hints on how much of the energy is used to propel the subject. Although this can be calculated in the kinetic energy equation, EK = 1/2mv2, and the power equation P = ?E/?t . However, I suspect that these energy fluctuations are taken into account via the VO2 max formula.\nBut even the formula itself isn’t very accurate in my opinion. As I mentioned earlier, if two subjects managed to sustain the same MAS (Maximum Aerobic Speed) but one of the subjects ran more laps, then logically that subject has a higher VO2 max. This logicality is not, in any way, included in the formula. On the website (See Note 1 of Reference) the tables show that if one of the subjects ran more laps during the same speed, that subject would have a higher VO2 max that if he/she managed to run a smaller number of laps. Therefore, I do not believe that the formula for calculating VO2 max on the Activity Sheet 26 gives a correct quantitative value of the subject’s VO2 max. Although the experiment has shown that some of the subjects are fitter that others (i.e. the experiment is correct qualitatively), it did not produce reliable figures with regard to the VO2 max of the subject.\nOn the first of the Activity Sheets, Figure A26.2 shows a line graph of VO2 max for boys and girls at different ages. According to the graph, 16-year-old boys should have a VO2 max of 52-53 mlO2 min-1 kg-1. Ops! (My VO2 max is nowhere near that, or at least the formula tells me that it is nowhere near that). It shows that I am not as healthy as I should be, considering that my mass is 85 kg! Unlike Robert, who is a very healthy person and managed to run at a high enough speed to get a high VO2 max. (Although I should stress again my doubt about the numbers given by the formula)\nThe Miscellaneous Calculations\nA velocity-time curve of the subject’s motion.\nNote the area under both of the curves should be equal since the same distance, 10 meters, is travelled. The distance is the speed multiplied by the time i.e. the area under the graph.\nA\nAs mentioned earlier, the energy needed to propel a subject can be calculated via the kinetic energy equation.\nEK = 1/2mv2 and P = ?E/?t\nMy mass is 85 kg. If I ran one lap at 8.5 kmh-1 (which is 2.36 ms-1, Refer to Table 1), the energy needed is:\nEK = 1/2mv2 so, EK = 1/2 x 85kg x (2.36ms-1)2\nEK = 236.7 J\nIf the above amount of energy were delivered by my muscles in one lap (10 meters, since that was the length each subject had to run) at 8.5 kmh-1, it would have taken 8.47/2 seconds (only half a lap, 10 meters). So:\nP = ?E/?t P = 236.7/4.235 P = 55.9 Js-1, W\nHowever, if I ran 20 meters, then the power is:\nP = 236.7/8.47 P = 27.9 Js-1, W\nThe amount of energy accumulated while running a lap is then dissipated towards the end (of each lap) as the subject must come to rest i.e. the velocity is zero. Notice also that because the subject has to accelerate at the beginning of every lap, some extra energy is needed for that acceleration. The subject must accelerate every 10 meters because he/she has stop and then run in the opposite direction.\nAs the lap distance decreases, the power transfer increases. This shortage (of lap distance) will also cause the subject to accelerate and decelerate more often. Therefore, the smaller the lap distance, the larger the error could be (due to the fact that some of the energy is used up in accelerating). Using a 10 meter track instead of a 20 meter track could have affected the results because this meant more energy used in accelerating. It is therefore, justifiable to say that had the track been longer (i.e. 20 meters), myself and all of the other subjects could have been able to sustain a higher speed instead of the one that was achieved.\n\nDon't use plagiarized sources. Get Your Custom Essay on\nVO2 Max and Aerobic Power\nJust from \\$13/Page\n\n## Calculate the price of your order\n\n550 words\nWe'll send you the first draft for approval by September 11, 2018 at 10:52 AM\nTotal price:\n\\$26\nThe price is based on these factors:\nAcademic level\nNumber of pages\nUrgency\nBasic features\n• Free title page and bibliography\n• Unlimited revisions\n• Plagiarism-free guarantee\n• Money-back guarantee\n• 24/7 support\nOn-demand options\n• Writer’s samples\n• Part-by-part delivery\n• Overnight delivery\n• Copies of used sources\n• Expert Proofreading\nPaper format\n• 275 words per page\n• 12 pt Arial/Times New Roman\n• Double line spacing\n• Any citation style (APA, MLA, Chicago/Turabian, Harvard)\n\n# Our guarantees\n\nDelivering a high-quality product at a reasonable price is not enough anymore.\nThat’s why we have developed 5 beneficial guarantees that will make your experience with our service enjoyable, easy, and safe.\n\n### Money-back guarantee\n\nYou have to be 100% sure of the quality of your product to give a money-back guarantee. This describes us perfectly. Make sure that this guarantee is totally transparent.\n\n### Zero-plagiarism guarantee\n\nEach paper is composed from scratch, according to your instructions. It is then checked by our plagiarism-detection software. There is no gap where plagiarism could squeeze in.\n\n### Free-revision policy\n\nThanks to our free revisions, there is no way for you to be unsatisfied. We will work on your paper until you are completely happy with the result.\n\n### Privacy policy\n\nYour email is safe, as we store it according to international data protection rules. Your bank details are secure, as we use only reliable payment systems.\n\n### Fair-cooperation guarantee\n\nBy sending us your money, you buy the service we provide. Check out our terms and conditions if you prefer business talks to be laid out in official language.\n\nOrder your essay today and save 20% with the discount code WELCOME" ]

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[ "# Excel TAKE function to extract rows or columns from array", null, "by , updated on\n\nMeet the new Excel TAKE function that can get a specified number of rows or columns from a range or array in a trice.\n\nWhen working with large arrays of data, you may sometimes need to extract a smaller part for closer examination. With the new dynamic array function introduced in Excel 365, it will be a walk in the park for you. Just specify how many rows and columns you want to take and hit the Enter key :)\n\n## TAKE function in Excel\n\nThe Excel TAKE function extracts the specified number of contiguous rows and/or columns from the array or range.\n\nThe syntax is as follows:\n\nTAKE(array, rows, [columns])\n\nWhere:\n\nArray (required) - the source array or range.\n\nRows (optional) - the number of rows to return. A positive value takes rows from the start of the array and a negative value from the end of the array. If omitted, columns must be set.\n\nColumns (optional) - the number of columns to return. A positive integer takes columns from the start of the array and a negative integer from the end of the array. If omitted, rows must be defined.\n\nHere's how the TAKE function looks like:", null, "Tips:\n\n• To return non-adjacent rows from a range, use the CHOOSEROWS function.\n• To pull non-adjacent columns, utilize the CHOOSECOLS function.\n• To get part of the array by removing a given number of rows or columns, leverage the DROP function.\n\n### TAKE function availability\n\nThe TAKE function is only supported in Excel for Microsoft 365 (Windows and Mac) and Excel for the web.\n\nIn earlier Excel versions, you can use an OFFSET formula as an alternative solution.\n\n## How to use TAKE function in Excel\n\nTo align expectations and reality when using the TAKE function in your worksheets, take notice of the following things:\n\n1. The array argument can be a range of cells or an array of values returned by another formula.\n2. The rows and columns arguments can be positive or negative integers. Positive numbers take a subset of data from the start of the array; negative numbers - from the end.\n3. The rows and columns arguments are optional, but at least one of them should be set in a formula. The omitted one defaults to the total number of rows or columns in the array.\n4. If the rows or columns value is greater than there are rows or columns in the source array, all rows / columns are returned.\n5. TAKE is a dynamic array function. You enter the formula in only one cell and it automatically spills into as many neighboring cells as needed.\n\n## Excel TAKE formula examples\n\nNow that you have a general understanding of how the TAKE function works, let's look at some practical examples to illustrate its real value.\n\n### Extract rows from a range or array\n\nTo return a given number of contiguous rows from the start of a 2D array or range, supply a positive number for the rows argument.\n\nFor example, to take the first 4 rows from the range A3:C14, the formula is:\n\n`=TAKE(A3:C14, 4)`\n\nThe formula lands in cell E3 and spills into four rows and as many columns as there are in the source range.", null, "### Take columns from an array or range\n\nTo get a certain number of contiguous columns from the start of a 2D array or range, provide a positive number for the columns argument.\n\nFor example, to pull the first 2 columns from the range A3:C14, the formula is:\n\n`=TAKE(A3:C14, ,2)`\n\nThe formula goes to cell E3 and spills into two columns and as many rows as there are in the supplied range.", null, "### Extract a certain number of rows and columns\n\nTo retrieve a given number of rows and columns from the beginning of an array, you provide positive numbers for both the rows and columns arguments.\n\nFor example, to take the first 4 rows and 2 columns from our dataset, the formula is:\n\n`=TAKE(A3:C14, 4, 2)`\n\nEntered in E3, the formula fills four rows (as set in the 2nd argument) and two columns (as defined in the 3rd argument).", null, "### Get last N rows\n\nTo pull a certain number of rows from the end of an array, provide a negative number for the rows argument. For example:\n\nTo take the last row, use -1:\n\n`=TAKE(A3:C14, -1)`\n\nTo get the last 3 rows, supply -3:\n\n`=TAKE(A3:C14, -3)`\n\nIn the screenshot below, you can observe the results.", null, "### Return last N columns\n\nTo extract some columns from the end of an array or range, use a negative number for the columns argument. For example:\n\nTo get the last column, set the 3rd argument to -1:\n\n`=TAKE(A3:C14, , -1)`\n\nTo pull the last 2 columns, set the 3rd argument to -2:\n\n`=TAKE(A3:C14, , -2)`\n\nAnd here are the results:", null, "Tip. To take rows and columns from the end of an array, provide negative numbers for both the rows and columns arguments.\n\n## How to take rows / columns from multiple ranges\n\nIn situation when you want to extract some columns or rows from several non-contiguous ranges, it takes two steps to accomplish the task:\n\n1. Combine multiple ranges into one vertically or horizontally using the VSTACK or HSTACK function.\n2. Return the desired number of columns or rows from the combined array.\n\nDepending on the structure of your worksheet, apply one of the following solutions.\n\n### Stack ranges vertically and take rows or columns\n\nLet's say you have 3 separate ranges like shown in the image below. To append each subsequent range to the bottom of the previous one, the formula is:\n\n`=VSTACK(A4:C6, A10:C14, A18:C21)`\n\nNest it in the array argument of TAKE, specify how many rows to return, and you will get the result you are looking for:\n\n`=TAKE(VSTACK(A4:C6, A10:C14, A18:C21), 4)`\n\nTo return columns, type the appropriate number in the 3rd argument:\n\n`=TAKE(VSTACK(A4:C6, A10:C14, A18:C21), ,2)`\n\nThe output will look something like this:", null, "### Stack ranges horizontally and take rows or columns\n\nIn case the data in the source ranges is arranged horizontally in rows, use the HSTACK function to combine them into a single array. For example:\n\n`=HSTACK(B3:D5, G3:H5, K3:L5)`\n\nAnd then, you place the above formula inside the TAKE function and set the rows or columns argument, or both, according to your needs.\n\nFor example, to get the first 2 rows from the stacked array, the formula is:\n\n`=TAKE(HSTACK(B3:D5, G3:H5, K3:L5), 2)`\n\nAnd this formula will bring the last 5 columns:\n\n`=TAKE(HSTACK(B3:D5, G3:H5, K3:L5), ,5)`", null, "## TAKE function alternative for Excel 2010 - 365\n\nIn Excel 2019 and earlier versions where the TAKE function is not supported, you can use OFFSET as an alternative. Though the OFFSET formula is not as intuitive and straightforward, it does provide a working solution. And here's how you set it up:\n\n1. For the 1st argument, supply the original range of values.\n2. The 2nd and 3rd arguments or both set to zero or omitted, assuming you are extracting a subset from the beginning of the array. Optionally, you can specify how may rows and columns to offset from the upper-left cell of the array.\n3. In the 4th argument, indicate the number of rows to return.\n4. In the 5th argument, define the number of columns to return.\n\nSumming up, the generic formula takes this form:\n\nOFFSET(array, , , rows, columns)\n\nFor instance, to extract 6 rows and 2 columns from the start of the range A3:C14, the formula goes as follows:\n\n`=OFFSET(A3:C14, , , 6, 2)`\n\nIn all versions except Excel 365 and 2021 that handle arrays natively, this only works as a traditional CSE array formula. There are two ways to enter it:\n\n• Select the range of cells the same size as the expected output (6 rows and 2 columns in our case) and press F2 to enter Edit mode. Type the formula and press Ctrl + Shift + Enter to enter it in all the selected cells at once.\n• Enter the formula in any empty cell (E3 in this example) and press Ctrl + Shift + Enter to complete it. After that, drag the formula down and to the right across as many rows and columns as needed.\n\nThe result will look similar to this:", null, "Note. Please be aware that OFFSET is a volatile function, and it may slow down your worksheet if used in many cells.\n\n## Excel TAKE function not working\n\nIn case a TAKE formula does not work in your Excel or results in an error, it's most likely to be one of the below reasons.\n\n### TAKE is not supported in your version of Excel\n\nTAKE is a new function and it has limited availability. If your version is other than Excel 365, try the alternative OFFSET formula.\n\n### Empty array\n\nIf either the rows or columns argument is set to 0, a #CALC! error is returned indicating an empty array.\n\n### Insufficient number of blank cells to fill with the results\n\nIn case there are not enough empty cells for the formula to spill the results into, a #SPILL error occurs. To fix it, just clear the neighboring cells below or/and to the right. For more details, see How to resolve #SPILL! error in Excel.\n\nThat's how to use the TAKE function in Excel to extract rows or columns from a range of cells. I thank you for reading and hope to see you on our blog next week!" ]

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[ "Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.\n\nA popular approach to deal with missing values is to impute the data to get a complete dataset on which any statistical method can be applied. Many imputation methods are available and provide a completed dataset in any cases, whatever the number of individuals and/or variables, the percentage of missing values, the pattern of missing values, the relationships between variables, etc.\n\nHowever, can we believe in these imputations and in the analyses performed on these imputed datasets?\n\nMultiple imputation generates several imputed datasets and the variance between-imputations reflects the uncertainty of the predictions of the missing entries (using an imputation model). In the missMDA package we propose a way to visualize the uncertainty associated to the predictions. The rough idea is to project all the multiple imputed datasets on the PCA graphical representations obtained from the “mean” imputed dataset.\n\nFor instance, for the incomplete orange data, the two following graphs read as follows: observation 6 has no uncertainty (there is no missing value for this observation) whereas there is more variability on the position of observation 10. For the variables, the clouds of points represent the uncertainties on the predictions. Ellipses as well as clouds are quite small and encourage to carry-on the analysis on the imputed dataset.", null, "", null, "The graphics above where obtained after performing multiple imputation with PCA simply be obtained using the function plot.MIPCA as follows:\n\nlibrary(missMDA)\ndata(orange)\nnbdim <- estim_ncpPCA(orange) # estimate the number of dimensions to impute\nres.comp <- MIPCA(orange, ncp = nbdim\\$ncp, nboot = 1000)\nplot(res.comp)\n\nNow we have hints to answer the famous questions: “I have a dataset with xx% of missing values, can I impute it with your method?” or “Is 30% of missing values too much or not?” or “What is the maximum percentage of missing values?” Indeed, the percentage of missing values impacts the quality of the imputation but not only! The structure of the data (i.e. the relationships between variables) is very important. It is indeed possible to have small ellipses with a high percentage of missing values and the other way around. That is why these graphs are useful. The following ones suggest that we must be very careful with subsequent analyses on the imputed dataset, and even it suggests stopping the analysis of this dataset. When there’s nothing good to do, it’s better to do nothing!", null, "", null, "This methodology is also available for categorical data with the functions MIMCA and plot.MIMCA to visualize the uncertainty around the prediction of categories.\n\n[email protected]       @JulieJosseStat\n[email protected]", null, "", null, "" ]

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[ "", null, "# 84.44 square kilometers to square centimeters\n\nto\n\n84.44 Square kilometers = 844,400,000,000 Square centimeters\n\nThis conversion of 84.44 square kilometers to square centimeters has been calculated by multiplying 84.44 square kilometers by 10,000,000,000 and the result is 844,400,000,000 square centimeters." ]

[ null, "https://unitconverter.io/img/area.png", null ]

[ "Document number: P2438R0 Date: 2021-09-14 Project: Programming Language C++, Library Working Group Reply-to:\n\n## 1. Before/After table\n\n without this proposal with this proposal ``````auto a = std::string(/* */); auto b = a.substr(/* */);`````` Value of `a` does not change Value of `a` does not change ``````auto foo() -> std::string; auto b = foo().substr(/* */);`````` `foo()` returns a temporary `std::string`. `.substr` creates a new string and copies the relevant content. At last the temporary string returned by foo is released. `foo()` returns a `std::string`. `.substr` implementation can reuse the storage of the string returned by foo and leave it in a valid but unspecified state. At last the temporary string returned by `foo()` is released. ``auto a = std::string(/* */).substr(/* */);`` A temporary `std::string` is created, on that instance `.substr` creates a new string and copies the relevant content. At last the temporary string is released. A temporary `std::string` is created, on that instance `.substr` implementation can reuse the storage and leave the temporary string in a valid but unspecified state. At last the temporary string is released. ``````auto a = std::string(/* */); auto b = std::move(a).substr(/* */);`````` Value of `a` does not change As `a` is casted to an xvalue, the implementation of `.substr` can reuse the storage and leave `this` string in a valid but unspecified state.\n\n## 2. Motivation\n\nSince C++11 the C++ language supports move semantic. All classes where it made sense where updated with move constructors and move assignment operators. This made it possible to take advantage of rvalues and \"steal\" resources, thus avoiding, for example, unnecessary costly copies.\n\nSome classes that came in later revisions of the language also take advantage of move semantic for member functions, like `std::optional::value` and `std::optional::value_or`.\n\nIn the case of `std::string::substr()`, it is possible to take advantage of move semantic to.\n\nConsider following two code snippets:\n\n``````// example 1\nbenchmark = std::string(argv[i]).substr(12);\n\n// example 2\nname_ = obs.stringValue().substr(0,32);``````\n\nIn the first example, `argv` is copied in a temporary string, then `substr` creates a new object. In this case one could use `string_view` to avoid the unnecessary copy, but changing already working code has a cost too.\n\nIn the second example, if `stringValue()` returns an `std::string` by value, the user of that API cannot use a `string_view` to avoid an unnecessary copy, like in the first case.\n\nIf `std::string` would have an overload for `substr() &&`, in both cases the standard library could avoid unnecessary work, and instead of copying the data \"steal\" it.\n\nIt is true that adding a new overload increases the already extremely high number of member functions of `std::string`.\n\nOn the other hand most users do not need to know it’s existence to take advantage of the provided optimization.\n\nThus this paper is not extending API surface, there is no names or behavior to be learned by user, and we just get extension that follows established language convection.\n\nFor users aware of the overload, they can move a string in order to \"steal\" it’s storage in a natural way:\n\n``````std::string foo = ...;\nstd::string bar = std::move(foo).substr(...);``````\n\n### 2.1. Couldn’t a library vendor provide such overload as QOI?\n\nNo, because it is a breaking change. Fur such library, following code would misbehave\n\n``````std::string foo = ...;\nstd::string bar = std::move(foo).substr(...);``````\n\n`[res.on.arguments]` says that a programmer can’t expect an object referred to by an rvalue reference to remain untouched. But there is currently no rvalue reference in `substr()`. This paper is proposing to add it.\n\n## 3. Design Decisions\n\nThis is purely a library extension.\n\nCurrently `substr` is defined as\n\n``constexpr basic_string substr(size_type pos = 0, size_type n = npos) const;``\n\nThis paper proposes to define following overloads\n\n``````constexpr basic_string substr(size_type pos = 0, size_type n = npos) const &;\nconstexpr basic_string substr(size_type pos = 0, size_type n = npos) &&;``````\n\nOther overloads (`constexpr basic_string substr(size_type pos = 0, size_type n = npos) const &&;` and `constexpr basic_string substr(size_type pos = 0, size_type n = npos) &;`) are not necessary.\n\nNotice that the current proposal is a breaking change, as following snippet of code might work differently if this paper gets accepted:\n\n``````std::string foo = ...;\nstd::string bar = std::move(foo).substr(...);``````\n\nUntil C++20, `foo` wont change it’s value, after this paper, the state of `foo` would be in a \"valid but unspecified state\".\n\nWhile a breaking change is generally bad:\n\n• I do not think there exists code like `std::move(foo).substr(…​)` in the wild\n\n• Even if such code exists, the intention of the author was very probably to tell the compiler that he is not interested in the value of `foo` anymore, as it is normally the case when using `std::move` on a variable. In other words, with this proposal the user is getting what he asked for.\n\nThe standard library proposes two way for creating a \"substring\" instance, either by calling \"substr\" method or via constructor that accepts (str, pos, len). We see both of them as different spelling of same functionality, and believe they behavior should remaining consistent. Thus we propose to add rvalue overload constructors.\n\n``````constexpr basic_string( basic_string&& other, size_type pos, const Allocator& alloc = Allocator() );\nconstexpr basic_string( basic_string&& other, size_type pos, size_type count, const Allocator& alloc = Allocator() );``````\n\n### 3.1. Note on the propagation of the allocator\n\n`basic_string` is one of the allocator-container, which means that any memory resource used by this class need to be acquired and released to from the associated allocator instance. This imposes some limitation on the behavior of the proposed overload. For example in:\n\n``````std::pmr::string s1 = ....;\nstd::pmr::string s2 = std::move(s1).substr();``````\n\nFor `s2` to be able to steal memory from `s1`, we need to be sure that the allocators used by both objects are equal (`s1.get_allocator() == s2.get_allocator()`). This is trivially achievable for the case of the for the allocators that are always equal (`std::allocator_traits<A>::is_always_equal::value` is true), including most common case of the stateless `std::allocator` and implementation can unconditionally steal any allocated memory in such situation.\n\nMoreover, the proposed overload can still provide some optimization in case of the stateful allocators, where `s2.get_allocator()` (which is required to be default constructed) happens to be the same as allocator of the source `s1`. In any remaining cases, behavior of this overload should follow existing const version, and as such it does not add any overhead.\n\nThis paper, recommends implementation to avoid additional memory allocation when possible (note if no-allocation would be performed, there is nothing to avoid), however it does not require so. This leave it free for implementation to decide, if the optimization should be guarded by:\n\n• compile time check of `std::allocator_traits<A>::is_always_equal`\n\n• runtime comparison of allocators instance (addition comparison cost).\n\n### 3.2. Overload with user supplied-allocator:\n\nWhile writing the paper, we have noticed that specification of the `substr()` requires returned object to use default constructed allocator. This means that invocation of this function is ill-formed for the `basic_string` instance with non-default constructing allocator, for example for invited `memory_pool_allocator<char>` that can be only constructed from reference to the pool, the following are ill-formed:\n\n``````memory_pool pool = ...;\nstd::basic_string<char, std::char_traits<char>, memory_pool_allocator<char>> s1(memory_pool_allocator<char>(pool));\nauto s2 = s1.substr();``````\n\nThis could be address by adding Allocator parameters to `substr()` overload that accepts allocator to be used as parameter:\n\n``````constexpr basic_string substr(size_type pos, const Allocator& alloc) const;\nconstexpr basic_string substr(size_type pos, size_type n, const Allocator& alloc) const;``````\n\nWhile the authors think that this additional feature is related to proposed changes, it is orthogonal to them and could be handled as separate paper. We seek LEWG guidance if that functionality should be included in the paper.\n\n### 3.3. Are they any other function of `std::string` that would benefit from a `&&` overload\n\nThe member function `append` and `operator+=` take `std::string` as const-ref parameter\n\n``````constexpr basic_string& operator+=( const basic_string& str );\n\nconstexpr basic_string& append(const basic_string& str);\nconstexpr basic_string& append(const basic_string& str, size_type pos, size_type n = npos);``````\n\nBut in this case, because of the interaction of two string instances, the benefits from stealing the resource of `str` are less clear. Supposing both string instances use the same allocator, an implementation should compare the capacity of `str` and `this`, and evaluate if moving `str.size()` elements is less costly than copying them. This would make the implementation of `append` less obvious, and the performance implications are difficult to predict.\n\nFor those reasons, the authors does not propose to add new overloads for `append` and `operator+`.\n\nThe authors are not aware of other functions that could benefit from a `&&` overload.\n\n### 3.4. Concerns on ABI stability\n\nChanging `basic_string substr(std::size_t pos, std::size_t len) const;` into `basic_string substr(std::size_t pos, std::size_t len) const&;` and `basic_string substr(std::size_t pos, std::size_t len) &&;` (the first change is required by the core language rules), can affect the mangling of the name, thus causing ABI break.\n\nFor a library it is possible to continue to define the old symbol, so that already existing code will continue to links and work without errors. For example, it is possible to use asm to define the old mangled name as an alias for the new `const&` symbol.\n\nThis is not a novel technique, as it has been explained by the ARG (ABI Review group), and similar breaks have already taken place for other papers, like P0408.\n\n## 4. Technical Specifications\n\nSuggested wording (against N4892):\n\nApply following modifications to definition of `basic_string class template in [basic.string.general] General`.\n\n```constexpr basic_string(const basic_string& str, size_type pos, const Allocator& a = Allocator());\nconstexpr basic_string(const basic_string& str, size_type pos, size_type n, const Allocator& a = Allocator());\nconstexpr basic_string( basic_string&& str, size_type pos, const Allocator& alloc = Allocator() );\nconstexpr basic_string( basic_string&& str, size_type pos, size_type n, const Allocator& alloc = Allocator() );```\n\nand\n\n```constexpr basic_string substr(size_type pos = 0, size_type n = npos) const &;\nconstexpr basic_string substr(size_type pos = 0, size_type n = npos) &&;```\n\nReplace the definition of the corresponding constructor `[string.cons] Constructors and assignment operators`\n\nWording note: We no longer define this constructors in terms of being equivalent to corresponding construction from `basic_string_view`, as that would prevent reuse of the memory, that we want to allow. The use of \"prior the call\", are not necessary for `const&`, but allow us to merge the wording.\n\n```constexpr basic_string(const basic_string& str, size_type pos, const Allocator& a = Allocator());\nconstexpr basic_string(const basic_string& str, size_type pos, size_type n, const Allocator& a = Allocator());\nconstexpr basic_string( basic_string&& str, size_type pos, const Allocator& alloc = Allocator() );\nconstexpr basic_string( basic_string&& str, size_type pos, size_type n, const Allocator& alloc = Allocator() );```\n\nEffects: Let `n` be `npos` for the first overload. Equivalent to: `basic_string(basic_string_view<charT, traits>(str).substr(pos, n), a)`.\nLet:\n\n• `s` be the value of `str` prior this call,\n\n• `rlen` be smaller of `n` and `s.size() - pos`, for overloads that define parameter `n`, and `s.size() - pos` otherwise.\n\nEffects: Constructs an object whose initial value is the range `[s.data() + pos, rlen)`\nThrows: `out_­of_­range` if `pos > s.size()`\nRemarks: The `str` is in valid but unspecified state, after invocation of either third or fourth overload.\nRecommended practice: For third and fourth overload implementations should avoid unnecessary copies and allocations, if `s.get_allocator() == a` is `true`.\n\nApply following changes to `[string.substr] basic_­string​::​substr`.\n\n`constexpr basic_string substr(size_type pos = 0, size_type n = npos) const &;`\n\nEffects: Determines the effective length `rlen` of the string to copy as the smaller of n and `size() - pos`.\nReturns: `basic_­string(data()+pos, rlen)`.\nThrows: `out_­of_­range` if `pos > size()`.\nEffects: Equivalent to: `return basic_string(*this, pos, n);`\n\n`constexpr basic_string substr(size_type pos = 0, size_type n = npos) &&;`\n\nEffects: Equivalent to: `return basic_string(std::move(*this), pos, n);`.\n\n## 5. Acknowledgements\n\nA big thank you to all those giving feedback for this paper." ]

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[ "Home »Self Introduction Email Template »Self Introduction Email Template\n\n# Self Introduction Email Template\n\nSelf Introduction Email Template from Sakamotoryouma.website one of ideas and inspiration for you to find any ideas for your Template Design Inspiration. this photo from Self Introduction Email Template\n\n## Self Introduction Email Template\n\n8 sample self introduction email to colleagues formal, 9 business introduction email templates sampletemplatess, self introduction letter to clients 2018 letter format, how to write an introduction email that wins you an in, 7 how to write letter of introduction for job, 5 self introduction mail introduction letter, 5 how to write a letter about yourself memo formats, 8 self introduction email sample introduction letter, writing a letter of introduction for yourself best, 5 self introduction letter introduction letter, 14 email introduction samples waa mood, 7 introduction email sample memo formats, self introduction quotes quotesgram, 8 basics of introduction email to client, 13 sample business introduction letters pdf doc, sample introduction letter new employee to customers, sample business introduction letter work related design, how to introduce yourself in an email to colleagues, sample self introduction letter for new employee, 8 sample self introduction email to colleagues, 4 sample self introduction email to client lease template, self introduction email, self introduction letter to customers sample, 9 self introduction to client email introduction letter, sample self introduction email to client lease template, 16 professional email examples pdf doc, 5 professional self introduction format introduction letter, 7 self introduction email sample to client introduction, 8 self introductory email sample writing a memo, introduction letter for new employee template, 5 self introduction letter sample memo formats, how to introduce yourself colleagues in email sample, self introduction email template job offer email welcome, how to introduce in email, 8 self introduction email sample for new employee, best photos of self introduction letter for employment, . many reference that we have, you can find other reference such as Apartment, Architecture, Bathroom, Bedroom, Furniture, Interior and more. you can find from gallery Self Introduction Email Template.\n\n Post name : 6 self introduction email sample memo formats Image Size width : 1277 px Image Size Height : 1652 px Date Post : 2018-08-31 17:45:46 Images Location : Desktop | Mobile" ]

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[ "» » Adding mixed fractions ideas\n\nIf you’re searching for adding mixed fractions images information related to the adding mixed fractions interest, you have pay a visit to the right blog. Our site always provides you with suggestions for seeking the maximum quality video and picture content, please kindly hunt and locate more enlightening video content and graphics that fit your interests.\n\nAdding Mixed Fractions. Add the numerators put that answer over the denominator Step 3. 31 4 2 1 4 1 10. Mixed fractions consist of three parts. A non-unit fraction is a fraction where the numerator is not 1 eg ⅘.", null, "Subtracting Mixed Numbers Same Denominators Adding Fractions Adding Improper Fractions Fractions Worksheets From pinterest.com\n\nAdd the numerators put that answer over the denominator Step 3. These are called the denominators. 110 11 1 6 11 38 11 3 5 11 9. Here are some examples of mixed number fractions. Its easy to add and subtract fractions when the numbers on the bottom are the same. Add to My Bitesize A unit fraction is any fraction where the numerator is 1 eg.\n\n### 31 4 1 5 6 17 12 1 5 12 2.\n\nThese are called the denominators. You could first convert each to an improper fraction. Adding and subtracting. When adding mixed numbers you can use a similar method to adding two fractions but this time you have to add whole numbers as well. 9 ⅔ 6 ⅓. 22022018 Two mazes where the aim is to get from start to finish only using the boxes with expressions that equal 10.", null, "Source: pinterest.com\n\n22022018 Two mazes where the aim is to get from start to finish only using the boxes with expressions that equal 10. Adding and subtracting fractions. If they dont have common denominators then find a common denominator and use it to rewrite each fraction. 11 6 1 1 7 1 42 12. Frac 20 7 20 div 7 2 remainder 6 so.", null, "Source: pinterest.com\n\nA non-unit fraction is a fraction where the numerator is not 1 eg ⅘. Here are some examples of mixed number fractions. Mixed fractions consist of three parts. 53 4 1 7 8 61 8 7 5 8 7. These are called the denominators.", null, "Source: pinterest.com\n\nAdding and Subtracting Mixed Fractions F Answers Find the value of each expression in lowest terms. Make sure the bottom numbers the denominators are the same Step 2. 1 8 11 1 1 4 21 44 11. A mixed fraction occurs when a whole number and a proper fraction combine. A non-unit fraction is a fraction where the numerator is not 1 eg ⅘.", null, "Source: pinterest.com\n\n101 2 3 2 3 41 6 6 5 6 5. When adding mixed number fractions you must first add the whole numbers together and then add the fractions. 101 2 3 2 3 41 6 6 5 6 5. Frac 20 7 20 div 7 2 remainder 6 so. These are called the denominators.", null, "Source: pinterest.com\n\n53 4 1 7 8 61 8 7 5 8 7. 21 2 1 4 7 13 14 8. Add the numerators put that answer over the denominator Step 3. Make sure the bottom numbers the denominators are the same Step 2. When adding mixed numbers you can use a similar method to adding two fractions but this time you have to add whole numbers as well.", null, "Source: pinterest.com\n\n31 4 1 5 6 17 12 1 5 12 2. 1 8 11 1 1 4 21 44 11. Its easy to add and subtract fractions when the numbers on the bottom are the same. 31 4 1 5 6 17 12 1 5 12 2. Make sure the bottom numbers the denominators are the same Step 2.", null, "Source: pinterest.com\n\nTo add fractions there are Three Simple Steps. 21 2 1 4 7 13 14 8. To add or subtract mixed numbers it is usually easiest to change them to improper fractions first and then change the answer back into a mixed number if needed. The whole number the numerator top half of the fraction and the denominator the bottom half of the fraction. 1 8 11 1 1 4 21 44 11.", null, "Source: pinterest.com\n\nIts easy to add and subtract fractions when the numbers on the bottom are the same. 110 11 1 6 11 38 11 3 5 11 9. The whole number the numerator top half of the fraction and the denominator the bottom half of the fraction. Frac 20 7 frac 7 7 frac 7 7 frac 6 7 2frac 6 7 Method 2. Mixed fractions are a mix of whole numbers.", null, "Source: pinterest.com\n\nAdd the numerators put that answer over the denominator Step 3. The whole number the numerator top half of the fraction and the denominator the bottom half of the fraction. A mixed fraction occurs when a whole number and a proper fraction combine. 101 2 3 2 3 41 6 6 5 6 5. 31 3 1 7 8 35 24 1 11 24 6.", null, "Source: pinterest.com\n\nMake sure the bottom numbers the denominators are the same Step 2. To add fractions there are Three Simple Steps. Add to My Bitesize A unit fraction is any fraction where the numerator is 1 eg. 72 3 2 1 2 31 6 5 1 6 4. 9 ⅔ 6 ⅓.", null, "Source: pinterest.com\n\n21 2 1 4 7 13 14 8. 1 8 11 1 1 4 21 44 11. These are called the denominators. Simplify the fraction if needed. Its easy to add and subtract fractions when the numbers on the bottom are the same.", null, "Source: pinterest.com\n\n21 2 1 4 7 13 14 8. 11 6 1 1 7 1 42 12. You could first convert each to an improper fraction. You can go vertically or horizontally but not diagonally. Add the numerators put that answer over the denominator Step 3.", null, "Source: pinterest.com\n\nWhen adding mixed number fractions you must first add the whole numbers together and then add the fractions. Frac 20 7 frac 7 7 frac 7 7 frac 6 7 2frac 6 7 Method 2. When adding mixed numbers you can use a similar method to adding two fractions but this time you have to add whole numbers as well. Simplify the fraction if needed. 9 ⅔ 6 ⅓.", null, "Source: pinterest.com\n\nIf they dont have common denominators then find a common denominator and use it to rewrite each fraction. 53 4 1 7 8 61 8 7 5 8 7. 9 ⅔ 6 ⅓. 72 3 2 1 2 31 6 5 1 6 4. Frac 20 7 20 div 7 2 remainder 6 so.", null, "Source: pinterest.com\n\nTo add or subtract mixed numbers it is usually easiest to change them to improper fractions first and then change the answer back into a mixed number if needed. 53 4 1 7 8 61 8 7 5 8 7. These are called the denominators. Simplify the fraction if needed. Add to My Bitesize A unit fraction is any fraction where the numerator is 1 eg.", null, "Source: pinterest.com\n\nFrac 20 7 frac 7 7 frac 7 7 frac 6 7 2frac 6 7 Method 2. Then add the fractions together and simplify. You could first convert each to an improper fraction. Frac 20 7 frac 7 7 frac 7 7 frac 6 7 2frac 6 7 Method 2. To add or subtract mixed numbers it is usually easiest to change them to improper fractions first and then change the answer back into a mixed number if needed.", null, "Source: pinterest.com\n\n1 8 11 1 1 4 21 44 11. 21 2 1 4 7 13 14 8. You could first convert each to an improper fraction. 15 6 10 1 2 37 3 12 1 3 3. 1 8 11 1 1 4 21 44 11.", null, "Source: pinterest.com\n\nWhen adding mixed numbers you can use a similar method to adding two fractions but this time you have to add whole numbers as well. 72 3 2 1 2 31 6 5 1 6 4. 11 6 1 1 7 1 42 12. 101 2 3 2 3 41 6 6 5 6 5. A mixed fraction occurs when a whole number and a proper fraction combine.\n\nThis site is an open community for users to do submittion their favorite wallpapers on the internet, all images or pictures in this website are for personal wallpaper use only, it is stricly prohibited to use this wallpaper for commercial purposes, if you are the author and find this image is shared without your permission, please kindly raise a DMCA report to Us.\n\nIf you find this site value, please support us by sharing this posts to your preference social media accounts like Facebook, Instagram and so on or you can also bookmark this blog page with the title adding mixed fractions by using Ctrl + D for devices a laptop with a Windows operating system or Command + D for laptops with an Apple operating system. If you use a smartphone, you can also use the drawer menu of the browser you are using. Whether it’s a Windows, Mac, iOS or Android operating system, you will still be able to bookmark this website." ]

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[ "×\n\nSearch anything:\n\n# Counter Objects in Python\n\n#### Software Engineering Python", null, "Get this book -> Problems on Array: For Interviews and Competitive Programming\n\nReading time: 15 minutes | Coding Time: 10 minutes\n\nA counter is basically a part of collections module and is a subclass of a dictionary which is used to keep a track off elements and their count. In Python, elements are stored as =Dict= keys and their counts are stored as dict values. These count elements can be positive, zero or negative integers. Simply speaking, if you add the value ‘hello’ thrice in the container, it will remember that you added it thrice. So, Counter counts hashable objects in Python.\n\nA counter class has no restrictions on its keys and values i.e. one can store anything in the value field.\n\n# Creating Python Counter Objects :\n\nCounters support three basic types of initialization. It can be called with iterable items or dictionary containing keys and counts or keyword args like mapping.\n\nFor creating a new empty counter:\n\n``````counter = Counter()\nprint(counter)\n--> Counter()\n``````\n\nFor creating a new counter with values:\n\n``````counter = Counter(['a','b','c','a','a'])\nprint(counter)\n--> Counter({'a':3, 'b':2, 'c':1})\n``````\n\nFor creating a new counter with iterable arguments:\n\n``````counter = Counter('abcdac')\nprint(counter)\n--> Counter({'a':2, 'b':1, 'c': 2, 'd':1})\n``````\n\nFor creating a new counter with Dictionary arguments:\n\n``````word = {'blue': 1, 'white': 3, 'black': 2}\ncounter = Counter(word)\nprint(counter)\n--> Counter({'white': 3, 'black': 2, 'blue': 1})\n``````\n\nFor creating a new counter with List arguments:\n\n``````list = ['Black', 'Blue', 'Blue', 'Red']\ncounter = Counter(list)\nprint(counter)\n--> Counter({'Blue': 2, 'Black': 1, 'Red': 1})\n``````\n\nNon-numeric data can also be used for counter values.\n\n# Accessing Counters in Python\n\nCounters are accessed like dictionaries. If any key is absent, it does not raise any KeyValue error, instead it shows the count as zero.\n\n``````c = collections.Counter('abcdefacc')\n\nfor letter in 'abcz':\nprint (letter, c[letter])\n\n--> a 2\nb 1\nc 3\nz 0\n``````\n\n# Getting Count of Elements in a Counter\n\nWe easily count the number of elements in a counter like shown in the example below. For a non-existing key, it will return zero.\n\n``````c = Counter({'a': 5, 'b': 3, 'c': 6})\nc1 = counter['b']\nprint(c1)\n\n--> 3\n``````\n\n# Setting Count of Elements in a Counter\n\nWe can set the count of an element in the counter as shown in the example below. For a non-existing key, it will get added to the counter.\n\n``````c = Counter({'a': 5, 'b': 3, 'c': 6})\nc['b']=0\nprint(c)\n\n--> Counter({'a': 5, 'b': 0, 'c': 6})\n``````\n\n# Deleting an element from a Counter\n\nAny element can easily be removed or deleted from a counter object by using =del=.\n\n``````c= Counter( {'A':3,'B':1,'C':5})\nprint(c)\n\ndel c['B']\nprint(c)\n\n--> Counter({'C':5, 'A':3, 'B':1})\nCounter({'C':5,'A':3})\n``````\n\n# Python Counter Methods\n\nCounter objects support three methods apart from those available to all dictionaries. Namely, elements(), most_common() and subtract().\n\n### elements()\n\nThis method returns the list of elements in the counter.\nOnly elements with positive counts are returned, if it is less than one, it gets ignored.\n\n``````c=Counter({'red':2, 'blue':-3, 'green':0})\n\ne= c.elements()\nfor value in e:\nprint(value)\n\n--> red\nred\n``````\n\nThe code above will print 'red' twice. This is because it's countg is 2. The other elements- 'blue' and 'green' will be ignored as they do not have a positive value like 'red'.\n\n### most_common(n)\n\nThis method returns the most common elements from the counter from most common to least common.\nIf the value of 'n' is not given, it returns all the elements in the counter.\n\n``````c=Counter({'red':2, 'blue':-1, 'green':0})\n\nm= c.most_common(1)\nprint(m)\n\n--> [('red' , 2)]\n``````\n\nThe code above has the output [('red' , 2)] as 'red' is the most common element in the counter with a value of 2.\n\n### subtract()\n\nCounter subtract() method is used to subtract element counts from another mapping (counter) or iterable. Inputs and outputs may be zero or negative.\n\n``````counter = Counter('abcabacaa')\nprint(counter)\nc = Counter('abc')\n\ncounter.subtract(c)\nprint(counter)\n\n--> Counter({'a':5, 'b':2, 'c':2})\nCounter({'a':4, 'b':1, 'c':1})\n``````\n\nIn the above code, using subtract() we get the output Counter({'a':4, 'b':1, 'c':1}) after subtracting 'abc' from the initial Counter containing elements 'abcabacaa'.\n\n### update()\n\nThis method is used to add counts from another mapping or counter.\n\n``````counter = Counter('abcabacaa')\nprint(counter)\nc = Counter('abc')\n\ncounter.update(c)\nprint(counter)\n\n--> Counter({'a':5, 'b':2, 'c':2})\nCounter({'a':6, 'b':3, 'c':3})\n``````\n\nIn the code above, update() is used to add 'abc' counts to the counter containing 'abcabacaa' elements. Hence, we get Counter({'a':6, 'b':3, 'c':3}) as the output.\n\n#### fromkeys(iterable)\n\nThis class method is not implemented for Counter objects.\n\n### Mathematical operations on Python Counters\n\n``````c = Counter(a=3, b=1)\nd = Counter(a=1, b=2)\nc + d # add two counters together: c[x] + d[x]\nCounter({'a': 4, 'b': 3})\nc - d # subtract (keeping only positive counts)\nCounter({'a': 2})\nc & d # intersection: min(c[x], d[x])\nCounter({'a': 1, 'b': 1})\nc | d # union: max(c[x], d[x])\nCounter({'a': 3, 'b': 2})\n``````\n\n### Miscellaneous operations on Python Counters :\n\n``````sum(c.values()) # total of all counts\nc.clear() # reset all counts\nlist(c) # list unique elements\nset(c) # convert to a set\ndict(c) # convert to a regular dictionary\nc.items() # convert to a list of (elem, cnt) pairs\nCounter(dict(list_of_pairs)) # convert from a list of (elem, cnt) pairs\nc.most_common()[:-n-1:-1] # n least common elements\nc += Counter() # remove zero and negative counts\n``````\n\n### Implementation:\n\nProgram that demonstrates accessing of Counter elements\n\n``````from collections import Counter\n\n# Create a list\nz = ['blue', 'red', 'blue', 'yellow', 'blue', 'red']\ncol_count = Counter(z)\nprint(col_count)\n\ncol = ['blue','red','yellow','green']\n\n# Here green is not in col_count\n# so count of green will be zero\nfor color in col:\nprint (color, col_count[color])\n``````\n\n#### Output\n\nCounter({'blue':3, 'red':2, 'yellow': 1})\nblue 3\nred 2\nyellow 1\ngreen 0\n\n# Applications\n\nCounter() creates a hash-map for data containers which is very helpful than manual processing of the elements. It has high processing and functioning tools and can work with a wide range of data. In general, we can conclude that counter objects and its functions are used for primarily for processing large amounts of data.\n\n## Question:\n\n#### Which of the following cannot be implemented for Counter objects?\n\nfromkeys()\nupdate()\nMost dictionary methods are available for Counter objects except for fromkeys() which cannot be implemented for the same.\n\nWith this article at OpenGenus, you must have the complete idea of Counter Objects in Python. Enjoy." ]

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[ "Variance computed using Taylor series does not agree with numerical experiment\n\nI would like to estimate an angle $\\theta\\in\\left(-\\frac{\\pi}{2},\\frac{\\pi}{2}\\right)$ given the noisy observations of its sine and cosine (this is related to my earlier question). My estimator is the inverse tangent of the ratio of the means of the observations of sine and cosine. Let's assume that the noise is additive and Gaussian. I am having trouble showing that the mean square error (MSE) of this estimator is decreasing in $n$, though numerical experiments (and intuition) suggest that it does.\n\nFormally, suppose independent sequences of i.i.d. zero-mean Gaussian random variables $(X_n)$ and $(Y_n)$ have variance that is inversely proportional to $n$. That is $X_n\\sim\\mathcal{N}\\left(0,\\frac{\\sigma^2}{n}\\right)$, $Y_n\\sim\\mathcal{N}\\left(0,\\frac{\\sigma^2}{n}\\right)$, and the joint p.d.f. is:\n\n$$f_{X_n,Y_n}(x_n,y_n)=\\frac{n}{2\\pi\\sigma^2}\\exp\\left[-\\frac{(x_n^2+y_n^2)n}{2\\sigma^2}\\right].$$\n\nI am interested in the mean squared error of the estimator $\\hat{\\theta}(X_n,Y_n)=\\tan^{-1}\\left(\\frac{\\sin(\\theta)+X_n}{\\cos(\\theta)+Y_n}\\right)$ as $n\\rightarrow\\infty$. Specifically, I would like to show that:\n\n$$\\lim_{n\\rightarrow\\infty}\\mathbb{E}_{X_n,Y_n}[(\\hat{\\theta}(X_n,Y_n)-\\theta)^2]=\\lim_{n\\rightarrow\\infty}\\mathbb{E}_{X_n,Y_n}\\left[\\left(\\tan^{-1}\\left(\\frac{\\sin(\\theta)+X_n}{\\cos(\\theta)+Y_n}\\right)-\\theta\\right)^2\\right]=0.$$\n\nFurthermore, I would like to show that $\\mathbb{E}_{X_n,Y_n}[(\\hat{\\theta}(X_n,Y_n)-\\theta)^2]=\\mathcal{O}\\left(\\frac{1}{n}\\right)$.\n\nI performed a numerical experiment in MATLAB (see code below), which suggests the inverse linear scaling of the MSE:", null, "I then tried to confirm it analytically using the Taylor series expansion of the function $\\tan^{-1}\\left(\\frac{\\sin(\\theta)+X_n}{\\cos(\\theta)+Y_n}\\right)$ at the origin $(x,y)=(0,0)$. However, utmost care in the control of the remainder must be exercised when taking the expectation of Taylor expansion--see this discussion (in particular, Mike McCoy's answer). With that in mind, I use Wolfram Mathematica (see code below) I obtain:\n\n\\begin{align}&\\tan^{-1}\\left(\\frac{\\sin(\\theta)+x}{\\cos(\\theta)+y}\\right) =\\theta+x \\cos (\\theta )-y \\sin (\\theta )\\\\ &\\qquad+\\frac{1}{2} \\left[\\sin (2 \\theta ) \\left(y^2-x^2\\right)-2 x y \\cos (2 \\theta )\\right] \\\\ &\\qquad+\\frac{1}{3} \\left[-y \\sin (3 \\theta ) \\left(y^2-3 x^2\\right)-x \\cos (3 \\theta ) \\left(x^2-3 y^2\\right)\\right] \\\\ &\\qquad+\\frac{1}{4} \\left[\\sin (4 \\theta ) \\left(x^4-6 x^2 y^2+y^4\\right)+4 x y \\cos (4 \\theta ) \\left(x^2-y^2\\right)\\right]\\\\ &\\qquad+\\frac{1}{5} \\left[-y \\sin (5 \\theta ) \\left(5 x^4-10 x^2 y^2+y^4\\right)+x \\cos (5 \\theta ) \\left(x^4-10 x^2 y^2+5 y^4\\right)\\right]\\\\ &\\qquad+\\frac{1}{6} \\left[-\\sin (6 \\theta ) \\left(x^6-15 x^4 y^2+15 x^2 y^4-y^6\\right)\\right. \\\\ &\\qquad\\qquad\\qquad\\left.+\\cos (6 \\theta ) \\left(-\\left(6 x^5 y-20 x^3 y^3+6 x y^5\\right)\\right)\\right] \\\\ &\\qquad+\\frac{1}{7}\\left[-y \\sin (7 \\theta ) \\left(-7 x^6+35 x^4 y^2-21 x^2 y^4+y^6\\right)\\right. \\\\ &\\qquad\\qquad\\qquad\\left.-x \\cos(7 \\theta ) \\left(x^6-21 x^4 y^2+35 x^2 y^4-7 y^6\\right)\\right]\\\\ &\\qquad+\\ldots\\end{align}\n\nThe zeroth-order term is of course $\\theta$. If we just plug $X_n$ and $Y_n$ into the first order term $x \\cos (\\theta )-y \\sin (\\theta )$ and compute the variance, we get the desired inverse dependence on $n$. However, per Mike McCoy's answer to the aforementioned post, I need to ensure that the error from the remainder $R_2(X_n,Y_n;\\theta)$ after the first order term goes to zero as well, where the remainder is defined as follows:\n\n$$R_2(x,y)=\\tan^{-1}\\left(\\frac{\\sin(\\theta)+x}{\\cos(\\theta)+y}\\right)-\\theta-x \\cos (\\theta )+y \\sin (\\theta ).$$\n\nRemainder is an arithmetic mess, however, subtracting $\\theta$ from the Taylor series expansion above, squaring, plugging in $X_n$ and $Y_n$, and finding expected value by integrating in Mathematica (see code below) yields:\n\n$$\\mathbb{E}_{X_n,Y_n}[(\\hat{\\theta}(X_n,Y_n)-\\theta)^2]=\\left(\\frac{\\sigma^2}{n}\\right)+\\left(\\frac{\\sigma^2}{n}\\right)^2+\\frac{8}{3} \\left(\\frac{\\sigma^2}{n}\\right)^3+12 \\left(\\frac{\\sigma^2}{n}\\right)^4+\\frac{384}{5}\\left(\\frac{\\sigma^2}{n}\\right)^5+640 \\left(\\frac{\\sigma^2}{n}\\right)^6+\\frac{46080}{7} \\left(\\frac{\\sigma^2}{n}\\right)^7+80640 \\left(\\frac{\\sigma^2}{n}\\right)^8+\\ldots$$\n\nFrom these first few terms one recognizes the following series:\n\n$$\\mathbb{E}_{X_n,Y_n}[(\\hat{\\theta}(X_n,Y_n)-\\theta)^2]=\\sum_{p=1}^\\infty\\frac{2^pp!\\sigma^{2p}}{pn^p},$$\n\nwhich diverges for any fixed $n$ (and $\\sigma^2$), implying that the MSE is infinite.\n\nWhat am I doing wrong? Is there any other way to show that the MSE converges to zero?\n\nCODE\n\nThis is the MATLAB code used in generating the figure:\n\nfor n=1:100\nx=randn(1,1000)/sqrt(n*10);\ny=randn(1,1000)/sqrt(n*10);\nv(n)=var(atan((sin(theta)+x)./(cos(theta)+y)));\nend\n\n\nThis is the Mathematica code to obtain the first seven terms of the Taylor series expansion:\n\nSum[FullSimplify[\nD[ArcTan[(Sin[\\[Theta]] + t*x)/(Cos[\\[Theta]] + t*y)], {t, i}] /.\nt -> 0, Assumptions -> \\[Theta] > -Pi/2 && \\[Theta] <\nPi/2]/(i!), {i, 0, 7}]\n\n\nand here is the code used to first the MSE to the first few terms:\n\nIntegrate[(Out - \\[Theta])^2/(2*Pi*s2)*\nExp[-(x^2 + y^2)/(2*s2)], {x, -Infinity, Infinity}, {y, -Infinity,\nInfinity},\nAssumptions -> \\[Theta] > -Pi/2 && \\[Theta] < Pi/2 && s2 > 0]\n\n\nmigrated from math.stackexchange.comJun 24 '16 at 22:11\n\nThis question came from our site for people studying math at any level and professionals in related fields.\n\n• Your messy remainder starts out as an expression that depends on $\\theta$. When you integrate that dependency out, you are including the area near $\\theta=\\frac{\\pi}{2}$. In that region, the remainder term grows rapidly as $\\theta \\to \\frac{\\pi}{2}$ so you would not expect the remainder to go to zero. If your region of integration were restricted to eliminate this, then the power series yo obtained would be messier but would assumedly converge. – Mark Fischler Jun 24 '16 at 21:50\n• Are you sure the single quadrant atan(z) and ArcTan[z] are adequate as opposed to atan2 and ArcTan[x,y]? – horchler Jun 24 '16 at 22:57\n• @Mark Could you please clarify your comment? I'm not integrating over $\\theta$. I think sines and cosines get squared and cancel out when one calculates variance. Then again, in my numerical experiments, as $\\theta$ gets closer to $\\pm\\pi/2$, the $n$ required to converge to small MSE gets larger. So, perhaps there is a problem there... However, I am not sure how to exclude $\\theta$ from the regions around $\\pm\\pi/2$... – M.B.M. Jun 25 '16 at 1:33\n• @horchler Using ArcTan[x,y] yields the same result... – M.B.M. Jun 25 '16 at 4:33\n• The Taylor series approach has little hope of succeeding, because the series converges only for arguments in absolute value less than $1$, whereas adding the Normal error assures there is positive probability of arguments greater than this. – whuber Jun 25 '16 at 18:31\n\nIt does not matter that whether the mean squared error is decreasing with $n$. All that matters is that it can be made arbitrarily small. Here is a simple demonstration.\n\nFixing $\\theta$, let $s=\\sin(\\theta)$ and $c=\\cos(\\theta)$. The estimator $\\hat\\theta$ is the slope of the ray through the origin and the point\n\n$$(c + Y_n, s+X_n).$$\n\nLet $\\epsilon \\gt 0$. Consider the disk of radius $\\sin\\delta$ around $(c,s)$. Euclidean geometry shows us this disk is contained within the wedge at the origin with angles $\\theta-\\delta$ to $\\theta+\\delta$. By choosing $n$ sufficiently large, the chance that $(c+Y_n, s+X_n)$ lies within that disk can be made to exceed $1-\\alpha$ for any tiny $\\alpha$. In fact, because $n(X_n^2 + Y_n^2)$ follows a $\\chi^2_2$ distribution,\n\n$$n = \\lceil\\frac{(\\chi^2_2)^{-1}(1-\\alpha)}{(\\sin\\delta)^2}\\rceil$$\n\nwill work.", null, "In this sketch, $(c,s)$ is at the red dot, the disk is drawn in black, the wedge in blue, and 10,000 simulated values of $(c+Y_n, s+X_n)$ are shown. $\\epsilon$ was equal to $1/50$ here, corresponding to a root mean error in the slope estimate of no more than $\\sqrt{1/50}\\approx 0.14$. With $\\alpha=0.05$, the value of $n$ was $300$, corresponding to a standard deviation of $\\sqrt{1/300} \\approx 0.058$ in the individual coordinates of the simulated points.\n\nNow, within that disk the angular error does not exceed $\\delta$ and outside that disk the angular error cannot exceed $\\pi$ under any circumstance (since the inverse tangent always produces values in the range $[-\\pi/2, \\pi/2]$). This bounds the expected squared error:\n\n$$\\mathbb{E}([\\hat\\theta - \\theta]^2) \\le (1-\\alpha)(\\sin(\\delta))^2 + \\alpha \\pi^2.$$\n\nBy choosing, say, $\\delta = \\arcsin{\\sqrt{\\epsilon}}$ and $\\alpha=\\epsilon/\\pi^2$, the right hand side will be less than $2\\epsilon$. Because $\\epsilon \\gt 0$ was arbitrary, the limiting mean square error is zero.\n\n(In the figure, the mean square error was $0.0033 \\ll 0.04 = 2\\epsilon$.)\n\nThe argument concerning angular error in the disk assumed the inverse tangent was continuous within a neighborhood of $\\theta$. That will not be the case for $\\theta=\\pm \\pi/2$, but a slight change in the definition of the inverse tangent in cases where $\\theta$ is close to these values will fix the problem." ]

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[ "Show whether the language with almost as many 0 as 1 in every prefix is regular [closed]\n\nThis is the exercise:\n\nLet A be a language defined over the alphabet Σ = {0, 1} composed by the strings with the property that in every prefix, the number of 0s and the number of 1s differ by at most 2. Determine if A is a regular language and prove it.\n\nOk, as @roi-divon commented, A is a regular language. I've tried to obtain de minimal DFA accepting A but I haven't got it... Someone knows?\n\nclosed as unclear what you're asking by David Richerby, Rick Decker, Juho, Luke Mathieson, ShaullNov 21 '14 at 15:01\n\nPlease clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.\n\n• This is a dump of an exercise problem, not a question. If you have a specific question regarding the wording of the problem or concrete steps in your own attempts at solving the problem, feel free to edit accordingly and we can reopen the question. See also here for our homework policy, and here for a relevant discussion. You may also want to check out our reference questions. – David Richerby Nov 18 '14 at 9:15\n\nThe language is regular.\n\nDenote\n\n$L^k = \\{ x \\ | \\ \\forall \\alpha \\in Prefix(x): |\\#_0(\\alpha)-\\#_1(\\alpha)| \\leq k \\}$\n\nYour language is $L = L^1$.\n\nWe shall prove, using Myhill-Nerode theorem that each $L^1$ is regular (actually for any $k,\\ L^k$ is regular).\n\nDefine equivalence classes of $R_L$ as follow:\n\n• for $-k \\leq i \\leq k: \\ S^i = \\{ x \\ | \\ \\#_0(\\alpha)-\\#_1(\\alpha) = i, \\forall \\alpha \\in Prefix(x): |\\#_0(\\alpha)-\\#_1(\\alpha)| \\leq k \\}$\n• $S_{out} = \\{ w | \\exists \\alpha\\in Prefix(x): |\\#_0(\\alpha)-\\#_1(\\alpha)| \\gt k\\}$\n\nWe can see that $L^k = \\bigcup_{-k \\leq i \\leq k}S^i$\n\nClearly, the equivalence classes are not empty, they don't intersect and the union is $\\Sigma^*$\n\nAssume $x,y \\in S_{out}$, we show that $x R_L y$: clearly from every $z \\in \\Sigma^*: \\ xz,yz \\notin L$ (have \"bad\" prefixes\")\n\nAssume $x,y \\in S^i$, and assume there is a $z \\in \\Sigma^*$ such that $xz \\in L$ but $yz \\notin L$. Then $yz$ has a prefix $y \\beta$ such that the difference between 0's and 1's is larger than k ($\\beta$ is prefix for z). By counting the number of 0's and 1's of $x \\beta$ we get that it's larger then k, in contradiction that $xz \\in L$\n\nNow we show that if $x,y$ in different classes, then $(x,y) \\notin R_L$.\n\nAssume $x \\in S^i, \\ y \\in S_{out}$: then for $z = \\epsilon: \\ xz = x \\in L, yz = y \\notin L$. Therefore $(x,y) \\notin R_L$\n\nAssume $x \\in S^i, \\ y \\in S^j$, and $j \\neq i$. w.l.o.g, $i > j$. Examine $z = 0^{k-i+1}$. Counting 0's and 1's, we get that $xz \\notin L$, and $yz \\in L$. Therefore $(x,y) \\notin R_L$\n\nBy that We've proved that $L = L^1$ is regular\n\n• And would you know how to obtain a minimal DFA accepting A? I've tried it, but... – eyhqtl Nov 18 '14 at 9:08\n• Also, rather than this lengthy proof via Myhill-Nerode, it's much easier to just produce a DFA for this: six states should suffice. – David Richerby Nov 18 '14 at 9:13\n• Please consider not to encourage undesirable posting behaviour. – Raphael Nov 18 '14 at 9:42\n• @DavidRicherby May i submit the DFA? – muradin Nov 18 '14 at 14:18" ]

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[ "Follow us on:", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "# Lcm of two numbers examples\n\nlcm of two numbers examples C. Let’s find the LCM of 120 and 45. The product of 6 and 4, for example, is 24, and 24 is a common multiple -- but it is not their lowest common multiple. Print the value returned. M. com/playlist?list=PL3JkVZVtegL9vWRCoXhNE71IWmSx3t2V2BASIC VOCABULARY- Use the LCM of two or more numbers Calculator to find the Least Common Multiple of numbers 44, 4, 3 i. Solution: The HCF of 18 and 48 based on the solution above is 6. e. In one example, we'll use the LCM to calculate the answer directly, and in the other we'll need it as a step to find the solution to the problem. Find the LCM of 3, 5, and 6. e. The LCM of Least Common Multiple of 2 numbers is the smallest number, divisible by the two numbers, evenly. Here’s a quick way to do this. 20 │0 1│1 2│2 3│3 4│4 5│5 6│6 7│7 8│8 9│9 10│10 11│11 12│12 13│13 14│14 15│15 16│16 17│17 18│18 19│ Take the LCM of each pair in a: *. Hence 12 is the least common multiple (LCM) of 3 and 4. Algorithm: Take two number’s as input. Find the prime factorization of the two numbers. Find the LCM of two Numbers using Command Line Language? It is highly advisable to go through Command Line Arguments Post before even looking at the code. Keep repeating the step 2 till no two numbers are divisible by the same number. com The LCM of two or more numbers is the smallest number that is evenly divisible by all numbers in the set. of fractions = (viii) If HCF (a, b) = H 1 and HCF (c, d) = H 2, then HCF (a, b, c, d) = HCF (H 1, H 2). The LCM of two natural numbers a and b is the smallest number y, which is a multiple of both a and b. Let ‘a’=48 Therefore 240*12=48*b 2880=48*b b=2880/48=60 Using the statement you can find any number. 24 = 2x2x2x3. Given LCM (a, b), the procedure for finding the LCM using GCF is to divide the product of the numbers a and b by their GCF, i. The least common multiple (L. The product of all is the LCM of the given set of numbers. The LCM of two integers n1 and n2 is the smallest positive integer that is perfectly divisible by both n1 and n2 (without a remainder). Example ☞ Consider the number 10 and 15. We can determine L. LCM (Least Common Multiple) of two numbers is the smallest number which can be divided by both numbers. (Also called Lowest Common Multiple) For example: The L. The product of any two numbers is equal to the product of their greatest common divisor and least common multiple. Let’s understand the concept of LCM with an example: Consider two numbers: 8 and 12. 15, Dec 20. h> int lcmCalculate(int a, int b); int main() { int n1, n2, lcmOf; printf(\" Recursion : Find the LCM of two numbers : \"); printf(\"----- \"); printf(\" Input 1st number for LCM : \"); scanf(\"%d\", &n1); printf(\" Input 2nd number for LCM : \"); scanf(\"%d\", &n2); // Ensures that first parameter of lcm must be smaller than 2nd if(n1 > n2) lcmOf = lcmCalculate(n2, n1);//call the function lcmCalculate for lcm calculation else lcmOf = lcmCalculate(n1, n2);//call the function If you have two numbers and one is a multiple of the other, then the multiple will be their least common multiple. Multiples of 5 - 5, 10, 15, 20, . Find least common multiple of this two numbers: 140 175. The common multiples of 3 and 4 are 0, 12, 24, . Program to find LCM of two numbers /** * C program to find LCM of any two numbers */ #include <stdio. Example 1 Greatest Common Factor: Find the greatest common factor of 12 and 18. 24 = 2 * 2 * 2 * 3. readData(); call. Just to clarify, these are the common multiples of 8 and 12 for their first ten multiples. If we multiply together all prime numbers used in these products in their highest power, we get the least common multiple. 10 is a multiple of 2. Disadvantages? Becomes tedious, as the number grows bigger, for example LCM (235, 512). This means that the 30 is the common number that can be divided on any of the numbers between 3,5 and 6 to get the answer in integers. Let the 2 numbers be ‘a’ and ‘b’. Lowest Common Multiple (LCM) : The Lowest Common Multiple or LCM (also sometimes referred to as the Least Common Multiple) is the smallest positive number that is a multiple of two or more given numbers. Therefore 72 is the Least Common Multiple of 24 and 36. 6 12. For instance, the LCM of 2, 3 and 7 is 42 because: 42 is a multiple of 2, 3 and 7. As regards the GCF, the greatest common factor must be a prime number –that is, a number that can only be divided by itself and 1. For example: Let us take three numbers such as 12, 15 and 18. The least common multiple (LCM) of two numbers is the smallest number (not zero) that is a multiple of both. x * c // gcd ( x, c) = y. a * b // gcd ( a, b) = x. For example, =LCM(25,40) returns 200. gcd ( y, d) = z. The Least Common Multiple of two nonzero numbers a and b, written LCM(a, b), is the least whole number (larger than zero), which is a multiple of both a and b. For example: We have two integers 4 and 6. Example. The LCM of $$2$$ or more numbers can not be less than the smallest numbers given in the set. The division method is another method to get the LCM of multiple numbers. knowledge book akbarpur kanpur classesThe Hindu Editorial Vocabularyhttps://www. 250 and 35. Free Least Common Multiplier (LCM) calculator - Find the lcm of two or more numbers step-by-step This website uses cookies to ensure you get the best experience. F. heart. For example: The LCM of 8 and 56 is 56 because 8 is a factor of 56 and can go into 56 seven times, and 56 is a factor of itself. Example. log(lcm_two_numbers(3,15)); console. डी. Step 2: Store the common multiple of A & B into the max variable. If a number cannot be divided it is copied down to the next step of division. The LCM is found by dividing the product a·b by the GCF. youtube. The test expression of while loop is always true. e. . Initialize the greater number to a variable say mp. Prerequisites: Proof that the prime factorization of an integer is unique; Maximum and minimum Solution . 2: Find the LCM of 12 and 18 12: 12, 24, 36, 48 18: 18, 36, 54 So the LCM is 36 because 36 is the lowest number in the list of multiples. If any number is multiplied by another number then the resulting number is called a multiple of the first number. The lcm of two or more than two numbers is the least number which is exactly divisible by each of the numbers. Following are the algorithm of LCM of two number, as follows: Step 1: Initialize the positive integer variables A and B. Use the LCM of two or more numbers Calculator to find the Least Common Multiple of numbers 44, 4, 3 i. To find the LCM, multiply the divisors and remaining quotients. It is also known as the Least Common Divisor or LCD. Sol: As mentioned above, take the differences between any two pairs of the three given numbers. There may be many multiples of a number. C plus plus Program to Find the GCD and LCM of two Numbers. In other words, a common multiple can be defined as a number that is a multiple of two or more numbers. or Least Common Multiple of two values is the smallest positive value which the multiple of both values. youtube. Let's find the LCM of 30 and 45. Factors of 18 = 2 x 3 x 3 = 2 x 32. Repetitive Division . The required number is the HCF of the two differences, i. It cannot be less than any of the two numbers, more specifically, it cannot be less than the greater number among the two. 12 = 2 * 2 * 3; 15 = 3 * 5; 18 = 2 * 3 * 3; So we take the common prime factors and then the rest of the ones and multiply them with each other which would give us the LCM. It is the smallest positive integer that is divisible by both \"a\" and \"b\". 72 = 2 3 x 3 2. C. The GCD of two positive integers is the largest integer that divides both of them without leaving a remainder (the GCD of two integers in general is defined in a more subtle way). So co-primes are numbers which do not share any divisors (other than 1) so the lcm would be the product of the two numbers. Integers: 10 15 lcm (10, 15) = ? step 2 Arrange the group of numbers in the horizontal form with space or comma separated format 10 and 15 Least Common Multiple (LCM)-The LCM of two or more given numbers is the least number which is exactly divisible by each of them. 5 = 5. Also, we know that if their is an LCM of two numbers, i. For example, express 36 + 8 as 4 (9 + 2). Given two numbers, the task is to find the LCM of two numbers using Command Line Arguments. 4. The least common multiple is the smallest positive integer that is a multiple of all supplied numbers. lcm of two number in c++; c++ lcm; lcm code; lcm using cpp; lcm of two numbers in c++; c++ code to find LCM - write a simple algorithm that finds the Least Common Multiple of two numbers A , B. With this factorization, finding the LCM is a cinch. October 22, 2020 . 1: Find the LCM of 10 and 30 10: 10, 20, 30, 40, 50 30: 30, 60, 90 So the LCM is 30 because 30 is the lowest number in the list of multiples Ex. 36 is a common multiple of 3 and 4. The LCM of two numbers 10 and 1100 is 1100. LCM of 3 and 7 = 3 x 7 = 21. 24 = 2 3 × 3 60 = 2 2 × 3 × 5. Apply and extend previous understandings of numbers to the system of rational numbers. Also try: Calculate HCF Online. input two number and displays it lcm; finding lcm in Least common multiple (LCM) of two numbers is the smallest number that they both divide. LCM of 9 and 21 = 63, Now multiply first fraction by 63/9 = 7. Simply, provide the inputs in the respective input field and click on the calculate button to avail the LCM of given numbers easily. Finding the LCM Using Listing Method List down the multiples of each number; take note of the common multiples that the numbers share and choose the lowest or least multiple. This Code To Calculate LCM of Two Integers makes use of While Loop, For Loop and Modulus Operator. 1) Write both numbers in a horizontal t-chart. C++; Least Common Denominator in c; lcm program in c; find lcm of two numbers using gcd; input two number and displays it lcm The first type of word problem that we will consider in this video lesson is the one that involves the greatest common factor, the greatest common factor that is shared between two or more numbers . Identify the common multiples. The least common multiple of two or more nonzero whole numbers is the smallest whole number that is divisible by each of the numbers. Next, they need to determine the LCM by finding the smallest common multiple that is divisible by both numbers. The product will be the LCM of the various numbers. Their lowest common multiple is 12. i. num1 = 4. 60. Enter two integers from the keyboard and find their lcm. Here’s the calculation to find the LCM given that we know already the GCF of the two integers and their product. So, LCM (12, 18) = 22 x 32 = 36. 132 smallest integer divisible by all numbers. factors: They have none common factor greater than . Now, the smallest number that can be divided by 10 and 4 evenly, will be 20. Find the LCM of two numbers using recursion Hint: You may assume that the first number is always smaller than the second number. The LCM of two numbers is the smallest number that is a multiple for both numbers. 30. 36 = 2 * 2 * 3 * 3. Enter two positive integers: 72 120 The LCM of 72 and 120 is 360. Another worksheet in which students determine the least common multiple for each number pair. 10 is a multiple of 2. The Output for the program has been displayed at the bottom. gcd ( x, c) = y. So now first fractions is (2 x 7)/ (9 x 7) = 14/63. \\) In a given set of two natural numbers if one number is the factor of the other number, then the LCM will be the bigger number. 6 = 2 × 3. Write a c program to find out L. To find the Least Common Multiple of two or more whole numbers, follow this procedure: Make a list of multiples for each whole number. First list all the multiples of 6. The least common multiple of two numbers is the lowest possible number that can be divisible by both numbers. The sum of the numbers is Formula. A faster way is to use repetitive division to find the least common multiple. Step 1: Rewrite the numbers in a row separated by commas. Example 1: The multiples of 3 and 5 are: Multiples of 3 - 3, 6, 9, 12, 15, 18, . The LCM must be determined before two fractions can be added or subtracted. . => x = 48 6 = 8 x = 48 6 = 8. This math video tutorial explains how to find the least common multiple of 3 numbers using prime factorization. lcm lets you find the least common multiple of symbolic rational numbers. Examples: Input: n1 = 10, n2 = 15 Output: 30 Input: n1 = 5, n2 = 10 Output: 10 Approach: How to make a method for finding the least common multiple (LCM). Example ☞ Consider the number 10 and 15. LCM factors: for 2, it is 2^3 = 8 for 3, it is 3^1 = 3 for 5, it is 5^1 = 5. Hence LCM of 2 x 2 x and 3 x 3 x is 6 x 6 x) Given that LCM of 2 x 2 x and 3 x 3 x is 48. com/playlist?list=PL3JkVZVtegL9vWRCoXhNE71IWmSx3t2V2BASIC VOCABULARY- Enter two numbers: 12 18 LCM = 36. Let’s find the LCM of 120 and 45. Formula To Find LCM of Two Integers. ) of two numbers is the smallest positive integer that is perfectly divisible by the two given numbers. एफ. For example, LCM of 15 and 20 is 60, and LCM of 5 and 7 is 35. Kirk. List the first few multiples of both numbers and stop as soon as you find a common multiple. The LCM of two numbers is the product of the two numberf sometimes; Sample examples: The LCM of 3 and 4 is 12 and 3 X 10, not 2 x IOor20. For example, 8 = 2 3 and 90 = 2 × 3 2 × 5. In the example given below, we will take two numbers and find their GCD and LCM. Also, the LCM of any two number is always greater or equal to one of them. The LCM of 4 and 3 is 12. . To continue, we are going to see how to calculate the least common multiple. Method 2: Find the LCM using prime factorization. 250 = 2*5*5*5. ab = \\left ( {72} \\right)\\left ( {90} \\right) = 6,480 ab = (72)(90) = 6,480. We can also find the least common multiple of three (or more) numbers. For example, the gcd of two numbers depends directly and simply on their factorizations, and this approach gives us significant new information. Least common multiple (LCM) of 44, 4, 3 is 132. or one prime and another composite like 3 & 8. Don't fret, any question you may have, will be answered. Example 8 : Find the LCM of numbers 12, 15, 20 & 27. Steps for Finding the LCM. Division Method. We store the value in a variable, and then, print the variable. Because LCM is the number that must be divisible by both the number (with leaving remainder 0). Theorem 17. Java program to find the HCF (Highest Common Factor) & LCM (Least Common Multiple) of two numbers using recursion. If two numbers have no common factors greater than , then their LCM will be their product. The LCM of two numbers 140 and 300 is 2100. See full list on calculatorsoup. C. For example, for the set of numbers 12, 24 and 36 the LCM = 72. We will be using this to find the result. By using this website, you agree to our Cookie Policy. Example 1. Ex. Follow the below steps to obtain the least common multiple using the division method. Use the LCM of two or more numbers Calculator to find the Least Common Multiple of numbers 44, 4, 3 i. Then find the LCM of c and q. youtube. In above program, the user is asked to integer two integers n1 and n2 and largest of those two numbers is stored in max. Let us use the above steps in the example given below: find the LCM of 15,30,90. integer(readline(prompt = \"Enter second number: \")) print(paste(\"The L. LCM of Two Numbers using Command Line Language. Many problems have appeared in various competitive exams based on this relationship. We follow the same method of prime factorization, with a few changes. Using prime factorization. Extremely fast when you’ve to find LCMs of two digit numbers for example 12,15,96. C program to calculate LCM. We would begin by listing the multiples of 4 and 6 until we find the smallest number in both lists, as shown below. There are two common methods for finding the least common A common use of the LCM function is to add fractions that have different denominators. Example 1: Find the LCM of 6 and 4 using prime factorization. Using the lists to find the LCM can be slow and tedious. In this method, the given numbers are divided by the common divisors until there is no possible further division by the common number. Check the prime factors page to learn how to find prime factors of an integer. For example, LCM $$(6,8)$$ can not be less than $$6$$ and $$8. It is a concept of arithmetic and number system. The least common multiple of two given natural numbers is the smallest positive integer that is a multiple of both given numbers. . gcd ( a, b) = x. 60 = 2x2x3x5. Hence numbers are 39 and 52 Ques: Find the largest number with which when 555, 1275, and 1635 are divided, the remainders are the same. Two numbers are said to be co-prime if their HCF is 1 HCF of fractions = HCF of numerators/LCM of denominators LCM of fractions = GCD of numerators/HCF of denominators; Questions and Solved Examples on LCM and HCF Problems. Logic: For GCD/HCF: We will take a number, check if it is perfectly divisible by both numbers. C. 3 = 3. Step 2 : Now, we have to draw two circles as shown above. Multiples that are shared by two or more numbers are common multiples 7. Solution. I. M. 132 smallest integer divisible by all numbers. Example: Let’s say we have to find the LCM of 2. Example – 11 : The GCD and LCM of two numbers are 66 and 384. Methods of finding the LCM To find the LCM of given expressions, we proceed as follows: Express each of the expressions as the product of its factors including prime factors where applicable. that mean's if we multiply them to get their product, we have a number with a factor of 9 which neither number has on it's own. of\", num1,\"and\", num2,\"is To find the GCF of two numbers, subtract the larger number from the smaller number while the two numbers are not equal. Example of LCM. Examples There is one very important relationship, given below, between two numbers and their HCF and LCM. 5 and 0. One of the quickest ways to find the LCM of two numbers is to use the prime factorization of each number and then the product of the least powers of the common prime factors According to Mathematics, the Least Common Multiple (LCM) of two or more integers is the smallest positive integer that is perfectly divisible by the given integer values without the remainder. Least common multiple (LCM) of 75, 42, 51 is 17850. Two Methods to Find the LCM Method 1: May be used for small numbers Examples: Find lowest common multiple of the numbers 6 and 8. For example, the L. This works in general, lcm(4,4)=4 not 16, lcm(6,3)=6 not 18, lcm(8,12)=24 not 96, etc. And usually in time speed work, pipe-cistern type questions have number in two digits (e. So for the above example, the LCM is 2 3 × 3 × 5 = 120. LCM = 144. For LCM: We use a formula here, LCM = Num1*Num2/GCD. Basically, in the list of two numbers’ respective list of multiples, the lowest number that the two numbers share is their lowest common multiple. The least common multiple (LCM) is equal to 1224 LCM = 2×2×2×3×3×17 = 1224 An example of a GCD and LCM Calculation Let's say you want to calculate the GCD and LCM of two numbers: 12 and 28. Multiply second fraction by 63/21 = 3. Continue your list until at least two multiples are common to all lists. EXAMPLE 3. g. The number 144 is the LCM of 18 and 48. Step 1 : Write down the standard form of numbers. 5 years ago. 24 is the common multiple of 8 and 648 is the c what is the least common multiple of 36 and 12 so when they say another way to say this is LCM in parentheses 36 and 12 and this is literally saying what's the least common multiple of 36 and 12 well this one might pop out at you because 36 itself is a multiple of 12 and 36 is also a multiple of 36 it's 1 times 36 so the smallest number that is both a multiple of 36 and 12 because 36 is a LCM stands for Lowest or Least Common Multiple. It is represented as LCM (a, b) or lcm (a, b). We have to only have factors the original numbers have. The numbers are in the ratio 2:3. 12, 15, 96)…so it is very easy to recall in which multiplication tables do they come. Least Common Multiple Use these worksheets when teaching students to calculate the least common multiple, or LCM, of a group of numbers. The sum of the numbers is To find the LCM, multiply the HCF by all the numbers in the products that have not yet been used. ☞ Multiples of 10 are 10, 20, 30, 40, 50, 60, 70, 80, 90, 100… ☞ Multiples of 15 are 15, 30, 45, 60, 75, 90, 105, 120, 135… Least Common Multiple Use these worksheets when teaching students to calculate the least common multiple, or LCM, of a group of numbers. C. According to Wikipedia - A common multiple is a number that is a multiple of two or more integers. Now the product of two numbers will always be a common multiple. let lcm = (n1, n2) => { //Find the gcd first let gcd = gcd(n1, n2); //then calculate the lcm return (n1 * n2) / gcd; } Finding Least Common Multiple of Decimals. 30 = 2 × 3 × 5. C. Decent website, although I want to be able to find the LCM for more than just two numbers Use the distributive property to express a sum of two whole numbers 1–100 with a common factor as a multiple of a sum of two whole numbers with no common factor. 12 = 2 × 2 × 3. org Example Input: 15 20 5 7 Output: 60 35 With using GCD. LCM = 144. Examples: C Program to find LCM of two numbers using Recursion. Find out the LCM of 3 and 7. Therefore, 10 is the LCM. 2. 27? Take the help of Free Online LCM of two or more numbers calculator and get the Least Common Multiple in the blink of an eye. For example, lcm of 6 and 15 is 30. Lowest Common Multiple (LCM): The smallest or lowest common multiple of any two or more than two integer numbers is termed as LCM. Question: LCM and HCF of two numbers are 2079 and 27 respectively. What is LCM? LCM (Lowest Common Multiple) of two numbers is the smallest positive integer that is perfectly divisible by the given numbers. HCF of 119 and 391 = 17 => Middle Number = 17 First Number =\\dfrac{119}{17}=7 Last Number =\\dfrac{391 Generate pairs of sequential integers from 0–19, store in a and list: ]a=. In the example shown above, we have the two numbers 24 and 60. Now we express those two numbers as a product of their prime factors. Example: Find the least common multiple of 4, 6, and 8 Multiples of 4 are: 4, 8, 12, 16, 20, 24 , 28, 32, 36, It is the smallest whole number that is multiple of those numbers. The first number they have in common is the LCM. » 4 Print this page. Today we're going to look at a couple of examples in which we use the Least Common Multiple (LCM) to solve math problems. For example, the LCM of 5 and 6 is 30, because Finding LCM using GCD is explained here but here the task is to find LCM without first calculating GCD. Example 2. 20 = 2 2 · 5. 36 and 0. Least Common Multiple (LCM): It is the smallest positive number that is a common multiple of two or more numbers. e. 4 8 12. Step 1: Write down the standard form of numbers. 10 × 1 = 10 and 2 × 5 = 10. x = np. Solution: step 1 Address input parameters & values. A multiple is a product of a number, if two numbers has one or more multiples in common they are called common multiples. C. Then take lowest common multiple, this lowest common multiple is LCM of numbers. Least Common Multiplier or LCM of any n numbers is the least value that is a multiple of the given numbers. 2) Next to each number list the first 10 products of that number and the whole numbers starting with 1. Example Questions (a) Find the highest common factor and lowest common multiple of 50 and 70 by following the steps below: Use the LCM of two or more numbers Calculator to find the Least Common Multiple of numbers 44, 4, 3 i. For example, LCM of 15 and 20 is 60 and LCM of 5 and 7 is 35. For example, the C LCM value of integer 2 and 3 is 12 because 12 is the smallest positive integer that is divisible by both 2 and 3 (the remainder is 0). 35. Important Results. GCF\\left ( {72,90} \\right) = 18 GC F (72,90) = 18. 72 is the least number which is divisible by both 24 and 36. Example: the Least Common Multiple of 3 and 5 is 15 Because 15 is a multiple of 3 and also a multiple of 5 and it is the smallest number like that. 6. This says \"the greatest common factor of 20 and 16 is 4. The common multiple which has the smallest in value is the least common multiple (LCM) of the given two numbers which are 8 and 12. e. Then the LCM of 2 and 3 is: 6. C. e. { 2,/\\i. Hence, if we take HCF of 119 and 391, we get the common middle number. Factors of 12 = 2 x 2 x 3 =22 x 3. Least Common Multiple (LCM) The Least Common Multiple of two numbers is the lowest number that is the multiple of both the numbers. For example, LCM of 12 and 18 is 36. Method #1: Set intersection method Example: Find LCM of 6 and 9. or 3,6,9. For example; Example: Find the LCM of 15 and 12 Determine the Greatest Common Factor of 15 and 12 which is 3 Either multiply the numbers and divide by the GCF (15*12=180, 180/3=60) OR - Divide one of the numbers by the GCF and multiply the answer times the other The smallest positive number that is a multiple of two or more numbers. The only way to teach this is by example, so that’s what I’ll do — by finding the LCM for 18 and 30. LCM stands for Least Common Multiple. 12 = 2 2 x 3 1 Example: Compute LCM in R # Program to find the L. The least common multiple (LCM) is the smallest number which is a multiple of two numbers. out. For example LCM of 15 and 20 is 60 and LCM of 5 and 7 is 35. if m & n are two numbers, mn = LCM x HCF mn = 6x210 = 1260 m + n = 72 (given) Now, 1/m + 1/n = (m+n)/mn = 72/1260 = 2/35 Math worksheets: Find the least common multiple (LCM) of 3 numbers (2-30) Below are six versions of our grade 6 math worksheet on finding the least common multiple of 3 numbers; all numbers are less than 30. . Let us multiply the two numbers in factored form : (2 x 3) x (2 x 2). 12 is a common multiple of 3 and 4. = 2 x + 3 x = 5 x = 2 x + 3 x = 5 x. . In the example above you clearly see that two numbers (2 and 3) are used and I have multiplied them with their same range number that is 2,4,6. What is the Least Common Multiple of 9 and 12? Least common multiple or lowest common denominator (lcd) can be calculated in two way; with the LCM formula calculation of greatest common factor (GCF), or multiplying the prime factors with the highest exponent factor. The sum of the numbers is Favorite Answer. c++ lcm and hcf . For 4 and 6, it would be: 3 • 2 • 2 = 12. ) . e. The least common multiple is sometimes referred to as lowest common multiple and abbreviated as (LCM). LCM = \\(12 \\times 2 \\times 3 \\times 5 = 360$$ Example: Considering the numbers 45, 60 and 90, the HCF is 15 since it is the highest number that divides 45, 60 and 90. LCM : Lowest Common Multiple of two or more integers is the smallest positive integer that is perfectly divisible by the given integers, without remainder I've always heard “mutually prime\" and am assuming that that is what co-primes are (after having looked it up). For example, the LCM value of integer 2 and 3 is 12 because 12 is the smallest positive integer that is divisible by both 2 and 3 (the remainder is 0). For example, 12, 18, 24 are all multiples of 6. *; class Hcf_Lcm { int n1, n2, hcf, lcm; public static void main( String args []) throws IOException { Hcf_Lcm call = new Hcf_Lcm (); call. LCM (Least Common Multiple) of two numbers is the smallest number which can be divided by both numbers. Beatles11. Hope that helps! Like Like This series walks you through the steps of finding the least common multiple in rational expressions. For example: Let’s say we have the following two numbers: 15 and 9. Example: LCM of 5 &amp; 6 is 30, which is the same as the Example 4: Find the LCM of 18 and 48. For example: Find the GCF of 6 and 10: 10 6 // 10 is larger than 6, so subtract 6 from 10; 4 6 // 6 is now larger than 4, so subtract 4 from 10; 4 2 // 4 is larger than 2, so subtract 2 from 4; 2 2 // the two numbers are now equal, and that's the GCF; The LCM in mathematics is simply the first number times the second number divided by their GCF. So the LCM of 60 and 72 is 2 × 2 × 2 × 3 × 3 × 5 which is 360. The Least Common Multiple (LCM) of two numbers is the smallest number that is a multiple of both. In this example, we will learn how to find the least common multiple (LCM) of two numbers in Python. Find hcf and lcm using recursive technique. One way to understand the least common multiple is by listing all whole numbers that are multiples of two given numbers, for example 3 and 5: The multiples of 3 are: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33 Python Program to Find LCM. e. C. How could you check to see if there are others that also work? On the other hand, the Lowest Common Multiple (or LCM) is the integer shared by two numbers that can be divided by both numbers. C. The numbers are in the ratio 2:3. g. M. ) × (their L. youtube. The common multiples of 3 and 4 are 0, 12, 24, . LCM using repeated division. 1. x and y. This product is the LCM. or both composite like 15 and 26. For example, LCM(6,10) = 30 because both 6 and 10 are factors of 30, and there is no smaller number than 30 that does this. M. • You have found a pair of numbers that work. Therefore, 10 is the LCM. gcd(a,b) print(\"The LCM of the two numbers is \", lcd) Therefore, the Least Common Multiple (LCM) is 6. display(); } public void readData() throws IOException { BufferedReader br = new BufferedReader (new InputStreamReader ( System. INFORMATION FINDING LCM USING PRIME FACTORIZATION. The least common multiple, or LCM, is another number that's useful in solving many math problems. HCF and LCM Questions & Answers for Bank Exams : The LCM of two numbers is 66. Let A B be their LCM. (a × b)/GCF (a,b). Find the LCM of the following two numbers: import numpy as np. For example, the LCM of 72 and 120 is 360. In order to do LCM of 2 x 2 x and 3 x 3 x = 6 x = 6 x (∵ LCM of 2 and 3 is 6. lcm (num1, num2) print(x) Try it Yourself ». Live Demo The least common multiple of two numbers, often written as the LCM, is the smallest number that can be divided evenly by those original two numbers. The multiples of 8 are: 8 X 1 = 8, 8 X 2 = 16, 8 X 3 = 24, 8 X 4 = 32, and so on… The multiples of 12 are: Highest Common Factor (HCF): The greatest common factor to any two or more than two integer numbers is known as HCF of these numbers. For rational numbers r i, LCM [r 1, r 2, …] gives the least rational number r for which all the r / r i are integers. Least Common Multiple (LCM): LCM(a,b) is the least (smallest) number that is a multiple of both a and b. h> int main() { int i, num1, num2, max, lcm=1; /* Input two numbers from user */ printf(\"Enter any two numbers to find LCM: \"); scanf(\"%d%d\", &num1, &num2); /* Find maximum between num1 and num2 */ max = (num1 > num2) ? num1 : num2; /* First multiple to be checked */ i = max; /* Run loop indefinitely till LCM is found */ while(1) { if(i%num1==0 && i%num2==0) { /* * If 'i' divides both 'num1 When two numbers have no common factors other than 1, then their Lowest Common Multiple is the product of these two numbers. What we’ve just found is the smallest number which is both a multiple of 40 and a multiple of 60. com/playlist?list=PL3JkVZVtegL9vWRCoXhNE71IWmSx3t2V2BASIC VOCABULARY- Use the distributive property to express a sum of two whole numbers 1-100 with a common factor as a multiple of a sum of two whole numbers with no common factor. import java. Let C++ program to find LCM of two numbers. Proof. Prime Factorization method. To find its LCM, we have to first check which one is greater. of 12 and 14 is 84. This is called the lowest common multiple or lcm. The largest number among n1 and n2 is stored in min. Vocab Vocabulary Check Fill in each blank with the correct word(s) to complete each sentence. Sum of the numbers. If you have two numbers and one is a multiple of the other, then the multiple will be their least common multiple. You can also find the least common multiple by using prime factorization. A common multiple is a number that is a multiple of two or more numbers. For example, to find the As you can see in the illustration below, the common multiples of 8 and 12 are 24, 48 and 72. For example, express 36 + 8 as 4 (9 + 2). The LCM of two numbers cannot be less than min. LCM of 5 and 9 = 5 x 9 = 45 Prime Factor Method Least Common Multiple (LCM) The Least Common Multiple of two numbers is the lowest number that is the multiple of both the numbers. 12 is the LCM of 4 and 6. In addition to the cake method, we can calculate the LCM of two numbers by using prime factorization. of two input number lcm <- function(x, y) { # choose the greater number if(x > y) { greater = x } else { greater = y } while(TRUE) { if((greater %% x == 0) && (greater %% y == 0)) { lcm = greater break } greater = greater + 1 } return(lcm) } # take input from the user num1 = as. To find the LCM (Least Common Multiple) of two or more numbers, make multiple of numbers and choose common multiple. Learners are required to list the first ten multiples for each set of two numbers featured in the questions. We take the answer and multiply it by the other number, 3 x 48 = 144. You will also learn to find LCM using GCD. 1. C. Then in the example above the Lowest Common Multiples between the numbers 2 and 3 are 6. The LCM of two integers a and b is denoted by LCM (a,b). readLine()); Write a c++ program to find LCM of two numbers. Example 7. For example: The LCM of 9 and 18 would be 18. lcm(sym([3/4, 7/3, 11/2, 12/3, 33/4])) L. Calculation of LCM 50=1 51=1x5 = 5 52=1x5x5=25. Find the least common multiple of these rational numbers, specified as elements of a symbolic vector. For example, the LCM of 2 and 3 is 6, as both numbers can evenly divide the number 6. For example, a multiple of 6 is 12. HCF and LCM Problems (Advanced) Q. knowledge book akbarpur kanpur classesThe Hindu Editorial Vocabularyhttps://www. A program to find the LCM of two numbers is given as follows − Example. Kaneppeleqw and 2 others learned from this answer. Examples: C Program to find LCM of two numbers using Recursion. Following with the previous example, if the common multiples of 2 and three 3 were 6 and 12, the least common multiple is 6, since it is smaller than 12. Must Read: Find LCM of N Numbers in C Programming HCF and LCM Questions & Answers for Bank Exams : The LCM of two numbers is 66. There are several methods for finding the least common multiple of two or more numbers. Let's say you have the numbers 2 and 10. You can use a factor tree, or whatever method you've learned. Find the side of the largest square slab which can be paved on the floor of a room 5 meters 44cm long and 3 meters 74 cm broad. For example, the numbers 10 and 15 are broken down as such: 10: 1, 2, 5 15: 1, 3, 5, 15 Least Common Multiple (LCM) The least common multiple of two or more numbers is the least number, except 0, that is a common multiple of both (or all) of the numbers. Watching this video will make you feel like your back in the classroom but rather comfortably from your home. 1275-555=720. Take a look at the following example. Now let's find the LCM of three numbers. knowledge book akbarpur kanpur classesThe Hindu Editorial Vocabularyhttps://www. G C F ( 7 2, 9 0) = 1 8. Prime Factorizations: 12 = 2^2 x 3 15 = 3 x 5 24 = 2^3 x 3. Let's say you have the numbers 2 and 10. If one of the numbers divided by 2 gives the results as 192, what is the second number? If one of the numbers divided by 2 gives the results as 192, what is the second number? To find the lcm of two numbers a a a and b b b, we use l c m (a, b) = (a ∗ b) / g c d (a, b) lcm(a,b) = (a*b)/gcd(a,b) l c m (a, b) = (a ∗ b) / g c d (a, b). 4. wikipedia. So, the LCM of 15 and 9 is 45. youtube. knowledge book akbarpur kanpur classesThe Hindu Editorial Vocabularyhttps://www. – user3026735 Aug 17 '14 LCM of the numbers = H × a × b = HCF × Product of the ratios. The Least Common Multiple (LCM) is the smallest of these common multiples. You can use two methods. Solution. #include <stdio. . e. I get the following result: 32 = 2 5. g. LCM works over Gaussian integers. Both may be primes like 5 and 7. For example, suppose we wanted to find the LCM of two numbers, 4 and 6. M. / each a │0│2│6│12│20│30│42│56│ Sometimes you are given numbers expressed as a product of prime factors. Product of two numbers = (their H. M. . of two numbers. However, =LCM (3,4,5) returns 60, since 60 is the smallest multiple of all three numbers. For example: Let’s say we have the following two numbers: 15 and 9. LCM: Least Common Multiple/ Lowest Common Multiple. If the two numbers do not have any factors in common (other than 1), then the LCM is the same as the product of the two numbers. Another Way to Find LCM. Least common multiple (LCM) of 44, 4, 3 is 132. 1635-555=1080. Divide the numbers by prime numbers. The sum of the numbers is 2/35 Product of the HCF and LCM of two numbers is the same as the product of the numbers themselves. 2100 = 2 3 x 3 3 x 5 2 x 7 1. For example, consider 2 numbers as 10 and 4. The LCM is therefore the product 2 × 3 × 5, which is 30. parseInt( br. Now using the statement: HCF and LCM Questions & Answers for Bank Exams : The LCM of two numbers is 66. 48 = 2 × 2 × 2 × 2 × 3; 180 = 2 × 2 × 3 × 3 × 5; The Excel LCM function returns the least common multiple of integers. and second fraction is (8 x 3)/ (21 x 3) = 24/63. HCF or Highest common factor, also known as GCD (Greatest common divisor) of two numbers is the highest number which is evenly divisible by both of those numbers. The least common multiple is the lowest number that two or more given If we take two numbers with product 12, but not co-prime, the HCF will not remain as 13) (Reference: Co-prime Numbers) Hence the numbers with HCF 13 and product 2028 = (13 × 1, 13 × 12) and (13 × 3, 13 × 4) = (13, 156) and (39, 52) Given that the numbers are 2 digit numbers. 6 = 2 x 3 4 = 2 x 2 A common multiple may be found by multiplying the two numbers 6 and 4. log(lcm_two_numbers(10,15 Free Least Common Multiplier (LCM) calculator - Find the lcm of two or more numbers step-by-step This website uses cookies to ensure you get the best experience. 15, Dec 20. 500 Formulas | 101 Functions Find the least common multiple of two numbers 10 & 15. Find the least common multiple of 6, 8, 15. Algorithm to find LCM of two number Find the prime factorization of each of the two numbers. For example, the LCM of two positive numbers, 72 and 120, is 360. LCM: Least Common Multiple of two numbers is the number that is a common multiple of the both the numbers. The Least Common Multiple (LCM) of two numbers is the smallest number that is a multiple of both. The first one is for \"24\" and the second one is for \"60\" Step 3 : LCM: Descriptive. (vi) H. +2. The numbers are in the ratio 2:3. A simple solution is to find all prime factors of both numbers, then find union of all factors present in both numbers. 20 is their LCM. C. etc. F of fractions = (vii) L. C. 12 is the least common multiple of 4 and 6. The easiest way to get the least common multiple of numbers is to express them as a product of prime numbers (factorize them). C. 4. Recursion : Find the LCM of two numbers : ----- Input 1st number for LCM : 4 Input 2nd number for LCM : 6 The LCM of 4 and 6 : 12 Flowchart: C Programming Code Editor: Example Formula Result; 1 =LCM(C5,D5) 15: 2 =LCM(C6,D6) 10: 3 =LCM(C7,D7) 156: 4 =LCM(C8,D8) #NUM! LCM, the least common multiple, is the smallest positives integers that is a multiple of two integers e. ☞ Multiples of 10 are 10, 20, 30, 40, 50, 60, 70, 80, 90, 100… ☞ Multiples of 15 are 15, 30, 45, 60, 75, 90, 105, 120, 135… Identify the greatest number of times any factor appears in either factorization and multiply those factors to get the least common multiple. There is more than one method to find the LCM of two numbers. Example: Relating this definition to the problem we solved on the previous page, we see that the two numbers n and m are the numbers 20 and 16. In this program, the integers entered by the user are stored in variable n1 and n2 respectively. com/playlist?list=PL3JkVZVtegL9vWRCoXhNE71IWmSx3t2V2BASIC VOCABULARY- Otherwise, it fails 15 and 3 for example have gcd of 3. It contains plenty of examples and practice In this example, you will learn two ways of calculating LCM (Lowest Common Multiple) of two integers entered by the user (C program to find LCM of two numbers). Least common multiple (LCM) of 44, 4, 3 is 132. LCM (250,35 Given an array of rational numbers, the task is to find the LCM of these numbers. Page content. in)); System. Example 1 Find the least common multiple of 4 and 10 by listing multiples. For example, the LCM of 72 and 120 is 360. io. Kindergarten-Grade 12 Standards for Mathematical Practice To calculate the GCD of more than two numbers you can do it as follows: For example: GCD of a, b, c, d. 2. For example, take numbers and . Solution . For example: LCM of two integers 2 and 5 is 10 since 10 is the smallest positive numbers which is divisible by both 2 and 5. Algorithm of LCM. First, we convert both numbers to like decimals i. The least common multiple (LCM) of two numbers is the smallest number that is a multiple of both numbers. Find out the LCM of 5 and 9. The LCM of two or more numbers can also be found using prime factorization. of 12, 16 and 30, then. To find the lcm of any two numbers, it is important to realise that if a divides b, then b needs to have all the prime factors of a (plus some more): 12. For example, HCF of 12 and 18 is 6. Notice that 2 is included twice, because it appears twice in the prime factorization of 4. These worksheets are pdf files. com/playlist?list=PL3JkVZVtegL9vWRCoXhNE71IWmSx3t2V2BASIC VOCABULARY- Least Common Multiple(LCM) The GCF of two natural numbers a and b is the greatest natural number x, which is a factor of both a and b. Now all the fractions must divide A B, that is, A/B ÷ 2/3, A/B ÷ 5/6 and A/B ÷ 7/9 all must be natural numbers. Least common multiple (LCM) of two numbers is the smallest number that they both divide. C plus plus Program to Find the GCD and LCM of two Numbers. Example, find the LCM for 12, 15, 24. The LCM would be the bigger number. \" The Least Common Multiple or the LCM is the lowest number you get when you compare the multiples of numbers. If two or more numbers have no common factor (other than 1),the LCM is their product. For example multiples of 3 and 4 are:3 → 3, 6, Two numbers are said to be coprime if they do not have common factor other than 1. . So, the LCM of 15 and 9 is 45. Returns: 12 because that is the lowest common multiple of both numbers (4*3=12 and 6*2=12). LCM = 2 × 2 × 3 × 3 × 2 × 2. Least common multiple (LCM) of 44, 4, 3 is 132. Example, $$\\text{LCM }(9,27) = 27$$ as $$9$$ is a factor of We have to decompose the given numbers in to prime factors. Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. Example 3: What is the greatest number which exactly divides 110, 154 and 242? Sol. Finding LCM using iterative method involves three basic steps: Initialize multiple variable with the maximum value among two given numbers. So, the least common multiple of 400, 500, and 550 is 22000. The following theorem is useful in finding the LCM of two numbers a and b when their prime factorization is not easy to find. GCF of two or more Numbers Calculator The GCF (GCD) of two or more numbers is the largest number that is evenly divisible by all numbers in the set with remainder zero. in sudocode - write a simple algorithm that finds the Least Common Multiple of two numbers A , B. 10 × 1 = 10 and 2 × 5 = 10. For example: HCF of 210, 45 is 20, HCF of 6, 18 is 6. Suppose we have to find the L. Find the GCF of 2 numbers: GCF of 21, 35 is 7: Find the GCF of 2 numbers (<100) GCF of 12, 100 is 4: Find the GCF of 3 numbers: GCF of 6, 21, 84 is 3: Least common multiples: Find the LCM of 2 numbers: LCM of 3, 4 is 12: Find the LCM of 2 numbers (<50) LCM of 4, 34 is 68: Find the LCM of 3 numbers: LCM of 4, 7, 14 is 28 C Examples; Find Lcm And Gcd Of Two Numbers; डाउनलोड पी. The LCM of these two numbers would be the number that is the multiple of the other. LCM or Least common multiple of two numbers is the smallest number that can be evenly divisible by both of those numbers: For example: LCM of 4, 10 is 20, LCM of 4, 3 is 12. . print(\"Enter 1st number : \"); n1 = Integer. . Note that 30 is a Using this method, first find the prime factorization of each number and write it in index form. <br>In its simplest form, Euclid's algorithm starts with a pair of positive integers, and forms a new pair that consists of the smaller number and the difference knowledge book akbarpur kanpur classesThe Hindu Editorial Vocabularyhttps://www. 15 = 5 * 3 9 = 3 * 3. e 2 3 x 3 1 x 5 2 x 7 1 = 37800( Here consider all prime numbers) Step 3: So LCM of the numbers 72, 108 & 2100 is 37800. We divide the HCF into either number. 35= 5*7. Below is a program to find LCM of two numbers using recursion. A program to find the LCM of two numbers is given as follows − Example. There is a proven statement LCM(a,b)*HCF(a,b)=product of the 2 numbers or ‘a’*’b’. M = (x*y)/G. Or A3/B2, A6/B5 and A9/B7 all must be natural numbers. 17850 smallest integer divisible by all numbers. num2 = 6. => Example: whenever you add fractions with different denominators you need to find the common denominator. 6. 132 smallest integer divisible by all numbers. Also, if we dont have any common Grade 6 » The Number System » Compute fluently with multi-digit numbers and find common factors and multiples. HCF and LCM Questions & Answers for Bank Exams : The LCM of two numbers is 66. Ques. Solution: When I want to find the LCM of two numbers, I generally start by doing a prime factorization of the two numbers. program to find LCM of two numbers using recursion How to find LCM of two numbers in C programming using recursion. The Greatest Common Divisor (GCD) or Highest Common Factor (HCF) of a given integer is the highest positive integer that divides the given integer without remainder. Once we find the prime factorization of the given numbers by using the factorization tree or the upside-down division method, we mark the common prime factors. integer(readline(prompt = \"Enter first number: \")) num2 = as. The least of the common multiples of two or more numbers is called the least common multiple or LCM. 24 is a common multiple of 3 and 4. Let 2 3, 5 6 and 7 9 be three fractions. D. e. Since the numbers are co-prime, their HCF = 1 Product of first two numbers = 119 Product of last two numbers = 391 The middle number is common in both of these products. Take common primes once and then take uncommon primes as below: LCM = 2 * 2 * 3 * 2 * 3 = 72. Example 7 : Find the LCM of numbers 72, 108 & 2100. For example, if 10 and 12 are the two numbers. One way to find the least common multiple of two numbers is to first list the prime factors of each number. 16 = 2 × 2 × 2 × 2. Example: Find the least common multiple of 40 and 50 with the division method? Amelia 2020-09-17 06:28:50. The HCF of two numbers is 8. See full list on en. = 5 × 8 = 40. Step 2 : Write each of the prime factors to their highest available power. Method #1 One method of finding the LCM is by listing multiples of each number. e. The least common multiple (LCM) of two numbers is the smallest number (not zero) that is a multiple of both. Find the product of each factor with the highest power which occurs in the given expressions. Implementation. The HCF of two or more numbers is the greatest number that divides each one of them exactly. You can easily compute the least common multiple using our LCM calculator on your mobile or desktop device. M of 3, 5 and 6 is 30. We find the GCF(a,b) using the Euclidean algorithm. 3) Go number by number in both columns and circle any numbers they have in common. 2 For any two nonzero whole numbers a and b we have LCM(a,b)GCF(a,b) = a·b. The answer is LCM (numerators)/LCM (b,d) Lets see this using example of 2/9 and 8/21. One way of computing LCM(a,b) is by prime factoring a and b. For the above example, the l c m lcm l c m turns out to be 6. Another worksheet in which students determine the least common multiple for each number pair. . The least common multiple(LCM) of two integers a and b, usually denoted by LCM (a, b), is the smallest positive integer that is divisible by both a and b. 3 and 5 have other common multiples such as 30, 45, etc, but they are all larger than 15. The LCM of two or more numbers is the smallest positive integer that is divisible by all the given numbers. 2. 2. A simple solution is to find all prime factors of both numbers, then find union of all factors present in both numbers. For example: Find LCM of 12 and 18 . The numbers are in the ratio 2:3. M. HCF × LCM = Product of the two numbers. There is a mathematical formula to find the LCM of two number using GCD (n1 x n2) / GCD(n1, n2). By using this website, you agree to our Cookie Policy. => 6 x = 48 6 x = 48. 30 = 2 × 3 × 5 LCM (Least Common Multiple) of two numbers is the smallest number which can be divided by both numbers. The least common multiple is the product of the each prime factors with the greatest power. When trying to determine the LCM of more than two numbers, for example LCM (a, b, c) find the LCM of a and b where the result will be q. The common multiples of 3 and 4 are 0, 12, 24, . if they have any common multiple except 1, then it will always be greater than both the numbers or equal to the greater number. The LCM of two numbers 100 and 152 is 3800. Codes #! /usr/bin/env python3 # -*- coding: utf-8 -*- import fractions a = int(input(\"Enter the first number:\")) b = int(input(\"Enter the second number:\")) lcd = a * b // fractions. L. It is checked whether max is divisible by n1 and n2, if it’s divisible by both numbers, max (which contains LCM) is printed and the loop is terminated. 108 = 2 2 x 3 3. In the intersection of the sets of common factors, it is the greatest value. What is LCM? A Least Common Multiple (LCM) of Two Integers is the Smallest Number which is a Multiple of Both the Numbers. Proof. Product of LCM factors: 8 x 3 x 5 = 120 So the LCM for the three numbers = 120. Since LCM is given by the product of the maximum exponent of each factor which has appeared in the prime factorisation of each of the given numbers. Ex: 10, 15, 20 (or) 24, 48, 96,45 (or) 78902, 89765, 12345 Use the LCM of two or more numbers Calculator to find the Least Common Multiple of numbers 75, 42, 51 i. Examples: Input: 7, 5 Output: 35 Input: 2, 6 Output: 6 The approach is to start with the largest of the 2 numbers and keep incrementing the larger number by itself till smaller number perfectly divides the resultant. C#; given a list of non negative integers return the longest number ; lcm coe; find lcm of two numbers in c; n which returns the LCM of N1 and N2. LCM = 2 × 2 × 3 × 4 × 3. C. That is the LCM. Many ideas in number theory can be interpreted profitably in terms of prime factorizations. However it may not be the lowest. The LCM (Least Common Multiple) of any two numbers is the smallest possible integer that is divisible by both. C++ program to find LCM and HCF of two. Least Common Multiple: the least common multiple is the smallest of the common multiples. Test Data: console. This excellent video shows you a clean board, with the instructors voice showing exactly what to do. Program to Compute LCM Examples of Finding LCM of Two numbers in Decimal Example 1: Find the least common multiple of 0. Least Common Multiple (LCM) In Mathematics, LCM of any two is the value which is evenly The least common multiple can be defined as the lowest positive integer that is a multiple of every number in a given set of numbers. of two given numbers by the following two methods: 1. . HCF of 720 and 1080 which is 360. The divisors and the remainders are multiplied together to obtain the LCM. The LCM, least common multiple, is the smallest value that two or more numbers multiply into. We divide 18 by 6 since it’s easier, 18÷6=3. For example, the LCM of 2 and 3 is 6, as both numbers can evenly divide the number 6. M. Live Demo In Mathematics, the Least Common Multiple (LCM) of two or more integers is the smallest positive integer that is perfectly divisible by the given integer values without the remainder. M. And to calculate the LCM of more than two numbers you can do it as follows: For example: LCM of a, b, c, d. Given an array of rational numbers, the task is to find the LCM of these numbers. For example, =LCM (3,4) returns 12, since 12 is the smallest multiple of both 3 and 4. ई-बुक्स This is a C program to find GCD of two numbers. We may write the solution symbolically as GCF (20, 16) = 4. If you want to find the LCM and HCF in an exam, we can use prime factor form In this example, you will about C++ program to find LCM (Lowest Common Multiple) using two different methods. M. 132 smallest integer divisible by all numbers. Lcm of three numbers What is the Lcm of 120 15 and 5; Lcm 2 Create the smallest possible number that is divisible by numbers 5,8,9,4,3; Decompose Decompose into primes and find the smallest common multiple n of (16,20) and the largest common divisor D of the pair of numbers (140,100) LCM • What is the least common multiple of these numbers? Assumes Moses’ numbers must be 2 and 60 For example: Student points out that the greatest common factor of 2 and 60 is 2 and the least common multiple of 2 and 60 is 60 (Q5). First we find the prime factorizations of these three numbers. 15 = 5 * 3 9 = 3 * 3. lcm of two numbers examples", null, "" ]

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[ "QA-Tf17: 1st QA-Hung’s Transformation\n\nQA-Tf17 maps a line into a line.\nLet P1, P2, P3, P4 be the defining Quadrangle Points.\nLet S1 = P1P2 ^ P3P4, S2 = P1P3 ^ P2P4 and S3 = P1P4 ^ P2P3.\nLine S2S3 meets lines P1P2 and P3P4 at S12 and S34.\nLine S1S2 meets lines P1P4 and P2P3 at S14 and S23.\nLine S3S1 meets lines P1P3 and P2P4 at S13 and S24.\nr is a random line.\nr meets lines S1S3, S3S2, S2S1 at T2, T1, T3.\nConstruct T1', T2', T3' such that\n• (S12,S34;T1,T1') = -1\n• (S14,S23;T3,T3') = -1\n• (S13,S24;T2,T2') = -1\nThen T1', T2', T3' are collinear on a line QA-Tf17(r).\nThis transformation was found by Tran Quang Hung. See Ref-34, QFG#3711.", null, "CT-coordinates\nof QA-Tf17(r), where r=(l:m:n)\n\n( q r (l2 p2  - l m p q - m2 q2 -  l n p r - m n q r  - n2 r2) :\n-p r (l2 p2 + l m p q - m2 q2 + l n p r + m n q r + n2 r2) :\n-p q (l2 p2 + l m p q + m2 q2 + l n p r + m n q r - n2 r2) )\n\nProperties:\nQA-Tf17(QA-Tf17( r )) = r. See Ref-34, QFG#3717.\nQA-Tf17(r) = QL-Tf2 of the dual QL as described at QA-8/QL-8. See Ref-34, QFG#3718.\n• For lines L of a line pencil wrt a point P, QA-Tf17(L) envelopes a conic tangent at QA-Tr1 in the cevian points of QA-Tf2(P). Example: Inscribed Steiner ellipse for QA-P16. See Ref-34, QFG#3713\nQA-Tf17(Infinity Line) = QA-Lx = line through the midpoints of (S12,S34), (S13,S24), (S14,S23). QA-Lx is a line // QA-P1.QA-Tf2(IP(QA-P1.QA-P16)) // QA-P5.QA-Tf2(IP(QA-P5.QA-P17)) // QA-P10.QA-Tf2(IP(QA-P10.QA-P16)) // QA-P19.QA-Tf2(IP(QA-P1.QA-P5)), where IP denotes “Infinity Point”. See Ref-34, QFG#3717.\n\n#### Plaats reactie", null, "Vernieuwen" ]

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[ "# 2634242\n\n## 2,634,242 is an even composite number composed of four prime numbers multiplied together.\n\nWhat does the number 2634242 look like?\n\nThis visualization shows the relationship between its 4 prime factors (large circles) and 16 divisors.\n\n2634242 is an even composite number. It is composed of four distinct prime numbers multiplied together. It has a total of sixteen divisors.\n\n## Prime factorization of 2634242:\n\n### 2 × 13 × 71 × 1427\n\nSee below for interesting mathematical facts about the number 2634242 from the Numbermatics database.\n\n### Names of 2634242\n\n• Cardinal: 2634242 can be written as Two million, six hundred thirty-four thousand, two hundred forty-two.\n\n### Scientific notation\n\n• Scientific notation: 2.634242 × 106\n\n### Factors of 2634242\n\n• Number of distinct prime factors ω(n): 4\n• Total number of prime factors Ω(n): 4\n• Sum of prime factors: 1513\n\n### Divisors of 2634242\n\n• Number of divisors d(n): 16\n• Complete list of divisors:\n• Sum of all divisors σ(n): 4318272\n• Sum of proper divisors (its aliquot sum) s(n): 1684030\n• 2634242 is a deficient number, because the sum of its proper divisors (1684030) is less than itself. Its deficiency is 950212\n\n### Bases of 2634242\n\n• Binary: 10100000110010000000102\n• Base-36: 1KGLE\n\n### Squares and roots of 2634242\n\n• 2634242 squared (26342422) is 6939230914564\n• 2634242 cubed (26342423) is 18279613522842900488\n• The square root of 2634242 is 1623.0348117031\n• The cube root of 2634242 is 138.1079111145\n\n### Scales and comparisons\n\nHow big is 2634242?\n• 2,634,242 seconds is equal to 4 weeks, 2 days, 11 hours, 44 minutes, 2 seconds.\n• To count from 1 to 2,634,242 would take you about six weeks!\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 2634242 cubic inches would be around 11.5 feet tall.\n\n### Recreational maths with 2634242\n\n• 2634242 backwards is 2424362\n• The number of decimal digits it has is: 7\n• The sum of 2634242's digits is 23\n• More coming soon!" ]

[ null ]

[ "## DEV Community is a community of 865,568 amazing developers\n\nWe're a place where coders share, stay up-to-date and grow their careers.", null, "Iulian Preda\n\nPosted on • Updated on\n\n# Codeforces - How to use Typescript/Javascript like a pro\n\nHi everybody,\n\nGetting that out of the way, let's get started. So, today I wanted to explore again some algorithmics problems and I returned to CodeForces.\n\nLately, I have used a lot of Typescript so naturally, I wanted to use it here too and I gladly discovered that you can actually use javascript in it.\n\nThen the next question was, but how can I run only a single file in my Typescript project that holds many other problems (or at least it will)?\n\nI tried many solutions but I came about creating an `npm package` that I can simply import or require since otherwise, I had issues with imports, paths, and more.\n\nThis is the package codeforces-io.\nIt is a simple mock for the `readline()` and `print()` functions so you do not have to modify your existing code while copying it to submission.\n\nThe documentation covers how to use it as a simple package, so I will write here a tutorial from start to finish with some advanced usages that you might like.\n\nI created this package for those who want their code to be as `close to the submission code` as possible without having to deal with the `boilerplate™` e.g.1,2.\n\n# CodeForces IO\n\nSmall package to mock Codeforces Javascript IO functions.\n\n# Install\n\n`npm i @ip-algorithmics/codeforces-io`\n\n# Intro\n\nCodeforces for Javascript/Typescript uses `readline()` and `print()` functions for input and output to the standard input/console.\n\n# How to use\n\nThis library exposes the functions in a manner that allows you to just copy the source code for the submission.\n\n## Importing the functions\n\nThe functions can be imported both as ES6 modules or using the `require` function. The difference lays in how you access it.\n\n```// using ES6 modules import - you need to have the \"module\" property set to \"commonjs\" in the package.json\nimport { readline, print } from '@ip-algorithmics/codeforces-io';\n// using require\nconst codeForcesIO = require('@ip-algorithmics/codeforces-io');\nconst print = codeForcesIO.print;\n// alternative\nconst { readline, print, console } = require('@ip-algorithmics/codeforces-io');```\n\nThis…\n\n# Too much talk, let's see the usage\n\nFirst, you have to have an `npm` project in the folder. So open up a terminal in your desired folder and just type `npm init`. You can use the default.\n\nThen `npm i @ip-algorithmics/codeforces-io`\n\nIf you do not use Typescript this is all you need to do, otherwise... just run this one after the other.\n\n• `npm i -g typescript ts-node`\n\nNote: if you want to configure Typescript you will additionally need to run\n\n• `tsc --init`\n\nRemember to `not` include the \"DOM\" library into the `lib` because the `print` function is already declared in the DOM library for printing files.\nYou can just require/import the `print` function for this package for output to the console. If need other functions from the console object you can require the entire `console`.\n\n## Usage with `ts-node` or `node`\n\nNow that you have a working environment you can create a `.js` or `.ts` file and get going.\nThis library uses modern ES6 import/export modules but that means declaring a few more things in the project configuration.\nSo let's stay to `require` for now.\nAt the top of your file just write\n\n``````const {nextLine, print, console} = require('@ip-algorithmics/codeforces-io')\n``````\n\nNow that we have our functions we can use them in the project.\n\nThe `readline` function reads from `input.txt` file that needs to be in the same folder.\nHere is a quick example\n\n``````// example.js or example.ts\nconst { readline, print } = require('@ip-algorithmics/codeforces-io');\n\nfor (let i = 0; i < numberOfLines; i++) {\n.trim()\n.split(',')\n.map((x) => parseInt(x, 10));\nif (x < 90 || x > 90 || x < 90 || x > 90) {\nprint(x);\nbreak;\n}\n// do something with x\n}\n``````\n\nAnd here is the `input.txt`\n\n``````3\n-47,23\n34,56\n91,82\n``````\n\nNow that we have our files in place we can just run\n`node example.js` or `ts-node example.ts`. You can use an absolute or relative path, this library will search for the place where the script is and then read the input file from there.\nSo you do not have to `cd <path>` you can just use `node <path>/<file>`.\n\n## Usage with jest", null, "For a simple reason. Paths can get really long and jest can use test names to find the file directly\n\nCompare these:\n\n• `ts-node src/1300-points/totally-real-problem/index.ts`\n• `jest -t 'totally-real-problem` I'd say it's a bit shorter. But how can you obtain that? Well, that is simple actually.\n\nJust run `npm i -g jest ts-jest` and `npm i @types/jest` and then in the root of the project you need to create a file named `jest.config.ts` and put the next lines in it:\n\n``````export default {\npreset: 'ts-jest',\ntestEnvironment: 'node',\ntransform: {\n'node_modules/variables/.+\\\\.(j|t)sx?\\$': 'ts-jest'\n},\ntransformIgnorePatterns: ['node_modules/(?!variables/.*)']\n};\n\n``````\n\nFor javascript you can just run `jest --init` and that will work by default.\n\nExcept for this, now all of your scripts will have to be a test\nso the naming should be .test.ts or .spec.ts or create a folder named `__tests__` and place your scripts in it (same things apply to javascript also).\n\nNow you have to convert your scripts to tests. This is as simple as wrapping everything in\n\n``````test('name of test', () => {\n})\n``````\n\nThe example from above will get transformed to this.\n\n``````test('example', () => {\nconst { readline, print, testOutput } = require('@ip-algorithmics/codeforces-io');\n\nfor (let i = 0; i < numberOfLines; i++) {\n.trim()\n.split(',')\n.map((y) => parseInt(y, 10));\n\nif ((x < 90 && x > -90) || (x < 90 && x > -90)) {\nprint(x + ',' + x);\nbreak;\n}\n}\n}\n``````\n\nTo run it just type `jest -t 'example'`.\n\n## Special mention\n\nFor typescript projects if you use the same function names without putting them in a class you will get an error that the name is taken, so you have to add `export {};` at the end of the file so the names will not conflict.\n\nIf something doesn't work do not hesitate to leave a comment, open an issue or even better create a pull request.\n\n# TL;DR\n\n• Created a library for Codeforces that reads `input.txt` from the folder the script was called.\n• `npm i @ip-algorithmics/codeforces-io`\n• Code example\n``````// example.js or example.ts\nconst { readline, print, testOutput } = require('@ip-algorithmics/codeforces-io');\n\nfor (let i = 0; i < numberOfLines; i++) {\n.trim()\n.split(',')\n.map((y) => parseInt(y, 10));\n\nif ((x < 90 && x > -90) || (x < 90 && x > -90)) {\nprint(x + ',' + x);\nbreak;\n}\n}\n\n``````\n\nAnd here is the input.txt\n\n``````3\n-47,23\n34,56\n91,82\n``````\n• `node <path to file>/file.js` or `ts-node <path to file>/file.ts`\n\nHappy codding!\n\n# Update\n\nI added the `testOutput` function. Now you can create your `output.txt` file and put the outputs in it.\nYou just have to run the function at the end of the script and it will prompt you with a `Passed` or `Failed` output in the console.\nRemember not to copy this function in the submission.\n\nInternally now the `print` caches everything so that the `testOutput` can check if what was printed is the same as what is in the `output.txt`\n\nExample:\n`input.txt`\n\n``````3\n-47,23\n34,56\n91,82\n``````\n\n`output.txt`\n\n``````-47,23\n``````\n``````const { readline, print, testOutput } = require('@ip-algorithmics/codeforces-io');\n\nfor (let i = 0; i < numberOfLines; i++) {\n.trim()\n.split(',')\n.map((y) => parseInt(y, 10));\n\nif ((x < 90 && x > -90) || (x < 90 && x > -90)) {\nprint(x + ',' + x);\nbreak;\n}\n}\n\ntestOutput(); // Result Passed\n``````\n\n## Update 2\n\nThe solution generator is completed. Now you can just create the `input.txt`, `output.txt`, `index.ts` and the folder with a simple command. It will also add the imports for you.\n\nYou can use this feature in 2 ways:\n\n• Without installing globally the library using `npx @ip-algorithmics/codeforces-io`.\n• Installing globally the library `npm i -g @ip-algorithmics/codeforces-io` and then using `cf` in the command line.\n\n### Parameters\n\n• `path` - the path to the solution. E.g. `./ProblemA`. Defaults to `./New Solution`.\n• `--f` or `--file` - the name of the js/ts file to create. E.g. `mainFile`. Defaults to `index`.\n• `--js` - Uses `.js` extension instead of `.ts`\n• `--cjs` - Uses `require` instead of ES6 `import`/`export`\n• `--c` or `--comment` - Adds a comment at the beginning of the file. E.g. the link to the problem.\n\n### Usage examples\n\n• `npx @ip-algorithmics/codeforces-io ./ProblemA --js --cjs`\n• `cf ./ProblemA --js --cjs`\n• `cf ./ProblemA`\n• `cf ./ProblemA --c http://link.to.problem`\n\n# Update 3 - Typescript ideas\n\nI noticed that Codforces uses ES5 internally in both node and V8 so it has problems with `const`, `let` and features like spreading an array.\n\nPersonally I found that the best solution is to:\n\n• add in the `package.json` the following script `\"build\": \"tsc --module es6 --moduleResolution node\"` using it like `npm run build <path>`\n• manually run `tsc --module es6 --moduleResolution node <path>`" ]

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[ "# 矩阵乘法的几何理解\n\n2x-y=0\n-x+2y=3", null, "$\\begin{bmatrix}2& -1\\\\-1& 2\\end{bmatrix}\\begin{bmatrix}x\\\\y\\end{bmatrix}=\\begin{bmatrix}0\\\\3\\end{bmatrix}$", null, "$x\\begin{bmatrix}2\\\\-1\\end{bmatrix}+y\\begin{bmatrix}-1\\\\2\\end{bmatrix}=\\begin{bmatrix}0\\\\3\\end{bmatrix}$", null, "$1\\begin{bmatrix}2\\\\-1\\end{bmatrix}+2\\begin{bmatrix}-1\\\\2\\end{bmatrix}=\\begin{bmatrix}0\\\\3\\end{bmatrix}$", null, "2x-y\n-x+2y", null, "$\\begin{bmatrix}2\\\\-1\\end{bmatrix}\\begin{bmatrix}-1\\\\2\\end{bmatrix}$不在同一个低维空间，也就是不在同一条直线上。\n\n2x+4y\n-x-2y\n\n## 1 对 “矩阵乘法的几何理解”的想法；\n\n1.", null, "autoxin说道：\n\n基于该理论，就可以将神经网络解释为高维空间中非常复杂的几何变换，就是把复杂的、高度折叠的数据流形找到简洁的表示" ]

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[ "# #0c000e Color Information\n\nIn a RGB color space, hex #0c000e is composed of 4.7% red, 0% green and 5.5% blue. Whereas in a CMYK color space, it is composed of 14.3% cyan, 100% magenta, 0% yellow and 94.5% black. It has a hue angle of 291.4 degrees, a saturation of 100% and a lightness of 2.7%. #0c000e color hex could be obtained by blending #18001c with #000000. Closest websafe color is: #000000.\n\n• R 5\n• G 0\n• B 5\nRGB color chart\n• C 14\n• M 100\n• Y 0\n• K 95\nCMYK color chart\n\n#0c000e color description : Very dark (mostly black) magenta.\n\n# #0c000e Color Conversion\n\nThe hexadecimal color #0c000e has RGB values of R:12, G:0, B:14 and CMYK values of C:0.14, M:1, Y:0, K:0.95. Its decimal value is 786446.\n\nHex triplet RGB Decimal 0c000e `#0c000e` 12, 0, 14 `rgb(12,0,14)` 4.7, 0, 5.5 `rgb(4.7%,0%,5.5%)` 14, 100, 0, 95 291.4°, 100, 2.7 `hsl(291.4,100%,2.7%)` 291.4°, 100, 5.5 000000 `#000000`\nCIE-LAB 0.993, 5.179, -4.36 0.231, 0.11, 0.424 0.302, 0.144, 0.11 0.993, 6.77, 319.907 0.993, 1.227, -1.995 3.315, 6.631, -5.272 00001100, 00000000, 00001110\n\n# Color Schemes with #0c000e\n\n• #0c000e\n``#0c000e` `rgb(12,0,14)``\n• #020e00\n``#020e00` `rgb(2,14,0)``\nComplementary Color\n• #05000e\n``#05000e` `rgb(5,0,14)``\n• #0c000e\n``#0c000e` `rgb(12,0,14)``\n• #0e0009\n``#0e0009` `rgb(14,0,9)``\nAnalogous Color\n• #000e05\n``#000e05` `rgb(0,14,5)``\n• #0c000e\n``#0c000e` `rgb(12,0,14)``\n• #090e00\n``#090e00` `rgb(9,14,0)``\nSplit Complementary Color\n• #000e0c\n``#000e0c` `rgb(0,14,12)``\n• #0c000e\n``#0c000e` `rgb(12,0,14)``\n• #0e0c00\n``#0e0c00` `rgb(14,12,0)``\n• #00020e\n``#00020e` `rgb(0,2,14)``\n• #0c000e\n``#0c000e` `rgb(12,0,14)``\n• #0e0c00\n``#0e0c00` `rgb(14,12,0)``\n• #020e00\n``#020e00` `rgb(2,14,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #0c000e\n``#0c000e` `rgb(12,0,14)``\n• #220028\n``#220028` `rgb(34,0,40)``\n• #380041\n``#380041` `rgb(56,0,65)``\n• #4e005b\n``#4e005b` `rgb(78,0,91)``\nMonochromatic Color\n\n# Alternatives to #0c000e\n\nBelow, you can see some colors close to #0c000e. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #09000e\n``#09000e` `rgb(9,0,14)``\n• #0a000e\n``#0a000e` `rgb(10,0,14)``\n• #0b000e\n``#0b000e` `rgb(11,0,14)``\n• #0c000e\n``#0c000e` `rgb(12,0,14)``\n• #0d000e\n``#0d000e` `rgb(13,0,14)``\n• #0e000e\n``#0e000e` `rgb(14,0,14)``\n• #0e000d\n``#0e000d` `rgb(14,0,13)``\nSimilar Colors\n\n# #0c000e Preview\n\nThis text has a font color of #0c000e.\n\n``<span style=\"color:#0c000e;\">Text here</span>``\n#0c000e background color\n\nThis paragraph has a background color of #0c000e.\n\n``<p style=\"background-color:#0c000e;\">Content here</p>``\n#0c000e border color\n\nThis element has a border color of #0c000e.\n\n``<div style=\"border:1px solid #0c000e;\">Content here</div>``\nCSS codes\n``.text {color:#0c000e;}``\n``.background {background-color:#0c000e;}``\n``.border {border:1px solid #0c000e;}``\n\n# Shades and Tints of #0c000e\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #0c000e is the darkest color, while #fef9ff is the lightest one.\n\n• #0c000e\n``#0c000e` `rgb(12,0,14)``\n• #1d0022\n``#1d0022` `rgb(29,0,34)``\n• #2e0035\n``#2e0035` `rgb(46,0,53)``\n• #3e0049\n``#3e0049` `rgb(62,0,73)``\n• #4f005c\n``#4f005c` `rgb(79,0,92)``\n• #600070\n``#600070` `rgb(96,0,112)``\n• #710084\n``#710084` `rgb(113,0,132)``\n• #820097\n``#820097` `rgb(130,0,151)``\n• #9300ab\n``#9300ab` `rgb(147,0,171)``\n• #a300bf\n``#a300bf` `rgb(163,0,191)``\n• #b400d2\n``#b400d2` `rgb(180,0,210)``\n• #c500e6\n``#c500e6` `rgb(197,0,230)``\n• #d600f9\n``#d600f9` `rgb(214,0,249)``\n• #dd0eff\n``#dd0eff` `rgb(221,14,255)``\n• #df22ff\n``#df22ff` `rgb(223,34,255)``\n• #e235ff\n``#e235ff` `rgb(226,53,255)``\n• #e549ff\n``#e549ff` `rgb(229,73,255)``\n• #e85cff\n``#e85cff` `rgb(232,92,255)``\n• #eb70ff\n``#eb70ff` `rgb(235,112,255)``\n• #ed84ff\n``#ed84ff` `rgb(237,132,255)``\n• #f097ff\n``#f097ff` `rgb(240,151,255)``\n• #f3abff\n``#f3abff` `rgb(243,171,255)``\n• #f6bfff\n``#f6bfff` `rgb(246,191,255)``\n• #f9d2ff\n``#f9d2ff` `rgb(249,210,255)``\n• #fbe6ff\n``#fbe6ff` `rgb(251,230,255)``\n• #fef9ff\n``#fef9ff` `rgb(254,249,255)``\nTint Color Variation\n\n# Tones of #0c000e\n\nA tone is produced by adding gray to any pure hue. In this case, #070608 is the less saturated color, while #0c000e is the most saturated one.\n\n• #070608\n``#070608` `rgb(7,6,8)``\n• #080608\n``#080608` `rgb(8,6,8)``\n• #080509\n``#080509` `rgb(8,5,9)``\n• #090509\n``#090509` `rgb(9,5,9)``\n• #09040a\n``#09040a` `rgb(9,4,10)``\n• #09040a\n``#09040a` `rgb(9,4,10)``\n• #0a030b\n``#0a030b` `rgb(10,3,11)``\n• #0a030b\n``#0a030b` `rgb(10,3,11)``\n• #0a020c\n``#0a020c` `rgb(10,2,12)``\n• #0b020c\n``#0b020c` `rgb(11,2,12)``\n• #0b010d\n``#0b010d` `rgb(11,1,13)``\n• #0c010d\n``#0c010d` `rgb(12,1,13)``\n• #0c000e\n``#0c000e` `rgb(12,0,14)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0c000e is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "", null, "", null, "### Home > INT3 > Chapter 1 > Lesson 1.1.3 > Problem1-46\n\n1-46.\n\nAlgebraically determine the y-intercept of the graph of each equation. Write each answer as an ordered pair.\n\nTo solve for the $y$-intercept, substitute $0$ for $x$ in the equation.\n\n1. $y = 3x + 6$\n\n1. $x = 5y – 10$\n\n1. $y = x^{2}$\n\n1. $y = 2x^{2} – 4$\n\n1. $y = \\left(x – 5\\right)^{2}$\n\n1. $y = 3x^{3} – 2x^{2} + 13$" ]

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SMAK/8hRgLL8S6SwvMcWDQzvascJkuopwm+szYqyA2SH3kRum89v6EE33NrjKLdwLy0Ffh2G4qUg32uVon3YtWxXrWXUEd8FCqftTH765n3cuqEC7zXUczvGyW8W5TzFrwvFmda1k/5wn0wEqelQJ7qWX/XlHC9Jr6z9hLrr0LRKws9tPhJS4FKutaTFjbUcSQcIhO48vcP7F9sZHWJhA58zshvpW/D9SoNNFAIMkRXQ27yHInWkL+ADa2LqTyGCXv+6ciz9GLs7aWfxLT3s4GIAxq8x5n2oALpQCB38X7PeXlw5bNM/2mmfdY59jz/38HjPr7BfFwVk4ejeXxG4NhHeN2XJJr/AOWJlfWOK/IO7D0v8fbv4z0Xnvlv3vNAfsf07+exh6ic+cR5Ae9jPVbYvijwbhDvMZv32jMmz0fy/FsK1P+TmZ9rCjz7VF7nm72ou7vElAfK6RGWq0/4tzL9PwJ1Au/04zH3QnDrLyRaCvkVvtvZRd7tRL7/13gOzv2l9OwGRPndXCBfuO8nipSFfbffKpBmBtNMLXKtk5gOsUTDlKYU/WmhZ2MIvbNCefqQ00BmaG3tE9Nozab2HCLoNY5G7Fp3owNp0T0wpgzFoFLYjB6Mnfn/VeYRDc6lEi0aM9GxEDZhwybcZxeoBfHbYMVT2ABZLX8bCqam/WlMPr4i+eF7Q4rkGaMbtuS76QqUWcJpxOud/HY69cfm91iS6IWedY38xgUsDuXxVd7+/VlvhrNsXmR5oSG+nedMi7EyJ/P4ZCoSqx2PyFjHE5Ry6ppb31c639P2tIirPCX4VxKtBgjMo/W1PZ/9Uzy2wrnODvRWYA6HCQEr3JbDigIWHIJGtyWxX0GPgA+U89Ysq3JRRyXGWrJZx1BA3vYyciiVsLWO8rgd03YG6vBRVODvcu6D7+MevosMFTYowntQcPw7Xt6+4xDnElrmyOsJLG8onU85dXIrJ1+2TXHzdQzzNTNG0Z1MRWwyvYAhq34sy+Ub/BbfiCnT8/jemjYy40PxHrTQQ+iqoFtoNK2PI9kQ7BtDtLDkf+6QiA806D8q4X7PsdFMDED5X83GaIFEa7uPpxxPUsAwv9O9cgZ+xgZ/R/4iNuA2ktN0yc++57pZz2BjEfIQuKMFisUjWCI7xcmDK+PZ+LrXQgO8k5Nmd8fC/j6f3ffQxE3qkw4QKkj8Jv7+kff6MJXDHzLNZVSQfNgpi4VKneuheJjPY8t5MvfPoQJkn/dwrx52eN/Dt0jYq1incc4H+X6XkbAv9JTmDsfrcEGJ5eBiJz4b0OwoE6FvN84zVgz2/UKp2I1ltAOf78tU9A/y6rDN77leHd6dym09CXGYo1TdSDKczfLYieV3GdOc79WhfRwyv5RpbZ14gG3M9Z4HzObrvJh81Xn58pXJcY6XZq8i3w6I+rSYNJ93PAgdou52xQAQ+kBgKt1icV6GIbRKFhS5DhqDtwcg/2igPsftMyVa/jXDjxgW5ZU8dnbAbbmazzWPv3B7TqIS00wLxMeOtH58wHrbtBf5X+TkwZW5bMh90niNx+fTMsJ8BLMc5aAv+CS9Bkv4PHNYlktIpo+wrp8ZOHcij83l/0nOsTbut+X8hkN+9nlej7G0xCGkE7l9Cb0IHSyTu0ggQqKPc69+m5ZoOTiGHoV5zO+kfqzLackHvM7n9g2S78I4WnpOKLXUq8OoEyfxnYEcd2G63aiItbKePM93i/7w7xm5m+lOdK5tn/XPVBiX8ZyX6alq4/UPCTwL7v8vL1+TuB+KcqhLwN77Nf6eUEKZTQ54C1EPz1JaUgw0oW/oRUlg2V5cJE2t89HH4T5q300DUPZoHBpp3TweOD6dpPftwHtKxlhLL3M7zl39TU8Bgqvwq45VWA7K6a6B5VoT2P9bx5rsSx3awfG2LA0cn0Kiv9Xb30yLKMuyWUhLb8uY+6Sc56ktMW9Qlmx/+gOB4w+R3DeR9fvdq0g8C3jfH5dxT6Q71lEGXqVC8MF+qstx5fG04wWqLaH+LCVxAkMdi1eoWL0WOOde/m7r7NveO+biLXrAzohRxEL5Wu7UK1/p2oyKwTpes4WK+ogSPJH+PBoHSnwMgULRL4Qeck03SnhseiXRzgbxMDZSxQjIRr+jEX8wcBxW0jkFnqm/Yee1XynhaG7sn0Fr3Y+E7o7xSNh+8IXesQdo2XzMs0pgOW1HC/8fZea/EjETbzl5b+jDdWwjG+dpQUAUgsf+GmhA4SlBlwC6CeBih2v1iAq+5yaSWafk+9r9et1CIqnzvrMsLbZVtCi/U+I94fL9AOsBvAD3U2Hqr9EdWQlH2u/rELVfx0PR+weQjLO08oHhzjUk5juxdci2aU1F6sPdVJifCRwL5etAyceCvOwd+yy/ZVjyCGJDtwCi8A8t0Hb+kt/w1x3FxSrcwEyJjw1SKCpiZbkNUKjRapJ8UE9fAGviSoeQYXku4wf+ai8UljQVgNmelfgTiSJJB7rsu6T8/stNaNW6VuC32OgsCxAXgv4w8c+1THc3G3jr3kMU9GllNN7AFWwwk16D9b2YhlJilCrrceiLhZ4sUDcLwbpGf+80pCdy/3SpzOp5SckPLQzFBXQ7+xMBJe0JiVzXeEfnUvF4usg9j3eIK81fBGIhIvxyqVwAq1uXMT/FWueZP8P8WgLzyxJW7OZMm6FX5EQqP4gHedF7t+uKKJZJpwxD9WFXfjdZJ13I6j/Cy9dYenf8fPllfadThw5mHZoRk2d8n2OoKEyi9wWWOUZ9wN3/fxLFZWj/uaLfCT2k9Q7nR+AT+v5s4NNO5QSp3sCPI4TFrNCVBAgGQTBnOhbs1AEue7dhKddDcDLFByL7vyw9o5mHsnFBfy2Gtu1GBeyjtDhmUukpB3EL8/y0DEJ3yyJbobIsFWioD2KjbUdVII5hCZ9tl148R2/ec7H3D+/Xj0jGu7Px372AEjhC8gFwv+bvoxL1Ce9A6/3+CtdlfP+PxRybwW/Px3HSc8hZG7/9s5xyK/ZuE166uHNQhhO8c690lA6LYwKeDHjIEIB7tqeYjGd5tku+L38W0+9PBXtujBJyNQkdVvr/UuGCAYKA1/kyMF5DxSAk9BcC+6C9fs2z8rDvssBHBFxVwPqp7qdnRV6OYkOOhV2WD3DZ9+WDfZtKSZKNACwjuPxulsi1HipTuG2voyJzjuOt+G82pMky84358Z+UvFswUaB+FPKgDFRZHk6yhJvddjesIrmfxkb9mQrlLdGH57CW4mkkzY+TBBbFXOMztEThfXrEsW7RdQOX/cR+IPRuWq7dfKcZEtmdjlLhA11hiB9AVx2i4D9EMjy1l+82UeQcxGu8QuPCkm1XgXwlWc7IF0ZOTAmktYGHs0jCwJtMj2NHSj641QW6l+5gvUM3GQJz0RXWQkLfSqlJsaEI/a8kR/+jQXAV+o7gEkRf4BdjyBxE9KCEg6T6E8v4cR0vPYOjBgJtzsddI4XXhk94FsgvJN//Xw5gZaCf7mj+XyDR+OjeAIQxu49lYPu+OyTvUrWKRZzClw4oA+scS7FURcK6SuGh2JPfQkbyoyKg/F1c5L2Ugg5aZPUSjhOwM9+JxA/Vs+WNbo6LJBri9ouYdLYb4SXvuawCcBjLaWUF6/JKWqpryzgHwai3OSQICxf90RjG+ZyTrt3xMoUwxClnW286vPplFVeLmwsQ+h+db+JNtmeH0ZvldtHVOJb8K3z+JOuntcqhPP1Qes7SZ2daRJ5ukXyA73S2Ux9QalL0Br2xkBBA9ZeYY0fzY/lpDJkDP6FLKjUAz3ujQ2YDjVX8qEfHNFZoQOACnik9I2t7a9kulfUnl7mOjXBvrldXgTKw0elLnEbYTuoyJuacTZ3ycz0WwLiYc6ZQibya/3eSfDQxJtV5lMdhrf+A+xE1vW8FnnEFSQllHJo2eRRJqU16Dvfzgbw9zXNs95Gr6CHP+3H7C95zXeeU38H94G0q1zho8Ej0CSo2/ph7G/W+eUybMc6rD1lHWdk65t7betcOKQhW6XhM8rP8uXBHDZxHb8iD/D2f+6Gc7FqgDOyshlYpvVYpSbGhCd0O8elNANzj1EIH0ipevJGU/Rx6K+okP3TMfS/Q2g8gma8ONKC9xfW0gEAMN/XhOi1lpE1Lz0AsDEeyE7Xc5+x/mL8TAoQKIjuJ2+5qfU84SpAfXTyWFu2+TkNvXaVv0Br7jSP4/6pDin3FUsfiDAUens73PUcKj2e3jf43aFmGukg+T6JEEOTtged6vsBztffxOftSJ9P0PgBwU3/CMyDWkZxPCNSHL3h1QBzP0XHSc6w3vAC7sx17rEi+YO3b2QWP8IwU6+GZS0+DW9b4P9/zBMV5by6nV+g6Cfe3KxQlo7f91a+wgt9awCoKWfbHSt9dmO8VrGUjdj01fFikGGJUS9I6hA3Kd6Uy0dYWi9lgurOR9QYns4FLBOoUvAovelb1+ZJ3PW5FTwkaW7g1f+aR80zWL/R7wmWJvkaMrf86FYGF9LZYPMWG9Bg2pldTYRlH5RPW3WtsNF1X6eUSng4XZT+Lv2OkbxMPZfme9yPBQIGzUd/HOXkBcZQy2uFJWuoXBAh1IrevlfA0txNIdgfwHSxwjkHhCc15kKLy9Eg/fw/38N1/gs/2WYcwf05FBvVkRyp9GP+Ncd8Y5vaW5GeNBG6gVwZu9XtZHkizN89JUZl9roR8WSt9Ar/FQ6lkH+5Y578LnIeI/RlUsnBea8z1URf+UKaCrFBUlNCFHzg+kMvYKMW5YGHJ3yzR0JvVXgPUHEhf7rKmdpUjH0PLuEbcilH93c8PMkFUMmaz+hLFAtbk2bJ+P7V1B5Y6ZrsupkxDQ4CaS3hmt6xPLZBuCQndXmszkqePZ+ideMuziibz3EMCxPQyFZ63A+ckaeH5i6y8SOsObtmjqBRkJD9TnY+H+Qyb0AK8xiub5hiLtNqpey4xoovqFF7ncIcMrKcDBHaHsy/pvOOQJY5vDv26OzvvAwqDndp2ZsxzQcnBzHbbsq5d6NxnP8m7631MjyF06wIfVoa3z9az2oCVPo1K7aFU6OxznMO6jzI8V9aPTH+ZyqXr3XiLRHozy+hG716/ooLgoqlIvv7A+ngg68WmrE9xAYb30usxjnVyRoF7rIkp16GiY9EVG4jQhZYSgt8QbIbpRnciQWXo9kODfZ/0nOjEupum8eNIO/mZ1wt33Q9oSaWdRnCJlD4U6kESjjseGNd4dgO8g8tpBdg5vrtpOaCBn+OlvZ3l83AZStc0elSKWZFX0QouZLV08nqjC3gNkpJ3f2Jq3qmyflBQgiSGYw9IeEz0clpoIL6DmS8ohugT/rX07IKwjeJRJDpEem9BpegR75x2PkMhFze8J6eTIBd75DGNhNEZ4/24hPfw83gTlbOJJJkEy+D2wPtZRpJHw7405tuBBXi8971cwW8t7n2jfqPvfU/nPFiIr0p+oZQQad8Xc715VC7WluF5g7W8jazvIreAgnUWyTLlKaCnsqxQJ7Zk+T7EfS0xyuIEltFeJMc3SMx/jsnXdgXydSYV03rWtWl8f3HBhVA4v0KPwhpHMYIy9XiRMprH72ZlActeoehpcWWz5Q3/3WrX0wZ7kUmiKjjC62w25NdrtVIoFJXG/KemayEo+tVCH3x0noiN/XlaCg87UigUCoVi47HQFQqFQqFQbHzQgAuFQqFQKJTQFQqFQqFQKKErFAqFQqGoCP4jwADQNvw20jA5ogAAAABJRU5ErkJggg==", null ]

[ "", null, "Practice Test | Kullabs.com\n0%\n\n(-3, -7)\n(-7, -3)\n(3, 7)\n(-7, 3)\n\n(3, 0)\n(-0, -3)\n(-0, 3)\n(0, 3)\n\n(7, 1)\n(1, 7)\n(7, -1)\n(-7, 1)\n\n(2, -4)\n(4, -2)\n(-4, -2)\n(2, 4)\n\n(2, -5)\n(2, 5)\n(-5, 2)\n(-2, 5)\n\n(7, -4)\n(-7, 4)\n(7, 4)\n(-4, 7)\n\n(0, -2)\n(-0, 2)\n(2, -0)\n(0, 2)\n\n(-9, -8)\n(8, 9)\n(9, 8)\n(-8, 9)\n\n(-0, 4)\n(0, 4)\n(4, 0)\n(-4, -0)\n\n(6, 1)\n(-6, -1)\n(-1, -6)\n(-1, 6)" ]

[ null, "https://www.facebook.com/tr", null ]

[ "# 微观经济学原理 Principles of microeconomics  ECON121\n\n0\n\n• $\\theta=\\lim {n \\rightarrow \\infty} \\bar{y}{n}$ must exist, and the marginal distribution (given $\\theta$ ) for each of the $y_{i}$ must be $p\\left(y_{i} \\mid \\theta\\right)=$ Bernoulli $(\\theta)=\\theta^{y_{i}}(1-\\theta)^{1-y_{i}}$;\n• $H(t)=\\lim {n \\rightarrow \\infty} P\\left(y{n} \\leq t\\right)$, the limiting cumulative distribution function (CDF) of the $\\bar{y}{n}$ values, must also exist for all $t$ and must be a valid CDF, where $P$ is Your joint probability distribution on $\\left(y{1}, y_{2}, \\ldots\\right)$; and\n• Your predictive distribution for the first $n$ observations can be expressed as\n$$p\\left(y_{1}, \\ldots, y_{n}\\right)=\\int_{0}^{1} \\prod_{i=1}^{n} \\theta^{\\gamma_{i}}(1-\\theta)^{1-y_{i}} d H(\\theta) .$$\nWhen (as will essentially always be the case in realistic applications) Your joint distribution $P$ is sufficiently regular that $H$ possesses a density (with respect to Lebesgue measure), $d H(\\theta)=p(\\theta) d \\theta,(1)$ can be written in a more accessible way as\n$$p\\left(y_{1}, \\ldots, y_{n}\\right)=\\int_{0}^{1} \\theta^{s_{n}}(1-\\theta)^{n-s_{n}} p(\\theta) d \\theta$$\n\n## ECON121 COURSE NOTES ：\n\nThe random effects model implies a homoskedastic variance var $\\left(u_{i t}\\right)=\\sigma_{\\mu}^{2}+\\sigma_{\\mathrm{v}}^{2}$ for all $i$ and $t$, and an equi-correlated block-diagonal covariance matrix which exhibits serial correlation over time only between the disturbances of the same individual. In fact,\n\\begin{aligned} \\operatorname{cov}\\left(u_{i t}, u_{j s}\\right) &=\\sigma_{\\mu}^{2}+\\sigma_{v}^{2} & \\text { for } & i=j, t=s \\ &=\\sigma_{\\mu}^{2} & & \\text { for } i=j, t \\neq s \\end{aligned}\nand zero otherwise. This also means that the correlation coefficient between $u_{i t}$ and $u_{j s}$ is\n\\begin{aligned} \\rho &=\\operatorname{correl}\\left(u_{i t}, u_{j s}\\right)=1 & & \\text { for } i=j, t=s \\ &=\\sigma_{\\mu}^{2} /\\left(\\sigma_{\\mu}^{2}+\\sigma_{\\mathrm{v}}^{2}\\right) & & \\text { for } i=j, t \\neq s \\end{aligned}\n\n# 微观经济学原理作业代写Principles of Microeconomics代考\n\n0\n\n## 代写微观经济学原理作业代写Principles of Microeconomics\n\n《微观经济学原理》是对教科书《微观经济学》的改编。D. Curtis和I. Irvine编写的《市场、方法和模型》，该书在加拿大和全球环境中提供了简明而完整的微观经济理论、应用和政策介绍。\n\n### 国际金融International finance代写\n\n• Global Economy全球经济\n• International political economy国际政治经济\n• International relations 国际关系学\n\n## 微观经济学原理的相关\n\nThis course begins with an introduction to supply and demand and the basic forces that determine an equilibrium in a market economy. Next, it introduces a framework for learning about consumer behavior and analyzing consumer decisions. We then turn our attention to firms and their decisions about optimal production, and the impact of different market structures on firms’ behavior.\n\n## 微观经济学原理的相关课后作业代写\n\nWhen studying changes in supply or demand in a market, one variable we often want to study is total revenue, the amount paid by buyers and received by sellers of the good. In any market, total revenue is $P \\times Q$, the price of the good times the quantity of the good sold. We can show total revenue graphically, as in Figure 5-2. The height of the box under the demand curve is $P$, and the width is $Q$. The area of this box, $P \\times Q$, equals the total revenue in this market. In Figure $5-2$, where $P=\\$ 4$and$Q=100$, total revenue is$\\$4 \\times 100$, or $\\$ 400\\$." ]

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[ "# Help with understanding op amp circuit using diodes at the output\n\nI'm trying to understand this circuit and having trouble figuring out how the diodes will affect it:", null, "I'm trying analyze this and understand how $V_{\\text{out}+}$ and $V_{\\text{out}-}$ will be affected by different values of $V_i$ but I am unsure. I believe $V_{\\text{out}+}$ will see a higher voltage with lower $V_i$ values and once $V_{\\text{out}+}$ is past 0.6V the current will go through both of the diodes and $V_{\\text{out}-}$ will begin to see an increase. I'm unsure of what to think about the op amp in the center though.\n\n• Start using \"ideal\" diodes with Von=0. When Vin is +ve, VOA-out = -ve. Top diode conducts with 0 drop so you get an inverting amploifier with Vout = Vin x R/R1 as OA- must be forced to remain at )A+ potential = ground. | Now perform the same analysis for negative input. – Russell McMahon Nov 5 '15 at 7:56\n\n## 3 Answers\n\nThe OpAmp is used in an inverting configuration, i.e. when you apply a positive Vi, the output Vo will be negative. So, the current flows through R3 and the upper path (R1, D1), while current through the lower path is blocked by D2.\nSo, for a positive input, Vo+ will be negative and Vo- will be zero.\nThe exact value on Vo+ is determined by the ratio of R3 and R1, as the output terminal Vo will be a value which cancels out the voltage at the negative input terminal.\n\nTo examine this a bit more, you can do a simulation, even here. I've drawn your circuit using the circuit designer of EE. Click on simulate this circuit and do a DC sweep for Vi.", null, "simulate this circuit – Schematic created using CircuitLab\n\nThe following picture shows the result, blue is Vo+, orange Vo-. I've also added Vo (brown) just to see the effect of the diodes. (Click the image to enlarge)", null, "Andy's answer is spot-on if this circuit operates in isolation.\n\nBut if there happens to be a positive bias current source connected to Vo+, and a negative current source connected to Vo-, then it is something different. (For simplicity, these bias current sources can be resistors connected to +V and -V respectively.)\n\nThen the outputs form three parallel lines, about 0.6V apart, with the central one (Vo = -Vi).\n\nVo+ and Vo- may be used to drive the bases of two complementary emitter followers, providing a crude power amplifier. Their emitters are connected together, and each base-emitter voltage is cancelled by the voltage across each diode. Therefore the final output (at the emitters) is approximately Vo = -Vi. But only approximately : errors here lead to crossover distortion.", null, "simulate this circuit – Schematic created using CircuitLab\n\nIn its current form, it's not very efficient : if you think about what happens when VO+ = V+/2 you'll see that it runs out of bias current and starts clipping (with the current resistor values), but this illustrates the basic principle, and you'll often see something like this in an audio power amplifier.\n\nFurther circuitry is usually added to fix the defects in this basic configuration.\n\nCompare this to the standard \"ideal diode\" op amp design.\n\nIf Vi is higher than A1+ (GND), the feedback loop will try to null the input voltage, driving the V+out line through the diode to \"pull\" A1- down.\n\nIf Vi is lower, the V-out line wil similarily be pulled up.\n\nThis circuit then behaves as a standard inverting amplifier, but with any negative output being delivered to the V+out line (for a positive input), and any positive to the V-out(for a negative input)" ]

[ null, "https://i.stack.imgur.com/ipGcB.png", null, "https://i.stack.imgur.com/Tlgnv.png", null, "https://i.stack.imgur.com/6DugN.png", null, "https://i.stack.imgur.com/MZGCo.png", null ]

[ "C Program\n\nObject:-Write a Program to check whether a number is even or odd\n\n````#include <stdio.h>````\n\nint main()\n{\nint number;\nprintf(\"Enter an integer: \");\nscanf(\"%d\", &number);\n\n// True if the number is perfectly divisible by 2\nif(number % 2 == 0)\nprintf(\"%d is even.\", number);\nelse\nprintf(\"%d is odd.\", number);\nreturn 0;\n}\n``````\n\n### Output\n\n```Enter an integer: 9\n9 is odd.\n```\n\n### Explanation\n\n```\nIf the number is perfectly divisible by 2, test expression `number%2 == 0` evaluates to 1 (true) and the number is even.\nHowever, if the test expression evaluates to 0 (false), the number is odd.\n\n```\n\nYou may also like:" ]

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[ "# Working with boolean variables\n\n`boolean` is a Java primitive data type with exactly 2 possible values: `true` and `false` (all lowercase, no quotes). Unlike in some other languages, it is NOT possible in Java to use `0` as `false` and `1` as `true`.\n\n`boolean` variables are declared, initialized, and used just like other primitive type variables in Java. `boolean` variables are often known as flag variables and are often used to identify whether a particular condition exists.\n\n## Example 1\n\n``````boolean childAteLunch = true;\n\nif(childAteLunch)\nSystem.out.println(\"Time for nap\");\nelse\nSystem.out.println(\"Time for lunch\");\n``````\n\nThe variable `childAteLunch` is initialized to `true`.\n\nThe conditional statement `if(childAteLunch)` checks if the value of `childAteLunch` is `true`. Since the value is `true`, the code segment prints `\"Time for nap\"`.\n\nAlthough it would work, it is not necessary to write `if(childAteLunch == true)`. The original condition already evaluates to `true` if the variable stores `true` and `false` if the variable stores `false`.\n\n## Example 2\n\n``````int grade = (int) (Math.random() * 101);\n// 0 <= grade <= 100\n\nboolean earnedHonorRoll = grade >= 90;\n\nif( ! earnedHonorRoll )\nSystem.out.println(\"Try again next marking period\");\nelse\nSystem.out.println(\"Congrats\");\n``````\n\nThe variable `grade` is initialized to a random value in the range `0 <= grade <= 100`. See Generate random numbers with Math.random() for additional details.\n\nThe variable `earnedHonorRoll` is initialized to the result of evaluating the `boolean` expression `grade >= 90`. The expression `grade >= 90` evaluates to either `true` or `false` depending on the value of `grade`. The variable `earnedHonorRoll` is set to either `true` or `false`.\n\nThe conditional statement `if( ! earnedHonorRoll )` checks if the value of `earnedHonorRoll` is `false` using the Java not operator (`!`).\n\nAlthough it would work, it is not necessary to write `if(earnedHonorRoll == false)`. The original condition already evaluates to `true` if the variable stores `false` and `false` if the variable stores `true`.\n\n## Why avoid `== true` and `== false`\n\n### Example with error 1\n\n``````boolean childAteLunch = false;\n\nif(childAteLunch = true) // mistake\nSystem.out.println(\"Time for nap\"); // prints\nelse\nSystem.out.println(\"Time for lunch\"); // does not print\n\nSystem.out.println(childAteLunch); // prints true\n``````\n\nThis code contains a mistake that is easy to make and easy to miss when reviewing code. The condition is:\n\n``````if(childAteLunch = true)\n``````\n\n``````if(childAteLunch == true)\n``````\n\nThe single equals sign (`=`) is an assignment operator. The code `childAteLunch = true` sets the value of `childAteLunch` to `true` rather than checking if it is `true`. When used as a condition, `childAteLunch = true` evalutes to `true`.\n\nThe code segment prints `\"Time for nap\"` instead of `\"Time for lunch\"`. The last print statement prints `true`, which is the value of `childAteLunch` at the end of the code segment.\n\n### Example with error 2\n\n``````boolean flag = true;\n\nif(flag = false) // mistake\nSystem.out.println(\"a\"); // does not print\nelse\nSystem.out.println(\"b\"); // prints\n\nSystem.out.println(flag); // prints false\n``````\n\nThe code `flag = false` sets the value of `flag` to `false`. When used as a condition, `flag = false` evaluates to `false`.\n\nThe code segment prints `\"b\"` instead of `\"a\"`. The last print statement prints `false`, which is the value of `flag` at the end of the code segment." ]

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[ "# Given order for every element in a symmetric group\n\nCompute the order of each of the elements in the symmetric group ##S_4##.\n\nIs the best way to do this just to write out each element's cycle decomposition, or is there a more efficient way?\n\nOrodruin\nStaff Emeritus\nHomework Helper\nGold Member\n2021 Award\nSince the order of the elements are the same within each conjugacy class, I would just take one representative of each conjugacy class. But yes, I would do it by looking at the cycles of that representative.\n\n•", null, "Mr Davis 97\nfresh_42\nMentor\n2021 Award\nCompute the order of each of the elements in the symmetric group ##S_4##.\n\nIs the best way to do this just to write out each element's cycle decomposition, or is there a more efficient way?\nYou can decompose the group:\n$$S_4 \\cong A_4 \\rtimes \\mathbb{Z}_2 \\cong (V_4 \\rtimes \\mathbb{Z_3}) \\rtimes \\mathbb{Z}_2 \\cong (\\mathbb{Z}_2^2 \\rtimes \\mathbb{Z_3}) \\rtimes \\mathbb{Z}_2$$" ]

[ null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null ]

[ "If you get a return of 9%, and inflation is 5%, what is your real return?\n\nYou might think it's (return - inflation)\n=> 9% - 5% = 4%\n\nWell, that's actually just an estimate. Here is how you calculate the real, real return (for real!)\n\nAnd let's say that, coincidentally, a widget also costs exactly R100\n\nUsing some advanced Maths, you can work out that your money can buy you precisely 1 widget.\nNow, let's say instead of buying a widget with your R100, you invested it and got a 9% return.\nAfter 1 year, you will have R109\n\nIn that same year, inflation (at 5%) will mean the price of the widget is now R105\nSo that means you can now buy R109/R105 = 1.038 widgets\n\nI.e. In widget terms (what your money can actually buy) you are 3.8% better off than a year ago\n\nSo your real return is 3.8%\n(Not quite the estimate of 9% - 5% = 4%)\nThe correct way to calculate real returns is as follows\nSo what's the big deal, why does the difference matter?\n\nWell, in low inflation, lower return environments like the US and Europe, it doesn't matter that much.\n\nE.g. with returns of 5% and inflation at 1%\nEstimated real return => 4%\nActual real return => 3.96%\nNo biggie\nBut in high inflation, high return environments (like the good old RS of A), the difference is bigger and can affect planning and projections (especially over the longer term)\n\nE.g. with returns of 12%, inflation at 8%\nEstimated Real Return=>4%\nActual Real Return=>3.7%" ]

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[ "# #01f296 Color Information\n\nIn a RGB color space, hex #01f296 is composed of 0.4% red, 94.9% green and 58.8% blue. Whereas in a CMYK color space, it is composed of 99.6% cyan, 0% magenta, 38% yellow and 5.1% black. It has a hue angle of 157.1 degrees, a saturation of 99.2% and a lightness of 47.6%. #01f296 color hex could be obtained by blending #02ffff with #00e52d. Closest websafe color is: #00ff99.\n\n• R 0\n• G 95\n• B 59\nRGB color chart\n• C 100\n• M 0\n• Y 38\n• K 5\nCMYK color chart\n\n#01f296 color description : Vivid cyan - lime green.\n\n# #01f296 Color Conversion\n\nThe hexadecimal color #01f296 has RGB values of R:1, G:242, B:150 and CMYK values of C:1, M:0, Y:0.38, K:0.05. Its decimal value is 127638.\n\nHex triplet RGB Decimal 01f296 `#01f296` 1, 242, 150 `rgb(1,242,150)` 0.4, 94.9, 58.8 `rgb(0.4%,94.9%,58.8%)` 100, 0, 38, 5 157.1°, 99.2, 47.6 `hsl(157.1,99.2%,47.6%)` 157.1°, 99.6, 94.9 00ff99 `#00ff99`\nCIE-LAB 84.848, -68.73, 31.149 37.267, 65.709, 39.571 0.261, 0.461, 65.709 84.848, 75.459, 155.62 84.848, -74.193, 54.803 81.061, -59.793, 27.799 00000001, 11110010, 10010110\n\n# Color Schemes with #01f296\n\n• #01f296\n``#01f296` `rgb(1,242,150)``\n• #f2015d\n``#f2015d` `rgb(242,1,93)``\nComplementary Color\n• #01f21d\n``#01f21d` `rgb(1,242,29)``\n• #01f296\n``#01f296` `rgb(1,242,150)``\n• #01d6f2\n``#01d6f2` `rgb(1,214,242)``\nAnalogous Color\n• #f21e01\n``#f21e01` `rgb(242,30,1)``\n• #01f296\n``#01f296` `rgb(1,242,150)``\n• #f201d6\n``#f201d6` `rgb(242,1,214)``\nSplit Complementary Color\n• #f29601\n``#f29601` `rgb(242,150,1)``\n• #01f296\n``#01f296` `rgb(1,242,150)``\n• #9601f2\n``#9601f2` `rgb(150,1,242)``\n• #5df201\n``#5df201` `rgb(93,242,1)``\n• #01f296\n``#01f296` `rgb(1,242,150)``\n• #9601f2\n``#9601f2` `rgb(150,1,242)``\n• #f2015d\n``#f2015d` `rgb(242,1,93)``\n• #01a667\n``#01a667` `rgb(1,166,103)``\n• #01bf77\n``#01bf77` `rgb(1,191,119)``\n• #01d986\n``#01d986` `rgb(1,217,134)``\n• #01f296\n``#01f296` `rgb(1,242,150)``\n• #0efea3\n``#0efea3` `rgb(14,254,163)``\n• #28feac\n``#28feac` `rgb(40,254,172)``\n• #41feb6\n``#41feb6` `rgb(65,254,182)``\nMonochromatic Color\n\n# Alternatives to #01f296\n\nBelow, you can see some colors close to #01f296. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #01f25a\n``#01f25a` `rgb(1,242,90)``\n• #01f26e\n``#01f26e` `rgb(1,242,110)``\n• #01f282\n``#01f282` `rgb(1,242,130)``\n• #01f296\n``#01f296` `rgb(1,242,150)``\n• #01f2aa\n``#01f2aa` `rgb(1,242,170)``\n• #01f2be\n``#01f2be` `rgb(1,242,190)``\n• #01f2d2\n``#01f2d2` `rgb(1,242,210)``\nSimilar Colors\n\n# #01f296 Preview\n\nThis text has a font color of #01f296.\n\n``<span style=\"color:#01f296;\">Text here</span>``\n#01f296 background color\n\nThis paragraph has a background color of #01f296.\n\n``<p style=\"background-color:#01f296;\">Content here</p>``\n#01f296 border color\n\nThis element has a border color of #01f296.\n\n``<div style=\"border:1px solid #01f296;\">Content here</div>``\nCSS codes\n``.text {color:#01f296;}``\n``.background {background-color:#01f296;}``\n``.border {border:1px solid #01f296;}``\n\n# Shades and Tints of #01f296\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000805 is the darkest color, while #f3fffa is the lightest one.\n\n• #000805\n``#000805` `rgb(0,8,5)``\n• #001b11\n``#001b11` `rgb(0,27,17)``\n• #002f1d\n``#002f1d` `rgb(0,47,29)``\n• #004229\n``#004229` `rgb(0,66,41)``\n• #005635\n``#005635` `rgb(0,86,53)``\n• #006941\n``#006941` `rgb(0,105,65)``\n• #017d4d\n``#017d4d` `rgb(1,125,77)``\n• #019059\n``#019059` `rgb(1,144,89)``\n• #01a466\n``#01a466` `rgb(1,164,102)``\n• #01b772\n``#01b772` `rgb(1,183,114)``\n• #01cb7e\n``#01cb7e` `rgb(1,203,126)``\n• #01de8a\n``#01de8a` `rgb(1,222,138)``\n• #01f296\n``#01f296` `rgb(1,242,150)``\n• #09fea0\n``#09fea0` `rgb(9,254,160)``\n• #1cfea8\n``#1cfea8` `rgb(28,254,168)``\n• #30feaf\n``#30feaf` `rgb(48,254,175)``\n• #43feb7\n``#43feb7` `rgb(67,254,183)``\n• #57febe\n``#57febe` `rgb(87,254,190)``\n• #6afec6\n``#6afec6` `rgb(106,254,198)``\n• #7efecd\n``#7efecd` `rgb(126,254,205)``\n• #91ffd5\n``#91ffd5` `rgb(145,255,213)``\n• #a5ffdc\n``#a5ffdc` `rgb(165,255,220)``\n• #b8ffe4\n``#b8ffe4` `rgb(184,255,228)``\n• #ccffeb\n``#ccffeb` `rgb(204,255,235)``\n• #e0fff3\n``#e0fff3` `rgb(224,255,243)``\n• #f3fffa\n``#f3fffa` `rgb(243,255,250)``\nTint Color Variation\n\n# Tones of #01f296\n\nA tone is produced by adding gray to any pure hue. In this case, #71827b is the less saturated color, while #01f296 is the most saturated one.\n\n• #71827b\n``#71827b` `rgb(113,130,123)``\n• #688b7e\n``#688b7e` `rgb(104,139,126)``\n• #5e9580\n``#5e9580` `rgb(94,149,128)``\n• #559e82\n``#559e82` `rgb(85,158,130)``\n• #4ca784\n``#4ca784` `rgb(76,167,132)``\n• #42b187\n``#42b187` `rgb(66,177,135)``\n• #39ba89\n``#39ba89` `rgb(57,186,137)``\n• #30c38b\n``#30c38b` `rgb(48,195,139)``\n• #26cd8d\n``#26cd8d` `rgb(38,205,141)``\n• #1dd68f\n``#1dd68f` `rgb(29,214,143)``\n• #14df92\n``#14df92` `rgb(20,223,146)``\n• #0ae994\n``#0ae994` `rgb(10,233,148)``\n• #01f296\n``#01f296` `rgb(1,242,150)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #01f296 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "# Mathematica Information\n\nPhysics students must become proficient in a computer algebra program. In the Paradigms program, we use Mathematica, by Wolfram. Similar programs are Maple and MathCad. After many years of using Maple, we have switched to Mathematica because we believe it is a better program and there is better support for it. Mathematica x is installed on the computers in Wngr 212, Wngr 304, Wngr 304F for your use. OSU students may also access Mathematica via OSU's Virtual Software Laboratory, Umbrella. There is also a free Mathmatica \"player\", which is sotware that allows you to run Mathematica programs, but not alter them. Mathematica 8 student version costs about \\$140.\n\nStudents need not have used Mathematica (or Maple or MathCad) before; we assue no previous knowledge. You will need to learn only very basic techniques, but we hope that you will explore on your own and become ever more proficient. It's a great tool to help you learn. We expect you to make extensive use of the built in Help features and the very impressive tutorials provided by Wolfram. Please collaborate with your fellow students; learn from them and help them to learn. We provide a template notebook with our favorite features (fonts, sizes, plot line thicknesses, etc.) already defined, so that you can start from this template each time.\n\nSite Description\nhttp://www.wolfram.com/broadcast/#Tutorials-GS Basic \"Getting Started\" Mathematica tutorials. The first four videos under \"Hands-on Start to Mathematica\" by Cliff Hastings are essential viewing. They refer to Mathematica 8, which has some new features, so they may have some extraneous info if you're using an earlier version. They are 4 min, 10 min, 7 min and 9 min respectively. The \"Mathematica Basics\" by Jon McLoone is also helpful.\nhttp://www.wolfram.com/support/learn/ Mathematica's learning center. This has a much wider range of information than the link above (which can be reached from this page). Very helpful with a mutlitude of ways to learn. Videos, tutorials, demos, etc.\nhttp://www.wolframalpha.com/ WolframAlpha - a Mathematica-based computation engine. Enter equations into the box in the window and get instant results. A remarkably powerful piece of sotware.\nhttp://www.wolfram.com/cdf-player/ Free Mathematica Player. You can run notebooks (.nb files) , but you can't create new files or new content.\nhttp://oregonstate.edu/is/mediaservices/scf/virtual-lab OSU's virtual computer lab \"Umbrella\". OSU students can use Mathematica via this interface.\nDownload the remote desktop client if you don't already have it, and log on to Umbrella according to the instructions on the page.\nFind Mathematica by clicking on: Start (windows logo) -> All programs -> Mathematics -> Mathematica -> Wolfram Mathematica 7\nYou're ready to use Mathematica and you can save notebooks to your onid account.\nBelow are Mathematica notebooks created for PH424. Control-click (Mac) or right-click (Windows) on the link and choose \"Save as ...\" Save the file on your computer, making sure it has a .nb extension and not .txt or anything else. Then open the file with Mathematica. (A normal click simply opens the raw code in the browser.)\nPH424 Mathematica template A mathematica notebook that I have created that has some useful default settings.\nCLASSICAL WAVES\nTraveling & standing waves Difference between \"v\" and \"d(psi)/dt\"\nMaterial velocity Difference between \"v\" and \"d(psi)/dt\"\nInitial conditions Shows how general solution to non-dispersivewave equation develops with different intiial conditions\nReflection and transmission Incident wave is reflected and transmitted at an abrupt boundary (animation)\nDisplacement and force (pressure) waves Illustration of longitudinal displacement and force waves (animation)\nEnergy density Potential, kinetic and total energy density in traveling and standing waves\nAnimate triangle wave Time development of triangle standing wave (hwk 2)\nQUANTUM WAVES\nFinite well eigenstates Manipulate the finite well to see effect of the change of the well depth on the eigenstates (note that the zero of potential energy is at the top of the well, not the bottom, as was the convention in class)\nAnimate Gaussian wave packet Time development of Gaussian wave packet that disperses but does not propagate\nAnimate Gaussian wave packet Time development of Gaussian wave packet that disperses and propagates", null, "" ]

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[ "# Lecture 40, Stack Top Caches\n\n## Implementing Stacks in Hardware\n\nThe Burroughs corporation, when they built the B5000 machine back in the 1960s, recognized that a pure stack architecture had an obvious performance problem: Consider the run-time cost of executing i=i+1 on a machine where the stack is entirely memory resident:\n\n```PUSHLA i -- sp = sp + 1; M[sp] = fp + i;\nPUSHLA i -- sp = sp + 1; M[sp] = fp + i;\nPUSHI 1 -- sp = sp + 1; M[sp] = 1;\nADD -- M[sp-1] = M[sp-1] + M[sp]; sp = sp - 1\nPOP -- M[M[sp-1]] = M[sp]; sp = sp - 1\n```\n\nIn sum, this requires 12 memory cycles (excluding instruction fetches). An obvious way to improve this is to use one register inside the CPU to hold the stack top; call it a. This offers a modest improvement:\n\n```PUSHLA i -- sp = sp + 1; M[sp] = a; a = fp + i;\nPUSHLA i -- sp = sp + 1; M[sp] = a; a = fp + i;\nPUSHI 1 -- sp = sp + 1; M[sp] = a; a = 1;\nADD -- a = M[sp] + a; sp = sp - 1;\nPOP -- M[M[sp]] = a; a = M[sp-1]; sp = sp - 2;\n```\n\nThis offers a modest improvement: 12 cycles are reduced to 8 cycles. But, if you think about building hardware, you realize that for any add operation, both operands must be held in registers while the ALU computes the result, and for any store operation, both the address and value being stored must be held in registers. The reads from the stack in the above would mostly be reading into this second register.\n\nThe most obvious way to do this is to keep the top two elements on the stack in registers at all times. Let's call the register used for the stack a and the one used for the element below the stack top b. The naive way to use this hardware to execute the above program would be:\n\n```PUSHLA i -- sp = sp + 1; M[sp] = b; b = a; a = fp + i;\nPUSHLA i -- sp = sp + 1; M[sp] = b; b = a; a = fp + i;\nPUSHI 1 -- sp = sp + 1; M[sp] = b; b = a; a = 1;\nADD -- a = a + b; b = M[sp]; sp = sp - 1;\nPOP -- M[b] = a; a = M[sp]; b = M[sp-1]; sp = sp - 2;\n```\n\nUnfortunately, this made no difference in the run time, since there are still 8 memory cycles. Note that we do not count the register to register data transfers as costing any time because, when this is done inside the CPU, these transfers can all be done in parallel.\n\n### A Brilliant Idea\n\nI suspect that Burroughs went through something like the above before someone on the design team came up with a brilliant idea: Instead of keeping the two registers full of data at all times, add two bits of state information, a bit to indicate whether a currently holds data, and a bit to indicate whether b currently holds data. As a result, the system has 3 potential states:\n\n• state 0: a and b both vacant\n• state 1: a vacant, b in use (the stack top)\n• state 3: a in use (the stack top), b in use\n\nNow, consider how the above program would run if the stack was initially in state 0:\n\n```PUSHLA i -- b = fp + i; state = 1;\nPUSHLA i -- a = fp + i; state = 3;\nLOAD -- a = M[a]; state = 3;\nPUSHI 1 -- sp = sp + 1; M[sp] = b; sp = sp + 1; b = a; a = 1; state = 3;\nADD -- b = a + b; state = 1;\nPOP -- a = b; b = M[sp]; sp = sp - 1; M[b] = a; state = 0;\n```\n\nThis reduces the run time to 4 memory cycles, a significant improvement. The Burroughs implementation simply included the current value of the stack state in its instruction decoding, so that each instruction had a different implementation in each potential stack state:\n\nPUSHLA i in state 0\nb = fp + i; state = 1;\nPUSHLA i in state 1\na = fp + i; state = 3;\nPUSHLA i in state 3\nM[sp] = b; sp = sp + 1; b = a; a = fp + i; state = 3;\n\nPUSHI i in state 0\nb = i; state = 1;\nPUSHI i in state 1\na = i; state = 3;\nPUSHI i in state 3\nM[sp] = b; sp = sp + 1; b = a; a = i; state = 3;\n\nb = M[sp]; sp = sp - 1; b = M[b]; state = 1;\nb = M[b]; state = 1;\na = M[a]; state = 3;\n\na = M[sp]; b = M[sp-1]; sp = sp - 2; b = a + b; state = 1;\na = b; b = M[sp]; sp = sp - 1; b = a + b; state = 1;\nb = a + b; state = 1;\n\nPOP in state 0\na = M[sp]; b = M[sp-1]; sp = sp - 2; M[b] = a; state = 0;\nPOP in state 1\na = b; b = M[sp]; sp = sp - 1; M[b] = a; state = 0;\nPOP in state 3\nM[b] = a; state = 0;\n\nIn effect, the a and b registers served as a cache for the top of stack. Later stack machines added additional registers, enlarging this cache and further reducing the likelihood that stack operations would result in actual memory cycles.\n\n## Stack Top Caches in Compilers\n\nThis idea can be directly applied to register allocation in a compiler, with one major change: In a compiler, we want to avoid using instructions for register-to-register transfers. As a result, we do not want to dedicate a particular register to holding the top item on the stack. As a result, our stack state is going to be far more complicated. If we have a block of n registers that is to be used as a stack-top cache, each of these registers has one of n+1 possible uses: It may be unused, it may hold the stack top, an item below the stack top, all the way up to the item n–1 items below the stack top.\n\nA first venture at this would attach, to each register, a counter indicating whether it was currently not part of the stack (0), or whether it represented the top element (1), the element below the top (2), and so on. This suggests that we might need to search the registers to find the stack top, and it suggests the need to update all of the registers with nonzero tags each time there is a push or pop. This is not appealing.\n\nFortunately, there is an alternative. If we arrange the set of cache registers used to cache the stack top as a circular buffer in the range min to max, where cache=(max-min)+1, we can use just one index to track which register holds the stack top, call it top, and a count to track how many stack elements are currently in registers, call it count. On the ARM, for example, min might be 3 and max might be 12. Here is the logic for generating code for the pushi c and add instructions in this context:\n\n```void pushi( int c ){\nint r = top + 1;\nif (r > max) r = min;\nif (count == cache) { /* we need to clear a register */\n-- emit code to push R[r] on the RAM stack --\ncount = count - 1;\n}\n-- emit code to load the constant c into R[r] --\ntop = r;\ncount = count + 1;\n}\nif (count == 0) { /* we to get the stack top into a register */\n-- emit code to pop the RAM stack into R[top] --\ncount = count + 1;\n}\nint r = top - 1;\nif (r < min) r = max;\nif (count == 1) { /* we to get the stack top into a register */\n-- emit code to pop the RAM stack into R[r] --\ncount = count + 1;\n}\n-- emit code to add R[r] plus R[top] and put the result in R[r]\ntop = r;\ncount = count - 1;\n}\n```\n\nThis scheme works well within a basic block, but whenever control converges on a label from two or more directions, it does not guarantee that the stack will be in the same state on each path into that label. As a result, at the end of each branch point in the code, the stack-top cache must be put into a standard state such as entirely in RAM. This was not a problem with the hardware stack-top cache, since the cache state there was a dynamic attribute. In general, stack-top caches are not a good stating point for developing more optimal code generators." ]

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[ "### Questions (2)\n\n 3 $O(n\\log n)$- algorithm for finding tree root 2 Find the nearest sum to a given number of two elements in sorted matrix\n\n### Reputation (161)\n\nThis user has no recent positive reputation changes\n\nThis user has not answered any questions\n\n### Tags (9)\n\n 0 algorithms × 2 0 matrices 0 trees 0 arrays 0 data-structures 0 optimization 0 algorithm-analysis 0 binary-trees 0 nearest-neighbour\n\n### Bookmarks (2)\n\n 3 $O(n\\log n)$- algorithm for finding tree root 2 Find the nearest sum to a given number of two elements in sorted matrix" ]

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[ "# Comparing variables from one loop to the next\n\nI'm trying to compare the value of a variable from the previous loop to the value in the current loop. Clearly, it's not working, I'm not sure what I'm doing wrong.\n\nThe variable in question is 'lastVal', the value is assigned by analogRead(0). In the if/else statement, the sketch only ever performs the 'else' portion:\n\n``````int x;\nint led = 13;\n\nvoid setup()\n{\nSerial.begin(9600);\npinMode(led, OUTPUT);\n}\n\nvoid loop()\n{\nint lastVal;\nif (x < lastVal)\n{\ndigitalWrite(led, HIGH);\nSerial.print(\"x: \");\nSerial.print(x, DEC);\nSerial.println(\". Light On.\");\n}\nelse\n{\ndigitalWrite(led, LOW);\nSerial.print(\"x: \");\nSerial.println(x, DEC);\n}\nx = lastVal;\ndelay(100);\n}\n``````\n\n``````int x;\nint lastVal = 0; //set initial value to 0\nint led = 13;\n\nvoid setup()\n{\nSerial.begin(9600);\npinMode(led, OUTPUT);\n}\n\nvoid loop()\n{\nif (x < lastVal)\n{\ndigitalWrite(led, HIGH);\nSerial.print(\"x: \");\nSerial.print(x, DEC);\nSerial.println(\". Light On.\");\n}\nelse\n{\ndigitalWrite(led, LOW);\nSerial.print(\"x: \");\nSerial.println(x, DEC);\n}\nlastVal = x; // 'save' the value of x to next iteration\ndelay(100);\n}\n``````\n\nNot tested. :]\n\nBrilliant! That worked just fine.\n\nThank you very much AlphaBeta\n\nI knew it had to be something simple, but you stare at something long enough, and it all starts to blur together." ]

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[ "## Information for Package authors\n\nThis section of the documentation describes how the external pointers produced by ast2ast can be used in other packages. This information is intended for package authors who want to use ast2ast to enable a simple user interface as it is not necessary anymore for the user to write C++. The R code is translated to a modified version of an expression template library called ETR (https://github.com/Konrad1991/ETR) which tries to mimic R.\nAs a side note all classes and functions can be found in the namespace etr.\n\n## Variables\n\nThe core class object is named VEC and can be initialized with any type. Furthermore, is a typedef defined for VEC called sexp alluding to the fact that all R objects are SEXP objects at C level. This class contains another class called STORE which manages memory. To do this a raw pointer of the form T* is used. Thus, all objects are located at the heap. Beyond that, it is important that no static methods are implemented or that memory is associated with functions or global variables. Therefore, the VEC objects can be used in parallel. However, the user has to take care that only one thread edits an object at a time. VEC a.k.a. sexp can be converted to Rcpp::NumericVectors, Rcpp::NumericMatrices, arma::vec and arma::mat or to SEXP. Moreover, it is also possible to copy the data from Rcpp::NumericVectors, Rcpp::NumericMatrices, arma::vec or arma::mat to a sexp variable. The constructors and the operator= are implemented for this conversions. It is possible to construct a sexp object by using a double* pointer and a size information (= int). The information about the Rcpp-, RcppArmadillo- and pointer-interface is explained in depth in the sections below.\n\nIf the user creates an R function all input arguments are of type SEXP. Furthermore, is the output always of type SEXP. Beyond that, there exists the possibility to create a function returning an external pointer to the C++ function. If using this interface the parameter passed to the function can be of one of the following types as long as an sexp object is defined as argument for the function.\n\n• double\n• SEXP\n• sexp\n• Rcpp::NumericVector or NumericVector\n• Rcpp::NumericMatrix or NumericMatrix\n• arma::vec or vec\n• arma::mat or mat\n\nThe variables are directly converted to sexp thereby copying the memory. Thus, the variable can not be used if an function expects a reference of type sexp If the aim is to modify a variable by the user function it is only possible to pass directly a sexp objects by reference. See the example below here a is modified whereas b is copied and thus not altered. The vector v can only be used for the second argument of the function foo.\n\n// [[Rcpp::depends(ast2ast)]]\n// [[Rcpp::plugins(cpp17)]]\n#include \"etr.hpp\"\nusing namespace Rcpp;\nusing namespace etr;\n\nvoid foo(sexp& a, sexp b) {\na = b + 1;\nb = 1;\n}\n\n// [[Rcpp::export]]\nvoid call_fct() {\nsexp a = coca(1, 2, 3);\nsexp b = coca(1, 2, 3);\nfoo(a, b);\nprint(a);\nprint(b);\n\nNumeric\n}\n\n\nBeyond that using “XPtr” as output argument of the function translate it is possible to specify ptr_vec or a ptr_mat as desired types. If these types are used sexp objects are created which form a shallow wrapper around these pointers. See an example below how this works.\n\n1. a Rcpp function is created which calls the translated R code:\n// [[Rcpp::depends(ast2ast)]]\n// [[Rcpp::plugins(cpp17)]]\n#include \"etr.hpp\"\n\n#include <Rcpp.h>\nusing namespace Rcpp;\n\ntypedef sexp (*fct_ptr) (double* a_double_ptr, int a_int_size);\n\n// [[Rcpp::export]]\nRcpp::NumericVector fcpp(XPtr<fct_ptr> f) {\nfct_ptr fct = *f;\n\nint size = 10;\ndouble* ptr = new double[size];\nfor(int i = 0; i < size; i++) {\nptr[i] = static_cast<double>(i);\n}\n\nRcpp::NumericVector ret = fct(ptr, size);\n\ndelete[] ptr;\n\nreturn ret;\n}\n1. a R function is defined and translated (not shown):\nf <- function(a) {\nd_db = 1\nret <- a + 2 + d_db\nreturn(ret)\n}\n1. The translated R function has the following code:\nSEXP f(double* a_double_ptr, int a_int_size ) {\n\ndouble d_db;\nsexp ret;\nsexp a(a_int_size, a_double_ptr, 2);\nd_db = etr::i2d(1);\nret = a + etr::i2d(2) + d_db;\nreturn(ret);\n}\n\n## How to use ast2ast\n\nIn this paragraph, a basic example demonstrates how to write the R code, translate it and call it from C++. Particular emphasis is placed on the C++ code. First of all, the R function is defined which accepts one argument called a, adds two to a and stores it into b. The variable b is returned at the end of the function. The R function called f is translated to an external pointer to the C++ function.\n\nf <- function(a) {\nb <- a + 2\nreturn(b)\n}\nlibrary(ast2ast)\nf_cpp <- translate(f, output = \"XPtr\", types_of_args = \"sexp\", return_type = \"sexp\")\n\nThe C++ function depends on RcppArmadillo and ast2ast therefore the required macros and headers were included. Moreover, ETR requires std=c++17 therefore the corresponding plugin is added. The function getXPtr is defined and is the function which is returned. In the last 5 lines, the translated code is depicted. The function f returns a sexp and gets one argument of type sexp called a. The body of the function looks almost identical to the R function. Except that the variable b is defined in the first line of the body with the type sexp. The function i2d converts an integer to a double variable. This is necessary since C++ would identify the 2 as an integer which is not what the user wants in this case.\n\n// [[Rcpp::depends(ast2ast, RcppArmadillo)]]\n#include <Rcpp.h>\n// [[Rcpp::plugins(cpp17)]]\nusing namespace Rcpp;\nusing namespace etr;\n#include <etr.hpp>\n// [[Rcpp::export]]\nSEXP getXPtr();\n\nsexp f(sexp a) {\nsexp b;\nb = a + i2d(2);\nreturn(b);\n}\n\nAfterwards, the translated R function has to be used in C++ code. This would be your package code for example. First, the macros were defined for RcppArmadillo and ast2ast. Subsequently, the necessary header files were included. As already mentioned ast2ast requires std=c++17 thus the required plugin is included. To use the function, it is necessary to dereference the pointer. The result of the dereferenced pointer has to be stored in a function pointer. Later the function pointer can be used to call the translated function. Therefore, a function pointer called fp is defined. It is critical that the signature of the function pointer matches the one of the translated function. Perhaps it would be a good idea to check the R function before it is translated. After defining the function pointer, a function is defined which is called later by the user (called call_package). This function accepts the external pointer. Within the function body, a variable f is defined of type fp and inp is assigned to it. Next, a sexp object called a is defined which stores a vector of length 3 containing 1, 2 and 3. The function coca is equivalent to the c function R. Afterwards a is printed. Followed by the call of the function f and storing the result in a. The variable a is printed again to show that the values are changed according to the code defined in the R function.\n\n// [[Rcpp::depends(RcppArmadillo, ast2ast)]]\n#include \"etr.hpp\"\n// [[Rcpp::plugins(\"cpp17\")]]\ntypedef sexp (*fp)(sexp a);\nusing namespace etr;\n\n// [[Rcpp::export]]\nvoid call_package(Rcpp::XPtr<fp> inp) {\nfp f = *inp;\nsexp a = coca(1, 2, 3);\nprint(a);\na = f(a);\nprint(\"a is now:\");\nprint(a);\n}\n\nThe user can call now the package code and pass the R function to it. Thus, the user only has to install the compiler or Rtools depending on the operating system. But it is not necessary to write the function in Rcpp.\n\ncall_package(f_cpp)\n## 1\n## 2\n## 3\n## a is now:\n## 3\n## 4\n## 5\n\n## Rcpp Interface\n\nIn the last section, the usage of ast2ast was described. However, only sexp variables were defined. Which are most likely not used in your package. Therefore interfaces to common libraries are defined. First of all, ast2ast can communicate with Rcpp which alleviates working with the library substantially. The code below shows that it is possible to pass a sexp object to a variable of type NumericVector or NumericMatrix and vice versa. Here, the data is always copied.\n\n// [[Rcpp::depends(ast2ast, RcppArmadillo)]]\n#include <Rcpp.h>\n// [[Rcpp::plugins(cpp17)]]\nusing namespace Rcpp;\n#include <etr.hpp>\nusing namespace etr;\n\n// [[Rcpp::export]]\nvoid fct() {\n// NumericVector to sexp\nNumericVector a{1, 2};\nsexp a_ = a;\nprint(a_);\n\n// sexp to NumericVector\nsexp b_ = coca(3, 4);\nNumericVector b = b_;\nRcpp::Rcout << b << std::endl;\n\n// NumericMatrix to sexp\nNumericMatrix c(3, 3);\nsexp c_ = c;\nprint(c_);\n\n// sexp to NumericMatrix\nsexp d_ = matrix(colon(1, 16), 4, 4);\nNumericMatrix d = d_;\nRcpp::Rcout << d << std::endl;\n}\ntrash <- fct()\n## 1\n## 2\n## 3 4\n## 0 0 0\n## 0 0 0\n## 0 0 0\n## 1.00000 5.00000 9.00000 13.0000\n## 2.00000 6.00000 10.0000 14.0000\n## 3.00000 7.00000 11.0000 15.0000\n## 4.00000 8.00000 12.0000 16.0000\n\nBesides Rcpp types, sexp objects can transfer data to RcppArmadillo objects and it is also possible to copy the data from RcppArmadillo types to sexp objects using the operator =. The code below shows that it is possible to pass a sexp object to a variable of type vec or mat and vice versa. Here the data is always copied.\n\n// [[Rcpp::depends(ast2ast, RcppArmadillo)]]\n#include <Rcpp.h>\n// [[Rcpp::plugins(cpp17)]]\nusing namespace arma;\n#include <etr.hpp>\nusing namespace etr;\n\n// [[Rcpp::export]]\nvoid fct() {\n// vec to sexp\narma::vec a(4, fill::value(30.0));\nsexp a_ = a;\nprint(a_);\n\n// sexp to vec\nsexp b_ = coca(3, 4);\nvec b = b_;\nb.print();\n\n// mat to sexp\nmat c(3, 3, fill::value(31.0));\nsexp c_ = c;\nprint(c_);\n\n// sexp to mat\nsexp d_ = matrix(colon(1, 16), 4, 4);\nmat d = d_;\nd.print();\n}\ntrash <- fct()\n## 30\n## 30\n## 30\n## 30\n## 3.0000\n## 4.0000\n## 31 31 31\n## 31 31 31\n## 31 31 31\n## 1.0000 5.0000 9.0000 13.0000\n## 2.0000 6.0000 10.0000 14.0000\n## 3.0000 7.0000 11.0000 15.0000\n## 4.0000 8.0000 12.0000 16.0000\n\n## Pointer Interface\n\nYou can pass the information of data stored on heap to a sexp object. The constructor for type vector accepts 3 arguments:\n\n• int defining the size of the data.\n• a pointer (T*) to the data\n• int called cob (copy, ownership, borrow).\n\nThe constructor for the type matrix accepts 4 arguments:\n\n• int defining number of rows\n• int defining number of cols\n• a pointer (T*) to the data\n• int called cob (copy, ownership, borrow).\n\nIf cob is 0 then the data is copied. Else if cob is 1 then the pointer itself is copied. Meaning that the ownership is transferred to the sexp object and the user should not call delete [] on the pointer. Be aware that only one sexp variable can take ownership of one vector otherwise the memory is double freed. Else if cob is 2 the ownership of the pointer is only borrowed. Meaning that the sexp object cannot be resized. The user is responsible for freeing the memory! The code below shows how the pointer interface works in general. Showing how sexp objects can be created by passing the information of pointers (double*) which hold data on the heap. Currently, only constructors are written which can use the pointer interface.\n\n// [[Rcpp::depends(ast2ast, RcppArmadillo)]]\n#include <Rcpp.h>\n// [[Rcpp::plugins(cpp17)]]\nusing namespace arma;\n#include <etr.hpp>\nusing namespace etr;\n\n// [[Rcpp::export]]\nvoid fct() {\nint size = 3;\n\n// copy\ndouble* ptr1;\nptr1 = new double[size];\nint cob = 0;\nsexp a(size, ptr1, cob);\ndelete [] ptr1;\na = vector(3.14, 5);\nprint(a);\n\nprint();\n\n// take ownership\ndouble* ptr2;\nptr2 = new double[size];\ncob = 1;\nsexp b(size, ptr2, cob);\nb = vector(5, 3);\nprint(b);\n\nprint();\n\n// borrow ownership\ndouble* ptr3;\nptr3 = new double[size];\ncob = 2;\nsexp c(size, ptr3, cob);\n//error calls resize\n//c = vector(5, size + 1);\nc = vector(4, size);\nprint(c);\n\nprint();\nsexp d(size, ptr3, cob);\nd = d + 10;\nprint(d);\nprint();\n\ndelete[] ptr3;\n}\n\ntrash <- fct()\n## 3.14\n## 3.14\n## 3.14\n## 3.14\n## 3.14\n##\n## 5\n## 5\n## 5\n##\n## 4\n## 4\n## 4\n##\n## 14\n## 14\n## 14\n\nThe pointer interface is particularly useful if the user function has to change the data of a vector or matrix of type NumericVector, vec, NumericMatrix or mat. Assuming that the user passes a function that accepts its arguments by reference it is easy to modify any variable which has a type that can return a pointer to its data. In the code below it is shown how sexp objects are constructed using the pointer interface. Thereby changing the content of variables which has an Rcpp type, a RcppArmadillo type, or is of type std::vector.\n\n// [[Rcpp::depends(ast2ast, RcppArmadillo)]]\n#include <Rcpp.h>\n// [[Rcpp::plugins(cpp17)]]\nusing namespace Rcpp;\nusing namespace arma;\n#include <etr.hpp>\nusing namespace etr;\n\ntypedef void (*fp)(sexp& a);\n\n// [[Rcpp::export]]\nvoid call_package(Rcpp::XPtr<fp> inp) {\nfp f = *inp;\n\n// NumericVector\nNumericVector a_rcpp{1, 2, 3};\nsexp a(a_rcpp.size(), a_rcpp.begin(), 2);\nf(a);\nRcpp::Rcout << a_rcpp << std::endl;\n\n//arma::vec\nvec a_arma(2, fill::value(30));\nsexp b(2, a_arma.memptr(), 2);\nf(b);\na_arma.print();\n\n// NumericMatrix\nNumericMatrix c_rcpp(2, 2);\nsexp c(2, 2, c_rcpp.begin(), 2);\nf(c);\nRcpp::Rcout << c_rcpp << std::endl;\n\n//arma::mat\nmat d_arma(3, 2, fill::value(30));\nsexp d(3, 2, d_arma.memptr(), 2);\nf(d);\nd_arma.print();\n}\n\nf <- function(a) {\na <- a + 2\n}\n\nlibrary(ast2ast)\nfa2a <- translate(f, reference = TRUE, output = \"XPtr\", types_of_args = \"sexp\", return_type = \"void\")\ntrash <- call_package(fa2a)\n## 3 4 5\n## 32.0000\n## 32.0000\n## 2.00000 2.00000\n## 2.00000 2.00000\n##\n## 32.0000 32.0000\n## 32.0000 32.0000\n## 32.0000 32.0000\n\n## Example r2sundials\n\nIn this section an example is shown to illustrate how ast2ast could be used in other R packages. to solve ODE-systems very efficiently. In order to do this the R package r2sundials is used. First a wrapper function is created which is then used by r2sundials. Probably it would be more efficient when the package code itself would take care of the transfer of data between Rcpp/RcppArmadillo and ast2ast.\n\nIn this example it is shown how ODE-systems can be solved by using r2sundials and ast2ast. The code below shows how r2sundials is used normally. Either using an R function or an external pointer to a C++ function.\n\nlibrary(RcppXPtrUtils)\nlibrary(Rcpp)\nlibrary(ast2ast)\nlibrary(r2sundials)\n## Loading required package: rmumps\nlibrary(microbenchmark)\n\n# R version\n# ==============================================================================\nti <- seq(0, 5, length.out=101)\np <- list(a = 2)\np <- c(nu = 2, a = 1)\ny0 <- 0\nfrhs <- function(t, y, p, psens) {\n-p[\"nu\"]*(y-p[\"a\"])\n}\n\nres1 <- r2cvodes(y0, ti,\nfrhs, param = p)\n\n# external pointer version\n# ==============================================================================\nptr_exp=cppXPtr(code='\nint rhs_exp(double t, const vec &y, vec &ydot, RObject &param, NumericVector &psens) {\nNumericVector p(param);\nydot = -p[\"a\"]*(y-1);\nreturn(CV_SUCCESS);\n}\nincludes=\"using namespace arma;\\n#include <r2sundials.h>\", cacheDir=\"lib\", verbose=FALSE)\n# For ease of use in C++, we convert param to a numeric vector instead of a list.\n\npv = c(a= 2)\n# new call to r2cvodes() with XPtr pointer ptr_exp.\nres3=r2sundials::r2cvodes(y0, ti, ptr_exp, param=pv)\n\nIn the code below is the wrapper function defined which is later called by r2sundials. This function is called rhs_exp_wrapper and has the correct function signature. Furthermore, a global function pointer named Fct is defined of type void (*user_fct) (sexp& y_, sexp& ydot_). Within rhs_exp_wrapper* the data of the vectors y and ydot are used to construct two sexp objects which are passed to Fct. Thus, the vector ydot is modified by the function passed from the user. Furthermore, another function called solve_ode is defined. Which calls the code from r2sundials, solves the ODE-system, and returns the output to R. In R the user defines the R function ode. Next, the function is translated and passed to solve_ode. Comparing the results shows that all three approaches (R, C++, ast2ast) generate the same result. Afterwards a benchmark is conducted showing that R is substantially slower than C++ and that the translated function is almost as fast as C++. Mentionable, it is possible to increase the speed of the ast2ast version by using the at function and the *_db* extension.\n\n// [[Rcpp::depends(RcppArmadillo)]]\n// [[Rcpp::depends(rmumps)]]\n// [[Rcpp::depends(r2sundials)]]\n// [[Rcpp::depends(ast2ast)]]\n// [[Rcpp::plugins(\"cpp17\")]]\n#include \"etr.hpp\"\n#include \"r2sundials.h\"\nusing namespace arma;\nusing namespace Rcpp;\nusing namespace etr;\n\ntypedef int (*fp)(double t, const vec &y,\nvec &ydot, RObject &param,\nNumericVector &psens);\n\ntypedef void (*user_fct)(sexp& y_,\nsexp& ydot_);\nuser_fct Fct;\n\nint rhs_exp_wrapper(double t, const vec &y,\nvec &ydot, RObject &param,\nNumericVector &psens) {\nNumericVector p(param);\nconst int size = y.size();\nsexp ydot_(size, ydot.memptr(), 2);\n\ndouble* ptr = const_cast<double*>(\ny.memptr());\nsexp y_(size, ptr, 2);\nFct(y_, ydot_);\nreturn(CV_SUCCESS);\n}\n\n// [[Rcpp::export]]\nNumericVector solve_ode(XPtr<user_fct> inp,\nNumericVector time,\nNumericVector y) {\nFct = *inp;\nXPtr<fp> ptr = XPtr<fp>(new fp(\n&rhs_exp_wrapper));\n\nEnvironment pkg =\nEnvironment::namespace_env(\"r2sundials\");\nFunction solve = pkg[\"r2cvodes\"];\nNumericVector output = solve(y, time,\nptr, time);\n\nreturn output;\n}\n# ast2ast XPtr version\n# ==============================================================================\nti <- seq(0, 5, length.out=101)\ny0 <- 0\n\nlibrary(ast2ast)\node <- function(y, ydot) {\nnu_db <- 2\na_db <- 1\nydot <- -nu_db*(y - a_db)\n}\npointer_to_ode <- translate(ode,\nreference = TRUE, output = \"XPtr\",\ntypes_of_args = \"sexp\",\nreturn_type = \"void\", verbose = TRUE)\n## // [[Rcpp::depends(ast2ast)]]\n## // [[Rcpp::plugins(cpp17)]]\n## #include \"etr.hpp\"\n## // [[Rcpp::export]]\n## SEXP getXPtr();\n## void ode ( sexp& y , sexp& ydot ) {double nu_db ; double a_db ;nu_db = etr::i2d(2);\n##\n## a_db = etr::i2d(1);\n##\n## etr::subassign(ydot, 1) = -nu_db * (etr::subset(y, 1) - a_db);\n##\n## }SEXP getXPtr() {\n## typedef void (*fct_ptr) ( sexp& y , sexp& ydot ); ;return Rcpp::XPtr<fct_ptr>(new fct_ptr(& ode ));}\n## Generated extern \"C\" functions\n## --------------------------------------------------------\n##\n##\n## #include <Rcpp.h>\n## #ifdef RCPP_USE_GLOBAL_ROSTREAM\n## Rcpp::Rostream<true>& Rcpp::Rcout = Rcpp::Rcpp_cout_get();\n## Rcpp::Rostream<false>& Rcpp::Rcerr = Rcpp::Rcpp_cerr_get();\n## #endif\n##\n## // getXPtr\n## SEXP getXPtr();\n## RcppExport SEXP sourceCpp_19_getXPtr() {\n## BEGIN_RCPP\n## Rcpp::RObject rcpp_result_gen;\n## Rcpp::RNGScope rcpp_rngScope_gen;\n## rcpp_result_gen = Rcpp::wrap(getXPtr());\n## return rcpp_result_gen;\n## END_RCPP\n## }\n##\n## Generated R functions\n## -------------------------------------------------------\n##\n## .sourceCpp_19_DLLInfo <- dyn.load('/tmp/Rtmp6tGawx/sourceCpp-x86_64-pc-linux-gnu-1.0.9/sourcecpp_caae5df1fe22/sourceCpp_20.so')\n##\n## getXPtr <- Rcpp:::sourceCppFunction(function() {}, FALSE, .sourceCpp_19_DLLInfo, 'sourceCpp_19_getXPtr')\n##\n## rm(.sourceCpp_19_DLLInfo)\n##\n## Building shared library\n## --------------------------------------------------------\n##\n## DIR: /tmp/Rtmp6tGawx/sourceCpp-x86_64-pc-linux-gnu-1.0.9/sourcecpp_caae5df1fe22\n##\n## /usr/lib/R/bin/R CMD SHLIB -o 'sourceCpp_20.so' 'filecaae7d7e1981.cpp'\nres4 <- solve_ode(pointer_to_ode,\nti, y0)\nres4 <- as.vector(res4)\n\n# Rfunction-ast2ast version\n# ==============================================================================\node <- function(t, y, p, psens) {\nnu_db <- 2\na_db <- 1\nreturn(-nu_db*(y - a_db))\n}\n\nodecpp <- translate(ode)\n\nti <- seq(0, 5, length.out=101)\ny0 <- 0\nres2 <- r2cvodes(y0, ti,\nodecpp, param = p)\n\n# check results and conduct benchmnark\n# ==============================================================================\ndf <- data.frame(time = rep(ti, 4),\nmethod = c(rep(\"R\", 101), rep(\"R ast2ast\", 101),\nrep(\"XPtr\", 101), rep(\"XPtr ast2ast\", 101) ),\ny = c(res1, res2, res3, res4))\n\nlibrary(ggplot2)\nggplot() +\ngeom_point(data = df, aes(x = time, y = y,\ncolour = method, shape = method,\ngroup = method, size = method) ) +\nscale_colour_manual(values = c(\"#FFDB6D\", \"#D16103\", \"#52854C\", \"#293352\") ) +\nscale_size_manual(values = c(6, 4, 2, 1.5)) +\nscale_shape_manual(values=c(19,17,4, 3)) +\nlabs(colour = \"\")", null, "r <- microbenchmark::microbenchmark(\nr2cvodes(y0, ti,\nfrhs, param = p),\nr2cvodes(y0, ti,\nodecpp, param = p),\nr2sundials::r2cvodes(y0, ti, ptr_exp, param=pv),\nsolve_ode(pointer_to_ode,\nti, y0))\nboxplot(r, names = c(\"pure R\", \"ast2ast R\", \"C++\", \"ast2ast XPtr\"))", null, "" ]

[ null, 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", null, 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", 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[ "Paul's Online Notes\nHome / Calculus II / Integration Techniques\nShow Mobile Notice Show All Notes Hide All Notes\nMobile Notice\nYou appear to be on a device with a \"narrow\" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.\n\n## Chapter 1 : Integration Techniques\n\nHere are a set of practice problems for the Integration Techniques chapter of the Calculus II notes.\n\n1. If you’d like a pdf document containing the solutions the download tab above contains links to pdf’s containing the solutions for the full book, chapter and section. At this time, I do not offer pdf’s for solutions to individual problems.\n2. If you’d like to view the solutions on the web go to the problem set web page, click the solution link for any problem and it will take you to the solution to that problem.\n\nNote that some sections will have more problems than others and some will have more or less of a variety of problems. Most sections should have a range of difficulty levels in the problems although this will vary from section to section.\n\nHere is a list of all the sections for which practice problems have been written as well as a brief description of the material covered in the notes for that particular section.\n\nIntegration by Parts – In this section we will be looking at Integration by Parts. Of all the techniques we’ll be looking at in this class this is the technique that students are most likely to run into down the road in other classes. We also give a derivation of the integration by parts formula.\n\nIntegrals Involving Trig Functions – In this section we look at integrals that involve trig functions. In particular we concentrate integrating products of sines and cosines as well as products of secants and tangents. We will also briefly look at how to modify the work for products of these trig functions for some quotients of trig functions.\n\nTrig Substitutions – In this section we will look at integrals (both indefinite and definite) that require the use of a substitutions involving trig functions and how they can be used to simplify certain integrals.\n\nPartial Fractions – In this section we will use partial fractions to rewrite integrands into a form that will allow us to do integrals involving some rational functions.\n\nIntegrals Involving Roots – In this section we will take a look at a substitution that can, on occasion, be used with integrals involving roots.\n\nIntegrals Involving Quadratics – In this section we are going to look at some integrals that involve quadratics for which the previous techniques won’t work right away. In some cases, manipulation of the quadratic needs to be done before we can do the integral. We will see several cases where this is needed in this section.\n\nIntegration Strategy – In this section we give a general set of guidelines for determining how to evaluate an integral. The guidelines give here involve a mix of both Calculus I and Calculus II techniques to be as general as possible. Also note that there really isn’t one set of guidelines that will always work and so you always need to be flexible in following this set of guidelines.\n\nImproper Integrals – In this section we will look at integrals with infinite intervals of integration and integrals with discontinuous integrands in this section. Collectively, they are called improper integrals and as we will see they may or may not have a finite (i.e. not infinite) value. Determining if they have finite values will, in fact, be one of the major topics of this section.\n\nComparison Test for Improper Integrals – It will not always be possible to evaluate improper integrals and yet we still need to determine if they converge or diverge (i.e. if they have a finite value or not). So, in this section we will use the Comparison Test to determine if improper integrals converge or diverge.\n\nApproximating Definite Integrals – In this section we will look at several fairly simple methods of approximating the value of a definite integral. It is not possible to evaluate every definite integral (i.e. because it is not possible to do the indefinite integral) and yet we may need to know the value of the definite integral anyway. These methods allow us to at least get an approximate value which may be enough in a lot of cases." ]

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[ "# Electron Lens\n\n(redirected from electron lenses)\nAlso found in: Dictionary, Thesaurus.\n\n## Electron lens\n\nAn electric or magnetic field, or a combination thereof, which acts upon an electron beam in a manner analogous to that in which an optical lens acts upon a light beam. Electron lenses find application for the formation of sharply focused electron beams, as in cathode-ray tubes, and for the formation of electron images, as in infrared converter tubes, various types of television camera tubes, and electron microscopes.\n\nAny electric or magnetic field which is symmetrical about an axis is capable of forming either a real or a virtual electron image of an object on the axis which either emits electrons or transmits electrons from another electron source. Hence, an axially symmetric electric or magnetic field is analogous to a spherical optical lens.\n\nThe lens action of an electric and magnetic field of appropriate symmetry can be derived from the fact that it is possible to define an index of refraction for electron paths in such fields. This index depends on the field distribution and the velocity and direction of the electrons.\n\nElectron lenses differ from optical lenses both in that the index of refraction is continuously variable within them and in that it covers an enormous range. Furthermore, in the presence of a magnetic field, the index of refraction depends both on the position of the electron in space and on its direction of motion. It is not possible to shape electron lenses arbitrarily. See Electrostatic lens, Magnetic lens\n\nMcGraw-Hill Concise Encyclopedia of Physics. © 2002 by The McGraw-Hill Companies, Inc.\nThe following article is from The Great Soviet Encyclopedia (1979). It might be outdated or ideologically biased.\n\n## Electron Lens\n\na device designed to shape and focus electron beams and to use such beams to produce electron-optical images of objects and parts of objects. Devices used to perform such operations on ion beams are called ion lenses. The action on the electron or ion beam is exerted by an electric or magnetic field, and the lenses are called electrostatic or magnetic lenses, respectively.", null, "Figure 1. Diaphragm with circular aperture (converging diaphragm): (1) electrode-diaphragm, (2) intersections of equipotential surfaces of the electrostatic field and the plane of the figure, (3) electron trajectories, (F) focal point of the lens. A uniform field borders the diaphragm on the left. Corresponding potential values are shown in arbitrary units next to the equipotential lines. It is assumed that the potential is equal to zero wherever the particle velocity is equal to zero. V = 30 is the potential of the electrode. The longitudinal component Ez, of the electric field E decelerates electrons, and the transverse component Er focuses electrons.\n\nElectron lenses are classified according to the type of field symmetry and other characteristic features of the field. In many cases the terminology used to describe electron lenses is borrowed from the classical optics of light rays. This is due to the consistent analogy between the optics of light rays and electron and ion optics as well as to considerations of convenience and clarity of presentation.\n\nThe simplest axially symmetric electrostatic lens consists of a diaphragm with a circular aperture whose field borders uniform electric fields on one or both sides (Figure 1). Depending on the potential distribution, the lens may be a converging lens (producing a beam of charged particles) or a diverging lens. If there is no field on either side of an axially symmetric electrostatic lens (that is, if the lens is bordered by regions of space with constant potentials V1 and V2) and if the potentials are different, then the lens is called a bipotential, or immersion, electron lens (Figure 2); if the potentials are the same, then the lens is called a unipotential", null, "Figure 2. Immersion electron lenses consisting of (a) two diaphragms and (b) two cylinders: (V1) and (V2) potentials of the electrodes. The thin lines are intersections of equipotential surfaces and the plane of the figure. The curved lines with arrows represent the trajectories of charged particles.\n\nlens; such a lens consists of three or more electrodes. An immersion lens changes the velocity of electrons passing through it; a unipotential lens does not alter electron velocities. Both immersion", null, "Figure 3. Cathode lens: (1) cathode, (2) focusing electrode, (3) anode, (O) one of the points on the cathode that emit electrons. The thin lines represent intersections of equipotential surfaces and the plane of the figure. The values of the potentials are entered on the upper scale (the potential of the cathode is assumed to be zero). The hatched area is a cross section of the region occupied by electron flow.\n\nand unipotential lenses always constitute converging lenses.\n\nIn some electrostatic lenses one of the electrodes is an electron-emitting cathode (cathode lenses). A lens of this type accelerates", null, "Figure 4. Magnetic lens with pole pieces: (1) exciting winding, (2) casing, (3) pole pieces. The casing serves as the magnetic circuit. The pole pieces concentrate the magnetic field in a small region near the optical axis z of the system.", null, "Figure 5. Cylindrical electrostatic lenses: (a) diaphragm with slit, (b) immersion lens consisting of two pairs of plates. In the region through which charged particles pass, the fields of the lenses are unaltered in the direction parallel to the diaphragm slits or to the gaps between the plates of neighboring electrodes.\n\nelectrons emitted by the cathode to form an electron beam. A cathode lens consisting of just two electrodes—a cathode and an anode—cannot focus an electron beam, and for this purpose an additional electrode, called a focusing electrode (Figure 3), is included in the lens design.\n\nAxially symmetric magnetic lenses are made in the form of coils of insulated wire, usually housed in an iron casing to amplify and concentrate the lens’s magnetic field. The construction of lenses with very small focal lengths requires that the extent of the field be reduced as much as possible. Pole pieces (Figure 4) are used for this purpose. The field of a magnetic lens can also be generated by a permanent magnet.", null, "Figure 6. Cross sections of electrodes of cylindrical electrostatic lenses in the plane passing through the axis z perpendicular to the midplane: (a) cylindrical-unipotential lens, (b) cylindrical immersion lens, (c) cylindrical single lens, (d) cylindrical cathode lens; (V1) and (V2) potentials of the corresponding electrodes\n\nThe electrodes of cylindrical electrostatic lenses (those exhibiting plane symmetry) are usually diaphragms with a slit or plates positioned symmetrically with respect to the midplane of the lens (Figure 5). Such lenses may be termed cylindrical, because they act on beams of charged particles just as cylindrical optical lenses act on light beams, focusing them in only one direction. The classification of cylindrical electron lenses is similar to that given for axially symmetric electron lenses; there are immersion, unipotential, cathode, and other types (Figure 6). Magnetic electron lenes, usually with an iron casing, may also be cylindrical.\n\nThe fields of transaxial electrostati lenses (Figure 7) have rotational symmetry with respect to an a is (the x-axis in the figure)", null, "Figure 7. Transaxial electrostatic lens with electrodes in the form of two coaxial tubes and with annular slits for the passage of particle beams: (1) tubular electrodes, (2) trajectories of charged particles, (V1) and (V2) potentials of the electrodes. A beam emerges from point A on the object, passes through the field of the lens, becomes astigmatic, and forms two line images B and B′. If the lens parameters are properly chosen, the lens will produce a stigmatic (point-for-point) image.\n\npositioned perpendicular to the optical axis z of the system. In cross sections parallel to the midplane yz of such a lens, the equipotential surfaces have the form of circles or, if the field is limited, arcs, analogous to cross sections of the spherical surfaces of optical lenses. The aberrations of a transaxial lens in a direction parallel to the midplane are therefore comparable in magnitude to the aberrations of optical lenses—that is, they are very small. The line image B’ of a point or rectilinear object perpendicular to the midplane will undergo practically no aberrational expansion.", null, "Figure 8. Quadrupole electron lenses in cross sections perpendicular to the direction of motion of a beam of charged particles: (a) electrostatic lens, (b) magnetic lens; (1) electrodes, (2) lines of force, (3) pole piece, (4) exciting winding\n\nQuadrupole electrostatic and magnetic lenses constitute a special class of electron lenses. Their fields have two planes of symmetry, and their field vectors in the region where the charged particles move are nearly perpendicular to the velocity vectors of the charged particles (Figure 8). Such lenses focus the beam in one direction and cause it to diverge in another perpendicular to the first, creating a linear image of a point object. Two quadrupole electron lenses whose fields are turned 90° to each other", null, "Figure 9. Doublet consisting of two quadrupole electrostatic lenses whose fields are turned 90° to each other about the optical axis z of the system\n\nabout their common optical axis can be mounted one behind the other to form a doublet (Figure 9). The resulting system converges a beam in two mutually perpendicular directions and gives a stigmatic image (one in which a point is represented by a point) when the electron lens parameters are properly selected. Quadrupole electron lenses can act on beams of charged particles with much higher energies and, in the case of magnetic lenses, with larger masses than can axially symmetric electron lenses.\n\n### REFERENCES\n\nSee references under .\n\nV. M. KEL’MAN and I. V. RODNIKOVA" ]

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[ "# calculating electric fields of charge distributions\n\nContinuous charge distributions usually involve a single body which is either uniformly charged, or the charge density across the body varies as a function of the location of the point. This function is given in the problem statement and what is required is the expression of electric field at some point outside/inside the charge distribution.\n\nElectric Charges and Fields 28. Introduction 29. Electric Charge 30. Conductors, Insulators, and Charging by Induction 31. Coulomb's Law 32. Electric Field 33. Calculating Electric Fields of Charge Distributions 34. Electric Field Lines 35. Electric Dipoles VI 36.\n\nFour Important Charge Distributions involving Potential point charges, a charged sphere, ring of charge, and parallel-plate capacitor Finding Potential of a continuous charge distribution is similar to calculating electric fields, but is easier because it is scalar ...\n\nPhys 232 – Lab 3 Ch 16 Electric Fields of Charge Distributions 5 Electric Field of a Uniformly Charged Ring You’ll use VPython to model a uniformly, positively charged ring. You’ll proceed much as you did for the rod – first get your program to evaluate the field at just one observation location, and ...\n\nOk, let's set up a numerical method for calculating the electric field due to the rod. Here is the recipe. Break the rod into N pieces (where you can change the value of N ).\n\n4/4/2013· calculating electric fields due to continuous charge distributions? a question I came across doing some electric field questions, and the answer was really confusing. Homework Statement Charge is distributed along a linear semicircular rod with a linear charge\n\nVideo introduction to electric field calculations and charge distributions for AP Physics C students. Home Books AP Physics 1 AP Physics 2 AP Physics C Honors Regents Regents Workbook iPad Book Courses Regents Book Calendar Course Notes Downloads\n\n1 Electric Charges and Fields Chapter Outline 1.1 Electric Charge 1.2 Conductors, Insulators, and Charging by Induction 1.3 Coulomb’s Law 1.4 Electric Field 1.5 Calculating Electric Fields of Charge Distributions 1.6 Electric Field Lines 1.7 Electric Dipoles Chapter\n\nElectric Fields – Integrating Continuous Charge Distributions I Due noon Thu. Jan. 25 in class Goals: 1. More practice with calculating electric fields, particularly with finding the vector from source points to field point and breaking vectors into components. 2.\n\n5.1 Electric Charge 5.2 Conductors, Insulators, and Charging by Induction 5.3 Coulomb's Law 5.4 Electric Field 5.5 Calculating Electric Fields of Charge Distributions 5.6 Electric Field …\n\nPhys222 Lab Activity 3: Electric Fields for Continuous Charge Distributions Problem 1: A thin, non-conducting rod of length L carries a line charge (x) that varies with distance according to (x)=Ax2 (in SI units) as shown in the figure.\n\nThe fields of nonsymmetrical charge distributions have to be handled with multiple integrals and may need to be calculated numerically by a computer. Exercise $$\\PageIndex{1}$$ How would the strategy used above change to calculate the electric field at a point a distance $$z$$ …\n\nContinuous charge distribution - definition, examples, and the method of dealing with electric charges in their continuous form. Explore more on other related concepts at BYJU'S We have dealt with the systems containing discrete charges such as Q1, Q2,…, Qn.\n\nElectric Field of a Continuous Charge Distribution We know that the E-field produced by a point charge q at a point P a distance r away is (kq/r^2)*r_hat, where r_hat is a unit vector along the direction from q to P. How can we use this knowledge to find the E-field for\n\nUniform charge density 5 6 calculating electric fields of charge distributions solved q4 a rod of length l has uniform linear charge solved there is an infinite line of charge with linear ch Uniform Charge Density 5 6 Calculating Electric Fields Of Charge Distributions ...\n\nGauss’ Law Gauss’ Law can be used as an alternative procedure for calculating electric fields. Gauss’ Law is based on the inverse-square behavior of the electric force between point charges. It is convenient for calculating the electric field of highly symmetric charge\n\nCalculating Electric Fields of Charge Distributions A spherical water droplet of radius 25 μm carries an excess 250 electrons. What vertical electric field is needed to balance the gravitational force on the droplet at the surface of the earth?\n\nCalculating Electric Fields of Charge Distributions A very large number of charges can be treated as a continuous charge distribution, where the calculation of the field requires integration. Common cases are: one-dimensional (like a wire); uses a line charge\n\nSo the electric field could be defined as Coulomb's constant times the charge creating the field divided by the distance squared, the distance we are away from the charge. So essentially, we've defined-- if you give me a force and a point around this charge anywhere, I can now tell you the exact force.\n\nGauss' law is a powerful tool for the calculation of electric fields when they originate from charge distributions of sufficient symmetry to apply it. Gauss' Law Gaussian surfaces If the charge distribution lacks sufficient symmetry for the application of Gauss' law, ...\n\n5.5 Calculating Electric Fields of Charge Distributions 5.5 Calculating Electric Fields of Charge Distributions\n\nElectric Charge and Force: Definition, Repulsion & Attraction Calculating Electric Forces, Fields & Potential What is Capacitance?\n\nDistribution of Charge Along a Curve Distribution of Charge Over a Surface Distribution of Charge in a Volume Contributors and Attributions The electric field intensity associated with $$N$$ charged particles is (Section 5.2):\\[{\\bf E}({\\bf r}) = \\frac{1}{4\\pi\\epsilon} \\sum ...\n\nElectric Fields In the previous unit on electricity and magnetism, we learned how an electric field E exists between two opposite charges q and –q at a distance r from each other. They attract each other, very much like gravity, with a force F given by .This force is called the Coulomb or the electric force, with units of Newtons like any other force.\n\nJOURNAL OF GEOPHYSICAL RESEARCH, VOL. 106, NO. D13, PAGES 14,191-14,205, JULY 16, 2001 On different approaches to calculating lightning electric fields Raj eev Thottappillil Angstr6m Laboratory, Uppsala University, Uppsala, Sweden Vladimir\n\n2.4 Electric Field of Charge Distributions So far we have considered point charges, we studied the forces that the point charges exert to one another. We also studied the electric field generated by the charge or system of the charges. Now, we’re going to study a\n\n25/9/2014· Ok, let's set up a numerical method for calculating the electric field due to the rod. Here is the recipe. Break the rod into N pieces (where you can change the value of N ).\n\nnet charge within the surface, and we use this relation to calculate the electric field for symmetric charge distributions. 22-1 Calculating From Coulomb’s Law Figure 22-1 shows an element of charge dq =r dV that is small enough to be con-sidered a point d at a\n\nThe electric potential at any point in space produced by a continuous charge distribution can be calculated from the point charge expression by integration since voltage is a scalar quantity. The continuous charge distribution requires an infinite number of charge elements to characterize it, and the required infinite sum is exactly what an integral does.\n\n15/9/2016· Electric Charges and Fields 28. Introduction 29. Electric Charge 30. Conductors, Insulators, and Charging by Induction 31. Coulomb's Law 32. Electric Field 33. Calculating Electric Fields of Charge Distributions 34. Electric Field Lines 35. Electric Dipoles VI 36.\n\nCalculating Electric Fields of Charge Distributions A proton enters the uniform electric field produced by the two charged plates shown below. ... Force experienced by the charge particle in an external electric field is expressed as, Here is particle’s charge and is …\n\nThis is a very common strategy for calculating electric fields. The fields of nonsymmetrical charge distributions have to be handled with multiple integrals and …\n\nAnswer to Calculating Electric Fields of Charge Distributions The charge per unit length on the thin semicircular wire shown.... The electric field due to point charge is expressed as follows: Here, is the permittivity of the free space, is the charge of the object, is the distance from the field point to the charged object, and is the unit directional vector.\n\nDifferent charge distributions produce different electric fields, but the individual bits of charge all produce fields in the same fashion. Electric Forces Have you ever walked across a carpet in ...\n\nAlso, note that Equation \\ref{m0104_eISC} is the electric field at any point above or below the charge sheet – not just on $$z$$ axis. This follows from symmetry. From the perspective of any point in space, the edges of the sheet are the same distance (i.e., infinitely far) away.\n\nlomb’s law, it is more convenient for calculating the electric fields of highly symmetric charge distributions and makes it possible to deal with complicated problems using qualitative reasoning. As we show in this chapter, Gauss’s law is important in understand ...\n\n5.6 Calculating Electric Fields of Charge Distributions A very large number of charges can be treated as a continuous charge distribution, where the calculation of the field requires integration. Common cases are: one-dimensional (like a wire); uses a line charge\n\nSummary: Gauss’s law is used to calculate the electric field surrounding a given charge distribution. The law, which is based on the concept of electric flux, is most convenient for calculating the electric fields of symmetrical charge distributions. Key Equations:" ]

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[ "", null, "# TimerWipy\n\n• Hi , can someone explain how the method\n''''timerchannel.irq(*, trigger, priority=1, handler=None)''' works?\nI have to use TimerWiPy.PERIODIC.\nI don't undestand also handler function.\n\n• @robert-hh Hi, sorry for my insistence on this topic.\nI 'm not able to use thread but I read about it and I think that it's better for my code if I use them . So do you know if is it possibile to use thread to read from two I2C through the code that I posted before?\nthank you.\n\n• @robert-hh I solved the problem creating a new function to compare data acquire by sensor with reference data and then it power on pwm. And in this way the period remain 50 ms.\nthank you\n\n• @vflorio94 For a consistent operation at a period of 0.02 or 50 ms it is required that the handler takes not longer than that, and that is executes the OWM changes at almost the same time within the handler. Looking at your code, I see at least two things that could affect timing:\na) You write to a file. File i/O is buffered. As long as the data is written into the buffer, it's fast. When the buffer will be written, it can be slow.\nb) You call lib_imu2.get_euler() several times in the timer handler. I do not know the complexity of that call, but if it includes access to external hardware, then it may take some time. Better call it once, store the result and use that in consecutive statements.\n\n• @robert-hh Thank you, I solved this problem. I put data as metod.\nNow I have a new problem because I want to read data from sensor and after a comparisono with reference data I want to activate pwm. I did it in this way.\nBut the problem is that the period change a lot, sample time decrease to 0.04.\nIs there another way to modify this code for activate pwm that can be better??\nI need if it is possible that period remain more possible like 0.02.\n\nthank you\n\n``````import machine\nimport pycom\nfrom machine import Timer\nfrom machine import idle\nimport time\nimport lib_imu\nimport lib_imu2\nfrom machine import I2C\nfrom math import sqrt\nfrom math import fabs\nfrom machine import PWM\nfrom machine import SD\nimport gc\nimport os\nsd = SD()\nos.chdir('/sd')\nfrom machine import idle\n\n#\n# add the other imports here that you need\n# fileIMU1 = open('/sd/dati_IMU1.txt', 'w')\n#\n# acquire values. The paramters of the constructor are:\n#\nclass Acquire:\ndef __init__(self, filename_imu1, data, period=0.02, runtime = 60):\nself.fileIMU1 = open(filename_imu1, \"w\")\nself.temp = Timer.Chrono()\nself.busy = True\nself.data = data\nself.runtime = runtime\nself.temp.start()\nself.alarm = Timer.Alarm(self.sample_and_write, period , periodic=True)\n\ndef stop(self):\nself.alarm.cancel()\npwm_c2.duty_cycle(0)\npwm_c5.duty_cycle(0)\nself.fileIMU1.close()\n#self.fileIMU2.close()\n\ndef sample_and_write(self, alarm):\n# check, if done, if the time for sampling is expired, then stop\nif passed >= self.runtime:\nself.stop()\nself.busy = False\nprint('end')\nelse:\n# get data and write\n#print(passed, fabs(lib_imu2.get_euler()))\nself.fileIMU1.write(repr(passed) + repr(lib_imu2.get_euler()) + \"\\n\")\n# avanti\nif fabs(lib_imu2.get_euler()) > ( fabs(dati_rif) + 0.2):\nif fabs(lib_imu2.get_euler()) < (fabs(dati_rif) + 0.4):\npwm2 = PWM(0, frequency=1000 )\npwm_c2.duty_cycle(0.5)\npwm_c5.duty_cycle(0)\nelif fabs(lib_imu2.get_euler()) >= (fabs(dati_rif) + 2):\npwm2 = PWM(0, frequency=2000 )\npwm_c2.duty_cycle(0.6)\npwm_c5.duty_cycle(0)\n# indietro\nelif fabs(lib_imu2.get_euler()) < fabs(dati_rif) + 0.2:\nif fabs(lib_imu2.get_euler()) > (fabs(dati_rif) + 0.4):\npwm5 = PWM(0, frequency=1000 )\npwm_c2.duty_cycle(0)\npwm_c5.duty_cycle(0.5)\nelif fabs(lib_imu2.get_euler())< (fabs(dati_rif) + 2):\npwm5 = PWM(0, frequency=2000 )\npwm_c2.duty_cycle(0)\npwm_c5.duty_cycle(0.5)\n# and what else you have to do.\n\n#\n# now here the skeleton main_\n#\nsd = SD()\nos.chdir('/sd')\n\n# set up the hardware\n''' SET UP IMU 0X29 '''\n# Scala Berg: imu L5 - lib_imu - acc\nlib_imu.set_acc()\ntime.sleep(0.01)\nlib_imu.set_configmode()\ntime.sleep(0.01)\nlib_imu.set_units()\ntime.sleep(0.01)\nlib_imu.set_mode('ndof')\ntime.sleep(0.01)\n''' SET UP IMU 0X28 '''\n# imu t1 - lib_imu2 - eul\nlib_imu2.set_configmode()\ntime.sleep(0.01)\nlib_imu2.set_units()\ntime.sleep(0.01)\nlib_imu2.set_mode('ndof')\ntime.sleep(0.01)\n''' PWM '''\npwm2 = PWM(0, frequency=500 )\npwm5 = PWM(0, frequency=500 )\n#This sets up the timer to\n#oscillate at the specified frequency. timer is an integer from 0 to 3.\n# frequency is an integer from 1 Hz to 78 KHz\npwm_c5 = pwm2.channel(5, pin='P12') #dietro\npwm_c2 = pwm5.channel(2, pin='P5') # avanti\npwm_c5.duty_cycle(0)\ntime.sleep(0.01)\npwm_c2.duty_cycle(0)\ntime.sleep(0.01)\n\n''' START'''\nwith open(\"/sd/Rif.txt\", \"r\") as f:\n# now you have a list of one string per line\ndati_rif = [float(el) for el in line.strip(\"()\\n \").split(\",\")] # reference data\nprint('start')\nacquire = Acquire('Acquisition_Vibration.txt', dati_rif, 0.02, 30)\n\n#\n# now get the data, until it's done\n#\nwhile acquire.busy:\nidle()\n#\n# whatever you have to do for cleanup, do it here\n#`````````\n\n• @vflorio94 That's a question of scope and visibility in Python.\nSo most simple but unelegant method is to have that data global. So it is created outside every function. Then at any part of your code you can declare is a global to access it. Example:\n\n``````# at the top level of your code:\ndata_buffer = bytearray(100)\ndata_ptr_in = 0\ndata_ptr_out = 0\n\n#\n#\nclass MyClass:\ndef my_func(self):\n# declare symbols as global\nglobal data_buffer, data_ptr_in, data_ptr_out\ndata_buffer[data_ptr_in] = some_data\ndata_ptr_in = (data_ptr_in + 1) % len(data:buffer)\n``````\n\nYou can also define functions (or better to say methods) to pass along data, and you can use property and setter and getter methods of classes, ...\n\n• @robert-hh sorry, I describe the problem in a bad way. I have another question.\nI have a data that I want to use inside the function 'sample_and_write', but I have some difficulties to pass it from main to the class '''Acquire''' and in the end to the function. Is it possible ? can you help me?\n\n• @robert-hh yes, I checked and you are right. And about command to activate pwm responding to data acquire by sensor?\n\n• @vflorio94 said in TimerWipy:\n\nself.pwm = PWM(0.02, frequency=500 )\n\nThe first argument of the class instantiation is a timer ID, not a time or the like. The value range is 0 - 3. See: https://docs.pycom.io/firmwareapi/pycom/machine/pwm\nSo it should be:\nself.pwm = PWM(0, frequency=500 )\n\n• @robert-hh Hello. I have to ask you a new thing correlate with this code that you wrote previously.\nI want to implement the code with another part in which I desire to activate PWM depending on the data sample and wrote by IMU. So, the question is:\nIs it correct if I introduced a control on the data that I wrote and the command to activate PWM in the fuction 'sample and write'?\nAnd have I also to complete class acquire like this?\n\n``````self.pwm = PWM(0.02, frequency=500 ) #This sets up the timer to\n#oscillate at the specified frequency. timer is an integer from 0 to 3.\n# frequency is an integer from 1 Hz to 78 KHz\n``````\n\nI know that I have also to set PWM channel where I put set-up Hardware parameters.\nI'll be newly grateful if you can help me.\n\n• @robert-hh I understand all. finally it works without any interrupt\nthan you a lot.\n\n• @vflorio94 Timer.Alarm() has to be called only once.\n\n• @vflorio94 You do not have to call alarm. Actually, it is not a function, it is a timer object. The callback function is called by the Timer periodically without further action by your main program. You main code has to prepare the data & objects needed by the callback and then start the timer. Typically, all of that would be enclosed in a class, and the main program would just instantiate this class. That could look like this:\n\n``````#\nfrom machine import Timer, idle\n#\n# add the other imports here that you need\n\n#\n# acquire values. The paramters of the constructor are:\n#\nclass Acquire:\ndef __init__(self, filename_imu1, period=0.2, runtime = 60):\nself.fileIMU1 = open(filename_imu1, \"w\") # or append\nself.temp = Timer.Chrono()\nself.busy = True\nself.runtime = runtime\nself.temp.start()\nself.alarm = Timer.Alarm(self.sample_and_write, period ,periodic=True)\n\ndef stop(self):\nself.alarm.cancel()\nself.fileIMU1.close()\n\ndef sample_and_write(self, alarm):\n# check, if done, if the time for sampling is expired, then stop\nif passed >= self.runtime:\nself.stop()\nself.busy = False\nelse:\n# get data and write\nprint(passed, self.runtime)\nself.fileIMU1.write(repr(passed) + \",\" + lib_imu.get_euler_int() + \"\\n\")\n#\n# and what else you have to do.\n\n#\n# now here the skeleton main_\n#\n\n# set up the hardware\n#\n\n# and start\nacquire = Acquire(\"my_data_file\", 0.2, 20)\n#\n# now get the data, until it's done\n#\nwhile acquire.busy:\nidle()\n#\n# whatever you have to do for cleanup, do it here\n#\n``````\n\n• @robert-hh the function can be like this?\n\n``````import machine\nimport pycom\nfrom machine import Timer\nimport time\nimport lib_imu\nimport lib_imu2\nfrom machine import I2C\nfrom math import sqrt\nfrom machine import PWM\nfrom machine import SD\nimport gc\nimport os\nsd = SD()\nos.chdir('/sd')\n\nPERIOD= 0.02 #sample time\n\nlib_imu.set_configmode()\ntime.sleep(0.01)\nlib_imu.set_units()\ntime.sleep(0.01)\nlib_imu.set_mode('ndof')\ntime.sleep(0.01)\n\nlib_imu2.set_configmode()\ntime.sleep(0.01)\nlib_imu2.set_units()\ntime.sleep(0.01)\nlib_imu2.set_mode('ndof')\ntime.sleep(0.01)\nfileIMU1 = open('/sd/dati_IMU1.txt', 'w')\n#fileT_IMU1 = open(\"/sd/datiT_IMU1.txt\", \"w\")\nfileIMU2 = open('/sd/dati_IMU2.txt', 'w')\n#fileT_IMU2 = open(\"/sd/datiT_IMU2.txt\", \"w\")\nfile_tempi = open('/sd/tempi.txt', 'w')\nlist_eul = [ ]\nlist_eul2 = [ ]\nlist_tempi = []\n\ndef sample_and_write(timer_obj): #timer_obj??\nlist_eul.append(lib_imu.get_euler_int())\nlist_eul2.append(lib_imu2.get_euler_int())\npass\n\ndef acquisition(sec):\nt = Timer.Chrono()\nt.start()\nwhile 1:\nif t <= sec :\nalarm = Timer.Alarm(sample_and_write, PERIOD, periodic = true)\nelse:\nalarm.cancel()\nfileIMU1.write( repr(list_eul) + '\\t')\nfileIMU2.write( repr(list_eul2) + '\\t')\ndel list_eul\ndel list_eul2\n``````\n\n• @robert-hh thank you, now i'm trying to modify my function.\n\n• so, is 'timer_obj' my time of acquisition?\n• is alarm the command that I have to use in main?\nthank you a lot\n\n• @vflorio94 With that method you can achieve, that a certain function, the handler, is called when the time expires. That is what you need for your project. The setting of timer.PERIODIC means, that it is called over & over again. Like:\n\n``````from machine import timer\nPERIOD = 0.2 # call every 0.2 seconds\n\ndef sample_and_write(timer_obj):\n# this is a function which reads form the sensor and writes\n# to disk. timer_obj identifies the timer, which calls this function\n# Can be ignored\npass\n\nalarm = Timer.Alarm(sample_and_write, PERIOD, periodic=True)\n``````\n\nAnd when you want to stop it, call\n\n``````alarm.cancel()\n``````\n\nEdit: It should be positively noted that the callback is no interrupt. This means you are free for almost any Python operation, especially allocating memory." ]

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[ "# 6 days or 14 Billion Years?", null, "## Time is relative.\n\nThe rate at which time passes is not the same everywhere. Einstein showed us this with his formula:\n\nHe showed us that changes in velocity or gravity will affect the passage of time. If fact, if we it were possible to travel at the speed of light, time would stop! We would exist simultaneously in the past, present and future.\n\nVisable light as well as all electromagnetic radiation has both frequency and velocity. The velocity is constant–the speed of light (300 million meters per second). The higher the freqency, the greater the wave engergy.\n\nOur universe is expanding and therefore the wavelength of light (actually all electromagnetic radiation) has streched. As the frequecy of light gets longer it turns red-hence the term redshift.\n\nJust as a train bell shifts lower in pich as it passes away from you, so time appears to go slower than its original rate.\n\nLight or radiation velocity is the cosmic clock . It exist in an “eternal now” state, a state were time does not pass. Absolute time is referenced from this cosmic clock. Earth time is relative.\n\nSo now we have earth time and absolute time.\n\nThe minimum frequency of radiation energy needed to turn energy into mass (The big bang) is known. It can be refered to as quark confinement. We know the temperature and hence frequency of radiated energy at quark confinement.\n\nAt present, the universe has a residue cosmic background radiation (CBR) observable from all directions. It is the residue energy leftover from the big bang. It’s temperature is also known (-270 C or 2.73 degrees Kelvin).\n\nThe energy at quark confinement is approximately a million million times greater than the CBR we measure today. Its redshift is 10 12. .\n\nThe universal streching of space alters our perception of time on earth as referenced from a universal clock.\n\n“The streching of light waves has slowed the frequency of the cosmic clock–it has expanded the perceived time between ‘ticks’ by a million million.”\n\nCosmological dating of the age of the universe is an approximation based on several different formulas and assumptions. It is also based on measurements of just one percent of the universe and assumes it represents the entire universe. Nevertheless, It is generally agreed that the universe has an apparent age of 10 to 20 billion years.\n\n“The cosmic clock records the passage of one minute while we on Earth experience a million million minutes.”\n\nSo to recap, the differnce of perceived time and cosmic time is a ratio of a millio n million to one. This is based on the known threshold temperature at which matter is formed and the measured temperature of the black of space (CBR).\n\nSo, what do you get if you divide 15 billion years by a million million? 6 days!! oh,oh I didn’t mean to get Biblical.\n\nreferences:\n\nby Schroeder, Gerald L. “The Science of God,” The Free Press 3:47-71 1997" ]

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[ "HomeCustomize Design Worksheet ➟ 20 20 Graphing Quadratics Review Worksheet\n\n# 20 Graphing Quadratics Review Worksheet", null, "Review solving quadratics by graphing from graphing quadratics review worksheet, graphing quadratics review worksheet answers page 2, graphing quadratics review worksheet answers page 4, graphing quadratics review worksheet key, graphing quadratics review worksheet algebra 1, graphing quadratics review worksheet kuta software, graphing quadratics review worksheet pdf answers, graphing quadratics review worksheet page 2, graphing quadratics review worksheet answers page 3, graphing quadratics review worksheet answer key, graphing quadratic functions review worksheet answers, image source: www.slideshare.net\n\n## 20 Homophones Worksheet 2nd Grade\n\nHomophones Worksheets from homophones worksheet 2nd grade, homophones worksheet for second grade, free homonyms worksheets for 2nd grade, homophones worksheet for 2nd graders, homophones worksheet 2nd grade pdf, free printable homophone worksheets for 2nd grade, free homophones worksheet 2nd grade, homophones worksheet for 2nd grade, homophones practice worksheets 2nd grade, homophones worksheet grade 2 pdf, […]\n\n## 20 Types Of Maps Worksheet\n\nTypes of Maps and Map Skills Pack Social Stu s grades 2 from types of maps worksheet, types of maps worksheets 6th grade, types of maps worksheet, types of maps worksheet 6th grade pdf, types of maps worksheet 5th grade, types of maps worksheet answers, types of maps worksheet 9th grade, types of maps worksheet […]\n\n## 20 Operations with Rational Numbers Worksheet\n\nAlgebra 1 Worksheets from operations with rational numbers worksheet, mixed operations with rational numbers worksheet pdf, order of operations with rational numbers worksheet pdf, operations with rational numbers worksheet pdf, adding and subtracting rational numbers worksheet kuta, operations with rational and irrational numbers worksheet pdf, adding rational numbers worksheet answer key, adding and subtracting rational […]" ]

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[ "# Efficient Recursive Diffusion Layers for Block Ciphers and Hash Functions\n\n## Abstract\n\nMany modern block ciphers use maximum distance separable (MDS) matrices as the main part of their diffusion layers. In this paper, we propose a very efficient new class of diffusion layers constructed from several rounds of Feistel-like structures whose round functions are linear. We investigate the requirements of the underlying linear functions to achieve the maximal branch number for the proposed 4×4 words diffusion layer, which is an indication of the highest level of security with respect to linear and differential attacks. We try to extend our results for up to 8×8 words diffusion layers. The proposed diffusion layers only require simple operations such as word-level XORs, rotations, and they have simple inverses. They can replace the diffusion layer of several block ciphers and hash functions in the literature to increase their security, and performance. Furthermore, it can be deployed in the design of new efficient lightweight block ciphers and hash functions in future.\n\n## Introduction\n\nBlock ciphers are one of the most important building blocks in many security protocols. Modern block ciphers are cascades of several rounds where every round consists of confusion and diffusion layers. In many block ciphers, while the confusion layer is often realized as a parallel application of non-linear substitution boxes (S-boxes), the diffusion layer is built from a linear transformation. The diffusion layer plays an efficacious role in providing resistance against the most well-known attacks on block ciphers, such as differential cryptanalysis (DC) , and linear cryptanalysis (LC) .\n\nWhen considering a word-based linear transformation, where the word size is equal to the input/output size of the S-box, the branch number provides a lower bound on the number of active S-boxes throughout the diffusion layer for differential and linear attacks. The goal for a designer is to maximize this number, in order to diffuse the non-linear properties of the S-Boxes faster to the subsequent rounds of the cipher. The faster this non-linearity spreads, the less number of rounds the cipher requires to become secure against linear and differential attacks. It has been shown that the maximal branch number for a linear transformation of s words is s+1 and diffusion layers with maximal branch number can be achieved by using MDS matrices .\n\nAn MDS matrix (Maximum Distance Separable) is a matrix representing a function with certain diffusion properties that have useful applications in cryptography. Technically, an m×n matrix A over a finite field K is an MDS matrix if it is the transformation matrix of a linear transformation f(x)=Ax from K n to K m such that no two different (m+n)-tuples of the form (x,f(x)) coincide in n or more components. Equivalently, the set of all (m+n)-tuples (x,f(x)) is an MDS code, i.e. a linear code that reaches the Singleton bound.\n\nIn 1994, Vaudenay [11, 12] suggested using MDS matrices in cryptographic primitives to produce what he called multipermutations, not-necessarily linear functions with the same property. These functions have what he called perfect diffusion: changing t of the inputs change at least mt+1 of the outputs. He showed how to exploit imperfect diffusion to cryptanalyze functions that are not multipermutations. MDS matrices were later used in many block ciphers such as Square, SHARK, AES, Twofish and Hierocrypt and in the stream cipher MUGI and the cryptographic hash function Whirlpool.\n\nThe common approach to construct MDS matrices is to extract them from MDS codes such as Reed–Solomon codes . However, constructing MDS diffusion layers with low-cost implementations is a challenge for designers. Another problem arises when MDS diffusion layers are exploited in substitution-permutation networks (SPN), where the MDS matrix is used in the encryption and its inverse is used in the decryption process. Thus, constructing MDS matrices with low-cost inverse is of great importance.\n\nIn this paper, we propose a new method to construct low-cost diffusion layers with an extra property that their inverse can also be implemented efficiently. We call the proposed layer a recursive diffusion layer. It is constructed from several rounds of Feistel-like structures whose round functions are linear. It consists of simple linear operations such as shift, rotation and XOR with very similar inversion operations. We are going to elaborate on the conditions for the underlying linear function to be an MDS matrix using one or multiple such linear functions by proposing a systematic method to find them. We believe that our proposed solution would be a rather simple recipe for designing a diffusion layer with maximal branch number and will be useful for future designs of cryptographic algorithms.\n\n### Notations\n\nLet x be an array of s n-bit elements x=[x 0(n),x 1(n),…,x s−1(n)]. The number of non-zero elements in x is denoted by w(x), also known as the Hamming weight of x. The following notations are used throughout this paper:\n\nFor a diffusion layer D applicable on x, we have the following definitions:\n\n### Definition 1\n\n()\n\nThe differential branch number of a linear diffusion layer D is defined as\n\n$$\\beta_{d}(D)=\\displaystyle\\min_{\\mathbf{x}\\not=0}\\bigl\\{ w(\\mathbf {x})+w\\bigl(D(\\mathbf{x})\\bigr)\\bigr\\}$$\n\nWe know that the linear function D can be shown as a binary matrix B, and D t is a linear function obtained from B t, where B t is the transposition of B.\n\n### Definition 2\n\n()\n\nThe linear branch number of a linear diffusion layer D is defined as:\n\n$$\\beta_{l}(D)=\\displaystyle\\min_{\\mathbf{x}\\not=0}\\bigl\\{ w(\\mathbf {x})+w\\bigl(D^{t}(\\mathbf{x})\\bigr)\\bigr\\}$$\n\nIt is well known that for a diffusion layer acting on s-word inputs, the maximal β d and β l are s+1 . A diffusion layer D taking its maximal β d and β l is called a perfect or MDS diffusion layer. Furthermore, a diffusion layer with β d =β l =s is called an almost perfect diffusion layer .\n\n### Our Contribution\n\nIn this paper, we define the notion of a recursive diffusion layer, and we propose a method to construct such perfect diffusion layers.\n\n### Definition 3\n\nA diffusion layer D with s words x i as the input and s words y i as the output is called a recursive diffusion layer if it can be represented in the following form:\n\n$$D: \\left\\{ \\begin{array}{l} y_{0}= x_{0} \\oplus F_0(x_{1}, x_{2},\\ldots, x_{s-1})\\\\ y_{1}= x_{1} \\oplus F_1(x_{2}, x_{3},\\ldots, x_{s-1}, y_0)\\\\ \\vdots\\\\ y_{s-1}= x_{s-1} \\oplus F_{s-1}(y_{0}, y_{1},\\ldots, y_{s-2}) \\end{array} \\right.$$\n(1)\n\nwhere F 0,F 1,…,F s−1 are arbitrary linear functions.\n\nAn advantage of this structure is that the inverse of D is very similar to D and does not require the inverse of F i functions. The inverse can be computed as:\n\n$$D^{-1}: \\left\\{ \\begin{array}{l} x_{s-1}= y_{s-1} \\oplus F_{s-1}(y_{0}, y_{1},\\ldots, y_{s-2})\\\\ x_{s-2}= y_{s-2} \\oplus F_{s-2}(x_{s-1}, y_{0},\\ldots, y_{s-3})\\\\ \\vdots\\\\ x_{0}= y_{0} \\oplus F_{0}(x_{1}, x_{2},\\ldots, x_{s-1}) \\end{array} \\right.$$\n(2)\n\nAs an example, consider a 2-round Feistel structure with a linear round function L as a recursive diffusion layer with s=2. The input–output relation for this diffusion layer is\n\n$$D: \\left\\{ \\begin{array}{l} y_{0}= x_{0} \\oplus L(x_{1})\\\\ y_{1}= x_{1} \\oplus L( y_0) \\end{array} \\right.$$\n\nThe quarter-round function of the stream cipher Salsa20 is an example of a non-linear recursive diffusion layer .\n\n$$D: \\left\\{ \\begin{array}{l} y_{1}= x_{1} \\oplus((x_0+x_3) \\lll7)\\\\ y_{2}= x_{2} \\oplus((x_0+y_1) \\lll9)\\\\ y_{3}= x_{3} \\oplus((y_1+y_2)\\lll13)\\\\ y_{0}= x_{0} \\oplus((y_2+y_3) \\lll18)\\\\ \\end{array} \\right.$$\n\nAlso, the lightweight hash function PHOTON and the block cipher LED use MDS matrices based on Eq. (1). In these ciphers, an m×m MDS matrix B m was designed based on the following matrix B for the performance purposes:\n\n$$\\mathbf{B} = \\left(\\begin{array}{c@{\\quad }c@{\\quad }c@{\\quad }c@{\\quad }c} 0 & 1 & 0 & \\ldots& 0\\\\ 0& 0 & 1 & \\ldots& 0\\\\ \\vdots& & \\ddots& \\\\ 0& 0 & 0 & \\ldots& 1\\\\ 1 & Z_{1} & Z_{2} & \\ldots& Z_{m-1} \\end{array} \\right)$$\n\nBy matrix B, one element of m inputs is updated and other elements are shifted. If we use B m, all inputs are updated, but we must check if this matrix is MDS. One example for m=4 is the PHOTON matrix working over GF(28):\n\n$$\\mathbf{B} = \\left(\\begin{array}{c@{\\quad}c@{\\quad}c@{\\quad}c} 0 & 1 & 0 & 0\\\\ 0& 0 & 1 & 0\\\\ 0& 0 & 0 & 1\\\\ 1& 2 & 1 & 4\\\\ \\end{array}\\right) \\quad \\Rightarrow \\quad \\mathbf{B}^4 = \\left(\\begin{array}{c@{\\quad}c@{\\quad}c@{\\quad}c} 1 & 2 & 1 & 4\\\\ 4& 9 & 6 & 17\\\\ 17& 38 & 24 & 66\\\\ 66& 149 & 100 & 11\\\\ \\end{array}\\right)$$\n\nIn this paper, we propose a new approach to design linear recursive diffusion layers with the maximal branch number in which F i ’s are composed of one or two linear functions and a number of XOR operations. The design of the proposed diffusion layer is based on the invertibility of some simple linear functions in GF(2). Linear functions in this diffusion layer can be designed to be low-cost for different sizes of the input words, thus the proposed diffusion layer might be appropriate for resource-constrained devices, such as RFID tags. Although these recursive diffusion layers are not involutory, they have similar inverses with the same computational complexity.\n\nThis paper proceeds as follows: In Sect. 2, we introduce the general structure of our proposed recursive diffusion layer. Then, for one of its instances, we systematically investigate the required conditions for the underlying linear function to achieve the maximal branch number. In Sect. 3, we propose some other recursive diffusion layers with less than 8 input words and only one linear function. We use two linear functions to have a perfect recursive diffusion layer for s>4 in Sect. 4. Finally, we conclude the paper in Sect. 5.\n\n## The Proposed Diffusion Layer\n\nIn this section, we introduce a new perfect linear diffusion layer with a recursive structure. The diffusion layer D takes s words x i for i={0,1,…,s−1} as input, and returns s words y i for i={0,1,…,s−1} as output. So, we can represent this diffusion layer as\n\n$$y_{0}||y_{1}||\\cdots||y_{s-1}=D(x_{0}||x_{1}|| \\cdots||x_{s-1})$$\n\nThe first class of the proposed diffusion layer D is represented in Fig. 1, where L is a linear function, α k ,β k ∈{0,1}, α 0=1 and β 0=0. This diffusion layer can be represented in the form of Eq. (1) in which the F i functions are all the same and can be represented as\n\n$$F_i(x_1,x_2,\\ldots,x_{s-1})= \\bigoplus_{j=1}^{s-1} \\alpha_{j}x_j \\oplus L \\Biggl(\\bigoplus_{j=1}^{s-1} \\beta_{j}x_j \\Biggr)$$", null, "Fig. 1.\n\nTo guarantee the maximal branch number for D, the linear function L and the coefficients α j and β j must satisfy some necessary conditions. Conditions on L are expressed in this section and those of α j ’s and β j ’s are expressed in Sect. 3. The diffusion layer described by Eq. (3) is an instance that satisfies the necessary conditions on α j , and β j with s=4. In the rest of this section, we concentrate on the diffusion layers of this form and show that we can find invertible linear functions L such that D becomes a perfect diffusion layer.\n\n$$D: \\left\\{ \\begin{array}{l} y_{0}= x_{0} \\oplus x_2 \\oplus x_3 \\oplus L(x_1 \\oplus x_3)\\\\ y_{1}= x_{1} \\oplus x_3 \\oplus y_0 \\oplus L(x_2 \\oplus y_0)\\\\ y_{2}= x_{2} \\oplus y_0 \\oplus y_1 \\oplus L(x_3 \\oplus y_1)\\\\ y_{3}= x_{3} \\oplus y_1 \\oplus y_2 \\oplus L(y_0 \\oplus y_2) \\end{array} \\right.$$\n(3)\n\nAs shown in Fig. 2, this diffusion layer has a Feistel-like (GFN) structure, i.e.,\n\n$$F_0(x_1,x_2,x_3)=x_2 \\oplus x_3 \\oplus L(x_1 \\oplus x_3)$$", null, "Fig. 2.\n\nThe inverse transformation, D −1, has a very simple structure and does not require the inversion of the linear function L. The inverse of D is\n\n$$D^{-1}: \\left\\{ \\begin{array}{l} x_{3}= y_{3} \\oplus y_1 \\oplus y_2 \\oplus L(y_0 \\oplus y_2)\\\\ x_{2}= y_{2} \\oplus y_0 \\oplus y_1 \\oplus L(x_3 \\oplus y_1)\\\\ x_{1}= y_{1} \\oplus x_3 \\oplus y_0 \\oplus L(x_2 \\oplus y_0)\\\\ x_{0}= y_{0} \\oplus x_2 \\oplus x_3 \\oplus L(x_1 \\oplus x_3) \\end{array} \\right.$$\n\nD and D −1 are different, but they have the same structure and properties. To show that D has the maximal branch number, first we introduce some lemmas and theorems.\n\nIf L(x) can be written as ax in a finite field, then Eq. (3) can be expressed as a matrix representation as below:\n\n$$\\mathbf{B} = \\left(\\begin{array}{c@{\\quad}c@{\\quad}c@{\\quad}c} 0 & 1 & 0 & 0\\\\ 0 & 0 & 1 & 0\\\\ 0 & 0 & 0 & 1\\\\ 1 & a & 1 & a+1\\\\ \\end{array}\\right) \\quad \\Rightarrow \\quad \\left(\\begin{array}{c} y_0 \\\\ y_1\\\\ y_2\\\\ y_3\\\\ \\end{array}\\right) =\\mathbf{B}^4 \\left(\\begin{array}{c} x_0 \\\\ x_1\\\\ x_2\\\\ x_3\\\\ \\end{array}\\right)$$\n(4)\n\nWe can construct MDS matrix similar to PHOTON matrix by the proposed diffusion layer. In Eq. (1), if F i (x 1,x 2,x 3)=F 0(x 1,x 2,x 3)=L(x 1)⊕x 2L 2(x 3), where L(x)=2x and xGF(28), PHOTON MDS matrix is obtained . If we change B to Eq. (3), and define L(x)=2x, we have\n\n$$\\mathbf{B} = \\left(\\begin{array}{c@{\\quad}c@{\\quad}c@{\\quad}c} 0 & 1 & 0 & 0\\\\ 0& 0 & 1 & 0\\\\ 0& 0 & 0 & 1\\\\ 1& 2 & 1 & 3\\\\ \\end{array}\\right) \\quad \\Rightarrow \\quad \\mathbf{B}^4 = \\left(\\begin{array}{c@{\\quad}c@{\\quad}c@{\\quad}c} 1 & 2 & 1 & 3\\\\ 3& 7 & 1 & 4\\\\ 4& 11 & 3 & 13\\\\ 13& 30 & 6 & 20\\\\ \\end{array}\\right)$$\n\n### Theorem 4\n\n()\n\nA Boolean function F has maximal differential branch number if, and only if it has maximal linear branch number.\n\nAs a result of Theorem 4, if we prove that the diffusion layer D represented in Eq. (3) has the maximal differential branch number, its linear branch number will be maximal too. Thus, in the following, we focus on the differential branch number.\n\n### Lemma 5\n\nA linear functions L(x) is invertible if, and only if for any non-zero value a, $$L(a)\\not=0$$.\n\n### Proof\n\nFor any linear function L(x), we have L(0)=0. If there exists $$a\\not =0$$ such that L(a)=0, then L(x) is not invertible. On the other hand, suppose a=0 is the unique zero of L(x), and L(x) is not invertible. So, there exist two values b and c ($$b\\not=c$$) such that L(b)=L(c). Since L(x) is a linear function, we have L(bc)=L(b)⊕L(c)=0, while $$b\\oplus c\\not=0$$. This contradicts the assumption that a=0 is the unique zero of L(x). □\n\n### Lemma 6\n\nAssume the linear operator $$\\mathcal{L}_{i}$$ corresponds to the linear function L i (x). If the linear operator $$\\mathcal{L}_{3}$$ can be represented as the multiplication of two operators $$\\mathcal{L}_{1}$$ and $$\\mathcal{L}_{2}$$, then the corresponding linear function L 3(x)=L 2(L 1(x)) is invertible if, and only if the linear functions L 1(x) and L 2(x) are invertible.\n\n### Proof\n\nIf L 1(x) and L 2(x) are invertible, clearly L 3(x) is invertible too. On the other hand, if L 3(x) is invertible then L 1(x) must be invertible, otherwise, there are distinct x 1, and x 2 such that L 1(x 1)=L 1(x 2). Thus, L 3(x 1)=L 2(L 1(x 1))=L 2(L 1(x 2))=L 3(x 2) which contradicts the invertibility of L 3(x). The invertibility of L 2(x) is proved in the same way. □\n\n### Example 1\n\nWe can rewrite the linear function L 3(x)=L 3(x)⊕x ($$\\mathcal{L}_{3}=\\mathcal{L}^{3} \\oplus I$$) as L 3(x)=L 2(L 1(x)), where L 1(x)=L(x)⊕x ($$\\mathcal {L}_{1}=\\mathcal{L} \\oplus I$$) and L 2(x)=L 2(x)⊕L(x)⊕x ($$\\mathcal{L}_{2}=\\mathcal{L}^{2} \\oplus \\mathcal{L} \\oplus I$$). Thus, the invertibility of L 3(x) is equivalent to the invertibility of the two linear functions L 1(x) and L 2(x).\n\n### Theorem 7\n\nFor the diffusion layer represented in Eq. (3), if the four linear functions L(x), xL(x), xL 3(x) and xL 7(x) are invertible, then this diffusion layer is perfect.\n\n### Proof\n\nWe show that the differential branch number of this diffusion layer is 5. First, the 4 words of the output are directly represented as functions of the 4 words of the input:\n\n$$D: \\left\\{ \\begin{array}{l} y_{0}= x_{0} \\oplus L(x_1 ) \\oplus x_2 \\oplus x_3 \\oplus L(x_3)\\\\ y_{1}= x_0 \\oplus L(x_0) \\oplus x_1 \\oplus L(x_1) \\oplus L^2(x_1) \\oplus x_2 \\oplus L^2(x_3)\\\\ y_{2}= L^2(x_0) \\oplus x_1 \\oplus L(x_1) \\oplus L^3(x_1) \\oplus x_2 \\oplus L(x_2)\\oplus x_3 \\oplus L^2(x_3) \\oplus L^3(x_3)\\\\ y_{3}= x_0 \\oplus L^2(x_0) \\oplus L^3(x_0) \\oplus L(x_1) \\oplus L^2(x_1) \\oplus L^3(x_1) \\oplus L^4(x_1) \\\\ \\phantom{y_{3}=} \\oplus L(x_2) \\oplus L^2(x_2) \\oplus L^2(x_3) \\oplus L^4(x_3) \\end{array} \\right.$$\n(5)\n\nIn the proof, we look at all different cases for the Hamming weight of the input. In other words, we show that if the Hamming weight of the input is m=1,2,3,4, then the Hamming weight of the output is greater than or equal to 5−m. Each case will pose different conditions on L which in the end can be summarized to the condition given in the theorem. The diffusion layer represented in Eq. (3) is invertible. Consider m=4, then all of the 4 words in the input are active, and we are sure at least one of the output words is active too. Thus, the theorem is correct for m=4. The remainder of the proof is performed for the 3 cases of w(Δ(x))=m, for m=1,2,3 separately. In each of these cases, some conditions are forced on the linear function L.\n\nCase 1: w(△x)=1\n\nTo study this case, first the subcase\n\n$$(\\triangle x_0\\not=0,\\triangle x_{1}=\\triangle x_{2}=\\triangle x_{3}=0 \\quad \\mbox{or}\\quad \\triangle \\mathbf{x}=\\triangle x_0|| 0|| 0||0 )$$\n\nis analyzed. For this subcase, Eq. (5) is simplified to:\n\n$$D: \\left\\{ \\begin{array}{l} \\triangle y_{0}= \\triangle x_{0}\\\\ \\triangle y_{1}=(I\\oplus L) (\\triangle x_0) \\\\ \\triangle y_{2}= L^2(\\triangle x_0)\\\\ \\triangle y_{3}= (I\\oplus L^2 \\oplus L^3)(\\triangle x_0 ) \\end{array} \\right.$$\n\nIf D is a perfect diffusion layer, then △y 0, △y 1, △y 2, and △y 3 must be non-zero. Clearly, △y 0 is non-zero and based on Lemma 5, the conditions for △y 1, △y 2, and △y 3 to be non-zero are that the linear functions IL, L 2, and IL 2L 3 must be invertible. Note that based on Lemma 6 the invertibility of L yields the invertibility of L 2. Considering Lemma 6, if the other three sub-cases are studied, it is induced that the linear functions xL(x)⊕L 2(x) and xL(x)⊕L 3(x) must also be invertible.\n\nCase 2: w(△x)=2\n\nIn this case, there exist exactly two active words in the input difference, and we obtain some conditions on the linear function L to guarantee the branch number 5 for D. In the following, we only analyze the subcase\n\n$$(\\triangle x_0,\\triangle x_{1}\\not=0 \\quad\\mbox{and}\\quad \\triangle x_{2}=\\triangle x_{3}=0 \\quad\\mbox{or}\\quad \\triangle\\mathbf{x}=\\triangle x_0|| \\triangle x_1|| 0||0 )$$\n\nWith this assumption, Eq. (5) is simplified to\n\n$$D: \\left\\{ \\begin{array}{l} \\triangle y_{0}= \\triangle x_{0} \\oplus L(\\triangle x_1 ) \\\\ \\triangle y_{1}=(I\\oplus L) (\\triangle x_0) \\oplus(I\\oplus L \\oplus L^2)( \\triangle x_1) \\\\ \\triangle y_{2}= L^2(\\triangle x_0) \\oplus(I\\oplus L \\oplus L^3)(\\triangle x_1) \\\\ \\triangle y_{3}= (I\\oplus L^2 \\oplus L^3)(\\triangle x_0 ) \\oplus (L\\oplus L^2\\oplus L^3\\oplus L^4)(\\triangle x_1) \\end{array} \\right.$$\n(6)\n\nTo show that w(△y) is greater than or equal to 3, we must find some conditions on L such that if one of the △y i ’s is zero, then the other three △y j ’s cannot be zero. Let △y 0=0, then\n\n$$\\triangle x_{0} \\oplus L(\\triangle x_1)=0\\quad \\Rightarrow\\quad \\triangle x_{0} = L(\\triangle x_1)$$\n\nIf △x 0 is replaced in the last three equations of Eq. (6), we obtain △y 1, △y 2 and △y 3 as follows:\n\n$$\\left\\{ \\begin{array}{l} \\triangle y_{1}=\\triangle x_1\\\\ \\triangle y_{2}=\\triangle x_1 \\oplus L(\\triangle x_1)\\\\ \\triangle y_{3}= L^2(\\triangle x_1)\\\\ \\end{array} \\right.$$\n\nObviously, △y 1 is not zero. Furthermore, considering Lemma 5, for △y 2 to be non-zero, we conclude that the function xL(x) must be invertible. For △y 1⇒△y 3, L 2(x) is invertible. This condition was already obtained in the Case 1. We continue this procedure for △y 1=0.\n\n$$\\begin{array}{l} \\triangle y_{1}= \\triangle x_0 \\oplus L(\\triangle x_0) \\oplus x_1 \\oplus L(\\triangle x_1) \\oplus L^2(\\triangle x_1)=0\\\\ \\quad\\Rightarrow\\quad \\triangle x_0 \\oplus L(\\triangle x_0) = x_1 \\oplus L(\\triangle x_1) \\oplus L^2(\\triangle x_1) \\end{array}$$\n\nFrom the previous subcase, we know that if △y 0=0, then $$\\triangle y_{1}\\not=0$$. Thus, we conclude that △y 0 and △y 1 cannot be simultaneously zero. Therefore, by contraposition, we obtain that if △y 1=0, then $$\\triangle y_{0}\\not=0$$. So, we only check △y 2 and △y 3. From the third equation in Eq. (6), we have\n\n$$\\begin{array}{lll} (I \\oplus L) (\\triangle y_{2})&=&L^2(\\triangle x_1) \\oplus L^3(\\triangle x_1) \\oplus L^4(\\triangle x_1) \\oplus\\triangle x_1\\\\ &&\\oplus\\, L^2(\\triangle x_1) \\oplus L^3(\\triangle x_1) \\oplus L^4(\\triangle x_1)\\\\ &=&\\triangle x_1 \\end{array}$$\n\nxL(x) is invertible, thus we conclude that with the two active words △x 0 and △x 1 in the input, △y 1 and △y 2 cannot be zero simultaneously. With the same procedure, we can prove that △y 1, and △y 3 cannot be zero simultaneously.\n\nHere we only gave the proof for the case ($$\\triangle x_{0},\\triangle x_{1}\\not=0$$, △x 2=△x 3=0). We performed the proof procedure for the other cases, and no new condition was added to the previous set of conditions in Case 1.\n\nCase 3: w(△x)=3\n\nIn this case, assuming three active words in the input, we show that the output has at least 2 non-zero words. Here, only the case\n\n$$(\\triangle x_{0},\\triangle x_{1},\\triangle x_{2}\\not=0 \\quad\\mbox{and}\\quad \\triangle x_{3}=0 \\quad \\mbox{or}\\quad \\triangle\\mathbf{x} = \\triangle x_0 || \\triangle x_1|| \\triangle x_{2} || 0 )$$\n\nis analyzed. The result holds for the other three cases with w(△x)=3. Let rewrite Eq. (5) for △x 3=0 as follows:\n\n$$D: \\left\\{ \\begin{array}{l} \\triangle y_{0}= \\triangle x_{0} \\oplus L(\\triangle x_1 ) \\oplus \\triangle x_2 \\\\ \\triangle y_{1}=(I\\oplus L) (\\triangle x_0) \\oplus(I\\oplus L \\oplus L^2)( \\triangle x_1) \\oplus\\triangle x_2 \\\\ \\triangle y_{2}= L^2(\\triangle x_0) \\oplus(I\\oplus L \\oplus L^3)(\\triangle x_1) \\oplus(I \\oplus L) (\\triangle x_2 )\\\\ \\triangle y_{3}= (I\\oplus L^2 \\oplus L^3)(\\triangle x_0 ) \\oplus (L\\oplus L^2\\oplus L^3\\oplus L^4)(\\triangle x_1) \\oplus (L\\oplus L^2)(\\triangle x_2) \\end{array} \\right.$$\n(7)\n\nWhen △y 0=△y 1=0, from the first two lines of Eq. (7), △x 0 and △x 1 are obtained as the function of △x 2.\n\n$$\\left\\{ \\begin{array}{lll} \\triangle y_{0}&=& \\triangle x_{0} \\oplus L(\\triangle x_1 ) \\oplus \\triangle x_2 =0\\\\ \\triangle y_{1}&=& \\triangle x_0 \\oplus L(\\triangle x_0) \\oplus \\triangle x_1 \\oplus L(\\triangle x_1)\\\\ &&\\oplus L^2(\\triangle x_1) \\oplus \\triangle x_2=0 \\end{array} \\right. \\quad \\Rightarrow \\quad \\left\\{ \\begin{array}{l} \\triangle x_{1} =L( \\triangle x_2)\\\\ \\triangle x_{0}= \\triangle x_2 \\oplus L^2(\\triangle x_2) \\end{array} \\right.$$\n\nNow, replacing △x 0=△x 2L 2(△x 2) and △x 1=L(△x 2) into △y 2 and △y 3 yields\n\n$$\\left\\{ \\begin{array}{rcl} \\triangle y_{2}&=& L^2(\\triangle x_0) \\oplus(I\\oplus L \\oplus L^3)(\\triangle x_1) \\oplus(I \\oplus L) (\\triangle x_2 )=\\triangle x_2\\\\ \\triangle y_{3}&=& (I\\oplus L^2 \\oplus L^3)(\\triangle x_0 ) \\oplus (L\\oplus L^2\\oplus L^3\\oplus L^4)(\\triangle x_1) \\oplus (L\\oplus L^2)(\\triangle x_2)\\\\ &=&(I\\oplus L)(\\triangle x_2) \\end{array} \\right.$$\n\nFrom Case 1, we know that the functions xL(x) are invertible. Therefore, △y 2, and △y 3 are non-zero. If the other sub-cases with three active words in the input are investigated, it is easy to see that no new condition is added to the present conditions on L. Finally, we conclude that the diffusion layer D presented in Fig. 1 is perfect if the linear functions\n\n$$\\left\\{ \\begin{array}{l} L_1(x)=L(x)\\\\ L_2(x)=x \\oplus L(x)\\\\ L_3(x)=x \\oplus L(x) \\oplus L^2(x)\\\\ L_4(x)=x \\oplus L(x) \\oplus L^3(x)\\\\ L_5(x)=x \\oplus L^2(x) \\oplus L^3(x) \\end{array} \\right.$$\n\nare invertible. We know that L 3(L 2(x))=xL 3(x) and L 5(L 4(L 2(x)))=xL 7(x). Thus, by Lemma 6, we can summarize the necessary conditions on the linear function L as the invertibility of L(x), (IL)(x), (IL 3)(x), and (IL 7)(x). □\n\nNext, we need a simple method to check whether a linear function L satisfies the conditions of Theorem 7 or not. For this purpose, we use the binary matrix representation of L. Assume that x i is an n-bit word. Hence, we can represent a linear function L with an n×n matrix A with elements in GF(2). As a result of Lemma 5, if L is invertible, A is not singular over GF(2) ($$|\\mathbf{A}|\\not=0$$). To investigate whether a linear function L satisfies the conditions of Theorem 7, we construct the corresponding matrix A n×n from L, and check the non-singularity of the matrices A, IA, IA 3, and IA 7 in GF(2).\n\nIn the following, we construct concrete functions L which are lightweight and satisfy the conditions mentioned in Theorem 7. For example, the functions L(x)=x, L(x)=xa and L(x)=xa are the examples of the most lightweight linear functions. However, they do not satisfy Theorem 7 conditions, because at least one of the two functions L(x) and xL(x) are not invertible. A set of candidates for lightweight linear functions can be expressed as:\n\n$$\\begin{array}{l} L(x_{(n)}) = (x_{(n)} \\ll a) \\oplus(x_{(n)} \\gg b) \\end{array} .$$\n(8)\n\nIf (a+b)|n, then L(x) is invertible . The remaining conditions xL(x), xL 3(x) and xL 7(x) have to be checked. Although the linear function in Eq. (8) has a complicated inverse, it does not require circular shift which is considered as an advantage for this function. Note that circular shift is not supported by some compilers. Another proposal for L(x) is\n\n$$\\begin{array}{l} L(x_{(n)}) = \\bigl(x_{(n)} \\oplus(x_{(n)} \\gg a)\\bigr) \\lll b \\end{array}$$\n(9)\n\nThe linear function in Eq. (9), for a>n/2, has a lightweight inverse L(x (n))=(x (n)b)⊕(x (n)b)≫a which will be used in diffusion layer proposed in Sect. 4.\n\nWe introduce some lightweight linear functions with n-bit inputs/outputs in Table 1 which satisfy the conditions of Theorem 7. Note that for n=8, there does not exist any linear function of the form Eq. (8) or Eq. (9) satisfying conditions of Theorem 7.\n\n### Application of the Proposed Diffusion Layer in Current Block Ciphers\n\nTogether with designing new lightweight block ciphers, the proposed diffusion layer can also be applied to diffuse the non-linearity of big-size S-boxes. One of these block ciphers is MMB that uses 32-bit S-boxes. Each round of MMB is composed of four transformations:\n\n• σ: bit-wise XOR of the intermediate value and the round key.\n\n• γ: modular multiplication of each 32-bit word of the intermediate value with a fixed 32-bit constant G i modulo 232−1.\n\n• η: an operation on two of the four input words.\n\n• θ: the only diffusion operation in MMB which is an involutory binary matrix as below:\n\n$$\\mathbf{B} = \\left(\\begin{array}{c@{\\quad}c@{\\quad}c@{\\quad}c} 1 & 1 & 0 & 1\\\\ 1& 1 & 1 & 0\\\\ 0 & 1 & 1 & 1\\\\ 1& 0 & 1 & 1\\\\ \\end{array}\\right)$$\n\nWe can use the proposed diffusion layer with L(x (32))=(x (32)≪3)⊕(x (32)≫1) instead of the diffusion layer used in the block cipher MMB. If we use the proposed diffusion layer in this cipher, it becomes stronger against differential and linear attacks, because branch number of the binary matrix of MMB is 4 while branch number of the proposed diffusion layer is 5. This change also prevents the attacks presented against this block cipher in . By computer simulations in C using a PC with CPU: 2.93 GHz and RAM: 2 GB, we observed that this modification reduces the performance of MMB by making it 30 % slower in the software implementations. This was achieved by comparing the running time of the protocol for 1 million encryptions.\n\nAnother block cipher where we can replace the diffusion layer by the proposed one is Hierocrypt . Hierocrypt does not explicitly use big-size S-boxes, but it constructs 32 bit S-boxes by using nested SPN structure together with four 8-bit S-boxes and the MDS L matrix. For diffusion within those 32-bit S-boxes, a 16×16 binary matrix called MDS H is used, which is MDS for four 32-bit inputs. If we use our proposed diffusion layer with the same L(x), instead of the MDS H , we can achieve a 2 times faster implementation with the same level of security.\n\nAES Mix-column layer has a simple implementation. As another comparison, we decided to replace the MDS H matrix in Hierocrypt with the MDS matrix of AES. But since MDS code of AES is over GF(28) and the inputs of MDS H are four 32-bit words, we modified the corresponding irreducible polynomial in AES and replaced it with x 32+x 7+x 5+x 3+x 2+x+1 to work over GF(232), which would still remain MDS. We call this new construction, sch 1. As another construction, we replaced the MDS H in Hierocrypt with the MDS code we proposed above in our solution and we called it sch 2. We observed that sch 2 still brings on 5 % better performance compared to sch 1.\n\n## Other Desirable Structures for the Proposed Diffusion Layer\n\nIn Sect. 2, the general form of the proposed diffusion layer was introduced in Fig. 1. Then, by assuming a special case of α i ’s and β i ’s, an instance of this diffusion layer was given in Eq. (3). In this section, we obtain all sets of α i ’s and β i ’s such that the diffusion layer of Fig. 1 becomes perfect. We know some properties of α i ’s and β i ’s; for instance if all the words of the output are directly represented as a function of input words, a function of each x i (0≤is−1) must appear in each equation. Another necessary condition is obtained for two active words of the input. Assume there exist only two indices i,j such that x i , $$x_{j}\\not= 0$$. If we write each two output words y p , y q in a direct form as a function of x i and x j , we obtain\n\n$$\\left\\{ \\begin{array}{l} y_{p}= L_{p_i}( x_i) \\oplus L_{p_j}( x_j)\\\\ y_{q}= L_{q_i}( x_i) \\oplus L_{q_j}( x_j)\\\\ \\end{array} \\right.$$\n\nIf\n\n$$\\frac{\\mathcal{L}_{p_i}}{\\mathcal{L}_{q_i}}=\\frac{\\mathcal {L}_{p_j}}{\\mathcal{L}_{q_j}} \\quad\\mbox{or}\\quad \\left\\lvert \\begin{array}{c@{\\quad }c} \\mathcal{L}_{p_i} & \\mathcal{L}_{p_j}\\\\ \\mathcal{L}_{q_i} & \\mathcal{L}_{q_j} \\end{array} \\right\\lvert=0$$\n\nthen, y p =0 is equivalent to y q =0. Thus, the minimum number of active words in the input and output is less than or equal to s and the branch number will not reach the maximal value s+1. This procedure must be repeated for 3, and more active words in the input. As an extension, we can use Lemma 3 of .\n\n### Lemma 8\n\n()\n\nAssume the diffusion layer has m inputs/outputs bits, and $$\\mathcal {L}$$ is the linear operator of L(x), and I is the linear operator of I(x). Moreover, ML D is an m×m matrix representation of the operator of the diffusion layer. If D is perfect, then all the submatrices of ML D are non-singular.\n\nIf we construct the ML D of Eq. (3), we have\n\n\\begin{aligned} \\mathbf{ML}_D= \\left(\\begin{array}{c@{\\quad}c@{\\quad}c@{\\quad}c} I & \\mathcal{L} & I & I\\oplus\\mathcal {L}\\\\ I\\oplus\\mathcal{L} & I\\oplus\\mathcal{L}\\oplus\\mathcal {L}^2 & I & \\mathcal{L}^2\\\\ \\mathcal{L}^2 & I\\oplus\\mathcal{L}\\oplus\\mathcal{L}^3 & I\\oplus\\mathcal{L} & I\\oplus\\mathcal{L}^2\\oplus \\mathcal{L}^3\\\\ I\\oplus\\mathcal{L}^2\\oplus\\mathcal{L}^3 & \\mathcal{L}\\oplus \\mathcal{L}^2\\oplus\\mathcal{L}^3\\oplus\\mathcal{L}^4 & \\mathcal{L}\\oplus\\mathcal{L}^2 & \\mathcal{L}^2\\oplus\\mathcal {L}^4 \\end{array}\\right) \\end{aligned}\n\nwhen calculating 69 sub-matrix determinants of ML D , we observe that these submatrices are non-singular only if L fulfills the condition of Theorem 7. However, by following this procedure, it is complicated to obtain all sets of α i ’s and β i ’s analytically. So, by systematizing the method based on Lemma 8, we performed a computer simulation to obtain all sets of α i ’s, and β i ’s in the diffusion layer in Fig. 1 that yield a perfect diffusion. We searched for all α i ’s and β i ’s that make the diffusion layer of Fig. 1 a perfect diffusion layer. This procedure was repeated for s=2,3,…,8. We found one set of (α i , β i ) for s=2, four sets for s=3, and four sets for s=4. The obtained diffusion layers along with the conditions on the underlying linear function L are reported in Table 2. We observed that for s=5, 6, 7, 8 the diffusion layer introduced in Fig. 1 cannot be perfect.\n\nNote that some linear functions in Table 1 such as L(x (64))=(x (64)≪15)⊕(x (64)≫1) are not suitable for diffusion layers, since x (64)L 15(x (64)) must be invertible.\n\nAs we can see in Fig. 1, and its instances presented in Table 2, there exists some kind of regularity in the equations defining y i ’s, in the sense that the form of y i+1 is determined by the form of y i , and vice versa (F i ’s are all the same in Eq. (1)). However, we can present some non-regular recursive diffusion layers with a more general form (F i ’s are different) as in Fig. 3, where A i,j ,B i,j ∈{0,1}.", null, "Fig. 3.\n\nIf A i,j =α (ji)mods , and B i,j =β (ji)mods , then Fig. 3 is equivalent to Fig. 1. The main property of this new structure is that it still has one linear function L, and a simple structure for the inverse. For example, if s=4, then, the diffusion layer D is\n\n$$\\left\\{ \\begin{array}{l} y_{0}= x_{0} \\oplus A_{0,1} \\cdot x_1 \\oplus A_{0,2} \\cdot x_2 \\oplus A_{0,3} \\cdot x_3 \\oplus L(B_{0,1} \\cdot x_1 \\oplus B_{0,2} \\cdot x_2 \\oplus B_{0,3} \\cdot x_3)\\\\ y_{1}= x_{1} \\oplus A_{1,0} \\cdot y_0 \\oplus A_{1,2} \\cdot x_2 \\oplus A_{1,3} \\cdot x_3 \\oplus L(B_{1,0} \\cdot y_0 \\oplus B_{1,2} \\cdot x_2 \\oplus B_{1,3} \\cdot x_3)\\\\ y_{2}= x_{2} \\oplus A_{2,0} \\cdot y_0 \\oplus A_{2,1} \\cdot y_1 \\oplus A_{2,3} \\cdot x_3 \\oplus L(B_{2,0} \\cdot y_0 \\oplus B_{2,1} \\cdot y_1 \\oplus B_{2,3} \\cdot x_3)\\\\ y_{3}= x_{3} \\oplus A_{3,0} \\cdot y_0 \\oplus A_{3,1} \\cdot y_1 \\oplus A_{3,2} \\cdot y_2 \\oplus L(B_{3,0} \\cdot y_0 \\oplus B_{3,1} \\cdot y_1 \\oplus B_{3,2} \\cdot y_2) \\end{array} \\right.$$\n\nWe searched the entire space for s=3 and s=4 (the order of search is 22s(s−1)). For s=3, we found 196 structures with branch number 4, and for s=4, 1634 structures with branch number 5. The conditions on linear functions that caused maximal branch number, are different for each structure. Among the 196 structures for s=3, the structure with the minimum number of operations (only 7 XORs, and one L evaluation) is the following:\n\n$$D: \\left\\{ \\begin{array}{l} y_{0}= x_{0} \\oplus x_1 \\oplus x_2 \\\\ y_{1}= x_{1} \\oplus x_2 \\oplus L( y_0 \\oplus x_2 )\\\\ y_{2}= x_{2} \\oplus y_0 \\oplus y_1 \\end{array} \\right.$$\n\nwhere L(x) and xL(x) must be invertible.\n\nThis relation is useful to enlarge the first linear function of the hash function JH for 3 inputs . For s=4, we did not find any D with the number of L evaluations less than four. However, the one with the minimum number of XORs is given as below:\n\n$$D: \\left\\{ \\begin{array}{l} y_{0}= x_{0} \\oplus x_1 \\oplus x_2 \\oplus L(x_3)\\\\ y_{1}= x_{1} \\oplus x_3 \\oplus y_0 \\oplus L( x_2 \\oplus y_0 )\\\\ y_{2}= x_{2} \\oplus x_3 \\oplus y_0 \\oplus L(x_3 \\oplus y_1)\\\\ y_{3}= x_{3} \\oplus y_1 \\oplus y_2 \\oplus L( y_0 ) \\end{array} \\right.$$\n\nSearching the whole space for s=5,6,… is too time consuming (note that for s=5 the order of search has complexity 240), and we could not search all the space for s≥5.\n\n## Increasing the Number of Linear Functions\n\nIn Sect. 3, we observed that for s>4 we cannot design a regular recursive diffusion layer in the form of Fig. 1 with only one linear function L. In this section, we increase the number of linear functions to overcome the regular structure of the diffusion layer of Eq. (3). A new structure is represented in Fig. 4, where α k ,β k ,γ k ∈{0,1}, k∈{0,1,…,s−1}, α 0=1,β 0=0 and γ 0=0.", null, "Fig. 4.\n\nIf L 1 and L 2 are two distinct linear functions, Fig. 4 is too complicated to easily obtain conditions on L 1 and L 2 that make it a perfect diffusion layer (the order of search for s input/output is 23(s−1)). To obtain simplified conditions for a maximal branch number, let L 1 and L 2 have a simple relation like $$L_{2}(x)=L_{1}^{2}(x)$$ or $$L_{2}(x)=L_{1}^{-1}(x)$$. For the linear functions in Eq. (8), L 2(x) is more complex in comparison to L(x). However, there exist some linear functions in the form of Eq. (9) such that L −1(x) is simpler than L 2(x). As an example, for L(x (32))=(x (32)x (32)≫31)⋘1, we have L −1(x (32))=((x (32)⋙1)⊕(x (32)⋙1)≫31), but L 2(x (32))=(x (32)⊕(x (32)≫31)⋘1)⊕((x (32)⋘1)≫31)⋘1.\n\nIn Table 3, we introduce some recursive diffusion layers with (L 1=L and L 2=L −1) or (L 1=L and L 2=L 2) that have maximal branch numbers. These diffusion layers are obtained similar to that of Table 2. In this table, for each case, only y 0 is presented. Other y i ’s can be easily obtained from Fig. 4, since the F i ’s are all the same.\n\nIf the 14 linear functions:\n\n$$\\begin{array}{l@{\\quad }l@{\\quad }l} L(x) & I\\oplus L(x) & I\\oplus L^3(x)\\\\ I\\oplus L^7(x) & I\\oplus L^{15}(x) & I\\oplus L^{31}(x)\\\\ I\\oplus L^{63}(x) & I\\oplus L^{127}(x) & I\\oplus L^{255}(x)\\\\ I\\oplus L^{511}(x) & I\\oplus L^{1023}(x) & I \\oplus L^{2047}\\\\ I\\oplus L^{4095}(x) & I\\oplus L^{8191}(x) \\end{array}$$\n\nare invertible (all irreducible polynomials up to degree 13), then all the diffusion layers introduced in Table 3 are perfect. One example for a 32-bit linear function satisfying these conditions is\n\n$$L(x_{(32)})=\\bigl(x_{(32)} \\oplus(x_{(32)} \\gg31)\\bigr) \\lll29$$\n\n## Conclusion\n\nIn this paper, we proposed a new family of efficient diffusion layers (recursive diffusion layers) which are constructed using several rounds of Feistel-like structures whose round functions are linear. The proposed diffusion layers are very efficient and have simple inverses, thus they can be deployed to improve the security or performance of some of the current block ciphers and hash functions and in the design of the future lightweight block ciphers and hash functions, even providing provable security against differential and linear attacks. For a fixed structure, we determined the required conditions for its underlying linear function to be perfectly secure with respect to linear and differential attacks. Then, for the number of words in input (output) less than eight, we extended our approach, and found all the instances of the perfect recursive diffusion layers with the general form described in Fig. 1. Also, we proposed some other diffusion layers with non-regular forms. Finally, diffusion layers with two linear functions were proposed. By using two linear functions, we designed perfect recursive diffusion layers for higher number of words.\n\n## References\n\n1. \n\nD.J. Bernstein, The Salsa20 Stream Cipher (2005). http://www.ecrypt.eu.org/stream/salsa20p2.html\n\n2. \n\nE. Biham, A. Shamir, Differential Cryptanalysis of DES-Like Cryptosystems, in CRYPTO’90. Lecture Notes in Computer Science, vol. 537 (Springer, Berlin, 1990), pp. 2–21\n\n3. \n\nJ. Daemen, Cipher and Hash function design strategies based on linear and differential cryptanalysis. Ph.D. thesis, Elektrotechniek Katholieke Universiteit Leuven, Belgium (1995)\n\n4. \n\nJ. Daemen, V. Rijmen, The Design of Rijndael: AES—The Advanced Encryption Standard (Springer, Berlin, 2002)\n\n5. \n\nJ. Guo, T. Peyrin, A. Poschmann, The PHOTON family of lightweight Hash functions, in CRYPTO’11. Lecture Notes in Computer Science, vol. 6841 (Springer, Berlin, 2011), pp. 222–239\n\n6. \n\nJ. Guo, T. Peyrin, A. Poschmann, M. Robshaw, The LED block cipher, in CHES’11. Lecture Notes in Computer Science, vol. 6917 (Springer, Berlin, 2011), pp. 326–341\n\n7. \n\nS. Lin, D. Costello, Error control coding: fundamentals and applications (Prentice Hall, New York, 2004)\n\n8. \n\nM. Matsui, Linear cryptanalysis method for DES cipher, in EUROCRYPT’93. Lecture Notes in Computer Science, vol. 765 (Springer, Berlin, 1993), pp. 386–397\n\n9. \n\nK. Ohkuma, H. Muratani, F. Sano, S. Kawamura, The block cipher hierocrypt, in SAC’01. Lecture Notes in Computer Science, vol. 2012 (Springer, Berlin, 2001), pp. 72–88\n\n10. \n\nM. Sajadieh, M. Dakhilalian, H. Mala, Perfect involutory diffusion layers based on invertibility of some linear functions. IET Inf. Secur. J. 5(1), 228–236 (2011)\n\n11. \n\nC. Schnorr, S. Vaudenay, Black box cryptoanalysis of Hash networks based on multipermutations, in EUROCRYPT’94. Lecture Notes in Computer Science, vol. 950 (Springer, Berlin, 1994), pp. 47–57\n\n12. \n\nS. Vaudenay, On the need for multipermutations: cryptanalysis of MD4 and SAFER, in FSE’94. Lecture Notes in Computer Science, vol. 1008 (Springer, Berlin, 1994), pp. 286–297\n\n13. \n\nM. Wang, J. Nakahara, Y. Sun, Cryptanalysis of the full MMB block cipher, in SAC’09. Lecture Notes in Computer Science, vol. 5867 (Springer, Berlin, 2009), pp. 231–248\n\n14. \n\nH. Wu, The Hash Function JH (2008). http://icsd.i2r.astar.edu.sg/staff/hongjun/jh/jh.pdf\n\n15. \n\nG. Zeng, K. He, W. Han, A Trinomial Type of σ-LFSR Oriented Toward Software Implementation. Science in China Series F-Information Sciences, vol. 50 (Springer, Berlin, 2007), pp. 359–372\n\nDownload references\n\n## Author information\n\nAuthors\n\n### Corresponding author\n\nCorrespondence to Mahdi Sajadieh.\n\n## Additional information\n\nThis paper was solicited by the Editors-in-Chief as one of the best papers from FSE 2012, based on the recommendation of the program committee.\n\n## Rights and permissions\n\nReprints and Permissions\n\n## About this article\n\n### Cite this article\n\nSajadieh, M., Dakhilalian, M., Mala, H. et al. Efficient Recursive Diffusion Layers for Block Ciphers and Hash Functions. J Cryptol 28, 240–256 (2015). https://doi.org/10.1007/s00145-013-9163-8\n\nDownload citation\n\n• Received:\n\n• Published:\n\n• Issue Date:\n\n### Key words\n\n• Block ciphers\n• Diffusion layer\n• Branch number\n• MDS matrix" ]

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[ "# #09f65c Color Information\n\nIn a RGB color space, hex #09f65c is composed of 3.5% red, 96.5% green and 36.1% blue. Whereas in a CMYK color space, it is composed of 96.3% cyan, 0% magenta, 62.6% yellow and 3.5% black. It has a hue angle of 141 degrees, a saturation of 92.9% and a lightness of 50%. #09f65c color hex could be obtained by blending #12ffb8 with #00ed00. Closest websafe color is: #00ff66.\n\n• R 4\n• G 96\n• B 36\nRGB color chart\n• C 96\n• M 0\n• Y 63\n• K 4\nCMYK color chart\n\n#09f65c color description : Vivid cyan - lime green.\n\n# #09f65c Color Conversion\n\nThe hexadecimal color #09f65c has RGB values of R:9, G:246, B:92 and CMYK values of C:0.96, M:0, Y:0.63, K:0.04. Its decimal value is 652892.\n\nHex triplet RGB Decimal 09f65c `#09f65c` 9, 246, 92 `rgb(9,246,92)` 3.5, 96.5, 36.1 `rgb(3.5%,96.5%,36.1%)` 96, 0, 63, 4 141°, 92.9, 50 `hsl(141,92.9%,50%)` 141°, 96.3, 96.5 00ff66 `#00ff66`\nCIE-LAB 85.372, -78.572, 58.929 34.998, 66.738, 21.162 0.285, 0.543, 66.738 85.372, 98.215, 143.13 85.372, -78.269, 86.482 81.694, -66.493, 41.827 00001001, 11110110, 01011100\n\n# Color Schemes with #09f65c\n\n• #09f65c\n``#09f65c` `rgb(9,246,92)``\n• #f609a3\n``#f609a3` `rgb(246,9,163)``\nComplementary Color\n• #2cf609\n``#2cf609` `rgb(44,246,9)``\n• #09f65c\n``#09f65c` `rgb(9,246,92)``\n• #09f6d3\n``#09f6d3` `rgb(9,246,211)``\nAnalogous Color\n• #f6092c\n``#f6092c` `rgb(246,9,44)``\n• #09f65c\n``#09f65c` `rgb(9,246,92)``\n• #d309f6\n``#d309f6` `rgb(211,9,246)``\nSplit Complementary Color\n• #f65c09\n``#f65c09` `rgb(246,92,9)``\n• #09f65c\n``#09f65c` `rgb(9,246,92)``\n• #5c09f6\n``#5c09f6` `rgb(92,9,246)``\nTriadic Color\n• #a3f609\n``#a3f609` `rgb(163,246,9)``\n• #09f65c\n``#09f65c` `rgb(9,246,92)``\n• #5c09f6\n``#5c09f6` `rgb(92,9,246)``\n• #f609a3\n``#f609a3` `rgb(246,9,163)``\nTetradic Color\n• #06ac40\n``#06ac40` `rgb(6,172,64)``\n• #07c54a\n``#07c54a` `rgb(7,197,74)``\n• #08dd53\n``#08dd53` `rgb(8,221,83)``\n• #09f65c\n``#09f65c` `rgb(9,246,92)``\n• #22f76c\n``#22f76c` `rgb(34,247,108)``\n• #3af87d\n``#3af87d` `rgb(58,248,125)``\n• #53f98d\n``#53f98d` `rgb(83,249,141)``\nMonochromatic Color\n\n# Alternatives to #09f65c\n\nBelow, you can see some colors close to #09f65c. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #09f621\n``#09f621` `rgb(9,246,33)``\n• #09f635\n``#09f635` `rgb(9,246,53)``\n• #09f648\n``#09f648` `rgb(9,246,72)``\n• #09f65c\n``#09f65c` `rgb(9,246,92)``\n• #09f670\n``#09f670` `rgb(9,246,112)``\n• #09f684\n``#09f684` `rgb(9,246,132)``\n• #09f697\n``#09f697` `rgb(9,246,151)``\nSimilar Colors\n\n# #09f65c Preview\n\nText with hexadecimal color #09f65c\n\nThis text has a font color of #09f65c.\n\n``<span style=\"color:#09f65c;\">Text here</span>``\n#09f65c background color\n\nThis paragraph has a background color of #09f65c.\n\n``<p style=\"background-color:#09f65c;\">Content here</p>``\n#09f65c border color\n\nThis element has a border color of #09f65c.\n\n``<div style=\"border:1px solid #09f65c;\">Content here</div>``\nCSS codes\n``.text {color:#09f65c;}``\n``.background {background-color:#09f65c;}``\n``.border {border:1px solid #09f65c;}``\n\n# Shades and Tints of #09f65c\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #011307 is the darkest color, while #ffffff is the lightest one.\n\n• #011307\n``#011307` `rgb(1,19,7)``\n• #01260e\n``#01260e` `rgb(1,38,14)``\n• #023915\n``#023915` `rgb(2,57,21)``\n• #034c1c\n``#034c1c` `rgb(3,76,28)``\n• #035f23\n``#035f23` `rgb(3,95,35)``\n• #04722a\n``#04722a` `rgb(4,114,42)``\n• #058432\n``#058432` `rgb(5,132,50)``\n• #069739\n``#069739` `rgb(6,151,57)``\n• #06aa40\n``#06aa40` `rgb(6,170,64)``\n• #07bd47\n``#07bd47` `rgb(7,189,71)``\n• #08d04e\n``#08d04e` `rgb(8,208,78)``\n• #08e355\n``#08e355` `rgb(8,227,85)``\n• #09f65c\n``#09f65c` `rgb(9,246,92)``\nShade Color Variation\n• #1cf769\n``#1cf769` `rgb(28,247,105)``\n• #2ff775\n``#2ff775` `rgb(47,247,117)``\n• #42f882\n``#42f882` `rgb(66,248,130)``\n• #55f98e\n``#55f98e` `rgb(85,249,142)``\n• #68f99b\n``#68f99b` `rgb(104,249,155)``\n• #7bfaa7\n``#7bfaa7` `rgb(123,250,167)``\n• #8dfbb4\n``#8dfbb4` `rgb(141,251,180)``\n• #a0fcc0\n``#a0fcc0` `rgb(160,252,192)``\n• #b3fccd\n``#b3fccd` `rgb(179,252,205)``\n• #c6fdd9\n``#c6fdd9` `rgb(198,253,217)``\n• #d9fee6\n``#d9fee6` `rgb(217,254,230)``\n• #ecfef2\n``#ecfef2` `rgb(236,254,242)``\n• #ffffff\n``#ffffff` `rgb(255,255,255)``\nTint Color Variation\n\n# Tones of #09f65c\n\nA tone is produced by adding gray to any pure hue. In this case, #7f807f is the less saturated color, while #09f65c is the most saturated one.\n\n• #7f807f\n``#7f807f` `rgb(127,128,127)``\n• #758a7c\n``#758a7c` `rgb(117,138,124)``\n• #6b9479\n``#6b9479` `rgb(107,148,121)``\n• #619e76\n``#619e76` `rgb(97,158,118)``\n• #57a874\n``#57a874` `rgb(87,168,116)``\n• #4eb171\n``#4eb171` `rgb(78,177,113)``\n• #44bb6e\n``#44bb6e` `rgb(68,187,110)``\n• #3ac56b\n``#3ac56b` `rgb(58,197,107)``\n• #30cf68\n``#30cf68` `rgb(48,207,104)``\n• #26d965\n``#26d965` `rgb(38,217,101)``\n• #1de262\n``#1de262` `rgb(29,226,98)``\n• #13ec5f\n``#13ec5f` `rgb(19,236,95)``\n• #09f65c\n``#09f65c` `rgb(9,246,92)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #09f65c is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "+0\n\n# I need help!\n\n0\n181\n2\n\nThree fair, six-sided dice are tossed. What is the probability that the product of the three rolls is 90?\n\nSep 2, 2021\n\n#1\n0\n\nLet's first find the ways that we could get products of 90.\n\nOne of the obvious first choices would be 6, as it's a factor of 90 and reduces it the most greatly. Let's start with it.\n\n90/6 = 20\n\nNow we can factor 20 into 5 and 4, so we now know that the only valid combination is 6-5-4.\n\nWe need the total number of permutations where we get a 6-5-4. In this case, our set is {6 5 4} and we are picking 3. This could be written as \"3 pick 3\", which happens to be equal to 3!. (3 factorial, not just an excited 3 :) )\n\nNow let's put this over the total number of perumations of dice:\n\n$$\\frac{3!}{6^3} = \\frac{6}{216} = \\frac{1}{36}$$\n\nThus, the chances of the dice having a product of 90 is 1/36.\n\nSep 2, 2021\n#2\n0\n\nGood technique, but there's a little hiccup:  90 / 6 = 15 not 20.  You get the same answer, though.\n\n.\n\nGuest Sep 2, 2021" ]

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[ "### Merge branch 'master' of https://gitlab.mpi-sws.org/iris/stdpp\n\nparents 705bb406 0b3d1e9f\nPipeline #19183 failed with stage\nin 0 seconds\n This file lists \"large-ish\" changes to the std++ Coq library, but not every API-breaking change is listed. ## std++ 1.2.1 (unreleased) This release of std++ received contributions by Michael Sammler, Paolo G. Giarrusso, Paulo Emílio de Vilhena, Ralf Jung, Robbert Krebbers, Rodolphe Lepigre, and Simon Spies. Noteworthy additions and changes: - Introduce `max` and `min` infix notations for `N` and `Z` like we have for `nat`. - Make `solve_ndisj` tactic more powerful. - Add type class `Involutive`. - Improve `naive_solver` performance in case the goal is trivially solvable. - Add a bunch of new lemmas for list, set, and map operations. - Rename `lookup_imap` into `map_lookup_imap`. ## std++ 1.2.0 (released 2019-04-26) Coq 8.9 is supported by this release, but Coq 8.6 is no longer supported. Use ... ...\n ... ... @@ -94,6 +94,9 @@ Proof. unfold bool_decide. destruct dec; [left|right]; assumption. Qed. Lemma bool_decide_decide P `{!Decision P} : bool_decide P = if decide P then true else false. Proof. reflexivity. Qed. Lemma decide_bool_decide P {Hdec: Decision P} {X : Type} (x1 x2 : X): (if decide P then x1 else x2) = (if bool_decide P then x1 else x2). Proof. unfold bool_decide, decide. destruct Hdec; reflexivity. Qed. Tactic Notation \"case_bool_decide\" \"as\" ident (Hd) := match goal with ... ...\n ... ... @@ -860,7 +860,7 @@ Proof. Defined. (** Properties of the imap function *) Lemma lookup_imap {A B} (f : K → A → option B) (m : M A) i : Lemma map_lookup_imap {A B} (f : K → A → option B) (m : M A) i : map_imap f m !! i = m !! i ≫= f i. Proof. unfold map_imap; destruct (m !! i ≫= f i) as [y|] eqn:Hi; simpl. ... ... @@ -876,6 +876,39 @@ Proof. rewrite elem_of_map_to_list in Hj; simplify_option_eq. Qed. Lemma map_imap_Some {A} (m : M A) : map_imap (λ _, Some) m = m. Proof. apply map_eq. intros i. rewrite map_lookup_imap. by destruct (m !! i). Qed. Lemma map_imap_insert {A B} (f : K → A → option B) (i : K) (v : A) (m : M A) : map_imap f (<[i:=v]> m) = match f i v with | None => delete i (map_imap f m) | Some w => <[i:=w]> (map_imap f m) end. Proof. destruct (f i v) as [w|] eqn:Hw. - apply map_eq. intros k. rewrite map_lookup_imap. destruct (decide (k = i)) as [->|Hk_not_i]. + by rewrite lookup_insert, lookup_insert. + rewrite !lookup_insert_ne by done. by rewrite map_lookup_imap. - apply map_eq. intros k. rewrite map_lookup_imap. destruct (decide (k = i)) as [->|Hk_not_i]. + by rewrite lookup_insert, lookup_delete. + rewrite lookup_insert_ne, lookup_delete_ne by done. by rewrite map_lookup_imap. Qed. Lemma map_imap_ext {A1 A2 B} (f1 : K → A1 → option B) (f2 : K → A2 → option B) (m1 : M A1) (m2 : M A2) : (∀ k, f1 k <\\$> (m1 !! k) = f2 k <\\$> (m2 !! k)) → map_imap f1 m1 = map_imap f2 m2. Proof. intros HExt. apply map_eq. intros i. rewrite !map_lookup_imap. specialize (HExt i). destruct (m1 !! i), (m2 !! i); naive_solver. Qed. (** ** Properties of the size operation *) Lemma map_size_empty {A} : size (∅ : M A) = 0. Proof. unfold size, map_size. by rewrite map_to_list_empty. Qed. ... ... @@ -1081,13 +1114,18 @@ Section map_filter. naive_solver. Qed. Lemma map_filter_insert_not m i x : (∀ y, ¬ P (i, y)) → filter P (<[i:=x]> m) = filter P m. Lemma map_filter_insert_not' m i x : ¬ P (i, x) → (∀ y, m !! i = Some y → ¬ P (i, y)) → filter P (<[i:=x]> m) = filter P m. Proof. intros HP. apply map_filter_lookup_eq. intros j y Hy. by rewrite lookup_insert_ne by naive_solver. intros Hx HP. apply map_filter_lookup_eq. intros j y Hy. rewrite lookup_insert_Some. naive_solver. Qed. Lemma map_filter_insert_not m i x : (∀ y, ¬ P (i, y)) → filter P (<[i:=x]> m) = filter P m. Proof. intros HP. by apply map_filter_insert_not'. Qed. Lemma map_filter_delete m i : filter P (delete i m) = delete i (filter P m). Proof. apply map_eq. intros j. apply option_eq; intros y. ... ... @@ -1106,6 +1144,16 @@ Section map_filter. Lemma map_filter_empty : filter P (∅ : M A) = ∅. Proof. apply map_fold_empty. Qed. Lemma map_filter_alt m : filter P m = list_to_map (filter P (map_to_list m)). Proof. apply list_to_map_flip. induction m as [|k x m ? IH] using map_ind. { by rewrite map_to_list_empty, map_filter_empty, map_to_list_empty. } rewrite map_to_list_insert, filter_cons by done. destruct (decide (P _)). - rewrite map_filter_insert by done. by rewrite map_to_list_insert, IH by (rewrite map_filter_lookup_None; auto). - by rewrite map_filter_insert_not' by naive_solver. Qed. End map_filter. (** ** Properties of the [map_Forall] predicate *) ... ...\n ... ... @@ -279,6 +279,13 @@ Section gset. - by rewrite option_guard_True by (rewrite elem_of_dom; eauto). - by rewrite option_guard_False by (rewrite not_elem_of_dom; eauto). Qed. Lemma dom_gset_to_gmap {A} (X : gset K) (x : A) : dom _ (gset_to_gmap x X) = X. Proof. induction X as [| y X not_in IH] using set_ind_L. - rewrite gset_to_gmap_empty, dom_empty_L; done. - rewrite gset_to_gmap_union_singleton, dom_insert_L, IH; done. Qed. End gset. Typeclasses Opaque gset.\nThis diff is collapsed.\n ... ... @@ -50,6 +50,26 @@ Inductive TlRel {A} (R : relation A) (a : A) : list A → Prop := Section sorted. Context {A} (R : relation A). Lemma elem_of_StronglySorted_app l1 l2 x1 x2 : StronglySorted R (l1 ++ l2) → x1 ∈ l1 → x2 ∈ l2 → R x1 x2. Proof. induction l1 as [|x1' l1 IH]; simpl; [by rewrite elem_of_nil|]. intros [? Hall]%StronglySorted_inv [->|?]%elem_of_cons ?; [|by auto]. rewrite Forall_app, !Forall_forall in Hall. naive_solver. Qed. Lemma StronglySorted_app_inv_l l1 l2 : StronglySorted R (l1 ++ l2) → StronglySorted R l1. Proof. induction l1 as [|x1' l1 IH]; simpl; [|inversion_clear 1]; decompose_Forall; constructor; auto. Qed. Lemma StronglySorted_app_inv_r l1 l2 : StronglySorted R (l1 ++ l2) → StronglySorted R l2. Proof. induction l1 as [|x1' l1 IH]; simpl; [|inversion_clear 1]; decompose_Forall; auto. Qed. Lemma Sorted_StronglySorted `{!Transitive R} l : Sorted R l → StronglySorted R l. Proof. by apply Sorted.Sorted_StronglySorted. Qed. ... ...\nMarkdown is supported\n0% or\nYou are about to add 0 people to the discussion. Proceed with caution.\nFinish editing this message first!" ]

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[ "# Spectrum of a Subspace of Matrices\n\nI conjecture the following:\n\nLet $$U \\subset \\Bbb C^{n \\times n}$$ be an affine subspace, and let $$S_U$$ denote the \"spectrum of $$U$$\", that is $$S_U = \\{\\lambda \\in \\Bbb C : \\det(A - \\lambda I) = 0 \\text{ for some } A \\in U\\}.$$ Then either all elements of $$U$$ have an identical spectrum, or $$S_U = \\Bbb C$$.\n\nIs this correct? Some simple examples of each case: for any fixed $$\\lambda_i$$, $$U_1 = \\left\\{\\pmatrix{\\lambda_1&t\\\\0&\\lambda_2}: t \\in \\Bbb C\\right\\}, \\quad U_2 = \\left\\{\\pmatrix{\\lambda_1&0\\\\0&t}: t \\in \\Bbb C\\right\\}.$$ Clearly, $$S_{U_1} = \\{\\lambda_1,\\lambda_2\\}$$ and $$S_{U_2} = \\Bbb C$$. Are there any other possibilities? I suspect that there is a quick algebraic-geometry approach here that I am missing.\n\nAn aside: I would also be interested in the case of real affine subspaces of $$\\Bbb C^{n \\times n}$$, if anyone has insights on what the possibilities are there. Note that the symmetric real matrices are an example of a subspace where we attain $$S_U^{(\\Bbb R)} = \\Bbb R \\subsetneq \\Bbb C$$. We can clearly attain any \"line\" in $$\\Bbb C$$, but I wonder if there are other possibilities.\n\nTo expand on Christian Remling's answer a bit: in his example, setting $$det(A(t) - \\lambda) = 0$$ becomes $$(1-\\lambda)t+\\lambda^3=0$$, and solving for $$t$$ gives $$t = x^3/(x-1)$$ -- so for any $$\\lambda \\in \\mathbb{C}$$, we have $$x \\in S$$ by choosing this $$t$$, with the exception of $$\\lambda=1$$.\n\nIn general for an affine subspace of dimension $$d$$, we can define the space by $$A(t_1,t_2,\\dots t_d)$$, and all entries in $$A$$ are linear in the $$t_i$$. Then $$\\det(A(t) - \\lambda)$$ is an $$n$$th degree polynomial in all the $$t_i$$ and $$\\lambda$$. It is possible that the determinant has no dependence on any $$t_i$$, in which case we have a finite spectrum. Otherwise, we have a dependence on the $$t_i$$. Then $$\\det(A(t)-\\lambda)=0$$ has a solution in $$t_1$$ whenever at least one of the non-constant coefficients is nonzero. (In the above example, as long as the linear coefficient $$1-\\lambda\\neq 0$$.) Since the leading coefficient (that is not a constant 0 polynomial) is an $$n-1$$th degree polynomial in $$\\lambda$$, it is zero at a most $$n-1$$ different places, and this is at most $$n-1$$ places that the spectrum can fail to be continuous. The spectrum only fails to be continuous when all of the coefficients in the $$t_1$$ polynomial are zero, so this is an upper bound. So, theorem:\n\nFor an affine subspace of $$n\\times n$$ matrices, the spectrum is either finite (of size at most $$n$$), or it is $$\\mathbb{C}\\setminus K$$, where $$K$$ is a finite set of exceptions (of size at most $$n-1$$).\n\n• Excellent! Thank you for a clear and thorough answer. Sep 19, 2019 at 3:01\n\nThis is not true. Consider $$A(t) = \\begin{pmatrix} 0 & t & 0 \\\\ 1 & 0 & 1 \\\\ 1 & 0 & 0 \\end{pmatrix} .$$ Then $$\\det (A(t)+1) = 1$$, so $$-1\\notin S$$, but clearly the spectrum of $$A(t)$$ is not constant (for example, $$0$$ is in the spectrum if and only if $$t=0$$).\n\n• Thank you for the example, I guess sometimes $2\\times 2$ isn't big enough :P Sep 19, 2019 at 3:06\n• Actually @Omnomnomnom, $2\\times 2$ can be enough. Consider [[t 1] [1 0]], which spectrum $\\mathbb{C}\\setminus\\{0\\}$. :) For any t, having an eigenvalue of 0 would imply a determinant of 0, but the determinant of that matrix is always -1. Sep 19, 2019 at 8:29\n• @AlexMeiburg Nice one, guess I really had no excuse here then Sep 19, 2019 at 8:34" ]

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[ "In this article, I will discuss and summarize the paper: “Structural Analysis of Criminal Network and Predicting Hidden Links using Machine Learning” by Emrah Budur, Seungmin Lee and Vein S Kong.\n\n# Background\n\nAs in most criminal network projects, data is key. However, data is not always available, and if so, only in limited amount. Therefore, link prediction can make sense from that perspective. In this paper, authors gathered large amount of data and turned link prediction in graphs (which is a hard task) into binary classification problems (which is much easier).\n\n# Data\n\nThe authors gathered a dataset of 1.5 millions nodes worldwide with 4 millions undirected edges from the Office of Foreign Asset Control. I think that they refer to a dataset available here.\n\nWith such amount of data, the authors propose to turn a link prediction (predict links between times T and T+1), to finding hidden links by a binary classifier (at time T). There are several steps in the project:\n\n• train a Gradient Boosting Model (supervised learning approach) on the current data for finding hidden links\n• remove 5 to 50% of the edges to create a test set and compute the accuracy of the model\n• distroy criminal networks using Weighted Pagerank index\n\nTo treat edge prediction as a classification problem, one should have features on each node and learn from them. The authors use as features:\n\n• the number of common neighbors\n• the jaccard index\n• the preferential attachment index\n• the hub index\n• the Leicht Holme index\n• the Salton index\n• the Sorensen index", null, "They build a training dataset (75%), and a test one in which they remove edges. To build the feature matrix, they compute combinations between all possible nodes. Therefore, they end up with more negative edges ($$E_{-}$$) than positive edges ($$E_{+}$$). To balance the training set, they apply a random undersampling on the negative edges.\n\n# Model and performance\n\nThe metric chosen is the prediction imbalance:\n\n$Pred_{Imbalance} = \\mid err_{+} − err_{-} \\mid$\n\nThe authors also compute the Area under the curve (AUC) criteria. They gradually remove from 5% up to 50% of the edges, and the findings are quite interesting:", null, "The light gray line shows the performance of the model when we build a train and a test set from a corrupted network. The more edges were removed before splitting in $$D_{train}$$ and $$D_{test}$$, the lower the model performance.\n\nThe dark line is the ability of the built model to find hidden links. And surprisingly, AUC increases with the proportion of links removed. Authors argue that it comes from the fact that the model might overfit on the structure of the graph when given too many links in training, and that removing some edges during training (typically anywhere close to 25-30%) could improve the model.\n\n# Destroying networks\n\nIn the paper “Disrupting Resilient criminal networks through data analysis”, authors identify the nodes with the highest betweenness centrality and remove them gradually in order to disrupt the network.\n\nIn this paper, authors argue that removing nodes with the highest pagerank score first improved the disruption of the network.\n\nThe data does not provide weights on edges. These weights are added by making the model predict the existing edges. A score closer to 1 will give a weight closer to 1, and a score closer to 0 will give a weight closer to 0.\n\nWeighted pagerank scores are then computed in the network, for each node. The index computed by the pagerank algorithm builds a “suspiciousness index”, and more suspicious nodes should be removed first. The authors compared several node removal strategies (Unweighted pagerank, weighted pagerank, node degree, jaccard index…).", null, "The removal of nodes having maximum Weighted PageRank score reduces the largest connected component (LCC) of the network much faster than removal of nodes based on any other metric, except for unweighted pagerank (no clear difference).\n\nAll pagerank methods reach a bottleneck after some time, since the pagerank is computed only once, and not after each iteration. Therefore, 2 options:\n\n• re-compute pagerank at each step, which can get quite expensive in terms of computation\n• use the hybrid method proposed by the authors to maximize the WCC score: parameters $$W_{hybrid} ≈ min(W_{Weighted Pagerank}, W_{Attack})$$\n\nThe first approach was not implemented by authors, but the second one seems to improve the percentage of the network removed where the pagerank reached a bottleneck.", null, "# Discussion\n\nThe work discussed in this paper is quite interesting by the volume of the data considered. A supervised machine learning model could be trained, and undersampling seemed to help. One should try other models and other features in future works.\n\nThe dataset, although massive, lacks the notion of weights and does not have a temporal notion, it’s only a snapshot.\n\nCategories:\n\nUpdated:" ]

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[ "###### Excel NPER function\n\nThe Excel NPER function calculates the number of periods, or amount of time, needed to payoff a loan or reach an investment objective.\n\n###### Excel IRR function\n\nThe Excel IRR function calculates the internal rate of return of an investment based upon a series of cashflows\n\n###### Excel NPV function\n\nThe Excel NPV function calculates the net present value of an investment based upon a discount rate and a series of cashflows.\n\n###### Loan Payment Calculator\n\nThis free easy to use Excel loan payment calculator for download will calculate loan payments with different terms.\n\n###### Excel FV function\n\nThe Excel FV function calculates the future value (e.g., the remaining amount) of a loan or investment at a point in time in the future.\n\n###### Excel PV function\n\nThe Excel PV function calculates the present value (or original amount) of a loan or investment.\n\n###### Excel RATE function\n\nThe Excel RATE function calculates the interest rate per period for a loan or investment.\n\n###### Excel PMT function\n\nThe Excel PMT function calculates the payment for a loan with a fixed interest rate. The function also assumes constant payments will be made.\n\nEnd of content" ]

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[ "# Paper Review: Polygonal Approximation of Digital Curves to Preserve Original Shapes\n\n## Keywords:\n\n• dominant points\n• consecutive vectors\n• toothbrush shape\n• distance metric\n• smallest perpendicular distance\n\n## Q1: How usual calculation of distance is done?\n\nMinor DPs are deleted in approximation. A minor DP is a DP where the perpendicular distance between the point and the straight line is minimum.\n\na a\nb\n\n\nHere b is deleted when its distance to the line a-a is minimum.\n\nThe perpendicular distance is calculated using\n\n[ d\\i = \\sqrt{\\frac{((xi - xa) (y:sub:b - ya) - (y:sub:i - ya) (x:sub:b - xa))2}{(xa - xb)2 + (y:sub:a - yb)2}} ]\n\nfor lines between points pa and pb and the point pi.\n\n## Q2: What is a toothbrush shape?\n\nIt's something like\n\naaaaaa\nbbbbbbbbbbbbbbbbbbbbbb\n\n\nHence the toothbrush.\n\nThough I don't get why is this particularly important.\n\n## Q3: Which information is included in distance metric?\n\nAngle acuteness is added to the information described above.\n\n## Q4: How the distance metric differs from others?\n\nIt includes angle acuteness in the metric and the more acute the angle, the less likely it's removed from DP set.\n\n## Q5: What's baseline for performance and how does this improve it?\n\nAs the number of DPs decrease, RMSE of the new metric decrease. For large number of DPs it doesn't matter much. (So performance penalty may not payoff)" ]

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[ "# Fluid Mechanics: Cone-Plate viscometer\n\n## Homework Statement", null, "Fluid Mechanics\n\nτ =μ (du/dy)\n\n## The Attempt at a Solution\n\nI got far enough to write down\n\ndM = μ (Ωr/tanθ) dA\n\nfrom just substitutions, easy enough.\n\nI get confused when I'm solving for the differential surface area. I somehow need the dA for a cone.\nFor a circle it's easy enough. A=πr^2 so then dA = 2πr dr.\n\nBut how do I go about getting this for a cone? the book lists it as (2πr/cosθ) dr, without any explanation, of course.\n\nTake a strip element 2πxdr, where x is the radius of circle formed by that strip element. Try relating x to r using similarity of triangles.\n\nNathanael\nHomework Helper\nWell you should first understand how to find the surface area of a \"truncated cone\" (ignoring the faces)\nIt is the average circumference of the cone multiplied by the side length (not the height)\n\nSo the surface area of a truncated cone (ignoring the faces) is $2\\pi r_{avg}L$ or you could say $2\\pi L\\frac{r_1+r_2}{2}$\n(Side note, this also applies to normal cones; just treat the tip as r2=0)\n\nSo with this understanding of truncated cones, look at the following picture I made:", null, "The differential surface area would be $2\\pi r_{avg}L$ but as you can see from the picture, $L=\\frac{dr}{\\cos\\theta}$ therefore the differential surface area is $\\frac{2\\pi R_{avg}}{\\cos\\theta}dr$\n\nThe two radii are r and (r+dr) but since dr is obviously infinitesimal, it suffices to say $r_{avg}=r$\n\nLast edited:\n•", null, "Feodalherren\nDoes it really require all that? Isn't using similarity to relate x and R, considering a string element of radius x, just enough?\n\nNathanael\nHomework Helper\nDoes it really require all that? Isn't using similarity to relate x and R, considering a string element of radius x, just enough?\nNo that is incorrect. The differential area element is $\\frac{2\\pi x}{\\cos\\theta}dx$\n\nMe and the OP were using r not to represent the radius of the base of the cone, but to represent what you called \"x\"\n\nNo that is incorrect. The differential area element is $\\frac{2\\pi x}{\\cos\\theta}dx$\n\nMe and the OP were using r not to represent the radius of the base of the cone, but to represent what you called \"x\"\n\nOh ok.\n\nWell you should first understand how to find the surface area of a \"truncated cone\" (ignoring the faces)\nIt is the average circumference of the cone multiplied by the side length (not the height)\n\nSo the surface area of a truncated cone (ignoring the faces) is $2\\pi r_{avg}L$ or you could say $2\\pi L\\frac{r_1+r_2}{2}$\n(Side note, this also applies to normal cones; just treat the tip as r2=0)\n\nSo with this understanding of truncated cones, look at the following picture I made:\nView attachment 78118\nThe differential surface area would be $2\\pi r_{avg}L$ but as you can see from the picture, $L=\\frac{dr}{\\cos\\theta}$ therefore the differential surface area is $\\frac{2\\pi R_{avg}}{\\cos\\theta}dr$\n\nThe two radii are r and (r+dr) but since dr is obviously infinitesimal, it is suffices to say $r_{avg}=r$\nExcellent explanation. Thank you very much, Sir." ]

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[ "A Practical Guide to Isotherms of ADSORPTION on Heterogeneous Surfaces\n\nA Practical Guide to\non Heterogeneous Surfaces\nby\n\n## An Overview\n\n### Preface\n\nThis WWW Guide reflects my own views on adsorption on heterogeneous surfaces and is \"phenomenologically oriented\". It is based mainly upon my own research (my papers, my PhD and unpublished results) as well as some papers/views of my co-workers (those from the past and current ones), generally agreed-upon views of an informal group of researchers and results available in other sources, e.g. books.\nMost of the text, equations and figures deal with adsorption from dilute solutions or gas adsorption. Where possible, the possibility of using equations formulated for gas (or vapour) adsorption in adsorption from dilute solutions - and vice versa is pointed out. The general approach to adsorption from liquid mixtures is also presented.\nThis Overview page presents a summary of problems this Guide tries to explain or help to solve in a simple, practical way. Each of the listed problems is shortly explained and more is available through series of hyperlinks. Most (though not all of the sub-pages) may be accessed directly by using links in the index (left frame). All external links (to other sites) within this WWW site as a default are opened in new windows. Internal links within right frame are opened by default in the same frame. Many of the sub-pages may be reached in several ways.\n\nGeneral Integral Equation / GL (Generalized Langmuir) / All equations (preview)\nAdsorption type ( Linear Langmuir plot / Graham plot / Consistency / Henry constant )\nPopular isotherms ( Mono-, Multilayer, Experimental, Micro-, Mesoporous )\nData analysis: ( LSq data fitting / Heterogeneity: Global , σE / Linear plots / φ-function / Pores )\nHeterogeneity and Molecular Size ( Theory and Prediction / Simple binary isotherm )\n\nHow to distinguish a type of isotherm your adsorption data (gas, dilute solution) fits?\n\n• Physical consistency\nPhysical consistency conditions require:\n• minimum and maximum adsorption energies\n• isotherm with Henry region (results from max. energy condition)\n\n•", null, "Henry constant\nHenry constant KH is defined as:\nKH = limp→0(KG)    or    KH = limc→0(KG)\nwhere:   KG = [θ/(1-θ)]/p   (gas adsorption) or   KG = [θ/(1-θ)]/c   (dilute solute adsorption) is Graham's equilibrium function .\nThis condition is in fact equivalent to:\nlimp→0(φ) = 1    or    limc→0(φ) = 1\nwhere:   φ = ∂ log(a)/ ∂ log(p)   or   φ = ∂ log(a)/ ∂ log(c)   is so-called φ-function.\nIt is usually believed, that the very existence of such a limit is a consequence of the existing maximum adsorption energy and is sometimes called a \"physical consistency condition\". However as it is easy to prove that it is enough that the energy distribution function defined in energy range (-∞ , +∞) behaves in a special way and such a limit is obtained (e.g. for Toth, RP or Gauss distribution-derived isotherm equations).", null, "Data analysis - dispersion of adsorption energy σE:\n\n• Calculation of adsorption energy dispersion σE allows to compare heterogeneities of various adsorption systems. This single value does not contain any information on energy distribution symmetry/asymmetry (in this aspect similar to global heterogeneity H), however if adsorption systems are similar and we may expect similar character of their energy distribution functions we may predict (or at least estimate) the behaviour of a mixed system.\n1. \"Unified Theoretical Description of Physical Adsorption from Gaseous and Liquid Phases on Heterogeneous Solid Surfaces and Its Application for Predicting Multicomponent Adsorption Equilibria\", A.W.Marczewski, A.Derylo-Marczewska and M.Jaroniec, Chemica Scripta, 28, 173-184 (1988) (pdf, hi-res pdf available upon e-mail request).\nQuite often, some adsorption data may be well described by more than 1 isotherm equation. If underlying energy distributions are of similar character (e.g. both are symmetrical quasi-gaussian), we may try to estimate parameters of another eqn. by using already determined parameters.\n1. \"Relationships Defining Dependence Between Adsorption Parameters of Dubinin-Astakhov and Generalized Langmuir Equations\", M.Jaroniec and A.W.Marczewski, J.Colloid Interface Sci., 101, 280-281 (1984), (doi).\n2. \"Correlations Among the Parameters of Dubinin-Radushkevich and Langmuir-Freundlich Isotherms for Adsorption from Binary Liquid Mixtures\", A.Derylo-Marczewska, M.Jaroniec, J.Oscik and A.W.Marczewski, J.Colloid Interface Sci., 117, 339-346 (1987), (doi).\n\n• For some isotherm equations the energy dispersion σE may be calculated very easily (see here).", null, "", null, "Data analysis - global heterogeneity/non-ideality:\nNOTE. Monolayer adsorption only - if multilayer effects are visible, the isotherm must be \"monolayerized\", i.e. multilayer part must be estimated and removed.\n\n•", null, "Global heteorgeneity, H, concept allows to estimate a single value characterizing the entire non-ideality of the adsorption system (i.e. both adsorbate and adsorbent). By using additional adsorption data measured on homogeneous surface the lateral interaction part of the non-ideality (Hint may be separated) and a part of non-ideality related to the system energetic heterogeneity may be determined (H = HE - Hint).\n\"A New Method for Characterizing the Global Adsorbent Heterogeneity by Using the Adsorption Data\", A.W.Marczewski, M.Jaroniec and A.Derylo-Marczewska, Mat.Chem.Phys., 14, 141-166 (1986), (doi)).\n• For several isotherm equations H may be calculated very easily (see here).\n\nData analysis - isotherm-specific methods:\n\n•", null, "Try simple linear dependencies using only experimental adsorption and concentration (or their functions, e.g. logarithms) - methods are equation-specific and you must decide what type of equation should be checked (Here are model pictures):\n\nPresented linear dependencies include: L, LF, BET, F, DR and DA isotherms.\n\nNOTE 1. Some of the methods may require adsorption monolayer (adsorption capacity), am or e.g. characteristic micropore filling concentration, co, to be know in advance (estimated by independent method or LSQ-fitted - in this case, linear plot is only a verification).\nNOTE 2. Replace concentration, c, by pressure, p, for gas adsorption. In dilute solute adsorption x=c/cs and in vapour adsorption x=p/ps, where index \"s\" indicates saturation concentration or pressure, respectively.\n\n•", null, "Use a so-called: φ-function method (where φ = ∂ log(a)/ ∂ log(c)) (derivative of adsorption isotherm in logarithmic co-ordinates). φ-function is a dimensionless (independent of adsorption and concentration/pressure units as well as independent of type of logarithm you use). Various plots of φ-function versus adsorption, pressure etc. allow to determine isotherm type and isotherm parameters (or sometimes at least to reject some existing choices) (Here are model pictures)\nPresented methods include analysis of: RP, Jossens, GL (partially), F, DR and DA isotherms.\nNOTE. This method allows to determine equation type and parameters very precisely, however your data should be very good - evenly spaced, smooth and in a wide range of relative adsorption φ) - method is sensitive to data scatter.\n\n• Isotherm analysis - separation of micropore and mesopore effects\n(Check here for isotherms: Micropores and Mesopores )\nMethods used for this purpose require the adsorption data of N2 (or sometimes C6H6) and include: t-plot (de Boer) (with HJ or Halsey/FHH isotherms), t/F-plot (Kadlec) (with DR isotherm) as well as αs-method (Singh).\n\n•", null, "Prediction of multicomponent adsorption for components with non-linearly correlated adsorption energy distributions. (Here are model pictures)\n\nProposed method assumes that a certain correlation (generally non-linear) exists between adsorption energies of various adsorbates on a solid surface. Generally no limit of no. of components exists. Prediction requires data for simple (single-component or binary) systems.\n1. \"Unified Theoretical Description of Physical Adsorption from Gaseous and Liquid Phases on Heterogeneous Solid Surfaces and Its Application for Predicting Multicomponent Adsorption Equilibria\", A.W.Marczewski, A.Derylo-Marczewska and M.Jaroniec, Chemica Scripta, 28, 173-184 (1988) (pdf, hi-res pdf available upon e-mail request).\nThis method may predict adsorption (or at least is able to estimate the magnitude of heterogeneity effects) in many situations:\n• Prediction of adsorption in gas mixtures (required: single gas adsorption data for all components; useful: binary gas mixture data - if e.g. prediction of tertiary-mixture adsorption is wanted; quality of prediction - very good).\n• Prediction of adsorption in multi-component dilute solutions (required: single solute adsorption data for all components; useful: binary dilute solute adsorption data - if e.g. prediction of tertiary mixture adsorption is wanted; quality of prediction - good or very good).\n• Prediction of adsorption in liquid mixtures (required: single gas/vapour adsorption data for all components; useful: binary gas mixture data - if e.g. prediction of tertiary-mixture adsorption is wanted; quality: heterogeneity parameters are nicely estimated, other parameters require correction) (for the prediction in e.g. tertiary systems, it is much better if binary liquid mixtures are available)\n\n•", null, "Prediction of adsorption in wastewater - unlimited no. of components and variable composition. (Here are model pictures)\n\nProposed method assumes that all adsorbates are similar in chemical properties and their distributions af adsorption energies have the same shape but different position on energy axis (same heterogeneities, different adsorption equilibrium constants). Generally no limit of no. of components exists. Prediction requires data for simple (single-component or binary) systems.\nIn the light of the above presented prediction method as well as a heterogeneity vs. molecular size and heterogeneity study it is crude oversimplification. However, the usefulness of this approach may overcome its limits.\nE.g. for wastewater treatment, where usually a large and only partially known no. and kind of pollutants is present. In such a case, if only an approximate mixture-profile is known, the adsorption of the entire mixture may be quite well predicted. In this way the right amount of adsorbent (e.g. active carbon) may be determined with a narrow margin error.\n1. \"A Simple Method for Describing Multi-Solute Adsorption Equilibria on Activated Carbons\", A.W.Marczewski, A.Derylo-Marczewska and M.Jaroniec, Chem.Engng.Sci., 45(1), 143-149 (1990), (doi).\n\n•", null, "Theoretical approach based on simple statistics.\nThe proposed theoretical approach does not assume any particular adsorption isotherm, though localised physical adsorption in monolayer is implied. It is assumed, that the observed heterogeneity comes from the entire system: i.e. specific properties of adsorbate and adsorbent as well as the topography of adsorption sites. (Here are model pictures)\n1. \"Energetic Heterogeneity and Molecular Size Effects in Physical Adsorption on Solid Surfaces\", A.W.Marczewski, A.Derylo-Marczewska and M.Jaroniec, J.Colloid Interface Sci., 109, 310-324 (1986), (doi).\n(This paper is probably my most-often cited one and I must say it gave me a lot of fun to find all this!)\nThis approach explained several observed facts and helped to reject some wide-spread (at that time) myths (here).\n\n• Simple equation of adsorption on heterogeneous solids from binary mixtures of gases or bi-component dilute solutions for components with different molecular sizes:\n• for quasi-gaussian energy distribution characterised by heterogeneity coefficient m and components with size ratio r12 = r1 / r2 in conditions where adsorbed layer is almost filled-up an equation is obtained (M.Jaroniec,1981; M.Jaroniec et.al. 1982):\n[a1 / a2 r12 ] = { K12 [c1 / c2 r12] } m\nMost useful forms of this equation are linear.\n\nAdsorption type ( Linear Langmuir plot / Graham plot / Consistency / Henry constant )\nPopular isotherms ( Mono-, Multilayer, Experimental, Micro-, Mesoporous )\nData analysis: ( LSq data fitting / Heterogeneity: Global , σE / Linear plots / φ-function / Pores )\nHeterogeneity and Molecular Size ( Theory and Prediction / Simple binary isotherm )\n\nGeneral Integral Equation / GL (Generalized Langmuir) / All equations (preview)\n\nTop", null, "", null, "" ]

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[ "# Normal subgroups of the semidihedral group of order 16?\n\nWhat are the normal subgroups of the semidihedral group of order 16? For such a fundamental result, there doesn't seem to be a reference to this online...\n\nThe semidihedral group of order 16 is given by $\\langle \\sigma, \\tau\\rangle$, where $\\sigma^8 = \\tau^2 = e$ and $\\sigma\\tau = \\tau\\sigma^3$.\n\n• Are you sure you mean $\\sigma^3$? – GPerez Jun 3 '15 at 11:20\n\nThere are 6 normal subgroups, namely, the trivial subgroup, $\\langle\\sigma^4\\rangle$, $\\langle\\sigma^4, \\tau\\rangle$, $\\langle\\sigma^2\\rangle$, $\\langle\\sigma^2, \\tau\\rangle$, and $\\langle\\sigma\\rangle$." ]

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[ "Solutions by everydaycalculation.com\n\n## Reduce 3919/168 to lowest terms\n\n3919/168 is already in the simplest form. It can be written as 23.327381 in decimal form (rounded to 6 decimal places).\n\n#### Steps to simplifying fractions\n\n1. Find the GCD (or HCF) of numerator and denominator\nGCD of 3919 and 168 is 1\n2. Divide both the numerator and denominator by the GCD\n3919 ÷ 1/168 ÷ 1\n3. Reduced fraction: 3919/168\nTherefore, 3919/168 simplified is 3919/168\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]

[ "# nLab Jones' theorem\n\nContents\n\ncohomology\n\n### Theorems\n\n#### Higher algebra\n\nhigher algebra\n\nuniversal algebra\n\n# Contents\n\n## Idea\n\nLet $X$ be a simply connected topological space.\n\nThe ordinary cohomology $H^\\bullet$ of its free loop space is the Hochschild homology $HH_\\bullet$ of its singular chains $C^\\bullet(X)$:\n\n$H^\\bullet(\\mathcal{L}X) \\simeq HH_\\bullet( C^\\bullet(X) ) \\,.$\n\nMoreover the $S^1$-equivariant cohomology of the loop space, hence the ordinary cohomology of the cyclic loop space $\\mathcal{L}X/^h S^1$ is the cyclic homology $HC_\\bullet$ of the singular chains:\n\n$H^\\bullet(\\mathcal{L}X/^h S^1) \\simeq HC_\\bullet( C^\\bullet(X) )$\n\n(Loday 11)\n\nIf the coefficients are rational, and $X$ is of finite type then this may be computed by the Sullivan model for free loop spaces, see there the section on Relation to Hochschild homology.\n\nIn the special case that the topological space $X$ carries the structure of a smooth manifold, then the singular cochains on $X$ are equivalent to the dgc-algebra of differential forms (the de Rham algebra) and hence in this case the statement becomes that\n\n$H^\\bullet(\\mathcal{L}X) \\simeq HH_\\bullet( \\Omega^\\bullet(X) ) \\,.$\n$H^\\bullet(\\mathcal{L}X/^h S^1) \\simeq HC_\\bullet( \\Omega^\\bullet(X) ) \\,.$\n\nThis is known as Jones' theorem (Jones 87)\n\nAn infinity-category theoretic proof of this fact is indicated at Hochschild cohomology – Jones’ theorem." ]

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[ "display | more...\nPower, Voltage and Resistance\n\nPower, measured in watts, can be calculated from the equation P=VI, or power=voltage times Amps (measures the current, full name Amperes, represented with an I in equations). If we use this relationship, along with V=I² R (see bellow), we can get:\n\nP= V²/R\n\nVoltage, measured in volts, represented by a v in equations, can be determined in the equation V²=PR\n\nResistance, measured in ohms, represented as a unit by the Greek symbol Omega and by an R in equations. Resistance is, surprisingly, the level of resistance of an object, meaning the resistance electricity faces when conducted through the object.\nUsing the following calculation, we can calculate resistance; R=P²/I\n\nThe following equations can also be used to determine the above:\n\nP=VI\nP=I²R\nP=I²R (Peter Is a Square Eyed Rabbit)\nR=P²/I\nR=VI\nV=P/I\n\nLog in or register to write something here or to contact authors." ]

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[ "# LCIS\n\nTime Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)\nTotal Submission(s): 440    Accepted Submission(s): 201\n\nProblem Description\nAlex has two sequences a1,a2,…,an and b1,b2,…,bm. He wants find a longest common subsequence that consists of consecutive values in increasing order.\n\nInput\n\nThere are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:The first line contains two integers n and m (1≤n,m≤100000) — the length of two sequences. The second line contains n integers: a1,a2,…,an (1≤ai≤106). The third line contains n integers: b1,b2,…,bm (1≤bi≤106).\n\nThere are at most 1000 test cases and the sum of n and m does not exceed 2×106.\n\nOutput\nFor each test case, output the length of longest common subsequence that consists of consecutive values in increasing order.\n\nSample Input\n3\n3 3\n1 2 3\n3 2 1\n10 5\n1 23 2 32 4 3 4 5 6 1\n1 2 3 4 5\n1 1\n2\n1\n\nSample Output\n1\n5\n0\n\nAC代码：\n\n#include<iostream>\n#include<algorithm>\n#include<stdio.h>\n#include<string.h>\n#include<cmath>\n#include<iostream>\nusing namespace std;\n#define MMAX 110000\nint a[MMAX],b[MMAX],dpa[MMAX],dpb[MMAX];\nint main()\n{\ncout.sync_with_stdio(false);\nint T;\ncin>>T;\nwhile(T--)\n{\nint n,m;\ncin>>n>>m;\nmemset(dpa,0,sizeof(dpa));\nmemset(dpb,0,sizeof(dpb));\nint maxx=0;\nfor(int i=1; i<=n; i++)\n{\ncin>>a[i];\ndpa[a[i]]=dpa[a[i]-1]+1;\nmaxx=max(maxx,a[i]);\n}\nfor(int i=1; i<=m; i++)\n{\ncin>>b[i];\ndpb[b[i]]=dpb[b[i]-1]+1;\nmaxx=max(maxx,b[i]);\n}\nint ans=0;\nfor(int i=1; i<=maxx; i++)\nans=max(ans,min(dpa[i],dpb[i]));\ncout<<ans<<endl;\n}\n}" ]

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[ "# Chapter 12 – Pure Monopoly example #30222\n\n## Unlock all answers in this set\n\nquestion\nPure monopoly refers to:\na single firm producing a product for which there are no close substitutes.\nquestion\nWhich of the following is correct?\nA purely competitive firm is a \"price taker,\" while a monopolist is a \"price maker.\"\nquestion\nA purely monopolistic firm:\nfaces a downsloping demand curve.\nquestion\nWhich of the following best approximates a pure monopoly?\nThe only bank in a small town.\nquestion\nLarge minimum efficient scale of plant combined with limited market demand may lead to:\nnatural monopoly.\nquestion\nFor an imperfectly competitive firm:\nthe marginal revenue curve lies below the demand curve because any reduction in price applies to all units sold.\nquestion\nRefer to the data. The marginal revenue obtained from selling the third unit of output is: <\\$6. <\\$1. <\\$3. <\\$5.\n\\$5\nquestion\nRefer to the data. At the point where 3 units are being sold, the coefficient of price elasticity of demand:\nis greater than unity (one).\nquestion\nA monopolistic firm has a sales schedule such that it can sell 10 prefabricated garages per week at \\$10,000 each, but if it restricts its output to 9 per week it can sell these at \\$11,000 each. The marginal revenue of the tenth unit of sales per week is: <-\\$1,000. <\\$9,000. <\\$10,000. <\\$1,000.\n\\$1,000.\nquestion\nThe demand curve faced by a pure monopolist:\nis less elastic than that faced by a single purely competitive firm.\nquestion\nThe marginal revenue curve for a monopolist:\nbecomes negative when output increases beyond some particular level.\nquestion\nWhich of the following is characteristic of a pure monopolist's demand curve?\nIt is the same as the market demand curve.\nquestion\nFor a pure monopolist the relationship between total revenue and marginal revenue is such that:\nmarginal revenue is positive when total revenue is increasing, but marginal revenue becomes negative when total revenue is decreasing.\nquestion\nFor a pure monopolist, marginal revenue is less than price because:\nwhen a monopolist lowers price to sell more output, the lower price applies to all units sold.\nquestion\nAssume a pure monopolist is currently operating at a price-quantity combination on the inelastic segment of its demand curve. If the monopolist is seeking maximum profits, it should:\ncharge a higher price.\nquestion\nSuppose a pure monopolist is charging a price of \\$12 and the associated marginal revenue is \\$9. We thus know that:\ntotal revenue is increasing.\nquestion\nWhich of the following is incorrect? Imperfectly competitive producers:\ndo not compete with one another.\nquestion", null, "Refer to the table. The monopolist will select its profit-maximizing level of output somewhere within the: <3-5 unit range of output. <1-3 unit range of output. <1-4 unit range of output. <2-4 unit range of output.\n1-3 unit range of output.\nquestion", null, "Refer to the table. The profit-maximizing monopolist will sell at a price:", null, "", null, "" ]

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[ "# How rates are calculated\n\nWe use a formula to work out how much your rates will be.\n\nThe formula for calculating rates is:\n\n• Capital Improved Value (CIV) x Rate in the dollar = Rate payable\n• add the Municipal charge, Fire Services Property Levy\n• add the cost of any waste services\n• subtract any concessions (if eligible)\n\n## Example\n\n• CIV of the residential property = \\$A\n• rate in the dollar for residential properties = B cents\n• A x B = \\$rates\n• \\$rates + levy charges + services – concessions = \\$total rates\n\nThe ‘rate in the dollar’ is determined by:\n\n• calculating the total amount of rate revenue we need to maintain our services\n• dividing this amount by the total value of the properties within the Frankston City Municipality (this number changes every year).\n\nPlease be aware that when you receive your rates notice, it is important not to add your capital improved value (CIV) and site value (SV) together, as this is not how rates are calculated." ]

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[ "# linear mixed model assumptions\n\nThe observations can be correlated. Linear mixed effects analyses - Mixed Effect Model Assumptions First review the Linear mixed effects analyses section.The same conditions we have in the fixed effect multivariate linear model apply to mixed and random effect models – co-linearity, influential data points, homoscedasticity, and lack of normality. The desire to extend the method to a linear mixed model I want to illustrate how to run a simple mixed linear regression model in SPSS. Finally, mixed model theory was incorporated, which led to generalized linear mixed models. What are the best methods for checking a generalized linear mixed model (GLMM) for proper fit?This question comes up frequently when using generalized linear mixed effects models.Unfortunately, it isn’t as straightforward as it is for a general linear model, where the requirements are easy to outline: linear relationships of numeric predictors to outcomes, normally … A G-side random effect in a mixed model is an element of , and its variance is expressed through an element in . 3 Overview As with any statistical manipulation, there are a specific set of assumptions under which we operate when conducting multilevel models (MLM). Details The glmmLasso algorithm is a gradient ascent algorithm designed for generalized linear mixed models, which incorporates variable selection by L1-penalized estimation. Unlike standard linear models (LMs), LMMs make assumptions not only about the distribution of2015 Formally, the assumptions of a mixed-effects model involve validity of the model, independence of the data points, linearity of the relationship between predictor and response, absence of mea - From the assumptions of the linear mixed model, each random effect specified is assumed to follow a normal distribution. If an effect, such as a medical treatment, affects the population mean, it … Given these assumptions, a heterogeneous linear mixed model can be specified as follows: Y i b i ∼ N X i ′ β + Z i ′ b i , R i , b i μ ~ ∼ N μ ~ , G , μ ~ ∈ μ ~ 1 , .... , μ ~ K . Therefore, these plots can be used to assess if this assumption is met. G eneralized Linear Model (GLM) is popular because it can deal with a wide range of data with different response variable types (such as binomial, Poisson, or multinomial). Nathaniel E. Helwig (U of Minnesota) Linear Mixed-Effects Regression Updated 04-Jan-2017 : Slide 13 One-Way Repeated Measures ANOVA Model Form and Assumptions … However, before we conduct linear regression, we must first make sure that four assumptions are met: 2. The standard linear mixed model (LMM) is thus represented by the following assumptions: The matrices and are covariance matrices for the random effects and the random errors, respectively. StATS: A simple example of a mixed linear regression model (October 18, 2006). Comparing to the non-linear models, such as the neural networks or tree-based models, the linear models may not be that powerful in terms of prediction. Linear Mixed Models in Linguistics and Psychology: A Comprehensive Introduction (DRAFT) 3.3 Checking model assumptions It is an assumption of the linear model that the residuals are (approximately) normally distributed, That is what … ects (i.e., the level-2 residuals) will not resemble the The target can have a non-normal distribution. Linear mixed effects model (xtmixed) relies on the assumption that the residuals are normally distributed. We have added 95% normal-theory These models are widely used in the biological and social sciences. These models describe the relationship between a response variable and independent variables, with coefficients that can vary with respect to one or more grouping variables. Learn about the assumptions and how to assess them for your model. If you are looking for help to make sure your data meets assumptions #4, #5, #6 and #7, which are required when using a mixed ANOVA and can be tested using SPSS Statistics, we show you how to do this in our enhanced AGR Summary. I will use some data on the plasma protein levels of turtles at baseline, after fasting 10 days, and after fasting 20 days. Linear mixed-effects model fit by maximum likelihood Data: data AIC BIC logLik 6425.735 6461.098 -3206.867 Random effects: Formula: ~1 | Time (Intercept) Residual StdDev: 0.07982052 0.7992555 Fixed effects To fit a mixed-effects model we are going to use the function lme from the package nlme . Moreover, usually approximations have not mixed designs) to then just use the lme package to streamline the model building process. In matrix notation, linear mixed models can be Linear Mixed-Effects Models Linear mixed-effects models are extensions of linear regression models for data that are collected and summarized in groups. Just to explain the syntax to use linear mixed-effects model in R for cluster data, we will assume that the factorial variable rep in our dataset describe some clusters in the data. We study robust designs for generalized linear mixed models (GLMMs) with protections against possible departures from underlying model assumptions. Generalized linear mixed models extend the linear model so that: The target is linearly related to the factors and covariates via a specified link function. However, if your model violates the assumptions, you might not be able to trust the results. In a linear mixed-effects model, responses from a subject are thought to be the sum (linear) of so-called fixed and random effects. Linear regression is a useful statistical method we can use to understand the relationship between two variables, x and y. This is the main page of the course and contains a course overview, schedule and learning outcomes. 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Of linear regression model in SPSS on the assumption that the residuals are normally distributed in. Glmmlasso algorithm is a gradient ascent algorithm designed for generalized linear mixed effects model xtmixed! For example a way of plotting the residuals against a normalcurve Summary four assumptions are met: 2 first." ]

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[ "M-x调出命令，输入：\n\n``````count-words\n``````\n\n## 老张说:\n\n``````(defun zjs-count-word ()\n(interactive)\n(let ((beg (point-min)) (end (point-max))\n(eng 0) (non-eng 0))\n(if mark-active\n(setq beg (region-beginning)\nend (region-end)))\n(save-excursion\n(goto-char beg)\n(while (< (point) end)\n(cond ((not (equal (car (syntax-after (point))) 2))\n(forward-char))\n((< (char-after) 128)\n(progn\n(setq eng (1+ eng))\n(forward-word)))\n(t\n(setq non-eng (1+ non-eng))\n(forward-char)))))\n(message \"English words: %d\\nNon-English characters: %d\"\neng non-eng)))\n``````" ]

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[ "", null, "# Section 2: Quantum Theory and the Atom\n\n## Presentation on theme: \"Section 2: Quantum Theory and the Atom\"— Presentation transcript:\n\nSection 2: Quantum Theory and the Atom\nElectrons in atoms Section 2: Quantum Theory and the Atom\n\nLearning Goals Compare the Bohr and quantum mechanical models of the atom. Explain the impact of de Broglie’s wave particle duality and the Heisenberg uncertainty principle on the current view of electrons in atoms. Identify the relationships among a hydrogen atom’s energy levels, sublevels, and atomic orbitals.\n\nBohr’s Model of the Atom\nEinstein’s theory of light’s dual nature accounted for several unexplainable phenomena, but it did not explain why atomic emission spectra of elements were discontinuous.\n\nBohr’s Model of the Atom\nIn 1913, Niels Bohr, a Danish physicist working in Rutherford’s laboratory, proposed a quantum model for the hydrogen atom that seemed to answer this question. This model correctly predicted the frequency lines in hydrogen’s atomic emission spectrum.\n\nBohr’s Model of the Atom\nThe lowest allowable energy state of an atom is called its ground state. When an atom gains energy, it is in an excited state.\n\nBohr’s Model of the Atom\nBohr suggested that an electron moves around the nucleus only in certain allowed circular orbits.\n\nBohr’s Model of the Atom\nEach orbit was given a number, called the quantum number.\n\nBohr’s Model of the Atom\nHydrogen’s single electron is in the n = 1 orbit in the ground state. When energy is added, the electron moves to the n = 2 orbit.\n\nBohr’s Model of the Atom\nThe electron releases energy as it falls back towards the ground state.\n\nBohr’s Model of the Atom\n\nBohr’s Model of the Atom\nBohr’s model explained the hydrogen’s spectral lines, but failed to explain any other element’s lines. The behavior of electrons is still not fully understood, but it is known they do not move around the nucleus in circular orbits.\n\nQuantum Mechanical Model\nLouis de Broglie (1892–1987) hypothesized that particles, including electrons, could also have wavelike behaviors. Electrons orbit the nucleus only in whole-number wavelengths.\n\nQuantum Mechanical Model\nIf electrons can only orbit in whole number wavelengths, then only certain frequencies and energies are possible. Not possible to have a continuous spectrum.\n\nQuantum Mechanical Model\nThe de Broglie equation predicts that all moving particles have wave characteristics.\n\nQuantum Mechanical Model\nHeisenberg showed it is impossible to take any measurement of an object without disturbing it. The Heisenberg uncertainty principle states that it is fundamentally impossible to know precisely both the velocity and position of a particle at the same time.\n\nQuantum Mechanical Model\nThe only quantity that can be known is the probability for an electron to occupy a certain region around the nucleus.\n\nQuantum Mechanical Model\nSchrödinger treated electrons as waves in a model called the quantum mechanical model of the atom. Schrödinger’s equation applied equally well to elements other than hydrogen (unlike Bohr’s model).\n\nQuantum Mechanical Model\nThe quantum mechanical model makes no attempt to predict the path of an electron around the nucleus.\n\nQuantum Mechanical Model\nInstead, Schrödinger’s wave function predicts a three-dimensional region around the nucleus called the atomic orbital in which an electron may be found.\n\nHydrogen’s Atomic Orbitals\nPrincipal quantum number (n) indicates the relative size and energy of atomic orbitals. n specifies the atom’s major energy levels, called the principal energy levels.\n\nHydrogen’s Atomic Orbitals\nEnergy sublevels are contained within the principal energy levels.\n\nHydrogen’s Atomic Orbitals\nEach energy sublevel relates to orbitals of different shape. s, p, d, f s, p, d s, p s\n\nHydrogen’s Atomic Orbitals\ns sublevel:\n\nHydrogen’s Atomic Orbitals\np sublevel:\n\nHydrogen’s Atomic Orbitals\nd sublevel:\n\nHydrogen’s Atomic Orbitals\nf sublevel:\n\nHydrogen’s Atomic Orbitals\n\nHydrogen’s Atomic Orbitals\nAt any given time, hydrogen’s electron can occupy just one orbital. When hydrogen is in the ground state, the electron occupies the 1s orbital. When the atom gains a quantum of energy, the electron is excited to one of the unoccupied orbitals.\n\nDownload ppt \"Section 2: Quantum Theory and the Atom\"\n\nSimilar presentations" ]

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