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https://nl.mathworks.com/matlabcentral/cody/problems/1054-what-s-your-bmi/solutions/1636849	[ "Cody\n\n# Problem 1054. What's Your BMI?\n\nSolution 1636849\n\nSubmitted on 2 Oct 2018 by Suraj Gurav\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\nh= 1; m= 1; y_correct = 703.06957964; assert(isequal(BMI(h,m),y_correct))\n\nans = []\n\n2 Pass\nh= 70; m= 90; y_correct = 12.91352289134694; assert(isequal(BMI(h,m),y_correct))\n\nans = []" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6247871,"math_prob":0.95250875,"size":532,"snap":"2019-51-2020-05","text_gpt3_token_len":161,"char_repetition_ratio":0.14015152,"word_repetition_ratio":0.024096385,"special_character_ratio":0.35902256,"punctuation_ratio":0.14150943,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9853797,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-08T21:19:14Z\",\"WARC-Record-ID\":\"<urn:uuid:f241392c-a376-40c5-841a-947cb044eed8>\",\"Content-Length\":\"71911\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a62a95ff-8812-40ed-986c-27435339b76f>\",\"WARC-Concurrent-To\":\"<urn:uuid:02f5c36b-a366-41a3-86f0-f1cdd3a48734>\",\"WARC-IP-Address\":\"23.50.228.199\",\"WARC-Target-URI\":\"https://nl.mathworks.com/matlabcentral/cody/problems/1054-what-s-your-bmi/solutions/1636849\",\"WARC-Payload-Digest\":\"sha1:P3PVDVFQ6HVJNUOJWBSCY5X7ZWUP5T3U\",\"WARC-Block-Digest\":\"sha1:NGAA2XQ7O5BPQJWIIG5BMLI6F2YJ2DLV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540514893.41_warc_CC-MAIN-20191208202454-20191208230454-00271.warc.gz\"}"}
https://whatisconvert.com/585-imperial-gallons-in-liters	[ "## Convert 585 Imperial Gallons to Liters\n\nTo calculate 585 Imperial Gallons to the corresponding value in Liters, multiply the quantity in Imperial Gallons by 4.54609 (conversion factor). In this case we should multiply 585 Imperial Gallons by 4.54609 to get the equivalent result in Liters:\n\n585 Imperial Gallons x 4.54609 = 2659.46265 Liters\n\n585 Imperial Gallons is equivalent to 2659.46265 Liters.\n\n## How to convert from Imperial Gallons to Liters\n\nThe conversion factor from Imperial Gallons to Liters is 4.54609. To find out how many Imperial Gallons in Liters, multiply by the conversion factor or use the Volume converter above. Five hundred eighty-five Imperial Gallons is equivalent to two thousand six hundred fifty-nine point four six three Liters.\n\n## Definition of Imperial Gallon\n\nThe imperial (UK) gallon, now defined as exactly 4.54609 litres (about 277.42 cubic inches), is used in some Commonwealth countries and was originally based on the volume of 10 pounds (approximately 4.54 kg) of water at 62 °F (17 °C). The imperial fluid ounce is defined as 1⁄160 of an imperial gallon; there are four quarts in a gallon, two pints in a quart, and 20 Imperial fluid ounces in an imperial pint.\n\n## Definition of Liter\n\nThe liter (also written \"litre\"; SI symbol L or l) is a non-SI metric system unit of volume. It is equal to 1 cubic decimeter (dm3), 1,000 cubic centimeters (cm3) or 1/1,000 cubic meter. The mass of one liter liquid water is almost exactly one kilogram. A liter is defined as a special name for a cubic decimeter or 10 centimeters × 10 centimeters × 10 centimeters, thus, 1 L ≡ 1 dm3 ≡ 1000 cm3.\n\n## Using the Imperial Gallons to Liters converter you can get answers to questions like the following:\n\n• How many Liters are in 585 Imperial Gallons?\n• 585 Imperial Gallons is equal to how many Liters?\n• How to convert 585 Imperial Gallons to Liters?\n• How many is 585 Imperial Gallons in Liters?\n• What is 585 Imperial Gallons in Liters?\n• How much is 585 Imperial Gallons in Liters?\n• How many L are in 585 uk gal?\n• 585 uk gal is equal to how many L?\n• How to convert 585 uk gal to L?\n• How many is 585 uk gal in L?\n• What is 585 uk gal in L?\n• How much is 585 uk gal in L?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8419184,"math_prob":0.9592188,"size":2196,"snap":"2023-14-2023-23","text_gpt3_token_len":584,"char_repetition_ratio":0.21578467,"word_repetition_ratio":0.07848101,"special_character_ratio":0.28688523,"punctuation_ratio":0.10561798,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.994896,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-01T19:33:28Z\",\"WARC-Record-ID\":\"<urn:uuid:e415a5bf-5aed-402f-aec9-7027417709ee>\",\"Content-Length\":\"30372\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fbf607f1-ea85-418d-a4d0-9c74e2963ad5>\",\"WARC-Concurrent-To\":\"<urn:uuid:919ec687-4bbf-44e7-a6ea-fd0020046301>\",\"WARC-IP-Address\":\"104.21.13.210\",\"WARC-Target-URI\":\"https://whatisconvert.com/585-imperial-gallons-in-liters\",\"WARC-Payload-Digest\":\"sha1:XBLB3UOCPN5CKSP25QN5UAC2ERBROKVQ\",\"WARC-Block-Digest\":\"sha1:KYUP4JGQPY6FIOST4IXUW5GM5C7SZKBR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224648000.54_warc_CC-MAIN-20230601175345-20230601205345-00244.warc.gz\"}"}
https://proofwiki.org/wiki/Definition:Convex_Real_Function/Definition_1/Strictly	[ "# Definition:Convex Real Function/Definition 1/Strictly\n\n## Definition\n\nLet $f$ be a real function which is defined on a real interval $I$.\n\n$f$ is strictly convex on $I$ if and only if:\n\n$\\forall x, y \\in I, x \\ne y: \\forall \\alpha, \\beta \\in \\R_{>0}, \\alpha + \\beta = 1: \\map f {\\alpha x + \\beta y} < \\alpha \\map f x + \\beta \\map f y$", null, "The geometric interpretation is that any point on the chord drawn on the graph of any convex function always lies above the graph.\n\n## Also defined as\n\nBy setting $\\alpha = t$ and $\\beta = 1 - t$, this can also be written as:\n\n$\\forall x, y \\in I, x \\ne y: \\forall t \\in \\openint 0 1 : \\map f {t x + \\paren {1 - t} y} < t \\map f x + \\paren {1 - t} \\map f y$" ]	[ null, "https://proofwiki.org/w/images/thumb/e/ef/ConvexFunction1.png/700px-ConvexFunction1.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.58952695,"math_prob":0.99997187,"size":717,"snap":"2021-31-2021-39","text_gpt3_token_len":236,"char_repetition_ratio":0.13464236,"word_repetition_ratio":0.093959734,"special_character_ratio":0.35006973,"punctuation_ratio":0.10191083,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999423,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-05T12:32:53Z\",\"WARC-Record-ID\":\"<urn:uuid:e80be7a3-1db2-4a81-bc47-d1a653d41741>\",\"Content-Length\":\"34019\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5a14d502-a775-4c9e-ab81-65fbae3f4b29>\",\"WARC-Concurrent-To\":\"<urn:uuid:77d2a717-5fa6-43af-9088-42808b5f4378>\",\"WARC-IP-Address\":\"172.67.198.93\",\"WARC-Target-URI\":\"https://proofwiki.org/wiki/Definition:Convex_Real_Function/Definition_1/Strictly\",\"WARC-Payload-Digest\":\"sha1:SMMETM3AYGP2DTGDPWL4APA4NOKBDXM3\",\"WARC-Block-Digest\":\"sha1:UDHXX4PVA4TNMCOXDIR37ZIRFZKFI6AG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046155529.97_warc_CC-MAIN-20210805095314-20210805125314-00047.warc.gz\"}"}
https://gourocklawnsllp.com/minimum-jumps-to-traverse-all-integers-in-range-1-n-such-that-integer-i-can-jump-i-steps/	[ "Tuesday, 19 Oct 2021\n\n# Minimum jumps to traverse all integers in range [1, N] such that integer i can jump i steps\n\nGiven an integer N, the task is to find the minimum steps to visit all integers in the range [1, N] by selecting any integer and jump i steps at every ith jump.Note: It is possible to revisit an integer more than once. Examples:Input: N = 6Output: 3Explanation: One of the following way is: First start at first number and visit the integers {1, 2, 4}.Now start at 2nd number and visit the integers as {2, 3, 5}Now start at the last number and visit it.Therefor, in total of 3 steps one can visit all the numbers in the range [1, 6]. And also it is the minimum number of steps needed.Input: N = 4Output: 2Naive Approach: The given problem can be solved based on the following observations: In each step the sizes of jumps increases therefore some numbers remains unvisited in a step.Starting from the first number and performing the jumps it can be observed that the maximum size of the jump is the total number of steps needed to visit every number. As in one move, one cannot visit each number between two unvisited numbers.Follow the below steps to solve the problem:Initialize two variables, say count = 1 and res = 0.Traverse over the range [1, N] and increment i by count and update res as res =max(res, count) and increment count by 1.After completing the above steps print the res.Below is the implementation of the above approach:C++#include using namespace std; int minSteps(int N){ int count = 1, res = 0; for (int i = 1; i" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8771824,"math_prob":0.9942948,"size":1526,"snap":"2021-43-2021-49","text_gpt3_token_len":373,"char_repetition_ratio":0.14914586,"word_repetition_ratio":0.0073260074,"special_character_ratio":0.24180865,"punctuation_ratio":0.12538226,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99333704,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-19T21:09:47Z\",\"WARC-Record-ID\":\"<urn:uuid:146c8420-fd13-4f37-9c42-4f3569a72377>\",\"Content-Length\":\"42625\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8b1232c0-ee52-4cf7-a740-44ae1b059649>\",\"WARC-Concurrent-To\":\"<urn:uuid:3471566d-c90d-4d66-beb8-f154b7b7a92d>\",\"WARC-IP-Address\":\"104.21.14.17\",\"WARC-Target-URI\":\"https://gourocklawnsllp.com/minimum-jumps-to-traverse-all-integers-in-range-1-n-such-that-integer-i-can-jump-i-steps/\",\"WARC-Payload-Digest\":\"sha1:C7LL3UXWUCV3UGNULOY2FIQN3CUZUKYM\",\"WARC-Block-Digest\":\"sha1:WR27WYOWXWOSQ3ZYKA2XBZGSQ5LPKW3N\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585281.35_warc_CC-MAIN-20211019202148-20211019232148-00563.warc.gz\"}"}
https://de.scribd.com/document/66806060/c-Notes-Complet	[ "Sie sind auf Seite 1von 144\n\n# C\n\n## #include <stdio.h> Function <printf Scanf\n\nC++\n#include <iostream.h> cout. in are objects cout << Hello; Insertion operator Or Insertors Cout << value = <<a; No need of format specefiers In C++ in >>a; Extraction operator Or Extractor In >>a >>b; Cascading of extractor; cout << Hello \\n User; Or Cout << Hello <<end <<User; in C++ default return type is an integer\n\n## In C, default return type Is void\n\nHISTORY OF C++\nYear 1982 Developed Bjarne stroustrap Lab Bell Labs Company At & T C is procedure oriented language Easy & fast programming in C. Logics can be easily developed Developed. C is object oriented language C++ closely models sear world problems\n\n## CLASSES AND OBJECTS\n\nIn structure of C :- only data can be member of structure and not functions All member of structure of are public by default In class of C++ data + functions accessing those data are member of class and All member of class are private by default Class stud { Int roll; Char grade; Float par; Public: Void get( ); Void show( ); }; Void stud : : get( ) { Cout << enter roll. Grade and per; Cin>>roll>> grade >> per; } Void stud : : show { Cout <<roll << << grade << <<per<< end1; } Void main( ) { Stud s; S. get ( ); s. show( ); } get show roll grade Per Function are never seplicated there is only one copy of function no matter now many objects are created only once memory is allocated to functions for all objects where as multiple copies of data are created for multiple objects.\n\n: : Scope resolution operator helps compiler to identify functions of which class if two classes have the same name. Q. 1 wap to add two numbers give by user Class add { Int a, b, c Public : void get( ); Void sum( ); Void show( ); }; Void add : : get ( ) { Cout << Enter no; Cin >> a >>b; } Void add : : sum( ) { C= a+b; } Void add : : show( ) { Cout << Numbers are = << a << << b; Cout << sum = <<c; } Void main( ) { Add obj; Obj. get( ); Obj.sum( ); Obj. show( ); Getch ( ); } C++ (Terminology) 1. objects 2. data members 3. member function 4. function call OOPs (Terminology) instances properties & attributes methods & behaviors message passing\n\n## BASIC PRINCIPLES OF OOP\n\nENCAPSULATION Word has been derived from a word capsule which means multiple medicines packed in one single unit. Similarly in a software there are two major unit data & functions acting on that data since functions and data are related entities it is advisable to score them within a single unit. Thus according to oops Encapsulation means building or wrapping up of data members and f n acting on those data members with in a single unit. Since a class allows us to hold data & functions within it are say that it supports the principle of encapsulation. POLYMORPHISM The word polymorphism is derived from combination of two words poly mcarning multiple and morph means the ability to have multiple forms. In other words if an entity can acquire multiple forms in different situation we say that its behaviors is polymorphic for eg in c++ it is possible for programmer to redife operator + in such a way that it can be used to add two integers as well as at the same time it can add two abject or two strings. So it a programmer define + to behave in above mentioned member we say that + behaves polymorphically . In C++, polymorphism is implemented in two way:(1). Compile time polymorphism :- function overloading and operator Overloading (ii). RunTime polymorphism:- Virtual functions, pure virtual functions Ads tract classes (iii). Inheritance :- to inherit means to acquire properties and fea of an Existing entity into a newly created entity Like a child acquire properties of his or has Parent, similary when designing software if a programmer wishes then he can acquire the fratures (data and member function) of an existing class in his own class with the help of inheritance. The class which gets inherited is known as bass class and the class. Thus by inheriting the members of base class it becomes possible to access them through the objects of derive class. The major advantage offered by principle of Inheritance is rcusability and reliability\n\nCREATING PARAMETERIZED FUNCTIONS WITHIN CLASSES Class Emp { Int age; Char name; Float salary; Public: Void set (int, char , float); Void show( ) }; Void Emp: : set (int I, char j, float K) { Age =I; Strcpy (name, j); Salary =k; } Void Emp : : show( ) { Cout <<age << <<name << salary; } ASSIGNMENT Wap to create a class c/a string. The class must contain a character array of size 20. class should have following fucnitons: (1). Getstring( ) which accept string as parameter + stores it in array str. (2). Show string( ) which display the string stored in str[]. (3). Reversestring ( ) which reverses the string stored in str[]. (4). Add string( ) which accepts string as parameter and if possible Concatenates in str. Write an object oriented program to calculate the factorial of a given by user. Provide separated function for initialization, inputting, calculating and display. Class Fact {\n\nInt I, n; Public: Void init( ); Void getno( ); Void calculate( ); Void display( ); }; Void fact : : init( ) { F=1; } Void fact : : getno( ) { Cout << Enter a number; Cin >> n; } Void fact : : calculate( ) { Int I; For (i=1; i<n; i++) F=f:; } Void fact : : display( ) { Cout << Number = << n; Cout << factorial= <<f; } Void main( ) { Fact obj ; Obj. init( ); Obj.getno( ); Obj.get calculate( ); Obj. display( ); }\n\nCONTRUCTOR\nConstructor :- are special member f n of a class with the following properties\n\n1. They have the same name as that of the class 2. They dont have any return type not even void 3. They are automatically called as soon as the object of class is created i.e. their calling is implicit 4. They cant be declared as static 5. They cant be declared as virtual Any class which does not contain any constructor then compiler from itself supplier a constructor but it is hidden. For programmer these constructors are default Class fact {\n\n} Constructors are automatically called even when not declared, at that time default constructors are called. Default contractors are destroyed as soon as we declare constructor Example :Class fact { Int I, b; Public: Fact( ) { F=1; } Void getno( ); Void calculate( ); Void display( ); }; Void fact : : getno( ) { cout << enter a no; an >> n; }\n\nVoid fact : : calculate( ) { For (i=1; i<=n; i++) F=fI; } Void fact : : display ( ) { Cout << no= <<n<<end1; Cout << factorial= <<f; } Void main( ) { Fact obj; Obj.getno( ); Obj.calculate( ); Obj.display( 0; }\n\nPARAMETERIZED CONSTRUCTOR\nClass Emp { Int age; Char name; Flaot sal; Public: Emp (int, char . Float); Void show( ); }; Emp : : Emp (int I, char j, float k) { Age =I; Strcpy (name, j); Sal=k; } Void Emp : : show( ) { Cout << age << << name<< <<sal; }\n\n1. No of Arguments 2. Type of Argument 3. Order of argument Int show ( ) Not allowed (return type differs) Void show( ) Compiler does not consider return type, other wise constructor can never be overloaded as they have no returns type. Function overloading allows some function name in same scope but there should be some different in functions. Void vol (int); Void vol (int, int, int ); Void main ( ) { Int choice; Cout << select a figure; Cout << (1) cube \\n (2) cuboid; cin >> choice ( ); Swith (choice) { Case 1: Int s; Cout << enter side of cube; Cin >> s; Vol (s); Break; Case 2: Int l.\\, b, h; Cout << Enter l, band h of alboid; Cin >> l >> b >> h; Vol (l,b, h); Breck; Default: Cout < wrong choice; Void show (int, int, int) void show (int) void show (double) void show (int, double) Void show (double, int)\n\n} Void vol (int s) { Cout << value of cube = << sss; } Void vol (int l, int b, int h) { Cout << value of alboid= << lbh; } ADVANTAGES OF FUNCTION OVERLODING 1. Overload of remembering the name is transformed from programmer to compiler 2. it develops symmetry & incrrases the readability of program CONSTRUCTURE OVERLOADING Class Box { Int l, b,h; Public: Box( ); //constructor for user defined Box Box (int); //constructor for albe }; Box : : Box( ) { Cout << enter l, b and h of box; Cin >> l >> b>> h; } Box : : Box(int s) { L=b=h=s; } Box : : Box (int I, int j, int k) { L=I; B=j; H=k; }\n\nVoid BOX : : show( ) { Cout << l << <<b<< << h; } Void main ( ) { Box B1; Box B2 (5,7,11); Box B2 (10); B1. show( ); B2. show ( ); B3. show ( ); }\n\nCOPY CONSTRUCTOR\nIt is a special constructor of a class which accepts the reference of the object of its our class as a parameter. It is called by c++ compiler in there situations 1. when ever programmer created an object and at the same time passes another object of the same class as a parameter 2. whenever a f n accepts an object as a parameter by value. 3. whenever a f n returns an object by value. Reference variable:Syntax:- <data type> & <sef-name> = <var-name>; Void main( ) { Int a = 10; Int p; P=&a; Cout <<p <<end1; } Void main ( ) { Int a=10; Int &p=a; prowbacks of pointer 1. occupy 2 bytes of memory 2. will be initialized by garbage 3. necessary to initialize before their use 4. very caseful in using indirection operator Advantage of Reference variable 1. we can have n reference variables of one variable\n\n## Cout <<a<< <<p; Cout << &a << <<&p;\n\n2. both variables get interlocked on each Other 3. does not require any memory space it Only reuse the memory of nay variable\n\nReference variable is a technology provided by c++ which allow a programmer to create anew variable which stores (not holds) the memory location of an existing variable. In other words, reference variable is a technique using which programmer can allot multiple mames to same memory location. POINTER TO REMEMBER 1. int &p = a; 2. int 7p =; multiple declarations for the variable p. 3. In case of array, always use pointer. Reference variable can not work with array 4. we can not make array of reference variable int &p C C pass by value pass by reference void swap (int, int) void swap (int, int) void main( ) void main( ) cout << Enter2 number; cout << Enter2no; cin >>a>>b; cin >> a >> b; swap (a,b); swap (&a, &b) cout << a << <<b; cout <<a<< <<b; } } Void swap(Int p,intq) void swap (intp, intq) } } Int temp; int temp; Temp =p; temp=p; P=q; p=q; Q=temp; q= temp; } } C++ pass by reference void swap (int7,int&) void main( ) { int a, b; cout << enter2no) cin>>a>>b; swap (a,b); cout <<a<< <<b <<end1; } void wsap(int&pint&q) { int temp; temp=p; P=q; Q=temp; }\n\nNote :- By call it is not possible to call whether it is call by value or call by reference Q. WAP to use a function call maximum which accepts an integer array of size pure as an argument & retirns the largest & smallest element of that array to main. Without changing the arcginal position of element of ht array. Maximum (int a[], int &, int&) Void main( ) { Int a, I, large, small; For (i=0; i<5; i++) { Cout << enter elements of array; Cin >>a[i]; } Maximum (a, large, small); Cout << maximum element= <<large; Cout << smallert element= <<small; } Void maximum (int a, int &max, int & main) { Int f; Max = a; Min=a; For 9j=1; j<5; i++) { If (* (a+j) >max) Max=* (a+j); Else if (* (a+j) <min) Min = (a+j); } } Class Box { Int l, b, h; Public: Box( );\n\nBox (int); Box (int, int, int); Box (Box &); Void show( ); }; Box : : Box( ) { Cout << enter l, b and h of Box; Cin >> l >> b >> h; } Box : : Box (int S) { L=b=h=s; } Box : : Box (Box &p) { L=p.l; B B=p.b; b1s l, b,h. H=p.h; } Box : : Box (int I, int j, int k) { L=I; B=j; H=k; } Void Box : : show( ) { Cout << L << << b << << h; } Void main ( ) B2. show( ) { Box B1; B3. show( ); Box B2 (10); B4. show( ); Box B2 (5,7,11); } Box B4 (B1); B1. show( 0; Box B2 (10) Box B4 (B1) Box B2 = 10; Box B4 = B;\n\nCall for copy constructor Box B4; B4 = B1 No call for copy constructor use of assignment operator other destination object is already made.\n\nDEFUALT ARGUMENTS\nVoid printline (char =, int=1) Void main ( ) { Printline ( ); // printline (, 1) Printline (#); // printline ( #, 1) Printline (!, 10) // printline (1, 10); } Void printline (char ch, int n) { For (int i=1 i<=n; i++) Cout <<ch; } Note;- Default arguments must always be trailing arguments. Printline(50); printline (ASCII of 50, 1); Class stud { Int age; Char grade; Float per; Public: Stud (int, char, float); // stud (int=o, char= ; float=0.0); Void get( ); Void show( ); }; Stud : : stud (int I, char j, float k) { Age =I; Char j; Per k;\n\n} Void std : : get( 0 { Cout << Enter age, grade & per ; Cin >> age >> grade >> per; } Void stud : : show( 0 { Stud main( ) { Stud t ( 15, A, 75); Stud p; p. get( ); t. show( ); p. show( 0; } Note:- In class at same time it is not possible to have default argument constructor and default constructor.\n\nDESTRUCTOR\nAre special member f n of class which have same name as that of the class but preceded with the symbol of field ( ). They are automatically called whenever an object goes out of scope & is about to be destroyed When an object is created first function which is automatically called is constructor & when object ends its life a last function is called to free occupied memory is destructor Class Emp { Int age; Char name ; Float sal; Public: Emp( ); Emp( ); Void show( 0; };\n\nEmp : : emp( ) { Cout << Enter age, name & sal; Cin >> age >> name >> sal; } Void emp : : show( ) Note { A class by default has 3 built in Cout <<age << name <<sal; fucntion } 1. Constructor Emp : : Emp( ) 2. copy constructor { 3. Destructor Cout << Object destroyed; } Note:- Destructor are always called in Void main( ) reverse order. { Emp e, f,; E. show( ); f. show( ); } Create a class c/o student having 3 data members (i) for storing Roll no. (ii) for storing name (iii) for storing marks in subjects The member for storing name should be char , member for storing marks should be int ; Create constructor of class which prompts user to enter length of name, allocates sufficient memory & accepts the name from user. The same constructor asks the user how many subject he want to enter, againallocate sufficient memory for that & accepts the marks given by user. Finally provide appropriate member f n which display % & grade of student. At the end define the destructor in proper manner to deallocate memory allocated by constructor. Class student { Int roll non; Char name, grade; Int marks; Float per;\n\nPublic: Student ( ); Void get( 0; Void calculate( ); Void show( 0 Student ( ) } Student : : student ( ) { Cout << how many letters; Cin >>n; Name = (int ) malloc (n+1) * size of (char)); If (p= = null) Exit(1); Get( ); } Void student : : get( ) { Cout << enter roll no; Cin >> roll ; Cout << enter name; Cin >> name; For (i=0; i<n; i++) { Cout << Enter marks; Cin >> * (q+i); } } Void student : : calculate ( ) { Per=0; For (int i=0; i<n; i++) Per+= * (q+i) Per / = n; If (per >=75) Grade = A; Else if (per >=60) Grade = B;\n\nElse grad = f; } Student : : student ( ) { Free (Q); Free(p); } Void student : : show( ) { } Void main( ) { Student s; s. calculate( ); s. show( ); Note In C++, cin dont accepts the space Void main( ) { Char str; Cout << enter name; Cin >> str; } Cin.getline (str, 80); //Enter Key /9. Prototype of get line ( ) Void get line (char , int ); Member of istream header file COMPARISION BETWEEN CONSTRUCTOR & DESTRUCTOR CONSTRUCTOR 1. are special member f n of a class having same name as that of the class. 2. constructors are automatically called as soon as object of class is created i.e. their calling is implicit 3. constructors can be parameterized. 4. since constructors accepts parameters they can be overloaded & thus a class can have multiple constructors.\n\n5. constructors are called in the orders in which objects are created. 6. constructors can not be inherited. 7. constructors can not be declared as virtual. DESTRUCTOR\n1. are special member fn of class having same name as that of the class but\n\n2. 3.\n4.\n\n5. 6. 7.\n\npreceded with the symbol of tilde a destructor is also automatically called but whenever an object is about to be destructor or goes out of scope thus these calling is implicit. we can not pass parameters to a destructor. as they dont accept parameters we can not overload then, thus a class can not have multiple destructor calling of destructor is always done in reverse or des of the creation of objects. inheriting of destructor is also not possible we can have virtual destructor in a class.\n\nINLINE FUCNTIONS\nInline functions Are those fn whose call is replaced by their body during compilation. Declaring a fn as inline has two major advantages.\n1. compiler does not has to leave the calling fn as it finds definition of fn\n\nbeing called their it self. 2. The overhead of maintaing the stack in belureen function call is seduced Thus declaring a function as inline increases the execution Speed, s\\reuces execution time & thus enhances the overall efficiency of program. But two ptr must be considered before declaring a fn as inline\n1. The definition of inline function must appear before its call i.e. if a non\n\nmember fn is to be made inline then its declaration & definition must appear about main as a single unit. 2. the body of inline fn must be short & small 3. it should not contain any complex keywords or keyword or statement like for, if, while, dowhile do,\n\nif any of the aboul sules are violated then compiler ignores the keyword inline an treats the fn as offline or normal one. Moreover a class can have two kinds of inline functions 1.Implicit Inline :- fn efined within the body of class 2. Explicit Inline :- fn declared within the class but defined outside the class preceded with the keyword inline. Thus at the end we can say, that declaring a fn as line is a request made by programmer which the later may agrel or deny. Class Emp { Char name ; Int age; Float sal; Public : Void get( ) Implicit { Inline cout << enter age, name and sal; Cin >> age >> name >> sal; } Void show( ); }; Explicit Inline inline void Emp : : show( ) { Cout << age << <<name << <<sal; }\n\nVoid main( ) { Emp E; Int I; For (i=0; i<5; i++) E[i]. get( ); For (i=0; i<5; i++) E[i] show( )(; }\n\nSTORAGE CLASSES\nStorage class decides the following 1. Default value 2. life (persistence) 3. scope (accessibility) storage default 1.auto garbage (automatic) 2. static 3. register 4. global Zero garbage zero li ti scope limited to limited to declaration block their Declaration Block throughout The program same as auto same as auto throughout throughout the program The program Static void display ( ); void main ( ) { display( ); display ( ); display( ); } void display( 0 { static int a; cout <<a<<end1; a++; } o/p o 1 2\n\nAuto Void display ( ); Void main( ) { Display( ); Display ( ); Display ( ); } Void display( ) { Cout <<a<<end1; A++; } o/p 3 garbage values will be generated\n\nSTATIC DATA\nStatic data members with in the class Class data { Int a ; Static int b; }; a int Data : : b; d1 D2 static variable dont wait for creation of object, before object creation memory is allocated for then of class 1. A static data member has a single copy shared amongst each object of that class. On other hand if a class has non static data then every object of that class has its own copy of non static data 2. static data members arrive in memory even before objects of the class get created. Because of their feature it becomes necessary to redefine them outside the class so that they can be given space in RAM without the help of object. 3. since they are not related to any particular object of the class & have a single copy in the entire class, a static data member never contributes in the size of object. In other words size of object is always calculated by summing up sizes of non static members. WAP to create a class c/o student having two data members roll & count. The member roll should keep the roll no. allocated to every student object while the members count should keep track of total no. of student objects currently in the memory. Finally provide appropriate members fn to initialize & display the values of Roll no. & count. Class student { Int roll; Static int count; Public: Student (int i) {\n\n0 a\n\nRoll=I; ++ count ; } Void show( ) { Cout << Roll no= <<roll<<end1; Cout << total objects alive = <<count; } Student ( ) { Count; } }; Int student : : count; Void main( ) { Student S=10; Student P=20; Student Q=30; S. show( ); P. show( ); Q. show( ); { Student X = 40; Student Y=50; X. show( ); Y. show( ); } }\n\nSTATIC FUCNITON\n<obj-name>. <function-name> Syntax <class-name> : : <function-name> DRAWBACKS OF STATIC FUNCTIONS\n1. static f n can never non static data numbers. But reverse is true.\n\n## 2. constructor & Destructor can never be made or declared as static.\n\nClass student { Int roll; Static int count; Public : Student (int i) { Roll =I; ++arunt; } Static void show( ) { Count<<Roll.X Count << total objects alive = <<count; } Student ( ) { Count; } }; Int student : : count; Void main( ) { Student S(10). P(20), Q(30); Student : : show( ); { Student X(40), Y(50); Student : : show( ); } Student : : show( ); } Student : : show ( ); Getch( ); } Program :- Create a class c/a employa having data members for storing age, name & id. Also provide another member which stores the next id which will be allocated to next incoming object. Provide a member f n to initialize\n\nthis variable with its initial value & also provide appropriate constructor & member f n for initialized all other members of class. Class Emp { Int age; Char name ; Int id; Static int n-id; Public: Static void init( ) { Id=n-id; } Emp( ) { Cout << enter age, name; Cin >> age >> name; Id=n-id; ++n-id; } Void show( ) { Cout <<age<<name<<id; } Solution :Class Emp { Int hae; Char name; Int id; Static int n-id; Public: Employce (int, char ); Static void init id( ); Void show( ); Static void get next id( );\n\nEmployce( ); }; Employa :: int next-id; Employce : : employce (int I, char * j) { Age =I; Strcpy (name if); Id = next=id; } Void employce : : init id( ) { Next-id=1; } Void employce :: show( ) { Cout <<age<< <name << << id<<end1; } Void employce : : get next id( ) { Cout << next object wil be given id= <<next id; } Employce : : employce( ) { Next id; } Void main( ) { Employce : : init id( ); Employce : : get next id( ); { Employce { (25, Rahul); Employce f (30, vipin); e. show( ); f. show( ); empoloyce : : get next id( ); } Emloyce : : get next id( ); }\n\nTHIS POINTER\nClass Emp { Int age; Char name; Float sal; Public : Void get( ) { Cout << enter age, name & sal: Cin >> age>> name>>sal; Cout << Add of calling object = this; } Void show( ) { Cout <<age<<name<<sal; Cout << Add of calling object= <<this; } } Void main( ) { Emp E, f; e.get( ); f.get( ); e.show( ); f.show( ); }\n1. Every member f n of class has this pointer.\n\n2. No need to declare & initialize this pointer. Compiler initialize it with base address of calling object.\n\nThis Pointer\n\nThis is a special pointer available in every member foundation of a class except static member f n of a class. Whenever the object of a class puts a call to any non static member f n of that class, the this pointer available in that member f n implicitly starts pointing to the address of calling object. Thus we can say that a this pointer always points or referr to the aurrent object. By default class has 3 this pointer. Constructor copy constructor Destructor Accessing Data Member Using this Class Box { Int l, b, h; Public: Box( ); Box (int, int, int( ); Box (Box &); Void show( ); }; Box : : Box( ) { Cout << Enter l, b, and h; Cin >> l >> b >> h; } Box : : Box (int I, int j, int k) { L=I; thisl=I; B=j; thisb=j; H=k; thish=k; } Box : ; Box (Box &p) { L=p.l; this=p; B=p.b; H=p.h }\n\nVoid Box : : show( ) { Cout << l << <<b<< <<h; Cout << this l << thisb << this h; } Void main( ) { Box B1; Box B2 (5, 7, 11); Box B3 = B1; B1. show( ); B2. show( ); B3. show( ); }\n\nLIMITATIONS OF THIS\nThis pointer always points to calling object thus it can not incremented or decremented. It is a constant pointer. Box = q; Q = this+1 valid Q=++this notvalid USING THE CONST KEYWORD 1. 2. 3. 4. Const variable pointer parameters C data members of class 5. member f n of a class\n\nC++\n\nConst Variables Void main ( ) { Const float pi=3.14 Const variable are initialized at the pt of declaration menns they are read only variables & their values can not be manipulated.\n\n## Cout <<pi++ X not valid.\n\nConst Pointer Const int p => pis a pointer a const int void main( ) { Int b=50; Int a =10; Const int p; or int const p P= &a; p=20 P=b; } This pointer comes in this category Int const P P is a const pointer to an integer means P cant be incremented or decremented. Selin they are initialized at the time of declaration. Void main( ) { Int b=50; Int a=10; Int * const P=&a; p=20; P=&b Const parameter Int strlen (const char p) { Int I; For (i=0; (p+i); i++) p= A X // value of parameter can not be changed } Box (const Box &p) // prototype of default copy constructor { L=p.l ++;\n\nB=p.b H=p.h } Const data members of class Class circle { Int rad; Const flaot pi X; // const var are initialized at the point of declaration thus this is not valid statement Circle (int r) : pie (3.14), rad(r) Initialiser Const Member Function:If we want that a f n do not change value of class member than we make them const. Class circle { Int r; Float a; Void get (int x) { R=x; } Void cal-area ( ) { A=r * r * r * 3.14; } Void show ( ) const { Count << Rad= <<r++ << Area= <<a; } If const is there then value of r will not be changed other wise at last it will be incremented.\n\nINITIALISES\nInialises It is a special syntax allotted by C++ used to initialize there thing . 1. const data members with in the class. 2. reference variables 3. calling parameterized constructor of the base class from derived class. They are always written in front of the constructors Orders of initializes must be same as that of the order of data members Class Data { Int x, y; Public: Data (int I, int j): x(i), y(j); { } Data (int I, int j): x(i+j), y(x+i); { } } Ist case Data D (s,10) X=5 Y=10 Data D(7,14) X=21 Y=28 PASSING OBEJCTS USING POINTER Class Box { Int l, b, h; Public: Void get( ) {\n\n// same as previous } Int compare (Box ) }; Int Box : : compare (Box p) { Int x, y; X= lbh; Y=P l* pb* ph; If (x= = y) Return (0); Else if (x>y) Return (1); Else Return (+) } Void main( ) { Box B1, B2; B1. get( ); B2. get( ); B1. show( ); B2. show( ); Int ans ; And = B1. compare (&B2);\n\nif (and= =0) cout << equal; else if (ans >=1) cout << B1 is greater; else cout << B2 is greater }\n\nPassing Objects by Reference Using Reference Variable Class Box { Int l, b, h; Public: Void get( ) { // same as previous; } Void show( ) { // same as previous; }\n\nInt compare (Box &); }; Int Box : : compare (const Box &P) { Int x, y; X = lbh; Y = p.lp.bp.h; If (x==y) Return (0); Else if (x > y) Return (1); Else Return (-1); } Void main( ) { Box B1, B2; B1.get( ); B2.get( ); B1.show( ); B2.show( ); Int and; Ans=B1. compare (B2); If (ans= =0) Cout << equal; else if (ans >1) cout << B1 is greater; else cout << B2 is greater; }\n\nFRIEND FUNCTIONS\nA friend function is a special function which despite of not being member f n of class has full access to the private, protected members of the class.\n\n## ATTRIBUTES OF FRIEND FUNCTIONS\n\n1. If a fn is to be made friend of a class then it has to be declared\n\nwithin the body of the class preceded with the keyword friend.\n\n## 2. whenever a friend f n is defined neither, the name of class nor scope\n\nresolution operator appears in its definition. Moreour the keyword friend also does not appear. 3. whenever a friend fn is called, neither the name of the object nor dot operator appears toward its left. It may however accept the object as a parameter whose members it wants to access. 4. It does not matter in which diction of class we declare a friend function as we can always access it from any portion of program 5. If a friend fn want to manipulate the values of data members of an object it veds the reference of object to be passed as parameter Example:- // Demonstrating the use of friend function Class student { Int roll; Char grade; Float per; Public: Void get( ); Friend void show (student); }; Void student : : get( ) { Cout << enter roll. Grade & per; Cin >> roll>> grade>>per; } Void show (student P) { Cout << p.roll<< << P.grade<< <<P.per; } Void main( ) { Student S; S. get( ); Show(S); } Demonstrating the use of friend function\n\nClass student { Int roll; Char grade; Float per; Public: Friend void get (student &); Void show( ); }; Void get (student & P) { Cout << enter roll, grade & per; Cin >> P.roll >> P.grade >> P.per; } Void student : : show( ) { Cout << roll << grade << per; } Void main( ) { Student S; Void get(S); s. show( ); } Class student { Int roll; Char grade; Float per; Public: Friend void get (student P); Void show( ); }; Void get (student q) { Cout << enter roll, grade & per; Cin >>q roll >> q grade >> q per; } Void student : : show ( )\n\n{ Cout << roll << grade << per; } Void main( ) { Student s; Get (&s0; s.show( ); } A FUNCTION BEING FRIEND OF TWO CLASSES Class Beta; // forward declaration Class Alpha { Int x; Public: Void get( ) { Cout << enter x=; Cin >> x; } Friend void compare (Alpha, Beta); }; Class Beta { Int y; Public: Void set( ) { Cout << enter y=; Cin >>y; } Friend void compare (Alpha, Beta); } Void compare (Alpha A, Beta B) { If (A, X > B. Y) Cout << greater= << A, X;\n\nElse if (B.Y> A.X) Cout << Greater= << B.Y; Else Cout << equal; } Void main( ) { Alpha Obj1; Beta Obj2; Obj1. get( ); Obj2. set( ); Compare (obj1, obj2); }\nADDING TWO OBJECTS OF A CLASS USING MEMEBR FUCNTIONS\n\nClass Distance { Int feet; Int in cher; Public: Void get( ) { Cout << enter feet & inches; Cin >> feet >> inches; } Distance add (Distance &) Void show( ) { Cout << feet= << feet<< inches= <<inches; } }; Distance Distance : : add (Distance &P) { Distance temp; Temp. feet = feet +P. feet; Temp.inches= inches + P.inches; If (temp. inches>=12) { temp. feer += temp. inches/12;\n\ntemp. inches % =12 } Return (temp); } Void main( ) { Distance D1, D2, D3; D1. get( ); D2, get( ); D3 = D1. add (d2); D3. show( ); } Class Distance { Int feet, inches; Public: Void get( 0 { // same as previous } Void add (distance &, distance &); Void show( ) { // same as previous; } }; Void distance : : add (distance &d1, distance &d2) { Feet = d1. feet + d2, feet; Inches = d1. inches + d2, inches If (inches >12) { Feet = feet + inches /12; Inches % = 12; } } Void main( ) { Distance d1, d2, d3; d1.get( );, d2.get( );\n\n## D3, add (d1, d2); D3. show( ); }\n\nADDING TWO AF CLASS UISNG PRIED FUNCTION Class distance { Int feet; Int inches; Public: Void get( ) { Void << enter feet & inches; Cin >>feet >> inches; } Void show( ) { //same } Friend distance add (distance, distance ): }; Distance add (Distance P, Distance Q) { Distance temp; Temp. feet = P.feet +Q.ffet; Temp. inches= P.inches + Q.inches; If (temp. inches>=12) { Temp.feet += temp. inches/12; Temp. inches % = 12; } Return (temp); } Void main( ) { Distance D1, D2, D3; D1.get( );\n\n## D2.get( ); D3 = add (D1, D2) D3.show( ); }\n\n{ Count = c } Void operator ++( ); Void show( ) { Cout << count <<end1; } }; Void counter : : operator ++ ( ) { ++ count; } Void main( ) { Counter a = 10; // Single parameterize constractor a.show( ); ++ a; // a. operator + +( ); a. show( ); } Class counter { Int count; Public: Counter( ) { Count = 0; } Counter (int c) { Count = c; } Counter operator ++ ( ); Void show( ) { Cout << count; } }; Counter &\n\nOr Counter counter : : operator ++ ( ) { Counter temp; ++ count; or Temp. count = count; ++ count Return (temp); return (* this); } Void main( ) { Counter c1 = 10, c2; C1. show( ); C2 = ++ c1; C1. show( ); C2. show ( ); } Overloading Of Post Increment Operator Using Member Function Of this class Class counter { Int count; Public: Counter( ) { Count=0; } Counter (int C) { Count = c; } Counter operator ++ (int); Void show( ) { Cout << count; } }; Counter counter : : operator ++ (int) {\n\nCounter temp; Temp. count = count++; Return (temp); } Void main( ) { Coun c1 = 10, c2; C2 = c1++; C1. show( ); C2. show( ); } Overloading Unary Operator As Friend Of The Class Class counter { Int count; Public: Counter( 0 { Count=0; } Counter (int i) { Count = I; } Void show( ) { Cout << count << end1; } Friend void operator ++ (counter &); }; Void operator ++ (counter &C) { ++ c.count Return (C); } Void main ( ) Or counter temp; temp. count = c.count++; return (temp);\n\n{ Counter C1 = 10, C2; C1. show ( ); C2. = ++(); C1. show ( ); C2. show( ); } Note:when unary operators are overloading using member function then dont accept any argument but when they are overloaded using friend fn then they have one argument of type object (class). Class counter { Int count; Public: Friend counter operator ++ (counter &); Counter { Count =0; } Count (int i) { Count = I; { Void show( ) { Cout << count << end1; } }; Void main( ) counter & operator ++(counter &c) { { Counter c1 =10, c2; counter temp; C1. show( 0; temp. count = C, count; C2= ++ C1; C1. show( ); return (temp); C2. show( ); } } Overloading Binary Operators As Member Functions Of The Class\n\nD3 = D1 + D2; D3 = D1. operator + (D2); Class Distance { Int feet, inches; Public: Void get( ) { Count << enter feet and inches; Cin >> feet >> inches; } Void show( ) { Cout << feet << inches; } Distance operator + (Distance); }; Distance Disttacne : : operator + (Distance P) { Distance t; t.feet = feet + P. feet t. inches = inches + P. inches; if (t.inches>= /12) { t.feet + = t.inches/12; t.inches % = 12; } Return (t); } Void main ( ) { Distance D1, D2, D3; D1, get( ); D2, get( ); D3 = D1+D2; D3. show( ); Getch( ); }\n\nOverloading Binary Operator Using Friend Function Class Distance { Int feet, inches; Public: Void get( ) { Cout << enter feet and inches: Cin >> feet>> inches; } Void show( ) { Cout << feet<< inches; } Friend Distance operator + (Distance, Distance); }; Distance Operator + (Distance p, Distance Q) { Distance t; t.feet = P.feet +Q.feet; t.inches = P.inches+ Q.inches; if (t. feet>=12) { t.feet = t.feet +t.inches/12; t.inches % = 12; } Return (t); } Void main ( ) { Distance D1, D2, D3; D1.get( ); D2.get( ); D3=D1+d2; D3.show( ); } Assignment:D2 = D1+n\n\nD3 = n + D1 Class Distance { Int feet, inches; Public: Void get( ) { // same as previous } Void show( ) { // same as previous } Distance Distance : : operator + (int n) { Distance temp; Temp feet = feet + n; Temp. inches = inches + n; If (temp. inches > = 12) { Temp. feet + = temp. inches / 12; Temp. inches % = 12; } Return (temp); } Distance operator + (int P, Distance Q) { Distance temp; Temp.feet = P+Q.feet; Temp. inches= P+Q. inches; If (temp. inches> = 12) { Temp.feet = temp. feet + temp. inches/12; Temp. inches % = 12; } Return (temp); } Void main( ) {\n\nDistance D1, D2, D3; Int n; Cout << enter an integer; Cin >> n; D1. get ( ); I D2 = D1+n; // can be done using member fn & friend fn D2. show( ); Cout << Enter an integer; Cin >>n; D3 = n+D1; // not possible through member fn D3. show( );\n\nII }\n\nNote:II can not be done using member function becoz n is an integer and only an object can call function not integer. So I call can be made using member function as well as friend function but II can be done only friend function Assignment :D1+=D2 using member fn & friend fn\n\nOverloading Relational Operators If (D1= = D2) Compiler if (D1.operator = = (D2) ) Class Distance { Int feet, inches; Public: Void get( ) { Cout << enter feet & inches; Cin >> feet >> inches; } Void show( ) { Cout << feet << << inches; }\n\nInt operator = = (distance D); }; Int Distance : : operator = = (Distance D) { Int x, y; X= feet &12 + inches; Y = D. feet 12 + D.inches; If (x= = y) Return (1); Else Return (0); } Void main( 0 { Distance D1, D2; D1. get( ); D2. get( ); D1. show( ); D2.show( ); If (D1= = D2) Cout << equal; Else Cout << not equal; } Assignment:Modify the above program so that now your cosle prents either of there message i. Object are equal. ii. Dl is greater. iii. D2 is greater Use minimum possible operator Overloading Binary Operator On String Class string { Char str; Public:\n\nString ( ) { Str = \\0; } String (char P); { Strcpy (str, P); } Void show( ) { Cout << str; } String operator + (string S) }; String string : : operator + (string S) { String t; Int I, j; For (i=0; str[i]; i++) t.str[i] = str[i]; for (j=0; s.str[j]; i++, j++) t.str[i] = s.str[j]; return (t); } Void main( ) { String s1= Hello; String s2 = How are you ?; String s3; S3 = s1 +s2; S1. show( ), s2. show( ); }\n\ns3. show( );\n\nAssignment :WAP which compares two string objects & checks which one is greater. Operators Which Can Not Be Overloaded\n1>\n\n:: (scope resolution operator) it already works on classes we overload only those operators which works on primitine and not on non prin 3> ?: (conditional Operator) It requires there arguments overloading of operators atmost can take 2 argument 4> >* Pointer to member operator 5> Size of operator\n2>\n\n## DYNAMIC MEMORY ALLOCATION (DMA)\n\nMALLOC 1. It is a function 2. Necessary to include the header file 3. It takes no, of bytes as parameter 4. brings memory from local heap 5. Maximum memory that malloc can allocate = 64KB 6. Malloc retirens the address of Allocate block in from of (void) Thus its return value has to be type Casted accordingly 7. when malloc fails it returns null 8. Malloc only creates a new block But not call the constructor of the Class for initialized the data member Of that block Such NEW It is a operator header file not required It takes no. of elements Required as parameter bring has no such memory store New has no such memory limitation New automatically converts according to the datatype given thus no need to explicitly type cast its return value. when new fails it returns zero on the other hand operator new can allocate memory as well as call the class for initializing the members of block created. Constructors are called dynamic Constructors. Syntax for new;(i) (ii) new <data type> new <data-type> [int];\n\nAllocations Of New :1. int p; P= new int (10); Cout << P; Delete P; new allows programmer to initialize the memory allocated but malloc does not provide nay such feature\n\n2000 P 2. int &P; P=new int ; Delete [ ] P; 3000 P Syntax for deletes .. 3000 2000\n\n10 2002\n\nHere value 10 denotes that programmer is interested in creating memory for elements\n\n1. Delete <ptr-name>; // deletes one block of data type 2. Delete [ ] <ptr-name>; // deletes whole allocated memory Eg:- float P; Delete P; // four bytes free\n\nDelete is a request to OS it does not work immediately where as new creates memory immediately WAP which creates a dynamic array of user defined size Accept values from user in that array & then find largest element in that array along with its position. Finally it should display the largest element & delete the array. Void main ( ) {\n\nInt n; Cout << How many integers; Cin >> n; Int P; P+ new int [n]; If (P= = 0) { Cout << Insufficient memory; Getch( ); Exit(1); } For (int i=0; i<n; i++) { Cout << enter element; Cin >> (p=i); } Int max = p; Int pos = 0; For (i=1; i<n; i++) { If ( (p+i) > max) { Max = * (p+i); Pos = I; } } Cout << largest element= << max; Cout << Position = << pos; Getch( ); Delete [ ] P; } WAP to create a class c/a string having a character pointer P. provide constructor in the class which accept int as parameter and allocates dynamic array of characters of special size. Now provide following member fn in the class. 1. get string which prompts users to enter string in dynamic array. 2. show string which display the string 3. Reverse string which accept an object as parameter and copies the\n\nReverse of string stored in calling object in object Passed as parameter. Class string { Char p; Public : String (int); String ( ); Void show str( ); Void reverse str ( str &); Void get str( ); }; String : : string (int size) { P = new char [size +1]; If (P= =0) Exit (1); N = size +1; } Void string : : get str( ) { Cout << enter string; Cin. Getline (P,n); } Void string : : reverse str (string &s) { Int I, j; For (i=0, j=strlen (p)-1; j>=0; j--, i++) { s.p[i] = p[j]; } S. P [i] = \\0; } Void string : : show str( ) { Cout << P << end1; }\n\nString : : string ( ) { Delete [ ]P; Cout << Array destroyed; } Void main ( ) { Int n; Cout << How may character; Cin >> n String S1 =n; String S2 =n; S1. getstr ( ); S1. reverse str(S2); S1. show str( ); S2. show str( ); }\n\nDYNAMIC OBJECTS\nAllocating memory for class with the help of new. Class Emp { Int age; Char name; Float sal; Public: Void get( ) { Cout << enter age, name & sal; Cin >> age >> name >> sal; } Void show( ) { Cout << age << << name<< <<sal; } }; Void main( ) {\n\nEmp P; P=new emp; If (p= =0) { Cout << Memory Insufficient; Exit (1); } p get( ); p show( ); getch ( ); delete p; }\n\n## ARRAY OF DYNAMIC OBJECTS\n\nModify the previous code so that now your program creates an array of a objects where n is given by user. Then it should accept values for each object and finally display them. Class Emp { Int age; Char anem; Float sal; Void get( ) { Cout << enter are, name and sal; Cin >> age>> name>> sal; } Void show( ) { Cout <<age<< name<<sal<<end1; } };\n\nVoid main( ) { Emp p; Int I, n; Cout << how many employees?; Cin >> n; New Emp [n]; If ( p = =0) { Cout << error in crating objects; Exit (1); } For (i=0; i<n; i++) (p+i) get( ); or p[i]. get( ); For (i=0; i<n; i++) (p+i) show( ); or p[i]. show( ); Getch( ); Delete[ ]p; } Enhance the above program so that after accepting records of n employees. Your program prompts the user to enter another name & display the record of that employee only. If employee name is not available then program should display the message record not found. Class Emp { Int age; Char name ; Float sal; Public: Void get( ) { Cout << enter age, name and sal; Cin >> age>> name>> sal; } Void show( ) { Cout << age<< name << sal; }\n\nInt compare (char ); }; Void main ( ) { Emp p; Int n; Cout << how many employees?; Cin >> n; P= new emp [n]; If (p = =0) { Cout << Insufficient Memory; Exit (1); } For (i=0; i<n; i++0 (p+i) get( ); or p[i]. get( ); Cout << enter name to search; Char str ; Cin. I gnore( ); Cin. Get line (str, 20); For (i=0; i<n; i++0 { If ( (p+i) compare (str) = = 0) { (p+i) show ( ); Break; } } If (I = = n) Cout << Record not found; Getch ( ); Delete [ ] p; } Int compare (char p) { Return (strcmp ( name.p) ); } Assignment\n\nWAP which creates an array of n objects. Accept values from user in then & display them Now sort the records on the basic of name in ascending order & again display the sorted record Class age; { Int age; Char name ; Float sal; Public: Void get( ) { Cout enter age, name and sal; Cin >> age >> name >> sal; } Void show( ) { Cout << age << name << sal; } Void sort (int); }; Void Emp : : sort (int r) { Int j, I, t; Char temp ; For (i=0; i<r; i++) { For (j=i+1; j<r; j++) { If (str cmp( (this +i) name, (this +j) name)>0) { Strcpy (temp, (this +i) name); Strcpy ((this +i) name, (this+jname); Strcpy ((this+j)naem, temp); T=(this+i) age; (this +i) age = t; T= (this+i) sal; (this +i)sal (this+j)sal; or Emp F; (this+i) = (this+j); (this+j) = F;\n\n## (this+j) sal=t; } //closing of it } // inner for } /outer for } // closing of function\n\nDYNAMIC COSNTRUCTORS\nClass Emp { Int age; Char age; Char name ; Float sal; Public; Emp( ) { Cout << enter age, name and sal; Cin >> age >> name >> sal; } Emp (int I, char j , float k) { Age = I; Strcpy (name, j); Sal = k; } Void show( 0 { Cout << age << name << sal; } Emp( ) { Cout << Object destroyed; } }; Void main ( ) { Emp P, q; P=new Emp;\n\nQ=new Emp (30, vineet, 200000); p show( ); q show( ); delete q; delete p; } Note:- The above Program will not call the destructor of the class even at the termination. This is because in C++ memory which is allocated using new can only be deal located using delete and calling of destructor is only made when memory gets deallocated. Since in above code, call for delete is not present so memory block still remains in RAM & might be collected in future through garbage collection. Thus if memory is freed for a dynamic object it can only be done through delete operator. Terms Used For Dynamic Objects Live Object :- Those object which are initialized using constructor are constructor are c/o live objects. Partially live objects\n\nINHERITANCE\n1. Single Inheritance A Base class 2. Multi level Inheritance A indirect base class of C\n\nDerived class\n\nB c\n\nbase class of C\n\n3. Multiple Inheritance A B\n\n4. Hierarchial Inheritance A\n\n## 5. Hybrid Inheritance or Multipath Inheritance A B C\n\nD Syntax for Inheritance:Class <class name>: public / private / protected < clas-name> Derived class Mode of Inheritance Class Box { Int l, b, h; Public: Void get( 0 { Cout << enter l, band; Cin >> l>>b>>h; } Void show( ) { Cout <<l<< << b<< << h; Base class\n\n} }; Class carton : public Box { Char type; Public: Void set( ) { Cout << enter material name; Cin. Get line (type, 20); } Void display ( ) { Cout << material = << type; } }; Void main( ) { Carton obj; Obj. get( ); Obj. set( ); Obj. show( ); Obj. display( ); } Accessability Rules When Mode Of Inheritance Is Public When a base class is inheritance in public mode then :1. All the public members of base class becomes public members of derived class i.e. they can be accessed through the function of derived class as well as by objects of derived class. 2. All the protected members of base class becomes protected members of derive class & thus can only be accessed through functions of derived class but not by the object (main) of derived class. 3. All private members of base remain private to their own class & thus can neither be accessed through functions of derive class nor by object of derive class. EARLY BINDING:-\n\nIt is a prours which is executed during compilation in which compiler binary the fn body with the fn call i.e. even before program starts executing decision regarding the fn call and fn body to be executed are taken by the compiler since this happens before execution, we can say it is early binding. Early binding is always an action until & unless keyword virtual is used infront of the fn return type. It is done on the basis of there criterias:(i) (ii) (iii) function name or calling object type or function parameter type in other words early binding is always in action in normal fn calls, overloaded fn calls and overloaded operators The major benefit of early binding is speed of execution i.e. function calls which are bound using early binding get executed at a fastest speed as compared to late binding fn calls. OVERRIING The function overriding is a term which is used when a derive class contains a function with the same prototype as its bass class. In other words, fn provided by base class has same prototype in derive class but in different body. Overriding Scope must be different By the classes related by Inheritance Prototype of functions Must be same . Overloading always in same class means at single level prototype must be different\n\nWhen Mode Of Inheritance Is Protected When base class is inheritance in protected mode then:-\n\n1. All public members of base class becomes protected member of derive class i.e. they can be accessed only through function of derived class but not by object of derive class. 2. All protected members of base class become protected members of derive class i.e. they two can only be accessed through functions of derived class but not by object of derive class. 3. All private members of base class remains private to their own class & thus neither accessible through the object nor through functions of derive class. Class Num { Protected: Int a, b; Public: Void get( ) { Cout << enter two numbers; Cin >> a >> b; } Void show( ) { Cout << Numbers are =; Cout <<a << << b; } }; Class Add Num : : protected Num { Protected : Int c; Public: Void set( ) { Get( ); } Void add( ) { C=a+b; }\n\nVoid display( ) { Show( ); Cout << sum= <<c; } }; Void main ( ) { Add Num obj; Obj. set( ); Obj/./ add( ); Obj. display( ); }\n\n## PRIVATE MODE OF INHERITANCE\n\nWhen a base class is inheritance in private mode then:1. All the public members of base class become private members of\n\nderive class i.e. they can be only accessed through fn of derive class but not by the objects of derive class. 2. All the protected members of base become private members of derive class i.e. they two can only be accessed through the function of derive class but not by objects of derive class. 3. All private members of base class remain private to their own class & thus can neither be accessed by fn nor by objects of derives class. Class Num { Protected : Int a, b; Public: Void get( ) { Cout << enter a and b; Cin >> a >> b; } Void show( ) { Cout << a = <<a<<end1;\n\nCout << b= << b<< end1; } }; Class Add Num: Private Num { Protected : Int c; Public: Void set( ) { Get( ); } Void add( ) { C=a+b; } Void display ( ) { Show( ); Cout << same = <<c; } }; Void main( ) { Add Num obj; Obj.set( ); Obj. add( ); Obj. display( ); } Note:- At single level Inheritance private & protected inheritance seems to be similar.\n\nMULTILEVEL INHERITANCE\nClass Num { Protected: Int a, b;\n\nPublic: Void get( ) { Cout << enter a nad b: Cin >> a >> b; } Void show( ) { Cout << a= <<a<<end1; Cout << b= <<b<<end1; } }; Class Add-Num : public Num { Protected : Int c; Public : Void set( ) { Get( ); } Void display( ) { Show( ); Cout << sum= <<c; } Void add( ) { C=a+b; } }; Class Diff Num : public Add Num { Int d; Public: Void accept( ) { Set( ); }\n\nVoid diff( ) { D= a b; } Void print( ) { Display ( ); Cout << Difference= <<d; } }; Void main( ) { Diff Num obj; Obj. accept( ); Obj. add( ); Obj. diff( ); Obj. print( ); } Program Class counter { Protected: Int count; Public: Void init (int a) { Count =a; } Void operator ++( ) { Count ++; } Void show( ) { Cout << count= <<count; } }; Class Dec Counter : public counter\n\n{ Public: Void operator - -( ) { - - count; } }; Void main( ) { Dec counter D; D.int(10); D.show( ); ++D; D.show( ); - - D; D. show( ); } Assignment :Create a class c/a array which contains an integer array of size 10. The class should have two member functions called get arr and showarr, which should accept values in the array & display its values respectively. Now create a derive class of array c/a sortarr, the class should accept a string as parameter if string contains asc then sorting should be done in ascending order & if it contains desc sorting should be done in descending order. Finally create the function main which should contain menu drive interface for the user having 5 options:(i) (ii) (iii) (iv) (v) input display sort in ascending sort in descending quit\n\n## Solution #include <iostream.h> #include <stdlib.h> Class Array { Protected:\n\nInt a; Public: Void get( ); Void display( ); }; Void Array : : get( ) { Int I; Cout << enter array elements; For (i=0; i<5; i++) Cin >>a[i]; } Void Array : : display ( ) { For (int i=0; i<5; i++) Cout << \\n elements are = << a[i]; } Class sortarr : public Array { Public: Void ascsort( ); Void descsort( ); }; Void sortarr : : arcsort ( ) { Int I, j, t; For (i=0; i<5; i++) { For (j=0; j<4; j++) { If (a [j] > a [j+1]) { T = a [j]; A [j] = a [j+1]; A [j+1] = j; } } } }\n\nVoid sortarr : : descsort( ) { Int I, j, t; For (i=0; i<5; i++) { For (j=0; j<4; j++) { If (a[j] <a [j+1]) { T=a[j]; A[j]=a[j+1] A[j+1] = t; } } } } Void main( ) { Clrscr( ); Sortarr sr; Int ch=0; Do { Cout << \\n \\t Enter (1) for Input data; Cout << \\n \\t Enter (2) for Display Data; Cout << \\n \\t Enter (3) sort in Ascending order; Cout << \\n \\t Enter (4) sort in Descending order; Cout << \\n \\t Enter (5) for quit; Cout << enter choice; Cin >> ch; Clrscr( ); Switch (ch) { Case (1): Sr.get( ); Break; Case (2): Sr.display( ); Break;\n\nCase (3): Case (4): Case (5): Exit(1) } } while (ch !=5) Getch( ); } Sr.ascsort( ); Break; sr.descsort( ); Break;\n\nMULTIPLE INHERITANCE\nClass Base 1 { Protected: Int a; Public: Void get( ) { Cout << enter a =; Cin >>a; } Void show( ) { Cout <<a << end1; } }; Class Base2 { Protected: Int b; Public: Void set( ) { Cout << enter b=; Cin >> b; } Void display( )\n\n{ Cout <<b << end1; } }; Class drv : public base1, public base2 { Int c; Public : Void accept ( ) { Get( ); Set( ); } Void add ( ) { C = a+d; } Void print( ) { Show( ); Display( ); Cout << sem = <<c; } }; Void main( ) { Drv obj; Obj. accept( ); Obj. add( 0; Obj. print( ); } Program :Base & derive having the function with same name & arguments. Class Base1 { Protected:\n\nInt a; Public: Void get( ) { Cout << enter a=; Cin >>a; } Void show( ) { Cout << a << end1; } }; Class Base 2 { Protected: Int b; Public: Void set( 0 { Cout << enter b=; Cin >> b; } Void show( ) { Cout <<b<<end1; } }; Class drv : public Base1, public Base2 { Int c; Public: Void accept( ) { Get( ); Set( ); } Void add( ) { C= a+b;\n\n} Void print( ) { Will show show( ); Error ambiguity error } }; Void main( ) { Drv obj; Obj. accept( ); Obj. add( ); Obj. print( ); }\n\n## base1: : show( ); Base2: : show( );\n\nRole Of Constructor & Destructor In Inheritance As a basic rule in inheritance, it a base class contains a constructor and destructor as well as derive class also contains constructor & destructor, then when object of derive class is created, constructor of base class gets executed first followed by constructor of derive class and the destructor is called in reverse order i.e. the destructor of derive class is called first followed by the destructor of base class. Thus we can say constructor are always called in the order of inheritance and destructor are called in reverse order. Class Base { Public : Base ( ) { Cout << In bases constructor <<end1; } Base( 0 { Cout << In bases destructor <<end1; } };\n\nClass drv: public name { Public: Drv( ) { Cout << In derives const <<end1; } Drv( ) { Cout << In derives destructor <<end1: } }; Void main( ) { { Drv obj; } Getch( ); } Class Base { Protected a, b; Public: Base (int I, int j) { A = I; B = j; } Void show( ) { Cout << a << << b; } }; Class drv : public base { Int c; Public: Drv ( ): base (10, 20)\n\n{ C=a+b; } Void show( ) { Base : : show( ); Cout << sum = << c; } }; Void main( ) { Drv obj1 Drv obj2 Obj1. show( ); Obj2. show( ); } Constructor Calling In Multiple Inheritance Base 1 Int a; Base1 (int); Void show( ); base2 int b; base2 (int); void display rv nt c;\n\nClass Base1 { Protected: Int a; Public: Base (int i) { A = I; } Void show( ) { Cout << a; } };\n\nClass Base2 { Protected : Int a; Public: Base2 (int j) { B = j; } Void display( ) { Cout << b; } }; Class drv: public base1, public base2 { Protected : Int c; Public: Drv (int p, int q) : base 1 (p), base2(q) { C=a+b; } Void print ( ) { Show ( ); Display( ); Cout << their sum = << c; } }; Void main( ) { Drv obj1 (10, 20); Drv obj2 (20, 70); Obj1. print ( ); Obj2. print( ); } Note :-\n\nIf the constructor of base class is parameterized then (i) (ii) derive class must have constructor Base class constructor should be called in derives constructor.\n\nConstructor Calling In Multilevel Inheritance Class Base1 { Protected: Int a ; Public: Base 1(int i) { A=1; } Void show( ) { Cout << a << end1; } }; Class Base2 { Protected: Int b; Public: Base2 (int I, int j): Base1 (i) { B=j; } Void display( ) { Cout << b << end 1; } }; Class drv : public Base1, public Base2 { Protected: Int c; Public:\n\nDrv (int x, int y) : base2 (x, y) { C=a+b; } Void print ( ) { Show( ); Display ( ); Cout << their sum= << c; } }; Void main ( ) { Drv obj(10, 20); Drv obj (50, 60); Obj1. print( ); Obj2. print( ); } Note Constructors can not be inherited :Constructors are special member fn which are exclusively used for initialing private data members of class. Now since the private data of class is not passed on to its derive classes, so the functions which explicitly initialize then (constructor) are not inherited. Same is the case with destructor,\n\nHIERARCHIAL INHERITANCE\nClass num { Protected : Int a, b: Public : Num (int I, int j) { A = I; B = j; }\n\nVoid show( ) { Cout << a = << a; Cout << b= << b; } }; Class Add Num : public Num { Int c; Public: Add num (int I, intj) : Num (I, j) { C=a+b; { Void show( ) { Num : : show( ); Cout << sum = <<c; } }; Class Diff num : public Num { Int d; Public : Diff Num (int x, int y) : Num (x, y) { b= a b; } Void show( ) { Num : : show( ); Cout << Difference = << d; } }; Void main( ) { Add Num addobj (10, 20); DiffNUm diffobj (30, 70); Add obj. show( );\n\nDiffobj. Show( ); }\n\nHYBRID INHERITANCE\nClass Base { Public: Int a; }; Class drv1 : virtual public base { Public: Int b; }; Class drv2: virtual pubic base { Public: Int c; }; Class drv3 : public drv1, public drv2 { Public: Int d; }; Void main( ) { Drv obj; Obj. a =10; Obj. b = 20 Obj. c 30; Obj.d= obj a + obj.b + obj. c; Cout << sum = << obj,d; }\n\nGRATING ACCESS\nClass Data\n\n{ Private : Int a; Protected : Int b; Public : Int c; }; Class drv : protected Data { Public: Data : : C // bring C in public mode instead of protected }; Foreg:Class Bank { Public : Void deposit( ); Void withdraw( ); Void int-cal( ); }; Class saving acct: Private Bank { Public : Bank : : deposit; Bank : : with draw; };\n\nPOLYMORPHISM\nClass Base { Public: Void show( ) { Cout << In base & show; } };\n\nClass drv : public Base { Public : Void show( ) { Cout << In drvs show; } }; Void main( ) { Base b, ptr; Drv d; Ptr = & b; Ptr show( ); Ptr = & d; Ptr show( ); } VIRTUAL FUNCTION & MULTILEVEL INHERITANCE Class Base { Public: Virtual void show( ) { Cout << In bases show; } }; Class drv1 : public Base { Public : Void show( ) { Cout << In drv1s show; } }; Class drv2: public drv1 { Public: Void show( )\n\n{ Cout << In drv2s show; } }; Void main( ) { Base ptr, b; Drv1 d1; Drv2 d2; Ptr = & b; Ptr show( ); Ptr = & d1; Ptr show( ); Ptr = &d2; Ptr show( ); } Note :Top level class ptr can access member of lower level class. Virtual is mandaroty in base show as function is originally from base. Virtual functions are those functions for which no decision regarding the call and the definition to be executed is token by the compiler during compilation. In other words, if a fn is preceded with keyword virtual then if never become the past of early binding & compiler delays its binding until runtime. Thus all the decisions regarding the call and the body to be executed are taken at runtime. These decisions are not based on the type of caller (as was the case with early binding) but on the bases of contents of the caller. If the calling pointer is storing the address of base class object then bases version of virtual function will be executed & if it pointing to an object of derive class then derive version of virtual fn is executed. But to fully uses the potential of virtual function the derive class while giving its own body /definition for virtual fn must keep its prototype same as the base class i.e. derive should override the virtual function of base class if it wants to place its own definition of virtual function. This is because pointer of base class can access only those function of derive class which are overridden in the derive class but not those which are hidden or added by derive class.\n\nInternal Working Of Virtual Function Class A { Int a; Public: Void f1( ) { } Virtual void f2( ) obj1 VPTR a { } }; Class B : public A { int x; Public: Void f4( ); Void f2( ); obj2 VPTR }; VTABLE FOR A &A : : f2( ) &A : : f3( 0\n\n## VTABLE FOR B &B : : f2( ) &A : f3( )\n\nVoid main( ) { size of class A = 4 bytes A obj1; 2 bytes for variable a Ptr = & obj1; 2 bytes for VPTR Ptr f2( ); Ptr = & obj2; Ptr f3( ); Ptr f3( ) } This will not be executed since this is not virtual & compiler will go to Early binding table for base class A where there is no function f4.\n\nVTABLE:For every class in C++ which contains at least one virtual function a special look up table for storing addresses of there virtual function is created which is known as VTABLE or virtual Table. Thus in short VTABLE is a table containing addresses of virtual functions of the class as well as virtual function of base class if they are not overridden. This table is then consulted by the compiler when a call for virtual function is encountered during runtime. VVPTR:(Virtual Void Pointer) For every class which contains a virtual function special pointer is created by the compiler known as VVPTR, which is used for pointing to the virtual table. Thus every object of class containing virtual function has its size incremented by2. Now whenever compiler encounters call for virtual function through a pointer, it first refers to the address of object to which pointer is pointing from their it reads the address contained in VVPTR of the object the which is the address of VTABLE of class. Lastly within the VATBLE it executes the virtual function called. Thus virtual function enhances the size of code as reduces the execution speed but provides a flexible way of designing program which can respond to the changes which accur at run time. Class Data { Int a ; Data (int ); Public: Static Data get Objects( ); Void show( ); }; Data : : Data (int i) { A = I; } Data Data : : get object( ) { Data D (10); Return (D); } Void Data : : show( )\n\n{ Cout << a; } Void main( ) { Data D = Data. Get object( ); D. show( ); } Note When only one object of class is created then that type of class is c/a single to n class. Polymorphism Compile Time Polymorphism 1. Function Overloading 2. Operator Overloading 3. Early Binding Virtual Inheritance Run Time Polymorphism 1. Function Overriding 2. Virtual Function 3. Late Binding avoids multiple copies Polymorphism (converts parly binding to late binding) Class Figure { Protected: Int dim1, dim2; Public : Void get( ) { Cout << enter 1st & 2nd dimension; Cin >> dim1>> dim2; } }; Class Rectangle : public Figure { Public:\n\nVoid area( ) { Cout << area of Rectangle=; Cout << dim1 * dim2; } }; Class Triangle : public Figure { Public: Void area( ) { Cout << area = <<5 * dim1 dim2; } }; Void main( ) { Figure p; Figure F; P = &F; p get( 0; p area( ); rectangle R; p=&R; p get( ); p area( ); triangle T; p = & T; p get( ); p area( ); }\n\n## PURE VIRTAUL FUNCTIONS\n\nDef:- If a virtual functions declarations is equated with zero then it is known as pure virtual fn. In other words, it we do not want to provide any definition for a virtual fn wee can equate with zero then it will be known as\n\npure virtual fn. Thus by definition a pure virtual fn is one which has no body define with in its own class or base class.\n\nABSTRACT CLASS\n1. If a class contain at least one pure virtual fn than it is known as\n\nabstract Base class. 2. we can never create any object of abstract class but we can always create its pointers. This is because whenever an object of a class containing virtual function is created a VTABLE is setup by the complete, which stores the address of virtual function has no body it can not have memory address & thus it can not places in VTABLE. So to prevent any accidental calls to a non-existing pure virtual function compiler prohibits the creation of an object of an abstract class. 3. An class which extends an Abstract Base class must provide its own definition of pure virtual fn available in its base class other wise the class itself would be created as an abstract class & then it too can not be instantiated. Note:- Constructors can not be virtual since link between VPTR VTABLE is made by constructor Virtual Destructor:Class Base { Protected: Public: Int p; base( ) { P=new int ; Cout << p constructed!; } base { Delete [ ] p; Cout << memory deallocated; }\n\nVirtual\n\n}; Class drv : public Base { Int q; Public : Drv( ) { Q = new int ; Cout << q constructed; } Drv( ) { Delete [ ] q; Cout << memory deallocated for q; } }; Void main( ) { Base ptr; Ptr = new drv; Delete ptr; } Destructor can not be declared as pure virtual. Since it is not possible to leave empty body of destructor since at the time of call of destructor same activity must be performed.\n\n## FILE HANDLING (STREAM I/O)\n\nFlow of data If flow of data is from program towards device then o/p stream eg. Cout . is used. If flow of data is from device to program then i/p stream eg cin is used. Monitor / Buffer Monitor\n\n## Input stream C++ Program Input stream\n\noutput stream\n\noutput stream\n\nHard-disk Classes Available In C++ For File Handling 1. Of Stream:2. It Stream:3. Fstream :-\n\nHard disk\n\nIf a class whose objects can be created for writing data to secondary memory of file. is a class whose objects can be created for reading data from file. whose object can read / write the data in a file.\n\nSteap Required For Writing Data In File Step 1 Create the object of Of stream class Eg:- Of stream obj; Step 2 Connect the object with file on your system. Step 3 write the data through the object in file. Step 4 Close the file. C++ Implement Of Above Steps Step 1 Step 2 Step 3 Step 4 (a) (a) (b) Of stream obj; obj. Open ( Data. Txt); obj << Hello; obj. put (H); obj.close ( );\n\n## static data File Opening Mode\n\n3 Of stream obj ( Data. Txt, ios : : app) Step For Creating A Program For Reading A File 1. Create the object of ifstream class. 2. Connect the object with the file on your system & check whether file is available or not. 3. Read the data through object from file. 4. class file. C++ IMPLEMENTATION 1. If stream obj 2. (a) Obj. Open ( Data.txt); Or If stream obj; Obj.open ( Data. Txt, ios : : in); Or If stream obj (Data.txt, ios: : in); Creating Object Of fstream Class 1. fstream obj; Obj.open ( Data.text, ios : : out \| ios : : in); 2. Using Constructor Fstream obj ( Data.txt, ios : : out \| ios : : in); WAP to create a file c/a message. text Accept a line of text from user & write it in file character by character #include <stdlib.h> #include <iostream.h> #include <fstream.h> #include <conio.h> Void main ( )\n\n{ Of stream out ( Message. Text); If (!out) { Cout << file can not be opened; Getch ( ); Exit(1); } Char str; Cout << enter a line of text; Cin.qetline (str, 80); Int i=0; While (str[i]) { Out.put (str[i]); I++; } Cout << file written successfully; Getch ( ); Out. Close( ); } 1 Note:isopen ( ) 0 1 Fail( ) 0 Que:- WAP to open the file created by the previous program. Read it on character by character basic & display its contents on the screen. Solution :Void main ( ) { If stream in ( message.txt); If (! in) { Cout << filoe can not be opened; Exit(1); Retruns I if not connected Returns 1 if conceted\n\n} Char ch; While (! In.enf( ) ) { Ch=in.get( ); Cout << ch; } Getch( ); In.close( ); } Que:- WAP to open a file c/a Data.txt. Accept a line of text from user & write it on file character by character basic and it in the same program read file and print its content on the screen . Void main( ) { Fstream Nisha ( Data, ios: : out \| ios: : in); If (! Nisha) { Cout << error opening file; Exit(1); } Char str ; Cout << enter a line of text; Cin.get line (str, 80); Int i=0; Char ch; While (str [i]) { Ch=str[i]; Obj.put (ch); I++; } Obj.seekg (0); While (! Obj.eof( ) ) { Ch=obj.get( ); Cout << ch; }\n\nGetch( ); Obj. close( ); } Que Assume there is a file c/a Message. Txt containing certain line of txt. Create a file c/a Message2.txt, and copy the contents of messages.txt into it. Before coping the character must be converted upper to lower & vice versa & spaces must be skipper. Finally display the contents of Message2 txt. Void main( ) { If stream obj1 ( Messages.txt); Stream obj2 (Message2. txt, ios : : out \| ios : : in); Char ch; If (! Obj1) { Cout << sourcl file cant be opened; Exit(1); } While (! Obj1, eof) { Ch = obj1. get( ); If (ch! = 32) { If (ch>=65 & & ch<=90) Ch=ch+32; Else if (ch > =97 && ch <=122) Ch=ch-32; } Obj2.put(ch); } Obj2.seekg(0); While ( ! obj2. eof( ) ) { Ch=obj2. get( ); Cout << ch; } Getch( ); Obj2.closs( );\n\nObj1. close( ); } READING AND WRITING STRINGS Void main ( ) { Fstream obj ( Data.txt, ios : : out \| ios : : in); If !(!obj) { Cout << error; Exit (1); } Char text ; Cout << how many lines; Int n; Cin >> n; Cout << Enter << n<< lines each terminated by enter : <<end1; For (in i=1; i<=n; i++) { Cin.get line (text, 80); Obj << text << end1; } Obj. seekg (0); Cout << File written press nay key to read; Gecth( 0; While (!obj.eof( ) ) { Obj.getline (text,80); Cout << text << end1; } Gecth( ); Obj.close( ); } Void main( ) { Fstream obj ( Data.txt, ios : : out \| ios : : in); If (!obj) { Cout << error;\n\nExit(1); } Char text ; Cout << enter lines and press enter on new line to stop; While (1) { Cin.getline (text, 80); If (strlen (text) = =0) Break; Obj << text << end1; } } FILE OPENING MODES 1. ios : : out (Default mode of ofstream) If file is not existing then it is created. Otherwise data gets Erased and pointer is placed at the beginning. If file is existing, pointer is placed a the beginning otherwise error is generated. 3. ios : : app It can not alter previous contents but can add new content at the End of file. 4. ios : : ate (ate stands for at the end) Allows updations as well as adding of new data. In this pointer can move forward and backward.\n\n2. ios : : in\n\n5. ios : : trunc Needed only in fstream. Used with ios : : out Truncate previous data & brings pointer to beginning. 6. ios : : nerplace Used with ios : : out if file is existing do not replace it otherwise create it. 7. ios : : nocreate\n\nUsed with ios : : out if file is existing overwrite it otherwise do not create it. 8. ios : : binary If we want to write data in binary form.\n\nBINARY I /O\nVoid write Of stream Member (char , address of variable whose data is to be Written int ) no of bytes to be written\n\nInt a = 23091 ; Obj, write (( char ) &a, size of (int) ); Char b = x; Obj.seekg(0); Int c; Obj. read ( (char ) &c, size of (int ) ); Cout << c; Char d; Obj. read (&d, size of (char) ); Cout <<d; Int read (char , int) Address of variable whose data is to be stored after reading From file On successfully reading from file it returns 1 on error it return 0. Reading and writing class objects Class Emp { Int age; Char name ; Float sal; Public: Void get( ) {\n\nCout << enter age, name and sal; Cin >> age >> name>> sal; } Void show( ) { Cout << age << } }; Void main ( ) { Emp E; Fstream obj ( Records.dat, ios : : out \| ios : : in \| ios : : trunc\| ios : : Binary ); If (! Obj) { Cout << error in opening file; Exit (1); } E.get( ); Obj.write ((char ) &E, size of (Emp) ); Obj.seekg(0); Emp F; Obj.read ((char ) &F, size of (Emp)); F.show( ); Getch( ); Obj.close( ); } Reading And Writing Multiple Objects Void main ( ) { Emp E; Fstream obj ( Record.dat, ios: : out\| ios : : in \| ios : : trunc \| ios : : Binary); If (! Obj) { Out << error in opening file; Exit (1);\n\n## <<name << sal <<end1;\n\n} Char choice; { e.get( ); obj.write ((char ) &E, size of (Emp)); cout << Any More (Y/N); cin.ignore( ); cin >> choice; } while ( ch = = y); Obj. seekg (0); While (opj.read (char ) &E, size of (emp)) E.show( ); getch( ); obj.close( ); } Typical Feature Of Read:Note :If in any ane program, we want to read file successively to eof we must clear the flag Obj. seekg (0); Obj. clear( );\n\nQ. WAP to write multiple records of type emp in a file. Accept a name from user & display the record of the employee by searching it with in the file and display the appropriate message Void main( ) { Int flat = 0 Emp E; Fstream obj1 ( Ekta, txt, ios : : out \| ios : : in \| ios : : trunc \| Ios : ; binary); If (! Obj) { Cout << error; Exit(1); } Char ch;\n\nDo { E.get( ); Obj.write ( ( char ) &E, size of (Emp)); Cout << Ant more (y/n); Cin . ignore ( ); } while (ch = = Y); Char name ; Cout << enter name to search; Cin >> name; Obj. seekg(0); While (obj.read (char ) &E, size of (emp)) { If ( ! ( E = = name) ) { E. show( ); Flag =1; Break; } } If ( ! flag) or if (flag = =0) { Cout << Record not found; Obj1. close( ); } Int operator = = (char ptr) { Return (strcmp (name, ptr) ); } } RANDOM I/O Q. Assume there is a file c/a Record.dat which contains several records of type emp. WAP to open this file & read the last record. Void main ( ) {\n\nIfstream in ( Records.dal, ios : : in \| ios : : binary); If ( ! in) { Cout << error; Exit (1); } In.seekg (-1 * size of (emp), ios : : end); Emp E; In. read ( ( char ) &E, size of (emp)); E.show( ); In.close( ); Getch( ); } Note:Prototype of seekg( ) Void seekg (int, int) No, of position of movement Bytes to ios : : beg Move ios : : cur Ios : : end WAP to accept a name from user. Search the record of that employee in Records.dat & Add a new record at its position by accepting it from user Void main( ) { Char temp; Emp E; Fstream in ( Records.dat, ios : : in \| ios : : ate\| ios : : binary); Cout << enter name to update; Cin >> temp; In.seekg(0); While (in.read ( (char ) &E, size of (emp) ) { If ( ( E = =temp) = =0) { E.get( ); In.seekg (-1 * size of (Emp), ios : : cur); In.write ( ( char ) &E, size of (emp));\n\nBreack; } } In.clear( ); In.seekg(0); While (in.read ( ( char ) &E, size of (Emp) ) E.show( ); Int operator = =(char n) { Return (strcmp (temp, ptr) ); } } Assignment :1. WAP to delete record given by user from file. 2. WAP to update just the salary of employee whose name given by user.\n\nTEMPLATES\nTemplates are a technique provide by C++ using which a programmer can define a single function in which last to be carried out is mentioned but the datatype is not mentioned. During runtime by looking at the call of the function & its parameter type the compiler generates specific version of that function to act according to parameter passed. Thus in other words, we can say templates are generic functions which at the time of creation are only told what to do & during run time they become aware on what type of argument it has to be done. Templates are of two type (i) Function Template (ii) Class Template Syntax Function Template\n\nTemplate <class <type-name> > <return type> < function-name> (< type-name> <arg-name>) Example Template <class T> Void display ( T n) { Cout << n << end1; } Void main ( ) { Int a =10; Char b = x; Float c= 11.5; Display (a); Display (b); Display (c); } Write a function template c/a swap which accepts two parameters & swaps then. Parameters passed can be two integer, two floats two class. Finally two swapped values must be displayed in the function main Template <class T> Void swap ( T &a, T &b) { T temp; Temp = a; A= b; B = temp; } Void main ( ) { Int a, b; Cout << enter two integers; Cin >> a >> b; Swap (a, b); Cout << a= <<a<< b= <<b<<end1; Cout << enter two character;\n\nChar p, q; Cin >> p >>q; Swap (p, q); Cout << p= <<p<<q = <<q <<end1; Float x, y; Cout << enter two float numbers; Cin >> x>>y; Swap (x, y); Cout << x= <<x<< y= <<y << end1; } Q. write a function template which accepts two values as an argument & return maximum amongst then. Template <class T> T greater ( T &P, T 7q) { If ( p > q) Return (p); Else Return (q); } Void main ( ) { Int a, b; Cout << enter two integers:; Cin >> a >> b; Cout << maximum = << greater (a, b); Cout << Enter two charr: Char x, y; Cin >> x >>y; Cout << maximum = << greater (x, y); } Write a function template c/a greatest which accepts an array as an argument & returns the largest element of that array. The array passed can be of different type and different size. Template <class T>\n\nT greatest ( T a, int n) { Int I; T max = a; For (i=1; i<n ; i++) { If ( (a+i) > max) Max = * (a+i); } Return (max); } Void main ( ) { Int arr[ ] = {7, 11, 2, 3, 4, 8}; Float brr[ ] = { 10.4, 11.3, 6.5, 8.2}; Cout << max int = << max (arr, 6); Cout max float= << max (brr, 4); }\n\nCLASS TEMPLATE\nTemplate <class T> Class Data { T a; T b; T c; Public: Void get( ) { Cin >> a >> b; } Void add ( ) { C = a+b; } Void show( )\n\n{ Cout << values are = << a << end1; Cout << b; Cout << sum= <<c; } }; Void main ( ) { Data <int> obj1; Data ,double> bj2; Cout << enter two int; Obj1.add( ); Cout << enter two doubles; Obj1, get( ); Obj2. add( ); Obj1. show( ); Obj2. show( ); } Class Template With Different Generic Type Template <class T1, class T2> Class Data { T1, a; T2, b; Public: Void get( ) { Cin >> a >>b; } Void show( ) { Cout << values are = << a << and << b<< end1; } }; Void main( ) { Data <int, float> obj1;\n\nData <double, char> obj2; Cout << enter an int and float; Obj1.get( ); Obj1.show( ); Cout << enter double and char; Obj2.get( ) Obj2.show( ); } Q. write a class template for a class called stack and implement three basic operations of stack push, pop and peek. The stack can be of int, char and flots. Template <class> Class stack { T arr [S]; Int tos; Public: Stack ( ) { Tos = -1; } T pop ( ); }; Template <class T> Void stack <T> : : push (T n) { If (tos = = 4) { Cout << stack overflow; Return; } ++ tos Arr [tos] = n; } Template <class T> T stack <T> : : pop ( ) { If (tos = = -1)\n\n{ Cout << stack under floaw; Return(-1) } Return (arr [tos - -]); } Void main ( ) { Stack <int> sa; Int n; Stack <char> s2; Char ch; For (int i=1; i<=6; i++) { Cout << enter int to be pushed; Cin >> n; S1.push(n); } For (int i=1; I,6; i++) Cout << element popped = << s1.pop( ); } Overloading Of Assignment Operator Class string { Char p; Public: String (int); Void set string (char ); Void resetstrign ( char ); Void display ( ); String( ); }; String : : string (int n) { P=new int [n+1]; } Void string : : setstring (char str)\n\n{ Strcpy (p, str); } Void string : : resetstring (char s) { Strcpy (p, s); } Void string : : display ( ) { Cout << p <<end1; } String : : string( ) { Delete [ ] p; Cout << Memory deallocated; } Void main ( ) { String s1 = 20; String s2 = 20; S11. setstring ( Hello User); S2=s1; S1. display ( ); S2. display ( ); S1. resetstring ( Welcome); S1. display ( ); S2. display( ); } Note:when ever the class is containing dynamic pointer then it is necessary to overload assignment operator At the time of overloading = it is necessary to pass parameter by reference. ( eg ( string & s) ). Note :For cascading of assignment operator it is necessary that return type must be string\n\n## void string : : operator (string &s) { strcpy 9p, s.p); x=s.x; }\n\nString string : : operator = (string &s) { String (p, s, p); X= s,x; Return ( this); } Note :Q. why do we overload assignment operator? Ans:- The default assignment operator provided by C++ is used for copying one object of class to another but it copies the value bit by bit i.e. the value contained in every bit of source object are copied in every bit of destination object. This behaviour is perfect if source cosle is not containing any pointer to dynamically alloveated block, but if it is so then the default equl to (=) operator will simply copy the address stored in the pointer with in source object to pointer in the destination object. This will make two the defferent pointers point to same location which may cause multiple problems like the destructor calling delete operator for the same memory to overcome this, a programmer should overload the equal (=) operator & provide his own definition for copying the data pointer by pointer of source object rather than address. Overloading Of Insertion And Extraction Operator Prototy Of << operator:Friend ostream & operator << (ostream & cout, strudent &p); Class Emp { Int age; Char name ; Float sal; Public: Friend istream & operator >> (istream 7, Emp&); Friend ostream & operator << (ostream &, Emp &); }; Istream & operator >> (istream & in, Emp & p) {\n\nIn >> P.age >> P.sal >> P.name; Return (in); } Ostream & operator << (ostream & out, Emp &P) { Out << P.age << << P.name<< << p.sal; Return out; } Void main( ) { Emp E, F; Cout << Enter age, name and sal; Cin >> E; Cout << Enter age, name & sal; Cin >> F; Cout << E << end1; Cout << F <<end1; Q. What is the need of overloading insertion and extraction operator? Ans:- In C++, all primitive types like int, float, char etc. are display on console using predefined object cout along with insertion operator (<<). Now if programmer wishes to use the same way of displaying objects of his class on screen then he has to overload insertion and extraction operator so that they can be derictly used with user defined objects. Q. Why dont we overload insertion and extraction operator as member function? Ans:- Insertion and Extraction operator are binary operator and if binary operator is overloaded as member function of class then a compulsion is their to keep the object of same class towards left of operator while calling. Thus if it is done then wll would become P << cout where P is object af class. Now this violotes the regular symmetry of insertion operator and so it must be overloaded as friend function. Q. Equal Operator can never be overloaded as friend function. Ans:- Since equal operator should return the reference of calling object (to support coscading). It has to be overloaded as member function as friend function do not have any calling object\n\nNote:Equal operator if defined by base class never inherited by derive class because it needs the same members in source as well as destination object.\n\nTYPE CONVERSION\nType conversion is the process using a programmer can convert value from primitive to non primitive and vice versa as well as from one object of class to another object of different class. Thus type conversion falls in three categories:1> Basic to User Defined 2> User Defined to Basic (Primitive ) 3> User Defined to User Defined Conversion Between Basic TO User Defined Constructor (Basic type { // steps to convert basic to user defined } Conversion Between User Defined TO Basic Operator primitive type of C++ ( ) return type { // steps to convert basic to user defined Return (<basic val>); } Example User To Basic & Basic To User Class Meter { Float length; Public : Meter ( ) { Length = 0.0;\n\n} Meter (float cm) { Length = CM/100; } Operator float( ) { Flaot ans= length * 100.0; Return (ans); } Void accept meters( ) { Cout << enter length in (int meters); Cin >> length ; } Void show meter( ) { Cout << In meter = << length ; } }; Void main ( ) { Meter M1 = 250; Meter M2; M2.accept Meter( ); Float cm = m2; // float CM= M2. operator float( ); Cout << 250 Meters when converted; M1. show Meter( ); M2. show Meter( ); Cout when converted to cms = << cm } Program :- Asignment Class String { Char str ; Public: String (int n) { Itoa (n, str, 10);\n\n} String( ) { Str = \\0; } Operator int( ) { K=1; I= strlen (str); While (i>=0) { Sum = sum + (str [i] -48)* k K * = 10; I - -; } Conversion From User Defined To User Defined 1. Conversion Routine In Source Object:Operator <destination_class-name> ( ) { //routines to convert source to destination Return ( <destination object>); } 2. Conversion Routine In Destination Object:Constructor ( <source class name>) { //routines to convert source to diction } Conversion From User Defined To User Defined By Placing Conversion Routine In Source Object Class Radian { Float rad; Public: Radian ( ) {\n\nRad = 0.0; } Radian (float r) { Rad = r; } Void show( ) { Cout << Radian = << rad << end1; } }; Class Degree { Float deg; Public: Degree( ) { Beg = 0.0; } Operator Radian ( ) { Float x = deg * PI/180; Radian obj = x; Return (obj ); } Void show( ) { Cout << Degree= << deg; } Void getdagree( ) { Cout << enter angle in degree=; Cin >> deg; } }; Void main ( ) { Degree D; d. getdegree( ); Radian R;\n\nR=D; D. show( ); R. show( ); } Class Degree { Float deg; Public: Degree( ) { Deg =0.0; } Void show( )_ { Cout << Degrees= << deg; } Void getdata( ) { Cout << enter angle in degrees; Cin >> deg; } Flaot getDree( ) { Return (deg); } }; Class Radian { Float rad; Public: Radian (Degree Obj) { Rad = obj.get Dece ( ) * PI/180; } Radian ( ) { Rad=0.0; } Void show( )\n\n{ Cout << Radian + << rad; } }; Void main( ) { Degree D; D.getData( ); Radian R = D; D. show( ); R.show( ); } Assignment:#include <iostream.h> #include <conio.h> Class hour { Float time; Public: Hour( ) { Time = 0.0; } Hour (double x) { Time =x/3600; } Operator float( ) { Float x = time * 3600; Return(x); } Void accept( ) { Cout << enter time in hours; Cin >> time; } Void show( )\n\n{ Cout << in hour = << time; } }; Void main( ) { Hour t1=3600; Hour t2; T2. accept( ); Float x = t2; Cout << 3600 when converted to hours=; T1.show( ); T2.shjow( ); Cout << when converted to sconds = <<x; Getch( ); }\n\nCONSOLE I/O OPERATIONS Unformatted 1. Unformatted I/O (a) istream & get (char &) (b) int get( ) (c) istream & get (char , int) Example :- cin.get(ch); Can read only characters Cin=cin.get( ); Istream & getline (char , int num, char delimiter); Istream & getline (char , int num); Example :Void main ( ) { Formatted can be called by cin\n\nChar ch; Cout << enter text and press enter to stop; Cin. Get (ch); While (ch ! =\\n) { Cout << ch; Cin.get(ch); } Getch( ); } Void main ( ) { Char country , capital ; Cout << enter country name:; Cin.get (country,20); cin.getline (country, 20); Cout << enter capital; Cin.getline (capital, 20); Cout << contry is << country << end1; Cout << its capital = << capital << end1; } Note:cin.getline (country, 20, )\n\nThis will terminate after 19 character or on pressing , nothing will happen on pressing enter rey. Functions Of Ostream Class (a) ostream & put (char) (b) ostream & write (char , int) This will start from base add upto no. of integers given. Example :Void main( ) { Char str[ ] = { programming }; Int I; For (i=0; i<strlen (str); i++) { Cout.put (str [i]);\n\nCout << end1; } For (i=1; i<=strlen(str); i++) { Cout.write (str, i); Cout << end1; } For (i=strlen (str) ; i>=1; i--) { Cout.write (str, i); Cout << end1; } }\n\nFORMATTED I/O\nFunction (i)width( ) (ii) precision( ) (iii) fill( ) (iv) setf( ) (v) unsetf( ) Description specifies required no of fields to be used for displaying the output. This fn is used for alignment. specifies the no of values to be displayed after decimal pt specifies a character to be used to fill the unused area of field. By default the fulling is done using spaces. sets the format flag to be used while displaying the output. clears the flogs set using setf & restones the default settling.\n\nDefining field Width Prototype:- int width( ) Returns the current width. Int width (int) Old width aurrent width Void main( )\n\n{ Cout. Width(4); Cout << 123; Cout.width(4); Cout. << 39; Cout.width(2); Cout << 2345 } Setting Precision Prototype :-int precision ( ) By default Int precision (int) precision is of 6 pt. Void main ( ) { Cout precision (2); Cout << 2.23 << end1; Cout << 5. 169 << end1; Cout << 4.003 << end1; } Output 2,23 5.17 4 Filling :Int fill( ) Int fill (char) Void main( ) { Cout.file ( ); Cout.precision(2); Cout.width(6); Cout << 12.53; Cout.width(6); Cout << 20.5; Cout. width(6); Cout << 2; }\n\no/p 12.53* * * 20.5 * * * * * 2 Note There is no need to set fill and precision again and again while width flag must be set again and again. Formatting With Flags & Bit Fields 1> setf long self (log-setbits, long field) Flag-value (1st argument) Bit field (2nd Arguments Description Ios : : left ios : : adjust field justifies the output on left side. Ios : : right ios : : adjeist field justifies the output in sight align mennes. Ios : : internal ios : : adjust field passing accurs between the sign or base indicator & the value when the value fails to fill the entire width ios : : dec ios : : base field display the data in decimal conversion ios : : oct display the data in actor ios : : hax display the data in hexadecimal ios : : scientific ios : : float field user exponential floating notation ios : : fixed user normal floating notation Example :Void main ( ) { Cout.self (ios : ; left, ios : : adjust field); Cout.fill (); Cout. precision (2); Cout.width (6); Cout << 12.53; Cout.width (6);\n\nCout << 20.5; Cout.witdth (6); Cout <<2; } 12.53 20.5 * * 2 * * * * * Void main ( ) { Cout.self (ios : : internal \| ios : : adjust field ); Cout.field (); Cout. precision (2); Cout. width (10); Cout << -420.53; } Output : - - * * 4 2 0 . 5 3 Displaying Trailing Zeros & Plus Sign Long self (long setbits) (i) ios : : show pint (for trailing zeros) (ii) ios : : show pos (for plus sign) void main ( ) { Cout.setf (ios : : show point); Cout.precision (2); Cout << 20.55 << end1; Cout << 55.55 << end1; Cout << 20.40 << end1; } Example :Void main ( ) { Cout.setf (ios : : showpoint); Cout.setf (ios : : show pos); Cout.setf (ios : : internal, ios : : adjust field); Cout.precison(3); Cout.width (10); Cout << 420.53;\n\n## output 20.55 55.55 20.40\n\n} Output + * 4 2 0 . 5 3 0\n\nFormatting Data Using Manipulators Non Parameterised Manipulators Manipulator 1. end1 2. dec 3. bex 4. oct 5. flush Description\n\nterminates the line and transfers the cursor to next row. set conversion base to 10. set the conversion field to 16. set the conversion field to 8 flushes the output screen\n\nExample :1. WAP to read a number in decimal and display it in hxadecimal. Void main ( ) { Cout << enter number; Cin a; Cin >> a; Cout << no is = << n << end; Cout << Its hexadecimal value= << hex<<n; Cout. self (ios : : show base); Cout << a; } Output:- no is = 64 Its hexaslccimal value = 40 0x 40 Example :- void main ( ) { Int n; Cout << enter number; Cin >> nex>> n;\n\nCout << number= << n; } Parameterised Manipulators Manipulator 1. setw (int) 2. setprecision (int) 3. setfil (char) 4. setbase (int) 5. setiosflag (long) 6. resetiosflag (long) Description sets the field width sets number of digits to be displayed after decimal pint sets the character to be field set the conversion base. Here possible value of int are 8.10 and 16. sets the format flag. resets the format flag.\n\nExample :Void main ( ) { Int n = 100; Cout << hex <<n << << dec <<n << end1; Float f = 122.3434; Cout << f << end1; Cout << setprecision (3); Cout << f << end1; Cout << setiosflag (ios : : internal } ios : : show base); Cout << hex << n << end1; Cout << setiosfloag (ios : : scientifie) << f << end1; } Output:64 100 122.34339 9 122. 343 0x0064 1. 2 2 3 c + 0. 2\n\nClass Array { Int P; Int n; Public : Array (int size) { P=new int [size]; N=size; } Int & operator [ ] (int i) { Return ( (p+i) ); } Void fil (int pos, int value) { * (p+pos) = value; } Array ( ) { Delete [ ]p; } }; Void main ( ) { Array obj(5); Int x; For (int i=0; i<5; i++) { Cin >> x; Obj.fill (I, x); or obj [i] = x } For (I = 0; i<5; i++) { Cout << obj [i]; } } Overloading Of Operator ( )\n\nClass Box { Int l, b, h; Public: Box operator ( ) (int, int, int); Void get( ) { Cout << enter l, b and h; Cin >> l >> b >> h; } Void show( ) { Cout << l, << << b << << h; } }; Box Box : : Operator (int l, int b, int h) { Box temp; Temp.l = l + this l; Temp.b = b + this b; Temp.h = h + this h; Return (temp); } Void main ( ) { Box B1; B1.get( ); Box B2; B2 = B1 (5, 3, 9); // B2 = B1. operator ( ) (5, 3, 9); B1. show( ); B2. show( 0; }\n\n## GRAPHICS MODE PROGRAMMING\n\nSteps Required For Designing Graphics Mode Program\n\n1> Concert the display monitor from text mode to graphics mode. 2> Perform the required graphics operation like filling coloring, drawing etc. 3> Finally close the graphics mode and restore character mode. Converting Character To Pixels 1. void init gragraph (int , int , char ) Driver resolution path of BGI file Or Mode 2. Drawing using built in functions. 3. void close graph( ); Void restorecrtmode( ); Program:- WAP to couvert monitor display mode from char to pixel and print welcome message. #include <graphics.h> #include <conio.h> #include <stdlib.h> Void main( ) { Int gd, gm, ec; Gd= DETECT; Inttgraph (&gd, &gm, C:\\\\TC\\\\BGI); Ec= graphresult( ); If (ec!=grOk) or (ec!=0) { Printf (error in initialising); Exit(1); } Cleardevice Outtext ( welcome to graphics); Getch( ); Closegraph( ); Restarectmode( ); } Note:-\n\n## Int getmaxx( ); Int getmaxy( ):Assignment:-\n\nreturns the maximum value of x coordinate. returns the minimum value of if coordinate. Modify the above code so that now code display the Message at the necter of screen.\n\nVoid main ( ) { Int gd, gm, ec; Gd = DETECT; Initgraph (&gd, & gm, C:\\\\TC\\\\BGI); Ec=graphresult( ); If (ec ! = grOk) { Printf ( error); Exit(1); } Cleardevice( ); Int a = getmaxx( ); Int b = getmaxy( ); Outtextry ( ( a/2), (b/2), welcome to graphics); Getch( ); Closegraph( ); Restorecrtmode( ); } Q. WAP to accept user name & then convert screen to graphics mode & display the name given by user at the center of the screen on char at a time. Void main ( ) { Char str Int gd, gm, ec; Gd = DETECT; Initgraph (&gd, &gm, c:\\\\TC\\\\BGI); Ec = graphresult( ); If (ec!=grOk) { Printf ( error); Exit(1);\n\n} Cleardevicl ( ); printf (enter name); Move to (getmaxx( ) /2, getmaxy( )/2); For (i=0; str[i]; i++) { Spritf (msg, %c, str[i]); Out text (msg); Delay (1000); } } Changing the font style and size 1. void settextstyle (int font, int dir, int charsize) description of parameters font default FONT (0) TRIPLEX FONT (1) SMALL FONT (2) SANS SERIT FONT (3) GOTHIC FONT (4) SCRIPT FONT (5) SIMPLEX FONT (6) PRIPLEX SCRIPT FONT (7) COMPLEX FONT (8) EUROPEAN FONT (9) BOLD FONT (10) horiz dir (0) Vert dir (1)\n\nDir\n\nCharsize :- 1 to 10 1 being size of 8 x 8 pixel. Program:Void main( ) { Char font style [ ] = { Defualt font ----, BOLDFONT); Int gd, gm, ec; char msg; Gd = DETECT;\n\nInitgraph (&gd, &gm, C: \\\\ TC\\\\ BGI); Ec = graphresult ( ); If (ec ! = 0) { Printf ( error); Exit (1); } Cleardevicl ( ) Move to (getmaxx( )/2, getmaxy( )/2); For (1=0; i<=10; i++) { Printf (msg = shiv - %s, font-style [i]); Settext style (I, 0, 1); Outtextry (getmaxx( )/2, getmaxy( )/2, msg); Getch( ); Cleardevicl ( ); } Cleardevicl( ); Restorecrtdevicl ( ); }\n\nDRAWING IN GRAPHICS\n1. For Drawing Line:a. void line (int x1, int y1, int x2, int y2) b. void linerel (int, int) c. void line to (int, int) Example :Void maiN ( ) { Int gd, gm, ec, x, y; Gd = DETECT; Initgraph ( &gd, &gm, c: \\\\TC\\\\BGI); Ec = graphresult( ); If (ec ! = 0) { Printf ( error);\n\nExit (1); } Cleardevicl ( ); X=getmaxx( )/2; Line (0, 0, x, y); Getch ( ); Closegraph( ); Restorertmode( ); y=getmaxy( )/2\n\nQ.WAP which draws a line from 20, 30 to 100 pixels further & display the wordinate of line at its end point. Void main ( ) { Int gd, gm, ec, x, y; Gd = DETECT; Initgraph (&gd, & gm, c: \\\\TC\\\\BGI); Ec = graphresult ( ); If (ec! = 0) { Printf ( error); Exit (1); } Cleardericl ( ); Move to (20, 30); Sprintf (msg, %d %d, get x( ), gety( ) ); Outtext (msg); Linerel (100, 100); Sprintf (msg, %d %d, getx( ), get y( ) ); Outtext (msg); } Drawing Linee In Specific Width & Style Void setline style (int style, int pattern, int thickness) Style:- No 0 1 2 Constant SOLID-LINE DOTTED LINE CENTER-LINE meaning solid line line withdot & dash\n\n3 4 Pattern:Thickners:-\n\nDASHED-LINE USERBIT-LINE\n\n## line with dashes\n\nIt is always zero except when first parameter is userbit line. 1 THICK-WIDTH 3 NORM-WIDTH\n\nVoid main( ) { Int gd, gm, ec, I; Char * style [ ] = { solidline}, Dotted-line,----, Dashedline}; Gd = DETECT; Initgraph ( &gd, & gm, C: \\\\TC\\\\BGI); Ec=graphresult( ); If (ec != grok) { Printf ( error); Exit(1); } Deardvice( ); For (i=0; i<4; i++) { Spritf (msg, %s in normal width, style[i]); Outtext (getmaxx( )/2-20, getmaxy( )/2-10, msg); Setlinestyle (I,0, NORM-WIDTH); Line (getmaxx( )/2-50, getmaxy( )/2 +20, getmaxx( )/2 +50, Getmaxy( )/2 +100); Getch( ); Cleardevicl( ); } Restorectdevicl( ); } Defining Pattens in User Defined Way Io define a user bit line wee have to build a sixbit pattern. In this pattern wheever a bit is one the curresposing pixel in the line is drawn in the current drawing colour. For eg:65535 or\n\nsetline style (4, OXFFFF, NORM-WIDTH); this will draw a solid line, setlinestyle (4, Ox3333, NORM-WIDTH) this will draw a dashed line. Drawing Arcs Or Circles Or Rectangles 1. void arc (int x, int y, intstangle, int endangle, int rad) 2. void piesline (int x, int y, int stangle, int endangle, int rad)\n\n3. void circle (int x, int y, int rad) 4. void rectangle (int left, int top, int right, int bottom) 5. void bar (int left, in top, int right, int bottom) Filling Image With Different Patterns Void set color (int) Void setfillstyle (int pattern, int style) Pattern:The pattern parameter signifies the pattern in which filing is to be made. value 0 1 2 3 Result Background color Solid Filling --------------------------\n\n## 5. SLASH-FILL 6 BKLASH-FILE 7. LTBKSLASH-FILE HATCH-FILE XHATCH-FILE INTERLEAVE-FILE WIDE-DOT-FILE CLOSE-DOT-FILE\n\n4 5 6 7 8 9 10 11\n\nWAP which draws a sectangle with white outline & red fill color & should display the fitling in all of the twelve patterns one at a time. Also name of pattern should be displayed with in the rectangle. Void main ( ) { Int gd, gm, ec, left, right, top, bottom, I; Char * style [ ] = { EMPTY-FILE, SOLID-FILE, ----, CLOSED-DOT-FILE}; Gd=DETECT Initgraph (&gd, &gm, C:\\\\TC\\\\BGI); Ec=grapgresult( ); If (ec !=0) { Printf (error); Exit (1); } Cleardevicl ( ); Left = getmaxx( )/2-100; Top=getmaxy( )/2-100 Right getmaxx( )/2+100; Bottom=getmaxy( )/2;\n\nFor(i=0; i<12; i++) { Setfill style (I, RED); Bar(left, top, right bottom); Rectangle (left, top, right, bottom); Outtextxy (left+50, top+50, style[i]); } }\n\n## FILLING CIRCLES & TRIANGLESS\n\n1. void floodfill (int x, int y, int boundary) (x, y) a point with in the figure which has to be filled using flodfill also known as sad. Boundary color:- Color at which filling should stop. Filling Circles With Different Pattern Void main ( ) { Char * pattern = { EMPTY-FILE, --------, CLOSE-DOT-FILE}; Int gd, gm, ec, x, y, 0; Gd=DETECT; Initgraph (&gd, &gm, C:\\\\TC\\\\BGI); Ec=graphresult( ); If (ec! = grOk) { Printf ( error); Exit(1); } Cleardevice( ); X=getmaxx( )/2; y=getmaxy( )/2; For (i=0; i<12; i++) { Setcolor (GREEN); Circle (x, y, 100); Setfill style (I, RED); Floodfill (x, y, GREER); Setcolor (WHITE);\n\nOuttextxy (x-50, y-50, patter[i]); Getch( ); Cleardivice( ); } Getch ( ); Restorecrtdevice( ); } Storing and Drawing Images On Screen 1. void getline (int, int, int, int, void ) 2. int imagesize (int, int, int, int) 3. void putimage (int, int, void , int option) void getimage (int, int, int, int, void ) copies the bit image of specified postion in memoery 1st parameter indicates left coordinates 2nd parameterer to rd 3 parameter right th 4 parameter bottom th 5 parameter pointer to an array large enough to store bit pattern. Void putimage [int x, int y, void are, int iption) Copies or outputs the image pattern form memoery to the specified portion on the screen. X = starting left coordinate Y = starting top coordinate Arr = maner un which color of the resultant pixel is to be decided, taking into consideration the pixels stored in memory & the pixels stores on screen. Int image size (int, int, int, int);Returns no. of bytes required to store the image, on any kind of error return -1 Void main ( ) {\n\nInt gd, gm, ec; Char * buffer, msg ; Int size -0f-image; Gd = DETECT; Initgraph ( &gd, &gm, C:\\\\TC\\\\BGI); Ec = graphresult( ); If (ec ! =0) { Printf (error); Exit (1); } Rectangle (150, 150, 200, 200); Size-of-image = image size (150, 150, 200, 200); If (size-of-image = = -1) { Outtextxy (getmaxx( )/2, getmaxy( )/2, Error); Getch( ); Close graph( ); Exit(1); } Buffer = (char ) malloc (size-of-image size of (char) ); If (buffer = = NULL) { Outtextxy (getmaxx( )/2, getmaxy( )/2, Can not allocate memory); Gecth( ); Exit(1); Close graph( ); } Getimage (150, 150, 200, 200, buffer); Line (200, 220, 220, 220); Putimage (175, 200, buffer, COPY-PUT); Getch( ); Closegraph( ); Restore crtdevice( ); } VALUES COPY-PUT CREEN ON OFF EMORY ON ON OUTPUT ON ON\n\nON OFF XOR-PUT ON OFF ON OFF ON OFF ON OFF ON OFF ON OFF THE END\n\nOR-PUT\n\nAND-PUT" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.65817374,"math_prob":0.8851826,"size":93672,"snap":"2019-51-2020-05","text_gpt3_token_len":27006,"char_repetition_ratio":0.18238886,"word_repetition_ratio":0.15148419,"special_character_ratio":0.34407294,"punctuation_ratio":0.21095529,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9852558,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-06T20:58:26Z\",\"WARC-Record-ID\":\"<urn:uuid:08fd6ec5-c384-4aad-8cdd-32bbf9ca6025>\",\"Content-Length\":\"426279\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ecb06f41-a987-44b1-9978-31e70ce14fdd>\",\"WARC-Concurrent-To\":\"<urn:uuid:19829dc5-2986-46d6-847a-6efe94a707f7>\",\"WARC-IP-Address\":\"151.101.250.152\",\"WARC-Target-URI\":\"https://de.scribd.com/document/66806060/c-Notes-Complet\",\"WARC-Payload-Digest\":\"sha1:JSNTFBNYWFT42YURYUUWAWKXLLKNXZ64\",\"WARC-Block-Digest\":\"sha1:GNOKVPTHUKTSZTVXC7AHXZBLTA3MKVRN\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540490972.13_warc_CC-MAIN-20191206200121-20191206224121-00177.warc.gz\"}"}
https://waseda.pure.elsevier.com/ja/publications/on-the-%CF%84-functions-of-the-reduced-ostrovsky-equation-and-the-a-su	[ "# On the τ-functions of the reduced Ostrovsky equation and the A (2) 2 two-dimensional Toda system\n\nBao Feng Feng, Ken Ichi Maruno, Yasuhiro Ohta\n\nこの研究の対応する著者\n\n8 被引用数 (Scopus)\n\n## 抄録\n\nThe reciprocal link between the reduced Ostrovsky equation and the A (2) 2 two-dimensional Toda (2D-Toda) system is used to construct the N-soliton solution of the reduced Ostrovsky equation. The N-soliton solution of the reduced Ostrovsky equation is presented in the form of pfaffian through a hodograph (reciprocal) transformation. The bilinear equations and the τ-function of the reduced Ostrovsky equation are obtained from the period 3-reduction of the B or C 2D-Toda system, i.e. the A (2) 2 2D-Toda system. One of the τ-functions of the A (2) 2 2D-Toda system becomes the square of a pfaffian which also becomes a solution of the reduced Ostrovsky equation. There is another bilinear equation which is a member of the 3-reduced extended BKP hierarchy. Using this bilinear equation, we can also construct the same pfaffian solution.\n\n本文言語 English 355203 Journal of Physics A: Mathematical and Theoretical 45 35 https://doi.org/10.1088/1751-8113/45/35/355203 Published - 2012 はい\n\n## ASJC Scopus subject areas\n\n• 統計物理学および非線形物理学\n• 統計学および確率\n• モデリングとシミュレーション\n• 数理物理学\n• 物理学および天文学（全般）\n\n## フィンガープリント\n\n「On the τ-functions of the reduced Ostrovsky equation and the A <sup>(2)</sup> <sub>2</sub> two-dimensional Toda system」の研究トピックを掘り下げます。これらがまとまってユニークなフィンガープリントを構成します。" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8807845,"math_prob":0.88496417,"size":2313,"snap":"2021-43-2021-49","text_gpt3_token_len":674,"char_repetition_ratio":0.17973149,"word_repetition_ratio":0.77437323,"special_character_ratio":0.2741029,"punctuation_ratio":0.08656036,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9808993,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-09T05:00:44Z\",\"WARC-Record-ID\":\"<urn:uuid:7be8f247-c4f5-436c-8740-956757a7aa7d>\",\"Content-Length\":\"48154\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aa1f085f-9d43-456b-8ef1-f237b6e063dc>\",\"WARC-Concurrent-To\":\"<urn:uuid:3350aeb3-afbb-4688-a02d-b38b4ee58468>\",\"WARC-IP-Address\":\"13.228.199.194\",\"WARC-Target-URI\":\"https://waseda.pure.elsevier.com/ja/publications/on-the-%CF%84-functions-of-the-reduced-ostrovsky-equation-and-the-a-su\",\"WARC-Payload-Digest\":\"sha1:YAXTXXL7SLDLIZHFCXECHV34EYIBMG7M\",\"WARC-Block-Digest\":\"sha1:KPEVD4LI7VJAXMM4TMMNHXFCHAZGCJK6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363659.21_warc_CC-MAIN-20211209030858-20211209060858-00457.warc.gz\"}"}
https://www.intechopen.com/books/topics-in-adaptive-optics/measurement-error-of-shack-hartmann-wavefront-sensor	[ "Open access peer-reviewed chapter\n\n# Measurement Error of Shack-Hartmann Wavefront Sensor\n\nBy Chaohong Li, Hao Xian, Wenhan Jiang and Changhui Rao\n\nSubmitted: March 8th 2011Reviewed: August 12th 2011Published: January 20th 2012\n\nDOI: 10.5772/29430\n\n## 1. Introduction\n\nA Shack-Hartmann sensor is one of the most important and popular wavefront sensors used in an adaptive optics system to measure the aberrations caused by either atmospheric turbulence, laser transmission, or the living eye [1-7]. Its design was based on an aperture array that was developed in 1900 by Johannes Franz Hartmann as a means to trace individual rays of light through the optical system of a large telescope, thereby testing the quality of the image. In the late 1960s Roland Shack and Platt modified the Hartmann screen by replacing the apertures in an opaque screen by an array of lenslets [9-10]. The terminology as proposed by Shack and Platt was “Hartmann-screen”. The fundamental principle seems to be documented even before Huygens by the Jesuit philosopher, Christopher Scheiner .\n\nThe schematic of a Shack-Hartman wavefront sensor is shown in Figure 1. It consists of an array of lenses (called lenslets, see Figure 1) of the same focal length. Each is focused onto a photon sensor (typically a CCD array or quad-cell). The local tilt of the wavefront across each lens can then be calculated from the position of the focal spot on the sensor. Any phase aberration can be approximated to a set of discrete tilts. By sampling an array of lenslets, all of these tilts can be measured and the whole wavefront can be approximated. Since only tilts are measured, the Shack-Hartmann can not measure the discontinuous steps of wavefront.\n\nTyler and Fried have obtained the theory expression, which evaluates the angular position error when a quadrant detector is used in the SHWFS . The formula they obtained, based on circular aperture diffraction, is shown in Eq. (1)\n\nσ=3π161SNRλDE1\n\nwhere SNR is defined as the ratio of the signal’s photoelectron counts to the noise’s fluctuation intensity within the detection area, λ is the wavelength, and D is the diameter of the aperture. Their analysis did not discuss the size of the incoming light spot on the detection area in detail. The formula was obtained based on a quadrant detector alone. Generally, the theory expression obtained by Tyler and Fried is not suitable for describing the angular position error when the scale of the discrete detector arrays is greater than 2×2 pixels.\n\nHardy has described formulas that can be used to evaluate the angular position error , under the conditions that the photon shot noise of signal is dominant. Although his formulas discussed the size of the diffraction-limited spot on the discrete detector arrays, it is reliable only under the approximation condition that l/f>>λ/Dor l/f<<λ/Dis satisfied, where l is the length of a pixel, f is the focal length, and Vsis the count of signal photoelectrons. Eq. (2) shows the formulas based on square aperture diffraction.\n\nσ={0.277λD/Vs1/2(whenl/f<<λ/D)0.500λD/Vs1/2(whenl/f>>λ/D)E2\n\nCao et al. have also analyzed the measurement error of a SHWFS. Their work emphasized the discrete sampling error of a CCD and obtained a formula which is used to describe the centroid position error induced by the readout noise of the CCD and the photon shot noise of the signal . Their research results are only the approximation of some cases discussed in this article. Jiang et al. have partially analyzed the measurement error of SHWFS and performed their method by setting a fixed threshold to suppress the impact of the random noise .\n\nIn this chapter, the wavefront error of a Shack-Hartmann wavefront sensor was analyzed in detail based on the research results of angular position error and wavefront error [16-17]. and the formula used to evaluate the wavefront error was derived, it concerns with the signal to noise ratio, number of photons and reconstruction matrix also.\n\n## 2. The angular position error caused by random noise\n\nThe wavefront to be measured is segmented into many subwavefronts by lenslet arrays, and the light spots at the focal plane of the subapertures are detected by the CCD. Particularly, the analysis is based on the notion that the wavefront is essentially flat over each subaperture and r0>D(r0is the coherent length of incoming wavefront). The centroid position can be calculated by Eq. (3) . The detection area of the subaperture is L1×L2pixels, and xnmand ynmare the (n,m)th pixel’s X coordinate and Y coordinates, respectively. Inmis the total intensity in the (n,m)th pixel, including the signal photons and all other noises.\n\nxi=m=1L1n=1L2xnmInmm=1L1n=1L2Inm,yi=m=1L1n=1L2ynmInmm=1L1n=1L2InmE3\n\nThe formulas which evaluate the centroid error associated with the signal’s photon shot noise and the readout noise of the detector, respectively, have been derived by Cao et al. When the photon shot noise of the signal is considered alone, the centroid fluctuation error was obtained by introducing the Gauss width of the signal. When the readout noise of the detector is considered alone, the centroid fluctuation error was also obtained, and its results are shown in Eq. (4) and Eq. (5), respectively . Nr is the rms error induced by the fluctuation of the readout noise in each pixel (with units of photoelectron counts). Vr is the sum of the readout noise’s photoelectron count among all of the pixels within the corresponding subaperture.\n\nϕcs2=Gs2VsE4\nϕcr2=Nr2Vr2L1L2L21112E5\nϕcs2is the variance of the centroid fluctuation in one direction ( X or Y), induced by the photon shot noise of signal itself, and ϕcr2is the variance of the centroid fluctuation in one direction ( X or Y), induced by the readout noise of the detector. ϕcr2and ϕcs2are both defined in pixel2 units. Gsis the equivalent Gauss width of the signal spot and is defined in pixel units by the expression:\nGs=ηflλDE6\n\nwhere η is the positive constant. η is 0.353 when the diffraction aperture is square and 0.431 when the diffraction aperture is circular.\n\nBased on Eq. (3), the centroid position in the X direction can be expressed by Eq. (7). The detailed derivation process of this expression is shown in appendix 1.1.\n\nxc=sbr1+sbrxcs+11+sbrxcbE7\n\nwhere xcis the calculated centroid position of the signal in the X direction, xcsis the real centroid position of the signal in the X direction, and xcbis the centroid position induced by the total noise, except for the signal in the X direction, where the total noise largely includes the readout noise of the detector and the heterogeneous light noise.sbr=<Sn,m>/<Bn,m>, <Snm>is the collective average of the signal intensity in the (n,m)th pixel, and <Bnm>is the collective average of the total noise intensity in the (n,m)th pixel (with units of ADU).\n\nBased on the error transition principles, the rms error of centroid measurement induced by random noise in the X direction can be written in Eq. (8) as:\n\nϕc=[(sbr1+sbr)2ϕcs2+(11+sbr)2ϕcb2]1/2E8\n\nwhere ϕcsis the rms error of centroid measurement in the X direction induced by the signal’s photon shot noise andϕcbis the rms error of centroid measurement in the X direction induced by all the other noise. The signal is mostly comprised of heterogeneous light and readout noise.\n\nIf there were no heterogeneous light and readout noise in the detection area, the signal’s photon shot noise should be the unique noise resource which affects centroid measurement. Based on Eq. (4) and Eq. (6), when the discrete sampling error of the detector is ignored, the rms error of angular position in the X direction caused by the photon shot noise of the signal can be written as:\n\nσ1=(Gs2Vs)1/2lf=ηλDVs1/2E9\n\nWhen the photon shot noise of the signal is small compared with the readout noise and the heterogeneous light noise, the heterogeneous light noise and the readout noise become the primary noise, which affects centroid calculation. When the heterogeneous light noise can be considered as a uniform noise, like the readout noise of the detector, it exists in each pixel and it has the same fluctuation characteristics among the pixels in the detection area. So, the noise in one pixel (including the heterogeneous light noise and the readout noise of the CCD) can be summed and described byNb. Nbis defined as the rms error of the heterogeneous light noise and the readout noise photoelectron count in one pixel, and it has the same fluctuation characteristics as the readout noise of the detector. Nbhas units of ADU. Subsequently, the rms error of centroid measurement in the X direction caused by the heterogeneous light noise and the readout noise of the CCD can be written as:\n\nϕc=11+sbrϕcb=(1+Snm/Bnm)1[L1L2(L121)/12]1/2Nb/Vb=(Bnm/Nb+C(λ,D,l,f)snr)1[L1L2(L121)/12]1/2E10\n\nwheresnr=max(Snm)/Nb, Vb=Bnm, and max(Snm)is the signal’s peak intensity. snr is defined as the ratio of the signal’s peak intensity to the rms error induced by the background noise, Bnmis the average intensity of noise in the (n, m)th pixel, which includes the heterogeneous light noise and the readout noise of the CCD. C(λ,D,l,f)is the light spot constant which is defined as the ratio of the total signal intensity to the signal’s peak intensity in the subaperture, and its value can be measured or calculated exactly byC(λ,D,l,f)=Snm/max(Snm).\n\nThe intensity distribution of the signal’s light spot at the focal plane of the subaperture can be calculated by circular or square aperture diffraction approximations. On the other hand, the Gauss distribution can also be used to approximately describe the intensity distribution of the light spot. The analytic expressions of C(λ,D,l,f)are described in Eq. (11) with different approximation conditions. The detailed derivation process is shown in appendix 2.1~2.3. The value of C(λ,D,l,f)can be calculated by Eq. (11).\n\nC(λ,D,l,f)={1/(1J02(r1)J12(r1))(circularaperturediffractionapproximation)[π2/(sin2x1x1+k=0+(1)k×(2x1)2k+1(2k+1)×(2k+1)!)]2(squareaperturediffractionapproximation)1/[1exp(n2l2D22πf2λ2)](Gaussdistributionapproximation)E11\n\nwherex1=πDλsin(θ)=πDλsin(l/2f)=πDl2λf,r14πx1=4ππDl2λf, n is a positive constant, and θ is the diffraction angle.\n\nWhen the direct current part of the noise (including the heterogeneous light noise and the readout noise of the CCD) is subtracted, it can be considered as white noise, andBnm=0. Then, the standard deviation of the angular position error in the X direction caused by noise can be described by Eq. (12):\n\nσ2=[L1L2(L121)/12]1/2C(λ,D,l,f)1snrlfE12\nWhenL1=L2=L,the Eq. (12) can also be expressed by Eq. (13)\nσ2=[L2(L21)/12]1/2Nb/(L2Nb+Vs1/2)C(λ,D,l,f)max(Snm)/(L2Nb+Vs1/2)lf=[(L21)/(12L2)]1/2L2Nb(L2Nb+Vs1/2)l/fλ/D1SNRλD=ω1SNRλDE13\n\nwhere SNR has the same definition as in Eq. (1). L2Nbis the sum of the rms error of total noise among all of the pixels within the detection area, and expresses the total intensity of noise fluctuation. Vs1/2expresses the photon shot noise induced only by the incoming signal. ω is the position error constant, and it is weighted by the intensity of background noise and the intensity of the signal’s photon shot noise (defined in Eq. (14)):\n\nω=[(L21)/(12L2)]1/2l/fλ/DL2Nb(L2Nb+Vs1/2)E14\n\nSubstituting Eq. (13) and Eq. (9) into Eq. (8), with the assumed condition that there are no correlations among the photon shot noise of the signal, the heterogeneous light noise, and the readout noise of CCD, then the total rms error of angular position in the X direction caused by random noise can be obtained:\n\nσ=(σ12+σ22)1/2ω1SNRλD+ηλDVs1/2E15\n\nEq. (15) is the desired result which can be used to precisely describe the angular position error of a Shack-Hartmann wavefront sensor caused by random noise, and therefore, the centroid algorithm is used to calculate the spot position of the incoming light. Generally, when the ideal detector with very small readout noise is used and there is no background light noise (ω0), the photon shot noise of the signal becomes the theoretical limits imposed on the angular position measurement. Eq. (9) showed this expression. In practice, the theoretical limits may not be achieved for the hardware and environment limitations. When the photon shot noise is small enough compared with the heterogeneous light noise and readout noise, it could be ignored in Eq. (15), and Eq. (13) could be used to describe the angular position error caused by the random noise approximately. Commonly, it has enough accuracy. The position-error constant ω described in Eq. (14) is concerned with the scale of the discrete detector arrays in the detection area, the noise characteristics of the detector, and the system parameters. Clearly, the formula based on a quadrant detector obtained by Tyler and Fried is only a special case in this article. On the other hand, the formula obtained in Eq. (13) is suitable to evaluate the angular position error for both a circular and square aperture.\n\n## 3. Wavefront measurement error caused by centroid position random error\n\nIn this chapter, Zernike modes are used as the basis for wavefront reconstruction. The wavefront measurement error can be written as [13, 19]\n\nΔϕ=ϕϕ'=j=1P(ajaj')Zj=j=1PΔajZjE16\n\nwhereΔϕis the wavefront measurement error induced by centroid position random error, ϕis the wavefront to be measured, ϕ'is the wavefront detected, P is the total number of Zernike modals, aj is the jth Zernike coefficient, and Z expresses the Zernike polynomial. Then, the mean-square of wavefront measurement error can be written as shown in Eq. (17) . The angle brackets denote a collective average.\n\nσϕ2=<ϕ2><ϕ'2>=j=1P<\|Δaj\|2>E17\n\nBased on the principles of the Zernike modal wavefront reconstruction algorithm , the Zernike-coefficients vector of a wavefront can be obtained:\n\nA=EHE18\n\nwhere E is the modal reconstruction matrix and H is the wavefront slope vector.\n\nTherefore, the variance of the modal Zernike coefficients that describe the wavefront measurement error can be written as:\n\n<\|Δaj\|2>=<\|k=12Qej,kΔhk\|2>=k=12Ql=12Qej,kej,l<ΔhkΔhl>E19\n\nwhere Q is the total number of subapertures, ej,kis the element of E, and Δhkis the error of the kth slope element.\n\nIn order to simplify analysis, we assume that there are no correlations among the different slope vectors in the corresponding subapertures and the intensity of the signals are uniform and isotropic among the different subapertures Subsequently, the following expression can be obtained:\n\n<ΔhkΔhl>=σc2f2δ(jkjl)E20\n\nwhere σc2is the variance of centroid position random error induced by random noise, f is the focal length of lenslets, δ(x,y)is the Kronecker delta function , and j and k are the subapertures which are connected with the slope hk and hl. Substituting Eq. (20) and Eq. (19) into Eq. (17), the mean-square of wavefront measurement error can be written as:\n\nσϕ2=j=1P<\|Δaj\|2>=j=1Pσg2K(j,Q)=j=1P(σcff0)2K(j,Q)E21\n\nwhere σgis the wavefront average slope of the corresponding subaperture in the unit circle,K(j,Q)=k=1Q(ej,2k1+ej,2k)2. It is concerned with the subaperture segmentation number and the distribution of subapertures. f0describes the normalized relationship between the real wavefront slope vector and the normalized wavefront slope vector in the unit circle, and is defined by the expression:\n\nf0=D2λE22\n\nwhere D is the diameter of the aperture and λ is the measuring wavelength.\n\nThen, the root mean square value of wavefront measurement error caused by centroid position random error is obtained:\n\nσϕ=σcD2λf[j=1Pk=1Q(ej,2k1+ej,2k)2]1/2E23\n\nEq. (23) is the desired expression used to evaluate the wavefront measurement error associated with the centroid position random error. σcis the standard deviation in pixels of centroid position random error caused by random noise. The formula described in Eq. (23) can help us to decide what the wavefront measurement error will be when the centroid position randomly fluctuates due to random noise, and it may be a factor that must be considered during the design of the SHWFS.\n\n## 4. Wavefront measurement error analysis based on Zernike modal reconstruction\n\nIn a Shack-Hartmann wavefront sensor, the angular position can be calculated from the centroid position in each subaperture and is proportional to the centroid position. The relationship between centroid and angular position can be described by\n\nσ=σcfE24\n\nIn Eq. (15), the angular position error caused by random noise was obtained. In Eq. (23), the wavefront error caused by random centroid error was obtained. Therefore, the total wavefront measurement error can be described by Eq. (25):\n\nσϕ=σfD2λf[j=1Pk=1Q(ej,2k1+ej,2k)2]1/2=12(1SNR+η/Vs)j=1Pk=1Q(ej,2k1+ej,2k)2E25\n\nIn this formula, we can determine the wavefront measurement error concerned with SNR (see the definition in Eq. (1)), aperture of lenslets (see the definition in Eq. (6)), counts of effective signal, and the reconstruction matrix parameters (see the definition in Eq. (19)).\n\n## 5. Conclusions\n\nIn this chapter, the exact formula (Eq. (25)), which evaluates the Shack-Hartmann wavefront sensor’s measurement error associated with the signal to noise ratio of effective signal, was derived in detail. This study was performed based on a modal wavefront reconstruction with Zernike polynomials, and provided an exact and universal formula to describe the wavefront measurement error of a Shack-Hartmann wavefront sensor with discrete detector arrays. It is critical to an adaptive optics system when the Shack-Hartmann sensor is used as the wavefront sensor, and it provides a reference when designing a Shack-Hartmann wavefront sensor and calculating its reconstruction matrix.\n\n## Acknowledgments\n\nWe would like to give our thanks to Shanqiu Chen, Li Shao, Daoai Dong, and Xuejun Zhang for their great discussion and assistance. We will also give our special thanks to Kevin M. Ivers for his great help in writing this chapter.\n\nchapter PDF\nCitations in RIS format\nCitations in bibtex format\n\n## More\n\n© 2012 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution 3.0 License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.\n\n## How to cite and reference\n\n### Cite this chapter Copy to clipboard\n\nChaohong Li, Hao Xian, Wenhan Jiang and Changhui Rao (January 20th 2012). Measurement Error of Shack-Hartmann Wavefront Sensor, Topics in Adaptive Optics, Robert K. Tyson, IntechOpen, DOI: 10.5772/29430. Available from:\n\n### chapter statistics\n\n1Crossref citations\n\n### Related Content\n\n#### This Book\n\nEdited by Robert Tyson\n\nNext chapter\n\n#### Acceleration of Computation Speed for Wavefront Phase Recovery Using Programmable Logic\n\nBy Eduardo Magdaleno and Manuel Rodríguez" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91842324,"math_prob":0.9600151,"size":10280,"snap":"2021-04-2021-17","text_gpt3_token_len":2124,"char_repetition_ratio":0.17127287,"word_repetition_ratio":0.0809838,"special_character_ratio":0.19747081,"punctuation_ratio":0.08803006,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9897733,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-21T03:21:17Z\",\"WARC-Record-ID\":\"<urn:uuid:edbbe7ff-09d4-47eb-bcf1-97a81cb36c3e>\",\"Content-Length\":\"512866\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2001166a-c382-40a6-93f3-626646901e6a>\",\"WARC-Concurrent-To\":\"<urn:uuid:2383fac6-cd00-47a2-8610-eb7c028b6a0d>\",\"WARC-IP-Address\":\"35.171.73.43\",\"WARC-Target-URI\":\"https://www.intechopen.com/books/topics-in-adaptive-optics/measurement-error-of-shack-hartmann-wavefront-sensor\",\"WARC-Payload-Digest\":\"sha1:LVQRM5MVC3JPFFVH4KYXILS7SM65TVMX\",\"WARC-Block-Digest\":\"sha1:YXEVEFL7C6GWLCVQ7PDRADUCNOBDCZQ5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703522150.18_warc_CC-MAIN-20210121004224-20210121034224-00226.warc.gz\"}"}
https://socratic.org/questions/58d2b088b72cff2f1f5fd5f0	[ "# Question #fd5f0\n\nJun 7, 2017\n\nI'll do the first one to show the general method of solving differential equations of this form.\n\nThese are all separable differential equations, meaning that all terms with $x$, including $\\text{d} x$, can be moved to one side of the equation, while all terms including $y$ can be moved to the other.\n\nFrom there, both sides can be integrated independently, and the constant of integration found using the initial condition.\n\nDepending on context, it may or may not be possible to solve for $y$ explicitly.\n\nIn the first example\n\n$\\left(\\text{d\"y)/(\"d} x\\right) = \\frac{x}{y} ^ 2$\n\nwe can cross multiply to isolate $x$ terms and $y$ terms as:\n\n${y}^{2} \\textcolor{w h i t e}{.} \\text{d\"y=xcolor(white).\"d} x$\n\nIntegrate both sides:\n\n$\\int {y}^{2} \\textcolor{w h i t e}{.} \\text{d\"y=intxcolor(white).\"d} x$\n\n$\\frac{1}{3} {y}^{3} = \\frac{1}{2} {x}^{2} + C$\n\nNote that $C$, an arbitrary constant of integration, has been added to only the right-hand side of the equation. This is arbitrary—we just as easily could have written $\\left(1 / 3\\right) {y}^{3} + {C}_{1} = \\left(1 / 2\\right) {x}^{2} + {C}_{2}$ or $\\left(1 / 3\\right) {y}^{3} + {C}_{3} = \\left(1 / 2\\right) {x}^{2}$. All are analogous forms.\n\nProceeding with the more standard form, which is to include the constant with the $x$ terms, we can solve for $C$ using the initial condition—that when $x = 1$, $y = 1$ as well.\n\n$\\frac{1}{3} {\\left(1\\right)}^{3} = \\frac{1}{2} {\\left(1\\right)}^{2} + C$\n\n$C = - \\frac{1}{6}$\n\nSo:\n\n$\\frac{1}{3} {y}^{3} = \\frac{1}{2} {x}^{2} - \\frac{1}{6}$\n\n$y = f \\left(x\\right) = {\\left(\\frac{3 {x}^{2} - 1}{2}\\right)}^{1 / 3}$" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.79859424,"math_prob":0.9999987,"size":828,"snap":"2021-43-2021-49","text_gpt3_token_len":265,"char_repetition_ratio":0.1092233,"word_repetition_ratio":0.0,"special_character_ratio":0.3031401,"punctuation_ratio":0.07909604,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99992836,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-07T16:12:38Z\",\"WARC-Record-ID\":\"<urn:uuid:13ec1b69-6ed1-4843-943d-9a10d63bdaad>\",\"Content-Length\":\"36628\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5eda3710-39e2-44ef-854f-3f5bc4448412>\",\"WARC-Concurrent-To\":\"<urn:uuid:f2f9e306-773c-4129-a3da-81ec689bb529>\",\"WARC-IP-Address\":\"216.239.36.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/58d2b088b72cff2f1f5fd5f0\",\"WARC-Payload-Digest\":\"sha1:27PA3DFYO5WQYTDPUN2467FY6JQALBFO\",\"WARC-Block-Digest\":\"sha1:RC56E5SR7KWX6VJIKB32HXVFAMJD2LHO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363400.19_warc_CC-MAIN-20211207140255-20211207170255-00589.warc.gz\"}"}
http://pubs.sciepub.com/education/6/11/12/index.html	[ " Influence of Problem Based Learning Model and Early Mathematics Ability to Mathematical Communication Skills and Self-Confidence in Junior High School", null, "Publications are Open\nAccess in this journal\nArticle Versions\nExport Article\n• Normal Style\n• MLA Style\n• APA Style\n• Chicago Style\nResearch Article\nOpen Access Peer-reviewed\n\nInfluence of Problem Based Learning Model and Early Mathematics Ability to Mathematical Communication Skills and Self-Confidence in Junior High School\n\nAmerican Journal of Educational Research. 2018, 6(11), 1539-1545. DOI: 10.12691/education-6-11-12\nReceived September 10, 2018; Revised October 30, 2018; Accepted November 23, 2018\n\nAbstract\n\nThis study aims to determine: (1) the effect of problem based learning model on students mathematical communication skills, (2) the effect of problem based learning model on students self-confidence, (3) the interaction between learning models (problem based learning and conventional learning) and early mathematics ability to students mathematical communication skills, and (4) the interaction between learning models (problem based learning and conventional learning) and early mathematics ability to student's self-confidence. This study is a quasi-experimental research. The population in this study were all students of class VIII junior high school 2 Tanjung Pura consisting of six classes. Samples were selected by simple random sampling as much as two classes. The selected class is VIII-2 as the experimental class (31 students) and VIII-1 as the control class (32 students). The instruments used are mathematical communication skills test, early mathematics ability test, self-confidence questionnaire. Data obtained from the research instruments were analyzed using two-way ANOVA by SPSS program. The results showed that: (1) there is significant effect of problem based learning model to students mathematical communication skill, (2) there is a significant effect of problem based learning model to students self-confidence, (3) there is interaction between learning models (problem based learning and conventional learning) and early mathematics ability to students' mathematical communication skills, (4) there is interaction between learning models (problem based learning and conventional learning) and early mathematics ability to student's self-confidence.\n\n1. Introduction\n\nIn general mathematics has five objectives in mathematics learning. These objectives include: developing mathematical attitudes, gaining proficiency in the use of mathematical language, gaining insight into applications from mathematics in other disciplines 1. One of the goals of mathematics learning has a purpose so students can communicate well. This is in accordance with the standard of the mathematics learning process formulated by the National Council of Teacher of Mathematics 2, namely: 1) mathematical problem solving, 2) mathematical reasoning, 3) mathematical communication, 4) mathematical connections, and 5) mathematical representation. Learn most of the communication and construct their own knowledge. In addition, with communication students can improve vocabulary, develop speaking skills, write ideas systematically and have better learning abilities 3, 4. Mathematical communication has an important role for students in formulating mathematical concepts and strategies, investing students in the completion of exploration and mathematical investigation, and means for students to communicate to obtain information, share ideas and findings 5, 6. However, the results of the study show that many students experience difficulties in mathematics. Arifin, Kartono, and Sutarto 7 that students' mathematical communication skills are still low. Students have difficulty identifying and conveying ideas contained in a problem 8.\n\nLearning objectives are also seen from the affective domain. In this case one of the affective domains that students need to have in mathematics learning is students' self-confidence attitude. Self-confidence can be interpreted as one part of feelings and thoughts about who we really are 9. In relation to mathematics learning, self-confidence can be built by erasing the impression that mathematics lessons place students as objects by accepting the theory and memorizing formulas. The ease in learning mathematics can make students appreciate and love mathematics 10, 11. From the results of other studies indicate that there is an influence between students 'confidence in students' mathematics learning outcomes 12, 13. In fact, students still have problems with confidence. Students always complain that they have no ability, especially in learning mathematics. When learning, students easily give up and complain that learning is difficult. If asked to work on a question in front of the class, students are overly afraid and feel uncertain about the answer 10, 14.\n\nMore specifically, the low mathematical communication skills and self-confidence of students cannot be separated from the teacher's view of the meaning and model of learning. This is in accordance with what was revealed by Slameto 15 that the role of teachers in the teaching and learning process is to encourage, guide, and provide learning facilities for students to achieve goals. One learning model that can be used to answer these problems is a problem based learning model. Rusman 16 suggests that problem based learning is useful to facilitate the success of problem solving, communication, group work and interpersonal skills better than other approaches. The results show that problem-based learning is effective to improve students' mathematical communication skills. When problem solving is used as a context in mathematics, the focus of learning activities is entirely on students, namely the process to understand a mathematical concept and procedure contained in the problem 17, 18. Amalia, Surya, and Syahputra 19 says that in the use of problem based learning, students were guided to find their own answers by following the steps of the PBL model. As in other studies problem based learning makes students more creative, dare to make decisions, think rationally and collaborate effectively with their classmates 20.\n\nThe success of students in taking mathematics lessons is also very influential on the early mathematical ability factor. Akramunnisa and Sulestry 21 revealed in the learning process, the teacher must pay attention to the students' early mathematical abilities in solving mathematical problems, because the mathematical concepts that are related to each other and form new concepts that are more complex. By knowing the early ability of each student's mathematics, the teacher will be easier in determining the method or strategy suitable for use in the classroom so that the learning carried out will be more effective and efficient 22. The expected end result, the problem based learning model can stimulate students to be more confident both in terms of communication verbally and in writing because each student always interacts between the teacher and other students. This is reinforced by the results of research which revealed that students who have high confidence in mathematics will easily answer mathematical communication problems 23.\n\n2. Literature Review\n\n2.1. Mathematical Communication Skills\n\nFiske 24 also states that communication as social interaction through messages and all communication involves signs and codes. Effendy 25 the communication process is essentially a process of delivering thoughts or feelings by a person (communicator) to others (communicant). The mind can be in the form of ideas, information, opinions, and others that emerge from his mind. Baroody 26 says that education should help children to communicate mathematical ideas with the representation, listening, reading, discussion and writing. Communication through mathematics, teachers can foster student engagement and participation while focusing on a deep conceptual understanding has been mentioned in the general standard of mathematics. Development the language of mathematics to note for the students to better understand mathematical concepts underlying 27. Yusra and Saragih 28 says that the mathematical communication skills is the ability to disclosure of mathematical ideas with symbols, tables, diagrams, or other media to clarify the issue of mathematics and delivered with a mathematical language in teaching and learning mathematics in the learning process of mathematics.\n\nIn this research study, mathematical communication skills of junior high school (SMP) has been expressed by Manitoba Education 29 that the communication in mathematics students daily delivery either orally, through diagrams and pictures, and writing about mathematics. Students need opportunities to read, present, view, write, listen, and discuss mathematical ideas. It is almost similar is said by NCTM 30 that the communication in mathematics for secondary level there are a few things to note are (1) related to mathematical ideas, (2) access some answers solution method, (3) allows multiple presentation of diverse, and (4) a chance in the interpretation, and hypothesizing. Mathematical communication skills Junior high school students are measured from four aspect, namely: 1) convert mathematical situations or ideas into drawing, 2) formulate mathematical ideas into language and mathematical symbols, 3) chage information from an image/table into mathematical language and symbols, and, 4) explain procedures settlement.\n\n2.2. Self-Confidence\n\nSelf-confidence is a positive mental attitude from a person individuals who position or condition themselves to be able to evaluate themselves and their environment so that they fell comfortable doing activities in an effort achieve planned goals 31, 32. Brewer and Barlow 33 also suggest that self-confidence is closely related to ability and knowledge in a particular domain and is influenced by the amount of information taken, the clarity and details of the target. As well as the influence of the domain of knowledge, self-confidence is also influenced by students’ beliefs about, awareness of themselves, general abilities in learning and strategy. Similar to this, Pierce and Stacey 34 defines mathematical beliefs, as students' perceptions of their ability to achieve good results and their assurance that they can handle difficulties in mathematics. Margono 35 divides ones’s confidence in mathematics into three components, namely as follows: 1) trust in understanding and self-awareness regarding mathematical abilities, 2) ability to realistically determine the goals to be achieved and develop an action plan as an effort to achieve the intended goals, and 3) trust in mathematics itself.\n\n2.3. Early Mathematics Ability\n\nAccording to Rusman 16 knowledge of students' early abilities is important for teachers to be able to provide the right portion of the lesson and it is useful to take the necessary steps. Early mathematical abilities describe the readiness of students in doing mathematics learning activities. Another thing explained by Astuti 36 is that initial knowledge is a framework in which students filter new information and find meaning about what is being learned in the learning process. Russeffendi 37 states that from a group of students who are randomly selected (not specifically chosen) there are always a number of children with high, moderate and low ability who are normally distributed. The student's initial mathematical abilities are important to know before he starts with his learning, because it can be known whether students already have or knowledge that is a prerequisite for learning. The extent to which students have known what material will be presented. By knowing this, the teacher will be able to design learning better. Because if students are given material that is already known, they will feel bored quickly. Classification of students' initial knowledge consists of groups consisting of smart groups, moderate and less 38. Akramunnisa and Sulestry 21 reveal the initial ability of mathematics is the level of students' ability to solve existing mathematical problems its relationship with the material underlying these questions. The students' initial mathematical abilities that vary greatly influence the achievement of the next mathematics learning outcomes.\n\n2.4. Model Problem Based Learning\n\nProblem based learning is learning that optimizes students' thinking skills through a systematic process of group work or teams, so students can empower, sharpen, test, and develop their thinking skills in a sustainable manner 16. Problem based learning will encourage students to think divergently. Divergent thought patterns will lead them to the formation of creativity. Learning by giving problems that contain many solutions or many ways of solving can improve students' mathematical creativity 39. Problem based learning has an effect on content knowledge which provides greater opportunities for students to learn with more involvement and increase student active participation, motivation and interest among students. This causes students to have a positive attitude towards mathematics and help them to improve their performance for the most part and which will cause long-term memory 26.\n\nProblem Based Learning is an effective approach to high level thinking. This learning helps students to process the information that has been formed in their minds and compiles their own knowledge about the social world and its surroundings. In this study the stages of the problem based learning model are: 1) Giving problems to students; 2) Reviewing the problem given; 3) Guiding individual and group mastery; 4) Develop and present the work; and 5) Analyze and evaluate the problem solving process. By using problem based learning that is able to develop inquiry and problem solving as well as collaborative, communicative, cooperative learning and learning processes, self-direction can be influenced by students' mathematical communication skills and self-confidence. From the various issues above, the problems that will be discussed in this study are:\n\n1. Is there an effect of problem based learning model on students' mathematical communication skills?\n\n2. Is there an effect of problem based learning model on student self-confidence?\n\n3. Are there interactions between the learning model and early mathematical abilities towards students' mathematical communication skills?\n\n4. Are there interactions between the learning model and the students early mathematical abilities towards students' self-confidence?\n\n3. Methods\n\nThe type of research used in this study is quasi-experimental. This research was conducted at junior high school 2 Tanjung Pura. The population in this study were all eighth grade students. While sample selection is done by cluster random sampling technique. The sample in this study was class, VIII-2 which was used as an experimental class with total 31 students and class VIII-1 was used as a control group totaling 32 students. This study involved two classes treated differently. The experimental group is treated by applying the problem based learning model, while the control group is treated by applying conventional learning. The instruments used are mathematical communication skills test consists of 4 essay, early mathematics ability test amounted to 6 essay, self-confidence questionnaire of 33 statements. Data obtained from research instruments were analyzed using two-way ANOVA through the SPSS program.\n\n4. Result\n\n4.1. Findings\n\nInitial description presented is the result of test early mathematical ability is given to know the average equality experimental class and control class. This test is also conducted to group students, namely high, medium, and low. The results of the summary presented in Table 1. below.\n\nTable 1. gives a conclusion that the average score of the early mathematical abilities for each sample class is relatively the same. For the problem based learning class obtained 73.26 while the conventional learning class was obtained 71.7. So if there is a difference in the results of students' abilities in each sample class is caused by different treatments, not because there are differences before learning. Next, the results of the calculation of mathematical communication skills and self-confidence of students can be seen in the following Table 2:\n\nIn Table 1, it can be seen the average mathematical communication skills and self-confidence of the two groups of students who are taught with the problem based learning and conventional learning models. The problem based learning model obtained an average mathematical communication ability of 76.63, while students who received regular learning obtained an average mathematical communication skills of 69.34. While the problem based learning model obtained an average self-confidence of 72.24, while students who received conventional learning gained an average self-confidence of 68.18. To determine the significance of the data in statistical testing with a two-way anava test, it was previously tested for data normality and homogeneity. The results of homogeneity and normality of the two classes from the tests of mathematical communication skills and self-confidence questionnaire indicated that the two sample groups had homogeneous variance and normal data distribution. The following are the results of the two-way anava calculation of mathematical communication skills presented in Table 3, namely.\n\nBased on the results of a two-way ANOVA in Table 3 above, the p-value was obtained for the study was 0,010 < 0,05, that is enough evidence to reject H0 and accept H1. Means that there is influence of problem based learning models on students' mathematical communication skills. In other words the influence of the problem based learning model on mathematical communication skills is better than conventional learning on mathematical communication skills. In the interaction of learning models (problem based learning and conventional learning) with early mathematical ability obtained p-value = 0.044 < 0.05 or in other words enough evidence to reject H0, thus significantly there is an influence of interaction between learning (problem based learning and conventional) with the early ability of mathematics towards students' mathematical communication skills. The following are the results of the two-way analysis of students' self-confidence in those presented in Table 4, namely.\n\nBased on the results of the two-way ANOVA test in Table 4 above, the p-value was obtained for the study was 0.028 < 0.05, that is enough evidence to reject H0 and accept H1. It means that there is the influence of the problem based learning model on students' self-confidence. In other words, the effect of the problem based learning model on self-confidence is better than conventional learning about self-confidence. In the interaction of learning models (problem based learning and conventional) with early mathematical abilities obtained p-value = 0.035 < 0.05 or in other words enough evidence to reject H0, thus significantly there is an influence of interaction between learning (problem based learning and conventional learning) with early mathematical abilities towards student self-confidence.\n\n4.2. Discussions\n\nThe research findings reveal that the average value of students 'mathematical communication skills taught with the problem based learning model is higher than the average value of students' mathematical communication skills taught by conventional learning. This proves that the problem based learning model is better than the usual learning done by the teacher in developing students' mathematical communication skills. The results of previous studies explain that mathematical communication skills in the experimental class (model problem based learning) is better than conventional learning by teachers 41, 42, 43, 44.\n\nSurya, Syahputra, and Juniati 43 during the process of PBL lead to interaction between students and teachers to learn is often the case, it will eventually make students brainstorm and to reflect the understanding that he had before. Khun-Inkeeree, Omar-Fauzee, and Othman 45 revealed confidence built by the ability to interact in the classroom. This is probably because the module or student activity sheet used in classroom interaction that creates opportunities for students to interact freely during class activities. The learning process is presented by the model problem based learning not only to transfer knowledge from teacher to student, but a process that is conditioned by the teacher so that students are active in a variety of ways to build his own knowledge. Kadir and Parman 46 gives a contextual problem students' attention and challenge students to solve it by means of a mathematical method or communicate their mathematical ideas. Kodariyati and Astuti 36 describes in more visible model of problem based learning students actively in learning. In a study using problem based learning models, communication skills can be developed in the form of questions at the beginning of learning. The use of student worksheets given to each group also influences the course of the learning process.\n\nThe study's findings prove that the use of problem-based student activity sheet structured nature of the inquiry was able to grow in the early mathematical concepts. With the activity sheet, students have no difficulty in answering math problems. students simply follow the patterns that have been provided in the sheet. It tersebur accordance with Kodariyati and Astuti 42 In a study using problem based learning models, communication skills can be developed in the form of questions at the beginning of learning. In addition, the presence of a class discussion makes the students are encouraged to bring an idea or ideas. Use of worksheets that are given to each group also influences the course of the learning process. Astriani, Surya, and Syahputra 47 explains when students are working on the student worksheet and test math skills through contextual problems would encourage students in learning activities to help each other, sharing, respect between the different learning abilities possessed by each student.\n\nThe above explanation is supported by learning theory that supports the mathematical communication skills contained in Runtukahu 48 the symbolic stage in the theory of child development from Bruner states that children manipulate symbols or symbols of certain objects. Students are able to use notation without relying on real objects. Other cognitive theories from Piaget say cognitive development as a child process actively builds systems of meaning and understanding of reality through their experiences and interactions 49. The study's findings also support previous research thatNurbaiti, Irawati and Lichteria 44 found that the model of problem based learning and expository able to jointly provide a positive influence on students' mathematical communication skills. Other results also indicate the interaction between early ability learning model mathematics to students' mathematical ability 50, 51. Problem based learning to students' attitudes that students use problem-based learning show a positive attitude in dealing with problems that are superior because of the element of interaction and constructivism were very prominent in problem-based learning 52.\n\nThe findings also showed that the average value of the self-confidence of students who are taught by a model problem based learning is higher than the average value of the self-confidence of students who are taught by conventional learning. This proved that the model of problem based learning is better than the usual lesson by the teacher in developing students' self-confidence. Results of previous studies also showed that the attitude of self-confidence of students who are taught by a model problem based learning is better than the expository 53, 54, 55. Nurqolbiah 52 the self-confidence of students who are built through problem-based learning with a scientific approach is superior in showing a positive attitude in dealing with problems.\n\nSelf-confidence can also be developed by doing rational and realistic learning in the student environment. This is in line with the problem based learning model which begins to present mathematical problems for students, so students are required to solve problems that are rich in mathematical concepts 56. Previous research by Cerezo 57 suggests that student responses are very positive by liking problem-based learning. Students are able to work in groups. Students feel problem-based learning makes them challenged to think differently, be open to new ideas and not judgmental, and to be supportive of each other. Rokhmawati, Djatmika, and Wardana 58 the application of PBL models can also improve students' positive attitudes. This shows that students have developed skills to adapt to the environment and can have self-control. This can be seen from the courage of students appearing to give opinions, the ability to think positively, and the confidence to communicate in class.\n\nFinally, this study shows that students who have the initial ability of high and moderate mathematics are more benefited than students who have low initial math skills. This means that the problem based learning model is well used for students who have high and moderate categories of early mathematical abilities in developing students' mathematical communication skills and self-confidence. Darkasyi, Johar, and Ahmad 59 explain learning models that enable students to improve students 'mathematical communication skills in terms of students' early mathematical abilities (low, medium, high). Utami and Misnasanti 60 if students can use the early knowledge well in understanding new material, it will affect the ability to solve mathematical problems. Students will be able to solve the problems faced by linking the knowledge they have with new knowledge.\n\n5. Conclusion\n\nBased on the research described in the previous section, can be summed up as follows:\n\n1. Mathematical communication ability of students taught using problem based learning models better than taught using conventional learning.\n\n2. The attitude of self-confidence of students showed better results in the classroom with a model of problem based learning than classroom with conventional learning. This study shows that there are positive effects of problem based learning model with self-confidence of students.\n\n3. There is an interaction between learning models (problem based learning and conventional learning) and early ability of mathematics to students' mathematical communication skills.\n\n4. There is an interaction between learning models (problem based learning and conventional learning) and early ability of mathematics to students' self-confidence.\n\n6. Suggestions for Future Studies\n\nFrom the findings obtained by feeding can be given to the next researcher who develops the learning model, first, the researcher can develop the learning components used in the problem-based learning model and computer-based model development. Second, before giving a learning model, the researcher should further examine students 'and teachers' perceptions of the learning model provided. So that students and teachers really understand the learning model use.\n\nAcknowledgements\n\nThe author would like to thank who have helped write this simple journal. I am really thankful to them.\n\nReferences", null, "This work is licensed under a Creative Commons Attribution 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by/4.0/" ]	[ null, "http://www.sciepub.com/common/showimages.aspx", null, "http://pubs.sciepub.com/images/icon-by.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.81056553,"math_prob":0.64032733,"size":51645,"snap":"2019-26-2019-30","text_gpt3_token_len":12972,"char_repetition_ratio":0.20479852,"word_repetition_ratio":0.51307684,"special_character_ratio":0.22819246,"punctuation_ratio":0.16604763,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.98477286,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-26T09:45:22Z\",\"WARC-Record-ID\":\"<urn:uuid:93abb510-53cc-4d0e-a840-da06a57d6ce7>\",\"Content-Length\":\"167290\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:63e22d68-5787-4236-b92e-65e064083a39>\",\"WARC-Concurrent-To\":\"<urn:uuid:6d4d573a-a6b1-499c-a1c8-bd33d7b212ce>\",\"WARC-IP-Address\":\"70.39.102.108\",\"WARC-Target-URI\":\"http://pubs.sciepub.com/education/6/11/12/index.html\",\"WARC-Payload-Digest\":\"sha1:XWYCBEGDGBBNVUQQI2JQUJKSZAXGLPCP\",\"WARC-Block-Digest\":\"sha1:IRJ3A4EQU5LLJVBA4BYS223XVESLPPBY\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560628000266.39_warc_CC-MAIN-20190626094111-20190626120111-00136.warc.gz\"}"}
https://bob.cs.sonoma.edu/IntroCompOrg-RPi/sec-codes.html	[ "## Section4.6Other Codes.\n\nThus far in this chapter we have used the binary number system to represent numerical values. It is an efficient code in the sense that each of the $2^{n}$ bit patterns represents a value. On the other hand, there are some limitations in the code. We will explore some other codes in this section.\n\n### Subsection4.6.1BCD Code\n\nOne limitation of using the binary number system is that a decimal number must be converted to binary before storing or performing arithmetic operations on it. And binary numbers must be converted to decimal for most real-world display purposes.\n\nThe Binary Coded Decimal (BCD) code is a code for individual decimal digits. Since there are ten decimal digits, the code must use four bits for each digit. The BCD code is shown in Table 4.6.1.\n\nFor example, in a 16-bit storage location the decimal number 1234 would be stored in the BCD code as\n\n\\begin{gather} \\binary{0001 \\; 0010 \\; 0011 \\; 0100} \\end{gather}\n\nand in binary as\n\n\\begin{gather} \\binary{0000 \\; 0100 \\; 1101 \\; 0010} \\end{gather}\n\nFrom Table 4.6.1 we can see that six bit patterns are “wasted.” The effect of this inefficiency is that a 16-bit storage location has a range of $0$ – $9999$ if we use BCD, but the range is $0$ – $65535$ if we use binary.\n\nBCD is important in specialized systems that deal primarily with numerical data. There are I/O devices that deal directly with numbers in BCD without converting to/from a character code {for example, ASCII). The COBOL programming language supports a packed BCD format where two digits (in BCD code) are stored in each 8-bit byte. The last (4-bit) digit is used to store the sign of the number as shown in Table 4.6.2. The specific codes used depend upon the particular implementation.\n\nFor example, $\\binary{0001 \\; 0010 \\; 0011 \\; 1010}$ would represent $+123\\text{,}$ $\\binary{0001 \\; 0010 \\; 0011 \\; 1011}$ would represent $-123\\text{,}$ and $\\binary{0001 \\; 0010 \\; 0011 \\; 1111}$ would represent $123\\text{.}$\n\n### Subsection4.6.2Gray Code\n\nOne of the problems with both the binary and BCD codes is that the difference between two adjacent values often requires that more than one bit be changed. For example, three bits must be changed when incrementing from $3$ to $4$ ($\\binary{0011}$ to $\\binary{0100}$). If the value is read during the time when the bits are being switched there may be an error. This is more apt to occur if the bits are implemented with, say, mechanical switches instead of electronic.\n\nThe Gray code is one where there is only one bit that differs between any two adjacent values. As you will see in Section 5.5, this property also allows for a very useful visual tool for simplifying Boolean algebra expressions.\n\n decimal Gray code $0$ $\\binary{0}$ $1$ $\\binary{1}$\n\nTo add a bit, first duplicate the existing pattern, but reflected:\n\n Gray code $\\binary{0}$ $\\binary{1}$ $\\binary{1}$ $\\binary{0}$\n\nThen add a zero to the beginning of each of the original bit patterns and a one to the beginning of each of the reflected set:\n\n decimal Gray code 0 $\\binary{00}$ 1 $\\binary{01}$ 2 $\\binary{11}$ 3 $\\binary{10}$\n\nThe Gray code for four bits is shown in Table 4.6.3. Notice that the pattern of only changing one bit between adjacent values also holds when the bit pattern “wraps around.” That is, only one bit is changed when going from the highest value ($15$ for four bits) to the lowest ($0$)." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84707373,"math_prob":0.9970338,"size":4182,"snap":"2020-34-2020-40","text_gpt3_token_len":1279,"char_repetition_ratio":0.21182384,"word_repetition_ratio":0.01904762,"special_character_ratio":0.37996173,"punctuation_ratio":0.088507265,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99880457,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-15T04:41:35Z\",\"WARC-Record-ID\":\"<urn:uuid:5fcd9691-a869-436c-992d-b5ab0a717eb9>\",\"Content-Length\":\"35171\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:43d4618a-8099-4cd8-b66d-934df829a59e>\",\"WARC-Concurrent-To\":\"<urn:uuid:38b8d671-9743-4d2d-9437-c18296075463>\",\"WARC-IP-Address\":\"130.157.166.29\",\"WARC-Target-URI\":\"https://bob.cs.sonoma.edu/IntroCompOrg-RPi/sec-codes.html\",\"WARC-Payload-Digest\":\"sha1:6CFX2MV7Z7TYNH4VS7A7ZLZDFAPI7YWM\",\"WARC-Block-Digest\":\"sha1:EZZYQGKEZ3G36V3QV6M3ADO3T3OTOM65\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439740679.96_warc_CC-MAIN-20200815035250-20200815065250-00158.warc.gz\"}"}
https://mathematica.stackexchange.com/questions/tagged/education?tab=newest&page=2	[ "# Questions tagged [education]\n\nQuestions about the use of Mathematica/WL in education and/or about teaching Mathematica, or learning it.\n\n127 questions\nFilter by\nSorted by\nTagged with\n58 views\n\n### Looking for introduction to equation solving [closed]\n\nI am a beginner with Mathematica. I wanna learn how to make a program to solve a system of algebraic equations related to my master [?], so I am looking for a way to learn this in as little time as ...\n78 views\n\n### Strain-stress relationship in spherical coordinates with Poisson's ratio [closed]\n\nI am wondering if there is a way to get the relationship between strain and stress in spherical coordinates with Poisson's ratio. I read the documentation here: https://reference.wolfram.com/...\n55 views\n\n### Two variable derivative function mathematica [closed]\n\nI am trying to find the point where the partial derivatives are equal to zero. 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Find the longest sequence of consecutive ...\n89 views\n\n### Understanding the question\n\nLet n be the integer shown below: ...\n46 views\n\n### Trying to find how to find how many prime factors have an odd exponent?\n\nHi I am doing a question for Mathematica and I'm having difficulties.The question is in the prime factorization of n, how many prime factors have an odd exponent? The given n is shown below (very ...\n938 views\n\n### Finding slope from a 2D Plot\n\nImagine I have a set of data, the plot is as following(just as an example consider a Gaussian curve): Is there any way to obtain the slop of this curve slop and plotted just using the initial data. ...\n166 views\n\n137 views\n\n### NonlinearModelFit understanding [closed]\n\nI have a trouble with NonlinearModelFit questions. Question: V is a function of T and ...\n114 views\n\n### Limit of an infinite nested radical?\n\nIs it possible to compute the limit of this infinite nested radical in Mathematica: Limit[Sqrt[n+Sqrt[n+Sqrt[n+Sqrt[n+...]]]], n->0]\n139 views\n\n### Mathematica Progression Path from Apprentice to Guru [closed]\n\nThere's an excellent set of responses to a question about learning Python on stack overflow. As there seems to be a large number of Mathematica experts here, it seems of value (to me at least) to ask ...\n43 views\n\n### Speak stops on first Hyphen with IntegerName\n\nIntegerName returns the text of an integer. However, when Speak is applied to the result it stops at the fist hyphen. For ...\n81 views\n\n### How do I use Mathematica to extract data points from DICOM images? [closed]\n\nI have few DICOM images whose data points I would like to extract and probably export (which would really be excellent) to an excel file.\n271 views\n\n### need help to enter the orthogonal sign, the upside down T.\n\nI have search on Google and never came across any idea how to enter this damn sign. would anyone give me some hint?\n332 views\n\n### Solving an ODE using the Taylor Method Help?\n\nNot sure how to proceed after this... ...\n506 views\n\n### How do I recursively calculate this equation and generate a list of iteration?\n\nHow do I write a recursive equation to compute a list of answers? I tried NestList, but it didn't work. ...\n269 views\n\n### Resources to learn about Neural Nets using Mathematica?\n\nI'm interested in learning about Neural Nets and I'd really like my instruction to be in Mathematica if possible. 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Could anyone help me to reproduce this picture with Mathematica?\n103 views\n\n### Graph not plotting any points [closed]\n\nf [x_] := (x + ln[x] - 3) Plot[f[x], {x, 0, 5}] How come this comes out with an empty graph? I tried this and same results: ...\n2k views\n\n### List of amazing things you can do with Mathematica [closed]\n\nWhat is the coolest (amazing) thing you can do with Mathematica? for instance there are my favorite things Automating xkcd Diagrams Extending Van Gogh’s Starry Night with Inpainting Making ...\n92 views\n\n### What syntax refers to “this notebook” object?\n\nI want a notebook to automatically save after any cell is evaluated. I found that I should use the command SetOptions[XXX, NotebookAutoSave -> True] to do ...\n933 views\n\n### Do[ ]s and Don't[ ]s when teaching Mathematica in undergraduate courses\n\nQuestion motivated by some horrific homework-related posts that I don't dare to link: What are the top Do's and Dont's when teaching Mathematica to undergrads? I don't ask for the design of a ...\n121 views" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8941707,"math_prob":0.63428336,"size":12297,"snap":"2020-10-2020-16","text_gpt3_token_len":3203,"char_repetition_ratio":0.14430977,"word_repetition_ratio":0.010466223,"special_character_ratio":0.26819548,"punctuation_ratio":0.1324,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99242395,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-09T04:48:59Z\",\"WARC-Record-ID\":\"<urn:uuid:33b1cffa-af72-44ce-9cf0-87f1542073a9>\",\"Content-Length\":\"248384\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3fe5102d-c264-4d41-b175-dabcd0a05670>\",\"WARC-Concurrent-To\":\"<urn:uuid:d5b94551-78d2-47f9-bc3e-563fb48fcac6>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://mathematica.stackexchange.com/questions/tagged/education?tab=newest&page=2\",\"WARC-Payload-Digest\":\"sha1:LVL53F7N2JRPRF7GAOKHBFOBJDODUW2G\",\"WARC-Block-Digest\":\"sha1:MY6RKKPHOKCRFEYDWOUAJWSCFYU7H4PM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585371829677.89_warc_CC-MAIN-20200409024535-20200409055035-00127.warc.gz\"}"}
https://cs50.stackexchange.com/questions/37163/ps2-substitution/37168	[ "# PS2 substitution\n\nI am trying to check that the substitution key contains all the alphabets but am getting error: expression result unused for this line of code: s, l ++; Can anyone help, this is my code as of now:\n\n`````` #include <cs50.h>\n#include <stdio.h>\n#include <ctype.h>\n#include <string.h>\n\nint main(int argc, string argv[])\n{\nint n = strlen(argv);\nint s = 97, l = 65, count = 0;\nif (argc != 2) //check 1 argument given only\n{\nprintf(\"Usage: ./substitution key\\n\");\nreturn 1;\n}\nelse if (n != 26) //check if key contains exactly 26 characters\n{\nprintf(\"Usage: ./substitution key\\n\");\nreturn 1;\n}\nelse //check key contains each alphabet exactly once\n{\nfor (int i = 0; i < n; i++)\n{\nfor (int j = 0; j < n; j++)\n{\nif (argv[i] == s \|\| argv[i] == l)\n{\ncount ++;\n}\n}\ns, l ++;\n}\n}\nif (count == 26)\n{\nprintf(\"OKAY!\");\n}\n}\n``````\n• If your intention is to increment `s` and `l` then use`s++, l++` – DinoCoderSaurus Apr 29 '20 at 21:14\n\nI'm going to suggest you an easier method. You can simply use the ctype.h library. The library provides you with a function called isalpha(). This function literally checks whether a string contains only alphabetical characters. And I do believe instead of s,l++ you better write them separately.\n\nAnother thing that I think I found is you could edit your code this way\n\n``````for (int i = 0; i < n; i++)\n{\nfor (int j = 0; j < n; j++)\n{\nif (argv[i] == s \|\| argv[i] == l)\n{\ncount ++;\n}\ns++;\nl++;\n}\n}\n``````\n• Thanks for your suggestion but in this case, i am not just checking if all the characters are alphabets but rather whether all 26 alphabets.- a,b,c,d etc are present. – Olivia Apr 30 '20 at 10:35\n\nIf you want to check whether all 26 alphabetical characters are present, you can check whether the characters are not repeated. It works the same way. This was my code. Take a look. Here I checked whether a character is repeated again. :)\n\n``````string repeated = key; //duplicating another equal string to check if same character repeats itself\nfor (int i = 0; i < len; i++) // loop which increases value of i after checking 26 times of k\n{\nfor (int k = 0; k < len; k++)\n{\nif (i != k) //when i is equal to k it does not check the conditition since that step must be skipped\n{\nif (key[i] == repeated[k]) //checking if same character is repeated\n{\nprintf(\"Key must not contain repeated characters\\n\");\nreturn 1;\n}\n}\n}\n}\n``````" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5922864,"math_prob":0.9573935,"size":865,"snap":"2021-04-2021-17","text_gpt3_token_len":278,"char_repetition_ratio":0.09872241,"word_repetition_ratio":0.048780486,"special_character_ratio":0.39884394,"punctuation_ratio":0.19251336,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97926366,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-17T22:17:57Z\",\"WARC-Record-ID\":\"<urn:uuid:378d5a70-6056-49a1-9ce1-2f0f3e9b85cf>\",\"Content-Length\":\"127486\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7dc2d86d-baf7-4e78-8c62-0df238eed691>\",\"WARC-Concurrent-To\":\"<urn:uuid:ae2143ce-b84d-40b8-8a18-0a36a99e8165>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://cs50.stackexchange.com/questions/37163/ps2-substitution/37168\",\"WARC-Payload-Digest\":\"sha1:SWDAPNZOSXALK3DYLV3BFUE3RMYV7BPC\",\"WARC-Block-Digest\":\"sha1:NXV4Z7CXUWKVGT7KKJO2YL3YJZ6XEMVD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703513194.17_warc_CC-MAIN-20210117205246-20210117235246-00102.warc.gz\"}"}
https://mne.tools/dev/generated/mne.decoding.SSD.html	[ "# mne.decoding.SSD¶\n\nclass mne.decoding.SSD(info, filt_params_signal, filt_params_noise, reg=None, n_components=None, picks=None, sort_by_spectral_ratio=True, return_filtered=False, n_fft=None, cov_method_params=None, rank=None)[source]\n\nM/EEG signal decomposition using the Spatio-Spectral Decomposition (SSD).\n\nSSD seeks to maximize the power at a frequency band of interest while simultaneously minimizing it at the flanking (surrounding) frequency bins (considered noise). It extremizes the covariance matrices associated with signal and noise 1.\n\nSSD can either be used as a dimensionality reduction method or a ‘denoised’ low rank factorization method 2.\n\nParameters\ninfo`mne.Info`\n\nThe `mne.Info` object with information about the sensors and methods of measurement. Must match the input data.\n\nfilt_params_signal`dict`\n\nFiltering for the frequencies of interest.\n\nfilt_params_noise`dict`\n\nFiltering for the frequencies of non-interest.\n\nreg`float` \| `str` \| `None` (default)\n\nWhich covariance estimator to use. If not None (same as ‘empirical’), allow regularization for covariance estimation. If float, shrinkage is used (0 <= shrinkage <= 1). For str options, reg will be passed to method to `mne.compute_covariance()`.\n\nn_components\n\nThe number of components to extract from the signal. If n_components is None, no dimensionality reduction is applied.\n\npicks\n\nThe indices of good channels.\n\nsort_by_spectral_ratiobool (default `False`)\n\nIf set to True, the components are sorted accordingly to the spectral ratio. See Eq. (24) in 1.\n\nreturn_filteredbool (default `True`)\n\nIf return_filtered is True, data is bandpassed and projected onto the SSD components.\n\nn_fft`int` (default `None`)\n\nIf sort_by_spectral_ratio is set to True, then the SSD sources will be sorted accordingly to their spectral ratio which is calculated based on `mne.time_frequency.psd_array_welch()` function. The n_fft parameter set the length of FFT used. See `mne.time_frequency.psd_array_welch()` for more information.\n\ncov_method_params\n\nAs in `mne.decoding.SPoC` The default is None.\n\nrank`None` \| `dict` \| ‘info’ \| ‘full’\n\nAs in `mne.decoding.SPoC` This controls the rank computation that can be read from the measurement info or estimated from the data. See Notes of `mne.compute_rank()` for details. We recommend to use ‘full’ when working with epoched data.\n\nReferences\n\n1(1,2,3)\n\nVadim V Nikulin, Guido Nolte, and Gabriel Curio. A novel method for reliable and fast extraction of neuronal EEG/MEG oscillations on the basis of spatio-spectral decomposition. NeuroImage, 55(4):1528–1535, 2011. doi:10.1016/j.neuroimage.2011.01.057.\n\n2(1,2)\n\nStefan Haufe, Sven Dähne, and Vadim V Nikulin. Dimensionality reduction for the analysis of brain oscillations. NeuroImage, 101:583–597, 2014. doi:https://doi.org/10.1016/j.neuroimage.2014.06.073.\n\nAttributes\nfilters_`array`, shape (n_channels, n_components)\n\nThe spatial filters to be multiplied with the signal.\n\npatterns_`array`, shape (n_components, n_channels)\n\nThe patterns for reconstructing the signal from the filtered data.\n\nMethods\n\n `__hash__`(/) Return hash(self). Remove selected components from the signal. `fit`(X[, y]) Estimate the SSD decomposition on raw or epoched data. `fit_transform`(X[, y]) Fit to data, then transform it. `get_params`([deep]) Get parameters for this estimator. `get_spectral_ratio`(ssd_sources) Get the spectal signal-to-noise ratio for each spatial filter. Not implemented yet. `set_params`(params) Set the parameters of this estimator. Estimate epochs sources given the SSD filters.\napply(X)[source]\n\nRemove selected components from the signal.\n\nThis procedure will reconstruct M/EEG signals from which the dynamics described by the excluded components is subtracted (denoised by low-rank factorization). See 2 for more information.\n\nNote\n\nUnlike in other classes with an apply method, only NumPy arrays are supported (not instances of MNE objects).\n\nParameters\nXarray, shape ([n_epochs, ]n_channels, n_times)\n\nThe input data from which to estimate the SSD. Either 2D array obtained from continuous data or 3D array obtained from epoched data.\n\nReturns\nXarray, shape ([n_epochs, ]n_channels, n_times)\n\nThe processed data.\n\nfit(X, y=None)[source]\n\nEstimate the SSD decomposition on raw or epoched data.\n\nParameters\nXarray, shape ([n_epochs, ]n_channels, n_times)\n\nThe input data from which to estimate the SSD. Either 2D array obtained from continuous data or 3D array obtained from epoched data.\n\ny`None` \| `array`, shape (n_samples,)\n\nUsed for scikit-learn compatibility.\n\nReturns\nselfinstance of `SSD`\n\nReturns the modified instance.\n\nExamples using `fit`:\n\nfit_transform(X, y=None, fit_params)[source]\n\nFit to data, then transform it.\n\nFits transformer to X and y with optional parameters fit_params and returns a transformed version of X.\n\nParameters\nX`array`, shape (n_samples, n_features)\n\nTraining set.\n\ny`array`, shape (n_samples,)\n\nTarget values.\n\nfit_params`dict`\n\nAdditional fitting parameters passed to `self.fit`.\n\nReturns\nX_new`array`, shape (n_samples, n_features_new)\n\nTransformed array.\n\nget_params(deep=True)[source]\n\nGet parameters for this estimator.\n\nParameters\ndeepbool, optional\n\nIf True, will return the parameters for this estimator and contained subobjects that are estimators.\n\nReturns\nparams`dict`\n\nParameter names mapped to their values.\n\nget_spectral_ratio(ssd_sources)[source]\n\nGet the spectal signal-to-noise ratio for each spatial filter.\n\nSpectral ratio measure for best n_components selection See 1, Eq. (24).\n\nParameters\nssd_sources`array`\n\nData projected to SSD space.\n\nReturns\nspec_ratio`array`, shape (n_channels)\n\nArray with the sprectal ratio value for each component.\n\nsorter_spec`array`, shape (n_channels)\n\nArray of indices for sorting spec_ratio.\n\nReferences\n\nExamples using `get_spectral_ratio`:\n\ninverse_transform()[source]\n\nNot implemented yet.\n\nset_params(params)[source]\n\nSet the parameters of this estimator.\n\nThe method works on simple estimators as well as on nested objects (such as pipelines). The latter have parameters of the form `<component>__<parameter>` so that it’s possible to update each component of a nested object.\n\nParameters\n**params`dict`\n\nParameters.\n\nReturns\ninstinstance\n\nThe object.\n\ntransform(X)[source]\n\nEstimate epochs sources given the SSD filters.\n\nParameters\nXarray, shape ([n_epochs, ]n_channels, n_times)\n\nThe input data from which to estimate the SSD. Either 2D array obtained from continuous data or 3D array obtained from epoched data.\n\nReturns\nX_ssdarray, shape ([n_epochs, ]n_components, n_times)\n\nThe processed data.\n\nExamples using `transform`:" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.60033226,"math_prob":0.82162255,"size":5539,"snap":"2021-43-2021-49","text_gpt3_token_len":1405,"char_repetition_ratio":0.11996387,"word_repetition_ratio":0.1281709,"special_character_ratio":0.2332551,"punctuation_ratio":0.16067654,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9700053,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-20T19:36:04Z\",\"WARC-Record-ID\":\"<urn:uuid:6e9cd5d2-c137-4c67-b535-7871bb67aff5>\",\"Content-Length\":\"139168\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:57111d26-c269-4a0a-93ed-095815e21512>\",\"WARC-Concurrent-To\":\"<urn:uuid:1757063c-46ed-4169-baa1-ca3e96c84456>\",\"WARC-IP-Address\":\"185.199.108.153\",\"WARC-Target-URI\":\"https://mne.tools/dev/generated/mne.decoding.SSD.html\",\"WARC-Payload-Digest\":\"sha1:PK3UGRWVEKQ6KH6BKOGYAPEOEL46GBZ3\",\"WARC-Block-Digest\":\"sha1:NCUFEQASTJN7TSXDTG2IHOPLJNTWMNEE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585348.66_warc_CC-MAIN-20211020183354-20211020213354-00166.warc.gz\"}"}
https://www.mathematicshomeworkhelp.com/how-to-solve-ordinary-differential-equation-homework/	[ "# Strategies for Solving Ordinary Differential Equation Homework Problems with Ease\n\nMay 16, 2023", null, "United Kingdom\nMathematics\nEmily Adams is a PhD candidate in Mathematics Education at the University of Cambridge. With years of experience teaching and tutoring students of all levels.\n\nSolving Ordinary Differential Equations (ODEs) can be difficult, especially for novices. Solving ODE homework problems, however, may be both a pleasurable and fruitful endeavor if you properly approach them. In this article, we'll go through nine tried-and-true methods for quickly and easily resolving ODE difficulties. These methods will aid you in comprehending the issue at hand, categorizing the ODE, selecting a suitable approach, checking for errors, gaining experience, making use of software, asking for assistance, pausing for reflection, and showing perseverance. These methods will help you master tackling ODE homework problems and increase your grasp of the material.\n\nIntroduction\n\nStudents fresh to the field may find it difficult to solve ordinary differential equations (ODEs). Solving ODE homework problems, however, may be both a pleasurable and fruitful endeavor if you properly approach them. This blog post will go over some of the most effective methods for quickly and easily resolving ODE difficulties at home.\n\n## Understand the Problem\n\nTo solve ODE homework problems, one must first and foremost comprehend the nature of the problem at hand. Reading the problem statement intently and picking out the crucial details is an important first step in solving the issue. The issue statement's wording, assumptions, and constraints must all make sense to you. The starting and boundary conditions are particularly important to consider because of the effect they can have on the final answer.\n\nThe ability to convert a verbal explanation of an ODE problem into a mathematical form is a crucial skill for solving such problems. To solve many problems, you must convert physical phenomena, such as motion or chemical reactions, into mathematical equations. To accomplish this, you must have a firm grasp of the mathematical principles at play.\n\nRecognizing the problem's classification is also crucial to getting a handle on it. Order, linearity, homogeneity, and boundary/initial conditions are some of the criteria that can be used to categorize ODEs. Knowing the type of ODE you're dealing with can aid in selecting the most effective strategy for solving it.\n\nFinally, make sure you fully understand the question posed by the difficulty. It is crucial to know what is expected of you in each step of an ODE problem. Keep in mind whether a precise solution, an approximate solution, or a solution presented in a series is required.\n\n## Classify the ODE\n\nThe first step in doing homework involving Ordinary Differential Equations (ODEs) is classifying the ODE. Order, linearity, homogeneity, and boundary/initial conditions are some of the criteria used to categorize ODEs. Identifying the correct strategies for tackling the problem is made easier by classifying the ODE first.\n\nTo determine the order of an ODE, we look at the largest derivative of the dependent variable. When y is the dependent variable, the form of an ODE is dy/dx = f(x,y). For example, d2y/dx2 = f(x,y,dy/dx) is an ODE of the second order. If you know the order of the ODE, you can figure out what method will work best to solve it.\n\nIf the dependent variable and its derivatives occur linearly in an ODE, we say that the ODE is linear. When both a(x) and b(x) are functions of x, then a linear ODE takes the form a(x)dy/dx + b(x)y = f(x). Where f(x,y) is a nonlinear function of x and y, we have a nonlinear ordinary differential equation of the type dy/dx = f(x,y). While linear ODEs can be solved with techniques like variable separation, integrating factors, and Laplace transforms, nonlinear ODEs cannot.\n\nWhether or not all terms in an ODE are of the same degree is what is meant by the term's homogeneity. When both f(x,y) and g(x,y) are functions of x and y of the same degree, then the resulting differential equation is said to be homogeneous. When x and y are functions of different degrees, the nonhomogeneous ODE takes the form dy/dx = f(x,y)/g(x,y) + h(x,y)/g(x,y), where h(x,y) is a function of x and y. Substitution methods and variable separation can be used to solve homogeneous ODEs, while more advanced approaches like the variation of parameters and the method of indeterminate coefficients are required to solve nonhomogeneous ODEs.\n\nThe final step in solving an ODE issue is to employ boundary and beginning conditions to identify a single solution. The dependent variable's values at discrete locations are defined by the boundary conditions, whereas the dependent variable's value and its derivative at a single point are defined by the initial conditions. The techniques utilized to solve the ODE may change depending on the boundary/initial circumstances.\n\n## Pick An approach\n\nTo solve problems involving Ordinary Differential Equations (ODEs), you must first understand the problem and categorize the ODE before moving on to selecting an acceptable strategy for solving the ODE. Using the appropriate strategy can streamline the process and guarantee a successful outcome.\n\nSolving ODEs can be done in a few different ways, each with its own set of benefits and drawbacks. Separation of variables, integrating factors, Laplace transforms, the power series approach and numerical methods are all frequently employed in the solution of ODEs.\n\nIf the dependent variable can be expressed as a product of functions of x and y, then the first-order ODE can be solved using the separation of variables approach. Integrating both sides of the equation concerning their variables is what this method is all about.\n\nFor first-order linear ODEs, the integrating factor technique is employed. The left-hand side of the equation becomes a product rule derivative by multiplying it by an appropriate integrating factor, which is incorporated into the multiplication of both sides of the equation. Integration can then be used to resolve the problem.\n\nFor constant-coefficient linear ODEs, the Laplace transform approach is utilized. The Laplace transform is used to convert the ODE into an algebraic equation, which can then be solved for the unknown function.\n\nWhen previous methods fail, the power series method is employed to resolve the underlying ODEs. By inserting the expanded power series into the ODE, we may determine the series' coefficients and so solve the unknown function.\n\nWhen analytic solutions to ODEs cannot be found, they are solved numerically. When solving an ODE numerically, we use numerical algorithms to approximate the solution as a series of discrete points.\n\n## Check for Errors\n\nAfter solving an Ordinary Differential Equation (ODE) problem, it is crucial to double-check it for accuracy. Error checking is a useful tool for catching flaws before submitting a solution.\n\nSeveral techniques exist for verifying the accuracy of an ODE solution. Substituting the answer back into the ODE and checking for satisfaction is one of the simplest approaches. Verification refers to this checking process.\n\nSolution verification requires both the solution and its derivative to be plugged back into the original ODE. The solution is valid if and only if it completely solves the equation. If the equation is not satisfied by the solution, then the solution is incorrect.\n\nThe solution can also be validated by contrasting it with the specified starting and ending points. If the answer is right, it will meet either the initial or boundary requirements. There is a problem with the solution if it fails to meet the starting or boundary requirements.\n\nMathematical flaws, such as algebraic blunders or improper differentiation or integration, should also be double-checked. Mathematical proofreading entails checking each calculation and solution step for accuracy.\n\n## Repeated Practice is Essential\n\nSolving Ordinary Differential Equation (ODE) homework problems is a skill that can only be developed by repeated practice. Solving ODE problems is like learning any other ability; the more you do it, the better you get.\n\nSolving ODE issues is a great way to hone your problem-solving abilities and expand your knowledge of the various approaches available for doing so. ODEs can be solved more quickly and easily if you can spot patterns and similarities between issues.\n\nWorking through textbook exercises and examples is a great method to gain experience in ODE problem-solving. Most textbooks include a variety of challenges, ranging in difficulty, which can be used to progressively hone your abilities.\n\nYou can also use online resources like math discussion boards and educational websites to practice answering ODE issues. You can hone your problem-solving abilities by working through the various scenarios presented in these materials.\n\nIt's also helpful to practice solving ODE problems with a study group or instructor. Working with others is a great way to expand your horizons and learn new ways to tackle difficult situations.\n\nKeeping tabs on your development is crucial for making the most of your practice time. You can gauge your progress and pinpoint trouble areas using practice puzzles of varied difficulties.\n\n## Use Software\n\nOrdinary Differential Equation (ODE) homework can also be tackled by turning to computer applications. Using software to solve problems is an easy way to save time and get better results.\n\nMATLAB, Mathematica, Maple, and Python are just some of the software packages that can be used to solve ODEs. These programs use numerical methods to solve ODEs, allowing for solutions to be obtained even when analytic ones are unavailable.\n\nThe behavior of the solution to ODEs over time can be visualized and explored with the use of the software. This can be very helpful when attempting to solve higher-order or nonlinear ODEs analytically.\n\nUsing software has the potential to lessen the likelihood of arithmetic mistakes. There is less room for error in software calculations because the algorithms used have been refined and tested repeatedly.\n\nThe software has the added benefit of being able to solve massive, complicated issues that would be impossible to tackle by hand. Since ODEs can simulate intricate systems and processes, they can prove extremely helpful in scientific and engineering contexts.\n\nWhile software can be helpful, it should never be utilized in place of a solid grasp of the topic at hand and the mathematics behind it. Before turning to software, it is still crucial to grasp the nature of the issue at hand, categorize the ODE, and settle on an approach.\n\n## Get Some Aid\n\nIf you're struggling with your Ordinary Differential Equation (ODE) homework, don't hesitate to ask for assistance. Seeking assistance when stuck with a problem might lead to fresh perspectives and ideas that can lead to a better resolution.\n\nHelp with ODE homework can be found in a variety of sources. One typical method is to seek help from a teacher or teaching assistant. They can offer advice on how to approach a subject and shed light on obscure ideas.\n\nJoining a study club or getting a tutor are two other options for those who need assistance. Working with others can foster an encouraging learning atmosphere and introduce you to different ways of thinking about ODE challenges.\n\nOnline resources, such as math discussion boards and educational websites, can also be useful. These materials help shed light on typical approaches to problems and answer frequently asked issues.\n\nWhen confronted with a challenging condition, it's also crucial to seek assistance as soon as possible. If you put off getting aid until the last minute, you can end up stressed out and with less-than-ideal results.\n\nSeeking assistance is not a show of weakness, which is something to keep in mind. Getting some outside assistance is a normal and healthy component of learning, especially while working on ODE homework issues. You can learn more about the issue and build better strategies for handling it by getting some outside input.\n\n## Take Breaks\n\nTaking a break is a crucial tactic for completing Ordinary Differential Equation (ODE) homework. Concentration and mental stamina tend to suffer when working on difficult problems for long periods. Taking a short break can assist restore mental energy and sharpen concentration, allowing for more efficient problem-solving once you return to work.\n\nTaking a break might help restore mental energy and boost performance. Taking short rests between problem-solving sessions has been demonstrated to boost performance, accuracy, motivation, and originality.\n\nWhen working on difficult problems that require sustained mental effort, frequent breaks are extremely vital. Taking a short break, even if only for ten or fifteen minutes, every hour may do wonders for the mind and make it easier to focus once you get back on the subject.\n\nRelaxing and enjoyable activities, such as going for a stroll, listening to music, or practicing mindfulness, can be quite beneficial during breaks. Resuming work on the issue after engaging in these pursuits may be approached with greater clarity and vigour.\n\nHaving a balanced schedule of work and play is also essential. Overwhelming and lack of interest might result from spending too much time on ODE homework. Maintaining a sense of equilibrium and developing better problem-solving skills over time requires setting reasonable goals and making self-care a top priority.\n\n## Be Patient\n\nThe solution to Ordinary Differential Equation (ODE) homework problems requires patience. ODEs are difficult mathematical problems to solve, especially for those who are just starting in the field. Therefore, working on ODE problems requires patience and perseverance.\n\nThe iterative procedure of solving ODE issues can be time-consuming and error-prone. Each challenge requires a curious mind and a thirst for knowledge. This entails having patience, both with oneself and the process of finding a solution.\n\nThe time required to resolve ODE difficulties should also be taken into account. When dealing with more complex issues, the time required to find a solution can easily exceed a day. This is frustrating, but keep in mind that even delayed growth is progress.\n\nBreaking down ODE problems into smaller, more manageable pieces can help you keep your patience levels up as you work on them. This may help the problem seem less daunting and more manageable. Setting attainable goals and rewarding progress along the way may keep the motivation and momentum going strong.\n\nLast but not least, keep in mind that patience is essential during the learning process. ODEs are a difficult topic that calls for dedicated study. A thorough comprehension of ODEs and enhanced problem-solving abilities can be attained with time, effort, and dedication.\n\n## Concluding Remarks\n\nFinally, ODE homework problems can be difficult to solve, but they are not insurmountable. You can improve your skill at solving ODE issues and your grasp of the subject by following the advice given in this blog. It's important to think through the problem, categorize the ODE, select the right approach, double-check your work, drill for errors, use software, get feedback, rest, and persevere. These methods will make solving ODE issues a breeze" ]	[ null, "https://www.mathematicshomeworkhelp.com/uploads/topic_files/Ordinary-Differential-Equation-Homework-Expert.webp", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9368012,"math_prob":0.91443497,"size":15090,"snap":"2023-40-2023-50","text_gpt3_token_len":2963,"char_repetition_ratio":0.13244067,"word_repetition_ratio":0.025427261,"special_character_ratio":0.18422796,"punctuation_ratio":0.09660767,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9578353,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-09T01:18:07Z\",\"WARC-Record-ID\":\"<urn:uuid:b6ff43cc-7c5a-4898-8ca3-00affefdc04b>\",\"Content-Length\":\"49903\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f26787a3-e7fe-487c-9772-d46f4194f763>\",\"WARC-Concurrent-To\":\"<urn:uuid:2c419098-c456-498b-abcd-12b1c0b00d11>\",\"WARC-IP-Address\":\"104.21.38.36\",\"WARC-Target-URI\":\"https://www.mathematicshomeworkhelp.com/how-to-solve-ordinary-differential-equation-homework/\",\"WARC-Payload-Digest\":\"sha1:BZUZ7O3CIRZY3DPSFSR65GM4RUUG5EFN\",\"WARC-Block-Digest\":\"sha1:2ZC7DPOJMDBDXRW2YEQDQMKCRO6FHDSQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100781.60_warc_CC-MAIN-20231209004202-20231209034202-00876.warc.gz\"}"}
https://labs.tib.eu/arxiv/?author=Sandro%20Wenzel	[ "• Particle physics has an ambitious and broad experimental programme for the coming decades. This programme requires large investments in detector hardware, either to build new facilities and experiments, or to upgrade existing ones. Similarly, it requires commensurate investment in the R&D of software to acquire, manage, process, and analyse the shear amounts of data to be recorded. In planning for the HL-LHC in particular, it is critical that all of the collaborating stakeholders agree on the software goals and priorities, and that the efforts complement each other. In this spirit, this white paper describes the R&D activities required to prepare for this software upgrade.\n• ### Vectorising the detector geometry to optimize particle transport(1312.0816)\n\nDec. 3, 2013 hep-ex, physics.comp-ph\nAmong the components contributing to particle transport, geometry navigation is an important consumer of CPU cycles. The tasks performed to get answers to \"basic\" queries such as locating a point within a geometry hierarchy or computing accurately the distance to the next boundary can become very computing intensive for complex detector setups. So far, the existing geometry algorithms employ mainly scalar optimisation strategies (voxelization, caching) to reduce their CPU consumption. In this paper, we would like to take a different approach and investigate how geometry navigation can benefit from the vector instruction set extensions that are one of the primary source of performance enhancements on current and future hardware. While on paper, this form of microparallelism promises increasing performance opportunities, applying this technology to the highly hierarchical and multiply branched geometry code is a difficult challenge. We refer to the current work done to vectorise an important part of the critical navigation algorithms in the ROOT geometry library. Starting from a short critical discussion about the programming model, we present the current status and first benchmark results of the vectorisation of some elementary geometry shape algorithms. On the path towards a full vector-based geometry navigator, we also investigate the performance benefits in connecting these elementary functions together to develop algorithms which are entirely based on the flow of vector-data. To this end, we discuss core components of a simple vector navigator that is tested and evaluated on a toy detector setup.\n• ### Zero-temperature Monte Carlo study of the non-coplanar phase of the classical bilinear-biquadratic Heisenberg model on the triangular lattice(1305.6418)\n\nSept. 6, 2013 cond-mat.str-el\nWe investigate the ground-state properties of the highly degenerate non-coplanar phase of the classical bilinear-biquadratic Heisenberg model on the triangular lattice with Monte Carlo simulations. For that purpose, we introduce an Ising pseudospin representation of the ground states, and we use a simple Metropolis algorithm with local updates, as well as a powerful cluster algorithm. At sizes that can be sampled with local updates, the presence of long-range order is surprisingly combined with an algebraic decay of correlations and the complete disordering of the chirality. It is only thanks to the investigation of unusually large systems (containing $\\sim 10^8$ spins) with cluster updates that the true asymptotic regime can be reached and that the system can be proven to consist of equivalent (i.e., equally ordered) sublattices. These large-scale simulations also demonstrate that the scalar chirality exhibits long-range order at zero temperature, implying that the system has to undergo a finite-temperature phase transition. Finally, we show that the average distance in the order parameter space, which has the structure of an infinite Cayley tree, remains remarkably small between any pair of points, even in the limit when the real space distance between them tends to infinity.\n• ### Evidence of columnar order in the fully frustrated transverse field Ising model on the square lattice(1207.1618)\n\nNov. 27, 2012 cond-mat.str-el\nUsing extensive classical and quantum Monte Carlo simulations, we investigate the ground-state phase diagram of the fully frustrated transverse field Ising model on the square lattice. We show that pure columnar order develops in the low-field phase above a surprisingly large length scale, below which an effective U(1) symmetry is present. The same conclusion applies to the Quantum Dimer Model with purely kinetic energy, to which the model reduces in the zero-field limit, as well as to the stacked classical version of the model. By contrast, the 2D classical version of the model is shown to develop plaquette order. Semiclassical arguments show that the transition from plaquette to columnar order is a consequence of quantum fluctuations.\n• ### Monte Carlo study of the critical properties of the three-dimensional 120-degree model(1106.3426)\n\nWe report on large scale finite-temperature Monte Carlo simulations of the classical $120^\\circ$ or $e_g$ orbital-only model on the simple cubic lattice in three dimensions with a focus towards its critical properties. This model displays a continuous phase transition to an orbitally ordered phase. While the correlation length exponent $\\nu\\approx0.665$ is close to the 3D XY value, the exponent $\\eta \\approx 0.15$ differs substantially from O(N) values. We also introduce a discrete variant of the $e_g$ model, called $e_g$-clock model, which is found to display the same set of exponents. Further, an emergent U(1) symmetry is found at the critical point $T_c$, which persists for $T<T_c$ below a crossover length scaling as $\\Lambda \\sim \\xi^a$, with an unusually small $a\\approx1.3$.\n• ### Unveiling the nature of three dimensional orbital ordering transitions: the case of $e_g$ and $t_{2g}$ models on the cubic lattice(1101.3259)\n\nMay 17, 2011 cond-mat.str-el\nWe perform large scale finite-temperature Monte Carlo simulations of the classical $e_g$ and $t_{2g}$ orbital models on the simple cubic lattice in three dimensions. The $e_g$ model displays a continuous phase transition to an orbitally ordered phase. While the correlation length exponent $\\nu\\approx0.66(1)$ is close to the 3D XY value, the exponent $\\eta \\approx 0.15(1)$ differs substantially from O(N) values. At $T_c$ a U(1) symmetry emerges, which persists for $T<T_c$ below a crossover length scaling as $\\Lambda \\sim \\xi^a$, with an unusually small $a\\approx1.3$. Finally, for the $t_{2g}$ model we find a {\\em first order} transition into a low-temperature lattice-nematic phase without orbital order.\n• ### Re-examining the directional-ordering transition in the compass model with screw-periodic boundary conditions(1002.3508)\n\nWe study the directional-ordering transition in the two-dimensional classical and quantum compass models on the square lattice by means of Monte Carlo simulations. An improved algorithm is presented which builds on the Wolff cluster algorithm in one-dimensional subspaces of the configuration space. This improvement allows us to study classical systems up to $L=512$. Based on the new algorithm we give evidence for the presence of strongly anomalous scaling for periodic boundary conditions which is much worse than anticipated before. We propose and study alternative boundary conditions for the compass model which do not make use of extended configuration spaces and show that they completely remove the problem with finite-size scaling. In the last part, we apply these boundary conditions to the quantum problem and present a considerably improved estimate for the critical temperature which should be of interest for future studies on the compass model. Our investigation identifies a strong one-dimensional magnetic ordering tendency with a large correlation length as the cause of the unusual scaling and moreover allows for a precise quantification of the anomalous length scale involved.\n• ### Finite-Temperature N\\'eel Ordering of Fluctuations in a Plaquette Orbital Model(0810.2378)\n\nAug. 10, 2009 cond-mat.stat-mech\nWe present a pseudospin model which should be experimentally accessible using solid-state devices and, being a variation on the compass model, adds to the toolbox for the protection of qubits in the area of quantum information. Using Monte Carlo methods, we find for both classical and quantum spins in two and three dimensions Ising-type Neel ordering of energy fluctuations at finite temperatures without magnetic order. We also readdress the controversy concerning the stability of the ordered state in the presence of quenched impurities and present numerical results which are at clear variance with earlier claims in the literature.\n• ### Comprehensive quantum Monte Carlo study of the quantum critical points in planar dimerized/quadrumerized Heisenberg models(0808.1418)\n\nWe study two planar square lattice Heisenberg models with explicit dimerization or quadrumerization of the couplings in the form of ladder and plaquette arrangements. We investigate the quantum critical points of those models by means of (stochastic series expansion) quantum Monte Carlo simulations as a function of the coupling ratio $\\alpha = J^\\prime/J$. The critical point of the order-disorder quantum phase transition in the ladder model is determined as $\\alpha_\\mathrm{c} = 1.9096(2)$ improving on previous studies. For the plaquette model we obtain $\\alpha_\\mathrm{c} = 1.8230(2)$ establishing a first benchmark for this model from quantum Monte Carlo simulations. Based on those values we give further convincing evidence that the models are in the three-dimensional (3D) classical Heisenberg universality class. The results of this contribution shall be useful as references for future investigations on planar Heisenberg models such as concerning the influence of non-magnetic impurities at the quantum critical point.\n• ### Evidence of Unconventional Universality Class in a Two-Dimensional Dimerized Quantum Heisenberg Model(0805.2500)\n\nThe two-dimensional $J$-$J^\\prime$ dimerized quantum Heisenberg model is studied on the square lattice by means of (stochastic series expansion) quantum Monte Carlo simulations as a function of the coupling ratio \\hbox{$\\alpha=J^\\prime/J$}. The critical point of the order-disorder quantum phase transition in the $J$-$J^\\prime$ model is determined as \\hbox{$\\alpha_\\mathrm{c}=2.5196(2)$} by finite-size scaling for up to approximately $10 000$ quantum spins. By comparing six dimerized models we show, contrary to the current belief, that the critical exponents of the $J$-$J^\\prime$ model are not in agreement with the three-dimensional classical Heisenberg universality class. This lends support to the notion of nontrivial critical excitations at the quantum critical point.\n• ### Monte Carlo simulations of the directional-ordering transition in the two-dimensional classical and quantum compass model(0804.2972)\n\nSept. 5, 2008 cond-mat.stat-mech\nA comprehensive study of the two-dimensional (2D) compass model on the square lattice is performed for classical and quantum spin degrees of freedom using Monte Carlo and quantum Monte Carlo methods. We employ state-of-the-art implementations using Metropolis, stochastic series expansion and parallel tempering techniques to obtain the critical ordering temperatures and critical exponents. In a pre-investigation we reconsider the classical compass model where we study and contrast the finite-size scaling behavior of ordinary periodic boundary conditions against annealed boundary conditions. It is shown that periodic boundary conditions suffer from extreme finite-size effects which might be caused by closed loop excitations on the torus. These excitations also appear to have severe effects on the Binder parameter. On this footing we report on a systematic Monte Carlo study of the quantum compass model. Our numerical results are at odds with recent literature on the subject which we trace back to neglecting the strong finite-size effects on periodic lattices. The critical temperatures are obtained as $T_\\mathrm{c}=0.1464(2)J$ and $T_\\mathrm{c}=0.055(1)J$ for the classical and quantum version, respectively, and our data support a transition in the 2D Ising universality class for both cases.\n• ### Percolation of Vortices in the 3D Abelian Lattice Higgs Model(0708.0903)\n\nJuly 9, 2008 hep-lat\nThe compact Abelian Higgs model is simulated on a cubic lattice where it possesses vortex lines and pointlike magnetic monopoles as topological defects. The focus of this high-precision Monte Carlo study is on the vortex network, which is investigated by means of percolation observables. In the region of the phase diagram where the Higgs and confinement phases are separated by a first-order transition, it is shown that the vortices percolate right at the phase boundary, and that the first-order nature of the transition is reflected by the network. In the crossover region, where the phase boundary ceases to be first order, the vortices are shown to still percolate. In contrast to other observables, the percolation observables show finite-size scaling. The exponents characterizing the critical behavior of the vortices in this region are shown to fall in the random percolation universality class.\n• ### Vortex Proliferation and the Dual Superconductor Scenario for Confinement: The 3D Compact U(1) Lattice Higgs Model(hep-lat/0510099)\n\nOct. 24, 2005 hep-lat\nIt is argued that the phase diagram of the 3D Compact U(1) Lattice Higgs Model is more refined than generally thought. The confined and Higgs phases are separated by a well-defined phase boundary, marked by proliferating vortices. It is shown that the confinement mechanism at work is precisely the dual superconductor scenario.\n• ### Kertesz Line in the Three-Dimensional Compact U(1) Lattice Higgs Model(cond-mat/0503599)\n\nMarch 24, 2005 cond-mat.str-el, hep-lat\nThe three-dimensional lattice Higgs model with compact U(1) gauge symmetry and unit charge is investigated by means of Monte Carlo simulations. The full model with fluctuating Higgs amplitude is simulated, and both energy as well as topological observables are measured. The data show a Higgs and a confined phase separated by a well-defined phase boundary, which is argued to be caused by proliferating vortices. For fixed gauge coupling, the phase boundary consists of a line of first-order phase transitions at small Higgs self-coupling, ending at a critical point. The phase boundary then continues as a Kertesz line across which thermodynamic quantities are nonsingular. Symmetry arguments are given to support these findings." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8723821,"math_prob":0.9759108,"size":12811,"snap":"2020-45-2020-50","text_gpt3_token_len":2739,"char_repetition_ratio":0.12727414,"word_repetition_ratio":0.07639979,"special_character_ratio":0.18905628,"punctuation_ratio":0.070935056,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97609127,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-12-04T20:53:01Z\",\"WARC-Record-ID\":\"<urn:uuid:28749a79-a985-431e-8b49-cdec2713ed7e>\",\"Content-Length\":\"116201\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cee4c9e1-f791-491f-a1f5-c60a529c0194>\",\"WARC-Concurrent-To\":\"<urn:uuid:86978a55-39e9-4757-9d90-b322ebcff13c>\",\"WARC-IP-Address\":\"194.95.114.13\",\"WARC-Target-URI\":\"https://labs.tib.eu/arxiv/?author=Sandro%20Wenzel\",\"WARC-Payload-Digest\":\"sha1:TLZ3SCB22QDZFDNZACYBETXPAYQP3D75\",\"WARC-Block-Digest\":\"sha1:Z2XNYHHYVAAWO264GUISSUFSVD3XXRXE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141743438.76_warc_CC-MAIN-20201204193220-20201204223220-00367.warc.gz\"}"}
https://jamesotto852.github.io/ggdensity/articles/method.html	[ "Almost every function in ggdensity accepts a method argument—this is true for geom_hdr() and other layer functions (geom_hdr_lines(), geom_hdr_points(), …), as well as get_hdr() and get_hdr_1d(). This vignette summarizes the many ways in which the method argument can be specified; first looking at it from a more basic perspective, then from the perspective of a developer wanting to implement additional estimators.\n\n## Using ggdensity’s method_() functions\n\nFirst, let’s load the necessary packages and generate some sample data.\n\nlibrary(\"ggdensity\"); theme_set(theme_minimal(8))\ntheme_update(legend.position = \"none\") # Suppressing legends for readability\nset.seed(1)\ndf <- data.frame(x = rnorm(500), y = rnorm(500))\np <- ggplot(df, aes(x, y))\np + geom_point()", null, "The easiest way to plot HDRs with geom_hdr() (or any other layer function from ggdensity) with a specified density estimator is to provide a character object to the method argument:\n\np + geom_hdr(method = \"kde\")\n\np + geom_hdr(method = \"mvnorm\")\n\np + geom_hdr(method = \"histogram\")\n\np + geom_hdr(method = \"freqpoly\")", null, "", null, "", null, "", null, "However, as of ggdensity v1.0.0 there is an alternative approach—providing a method_() function call:\n\np + geom_hdr(method = method_kde())\n\np + geom_hdr(method = method_mvnorm())\n\np + geom_hdr(method = method_histogram())\n\np + geom_hdr(method = method_freqpoly())", null, "", null, "", null, "", null, "The default behaviors of these two approaches are the same and always will be—in this way, they are completely interchangeable. However, the method_() function call is required to estimate HDRs with non-default estimator parameters. For example, we can set the adjust parameter to apply a multiplicative adjustment to the heuristically determined bandwidth in method_kde() (which itself uses the one computed by MASS::bandwidth.nrd()):\n\np + geom_hdr(method = method_kde(adjust = 1/2))", null, "The relevant parameters for each method are documented in their respective ?method_ help pages. Note that these parameters can not be provided to geom_hdr() or stat_hdr() and thus are not accessible if a character value is provided to method.\n\nThe method argument of get_hdr() functions in the same way:\n\nres <- get_hdr(df, method = method_kde(adjust = 1/2))\n\nstr(res)\n#> List of 3\n#> $df_est:'data.frame': 10000 obs. of 5 variables: #> ..$ x : num [1:10000] -3.01 -2.94 -2.87 -2.8 -2.73 ...\n#> ..$y : num [1:10000] -3 -3 -3 -3 -3 ... #> ..$ fhat : num [1:10000] 4.72e-17 1.30e-15 2.88e-14 5.16e-13 7.44e-12 ...\n#> ..$fhat_discretized: num [1:10000] 2.00e-19 5.50e-18 1.22e-16 2.18e-15 3.15e-14 ... #> ..$ hdr : num [1:10000] 1 1 1 1 1 1 1 1 1 1 ...\n#> $breaks: Named num [1:5] 0.00422 0.01273 0.03024 0.07544 Inf #> ..- attr(, \"names\")= chr [1:5] \"99%\" \"95%\" \"80%\" \"50%\" ... #>$ data :'data.frame': 500 obs. of 3 variables:\n#> ..$x : num [1:500] -0.626 0.184 -0.836 1.595 0.33 ... #> ..$ y : num [1:500] 0.0773 -0.2969 -1.1832 0.0113 0.9916 ...\n#> ..$hdr_membership: num [1:500] 0.5 0.5 0.8 0.8 0.5 0.95 0.8 0.5 0.5 0.5 ... For details on the output of get_hdr(), see ?get_hdr. ### method__1d() functions In ggdensity, it is possible to estimate and plot 1-dimensional HDRs with geom_hdr_rug() and get_hdr_1d(). These functions also accept a method argument, but they do not accept the previously discussed method_() functions. Instead they accept the 1-dimensional analogues: method__1d(). p + geom_point() + geom_hdr_rug(method = method_kde_1d()) p + geom_point() + geom_hdr_rug(method = method_norm_1d()) p + geom_point() + geom_hdr_rug(method = method_histogram_1d()) p + geom_point() + geom_hdr_rug(method = method_freqpoly_1d())", null, "", null, "", null, "", null, "Just like we saw with geom_hdr(), geom_hdr_rug() also accepts character values for method: p + geom_point() + geom_hdr_rug(method = \"kde\") p + geom_point() + geom_hdr_rug(method = \"norm\") p + geom_point() + geom_hdr_rug(method = \"histogram\") p + geom_point() + geom_hdr_rug(method = \"freqpoly\")", null, "", null, "", null, "", null, "Because the return values of the method_() functions are incompatible with the 1-dimensional HDR estimation procedure, if a 2-dimensional method is specified the following error message is issued: p + geom_point() + geom_hdr_rug(method = method_kde()) #> Warning: Computation failed in stat_hdr_rug() #> Caused by error in get_hdr_1d(): #> ! Invalid method argument -- did you forget the _1d()? Lastly, we see that the method argument of get_hdr_1d() behaves similarly. res <- get_hdr_1d(df$x, method = method_kde_1d())\n\nstr(res)\n#> List of 3\n#> $df_est:'data.frame': 512 obs. of 4 variables: #> ..$ x : num [1:512] -3.01 -2.99 -2.98 -2.97 -2.95 ...\n#> ..$fhat : num [1:512] 0.00728 0.00748 0.00769 0.00789 0.00809 ... #> ..$ fhat_discretized: num [1:512] 9.73e-05 1.00e-04 1.03e-04 1.05e-04 1.08e-04 ...\n#> ..$hdr : num [1:512] 1 1 1 1 1 1 1 1 1 1 ... #>$ breaks: Named num [1:5] 0.0188 0.0562 0.1601 0.3146 Inf\n#> ..- attr(, \"names\")= chr [1:5] \"99%\" \"95%\" \"80%\" \"50%\" ...\n#> $data :'data.frame': 500 obs. of 2 variables: #> ..$ x : num [1:500] -0.626 0.184 -0.836 1.595 0.33 ...\n#> ..$hdr_membership: num [1:500] 0.5 0.5 0.8 0.95 0.5 0.8 0.5 0.8 0.5 0.5 ... Again, for details on the above output of get_hdr_1d(), see ?get_hdr_1d. ## A detailed look at method_() functions Now that we understand the ways in which method can be specified let’s look at the internals of the method_() functions. Note: the implementations discussed in this section depend heavily on topics in functional programming, especially closures and function factories. While not necessary, a good understanding of these ideas is helpful—the linked chapters from Hadley Wickham’s Advanced R are a great place to start. Looking at the definition of method_kde(), we see that it is a function of h and adjust, returning a closure with arguments data, n, rangex, and rangey. The closure passes the x and y columns of data to MASS::kde2d(), returning the estimated density evaluated on a grid with columns x, y, and fhat. This closure is what geom_hdr() expects as its method argument, and is how the HDRs are estimated (via get_hdr()). method_kde function (h = NULL, adjust = c(1, 1)) { function(data, n, rangex, rangey) { if (is.null(h)) { h <- c(MASS::bandwidth.nrd(data$x), MASS::bandwidth.nrd(data$y)) } h <- h * adjust kdeout <- MASS::kde2d(x = data$x, y = data$y, n = n, h = h, lims = c(rangex, rangey)) df <- with(kdeout, expand.grid(x = x, y = y)) df$fhat <- as.vector(kdeout$z) df } } <bytecode: 0x560c272be498> <environment: namespace:ggdensity> Both method_histogram() and method_freqpoly() behave similarly, accepting parameters governing the density estimation procedure and returning a closure with arguments data, n, rangex, and rangey. However, these functions are significantly more complicated as the density estimation procedures are implemented entirely in ggdensity. method_mvnorm() is different in a few ways. The closure it returns is a function of just one argument: data. This is because it does not return the estimated density evaluated on a grid. Instead, it returns yet another closure with (vectorized) arguments x and y. As in method_kde(), the return value of the closure is a representation of the estimated pdf. The difference is the manner in which the pdf is represented. Whereas before we had a pdf defined by a discrete approximation on a grid, we now have an explicit definition of the pdf in terms of x and y. method_mvnorm function () { function(data) { data_matrix <- with(data, cbind(x, y)) mu_hat <- colMeans(data_matrix) R <- chol(cov(data_matrix)) function(x, y) { X <- cbind(x, y) tmp <- backsolve(R, t(X) - mu_hat, transpose = TRUE) logretval <- -sum(log(diag(R))) - log(2 * pi) - 0.5 * colSums(tmp^2) exp(logretval) } } } <bytecode: 0x560c26eab8c0> <environment: namespace:ggdensity> To summarize each of the above cases: in the first example, the method_() function returned a closure with arguments data, n, rangex, and rangey which itself returned the estimated density evaluated on a grid; in the second, the method_() function returned a closure with a single argument, data, which itself returned a closure with arguments x and y, representing the estimated density explicitly. In both cases, the method_() function can have any number of parameters governing the density estimation procedure. These are the two ways the method argument may be specified. The first is necessary for cases in which an explicit definition of the estimated density is not computationally feasible (for example, KDEs). The second is an easier option for the cases in which a closed form of the estimated density is available (for example, parametric estimators). Let’s look at how we might define our own method_() functions in each case, beginning with a simple parametric estimator. ### Implementing a method returning a PDF In ggdensity, method_mvnorm() estimates HDRs based on the parametric multivariate normal model. If we wanted to fit a simpler model in which the data is further assumed to be independent, we could implement method_mvnorm_ind(). method_mvnorm_ind <- function() { function(data) { xbar <- mean(data$x)\nybar <- mean(data$y) sx <- sd(data$x)\nsy <- sd(data$y) # joint pdf is simply the product of the marginals function(x, y) dnorm(x, xbar, sx) * dnorm(y, ybar, sy) } } To use our method_mvnorm_ind(), we just need to supply it to geom_hdr()’s method argument. ggplot(df, aes(x, y)) + geom_hdr(method = method_mvnorm_ind())", null, "If we transform our data to have non-zero covariance we still see the major and minor axes of the contours coincide with the plot axes—exactly what we would expect with this (incorrectly) constrained model. A <- matrix(c( 2cos(pi/6), -2sin(pi/6), 1sin(pi/6), 1cos(pi/6) ), byrow = TRUE, ncol = 2) df_rot <- as.data.frame(as.matrix(df) %% A) colnames(df_rot) <- c(\"x\", \"y\") ggplot(df_rot, aes(x, y)) + geom_hdr(method = method_mvnorm_ind()) + geom_point(size = .4) + coord_fixed(xlim = c(-6, 6), ylim = c(-6, 6))", null, "Notice, method_mvnorm_ind() accepts no arguments. The density estimation procedure is so simple that there are no parameters to govern it. To allow for circular models in which the fitted variances are required to be equal, we can implement a circular argument. method_mvnorm_ind <- function(circular = FALSE) { function(data) { xbar <- mean(data$x)\nybar <- mean(data$y) if (circular) { sx <- sd(c(data$x - xbar, data$y - ybar)) sy <- sx } else { sx <- sd(data$x)\nsy <- sd(data$y) } function(x, y) dnorm(x, xbar, sx) dnorm(y, ybar, sy) } } Now, the contours are perfectly circular. ggplot(df_rot, aes(x, y)) + geom_hdr(method = method_mvnorm_ind(circular = TRUE)) + geom_point(size = .4) + coord_fixed(xlim = c(-6, 6), ylim = c(-6, 6))", null, "In the above plot, the upper and lower portions of the HDRs are cut off. This is because the default behavior of ggdensity is to not draw HDRs outside of the “bounding box” around observed data. This is not because we are using a custom method_() function. To fix this, we need to either set a better ylim value for geom_hdr() or specify a larger range in scale_y_continuous(). ggplot(df_rot, aes(x, y)) + geom_hdr(method = method_mvnorm_ind(circular = TRUE), ylim = c(-6, 6)) + geom_point(size = .4) + coord_fixed(xlim = c(-6, 6), ylim = c(-6, 6)) ggplot(df_rot, aes(x, y)) + geom_hdr(method = method_mvnorm_ind(circular = TRUE)) + geom_point(size = .4) + scale_y_continuous(limits = c(-6, 6)) + coord_fixed(xlim = c(-6, 6), ylim = c(-6, 6))", null, "", null, "Notice, neither of these approaches involve arguments to method_mvnorm_ind(). Internally, the closure returned by method_mvnorm_ind() is used by get_hdr(), along with information from the scales associated with the ggplot object. It is the scales that need adjusting, not anything related to the method argument. ### Implementing a method returning an evaluated PDF To illustrate the other case, in which the object returned by the closure is the estimated density evaluated on a grid, we implement method_mvnorm_ind_grid(). This estimates the same independent normal density as method_mvnorm_ind(), the only difference is the behavior of the returned closure. method_mvnorm_ind_grid <- function() { function(data, n, rangex, rangey) { # First, we estimate the density ----------------------------- xbar <- mean(data$x)\nybar <- mean(data$y) sx <- sd(data$x)\nsy <- sd(data$y) f_est <- function(x, y) dnorm(x, xbar, sx) dnorm(y, ybar, sy) # Return the density evaluated on a grid --------------------- # df_grid defined by rangex, rangey, and n df_grid <- expand.grid( x = seq(rangex, rangex, length.out = n), y = seq(rangey, rangey, length.out = n) ) df_grid$fhat <- f_est(df_grid$x, df_grid$y)\n\ndf_grid\n}\n\n}\n\nSee that returned closure has additional arguments n, rangex, and rangey which define the grid. Also, the grid is represented a data.frame with columns x, y, and fhat, where fhat is the (potentially unnormalized) density estimate.\n\nAgain, to use our method_mvnorm_ind_grid() we provide it to geom_hdr()’s method argument.\n\nggplot(df, aes(x, y)) +\ngeom_hdr(method = method_mvnorm_ind_grid())", null, "Like we saw in the previous example, we could prevent the HDRs from being “cut off” by specifying either the x/ylim arguments in geom_hdr() or by setting a larger range in scale_x/y_continuous().\n\n## The method__1d() functions\n\nWe saw before that ggdensity uses method__1d() functions for the estimation of 1-dimensional densities. The internals of these functions are very similar to the method_() functions, the only differences are slight changes to the arguments and return values of the returned closures.\n\nLooking at the definition of method_kde_1d(), we see the returned closure has arguments x, n, and range. This is very similar to method_kde(), the only difference is we are now dealing with univariate data: the vector argument x is used instead of data, and we have a single range parameter instead of rangex and rangey. Similarly, the closure now returns the estimated density evaluated on a univariate grid, with columns x and fhat instead of the bivariate grid with columns x, y, and fhat. Finally, see that method_kde_1d() accepts several arguments governing the density estimation procedure just like method_kde().\n\nmethod_kde_1d\nfunction (bw = \"nrd0\", adjust = 1, kernel = \"gaussian\", weights = NULL,\nwindow = kernel)\n{\nfunction(x, n, range) {\nnx <- length(x)\nif (is.null(weights)) {\nweights <- rep(1/nx, nx)\n}\nelse {\nweights <- normalize(weights)\n}\ndens <- stats::density(x, bw = bw, adjust = adjust, kernel = kernel,\nweights = weights, window = window, n = n, from = range,\nto = range)\ndata.frame(x = dens$x, fhat = dens$y)\n}\n}\n<bytecode: 0x560c28d2c850>\n<environment: namespace:ggdensity>\n\nEstimated univariate densities can also be represented explicitly, as illustrated by method_norm_1d(). Comparing this to the previously discussed method_mvnorm() we see that little has changed: the closure is now a function of a vector x instead of data and returns a function of one variable (x) instead of two (x and y).\n\nmethod_norm_1d\nfunction ()\n{\nfunction(x) {\nmu_hat <- mean(x)\nsigma_hat <- sd(x)\nfunction(x) dnorm(x, mu_hat, sigma_hat)\n}\n}\n<bytecode: 0x560c2edd98b0>\n<environment: namespace:ggdensity>\n\nAdditional method__1d() functions can be implemented in the same way as the 2-dimensional method_*() functions, so long as the returned closure is structured in one of the two ways we have seen here." ]	[ null, "https://jamesotto852.github.io/ggdensity/articles/method_files/figure-html/unnamed-chunk-2-1.png", null, "https://jamesotto852.github.io/ggdensity/articles/method_files/figure-html/unnamed-chunk-3-1.png", null, "https://jamesotto852.github.io/ggdensity/articles/method_files/figure-html/unnamed-chunk-3-2.png", null, "https://jamesotto852.github.io/ggdensity/articles/method_files/figure-html/unnamed-chunk-3-3.png", null, "https://jamesotto852.github.io/ggdensity/articles/method_files/figure-html/unnamed-chunk-3-4.png", null, "https://jamesotto852.github.io/ggdensity/articles/method_files/figure-html/unnamed-chunk-4-1.png", null, "https://jamesotto852.github.io/ggdensity/articles/method_files/figure-html/unnamed-chunk-4-2.png", null, "https://jamesotto852.github.io/ggdensity/articles/method_files/figure-html/unnamed-chunk-4-3.png", null, "https://jamesotto852.github.io/ggdensity/articles/method_files/figure-html/unnamed-chunk-4-4.png", null, "https://jamesotto852.github.io/ggdensity/articles/method_files/figure-html/unnamed-chunk-5-1.png", null, "https://jamesotto852.github.io/ggdensity/articles/method_files/figure-html/unnamed-chunk-7-1.png", null, "https://jamesotto852.github.io/ggdensity/articles/method_files/figure-html/unnamed-chunk-7-2.png", null, "https://jamesotto852.github.io/ggdensity/articles/method_files/figure-html/unnamed-chunk-7-3.png", null, "https://jamesotto852.github.io/ggdensity/articles/method_files/figure-html/unnamed-chunk-7-4.png", null, "https://jamesotto852.github.io/ggdensity/articles/method_files/figure-html/unnamed-chunk-8-1.png", null, "https://jamesotto852.github.io/ggdensity/articles/method_files/figure-html/unnamed-chunk-8-2.png", null, "https://jamesotto852.github.io/ggdensity/articles/method_files/figure-html/unnamed-chunk-8-3.png", null, "https://jamesotto852.github.io/ggdensity/articles/method_files/figure-html/unnamed-chunk-8-4.png", null, "https://jamesotto852.github.io/ggdensity/articles/method_files/figure-html/unnamed-chunk-14-1.png", null, "https://jamesotto852.github.io/ggdensity/articles/method_files/figure-html/unnamed-chunk-15-1.png", null, "https://jamesotto852.github.io/ggdensity/articles/method_files/figure-html/unnamed-chunk-17-1.png", null, "https://jamesotto852.github.io/ggdensity/articles/method_files/figure-html/unnamed-chunk-18-1.png", null, "https://jamesotto852.github.io/ggdensity/articles/method_files/figure-html/unnamed-chunk-18-2.png", null, "https://jamesotto852.github.io/ggdensity/articles/method_files/figure-html/unnamed-chunk-20-1.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6423976,"math_prob":0.994042,"size":15200,"snap":"2023-14-2023-23","text_gpt3_token_len":4505,"char_repetition_ratio":0.15569887,"word_repetition_ratio":0.12542808,"special_character_ratio":0.31822369,"punctuation_ratio":0.18085831,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99746805,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-01T05:36:32Z\",\"WARC-Record-ID\":\"<urn:uuid:21caf1d5-8569-4ef4-a743-b42230a9d0b9>\",\"Content-Length\":\"75044\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:31beea7a-9d34-45a8-bc0b-c026e6a3c7f3>\",\"WARC-Concurrent-To\":\"<urn:uuid:421ea940-3d01-4cfd-8646-e067c731f0b3>\",\"WARC-IP-Address\":\"185.199.110.153\",\"WARC-Target-URI\":\"https://jamesotto852.github.io/ggdensity/articles/method.html\",\"WARC-Payload-Digest\":\"sha1:XCN6CKSYYWF3O25AOJBNNM6ZHOTHUPAC\",\"WARC-Block-Digest\":\"sha1:4J5HREONAOABFP4EAMAHOQFWG7QPHQML\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224647614.56_warc_CC-MAIN-20230601042457-20230601072457-00312.warc.gz\"}"}
https://arguablywrong.home.blog/2019/10/17/selection-pressure-on-prevalence-of-a-dominant-mutation/	[ "Design a site like this with WordPress.com\n\n# Selection pressure on prevalence of a dominant mutation\n\nSuppose we have a dominant mutation in the population with some prevalence", null, "$p$. That means that a fraction", null, "$p$ of the alleles at that locus in the population are our mutation, while the rest are wild-type. We’d like to know how that prevalence changes over time. Some of those changes will be random, but we can average out to ask what the expectation of the change is at any particular point. We’ll have to make some assumptions first:\n\n• Generations will be single events. We’ll start off with some generation, apply a selection model, and use that to generate the next generation, ad infinitum.\n• Individuals will randomly mate to produce the next generation. This means there won’t be any population substructure and we can calculate the fraction of homozygotes and heterozygotes using only the mutation prevalence", null, "$p$.\n• The mutation confers a selective advantage", null, "$s$ that is equal in both homozygote and heterozygote carriers.\n\nGiven this, we can work out a table comparing different parts of the population, where", null, "$q = 1 - p$ is the probability of the wild-type allele:\n\nRelative fitness here means that parents with the allele will have on average", null, "$1 + s$ children if wild type parents have", null, "$1$. That lets us calculate the fraction of the parents of the next generation, adjusting for increased reproduction of the carriers.\n\nNow that we know what fraction of the parental population has each status, we can calculate the expected prevalence of the allele after one generation. The next generation will have the same prevalence as in the parental population.", null, "$p' = \\frac{pq(1+s) + p^2(1+s)}{1 + sp + spq}$.\n\nThat is, we sum up half the fraction of heterozygotes and the fraction of homozygotes. This simplifies to", null, "$p' = \\frac{p (1 + s)}{1 + sp + spq}$.\n\nNote that for the case where", null, "$s=0$, this reduces to", null, "$p' = p$, indicating no expected change in allele prevalence when there is no selective pressure, as we would expect. We can then calculate the change in prevalence as", null, "$\\Delta p = p' - p$, which simplifies down to", null, "$\\Delta p = \\frac{spq^2}{1 + sp + spq}$.\n\nThis change is zero if and only if", null, "$spq^2$ is zero. That is, if any of", null, "$s$,", null, "$p$, or", null, "$q$ are zero. This means that the allele changes frequency unless there is no selective pressure or the frequency is fixed at 0 or 1.\n\nWe can get a better handle on the behavior of this by treating it as a differential equation and solving. We can rewrite it as", null, "$\\frac{dp}{dt} \\frac{-p^2 + 2p + \\frac{1}{s}}{p^3 - 2p^2 + s} = 1$.\n\nThen we integrate both sides with respect to t:", null, "$\\int \\frac{-p^2 + 2p + \\frac{1}{s}}{p^3 - 2p^2 + s} \\frac{dp}{dt} dt = \\int 1 dt$.\n\nSimplify the left side by partial fraction decomposition and solve the right side:", null, "$\\int \\left [\\frac{1}{sp} + \\frac{1}{1-p} + \\frac{1}{s(1-p)} + \\frac{1}{(1-p)^2} + \\frac{1}{s(1-p)^2} \\right] dp = t + c$.\n\nThen we simply integrate each piece of the left integral separately and combine the constants of integration.", null, "$\\frac{1 + s}{sq} + \\frac{1}{s}ln(\\frac{p}{q}) - ln(q) = t + c$.\n\nThere’s no way to solve explicitly for", null, "$p$ at any particular number of generations. Instead what this gives us is a solution for", null, "$t$, the number of generations it takes to reach any particular prevalence, where the constant of integration is provided by the initial prevalence of the allele.", null, "$t(p \| s) = \\frac{1+s}{s} \\left ( \\frac{q_0 - q}{q_0q} - ln \\frac{q}{q_0} \\right ) + \\frac{1}{s} ln \\frac{p}{p_0}$.\n\nThis lets us answer the question of how changes in the selection pressure affects the speed of selection. Suppose we keep the same starting prevalence and we wish to know how fast the population will reach some ending prevalence. The time needed is of the form", null, "$t(p \| s) = \\frac{1}{s}(s K(p) + Q(p))$", null, "$K(p) = \\frac{q_0 - q}{q_0q} - ln \\frac{q}{q_0}$", null, "$Q(p) = \\frac{q_0 - q}{q_0q} - ln \\frac{q}{q_0} + ln\\frac{p}{p_0}$\n\nIf", null, "$K(p) >> Q(p)$ then the time becomes approximately independent of", null, "$s$. If", null, "$K(p) << Q(p)$ then the time becomes approximately inversely proportional to the selection pressure. This gives us a bound on how fast increased selection can work: at most, doubling the selection pressure will half the time needed to increase allele prevalence to a given point." ]	[ null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89112264,"math_prob":0.9998037,"size":3509,"snap":"2022-40-2023-06","text_gpt3_token_len":714,"char_repetition_ratio":0.15378031,"word_repetition_ratio":0.006779661,"special_character_ratio":0.19321744,"punctuation_ratio":0.093603745,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99990976,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-28T04:52:35Z\",\"WARC-Record-ID\":\"<urn:uuid:98f0bed5-4eb8-4e75-81e6-44cc770b398a>\",\"Content-Length\":\"111204\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:52815365-4c39-4102-89d0-4d63f04f2b8f>\",\"WARC-Concurrent-To\":\"<urn:uuid:02219ec9-3331-48ec-86db-26fc07d3cdd8>\",\"WARC-IP-Address\":\"192.0.78.31\",\"WARC-Target-URI\":\"https://arguablywrong.home.blog/2019/10/17/selection-pressure-on-prevalence-of-a-dominant-mutation/\",\"WARC-Payload-Digest\":\"sha1:AZ7JEW3MD6AK3YVVVKBTXXJEH5U7IST5\",\"WARC-Block-Digest\":\"sha1:XBSHZE6PZFQAOVCCBTUYF2TF35BFTIXA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499470.19_warc_CC-MAIN-20230128023233-20230128053233-00212.warc.gz\"}"}
https://thtsearch.com/content/Arc_(geometry)/	[ "You Might Like\n\nIn Euclidean geometry, an arc (symbol: ) is a closed segment of a differentiable curve. A common example in the plane (a two-dimensional manifold), is a segment of a circle called a circular arc. In space, if the arc is part of a great circle (or great ellipse), it is called a great arc.\n\nEvery pair of distinct points on a circle determines two arcs. If the two points are not directly opposite each other, one of these arcs, the minor arc, will subtend an angle at the centre of the circle that is less than π radians (180 degrees), and the other arc, the major arc, will subtend an angle greater than π radians.\n\n# Circular arcs\n\nThe length (more precisely, arc length) of an arc of a circle with radius r and subtending an angle θ (measured in radians) with the circle center — i.e., the central angle — is\n\nThis is because\n\nSubstituting in the circumference\n\nand, with α being the same angle measured in degrees, since θ = α/180π, the arc length equals\n\nA practical way to determine the length of an arc in a circle is to plot two lines from the arc's endpoints to the center of the circle, measure the angle where the two lines meet the center, then solve for L by cross-multiplying the statement:\n\nFor example, if the measure of the angle is 60 degrees and the circumference is 24 inches, then\n\nThis is so because the circumference of a circle and the degrees of a circle, of which there are always 360, are directly proportional.\n\nThe upper half of a circle can be parameterized as\n\nThe area of the sector formed by an arc and the center of a circle (bounded by the arc and the two radii drawn to its endpoints) is\n\nThe area A has the same proportion to the circle area as the angle θ to a full circle:\n\nWe can cancel π on both sides:\n\nBy multiplying both sides by r2, we get the final result:\n\nUsing the conversion described above, we find that the area of the sector for a central angle measured in degrees is\n\nThe area of the shape bounded by the arc and the straight line between its two end points is\n\nUsing the intersecting chords theorem (also known as power of a point or secant tangent theorem) it is possible to calculate the radius r of a circle given the height H and the width W of an arc:\n\nConsider the chord with the same endpoints as the arc. Its perpendicular bisector is another chord, which is a diameter of the circle. The length of the first chord is W, and it is divided by the bisector into two equal halves, each with length W/2. The total length of the diameter is 2r, and it is divided into two parts by the first chord. The length of one part is the sagitta of the arc, H, and the other part is the remainder of the diameter, with length 2r − H. Applying the intersecting chords theorem to these two chords produces\n\nwhence\n\nso" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92033637,"math_prob":0.99274236,"size":2742,"snap":"2022-40-2023-06","text_gpt3_token_len":634,"char_repetition_ratio":0.16325785,"word_repetition_ratio":0.007736944,"special_character_ratio":0.22428884,"punctuation_ratio":0.08596491,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9995291,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-25T09:04:20Z\",\"WARC-Record-ID\":\"<urn:uuid:9ab1167e-c5db-4788-b3e1-872057e55d1f>\",\"Content-Length\":\"20048\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ca6b56ae-3a8c-4650-a732-14d0a9fc5d1a>\",\"WARC-Concurrent-To\":\"<urn:uuid:d3013bc3-e713-4a99-9877-cc2d9ed74b9b>\",\"WARC-IP-Address\":\"104.21.23.82\",\"WARC-Target-URI\":\"https://thtsearch.com/content/Arc_(geometry)/\",\"WARC-Payload-Digest\":\"sha1:U3UULORRKMYHJCFGKJJ32MP6CAA3VTIH\",\"WARC-Block-Digest\":\"sha1:V4HT6ULNA4F3KQWOKG3VERNR53RYP5ED\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030334515.14_warc_CC-MAIN-20220925070216-20220925100216-00345.warc.gz\"}"}
https://adaic.org/resources/add_content/standards/95lrm/ARM_HTML/RM-4-4.html	[ "# 4.4 Expressions\n\n1\nAn expression is a formula that defines the computation or retrieval of a value. In this International Standard, the term ``expression'' refers to a construct of the syntactic category expression or of any of the other five syntactic categories defined below.\n\n#### Syntax\n\n2\nexpression ::=\nrelation {and relation} \| relation {and then relation}\n\| relation {or relation} \| relation {or else relation}\n\| relation {xor relation}\n3\nrelation ::=\nsimple_expression [relational_operator simple_expression]\n\| simple_expression [notin range\n\| simple_expression [notin subtype_mark\n4\n5\nterm ::= factor {multiplying_operator factor}\n6\nfactor ::= primary [** primary] \| abs primary \| not primary\n7\nprimary ::=\nnumeric_literal \| null \| string_literal \| aggregate\n\| name \| qualified_expression \| allocator \| (expression)\n\n#### Name Resolution Rules\n\n8\nA name used as a primary shall resolve to denote an object or a value.\n\n#### Static Semantics\n\n9\nEach expression has a type; it specifies the computation or retrieval of a value of that type.\n\n#### Dynamic Semantics\n\n10\nThe value of a primary that is a name denoting an object is the value of the object.\n\n#### Implementation Permissions\n\n11\nFor the evaluation of a primary that is a name denoting an object of an unconstrained numeric subtype, if the value of the object is outside the base range of its type, the implementation may either raise Constraint_Error or return the value of the object.\n\n#### Examples\n\n12\nExamples of primaries:\n13\n4.0 -- real literal\nPi -- named number\n(1 .. 10 => 0) -- array aggregate\nSum -- variable\nInteger'Last -- attribute\nSine(X) -- function call\nColor'(Blue) -- qualified expression\nReal(MN) -- conversion\n(Line_Count + 10) -- parenthesized expression\n14\nExamples of expressions:\n15\nVolume -- primary\nnot Destroyed -- factor\n2Line_Count -- term\n-4.0 -- simple expression\n-4.0 + A -- simple expression\nB*2 - 4.0AC -- simple expression\nPassword(1 .. 3) = \"Bwv\" -- relation\nCount in Small_Int -- relation\nCount not in Small_Int -- relation\nIndex = 0 or Item_Hit -- expression\n(Cold and Sunny) or Warm -- expression (parentheses are required)\nA(B*C) -- expression (parentheses are required)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.54017705,"math_prob":0.9290947,"size":2325,"snap":"2020-45-2020-50","text_gpt3_token_len":583,"char_repetition_ratio":0.17277035,"word_repetition_ratio":0.06970509,"special_character_ratio":0.2627957,"punctuation_ratio":0.09090909,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98522025,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-27T20:26:12Z\",\"WARC-Record-ID\":\"<urn:uuid:aba2347e-6724-4cd3-9146-1ae40d7ddaa3>\",\"Content-Length\":\"18335\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:25754b79-c831-4989-b26f-8c6e3dd124c0>\",\"WARC-Concurrent-To\":\"<urn:uuid:24201d2b-0ea2-4751-a8dc-369c3b38004c>\",\"WARC-IP-Address\":\"35.214.219.108\",\"WARC-Target-URI\":\"https://adaic.org/resources/add_content/standards/95lrm/ARM_HTML/RM-4-4.html\",\"WARC-Payload-Digest\":\"sha1:WSULSUQMWGVS4RWITXOAW4WWSYXNWA2Q\",\"WARC-Block-Digest\":\"sha1:EPNPDHK5K3DI4XBPCO4NRWDP7NBN6KP6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141194171.48_warc_CC-MAIN-20201127191451-20201127221451-00395.warc.gz\"}"}
http://oeis.org/A165726/internal	[ "The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.", null, "Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)\n A165726 Number of reduced words of length n in Coxeter group on 50 generators S_i with relations (S_i)^2 = (S_i S_j)^9 = I. 0\n\n%I\n\n%S 1,50,2450,120050,5882450,288240050,14123762450,692064360050,\n\n%T 33911153642450,1661646528478825,81420679895402400,\n\n%U 3989613314871777600,195491052428573042400,9579061568993020137600,469374016880312098682400\n\n%N Number of reduced words of length n in Coxeter group on 50 generators S_i with relations (S_i)^2 = (S_i S_j)^9 = I.\n\n%C The initial terms coincide with those of A170769, although the two sequences are eventually different.\n\n%C Computed with MAGMA using commands similar to those used to compute A154638.\n\n%H <a href=\"/index/Rec#order_09\">Index entries for linear recurrences with constant coefficients</a>, signature (48, 48, 48, 48, 48, 48, 48, 48, -1176).\n\n%F G.f. (t^9 + 2t^8 + 2t^7 + 2t^6 + 2t^5 + 2t^4 + 2t^3 + 2t^2 + 2t +\n\n%F 1)/(1176t^9 - 48t^8 - 48t^7 - 48t^6 - 48t^5 - 48t^4 - 48t^3 -\n\n%F 48t^2 - 48*t + 1)\n\n%K nonn\n\n%O 0,2\n\n%A _John Cannon_ and _N. J. A. Sloane_, Dec 03 2009\n\nLookup \| Welcome \| Wiki \| Register \| Music \| Plot 2 \| Demos \| Index \| Browse \| More \| WebCam\nContribute new seq. or comment \| Format \| Style Sheet \| Transforms \| Superseeker \| Recent\nThe OEIS Community \| Maintained by The OEIS Foundation Inc.\n\nLast modified April 19 17:46 EDT 2021. Contains 343117 sequences. (Running on oeis4.)" ]	[ null, "http://oeis.org/banner2021.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6269025,"math_prob":0.92813677,"size":911,"snap":"2021-04-2021-17","text_gpt3_token_len":376,"char_repetition_ratio":0.13340683,"word_repetition_ratio":0.022727273,"special_character_ratio":0.6026345,"punctuation_ratio":0.17073171,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9640882,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-19T23:09:32Z\",\"WARC-Record-ID\":\"<urn:uuid:33e4d5f3-23ea-4f9a-9ea5-0e0ee0af48ff>\",\"Content-Length\":\"8321\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fbeb8ff0-60c7-4ded-a60b-23ae79db5d2c>\",\"WARC-Concurrent-To\":\"<urn:uuid:3a9dd96f-5513-48f1-9ed2-639fde71f456>\",\"WARC-IP-Address\":\"104.239.138.29\",\"WARC-Target-URI\":\"http://oeis.org/A165726/internal\",\"WARC-Payload-Digest\":\"sha1:CVQOR3I3XWZKJR6VG6OHFFRDEBIDRTQ3\",\"WARC-Block-Digest\":\"sha1:E3GMGMZ5GT5IUEKF4D2LJC4EGBPXHJQI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038917413.71_warc_CC-MAIN-20210419204416-20210419234416-00520.warc.gz\"}"}
https://www.jiskha.com/questions/514200/perimeter-of-a-equilateral-triangle-is-given-as-42-cm-find-the-perimeter-of-a-shaded-part	[ "# maths\n\nperimeter of a equilateral triangle is given as 42 cm. Find the perimeter of a shaded part of triangle if the 3 sectors of a circle marked are identical\n\n1. 👍 0\n2. 👎 0\n3. 👁 206\n1. Sorry, we do not see the sectors or the shading. You will need to describe the figure or try to post a link to the figure.\n\n1. 👍 0\n2. 👎 0\n\n## Similar Questions\n\n1. ### Math\n\nThe formula for finding the perimeter of an equilateral triangle is P = 3s. How can you transform the formula so that it can be used to determine the length of one side of the triangle when given the perimeter?\n\n2. ### Algebra\n\nA regular pentagon and an equilateral triangle have the same perimeter. The perimeter of the pentagon is 5 (1/2x +2) inches. The perimeter of the triangle is 4(x-2) inches. What is the perimeter of each figure?\n\n3. ### Maths\n\nThe sides of the equilateral triangle is 3x+2,2y-x and y+3 find x and y.and the perimeter of the triangle\n\n4. ### geometry\n\nhow do you find if the perimeter of a triangle EFG is 32, triangle EFG scalene, isosceles, or equilateral?\n\n1. ### Algebra\n\nA square and an equilateral triangle have the same perimeter.Each side of the triangle is 5 inches longer than each side of the square. What is the perimeter of the square.\n\n2. ### math\n\nAn equilateral triangle has a perimeter of 15x +18. What is the length of each side of the triangle?\n\n3. ### geometry\n\none side of an equilateral triangle is 7 dm shorter than one side of square.the sum of the perimeter of the two figure is 49 dm . find the perimeter of each figure.\n\n4. ### geometry\n\nfind the perimeter of an equilateral triangle with one side that measures 9.6 centimeters\n\n1. ### Geometry\n\nthe perimeter of an equilateral triangle is 32 centimeters. find the length of an altitude of the triangleto the nearest tenth of a centimeter.\n\n2. ### Geometry\n\nAn equilateral triangle has a perimeter of 120 inches. What is the area of the triangle? Express your answer in simplest radical form.\n\n3. ### Centroids and Triangles - determining perimeter?\n\nLet G denote the centroid of triangle ABC. If triangle ABG is equilateral with side length 2, then determine the perimeter of triangle ABC. I drew the diagram, but it doesn't really help....\n\n4. ### Algebra\n\nThe perimeter of an equilateral triangle is 7 in. more than the perimeter of a square. The side of the triangle is 5 in. longer than the side of the square. Find the length of each side of the triangle. (Note: An equilateral" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84280735,"math_prob":0.9972501,"size":2158,"snap":"2020-34-2020-40","text_gpt3_token_len":545,"char_repetition_ratio":0.2650882,"word_repetition_ratio":0.1043257,"special_character_ratio":0.23586655,"punctuation_ratio":0.0990566,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999361,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-27T00:17:25Z\",\"WARC-Record-ID\":\"<urn:uuid:88d8c68e-1c14-4f7b-b985-ac04f498a4f2>\",\"Content-Length\":\"17143\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7a665541-1f61-4f79-9309-2e64b99941d3>\",\"WARC-Concurrent-To\":\"<urn:uuid:d31eef8b-6ef7-414e-8808-e8cbbaed86ed>\",\"WARC-IP-Address\":\"66.228.55.50\",\"WARC-Target-URI\":\"https://www.jiskha.com/questions/514200/perimeter-of-a-equilateral-triangle-is-given-as-42-cm-find-the-perimeter-of-a-shaded-part\",\"WARC-Payload-Digest\":\"sha1:E4Y62KP7EFAKIAESGH3FHR6XTZJIDISK\",\"WARC-Block-Digest\":\"sha1:KHQSHE265OJHHXDPSHW4NBCTFELPAYLP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400249545.55_warc_CC-MAIN-20200926231818-20200927021818-00480.warc.gz\"}"}
https://in.mathworks.com/help/robotics/ug/design-position-controlled-manipulator-using-simscape.html	[ "# Design Position Controlled Manipulator Using Simscape\n\nThis example shows you how to use Simulink® with Robotics System Toolbox™ to design a position controller for a manipulator and compute joint position required to drive the Simscape™ Multibody™ model of the manipulator.\n\nRobotics System Toolbox, Simscape Multibody, and Robotics System Toolbox Robot Library Data support package are required to run this example.\n\n### Introduction\n\nIn this example, you will load an included robot model using `loadrobot` as a `rigidBodyTree` object, then create a Simscape Multibody model of the robot using `smimport`. Configure the model to accept joint torque and return the computed joint position and velocity. Implement a computed torque controller with joint position and velocity feedback using manipulator algorithm blocks. The controller receives joint position and velocity information from the robot model and sends torque commands to drive the robot to the desired joint position computed using Inverse Kinematics (IK).\n\n### Load Robot Model in Workspace\n\nThis example uses a model of the KINOVA® Gen3, a 7 degree-of-freedom robot manipulator. Call `loadrobot` to generate a `rigidBodyTree` model of the robot. Set the `DataFormat` properties to be consistent with Simscape.\n\n`robot = loadrobot(\"kinovaGen3\",DataFormat=\"column\");`\n\n### Generate Simscape Multibody Model from Rigid Body Tree\n\nImport the `robot` object into Simscape Multibody and get the model parameters.\n\n```robotSM = smimport(robot,ModelName=\"ManipulatorPositionControl_Subsystem\"); sm_mdl = get_param(robotSM,\"Name\");```", null, "### Configure Simscape Multibody Model\n\nPrepare the Simscape Multibody model to accept the joint torque inputs and return the joint positions and velocities. You can follow the steps below to manually configure the model or use the `helperInstrumentSMModels` helper function to automatically configure the model.\n\n#### Manual Configuration of Simscape Multibody Model\n\n1. In your model, double-click a Joint block. The Property Inspector dialog box opens.\n\n2. In the Property Inspector dialog box, select Z Revolute Primitive (Rz) > Actuation > Torque > Provided by Input, and select Z Revolute Primitive (Rz) > Actuation > Motion > Automatically computed. The block exposes a physical signal input port, labeled `t`.", null, "3. Select Z Revolute Primitive (Rz) > Sensing and enable Position, Velocity, and Acceleration. The block exposes a physical signal output ports, labeled `q`, `w`, and `b`.", null, "4. Add a Simulink-PS Converter block from the Simscape > Utilities library, connect the Simulink-PS Converter block to physical signal input port `t` of the Joint block.\n\n5. Add a From block from the Simulink > Signal Routing library to the input port of the Simulink-PS Converter block.\n\n6. Add three PS-Simulink Converter blocks from the Simscape > Utilities library, connect the PS-Simulink Converter blocks to physical signal output ports `q`, `w`, and `b` of the Joint block.\n\n7. Add three Goto blocks from the Simulink > Signal Routing library to the output port of the PS-Simulink Converter blocks.", null, "8. Repeat these steps for all the Joint blocks.\n\n9. Add a Demux block from the Simulink > Signal Routing library and connect the joint torque Goto blocks related to the respective joint torque From blocks.\n\n10. Add three Mux blocks from the Simulink > Signal Routing library and connect the joint motions From blocks related to the respective joint motions Goto blocks.\n\n11. Create a subsystem of the Simscape Multibody model.\n\n#### Configure Simscape Multibody Model Using Helper Function\n\nUse the `helperInstrumentSMModels` helper function to automatically configure the model.\n\nCall the helper function to automatically configure the Simscape Multibody model to accept torque input.\n\n`helperInstrumentSMModels.instrumentRBTSupportedJointInputs(sm_mdl,robot,\"torque\")`\n\nCall the helper function again to configure the Simscape Multibody model to enable position, velocity, and acceleration sensing at each joint.\n\n`helperInstrumentSMModels.instrumentRBTSupportedJointOutputs(sm_mdl,robot,\"motion\")`\n\nCreate a subsystem of the Simscape Multibody model.\n\n`helperInstrumentSMModels.convertToSubsystem(sm_mdl)`", null, "#### Set Up Variables in Model Workspace\n\nSet up the variables in the model workspace that specify the start and end waypoints, and the joint starting position and velocity.\n\n```mdlWks = get_param(robotSM,\"ModelWorkspace\"); assignin(mdlWks,\"robotToTest\",robot) assignin(mdlWks,\"q0\",robot.homeConfiguration) assignin(mdlWks,\"dq0\",zeros(size(robot.homeConfiguration)))```\n\n### Computed Torque Controller\n\nThe Computed Torque Controller subsystem is built using three robotics manipulator blocks: Joint Space Mass Matrix, Velocity Product Torque, and Gravity Torque. The `rigidBodyTree` model, `robotToTest`, is assigned in all those blocks.", null, "The Computed Torque Controller subsystem accepts Measured Configuration, Measured Velocities and Desired Configuration and returns Applied Torque for each joint of the manipulator.\n\n### Set Up Controller Input\n\n1. Add a Coordinate Transformation Conversion block from the Robotics System Toolbox > Utilities library to the model. Set the input representation as Translation Vector and the output representation as Homogeneous Transformation.", null, "2. Add a Constant block and set the value as `[0.5 0.5 0.5]`. Connect the Constant block to the input port of the Coordinate Transformation Conversion block.\n\n3. Add a Inverse Kinematics block from the Robotics System Toolbox > Manipulator Algorithms library to the model.\n\n4. In the Inverse Kinematics block, specify the Rigid body tree model as `robotToTest`, then click Select body next to the End effector to select the end effector body.", null, "5. Connect the output port of Coordinate Transformation Conversion block to the Pose port of the Inverse Kinematics block.\n\n6. Add another Constant block and set the value as `[0 0 0 1 1 1]`. Connect the Constant block to the Weights port of the Inverse Kinematics block.\n\n7. Connect a Delay block to the Config port of the Inverse Kinematics block and specify the Initial condition as `q0`.", null, "8. Connect the output of the Delay block to the InitialGuess port of the Inverse Kinematics block.\n\n### Final Setup\n\nConnect the Simscape Multibody Model subsystem, Computed Torque Controller subsystem, Controller Input blocks, and a Scope block as shown in figure.", null, "### Simulate Model\n\nOpen the provided ManipulatorPositionControl.slx model and replace the Robot subsystem with the subsystem created in ManipulatorPositionControl_Subsystem model above, for it to be able to fetch the meshes correctly.\n\n```open_system(\"ManipulatorPositionControl.slx\") ```\n\nSave the model and simulate it.\n\n```sim(\"ManipulatorPositionControl.slx\",\"StopTime\",\"5\") ```\n\nVisualize the multibody model in the Mechanics Explorer.", null, "Visualize the joint positions in the Scope.", null, "" ]	[ null, "https://in.mathworks.com/help/examples/robotics/win64/DesignPositionControlledManipulatorUsingSimscapeExample_01.png", null, "https://in.mathworks.com/help/examples/robotics/win64/DesignPositionControlledManipulatorUsingSimscapeExample_02.png", null, "https://in.mathworks.com/help/examples/robotics/win64/DesignPositionControlledManipulatorUsingSimscapeExample_03.png", null, "https://in.mathworks.com/help/examples/robotics/win64/DesignPositionControlledManipulatorUsingSimscapeExample_04.png", null, "https://in.mathworks.com/help/examples/robotics/win64/DesignPositionControlledManipulatorUsingSimscapeExample_05.png", null, "https://in.mathworks.com/help/examples/robotics/win64/DesignPositionControlledManipulatorUsingSimscapeExample_06.png", null, "https://in.mathworks.com/help/examples/robotics/win64/DesignPositionControlledManipulatorUsingSimscapeExample_07.png", null, "https://in.mathworks.com/help/examples/robotics/win64/DesignPositionControlledManipulatorUsingSimscapeExample_08.png", null, "https://in.mathworks.com/help/examples/robotics/win64/DesignPositionControlledManipulatorUsingSimscapeExample_09.png", null, "https://in.mathworks.com/help/examples/robotics/win64/DesignPositionControlledManipulatorUsingSimscapeExample_10.png", null, "https://in.mathworks.com/help/examples/robotics/win64/DesignPositionControlledManipulatorUsingSimscapeExample_11.png", null, "https://in.mathworks.com/help/examples/robotics/win64/DesignPositionControlledManipulatorUsingSimscapeExample_12.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.677656,"math_prob":0.7258762,"size":6804,"snap":"2022-27-2022-33","text_gpt3_token_len":1460,"char_repetition_ratio":0.1525,"word_repetition_ratio":0.1138976,"special_character_ratio":0.18239272,"punctuation_ratio":0.11722913,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9633154,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-11T21:31:55Z\",\"WARC-Record-ID\":\"<urn:uuid:ffe2c8aa-bb0a-4940-bdf0-adf6e296d09a>\",\"Content-Length\":\"89689\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:61911300-1938-4da0-92d5-d00c8aa07b10>\",\"WARC-Concurrent-To\":\"<urn:uuid:044a8b7a-2449-4fe5-9253-5c5f10bb3f2e>\",\"WARC-IP-Address\":\"23.39.174.83\",\"WARC-Target-URI\":\"https://in.mathworks.com/help/robotics/ug/design-position-controlled-manipulator-using-simscape.html\",\"WARC-Payload-Digest\":\"sha1:GXJMEZB4ZNOBMVXCDOLYSBY2FQMGNDJT\",\"WARC-Block-Digest\":\"sha1:ER5ZFDXZNQIOLQFWFETD2PVU76WFUKW6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882571502.25_warc_CC-MAIN-20220811194507-20220811224507-00680.warc.gz\"}"}
https://wizardcalc.com/what-is/7-of-120	[ "", null, "Percent Calculator\n\n## 7% of 120 is 8.4\n\nCalculate percent\nWhat is% of?\n\n## How to calculate 7 percent of 120\n\n• step 1: 7%120 =\n• step 2: (7:100)120 =\n• step 3: (7120):100 =\n• step 4: 840.0:100=8.4\nAnswer: 7 of 120 is 8.4\n\n## 7% of other values:\n\n• 7% of 132 = 9.24\n• 7% of 144 = 10.08\n• 7% of 156 = 10.92\n• 7% of 168 = 11.76\n• 7% of 180 = 12.6\n• 7% of 192 = 13.44\n• 7% of 204 = 14.28\n• 7% of 216 = 15.12\n• 7% of 228 = 15.96\n• 7% of 240 = 16.8\n• 7% of 252 = 17.64\n• 7% of 264 = 18.48\n• 7% of 276 = 19.32\n• 7% of 288 = 20.16\n• 7% of 300 = 21\n• 7% of 312 = 21.84\n• 7% of 324 = 22.68\n\n## How much will you save if product costs 120 and discount is 7%\n\nSavings: Original price percentage off / 100\nAmount saved: (7 * 120) / 100 =\nSavings: \\$8.4\n\nThat means for an original price of 120 and a 7% discount, you would pay \\$111.6 and save \\$8.4\n\n## What is 7 percent (calculated percentage %) of number 120?\n\n7% of 120 is equal to their multiplication: 7% * 120.\n7 percent of 120:\n• 7% of 120 =\n• 7/100 * 120 =\n• = 8.4" ]	[ null, "https://wizardcalc.com/backArrow.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91163146,"math_prob":0.9963641,"size":1029,"snap":"2021-31-2021-39","text_gpt3_token_len":458,"char_repetition_ratio":0.21658537,"word_repetition_ratio":0.008196721,"special_character_ratio":0.6122449,"punctuation_ratio":0.16356878,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998766,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-18T13:24:41Z\",\"WARC-Record-ID\":\"<urn:uuid:120b2dec-d16d-4b91-86d9-6089db0c9a00>\",\"Content-Length\":\"11288\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a9b9b082-d478-4a2e-ae12-97ef37f511d0>\",\"WARC-Concurrent-To\":\"<urn:uuid:0abe13f7-1617-4f1f-975f-26b55e5757df>\",\"WARC-IP-Address\":\"76.76.21.21\",\"WARC-Target-URI\":\"https://wizardcalc.com/what-is/7-of-120\",\"WARC-Payload-Digest\":\"sha1:7E5U5757AOZQNX63RD6FDKHLLMQ2Q4VV\",\"WARC-Block-Digest\":\"sha1:VIDBTJ4F4PFZ6LQTORO6OEN5KAEGXL72\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780056476.66_warc_CC-MAIN-20210918123546-20210918153546-00487.warc.gz\"}"}
http://rportal.lib.ntnu.edu.tw/items/f7c101a0-3ac4-47c9-9ece-8ea6c3b84e17	[ "# 完全圖上最大權重配對問題之自我穩定演算法的設計及分析\n\nNo Thumbnail Available\n\n2000-04-??\n\n## Publisher\n\nOffice of Research and Development\n\n## Abstract\n\nIn 1974, Dijsktra defined a self-stabilizing system as a system which is guaranteed to arrive at a legitimate state in a finite number of steps regardless of its initial state. Since his introduction, self-stabilizing algorithms gained wide-spread research interest. The objectives of this research are to design and analyze self-stabilizing algorithms for maximal weight matching problem. Firstly, Hsu and Huang proved that the time complexity of their self-stabilizing algorithm for finding a maximal matching in distributed networks is O(n3), where n is the number of nodes in the graph. In 1994, Tel introduced a variant function to show that the time complexity of Hsu-Huang's algorithm is O(n2). In this paper, we design a self-stabilizing algorithm for maximal weight matching of the complete graph and prove its correctness. The maximal weight matching problem is defined not only to find the maximal matching of the complete graph, but also to let the total weight of the matching edges be maximal. We combine Hsu-Huang's maximal matching alogrithm and new swapping rules. This system possesses the properties of fault tolerance and self-stabilization and has a time complexity O(n2+nk), where k is the largest weight over all edges in the graph." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7420317,"math_prob":0.9627091,"size":3182,"snap":"2023-40-2023-50","text_gpt3_token_len":1194,"char_repetition_ratio":0.14348647,"word_repetition_ratio":0.93012047,"special_character_ratio":0.17850408,"punctuation_ratio":0.06990291,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98091626,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-10T19:44:11Z\",\"WARC-Record-ID\":\"<urn:uuid:38e4caa4-4691-4ed0-baed-6b2f320cc17b>\",\"Content-Length\":\"433778\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:59da1f87-cf89-4b5a-aca7-f5fa7b631798>\",\"WARC-Concurrent-To\":\"<urn:uuid:c1276a6c-d45c-4711-adcb-a842efe38ffd>\",\"WARC-IP-Address\":\"140.122.127.138\",\"WARC-Target-URI\":\"http://rportal.lib.ntnu.edu.tw/items/f7c101a0-3ac4-47c9-9ece-8ea6c3b84e17\",\"WARC-Payload-Digest\":\"sha1:O34QWTFBB2QCCMKBHWPN53UQFTV5KI6J\",\"WARC-Block-Digest\":\"sha1:7QLZX4LVNPR7QKIAOMKOI5XYDJ5DLJ7Y\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679102637.84_warc_CC-MAIN-20231210190744-20231210220744-00270.warc.gz\"}"}
https://www.physicsforums.com/threads/finding-critical-numbers.388758/	[ "# Finding critical numbers\n\n## Homework Statement\n\nFind all critical numbers of:\ng(x)= sqrt(x2-4)\nand\nf(x)= (1)/(x2-9)\n\nn/a\n\n## The Attempt at a Solution\n\n1) sqrt(x2-4)\nand got zeroes as x=0, x=2, x=-2\nand I got confused because if you do g(2) and g(-2) it equals zero. For some reason I can't tell if they are defined or undefined. x=0 works, so that is a critical number. The other two are throwing me off.\n\n2) (1)/(x2-9)\nzeroes were x=-3, x=3, x=0. Plugging 3 and -3 back into f(x) gave me undefined, so I'm pretty sure 0 is the only critical number.\n\nIf you could please check the second and help with the first that would be great. Thank you\n\nHomework Helper\n\n\"1) sqrt(x2-4)\nsimplified to x(x-2)(x+2)-1/2\"\n\n- is that supposed to be the derivative of $$\\sqrt{\\,x^2 -4}$$? if so, it isn't correct.\n\nwhat is your definition of a critical value? (writing it out can help you see the appropriate path)\n\nI'm honestly not sure what I did there... I re-did my derivative and found\n\nx/((x2-4)1/2)\n\nwhich would go to x/(((x+2)(x-2))1/2)\n\nand give zeroes of -2,2,0. -2 and 2 would make the denominator 0 and be undefined, leaving 0 as the only CP. Sound right? Sorry if any errors I did that really quick.\n\nHomework Helper\nyes, the derivative is\n\n$$\\frac{x}{\\sqrt{\\, x^2 - 4}}$$\n\nand this is zero or undefined for $0 \\text{ and } \\pm 2$.\n\nAgain, what is your definition of a critical number? (same as critical value)\n\nDefinition is:\nx=c is a critical number for f(x) if f(c) is defined and f'(c)=0 or f'(c) is undefined.\n\nSo that means 2 and -2 WOULD be CPs?\n\nHomework Helper\nDefinition is:\nx=c is a critical number for f(x) if f(c) is defined and f'(c)=0 or f'(c) is undefined.\n\nSo that means 2 and -2 WOULD be CPs?\n\nyes.\n\nAnd 0 would not be one because it is undefined in f(x) (sqrt of a negative makes it undefined), correct?\n\nAnd for the second problem, zero was the only one defined in f(x). Therefore it IS indeed a CP." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.93120223,"math_prob":0.99545145,"size":686,"snap":"2021-21-2021-25","text_gpt3_token_len":228,"char_repetition_ratio":0.09530792,"word_repetition_ratio":0.0,"special_character_ratio":0.33965015,"punctuation_ratio":0.085365854,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99873143,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-14T14:36:15Z\",\"WARC-Record-ID\":\"<urn:uuid:b44f3388-e36e-455f-afec-bb84981535cd>\",\"Content-Length\":\"74415\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aaec179b-6a92-4d57-a7f2-a53300d66a9d>\",\"WARC-Concurrent-To\":\"<urn:uuid:6cab115b-2790-4155-813a-c343aa98fb04>\",\"WARC-IP-Address\":\"104.26.15.132\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/finding-critical-numbers.388758/\",\"WARC-Payload-Digest\":\"sha1:RWVKU2EKKT7N3BQHVDDCFBDC53I523VM\",\"WARC-Block-Digest\":\"sha1:OYHLTXTCB44FTJI544QAY5QVYOLV7CUA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243989526.42_warc_CC-MAIN-20210514121902-20210514151902-00138.warc.gz\"}"}
https://grinebiter.com/Coin/CoinConverter/Nickels-in-Dimes/How-many-Nickels-are-in-1-Dimes.html	[ "How many nickels are in 1 dimes?", null, "Here, we will show you how to calculate how many nickels there are in 1 dimes.\n\nFirst, calculate how many cents there are in 1 dimes by multiplying 1 by 10, and then divide that result by 5 cents to get the answer.\n\nHere is the math to illustrate better:\n\n1 dimes x 10 cents\n= 10 cents\n\n10 cents / 5 cents\n= 2 nickels\n\nThus, the answer to the question \"How many nickels are in 1 dimes?\" is as follows:\n\n2 nickels\n\nNote: We multiplied 1 by 10, because there are 10 cents in a dime, and we divided 10 by 5, because there are 5 cents in a nickel.\n\nCoin Converter\nFill out the form below or go here if you need to convert another coin denomination.\n\nHow many\n\nare in\n\nHow many nickels are in 2 dimes?\nHere is the next number of coins we converted." ]	[ null, "https://grinebiter.com/Images/CoinConverter.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.932305,"math_prob":0.9781114,"size":766,"snap":"2021-31-2021-39","text_gpt3_token_len":205,"char_repetition_ratio":0.15223098,"word_repetition_ratio":0.0,"special_character_ratio":0.26762402,"punctuation_ratio":0.1030303,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99979085,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-19T08:46:51Z\",\"WARC-Record-ID\":\"<urn:uuid:a4f29f8d-d5f2-468b-9176-7a2740839498>\",\"Content-Length\":\"8170\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:17ae92f5-2cdd-47d7-959a-58730605d4ba>\",\"WARC-Concurrent-To\":\"<urn:uuid:5901371d-92e8-4587-be0b-59f7dbc9ecf2>\",\"WARC-IP-Address\":\"99.86.230.8\",\"WARC-Target-URI\":\"https://grinebiter.com/Coin/CoinConverter/Nickels-in-Dimes/How-many-Nickels-are-in-1-Dimes.html\",\"WARC-Payload-Digest\":\"sha1:5APHWTAKFNDDSZHJFIR7CTFUGN7XKKMS\",\"WARC-Block-Digest\":\"sha1:GAACLELHU2HIP4OJN2RVDOHJ4VY7NH4X\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780056752.16_warc_CC-MAIN-20210919065755-20210919095755-00079.warc.gz\"}"}
https://www.physicsforums.com/threads/net-thermal-radiation-from-an-object-outside-at-night.546962/	[ "# Net thermal radiation from an object outside at night\n\nI am trying to model the cooling of an object (for example, a sheet of glass) placed outside at night. At the moment I am only considering heat loss by radiation.\n\nI know that the net radiation from the object will be:\n\nRnet = Robj - Rsky\n\nwhere:\nRnet = the net radiation from the object\nRobj = the total thermal radiation from the object\nRsky = the thermal downwelling radiation from the sky\n\nI have come across a formula which does the above, but also takes into consideration the absorptivity of the sheet of glass at the 8-14μm wavelength (I have only considered emissivity of the glass in this wavelength as I'm only concerned with thermal radiation heat loss to the sky).\n\nThe formula is:\n\nRnet = A((εobj2obj2)σTobj4 - εskyσTamb4)\n\nwhere:\nεobj = the emissivity of the object at 8-14μm\nαobj = the absorptivity of the object at 8-14μm\nεsky = the emissivity of the sky\nTobj = the temperature of the object\nTamb4 = the ambient air temperature\nA = the area of the object\n\nWhat I would like to know is where does the εobj2obj2 bit come from? I know if I wasn't considering absorptivity then it would just be εobj, but why are εobj and αobj now squared? I have since lost where I saw it, and I'm pretty sure there wasn't an explanation there anyway. I have searched all over for a derivation of it but have had no luck.\n\nCan anyone help?\n\n## Answers and Replies\n\nI am not familiar with this, but I would agree that something is off here:\n\nSince the eqn is Rnet= Robj - Rsky\n\nthese terms should have equal forms.\n\nBut as you pointed out there is a squared ratio only on one term.\nMy guess is that it may be simplified, but who knows.\n\nSo I can't really answer your question, but I can say that I agree with it.\nPerhaps try to find a different formula / approach to use?\n\nI did find this though - a program for modelling thermal radiation. Looks like it is free to use, and mentions finding radiation from any given objects of given temperature. Maybe it will help. Here's the link http://www.fire-engineering-software.com/tra.html\n\ngood luck\n\nLast edited:\nWithout knowing how the terms are defined, and what units they have, it's impossible to say what is going on - \"I found it on the internet somewhere\" isn't very helpful.", null, "What units does each term in this equation have, and does the result (\"net radiation\" is not precise either) make sense?\n\nThanks for the link, elegysix.\n\nJeffKoch - my problem in finding the source of the equation stems from the fact that it was found in a Solar Energy journal somewhere but I can no longer gain access to Solar Energy journals via my university account. Hence why I'm trying to find it by other means. I thought perhaps it may have been used elsewhere but now I'm beginning to realise that perhaps the assumptions and simplifications made in the particular journal article are more important than I'd first thought!\n\nAs for the units, they are all SI units, i.e. T in Kelvin, Stefan-Boltzmann in W(m^−2)(K−4). Emissivity and absorptivity are dimensionless. Net radiation is therefore in units of W. So the equation is dimensionally correct as far as I can tell.\n\nWhat do you mean about \"net radiation\" not being precise? By net radiation, I mean the net thermal radiation (i.e. primarily in the 8-14micro m wavelength band) leaving the surface of the glass, as opposed to the thermal radiation leaving the glass before one has taken into account the downwelling thermal radiation from the sky.\n\n256bits\nGold Member\nThe formula is:\n\nRnet = A((εobj2/αobj2)σTobj4 - εskyσTamb4)\n\nwhere:\nεobj = the emissivity of the object at 8-14μm\nαobj = the absorptivity of the object at 8-14μm\nεsky = the emissivity of the sky\nTobj = the temperature of the object\nTamb4 = the ambient air temperature\nA = the area of the object\n\nYour equation is most likely derived for a particluar situation since it has a ratio of emissivity/ absortivity, which is what I have no idea, and the ambiant air temperature rather than the temperature of the sky ( which is different than the sky temperature ),\n\nWhat do you mean about \"net radiation\" not being precise? By net radiation, I mean the net thermal radiation (i.e. primarily in the 8-14micro m wavelength band) leaving the surface of the glass, as opposed to the thermal radiation leaving the glass before one has taken into account the downwelling thermal radiation from the sky.\n\nThink, man. Net thermal radiation per unit time? Per unit area? Per Hz? Per Sr? Integrated over any of these quantities? Photons or joules? What units do each of the terms have, and does the result make sense? Find a book on this stuff, there are many - for example the very useful one by Rybicki and Lightman." ]	[ null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.95843196,"math_prob":0.8919592,"size":2515,"snap":"2021-04-2021-17","text_gpt3_token_len":671,"char_repetition_ratio":0.1668658,"word_repetition_ratio":0.9102564,"special_character_ratio":0.23777336,"punctuation_ratio":0.059793815,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9811241,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-15T12:05:00Z\",\"WARC-Record-ID\":\"<urn:uuid:c10fa7df-8765-4e5d-b406-72a3f832626e>\",\"Content-Length\":\"78660\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:931ca4f7-a33c-4a8a-9fb4-01e4fd05338c>\",\"WARC-Concurrent-To\":\"<urn:uuid:049e0b46-df82-4ce3-960d-b4c739d9b9bb>\",\"WARC-IP-Address\":\"172.67.68.135\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/net-thermal-radiation-from-an-object-outside-at-night.546962/\",\"WARC-Payload-Digest\":\"sha1:B2Z2RR567IR4IHXUJV7M5S3EA5E6LWEM\",\"WARC-Block-Digest\":\"sha1:JG4ZUYGXN7GSAPR7FBED3PAD4MBVWRP3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038084765.46_warc_CC-MAIN-20210415095505-20210415125505-00248.warc.gz\"}"}
https://vincenttam.github.io/blog/2014/11/13/plot-polar-coordinates-graphs/	[ "# Plot Polar Coordinates Graphs\n\n\| Comments \|\n\nGoogling “tikz tutorial”, one can find several useful PDF documents which are easy to follow, such as the first two results.\n\n1. PGF/TikZ - Graphics for $\\rm \\LaTeX$ —A tutorial by Meik Hellmund\n2. A very minimal introduction to TikZ by Jacques Crémer\n\nThe second PDF document covers plotting graphs in rectangular coordinates. How about polar coordinates?\n\nIt’s easy to find a solution online, but understanding the code and adapting them according to your needs are much harder. One has to make use of “cs”, which stands for “coordinate system”. If one is too lazy to read more webpages, then one may use the code below.\n\nWhen I drew figures with TikZ a recent post on limits of composite functions on $\\R^n$, I found a way to construct random region with TikZ in a $\\rm \\TeX$–$\\rm \\LaTeX$ Stack Exchange question.1 This inspired me to come up with the following graph.\n\n## Output", null, "## Source code\n\n• I got the idea of for loop in TikZ from TikZ and PGF Manual.\n• I learnt the syntax for plotting functions from the PDF in item (2).\n\n1. Refer to TikZ: Arbitrary shapes and filling? for details." ]	[ null, "https://vincenttam.github.io/images/posts/TikZPolar/graph.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8695698,"math_prob":0.65550834,"size":1125,"snap":"2019-13-2019-22","text_gpt3_token_len":296,"char_repetition_ratio":0.088314004,"word_repetition_ratio":0.0,"special_character_ratio":0.25955555,"punctuation_ratio":0.083333336,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98940873,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-19T08:46:57Z\",\"WARC-Record-ID\":\"<urn:uuid:ff2fd306-9f14-4ee6-a2d7-073777da1bc0>\",\"Content-Length\":\"27206\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a70f0cf4-b144-41e6-b748-a12a4e819477>\",\"WARC-Concurrent-To\":\"<urn:uuid:9575d6ff-eba3-45f3-9ef9-8fb65f6240cb>\",\"WARC-IP-Address\":\"185.199.110.153\",\"WARC-Target-URI\":\"https://vincenttam.github.io/blog/2014/11/13/plot-polar-coordinates-graphs/\",\"WARC-Payload-Digest\":\"sha1:LZRXLL2WFMWAQMK4TWB72HR6NAIJXQUW\",\"WARC-Block-Digest\":\"sha1:HWKSQJSXHIJABYQSSMXYVRVRXMTDTVHA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232254731.5_warc_CC-MAIN-20190519081519-20190519103519-00283.warc.gz\"}"}
http://computerchess.org.uk/ccrl/404/cgi/engine_details.cgi?match_length=30&print=Details&each_game=1&eng=Asterisk%200.6b	[ "Contents: CCRL Blitz Downloads and Statistics January 9, 2021 Testing summary: Total: 2'437'048 games played by 2'898 programs White wins: 937'753 (38.5%) Black wins: 752'816 (30.9%) Draws: 746'479 (30.6%) White score: 53.8%\n\n## Engine Details\n\n Options Show each game results\nAsterisk 0.6b (2316+36\n−36\n)Quote\n Author: Peter Horvath (Hungary) Link: Homepage\nThis is one of the 2 Asterisk versions we tested: Compare them!\n Opponent Elo Diff Results Score LOS Perf – RESP 0.19 64-bit 2363 +20−20 (+47) 10.5 − 21.5(+8−19=5) 32.8%10.5 / 32 1.1% −85 = 0 1 = 1 1 0 1 0 1 = 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 = 0 0 = 0 0 – Wasabi 1.3.0 64-bit 2347 +20−20 (+31) 15.5 − 15.5(+14−14=3) 50.0%15.5 / 31 6.5% +34 = 1 0 0 0 1 = 1 0 0 1 0 1 0 0 0 1 1 1 1 1 1 0 = 1 0 0 0 1 0 1 – Halogen 5 64-bit 2340 +20−20 (+24) 15.5 − 15.5(+10−10=11) 50.0%15.5 / 31 11.6% +24 1 0 = = 1 = 0 0 0 = 1 1 0 1 0 = 0 0 1 = 1 1 = = 1 0 = 0 = = 1 – Ceibo 0.5 64-bit 2329 +17−17 (+13) 6 − 12(+3−9=6) 33.3%6.0 / 18 25.2% −102 1 = 0 1 0 0 0 = 0 0 1 = 0 0 = = 0 = – FoxSEE 5.0.1 64-bit 2321 +21−22 (+5) 12.5 − 19.5(+10−17=5) 39.1%12.5 / 32 40.5% −82 1 = 1 0 1 = 0 0 1 0 0 0 1 0 = 0 0 0 0 0 1 0 1 0 0 = 1 = 1 0 0 1 – Fornax 1-2 64-bit 2292 +19−19 (−24) 16.5 − 15.5(+13−12=7) 51.6%16.5 / 32 87.1% −11 0 0 1 0 = 1 = 1 = 0 1 = 1 1 0 0 0 = 1 0 = 1 0 1 1 0 1 0 1 = 1 0 – FoxSEE 5.0.0 64-bit 2283 +19−19 (−33) 19 − 13(+15−9=8) 59.4%19.0 / 32 94.1% +35 1 1 0 0 1 1 1 0 1 1 = 1 = = 0 1 1 0 1 0 = 1 1 1 0 = 1 = = 0 = 0 – Wasabi 1.2.1 64-bit 2265 +21−21 (−51) 45.5 − 24.5(+40−19=11) 65.0%45.5 / 70 99.4% +65 1 1 = 1 0 = = 1 = 1 1 = = 0 = 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 1 0 1 0 0 0 1 1 1 1 1 1 1 1 1 0 0 1 0 1 0 1 1 1 0 1 1 0 0 0 = 1 1 1 = 1 1 1 = =\n\n### Rating changes by day", null, "### Rating changes with played games", null, "Created in 2005-2013 by CCRL team Last games added on January 9, 2021" ]	[ null, "http://computerchess.org.uk/ccrl/404/rating-history-by-day-graphs/Asterisk_0_6b.png", null, "http://computerchess.org.uk/ccrl/404/rating-history-by-day-graphs-2/Asterisk_0_6b.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5352794,"math_prob":0.99983555,"size":1348,"snap":"2021-04-2021-17","text_gpt3_token_len":992,"char_repetition_ratio":0.34598213,"word_repetition_ratio":0.39529413,"special_character_ratio":0.92284864,"punctuation_ratio":0.098712444,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98430216,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-15T23:35:20Z\",\"WARC-Record-ID\":\"<urn:uuid:39d81593-bf40-4fc6-956e-bf94c73a57cd>\",\"Content-Length\":\"14128\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1549d477-516c-4b0f-a53e-0c42fcabe198>\",\"WARC-Concurrent-To\":\"<urn:uuid:ad680508-639d-40fd-b407-eb8bd6fd4b86>\",\"WARC-IP-Address\":\"185.45.66.155\",\"WARC-Target-URI\":\"http://computerchess.org.uk/ccrl/404/cgi/engine_details.cgi?match_length=30&print=Details&each_game=1&eng=Asterisk%200.6b\",\"WARC-Payload-Digest\":\"sha1:KJ7XH4JWNFB2XWY5HPKEU4XLUDOWLUNS\",\"WARC-Block-Digest\":\"sha1:PESJ37B6CQ5GHWXSMDRCSCWK7WTRPQCH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703497681.4_warc_CC-MAIN-20210115224908-20210116014908-00504.warc.gz\"}"}
http://conversion.org/volume/fluid-dram-us/bushel-us-dry-level	[ "# fluid dram (US) to bushel (US dry level) conversion\n\nConversion number between fluid dram (US) [fl dr] and bushel (US dry level) [bu (US lvl)] is 0.00010490319914249. This means, that fluid dram (US) is smaller unit than bushel (US dry level).\n\n### Contents [show][hide]", null, "Switch to reverse conversion:\nfrom bushel (US dry level) to fluid dram (US) conversion\n\n### Enter the number in fluid dram (US):\n\nDecimal Fraction Exponential Expression\n [fl dr]\neg.: 10.12345 or 1.123e5\n\nResult in bushel (US dry level)\n\n?\n precision 0 1 2 3 4 5 6 7 8 9 [info] Decimal: Exponential:\n\n### Calculation process of conversion value\n\n• 1 fluid dram (US) = (exactly) (3.696691195312510^-06) / (0.03523907016688) = 0.00010490319914249 bushel (US dry level)\n• 1 bushel (US dry level) = (exactly) (0.03523907016688) / (3.696691195312510^-06) = 9532.5977489177 fluid dram (US)\n• ? fluid dram (US) × (3.6966911953125*10^-06 (\"m³\"/\"fluid dram (US)\")) / (0.03523907016688 (\"m³\"/\"bushel (US dry level)\")) = ? bushel (US dry level)\n\n### High precision conversion\n\nIf conversion between fluid dram (US) to cubic-metre and cubic-metre to bushel (US dry level) is exactly definied, high precision conversion from fluid dram (US) to bushel (US dry level) is enabled.\n\nDecimal places: (0-800)\n\nfluid dram (US)\nResult in bushel (US dry level):\n?\n\n### fluid dram (US) to bushel (US dry level) conversion chart\n\n Start value: [fluid dram (US)] Step size [fluid dram (US)] How many lines? (max 100)\n\nvisual:\nfluid dram (US)bushel (US dry level)\n00\n100.0010490319914249\n200.0020980639828499\n300.0031470959742748\n400.0041961279656997\n500.0052451599571247\n600.0062941919485496\n700.0073432239399745\n800.0083922559313994\n900.0094412879228244\n1000.010490319914249\n1100.011539351905674\nCopy to Excel\n\n## Multiple conversion\n\nEnter numbers in fluid dram (US) and click convert button.\nOne number per line.\n\nConverted numbers in bushel (US dry level):\nClick to select all\n\n## Details about fluid dram (US) and bushel (US dry level) units:\n\nConvert Fluid dram (US) to other unit:\n\n### fluid dram (US)\n\nDefinition of fluid dram (US) unit: ≡ 1⁄8 US fl oz. Fluid dram (or Drachm in UK spelling) = US fl oz / 8 = 3.6966911953125×10−6 m³\n\nConvert Bushel (US dry level) to other unit:\n\n### bushel (US dry level)\n\nDefinition of bushel (US dry level) unit: ≡ 2150.42 cu in. = 2150.42 × 0.0254³ = 0.03523907016688 m³", null, "" ]	[ null, "http://conversion.org/images/switch.png", null, "http://conversion.org/menufiles/top.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7180213,"math_prob":0.57854134,"size":1213,"snap":"2019-43-2019-47","text_gpt3_token_len":429,"char_repetition_ratio":0.2158809,"word_repetition_ratio":0.021857923,"special_character_ratio":0.5012366,"punctuation_ratio":0.13304721,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9759735,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-13T23:36:33Z\",\"WARC-Record-ID\":\"<urn:uuid:be40276a-17f4-4152-8602-53c655fcd61f>\",\"Content-Length\":\"35490\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:858dec35-68f9-420d-85c6-e1d9033f658c>\",\"WARC-Concurrent-To\":\"<urn:uuid:9438cb71-33b1-45d7-8f2d-e3ef9af0ae1a>\",\"WARC-IP-Address\":\"142.54.171.162\",\"WARC-Target-URI\":\"http://conversion.org/volume/fluid-dram-us/bushel-us-dry-level\",\"WARC-Payload-Digest\":\"sha1:FGKPMTSMKTC2FM6AXXPDVLCWX3RI3GLW\",\"WARC-Block-Digest\":\"sha1:7DJTBWSM6XLYDHQMKDCRDGLJ37FAKI3J\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496667442.36_warc_CC-MAIN-20191113215021-20191114003021-00285.warc.gz\"}"}
https://www.expertsmind.com/library/calculate-the-daily-market-return-over-the-last-five-years-51169072.aspx	[ "### Calculate the daily market return over the last five years\n\nAssignment Help Financial Management\n##### Reference no: EM131169072\n\nAssignment\n\nQuestion 1 Risk and Return\n\nCalculate the following using the data from Yahoo Finance (https://au.finance.yahoo.com/) for the company you selected for Question 1 of Assignment .\n\n1. Calculate the daily market return over the last five years from the daily prices, calculate the monthly returns from the daily returns, and calculate the yearly returns from the monthly returns.\n\n2. Calculate the total risk (i.e. yearly standard deviation of the daily returns).\n\n3. Calculate the yearly systematic / market risk using the daily returns of the stock and daily return of the market index.\n\n4. Calculate the unsystematic risk / firm specific risk. Suggest whether this company is a good investment. Answer the following questions while making your suggestion.\n\na) What is the basis for selection of this stock if you suggest this as a good investment?\n\nb) Would you invest all your money into this stock? If not, why not? How will you address this concern?\n\nQuestion 2 Capital Budgeting\n\nABC Ltd. would like to set up a new expansion plant. Currently, ABC has an option to buy an existing building at a cost of AUD 24 000. Necessary equipment for the plant will cost AUD 16 000, including installation costs. The economic life of the equipment and building are 5\nand 40 years, respectively. The project also requires an initial investment of AUD 12 000 in net working capital. The initial working capital investment will be made at the time of the purchase of the building and equipment.\n\nThe project's estimated economic life is four years. At the end of that time, the building is expected to have a market value of AUD 15 000 and a book value of AUD 21 600, whereas the equipment is expected to have a market value of AUD 4 000 and a book value of AUD 3\n200.\n\nAnnual sales will be AUD 80 000. The production department has estimated that variable manufacturing costs will total 60% of sales and that fixed overhead costs, excluding depreciation, will be AUD 10 000 a year. Depreciation expense will be determined for the year using straight line depreciation method.\n\nABC's tax rate is 40%; its cost of capital is 12%; and, for capital budgeting purposes, the company's policy is to assume that operating cash flows occur at the end of each year. The plant will begin operations immediately after the investment is made, and the first operating cash flows will occur exactly one year later.\n\nRequirements:\n\n1. Compute the initial investment outlay, operating cash flow over the project's life, and the terminal-year cash flows for ABC's expansion project.\n\n2. Determine whether the project should be accepted using NPV analysis.\n\n3. Do the sensitivity analysis using different levels of change (e.g. 2%, 5% and 10% increase and decrease) of each of the key inputs (e.g., sales, variable costs and cost of capital)\n\n4. Identify the most sensitive factor\n\n5. Perform the scenario analysis\n\nReturn and Risk Calculation\n\n1. Calculation of daily market returns for the company selected for Module 1 assignment for over 5 years What is daily market return?\nA: Percentage change in prices over one day holding period.\nHow to calculate?\n\nRt=((Pt-Pt-1)/Pt-1)\nWhere,\n\nPt = Price today\nPt-1= Price yesterday\n\nAlternatively:\n\nRt= ln(Pt/Pt-1)\n\nWhere,\n\nln = Natural logarithm\n\nWhat is the source of market price data?\nA: yahoo finance\n\n2. Calculation of monthly market returns from the daily market returns.\n\nHow to calculate monthly returns from daily returns?\n\nA: Demonstrated in the \"return calculation.xlsx\" file available under key resources.\n\nStep 1: Add 1 to the daily returns calculated using either Equation (1) or Equation (2)\n\nStep 2: Use the product function in Excel (i.e., = PRODUCT (select the daily returns in a month)\n\nStep 3: Subtract 1 from the product (I have combined step 2 and step 3 in my calculations)\n\n3. Calculation of yearly market returns from the monthly market returns.\n\nHow to calculate yearly market returns from the monthly market returns?\n\nStep 1: Add 1 to the monthly returns\n\nStep 2: Use the product function in Excel (i.e., = PRODUCT (select the 12 monthly returns in a year)\n\nStep 3: Subtract 1 from the product (I have combined step 2 and step 3 in my calculations)\n\n4. Should I do calculations for five years in one file?\n\nA: For simplicity, it is better to do the calculations for each year separately (so five files for five years). I am, however, happy if you can do all calculations in one file.\n\n5. Calculate yearly standard deviation (i.e. total risk) of the daily returns.\n\nHow to calculate standard deviation of the daily returns?\n\nA: Please use the Excel formula (=STDEV.S (select the range of daily returns))\n\n6. Calculate yearly market risk (or systematic risk).\n\nA: It is a yearly calculation using the daily returns of the company and daily returns of the market\n\nSteps in Excel:\n\ni) Data>Analysis>Regression (if you do not find analysis tab under Data, please add the analysis tool pack from options)\n\nii) Select the range of stock return as Y inputs\n\niii) Select the range of market returns as X inputs\n\niv) Tick the label box\n\nv)Select a cell for the output range\n\nvi) Click OK\n\nvii) The coefficient of the market return is the systematic risk (commonly known as beta). Please ignore the other statistics.\n\n SUMMARY OUTPUT Regression Statistics Multiple R 0.2154 R Square 0.0464 Adjusted R Square 0.0426 Standard Error 0.0265 Observations 252\n\n ANOVA df SS MS F Significance F Regression 1 0.0085 0.0085 12.1683 0.0006 Residual 250 0.1753 0.0007 Total 251 0.1838 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept -0.0015 0.0017 -0.8779 0.3808 -0.0048 0.0018 Market return 0.5428 0.1556 3.4883 0.0006 0.2364 0.8493\n\n7. How to calculate the daily market returns?\n\nStep 1: Please use the daily price of a market index (i.e., ASX All Ordinaries or ASX S&P200)\n\nStep 2: Use either Equation (1) or Equation (2) to calculate the daily market retruns.\n\n8. How to calculate the unsystematic risk (or firm specific risk)?\n\nStep 1: Use the following model to calculate the daily fitted returns (or forecasted returns) of the stock of the company\n\nE(Ri)= α +β Rm\n\nWhere,\n\nα = Intercept\nβ = Coefficient of X Variable 1 (i.e., beta)\nRm = Daily market return\n\nStep 2: Calculate the residuals (i.e., actual return minus expected return) for every day.\nε=Ri-E( Ri )\n\nStep 3: Calculate the yearly standard deviation of the daily residuals. This is defined as the unsystematic risk.\n\n### Write a Review\n\n#### Unsalable inventory-uncollectible accounts receivable\n\nFinancial statement analysis can be used to identify weaknesses in a firm’s operations. Uncollectible accounts receivable – you could compare the firm’s average collection period to other peer firms and look for trends over time that might suggest de..\n\n#### What is the macaulay duration\n\nThere is a 9 percent coupon bond with six years to maturity and a current price of \\$958.50. What is the dollar value of an 01 for the bond? You find a bond with 14 years until maturity that has a coupon rate of 8.2 percent and a yield to maturity of..\n\n#### Development and testing of the new app will take four months\n\nTake Five Systems, a new start-up, is developing a new iPhone application (“app”) and provides you with the following assumptions: Development and testing of the new app will take four months. Month five is the first month of revenue generation. Init..\n\n#### Basic trade-off between efficiency and equity\n\nThere is a basic trade-off between efficiency and equity because A. Income redistribution tends to reduce incentives for efficient behavior. B. People who are efficient dislike equity. C. Pareto improvements can only be made by sacrificing efficiency..\n\n#### What was the arithmetic average return on the stock\n\nYou’ve observed the following returns on Doyscher Corporation’s stock over the past five years: –12 percent, 21 percent, 27 percent, 6 percent, and 17 percent. What was the arithmetic average return on the stock over this five-year period? What was ..\n\n#### Declining growth stock valuation\n\nDeclining Growth Stock Valuation Brushy Mountain Mining Company's coal reserves are being depleted, so its sales are falling. Also, environmental costs increase each year, so its costs are rising. As a result, the company's earnings and dividends are..\n\n#### Capital asset pricing model and wacc-beta coefficient\n\nThe “beta” coefficient for ABC is 1.25 based on the past information. The 30-day T-bill rate is 1.5%, The 5-year average market return of (say, S&P 500 index) in the same period is 11%. Suppose that ABC's current dividend is \\$1.24 per share with poss..\n\n#### How reasoning errors influence financial decisions\n\nThe textbook describes the field of Behavioral Finance as the study of “how reasoning errors influence financial decisions.” In this context, explain the difference between biases, framing effects and heuristics with examples.\n\n#### Explain what is the price of the coupon bond\n\nWhat is the price of the coupon bond. What is the yield to maturity of the coupon bond. Under the expectations hypothesis, what is the expected realized compound yield of the coupon bond\n\n#### What would be the maintenance margin if a margin call\n\nAssume you sold short 100 shares of common stock at \\$50 per share. The initial margin is 60%. What would be the maintenance margin if a margin call is made at a stock price of \\$60?\n\n#### What annual rate of return is analyst assuming you can earn\n\nA financial analyst tells you that investing in stocks will allow you to double your money in 7 years. What annual rate of return is the analyst assuming you can earn?\n\n#### What will the spot rate be in one year according to the ife\n\nThe one-year interest rate is 11% in the United Kingdom and 7% in Singapore. What will the spot rate be in one year according to the IFE?", null, "" ]	[ null, "https://www.expertsmind.com/prostyles/images/3.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84803194,"math_prob":0.9208385,"size":6606,"snap":"2023-40-2023-50","text_gpt3_token_len":1656,"char_repetition_ratio":0.17540139,"word_repetition_ratio":0.08732877,"special_character_ratio":0.27066302,"punctuation_ratio":0.12930375,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98398983,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-07T06:01:49Z\",\"WARC-Record-ID\":\"<urn:uuid:aa385994-cff1-4a94-9639-68c829c10856>\",\"Content-Length\":\"69774\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:286bd6d8-65da-4c37-9788-9a6b6fa6669c>\",\"WARC-Concurrent-To\":\"<urn:uuid:c43f291b-725b-4968-828f-7c506bd9a474>\",\"WARC-IP-Address\":\"104.21.1.109\",\"WARC-Target-URI\":\"https://www.expertsmind.com/library/calculate-the-daily-market-return-over-the-last-five-years-51169072.aspx\",\"WARC-Payload-Digest\":\"sha1:44XM7LMXF64IB3OTPT6A7D54N2FK2P6I\",\"WARC-Block-Digest\":\"sha1:TDKZZAXQSSFYBUOKLLJJXWFKITJE3M4Q\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100650.21_warc_CC-MAIN-20231207054219-20231207084219-00076.warc.gz\"}"}
https://silosy-zbozowe.com.pl/coal/Oct-4755/	[ " how to calculate mill copper recovery\n\n## how to calculate mill copper recovery", null, "### Calculate Copper Wire Recovery Rates Gardner Metal\n\n2021-3-15 Fortunately, calculating the copper wire recovery percentage is a fairly simple process. Simply take a small segment of cable and weigh it. Take note of that number and then strip the casing so that you are left with only the copper wire at the core of the cable. Weigh the copper by itself, then divide that number by the overall weight of the ...\n\nMore", null, "### How to Calculate the Recovery Rate of your Scrap Copper ...\n\n2021-5-21 You calculate the recovery rate of your scrap copper wire by dividing the weight of the piece of copper before and after the rubber or plastic casing that encloses the wire itself has been removed. It’s also a good idea to have a piece of paper and a pen to help you record the weights. Then, you’re going to want to have a calculator close ...\n\nMore", null, "### How To Calculate Copper Wire Recovery Rates » Super\n\n2021-9-20 Fortunately, calculating the copper wire recovery percentage is a fairly simple process. Simply take a small segment of cable and weigh it. Take note of that number and then strip the casing so that you are left with only the copper wire at the core of the cable. Weigh the copper by itself, then divide that number by the overall weight of the ...\n\nMore", null, "### Metallurgical Accounting Formulas Concentration and ...\n\n2013-9-8 R = 100 c/Kf = recovery % By weights F and C, plus assays c and t; R = 100 Cc / (Cc+t(F—C)) = recovery % A copper concentrator is milling 15,000 tons/day of a chalcopyrite ore assaying 1.15% copper. The concentrate and tailings produced average 32.7% and 0.18% copper\n\nMore", null, "### Common Basic Formulas for Mineral Processing\n\n2016-3-20 Concentration and Recovery Formulas. These are used to compute the production of concentrate in a mill or in a particular circuit. The formulas are based on assays of samples, and the results of the calculations are generally accurate— as\n\nMore", null, "### Recovery of Copper - Valencia College\n\n2009-10-27 After our product has been weighed we will need to calculate the percent recovery. The equation for percent recovery can be seen in equation 7. (7) Percent Recovery = (mass of Cu recovered / mass of initial Cu) 100%. Procedure . 1. Weigh 0.35-0.40 g of copper wire and place into a 250 mL graduated beaker. 2.\n\nMore", null, "### How to Calculate Percent Recovery - Science Struck\n\nPercent recovery = (8.67 ÷ 11.23) × 100 = 77.20 %. 77.20% of zinc is recouped in this process. Problem II: 14.18 gm of copper is used for a recrystallization experiment. The amount of copper recovered at the end of the purification process is 18.29 gm. Calculate the percentage of copper\n\nMore", null, "### Solved Calculate the percent recovery of copper. Use the ...\n\nThis problem has been solved! See the answer. See the answer See the answer done loading. Calculate the percent recovery of copper. Use the formula % recovery= ending mass of copper (g)/inital mass of copper x 100%. Mass of empty centrifuge tube- 2.639 g. Mass of centrifuge + starting copper- 2.868 g. MAss of centrifuge + dried copper- 3.125 g.\n\nMore", null, "### How to Calculate Percent Recovery.\n\nFormula to calculate percent recovery. Example: Suppose you had 15g of blue Copper (II) sulfate, after heating it, you were left with 12.8g of white Copper (II) sulfate, Calculate the percent recovery of the compound. Thus, the percent recovery of the substance is 85.3%. Prev Article.\n\nMore", null, "### Calculate Copper Wire Recovery Rates Gardner Metal\n\n2021-3-15 Fortunately, calculating the copper wire recovery percentage is a fairly simple process. Simply take a small segment of cable and weigh it. Take note of that number and then strip the casing so that you are left with only the copper wire at the core of the cable. Weigh the copper by itself, then divide that number by the overall weight of the ...\n\nMore", null, "### Calculating Copper Recovery Rate For Scrap Cables\n\n2021-6-8 Once you have the weight of your copper wire on the inside of the cable, take that number and divide it by the overall weight of the sample and the result will be the percentage of copper recovery rate from your scrap cable or wire. You can bring that\n\nMore", null, "### OneClass: Calculate the percent recovery of copper. Use ...\n\nCalculate the percent recovery of copper. Use the formula % recovery= ending mass of copper (g)/inital mass of copper x 100%. Mass of empty centrifuge tube- 2.639 g. Mass of centrifuge + starting copper- 2.868 g. MAss of centrifuge + dried copper- 3.125 g. Would it be 2.12 or 212 %?\n\nMore", null, "### Recovery of Copper - Valencia College\n\n2009-10-27 After our product has been weighed we will need to calculate the percent recovery. The equation for percent recovery can be seen in equation 7. (7) Percent Recovery = (mass of Cu recovered / mass of initial Cu) 100%. Procedure . 1. Weigh 0.35-0.40 g of copper wire and place into a 250 mL graduated beaker. 2.\n\nMore", null, "### How to Calculate Percent Recovery - Science Struck\n\nPercent recovery = (8.67 ÷ 11.23) × 100 = 77.20 %. 77.20% of zinc is recouped in this process. Problem II: 14.18 gm of copper is used for a recrystallization experiment. The amount of copper recovered at the end of the purification process is 18.29 gm. Calculate the percentage of copper\n\nMore", null, "### Metal Recovery Rate - How to Interpreted the Mineral ...\n\nThe metal recovery rate can be found in a mining company’s National Instrument 43-101 or in a similar applicable international reporting standard. Example: When you read a press release in which a mining company announces a resource of 500 million tonnes of 1% copper and a mineral recovery\n\nMore", null, "### Estimation of Open Cut Mining Recovery and Mining\n\n2013-9-27 Mine to mill is a process for optimisation of mining and processing recovery and costs. The choice of either selective mining or bulk mining techniques is a major consideration, however evaluations rarely undertake careful consideration of Mining Recovery and Mining Dilution in the context of mine to mill optimisation.\n\nMore", null, "### Solved Calculate the percent recovery of copper. Use the ...\n\nThis problem has been solved! See the answer. See the answer See the answer done loading. Calculate the percent recovery of copper. Use the formula % recovery= ending mass of copper (g)/inital mass of copper x 100%. Mass of empty centrifuge tube- 2.639 g. Mass of centrifuge + starting copper- 2.868 g. MAss of centrifuge + dried copper- 3.125 g.\n\nMore", null, "### Copper recovery using leach/solvent extraction ...\n\n2009-8-27 Copper recovery by L/SX/EX in 1968 In 1968 there were only two widely practiced copper leaching processes using dilute sulphuric acid. The first process, vat leaching of high-grade copper oxide ore followed by EW of copper from the leach solution, produced low quality copper cathode at relatively high cost. In 1968 the tonnage of high-\n\nMore", null, "### What is a theoretical grade-recovery curve? An example ...\n\n2009-10-26 The theoretical grade-recovery curve for an ore is a definition of the maximum expected recovery by flotation of a mineral or element at a given grade. This is defined by the surface area liberation of the value minerals and is consequently directly related to the grind size utilised in the process. The theoretical grade-recovery can be readily used to quickly identify potential recovery ...\n\nMore", null, "### Calculate Copper Wire Recovery Rates Gardner Metal\n\n2021-3-15 Fortunately, calculating the copper wire recovery percentage is a fairly simple process. Simply take a small segment of cable and weigh it. Take note of that number and then strip the casing so that you are left with only the copper wire at the core of the cable. Weigh the copper by itself, then divide that number by the overall weight of the ...\n\nMore", null, "### The electrolytic recovery of copper from rod mill pickling ...\n\n2020-10-13 COPPER AVAILABLE FOR RECOVERY. The. two. fao. tors governing. the. amount of copper entering the piokling solution are (ll. the rod mill production rate, and (2) the amount of oopper oxide coating dissolved from each coil of rod. Rod Kl11. Production Rate; In es. t. imatlng the quant. i. ty. ot. oopper available. for. recovery annually. the ...\n\nMore", null, "### OneClass: Calculate the percent recovery of copper. Use ...\n\nCalculate the percent recovery of copper. Use the formula % recovery= ending mass of copper (g)/inital mass of copper x 100%. Mass of empty centrifuge tube- 2.639 g. Mass of centrifuge + starting copper- 2.868 g. MAss of centrifuge + dried copper- 3.125 g. Would it be 2.12 or 212 %?\n\nMore", null, "### Recovery of Copper - Valencia College\n\n2009-10-27 After our product has been weighed we will need to calculate the percent recovery. The equation for percent recovery can be seen in equation 7. (7) Percent Recovery = (mass of Cu recovered / mass of initial Cu) 100%. Procedure . 1. Weigh 0.35-0.40 g of copper wire and place into a 250 mL graduated beaker. 2.\n\nMore", null, "### How to Calculate Percent Recovery - Science Struck\n\nPercent recovery = (8.67 ÷ 11.23) × 100 = 77.20 %. 77.20% of zinc is recouped in this process. Problem II: 14.18 gm of copper is used for a recrystallization experiment. The amount of copper recovered at the end of the purification process is 18.29 gm. Calculate the percentage of copper\n\nMore", null, "### Copper recovery using leach/solvent extraction ...\n\n2009-8-27 Copper recovery by L/SX/EX in 1968 In 1968 there were only two widely practiced copper leaching processes using dilute sulphuric acid. The first process, vat leaching of high-grade copper oxide ore followed by EW of copper from the leach solution, produced low quality copper cathode at relatively high cost. In 1968 the tonnage of high-\n\nMore", null, "### Dilution and ore recovery - QueensMineDesignWiki\n\n2019-6-28 The generalized equation for recovery is given in the following equation: More specifically, ore recovery can be defined by the percentage of minable reserves extracted in the mining process. The issue of balancing dilution and ore recovery is a challenging one as profitability is to be optimized while not effecting the efficiency of operation.\n\nMore", null, "### Copper Recovery - INDUSTRIAL SCRAP RECYCLING\n\nINDUSTRIAL SCRAP. RECYCLING EQUIPMENT. FOR RESOURCE RECOVERY. Copper Recovery is a manufacturer of wire and cable recycling equipment, offering sales and service worldwide. We also act as agent or representative for some of the finest European manufacturers of recycling machinery. Most recently, we infused our knowledge and know-how gained over ...\n\nMore", null, "### Recommended machining parameters for copper and\n\n2021-8-12 for copper and copper alloys” contin-ues a long tradition established by the German Copper Institute (DKI). The publication “Processing Copper and Copper Alloys” (“Das Bearbeiten von Kupfer und Kupferlegierungen”) first appeared in 1938 and again in 1940. The handbook “Metal cutting tech-niques for copper and copper alloys”\n\nMore", null, "### Percent Recovery Calculator - Calculator Academy\n\n2021-10-13 Percent Recovery Formula. The following equation is used to calculate the percent recovery from a purification process. PR = SAP / SBP *100. Where PR s the percent recovery. SAP is the substance amount after purification. SBP is the\n\nMore" ]	[ null, "https://silosy-zbozowe.com.pl/pic/yt/69.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/109.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/94.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/7.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/161.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/98.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/124.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/134.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/177.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/69.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/166.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/173.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/98.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/124.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/43.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/150.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/134.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/7.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/189.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/69.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/56.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/173.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/98.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/124.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/7.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/3.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/131.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/142.jpg", null, "https://silosy-zbozowe.com.pl/pic/yt/25.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.878884,"math_prob":0.94997233,"size":10920,"snap":"2022-27-2022-33","text_gpt3_token_len":2576,"char_repetition_ratio":0.19961524,"word_repetition_ratio":0.5512465,"special_character_ratio":0.24624541,"punctuation_ratio":0.124824025,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9517807,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58],"im_url_duplicate_count":[null,5,null,1,null,3,null,3,null,1,null,5,null,5,null,2,null,6,null,5,null,1,null,5,null,5,null,5,null,1,null,1,null,2,null,3,null,1,null,5,null,1,null,5,null,5,null,5,null,3,null,1,null,4,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-05T05:54:41Z\",\"WARC-Record-ID\":\"<urn:uuid:ef320ea5-f645-48ea-8a3c-ce5f949b1a62>\",\"Content-Length\":\"23856\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:95032fdb-4c32-45f5-9b1e-f176090fe890>\",\"WARC-Concurrent-To\":\"<urn:uuid:84ac8f1a-ad3f-48fa-a8de-cd75e3b2a5ad>\",\"WARC-IP-Address\":\"172.67.157.221\",\"WARC-Target-URI\":\"https://silosy-zbozowe.com.pl/coal/Oct-4755/\",\"WARC-Payload-Digest\":\"sha1:VZKVEJMOYF5EKFMHBGJDV3OTJIFPPR7Q\",\"WARC-Block-Digest\":\"sha1:PTXOEYXYQ2RT2TLCM6TPMVTYF57BDAN6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104514861.81_warc_CC-MAIN-20220705053147-20220705083147-00779.warc.gz\"}"}
https://codereview.stackexchange.com/questions/187283/aoc-day-22-sporifica-virus-part-1-a-solution-in-beginners-haskell	[ "# AoC Day 22: Sporifica Virus Part 1, a Solution in Beginner's Haskell\n\nUntil recently, I was very much only devoted to imperative languages (mainly C++ and C, to be precise), when I decided to venture into unknown waters by picking up a new, completely different language, which happened to be Haskell, a decision which happened to be influenced by the fact that I owned a copy of \"Learn You a Haskell for Great Good!\", a book which I very much enjoyed learning from.\n\nSome days ago, I finished said book, and, wanting to apply my newly acquired knowledge, ventured our to find some programming exercises. I quickly remembered Advent of Code, which offers a whole pre-Christmas period's worth of easy to mildly difficult programming problems, a few of which I had already solved during the holidays.\n\nI skimmed through the exercises, looking for one simple enough to be conquerable with my still very inadequate and shaky Haskell skills, and finally chose the task of Day 22.\n\n# Problem Description\n\nDiagnostics indicate that the local grid computing cluster has been contaminated with the Sporifica Virus. The grid computing cluster is a seemingly-infinite two-dimensional grid of compute nodes. Each node is either clean or infected by the virus.\n\nTo prevent overloading the nodes (which would render them useless to the virus) or detection by system administrators, exactly one virus carrier moves through the network, infecting or cleaning nodes as it moves. The virus carrier is always located on a single node in the network (the current node) and keeps track of the direction it is facing.\n\nTo avoid detection, the virus carrier works in bursts; in each burst, it wakes up, does some work, and goes back to sleep. The following steps are all executed in order one time each burst:\n\n• If the current node is infected, it turns to its right. Otherwise, it turns to its left. (Turning is done in-place; the current node does not change.)\n• If the current node is clean, it becomes infected. Otherwise, it becomes cleaned. (This is done after the node is considered for the purposes of changing direction.)\n• The virus carrier moves forward one node in the direction it is facing. Diagnostics have also provided a map of the node infection status (your puzzle input).\n\nClean nodes are shown as .; infected nodes are shown as #. This map only shows the center of the grid; there are many more nodes beyond those shown, but none of them are currently infected.\n\nThe virus carrier begins in the middle of the map facing up.\n\n(The full puzzle description, including examples, is available on the official AoC website.)\n\nThe mentioned puzzle input consists of a file containing a grid of . and #.\n\n# My Solution\n\nimport Data.List.Index\nimport qualified Data.Set as Set\nimport System.Environment\nimport System.IO\nimport qualified System.IO.Strict as IOS\n\ntype Position = (Int, Int)\ntype Dimensions = (Int, Int)\ntype NodeMap = Set.Set Position\n\ndata Rotation = Clockwise \| Counterclockwise deriving (Eq, Show)\ndata Direction = North \| East \| South \| West deriving (Eq, Show)\n\nnextDirection :: Rotation -> Direction -> Direction\nnextDirection rot dir\n\| rot == Clockwise && dir == West = North\n\| rot == Clockwise && dir == East = South\n\| rot == Clockwise && dir == North = East\n\| rot == Clockwise && dir == South = West\n\| rot == Counterclockwise && dir == West = South\n\| rot == Counterclockwise && dir == East = North\n\| rot == Counterclockwise && dir == North = West\n\| rot == Counterclockwise && dir == South = East\n\nparseInput :: String -> (Dimensions, NodeMap)\nparseInput s = ((length . head . lines $s, length . lines$ s),\nfoldl (\\set (index, char) ->\nif char == '#'\nthen Set.insert index set\nelse set)\nSet.empty\n. ifoldl (\\ls index line ->\nls ++ zipWith (\\ix (i, char) ->\n((i, ix), char))\n(repeat index) (indexed line))\n[]\n. lines $s) simulateNBurstsImpl :: Int -> (Int, Position, Direction, NodeMap) -> (Int, Position, Direction, NodeMap) simulateNBurstsImpl 0 x = x simulateNBurstsImpl i (count, pos, dir, map) = simulateNBurstsImpl (i - 1) transitionFunction where step (a, b) dir \| dir == West = (a - 1, b) \| dir == East = (a + 1, b) \| dir == North = (a, b - 1) \| dir == South = (a, b + 1) transitionFunction \| Set.member pos map == True = let nextDir = nextDirection Clockwise dir in (count, step pos nextDir, nextDir, Set.delete pos map) \| Set.member pos map == False = let nextDir = nextDirection Counterclockwise dir in (count + 1, step pos nextDir, nextDir, Set.insert pos map) simulateNBursts :: Int -> Dimensions -> NodeMap -> Int simulateNBursts i (width, height) map = first (simulateNBurstsImpl i (0, startingPos, North, map)) where startingPos = (width quot 2, height quot 2) first (a, _, _, _) = a main = getArgs >>= \\(filename : _) -> withFile filename ReadMode IOS.hGetContents >>= \\content -> let parsed = parseInput content in print . simulateNBursts 10000 (fst parsed) . snd$ parsed\n\n\nNotes:\n\n• The program takes a single argument on the commandline; the path to the file containing the starting grid.\n• I decided on using a set as the underlying data structure as it makes it easy to work with growing grids. My first attempt used a two-dimensional sequence, but extending the grid turned out to be too much of a hassle for my taste.\n• The code assumes that all inputs are valid, including the fact that a commandline parameter was passed and points to a valid file.\n\n# Review Requests\n\nPlease feel free to review anything and everything that comes to mind! That said, I do have a few concrete questions:\n\n1. Having simulateNBursts as a beautified interface to simulateNBurstsImpl seems kind of ugly to me. Is there a way to clean this up and join the two functions? Or is this a common pattern?\n2. How readable is this code? As its author, I find it hard to judge how (un-)pleasant this code is to the eye of a third person, especially since I have next to no experience in reading and writing Haskell code. What can I do to improve readability?\n3. nextDirection seems very verbose to me. Is there a more concise way to implement it?\n\n## 1 Answer\n\nYou can inline simulateNBurstsImpl if you get rid of the recursion:\n\nsimulateNBurstsImpl n = foldr (.) id $replicate n transitionFunction Combining zipWith with repeat is a fool's errand. (I don't know where you get ifoldl and indexed, so I'll assume they start at 0.) parseInput :: String -> (Dimensions, NodeMap) parseInput s = ((length . head$ lines s, length $lines s), Set.fromList [ (x, y) \| (y, line) <- zip [0..]$ lines s\n, (x, char) <- zip [0..] line\n, char == '#'\n])\n\n\nDirection and Rotation can be directly represented as offsets and functions on offsets.\n\nimport Data.NumInstances.Tuple\n\ntype Rotation = Direction -> Direction\ntype Direction = (Int, Int)\n\ntransitionFunction (count, pos, (dx,dy), map) = if Set.member pos map\nthen let nextdir = (-dy,dx)\nin (count , pos + nextDir, nextDir, Set.delete pos map)\nelse let nextdir = (dy,-dx)\nin (count + 1, pos + nextDir, nextDir, Set.insert pos map)\n\n\nOptionally: lens and State specialize in this fiddly sort of stuff.\n\ndata S = S\n{ _count :: Int\n, _pos :: Position\n, _dir :: Int\n, _map :: NodeMap\n}\n\nmakeLenses ''S\n\nsimulateNBursts i (width, height) map =\n(evalState (0, (width quot 2, height quot 2), (0,-1), map)) $do replicateM_ i$ do\nhashtagged <- map . contains pos <<%= negate\nnextDir <- dir <%= \\(x,y) -> if hashtagged then (-y,x) else (y,-x)\npos += nextDir\nunless hashtagged \\$ count += 1\nuse count\n\n• Thank you for the answer. ifoldl and indexed are from the ilist package (from Data.List.Index and start, as you guessed, from 0. – Ben Steffan Feb 11 '18 at 16:23" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89705175,"math_prob":0.89612436,"size":5957,"snap":"2019-35-2019-39","text_gpt3_token_len":1421,"char_repetition_ratio":0.10431715,"word_repetition_ratio":0.05167464,"special_character_ratio":0.25029376,"punctuation_ratio":0.13748854,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9543783,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-21T01:47:36Z\",\"WARC-Record-ID\":\"<urn:uuid:736ee9d1-a3f4-43e5-a30b-9be3934cabd5>\",\"Content-Length\":\"143965\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:435ad3f4-e415-47d3-ae1f-5cbf77783f53>\",\"WARC-Concurrent-To\":\"<urn:uuid:8ebb9ebb-f4e5-4882-9d90-5d49062351c5>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://codereview.stackexchange.com/questions/187283/aoc-day-22-sporifica-virus-part-1-a-solution-in-beginners-haskell\",\"WARC-Payload-Digest\":\"sha1:J7TJ2L7RYLYMNIAJKLYV4R2JR5RGR5QS\",\"WARC-Block-Digest\":\"sha1:QRJMNSS6VRGY7BCOQ3KO5QJNL3A2UWBA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027315695.36_warc_CC-MAIN-20190821001802-20190821023802-00307.warc.gz\"}"}
https://docs.derivative.ca/LSystem_SOP	[ "# LSystem SOP\n\n## Summary\n\nThe Lsystem SOP implements L-systems (Lindenmayer-systems, named after Aristid Lindenmayer (1925-1989)), allow definition of complex shapes through the use of iteration. They use a mathematical language in which an initial string of characters is evaluated repeatedly, and the results are used to generate geometry. The result of each evaluation becomes the basis for the next iteration of geometry, giving the illusion of growth.\n\nYou begin building an L-system by defining a sequence of rules which are evaluated to produce a new string of characters. Each character of the new string represents one command which affects an imaginary stylus, or \"turtle\". Repeating this process will grow your geometry.\n\nYou can use L-systems to create things such as:\n\n• Create organic objects such as trees, plants, flowers over time.\n• Create animated branching objects such as lightning and snowflakes.\n\nThe file can be read in from disk or from the web. Use http:// when specifying a URL.\n\n### The Algorithmic Beauty of Plants\n\nThe descriptions located here should be enough to get you started in writing your own L-system rules, however, if you have any serious interests in creating L-systems, you should obtain the book:\n\n```The Algorithmic Beauty of Plants\nPrzemyslaw Prusinkiewicz & Aristid Lindenmayer\nSpringer-Verlag, New York, Phone: 212.460.1500\nISBN: 0-387-94676-4, 1996.\n```\n\nwhich is the definitive reference on the subject. It contains a multitude of L-systems examples complete with descriptions of the ideas and theories behind modelling realistic plant growth.\n\n## Parameters - Geometry Page\n\nType `type` - - Provides two options for output geometry:\n\n• Skeleton `skel` - Creates wire frame geometry. This option is ideal for geometry that is stiff and jagged like lightning or snowflakes. It is also useful to reduce SOP cooking time.\n• Tube `tube` - Creates tube geometry. This option can be used with solid geometry that would need smooth curves, like trees or shrubs. Parameters on the Tube Page are only enabled when this Type is selected.\n\nGenerations `generations` - Determines the number of times to apply the rules to the initial string. This value controls the growth of the L-system. Place a time-based function here to animate the L-system growth.\n\nRandom Scale `randscale` - Random Scale as a percentage. This will apply a random scale to the changing geometry's lengths, angles and thickness.\n\nRandom Seed `randseed` - Random Seed for the SOP. This value can be used to select different sequences of random values.\n\nContinuous Angles `contangl` - Calculates the incremental angles of branches, if a non-integer generational value is used. If the Generations field is animating, this should be set to ensure smooth growth.\n\nContinuous Length `contlength` - Calculates the incremental lengths of the geometry points if a non-integer generational value is used. As with Continuous Angles, if the Generations field is animating, this should be set to ensure smooth, continuous growth. The Continuous Width field applies to tube thickness.\n\nContinuous Width `contwidth` - Calculates the incremental lengths of the geometry points if a non-integer generational value is used. As with Continuous Angles, if the Generations field is animating, this should be set to ensure smooth, continuous growth. The Continuous Width field applies to tube thickness.\n\nApply Color `docolor` - Use a TOP to apply color to the L-system as it grows.\n\nImage File `colormap` - Defines a TOP to use when the Apply Color button is selected. Also see the ` and # turtle operators.\n\nUV Increment `inc` - - Defines the default color U, V index increments when the turtle symbols ` or # are used.\n\n• `incu` -\n• `incv` -\n\nPoint Width Attribute `pointwidth` - Adds a point `width` attribute to each point in the geometry. This width is effected by the Thickness and Thickness Scale parameters on the Tube Page.\n\n## Parameters - Tube Page\n\nThe parameters on this page are active only if Geometry Page > Type has been set to the Tube type.\n\nRows `rows` - The first option sets the number of tube sides and the second sets the number of divisions per step length if tube geometry is selected.\n\nColumns `cols` - The first option sets the number of tube sides and the second sets the number of divisions per step length if tube geometry is selected.\n\nTension `tension` - Tension defines the smoothness of branching corners.\n\nBranch Blend `smooth` - Enabling this option allows a child branch to be continuously joined to its parent branch.\n\nThickness `thickinit` - This number defines the default tube thickness.\n\nThickness Scale `thickscale` - This number is the scale factor used with the ! or ? operator.\n\nApply Tube Texture Coordinates `dotexture` - When enabled, UV texture coordinates are applied to the tube segments, such that the texture wraps smoothly and continuously over branches.\n\nVertical Increment `vertinc` - Defines the vertical spacing of texture coordinates over tube geometry when tube texture is applied.\n\n## Parameters - Values Page\n\nStep Size `stepinit` - Step Size allows you to define the default length of the edges when new geometry is generated.\n\nStep Size Scale `stepscale` - Step Size Scale defines the scale by which the geometry will be modified by the \" or _ (double quote, or underscore) turtle operators.\n\nAngle `angleinit` - Angle defines the default turning angle for turns, rolls and pitches.\n\nAngle Scale `anglescale` - Angle Scale allows you to enter the scaling factor to be employed when the ; or @ operators are used.\n\nVariable b `varb` - Substitutes user-defined b, c and d variables in rules or premise. These variables are expanded and so may include system variables such as `\\$F` and `\\$T`.\n\nVariable c `varc` - Substitutes user-defined b, c and d variables in rules or premise. These variables are expanded and so may include system variables such as `\\$F` and `\\$T`.\n\nVariable d `vard` - Substitutes user-defined b, c and d variables in rules or premise. These variables are expanded and so may include system variables such as `\\$F` and `\\$T`.\n\nGravity `gravity` - This parameter determines the amount of gravity applied to the geometry via the T (tropism vector) turtle operator. Tropism is when a plant bends or curves in response to an external stimulus. L-systems employ a tropism vector to simulate this behaviour. The bending is characterised by the fact that the thicker or shorter parts bend less than the longer or thinner parts.\n\n## Parameters - Funcs Page\n\nThe parameters on this page allow you to stamp your leaf geometry (each copy can be different) as opposed to simply copying them. See the example in Example - Stamping L-system Leaves.\n\nPic Image TOP `pictop` - This is the TOP which the pic() function uses. See #Expressions L-system Specific Expression Functions below.\n\nGroup Prefix `grpprefix` - If the production g(n) is encountered, all subsequent geometry is included in a primitive group prefixed with this label and ending with the ascii value of n. See #CreateGroup Creating Groups within L-systems below for an example.\n\nChannel Prefix `chanprefix` - If the expression chan(n) is encountered, it is replaced with the local channel prefixed with this label and ending with the ascii value of n.\n\nLeaf Param A `stampa` - You can determine which parameters are used by leaves.\nSee #CreateGroup Creating Groups within L-systems below for an example.\n\nLeaf Param B `stampb` - You can determine which parameters are used by leaves.\nSee #CreateGroup Creating Groups within L-systems below for an example.\n\nLeaf Param C `stampc` - You can determine which parameters are used by leaves.\nSee #CreateGroup Creating Groups within L-systems below for an example.\n\nRules DAT `rules` - Path to the DAT defining the rules for the LSystem.\n\n• Context Ignore `context_ignore:` - Defining this in the Rules DAT specifies all characters which are to be skipped when testing context sensitivity in the rules below.\n• Premise `premise:` - Define an initial string of characters to which the substitution rules are applied.\n• Rules - This is where the turtle substitution rules are defined.\n\n## Rule Substitution\n\nYou create the highly structured organic and branching objects using L-systems grammar. An L-system is a process in which a sequence of rules are applied to an initial string of characters to create a new string. To build the geometry, each character of the final string represents one command which affects an imaginary stylus, or \"turtle\".\n\nThe process begins by examining the first character of the premise string. All sixteen rules are searched sequentially until an applicable rule is found. The current character is then replaced with one or more characters as defined by the rule. The remaining characters in the premise string are replaced in a similar fashion. The entire process is repeated once for each generation.\n\n### Limitations to Rules\n\n• Polygon {} and branch [] operators can be nested 30 levels deep\n• Rules can be 256 characters in length\n• Variables can have up to 5 parameters\n• Up to 25 rules can be defined\n\n### Turtle Operators\n\n`F` Move forward (creating geometry)\n\n`H` Move forward half the length (creating geometry)\n\n`G` Move forward but don't record a vertex\n\n`f` Move forward (no geometry created)\n\n`h` Move forward a half length (no geometry created)\n\n`J K M` Copy geometry source J, K or M at the turtle's position\nafter rescaling and reorienting the geometry.\n\n`T` Apply tropism vector\n\n`+` Turn right\n\n`-` Turn left (minus sign)\n\n`&` Pitch up\n\n`^` Pitch down\n\n`\\` Roll clockwise\n\n`/` Roll counter-clockwise\n\n`\|` Turn 180 degrees\n\n`` Roll 180 degrees\n\n`~` Pitch / Roll / Turn random amount\n\n`\"` Multiply current length\n\n`!` Multiply current thickness\n\n`;` Multiply current angle\n\n`_` Divide current length (underscore)\n\n`?` Divides current width\n\n`@` Divide current angle\n\n`'` Increment color index U (single quote)\n\n`#` Increment color index V\n\n`%` Cut off remainder of branch\n\n`\\$` Rotate `up' towards the sun about heading\n\n`[` Push turtle state (start a branch)\n\n`]` Pop turtle state (end a branch)\n\n`{` Start a polygon\n\n`.` Make a polygon vertex\n\n`}` End a polygon\n\n`g` Create a new primitive group to which subsequent geometry is added\n\n## Production Rule Syntax\n\nA Touch L-system rule is specified as:\n\n[lc<] pred [>rc] [:cond]=succ [:prob]\n\nwhere:\n\n• lc - Optional left context\n• pred - predecessor symbol to be replaced\n• rc - Optional right context\n• cond - Condition expression (optional)\n• succ - Replacement string\n• prob - Probability of rule execution (optional)\n\n### Context Sensitivity\n\nThe most basic type of rule is:\n\npred = succ\n\nIn this case, a character is replaced with the characters of succ if, and only if, it matches pred.\n\nFor example:\n\nPremise: `ABC`\nRule 1: `B=DOG`\n\nwill result in `ADOGC`\n\npred can only specify one letter, but left and right context symbols can be specified. The general syntax is [lc<] pred [>rc] = succ.\n\nFor example:\n\nPremise: `ABC`\nRule 1: `A<B=DOG`\n\nagain results in `ADOGC` because `B` is preceded by `A`. If the rule were: `Z<B=DOG` or `B>A=DOG` the rule would not be applied.\n\n## Parameter Symbols\n\nEach symbol can have up to five user-defined variables associated with it which can be referenced or assigned in expressions. Variables in the predecessor are instanced while variables in the successor are assigned.\n\nFor example:\n\nThe rule `A(i, j) = A(i+1, j-1)`, will replace each `A` with a new `A` in which the first parameter has been incremented and the second parameter decremented.\n\nNote that the variables in the predecessor can also be referenced by the condition or probability portions of the rule.\n\nFor example:\n\nThe rule `A(i):i<5 = A(i+1) A(i+1)`, will double each `A` a maximum of five times (assuming a Premise of `A(0)`).\n\nParameters assigned to geometric symbols (e.g. `F`, `+`, or `!`) are interpreted geometrically.\n\nFor example:\n\nThe rule `F(i, j) = F(0.5i, 2j)`, will again replace each `F` with a new `F` containing modified parameters. In addition to this, the new `F` will now be drawn at half the length and twice the width.\n\n## Operator Override\n\nNormally turtle symbols use the current length/angle/thickness etc. to determine their effect. By providing a turtle operator with an explicit parameter, it will override the value normally used by the turtle operator.\n\nOverride parameters for `F`, `f`, `G`, `h`, `H` take the form of:\n\n`F(i,j,k,l,m)`\n\n• `i` - Override Length.\n• `j` - Override Thickness.\n• `k` - Override # Tube Segments.\n• `l` - Override # Tube Rows.\n• `m` - User parameter.\n\nThe k and l override parameters allow dynamic resolution of tube segments.\n\n### Examples\n\n`F` - Moves forward current length creating geometry.\n\n`H` - Moves forward half current length creating geometry.\n\n`F(i, j)` - Moves forward a distance of `i,` creating geometry of thickness `j`.\n\n`H(i, j)` - Move forward half the distance of `i`, creating geometry of thickness half of `j`.\n\n`+` - Turn by current angle amount.\n`~` - Rotate by random angle.\n\n`+(i)` - Turn by i degrees.\n`~(i)` - Override random angle with value of `i`.\n\n`\\$(x0,y0,z0)` - Points the turtle to location `(x0,y0,z0)`.\n\nGiven the above, the Premise:\n\n`F(1) +(90) F(1) +(90) F(1) +(90) F(1)`\n\ngenerates a unit box regardless of the default Step Size or Angle settings.\n\n### List of Operator Overrides\n\nThe following list describes the geometric interpretation of parameters assigned to certain turtle symbols:\n\n## Edge Rewriting\n\nIn The Algorithmic Beauty of Plants, many examples use a technique called Edge Rewriting which involve left and right subscripts. A typical example might look like:\n\nGenerations `10`\nAngle `90`\nPremise `F(l)`\nRules\n`F(l) = F(l)+F(r)+ F(r) = -F(l)-F(r)`\n\nHowever, Touch doesn't know what `F(l)` and `F(r)` are. In this case, we can modify the rules to use parameter passing. For the `F` turtle symbol, the first four parameters are length, width, tubesides, and tubesegs, leaving the last parameter user-definable. We can define this last parameter such that `0` is left, and `1` is right:\n\nGenerations `10`\nAngle `90`\nPremise `F(1,1,3,3,0)`\nRules\n\nAfter two generations this produces: `Fl+Fr+-Fl-Fr`. There should not be any difference between this final string and `F+F+-F-F.`\n\nAnother approach is to use two new variables, and use a conditional statement on the final step to convert them to `F`:\n\nVariable b `ch(\"generations\")`\nPremise `l`\nRules\n`l:t<b=l+r+ r:t<b=-l-r l=F r=F`\nOutput `l`\n`F`\n`F+F+`\n`F+F++-F-F+`\n\n## Expressions\n\nIn the earlier example, the expressions `0.5i` and `2j` are used. In fact, expressions can be used anywhere a numeric field is expected. Currently the following symbols can be used in expressions:\n\n`( )` - brackets for nesting priority\n\n`^ + - / %` - arithmetic operators\n\n`min() max() sin() cos() asin() acos() pic() in()` - supported functions\n\n`== != = < <= > >=` - logical operators\n\n`& \| !` - logical operators: and, or, not\n\n`b c d` - SOP b, c, d parameters after expansion\n\n`x y z` - current turtle position\n\n`g` - age of symbol\n\n`t` - time (generations) of L-system\n\n`a` - SOP angle parameter\n\n`T` - SOP tropism (gravity) parameter\n\nThe pre-defined variables above should not be used in the arguments of the predecessor.\n\nFor example: `A(a,b) = B(a2,b2)` is wrong (a is the SOP Angle parameter). `a<b (A,B) = b(A+1,B)` is right.\n\nThe last statement is correct because `a` and `b` are used as symbols and not variable names. `A` and `B` are correct because variable names are case sensitive.\n\n### L-System Specific Expression Functions\n\n• `pic(u, v, c)` - Using the image specified with Pic Image File, this function returns a normalized value (between `0` and `1`) of the pixel at the normalized coordinates `(u,v)`; `c` selects one of four channels to examine:\n`0`\n`1`\n`2`\n`3`\n• `in(q, r, s)` - Given a MetaTest input source containing a metaball geometry, this function returns a `1` if the point `(q, r, s)` is contained within the metaball, and `0` if not. Use `in(x, y, z)` (the letters `x`, `y`, and `z` are special and contain the X, Y, and Z location of the turtle) to test whether or not the turtle is currently inside the metaball to create pruned outputs.\n\n### Conditions\n\nEach rule may have an optional condition specified. The syntax is:\n\n[lc<] pred [>rc]:cond = succ\n\nFor example:\n\nThe Rule1 `A:y>2=J` includes source `J` at all `A`'s above the height of `2`.\n\n### Probability\n\nEach rule can specify the probability that it is used (provided it is otherwise applicable). The syntax is:\n\n[lc<] pred [>rc]:cond=succ=prob\n\nFor example:\n\nRule1 `A=B:0.5`\nRule2 `A=C:0.5 `\n\nwill replace `A` with either `B` or ` C` with equal probability.\n\n## Creating Groups with L-Systems\n\nThere is a group operator '`g`' which lumps all geometry currently being built into group g.\n\nFor example: `g[F]` lumps geometry from F into a group called `lsys0`. You can set the lsys prefix from the Funcs page.\n\n### Optional Group Parameters\n\n`g` takes an optional parameter as well.\n\nFor example: `g(1)[F]` lumps geometry from `f` into a group called `lsys1`. If no parameter is given, the default index is bumped up appropriately.\n\nThe current group container is pushed/popped with the turtle state so you can do things like:\n\n`gF [ gFF] F `- The first and last `F`'s are put into group `0`, and the middle `FF`'s are put into group `1`.\n\n`gF [ FF] F` - The geometry from all four `F`'s are put into group `0`, (pushing the turtle adopts the parent's group).\n\nTo exclude the middle `FF` from the parent's group type: `gF [ g(-1)FF ] F`\n\n## Controlling the Length over Time\n\nTo create an L-system which goes forward X percent less for each iteration, you need to start your Premise with a value, and then within a rule, multiply that value by the percentage you want to remain:\n\nPremise `A(1)`\nRule `A(i) = F(i)A(i*0.5)`\n\nThis way \"`i`\" is scaled before `A` is again evaluated. The important part is the Premise. You need to start with a value to be able to scale that value.\n\n## Example\n\nStep 1) Place a Circle SOP, and set the Number of Divisions to`: param(\"lsys\", 3`)\n\nIt then displays a triangle (3 is default value).\n\nStep 2) Pipe this into the J input of a L-System SOP. If the L-system Premise is:\n\n`J A`\n`J(,,4) A`\n`J(,,5) A`\n\nThis way, you can customize each leaf before it gets copied.\n\nStep 3) Change the Premise and Rule to:\n\nPremise`A(0)`\nRule1`A(i)=FJ(,,i+3)A(i+1)`\n\nThis creates a line of increasing-order polygons.\n\nStep 4) Finally, we will want to create 20 leaves, and put them all into a Switch SOP. Do this by entering the following expression into the Switch SOP's Select Input: `param(\"lsys\",0)`\nStep 5) Then in your L-system, J(,,0) gives you the first SOP, J(,,1) gives you the second, and so on. This solves the problem of a limited number of leaves using only JKM.\n\nAlso note that these examples use only the first stamp parameter, you can use up to three parameters: e.g. J(,,1,2,3)\n\nThe first two parameters of J, K, M are used to override length and width, like symbol F.\n\n• Input 0 -\n• Input 1 -\n• Input 2 -\n• Input 3 -\n\n## Info CHOP Channels\n\nExtra Information for the LSystem SOP can be accessed via an Info CHOP.\n\n### Common SOP Info Channels\n\n• num_points - Number of points in this SOP.\n• num_prims - Number of primitives in this SOP.\n• num_particles - Number of particles in this SOP.\n• last_vbo_update_time - Time spent in another thread updating geometry data on the GPU from the SOP's CPU data. As it is part of another thread, this time is not part of the usual frame time.\n• last_meta_vbo_update_time - Time spent in another thread updating meta surface geometry data (such as metaballs or nurbs) on the GPU from the SOP's CPU data. As it is part of another thread, this time is not part of the usual frame time.\n\n### Common Operator Info Channels\n\n• total_cooks - Number of times the operator has cooked since the process started.\n• cook_time - Duration of the last cook in milliseconds.\n• cook_frame - Frame number when this operator was last cooked relative to the component timeline.\n• cook_abs_frame - Frame number when this operator was last cooked relative to the absolute time.\n• cook_start_time - Time in milliseconds at which the operator started cooking in the frame it was cooked.\n• cook_end_time - Time in milliseconds at which the operator finished cooking in the frame it was cooked.\n• cooked_this_frame - 1 if operator was cooked this frame.\n• warnings - Number of warnings in this operator if any.\n• errors - Number of errors in this operator if any.\n\nTouchDesigner Build:\n\nAdd • Alembic • Align • Arm • Attribute Create • Attribute • Basis • Blend • Bone Group • Boolean • Box • Experimental:Box • Bridge • Cache • Cap • Capture Region • Capture • Carve • CHOP to • Circle • Experimental:Circle • Clay • Clip • Convert • Copy • CPlusPlus • Creep • Curveclay • Curvesect • DAT to • Deform • Delete • Divide • Extrude • Face Track • Facet • File In • Fillet • Fit • Font • Force • Fractal • Grid • Experimental:Grid • Group • Hole • Import Select • In • Introduction To s Vid • Inverse Curve • Iso Surface • Join • Joint • Kinect • Lattice • Limit • Line • Line Thick • LOD • LSystem • Magnet • Material • Merge • Metaball • Model • Noise • Null • Object Merge • Oculus Rift • OpenVR • Out • Particle • Point • Polyloft • Polypatch • Polyreduce • Polyspline • Polystitch • Primitive • Profile • Project • Rails • Raster • Ray • Rectangle • Experimental:Rectangle • Refine • Resample • Revolve • Script • Select • Sequence Blend • Skin • Sort • Sphere • Experimental:Sphere • Spring • Sprinkle • Sprite • Stitch • Subdivide • Superquad • Experimental:Superquad • Surfsect • Sweep • Switch • Text • Texture • Torus • Experimental:Torus • Trace • Trail • Transform • Trim • Tristrip • Tube • Experimental:Tube • Twist • Vertex • Wireframe • ZED" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7875646,"math_prob":0.87549835,"size":20031,"snap":"2022-05-2022-21","text_gpt3_token_len":4818,"char_repetition_ratio":0.12872621,"word_repetition_ratio":0.14794755,"special_character_ratio":0.23977834,"punctuation_ratio":0.10526316,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98141855,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-17T14:00:21Z\",\"WARC-Record-ID\":\"<urn:uuid:c1ed4b8d-ba35-4704-a2a6-dc20b12474a3>\",\"Content-Length\":\"83161\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ebd4c70f-9b55-4abe-9b97-395f4be17fa8>\",\"WARC-Concurrent-To\":\"<urn:uuid:4465b195-cb42-4168-8c6f-e92ec83ecb69>\",\"WARC-IP-Address\":\"3.15.54.19\",\"WARC-Target-URI\":\"https://docs.derivative.ca/LSystem_SOP\",\"WARC-Payload-Digest\":\"sha1:ESJXSMBNXJRK22744UYZJKI5NPWOENHQ\",\"WARC-Block-Digest\":\"sha1:RLYN36FPZOKL2IAD5EH3KB4LXVSWZR2F\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662517485.8_warc_CC-MAIN-20220517130706-20220517160706-00419.warc.gz\"}"}
https://aalopes.com/blog/?p=291	[ "### Implementing mod function in C\n\nSomething some people don’t realize is that C and several other programming languages don’t offer you the standard modulo operation but simply a remainder operation (this issue has been in any case discussed countless times, e.g. see, although most of the times without offering a solution).\n\nThe problem arises when people try to use the remainder operator, % or", null, "$\\text{rem}$, for the modulo without realizing the implications of doing so.\nI recall when a programmer I knew tried to write C code for a steering angle sensor system using the % operator for the mod operation (which was needed by the algorithm), leading to erratic and difficult to debug results (in this case, a function that was supposed to be smooth exhibited erratic jumps), leaving me the job to understand the algorithm and then find the bug.\nNote that the C standard defines the remainder of an integer division via the % operator for positive integers. For negative integers the operation is implementation dependent and only fixed by the C99 standard or newer. Note that the remainder can be negative but the modulo can’t, and that", null, "$\\mod(a,b)$ is contained in", null, "$[0,b[$:\n\nThe question is then, if the remainder operation, % here also denoted as rem is available, how to leverage this to calculate the modulo operation? And here’s one way of doing it:", null, "$\\mod (a,b) = \\begin{cases} 0, & a = 0 \\vee b = 0\\\\ b - \\text{rem}(-a -1, b) -1, &a<0 \\\\ \\text{rem} (a,b), &a>0\\end{cases}$.\nNote that this is only valid for a non-negative divisor", null, "$b$, however the generalization is rather trivial and as is common practice to say, left as an exercise for the reader :)." ]	[ null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9398961,"math_prob":0.98257655,"size":1507,"snap":"2021-21-2021-25","text_gpt3_token_len":297,"char_repetition_ratio":0.13040586,"word_repetition_ratio":0.0,"special_character_ratio":0.20238885,"punctuation_ratio":0.09122807,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9865527,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-18T14:13:38Z\",\"WARC-Record-ID\":\"<urn:uuid:0c6f3db5-c7fe-4b25-8f40-d98eed960328>\",\"Content-Length\":\"19140\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8e9fe2c1-3a5d-4759-a4c2-10d6691531a7>\",\"WARC-Concurrent-To\":\"<urn:uuid:bed3ed01-d7ee-4e4b-8e50-05ab6f989b59>\",\"WARC-IP-Address\":\"188.93.230.163\",\"WARC-Target-URI\":\"https://aalopes.com/blog/?p=291\",\"WARC-Payload-Digest\":\"sha1:MJHFZMROIWNVIFD5K7OJZQL3DJZMZTQI\",\"WARC-Block-Digest\":\"sha1:3BMWD57PUW32MDDBECKH6WZDAIU6W4NL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243989637.86_warc_CC-MAIN-20210518125638-20210518155638-00346.warc.gz\"}"}
https://www.enoisecontrol.com/reverberation-time-gymnasiums-multi-purpose-rooms/	[ "## Reverberation Time for Gymnasiums and Multi-Purpose Rooms\n\nThe below excerpt from a booklet on classroom acoustics published by the Acoustical Society of America can help understand reverberation time for community and industrial applications. Although this was written about classrooms, the principle is the same for all indoor spaces.\n\nSource: Acoustical Society of America, A publication of the technical committee on architectural acoustics of the Acoustical Society of America. https://asa.aip.org/classroom/booklet.html\n\nOver 100 years ago, a Harvard physics professor named Wallace Clement Sabine developed the first equation for reverberation time, which has since been named after him and is still used today. Reverberation time is defined as the length of time required for sound to decay 60 dB from its initial level. Sabine’s simple formula is", null, "where:\n\nRT (60) = reverberation time (seconds)\n\nV = room volume (cubic feet)\n\nS = surface area (square feet)\n\n? = absorption coefficient of material(s) at given frequency\n\n? indicates the summation of S times ? for all room surfaces\n\nTo use this formula, the volume of the room, surface area of each material in the room, and absorption coefficients for those materials must be known. Absorption coefficients are measured in specialized laboratories, and represent the fraction of sound energy (not sound level – dB) the material will absorb as a decimal from 0 to 1. Figure 15 gives absorption coefficients for common classroom materials.", null, "A commonly used one-number rating called NRC, noise reduction coefficient, is simply the average of the absorption coefficients at 250, 500, 1000, and 2000 Hz. This simple, one-number rating can be useful for comparing the relative absorption of two materials; however, examining absorption coefficients in each octave band gives a better idea of the performance of a material at various frequencies.\n\nReverberation time is often calculated with the room unoccupied. Since people and their clothing provide additional sound absorption, an unoccupied room is the worst-case scenario, though not an unreasonable one, since occupancy of most classrooms varies. In a complete analysis, this calculation should be performed for each octave band, as the RT can vary widely at different frequencies. However, for a quick estimate, the RT of a classroom can be calculated for just one octave band representative of speech frequencies, such as 1000 Hz. If this RT is acceptable, then the RT throughout the speech range will likely be acceptable." ]	[ null, "https://www.enoisecontrol.com/wp-content/uploads/2015/01/sabine-formula.jpg", null, "https://www.enoisecontrol.com/wp-content/uploads/2015/01/absorption-coefficients.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9296257,"math_prob":0.91018623,"size":2522,"snap":"2023-40-2023-50","text_gpt3_token_len":513,"char_repetition_ratio":0.12986498,"word_repetition_ratio":0.0,"special_character_ratio":0.19547978,"punctuation_ratio":0.11060948,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96365225,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,5,null,5,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-26T22:34:02Z\",\"WARC-Record-ID\":\"<urn:uuid:89fb727c-8949-45e9-994a-5f444b0a407d>\",\"Content-Length\":\"188038\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a636a187-4fa9-41a8-80fd-5a66e0383708>\",\"WARC-Concurrent-To\":\"<urn:uuid:f6839a1b-b3f8-4666-a847-1f191b7ed916>\",\"WARC-IP-Address\":\"64.255.252.21\",\"WARC-Target-URI\":\"https://www.enoisecontrol.com/reverberation-time-gymnasiums-multi-purpose-rooms/\",\"WARC-Payload-Digest\":\"sha1:T54NGXIGWEUHHL4ESWSNH6OCU55QUDSY\",\"WARC-Block-Digest\":\"sha1:BHHQ7E7TIBTNMK66MBH2ZC7EAKGV646X\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510225.44_warc_CC-MAIN-20230926211344-20230927001344-00686.warc.gz\"}"}
https://wintertalesevents.com/ounces-in-a-pound/	[ "# How many ounces in a pound\n\nA pound is a unit of weight in the United States customary measurement system. One of the most common questions that people have when purchasing items by weight is, “How many ounces in a pound?” The answer to this question is particularly helpful when calculating the price per pound. It is equivalent to sixteen ounces. Therefore, one pound is equal to sixteen ounces. In other words, there are sixteen ounces in one pound. One pound is a unit of weight in the United States customary measurement system. It equals sixteen ounces, the same as four-quarters, eight-eighths, or sixteen sixteenths.\n\nIn other words, there are sixteen ounces in a pound. A pound is a unit of weight equal to sixteen ounces in the United States standard measurement system. An ounce is 16 ounces, therefore a pound is 16 ounces. This can also be expressed as four-quarters, eight-eighths, or sixteen-sixteenths. The traditional system of measurement used in the United States defines a pound as a unit of weight equal to sixteen ounces. Consequently, one pound contains sixteen ounces. This can also be expressed as four-quarters, eight-eighths, or sixteen-sixteenths.\n\nWhat are ounces?\n\nOunces are a unit of weight measurement used in the imperial and United States customary systems. The weight of an ounce is approximately 28.35 grams. Ounces are commonly used to measure the weight of food, liquids, and other small objects.\n\nWhat is a pound?\n\nThe Imperial system of measurement uses pounds to measure mass or weight. It is equal to 16 ounces and is commonly used to measure a variety of items, including food, medicines, and other materials. The pound is widely used in the United States and around the world in a variety of industries, including retail, health care, and manufacturing. It is also used to measure the weight of people, animals, and other objects. In the United States, the pound is often abbreviated as “lb”.\n\nIs there a difference between imperial and metric units?\n\n1. The Imperial System is based on units of measurement like feet, inches, ounces, pounds, and gallons. The Metric System is based on units of measurement like meters, liters, grams, and kilograms.\n\n2. The Imperial System uses ounces to measure a pound, while the Metric System uses grams to measure a kilogram.\n\n3. The Imperial System is used mainly in the United States, while the Metric System is used by most of the rest of the world.\n\n4. The Imperial System is more complicated and less precise than the Metric System.\n\n5. The Imperial System is based on the British Imperial System of measurement, while the Metric System is based on the metric system of measurement.\n\nOunces in a pound conversion table" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9306485,"math_prob":0.9786085,"size":2915,"snap":"2023-40-2023-50","text_gpt3_token_len":702,"char_repetition_ratio":0.16901408,"word_repetition_ratio":0.14481409,"special_character_ratio":0.24493997,"punctuation_ratio":0.123333335,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9606183,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-01T15:48:27Z\",\"WARC-Record-ID\":\"<urn:uuid:e0b14a78-c838-48e5-b081-ae15af429c3d>\",\"Content-Length\":\"86110\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2e9c987c-f9b4-4eb4-b145-fd2edf1bc7ff>\",\"WARC-Concurrent-To\":\"<urn:uuid:cbaa9350-f795-4d9f-b2e9-4345546d8549>\",\"WARC-IP-Address\":\"172.67.203.230\",\"WARC-Target-URI\":\"https://wintertalesevents.com/ounces-in-a-pound/\",\"WARC-Payload-Digest\":\"sha1:CEKMOXIE5DFT6WXLB5M4WTRGIF7FHMOG\",\"WARC-Block-Digest\":\"sha1:BUTDTEQXYACHPE22V7QFCIS4N6FW72TH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510903.85_warc_CC-MAIN-20231001141548-20231001171548-00834.warc.gz\"}"}
https://www.etoolsage.com/Calculator/MolarMass.asp?CalText=c6h5n2	[ "Home\|Add to Favorites", null, "\|Add to My Toolkit Petroleum & Chemicals", null, "Molar Mass Calculator", null, "\|Register\|Sign in\|Customization", null, "# Calculate molar mass\n\nEnter a chemical formula in the field below, pay attention to formula is case sensitive!!!\n\nThe formula should be entered as:\nH2SO4= H2SO4\nCuSO4.5H2O=CuSO4.5H2O\n3BeO.Al2O3.6SiO2= (BeO)3.Al2O3.6SiO2\n Calculations:Formula: c6h5n2Molar Mass: 105.1195 g/mol 1g=9.51298284333544E-03 molPercent composition (by mass):Element Count Atom Mass %(by mass)c 6 12.011 68.56%h 5 1.0079 4.79%n 2 14.007 26.65%\n Top Use:", null, "Molar Mass Calculator Formula: Co2(Cr2O7)3Recent user inquiry:2020/6/2 15:08Molar Mass Calculator Formula: Kr2020/6/2 15:07Molar Mass Calculator Formula: n2h42020/6/2 15:06Molar Mass Calculator Formula: (NH4)6.Mo7.O24.4H2O2020/6/2 15:06Molar Mass Calculator Formula: C6H6O32020/6/2 15:06Molar Mass Calculator Formula: (K)2.MnO4" ]	[ null, "https://www.etoolsage.com/images/noteD.gif", null, "https://www.etoolsage.com/images/arrowRight.gif", null, "https://www.etoolsage.com/images/DownArrow.gif", null, "https://www.etoolsage.com/images/eTools.gif", null, "https://www.etoolsage.com/images/rightArrow.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.54526174,"math_prob":0.9585135,"size":423,"snap":"2020-24-2020-29","text_gpt3_token_len":152,"char_repetition_ratio":0.13365155,"word_repetition_ratio":0.0,"special_character_ratio":0.38297874,"punctuation_ratio":0.1724138,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9916902,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-02T07:08:23Z\",\"WARC-Record-ID\":\"<urn:uuid:70206f8a-6859-4677-a279-ad039c1a9ad9>\",\"Content-Length\":\"26152\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9c2024c3-fdea-4e98-b255-963ad20e78de>\",\"WARC-Concurrent-To\":\"<urn:uuid:b63f1225-99ad-483d-a68c-a43d833f798d>\",\"WARC-IP-Address\":\"172.67.157.159\",\"WARC-Target-URI\":\"https://www.etoolsage.com/Calculator/MolarMass.asp?CalText=c6h5n2\",\"WARC-Payload-Digest\":\"sha1:YVSXOAHW3X3NNUOBOSNMH2XU3WPALPUN\",\"WARC-Block-Digest\":\"sha1:B4CKCIK4AYPBCEO3R46DYF62FIFZA3TN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347423915.42_warc_CC-MAIN-20200602064854-20200602094854-00353.warc.gz\"}"}
https://alien.ren/article/7.html	[ "## ES6 对数组的扩展\n\nAlien\| 阅读：1795 发表时间：2018-04-08 10:37:08 JavaScript\n\n``````var arr=Array.of(1,2,3,4)\nconsole.log(arr) // [1,2,3,4]``````\n\n1.可以将类似数组的对象或者可遍历的对象转换成真正的数组。\n\n``````let ele = document.getElementsByTagName('a');\nele instanceof Array; //结果：false，不是数组\nele instanceof Object; //结果:true，是对象\nArray.from(ele) instanceof Array; //结果：true，是数组``````\n\n2.将字符串转换成数组\n\n`````` let str='hello';\nlet newarr=Array.from(str)\nconsole.log(newarr) // ['h','e','l','l','o']``````\n\n``````let arr=[1,3,5,6,7]\nlet x=arr.find(function(num){\nreturn num<8 // 1\nnum>2 //3\nnum>9 // underfined\n})``````\n\n``````let arr=[1,3,5,6,7];\nlet x=arr.findIndex(function(num){\nreturn num<8 // 0\nnum>2 // 1\nnum>9 // -1\n})``````\n\n``````let arr=[1,3,5,6,7]\narr.fill(10,0,1)\nconsole.log(arr) // [10,3,5,6,7]``````\n\n``````for(let[key,value] of ['aa','bb'].entries()){\nconsole.log(key,value); // 0 'aa' 1 'bb'\n}``````\n\n``````for(let index of ['aa','bb'].keys()){\nconsole.log(index); // 0 1\n}``````\n\n``````for(let value of ['aa','bb'].values()){\nconsole.log(value); // aa bb\n}``````\n\n*本文由Alien发表并编辑，转载此文章请附上出处及本页链接。如有侵权，请联系本站删除。" ]	[ null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.5623868,"math_prob":0.9792599,"size":1344,"snap":"2020-45-2020-50","text_gpt3_token_len":669,"char_repetition_ratio":0.11492537,"word_repetition_ratio":0.0,"special_character_ratio":0.34151787,"punctuation_ratio":0.22408026,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98479545,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-21T11:01:58Z\",\"WARC-Record-ID\":\"<urn:uuid:19cf1916-2762-46a7-850c-066f433cc97e>\",\"Content-Length\":\"17202\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a03d68e2-7731-4ac6-94bc-071cbe63c9f5>\",\"WARC-Concurrent-To\":\"<urn:uuid:b3c1c4f2-3d52-4fbf-8eb9-6a5783069a4d>\",\"WARC-IP-Address\":\"47.106.69.132\",\"WARC-Target-URI\":\"https://alien.ren/article/7.html\",\"WARC-Payload-Digest\":\"sha1:ITYNWYO4KRJQYLO4YAFWPSDSKSASAN4Z\",\"WARC-Block-Digest\":\"sha1:ZZBTPDMUJNDZCRT7HFLJW6M74TC544GW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107876307.21_warc_CC-MAIN-20201021093214-20201021123214-00182.warc.gz\"}"}
https://itectec.com/matlab/matlab-multiple-plot-lines-on-a-single-plot/	[ "# MATLAB: Multiple Plot lines on a single plot\n\nMATLABmultiple plotsplot linesplotting coordinates\n\nHi\nI would like to join x,y coordinates for 2 points together on a plot. But also plot all 361 lines.\nI have all coordinates saved as 361 x 1\nx coordinate left = xl\ny coordinate left = yl\nx coordinate right = xr\ny coordinate right = yr\nI can combine into 2 columns if required so which will determine a point on a plot a(1), a(2), b(1), b(2) etc to 361\na = [xl, yl]\nb = [xr, yr]\nBut I basically need to plot a line between\na(1) and b(1)\na(2) and b(2)\nand so on for 361 seperate lines on 1 plot.\nIs this possible and is there a quicker way than using hold on 361 times?\n``x = [xl.'; xr.']; % xl, xr, yl and yr should be column vectorsy = [yl.'; yr.'];plot(x, y)``" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.81053454,"math_prob":0.99893636,"size":609,"snap":"2020-34-2020-40","text_gpt3_token_len":186,"char_repetition_ratio":0.14214876,"word_repetition_ratio":0.0,"special_character_ratio":0.3136289,"punctuation_ratio":0.07042254,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9986536,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-09T21:02:51Z\",\"WARC-Record-ID\":\"<urn:uuid:b46bb214-0c68-4152-a9ab-d75f1f9c29f3>\",\"Content-Length\":\"12672\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:98c9956d-1e9f-483e-baf1-48ebebecd833>\",\"WARC-Concurrent-To\":\"<urn:uuid:f94a66de-d76c-44e3-8c72-16e4473d9b3b>\",\"WARC-IP-Address\":\"45.76.174.189\",\"WARC-Target-URI\":\"https://itectec.com/matlab/matlab-multiple-plot-lines-on-a-single-plot/\",\"WARC-Payload-Digest\":\"sha1:57R53LSVEZYNK4JKE2FMLYMBXCF2CZOT\",\"WARC-Block-Digest\":\"sha1:KXHO7OQKSBZDA2Z2TOUGS32ONWTRF6KZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738573.99_warc_CC-MAIN-20200809192123-20200809222123-00485.warc.gz\"}"}