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http://rosettacode.org/wiki/Draw_a_rotating_cube | Draw a rotating cube | Task
Draw a rotating cube.
It should be oriented with one vertex pointing straight up, and its opposite vertex on the main diagonal (the one farthest away) straight down. It can be solid or wire-frame, and you can use ASCII art if your language doesn't have graphical capabilities. Perspective is optional.
Related tasks
Draw a cuboid
write language name in 3D ASCII
| #JavaScript | JavaScript | <!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<style>
canvas {
background-color: black;
}
</style>
</head>
<body>
<canvas></canvas>
<script>
var canvas = document.querySelector("canvas");
canvas.width = window.innerWidth;
canvas.height = window.innerHeight;
var g = canvas.getContext("2d");
var nodes = [[-1, -1, -1], [-1, -1, 1], [-1, 1, -1], [-1, 1, 1],
[1, -1, -1], [1, -1, 1], [1, 1, -1], [1, 1, 1]];
var edges = [[0, 1], [1, 3], [3, 2], [2, 0], [4, 5], [5, 7], [7, 6],
[6, 4], [0, 4], [1, 5], [2, 6], [3, 7]];
function scale(factor0, factor1, factor2) {
nodes.forEach(function (node) {
node[0] *= factor0;
node[1] *= factor1;
node[2] *= factor2;
});
}
function rotateCuboid(angleX, angleY) {
var sinX = Math.sin(angleX);
var cosX = Math.cos(angleX);
var sinY = Math.sin(angleY);
var cosY = Math.cos(angleY);
nodes.forEach(function (node) {
var x = node[0];
var y = node[1];
var z = node[2];
node[0] = x * cosX - z * sinX;
node[2] = z * cosX + x * sinX;
z = node[2];
node[1] = y * cosY - z * sinY;
node[2] = z * cosY + y * sinY;
});
}
function drawCuboid() {
g.save();
g.clearRect(0, 0, canvas.width, canvas.height);
g.translate(canvas.width / 2, canvas.height / 2);
g.strokeStyle = "#FFFFFF";
g.beginPath();
edges.forEach(function (edge) {
var p1 = nodes[edge[0]];
var p2 = nodes[edge[1]];
g.moveTo(p1[0], p1[1]);
g.lineTo(p2[0], p2[1]);
});
g.closePath();
g.stroke();
g.restore();
}
scale(200, 200, 200);
rotateCuboid(Math.PI / 4, Math.atan(Math.sqrt(2)));
setInterval(function() {
rotateCuboid(Math.PI / 180, 0);
drawCuboid();
}, 17);
</script>
</body>
</html> |
http://rosettacode.org/wiki/Element-wise_operations | Element-wise operations | This task is similar to:
Matrix multiplication
Matrix transposition
Task
Implement basic element-wise matrix-matrix and scalar-matrix operations, which can be referred to in other, higher-order tasks.
Implement:
addition
subtraction
multiplication
division
exponentiation
Extend the task if necessary to include additional basic operations, which should not require their own specialised task.
| #Phix | Phix | constant m = {{7, 8, 7},{4, 0, 9}},
m2 = {{4, 5, 1},{6, 2, 1}}
?{m,"+",m2,"=",sq_add(m,m2)}
?{m,"-",m2,"=",sq_sub(m,m2)}
?{m,"*",m2,"=",sq_mul(m,m2)}
?{m,"/",m2,"=",sq_div(m,m2)}
?{m,"^",m2,"=",sq_power(m,m2)}
?{m,"+ 3 =",sq_add(m,3)}
?{m,"- 3 =",sq_sub(m,3)}
?{m,"* 3 =",sq_mul(m,3)}
?{m,"/ 3 =",sq_div(m,3)}
?{m,"^ 3 =",sq_power(m,3)}
|
http://rosettacode.org/wiki/Draw_a_sphere | Draw a sphere | Task
Draw a sphere.
The sphere can be represented graphically, or in ASCII art, depending on the language capabilities.
Either static or rotational projection is acceptable for this task.
Related tasks
draw a cuboid
draw a rotating cube
write language name in 3D ASCII
draw a Deathstar
| #BASIC | BASIC | clg
color white
rect 0,0,graphwidth, graphheight
for n = 1 to 100
color rgb(2*n,2*n,2*n)
circle 150-2*n/3,150-n/2,150-n
next n |
http://rosettacode.org/wiki/Doubly-linked_list/Element_insertion | Doubly-linked list/Element insertion | Doubly-Linked List (element)
This is much like inserting into a Singly-Linked List, but with added assignments so that the backwards-pointing links remain correct.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack | #FreeBASIC | FreeBASIC | #define FIL 1
#define DATO 2
#define LPREV 3
#define LNEXT 4
Dim Shared As Integer countNodes, Nodes
countNodes = 0
Nodes = 10
Dim Shared As Integer list(Nodes, 4)
list(0, LNEXT) = 1
Function searchNode(node As Integer) As Integer
Dim As Integer Node11
For i As Integer = 0 To node-1
Node11 = list(Node11, LNEXT)
Next i
Return Node11
End Function
Sub insertNode(node As Integer, newNode As Integer)
Dim As Integer Node11, i
If countNodes = 0 Then node = 2
For i = 1 To Nodes
If Not list(i, FIL) Then Exit For
Next
list(i, FIL) = True
list(i, DATO) = newNode
Node11 = searchNode(node)
list(i, LPREV) = list(Node11, LPREV)
list(list(i, LPREV), LNEXT) = i
If i <> Node11 Then list(i, LNEXT) = Node11 : list(Node11, LPREV) = i
countNodes += 1
If countNodes = Nodes Then Nodes += 10 : Redim list(Nodes, 4)
End Sub
Sub removeNode(n As Integer)
Dim As Integer Node11 = searchNode(n)
list(list(Node11, LPREV), LNEXT) = list(Node11, LNEXT)
list(list(Node11, LNEXT), LPREV) = list(Node11, LPREV)
list(Node11, LNEXT) = 0
list(Node11, LPREV) = 0
list(Node11, FIL) = False
countNodes -= 1
End Sub
Sub printNode(node As Integer)
Dim As Integer Node11 = searchNode(node)
Print list(Node11, DATO);
Print
End Sub
Sub traverseList()
For i As Integer = 1 To countNodes
printNode(i)
Next i
End Sub
insertNode(1, 1000)
insertNode(2, 2000)
insertNode(2, 3000)
traverseList()
removeNode(2)
Print
traverseList()
Sleep |
http://rosettacode.org/wiki/Doubly-linked_list/Traversal | Doubly-linked list/Traversal | Traverse from the beginning of a doubly-linked list to the end, and from the end to the beginning.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Delphi | Delphi | uses system ;
type
// declare the list pointer type
plist = ^List ;
// declare the list type, a generic data pointer prev and next pointers
List = record
data : pointer ;
prev : pList ;
next : pList ;
end;
// since this task is just showing the traversal I am not allocating the memory and setting up the root node etc.
// Note the use of the carat symbol for de-referencing the pointer.
begin
// beginning to end
while not (pList^.Next = NIL) do pList := pList^.Next ;
// end to beginning
while not (pList^.prev = NIL) do pList := pList^.prev ;
end; |
http://rosettacode.org/wiki/Doubly-linked_list/Traversal | Doubly-linked list/Traversal | Traverse from the beginning of a doubly-linked list to the end, and from the end to the beginning.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #E | E | def traverse(list) {
var node := list.atFirst()
while (true) {
println(node[])
if (node.hasNext()) {
node := node.next()
} else {
break
}
}
while (true) {
println(node[])
if (node.hasPrev()) {
node := node.prev()
} else {
break
}
}
} |
http://rosettacode.org/wiki/Doubly-linked_list/Element_definition | Doubly-linked list/Element definition | Task
Define the data structure for a doubly-linked list element.
The element should include a data member to hold its value and pointers to both the next element in the list and the previous element in the list.
The pointers should be mutable.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #C.23 | C# | class Link
{
public int Item { get; set; }
public Link Prev { get; set; }
public Link Next { get; set; }
//A constructor is not neccessary, but could be useful
public Link(int item, Link prev = null, Link next = null) {
Item = item;
Prev = prev;
Next = next;
}
} |
http://rosettacode.org/wiki/Doubly-linked_list/Element_definition | Doubly-linked list/Element definition | Task
Define the data structure for a doubly-linked list element.
The element should include a data member to hold its value and pointers to both the next element in the list and the previous element in the list.
The pointers should be mutable.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #C.2B.2B | C++ | template <typename T>
struct Node
{
Node* next;
Node* prev;
T data;
}; |
http://rosettacode.org/wiki/Dutch_national_flag_problem | Dutch national flag problem |
The Dutch national flag is composed of three coloured bands in the order:
red (top)
then white, and
lastly blue (at the bottom).
The problem posed by Edsger Dijkstra is:
Given a number of red, blue and white balls in random order, arrange them in the order of the colours in the Dutch national flag.
When the problem was first posed, Dijkstra then went on to successively refine a solution, minimising the number of swaps and the number of times the colour of a ball needed to determined and restricting the balls to end in an array, ...
Task
Generate a randomized order of balls ensuring that they are not in the order of the Dutch national flag.
Sort the balls in a way idiomatic to your language.
Check the sorted balls are in the order of the Dutch national flag.
C.f.
Dutch national flag problem
Probabilistic analysis of algorithms for the Dutch national flag problem by Wei-Mei Chen. (pdf)
| #Ceylon | Ceylon | import ceylon.random {
DefaultRandom
}
abstract class Colour(name, ordinal) of red | white | blue satisfies Comparable<Colour> {
shared String name;
shared Integer ordinal;
string => name;
compare(Colour other) => this.ordinal <=> other.ordinal;
}
object red extends Colour("red", 0) {}
object white extends Colour("white", 1) {}
object blue extends Colour("blue", 2) {}
Colour[] allColours = `Colour`.caseValues;
shared void run() {
function ordered({Colour*} colours) =>
colours.paired.every(([c1, c2]) => c1 <= c2);
value random = DefaultRandom();
function randomBalls(Integer length = 15) {
while (true) {
value balls = random.elements(allColours).take(length);
if (!ordered(balls)) {
return balls.sequence();
}
}
}
function dutchSort({Colour*} balls, Colour mid = white) {
value array = Array { *balls };
if (array.empty) {
return [];
}
variable value i = 0;
variable value j = 0;
variable value n = array.size - 1;
while (j <= n) {
assert (exists ball = array[j]);
if (ball < mid) {
array.swap(i, j);
i ++;
j ++;
}
else if (ball > mid) {
array.swap(n, j);
n --;
}
else {
j ++;
}
}
return array;
}
function idiomaticSort({Colour*} balls) =>
balls.sort(increasing);
value initialBalls = randomBalls();
"the initial balls are not randomized"
assert (!ordered(initialBalls));
print(initialBalls);
value sortedBalls1 = idiomaticSort(initialBalls);
value sortedBalls2 = dutchSort(initialBalls);
"the idiomatic sort didn't work"
assert (ordered(sortedBalls1));
"the dutch sort didn't work"
assert (ordered(sortedBalls2));
print(sortedBalls1);
print(sortedBalls2);
} |
http://rosettacode.org/wiki/Dutch_national_flag_problem | Dutch national flag problem |
The Dutch national flag is composed of three coloured bands in the order:
red (top)
then white, and
lastly blue (at the bottom).
The problem posed by Edsger Dijkstra is:
Given a number of red, blue and white balls in random order, arrange them in the order of the colours in the Dutch national flag.
When the problem was first posed, Dijkstra then went on to successively refine a solution, minimising the number of swaps and the number of times the colour of a ball needed to determined and restricting the balls to end in an array, ...
Task
Generate a randomized order of balls ensuring that they are not in the order of the Dutch national flag.
Sort the balls in a way idiomatic to your language.
Check the sorted balls are in the order of the Dutch national flag.
C.f.
Dutch national flag problem
Probabilistic analysis of algorithms for the Dutch national flag problem by Wei-Mei Chen. (pdf)
| #Clojure | Clojure | (defn dutch-flag-order [color]
(get {:red 1 :white 2 :blue 3} color))
(defn sort-in-dutch-flag-order [balls]
(sort-by dutch-flag-order balls))
;; Get a collection of 'n' balls of Dutch-flag colors
(defn random-balls [num-balls]
(repeatedly num-balls
#(rand-nth [:red :white :blue])))
;; Get random set of balls and insure they're not accidentally sorted
(defn starting-balls [num-balls]
(let [balls (random-balls num-balls)
in-dutch-flag-order? (= balls
(sort-in-dutch-flag-order balls))]
(if in-dutch-flag-order?
(recur num-balls)
balls))) |
http://rosettacode.org/wiki/Draw_a_cuboid | Draw a cuboid | Task
Draw a cuboid with relative dimensions of 2 × 3 × 4.
The cuboid can be represented graphically, or in ASCII art, depending on the language capabilities.
To fulfill the criteria of being a cuboid, three faces must be visible.
Either static or rotational projection is acceptable for this task.
Related tasks
draw a sphere
draw a rotating cube
write language name in 3D ASCII
draw a Deathstar
| #C | C |
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
const char *shades = ".:!*oe&#%@";
void vsub(double *v1, double *v2, double *s) {
s[0] = v1[0] - v2[0];
s[1] = v1[1] - v2[1];
s[2] = v1[2] - v2[2];
}
double normalize(double * v) {
double len = sqrt(v[0]*v[0] + v[1]*v[1] + v[2]*v[2]);
v[0] /= len; v[1] /= len; v[2] /= len;
return len;
}
double dot(double *x, double *y) {
return x[0]*y[0] + x[1]*y[1] + x[2]*y[2];
}
double * cross(double x[3], double y[3], double s[3]) {
s[0] = x[1] * y[2] - x[2] * y[1];
s[1] = x[2] * y[0] - x[0] * y[2];
s[2] = x[0] * y[1] - x[1] * y[0];
return s;
}
double* madd(double *x, double *y, double d, double *r) {
r[0] = x[0] + y[0] * d;
r[1] = x[1] + y[1] * d;
r[2] = x[2] + y[2] * d;
return r;
}
double v000[] = { -4, -3, -2 };
double v100[] = { 4, -3, -2 };
double v010[] = { -4, 3, -2 };
double v110[] = { 4, 3, -2 };
double v001[] = { -4, -3, 2 };
double v101[] = { 4, -3, 2 };
double v011[] = { -4, 3, 2 };
double v111[] = { 4, 3, 2 };
typedef struct {
double * v[4];
double norm[3];
} face_t;
face_t f[] = {
{ { v000, v010, v110, v100 }, { 0, 0, -1 } },
{ { v001, v011, v111, v101 }, { 0, 0, 1 } },
{ { v000, v010, v011, v001 }, { -1, 0, 0 } },
{ { v100, v110, v111, v101 }, { 1, 0, 0 } },
{ { v000, v100, v101, v001 }, { 0, -1, 0 } },
{ { v010, v110, v111, v011 }, { 0, 1, 0 } },
};
int in_range(double x, double x0, double x1) {
return (x - x0) * (x - x1) <= 0;
}
int face_hit(face_t *face, double src[3], double dir[3], double hit[3], double *d)
{
int i;
double dist;
for (i = 0; i < 3; i++)
if (face->norm[i])
dist = (face->v[0][i] - src[i]) / dir[i];
madd(src, dir, dist, hit);
*d = fabs(dot(dir, face->norm) * dist);
if (face->norm[0]) {
return in_range(hit[1], face->v[0][1], face->v[2][1]) &&
in_range(hit[2], face->v[0][2], face->v[2][2]);
}
else if (face->norm[1]) {
return in_range(hit[0], face->v[0][0], face->v[2][0]) &&
in_range(hit[2], face->v[0][2], face->v[2][2]);
}
else if (face->norm[2]) {
return in_range(hit[0], face->v[0][0], face->v[2][0]) &&
in_range(hit[1], face->v[0][1], face->v[2][1]);
}
return 0;
}
int main()
{
int i, j, k;
double eye[3] = { 7, 7, 6 };
double dir[3] = { -1, -1, -1 }, orig[3] = {0, 0, 0};
double hit[3], dx[3], dy[3] = {0, 0, 1}, proj[3];
double d, *norm, dbest, b;
double light[3] = { 6, 8, 6 }, ldist[3], decay, strength = 10;
normalize(cross(eye, dy, dx));
normalize(cross(eye, dx, dy));
for (i = -10; i <= 17; i++) {
for (j = -35; j < 35; j++) {
vsub(orig, orig, proj);
madd(madd(proj, dx, j / 6., proj), dy, i/3., proj);
vsub(proj, eye, dir);
dbest = 1e100;
norm = 0;
for (k = 0; k < 6; k++) {
if (!face_hit(f + k, eye, dir, hit, &d)) continue;
if (dbest > d) {
dbest = d;
norm = f[k].norm;
}
}
if (!norm) {
putchar(' ');
continue;
}
vsub(light, hit, ldist);
decay = normalize(ldist);
b = dot(norm, ldist) / decay * strength;
if (b < 0) b = 0;
else if (b > 1) b = 1;
b += .2;
if (b > 1) b = 0;
else b = 1 - b;
putchar(shades[(int)(b * (sizeof(shades) - 2))]);
}
putchar('\n');
}
return 0;
} |
http://rosettacode.org/wiki/Dynamic_variable_names | Dynamic variable names | Task
Create a variable with a user-defined name.
The variable name should not be written in the program text, but should be taken from the user dynamically.
See also
Eval in environment is a similar task.
| #PicoLisp | PicoLisp | (de userVariable ()
(prin "Enter a variable name: ")
(let Var (line T) # Read transient symbol
(prin "Enter a value: ")
(set Var (read)) # Set symbol's value
(println 'Variable Var 'Value (val Var)) ) ) # Print them |
http://rosettacode.org/wiki/Dynamic_variable_names | Dynamic variable names | Task
Create a variable with a user-defined name.
The variable name should not be written in the program text, but should be taken from the user dynamically.
See also
Eval in environment is a similar task.
| #PowerShell | PowerShell | $variableName = Read-Host
New-Variable $variableName 'Foo'
Get-Variable $variableName |
http://rosettacode.org/wiki/Dynamic_variable_names | Dynamic variable names | Task
Create a variable with a user-defined name.
The variable name should not be written in the program text, but should be taken from the user dynamically.
See also
Eval in environment is a similar task.
| #ProDOS | ProDOS | editvar /newvar /value=a /userinput=1 /title=Enter a variable name:
editvar /newvar /value=b /userinput=1 /title=Enter a variable title:
editvar /newvar /value=-a- /title=-b- |
http://rosettacode.org/wiki/Draw_a_pixel | Draw a pixel | Task
Create a window and draw a pixel in it, subject to the following:
the window is 320 x 240
the color of the pixel must be red (255,0,0)
the position of the pixel is x = 100, y = 100 | #Oberon | Oberon | MODULE pixel;
IMPORT XYplane;
BEGIN
XYplane.Open;
XYplane.Dot(100, 100, XYplane.draw);
REPEAT UNTIL XYplane.Key() = "q"
END pixel.
|
http://rosettacode.org/wiki/Draw_a_pixel | Draw a pixel | Task
Create a window and draw a pixel in it, subject to the following:
the window is 320 x 240
the color of the pixel must be red (255,0,0)
the position of the pixel is x = 100, y = 100 | #Objeck | Objeck | use Game.SDL2;
use Game.Framework;
class DrawPixel {
@framework : GameFramework;
function : Main(args : String[]) ~ Nil {
DrawPixel->New()->Run();
}
New() {
@framework := GameFramework->New(320, 240, "RGB");
@framework->SetClearColor(Color->New(0, 0, 0));
}
method : Run() ~ Nil {
if(@framework->IsOk()) {
e := @framework->GetEvent();
quit := false;
while(<>quit) {
@framework->FrameStart();
@framework->Clear();
while(e->Poll() <> 0) {
if(e->GetType() = EventType->SDL_QUIT) {
quit := true;
};
};
@framework->GetRenderer()->SetDrawColor(255, 0, 0, 255);
@framework->GetRenderer()->DrawPoint(100, 100);
@framework->Show();
@framework->FrameEnd();
};
}
else {
"--- Error Initializing Game Environment ---"->ErrorLine();
return;
};
leaving {
@framework->Quit();
};
}
}
|
http://rosettacode.org/wiki/Egyptian_division | Egyptian division | Egyptian division is a method of dividing integers using addition and
doubling that is similar to the algorithm of Ethiopian multiplication
Algorithm:
Given two numbers where the dividend is to be divided by the divisor:
Start the construction of a table of two columns: powers_of_2, and doublings; by a first row of a 1 (i.e. 2^0) in the first column and 1 times the divisor in the first row second column.
Create the second row with columns of 2 (i.e 2^1), and 2 * divisor in order.
Continue with successive i’th rows of 2^i and 2^i * divisor.
Stop adding rows, and keep only those rows, where 2^i * divisor is less than or equal to the dividend.
We now assemble two separate sums that both start as zero, called here answer and accumulator
Consider each row of the table, in the reverse order of its construction.
If the current value of the accumulator added to the doublings cell would be less than or equal to the dividend then add it to the accumulator, as well as adding the powers_of_2 cell value to the answer.
When the first row has been considered as above, then the integer division of dividend by divisor is given by answer.
(And the remainder is given by the absolute value of accumulator - dividend).
Example: 580 / 34
Table creation:
powers_of_2
doublings
1
34
2
68
4
136
8
272
16
544
Initialization of sums:
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
8
272
16
544
0
0
Considering table rows, bottom-up:
When a row is considered it is shown crossed out if it is not accumulated, or bold if the row causes summations.
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
8
272
16
544
16
544
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
8
272
16
544
16
544
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
16
544
8
272
16
544
powers_of_2
doublings
answer
accumulator
1
34
2
68
16
544
4
136
8
272
16
544
powers_of_2
doublings
answer
accumulator
1
34
17
578
2
68
4
136
8
272
16
544
Answer
So 580 divided by 34 using the Egyptian method is 17 remainder (578 - 580) or 2.
Task
The task is to create a function that does Egyptian division. The function should
closely follow the description above in using a list/array of powers of two, and
another of doublings.
Functions should be clear interpretations of the algorithm.
Use the function to divide 580 by 34 and show the answer here, on this page.
Related tasks
Egyptian fractions
References
Egyptian Number System
| #Racket | Racket | #lang racket
(define (quotient/remainder-egyptian dividend divisor (trace? #f))
(define table
(for*/list ((power_of_2 (sequence-map (curry expt 2) (in-naturals)))
(doubling (in-value (* divisor power_of_2)))
#:break (> doubling dividend))
(list power_of_2 doubling)))
(when trace?
(displayln "Table\npow_2\tdoubling")
(for ((row table)) (printf "~a\t~a~%" (first row) (second row))))
(define-values (answer accumulator)
(for*/fold ((answer 0) (accumulator 0))
((row (reverse table))
(acc′ (in-value (+ accumulator (second row)))))
(when trace? (printf "row:~a\tans/acc:~a ~a\t" row answer accumulator))
(cond
[(<= acc′ dividend)
(define ans′ (+ answer (first row)))
(when trace? (printf "~a <= ~a -> ans′/acc′:~a ~a~%" acc′ dividend ans′ acc′))
(values ans′ acc′)]
[else
(when trace? (printf "~a > ~a [----]~%" acc′ dividend))
(values answer accumulator)])))
(values answer (- dividend accumulator)))
(module+ test
(require rackunit)
(let-values (([q r] (quotient/remainder-egyptian 580 34)))
(check-equal? q 17)
(check-equal? r 2))
(let-values (([q r] (quotient/remainder-egyptian 192 3)))
(check-equal? q 64)
(check-equal? r 0)))
(module+ main
(quotient/remainder-egyptian 580 34 #t)) |
http://rosettacode.org/wiki/Egyptian_division | Egyptian division | Egyptian division is a method of dividing integers using addition and
doubling that is similar to the algorithm of Ethiopian multiplication
Algorithm:
Given two numbers where the dividend is to be divided by the divisor:
Start the construction of a table of two columns: powers_of_2, and doublings; by a first row of a 1 (i.e. 2^0) in the first column and 1 times the divisor in the first row second column.
Create the second row with columns of 2 (i.e 2^1), and 2 * divisor in order.
Continue with successive i’th rows of 2^i and 2^i * divisor.
Stop adding rows, and keep only those rows, where 2^i * divisor is less than or equal to the dividend.
We now assemble two separate sums that both start as zero, called here answer and accumulator
Consider each row of the table, in the reverse order of its construction.
If the current value of the accumulator added to the doublings cell would be less than or equal to the dividend then add it to the accumulator, as well as adding the powers_of_2 cell value to the answer.
When the first row has been considered as above, then the integer division of dividend by divisor is given by answer.
(And the remainder is given by the absolute value of accumulator - dividend).
Example: 580 / 34
Table creation:
powers_of_2
doublings
1
34
2
68
4
136
8
272
16
544
Initialization of sums:
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
8
272
16
544
0
0
Considering table rows, bottom-up:
When a row is considered it is shown crossed out if it is not accumulated, or bold if the row causes summations.
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
8
272
16
544
16
544
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
8
272
16
544
16
544
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
16
544
8
272
16
544
powers_of_2
doublings
answer
accumulator
1
34
2
68
16
544
4
136
8
272
16
544
powers_of_2
doublings
answer
accumulator
1
34
17
578
2
68
4
136
8
272
16
544
Answer
So 580 divided by 34 using the Egyptian method is 17 remainder (578 - 580) or 2.
Task
The task is to create a function that does Egyptian division. The function should
closely follow the description above in using a list/array of powers of two, and
another of doublings.
Functions should be clear interpretations of the algorithm.
Use the function to divide 580 by 34 and show the answer here, on this page.
Related tasks
Egyptian fractions
References
Egyptian Number System
| #Raku | Raku | sub egyptian-divmod (Real $dividend is copy where * >= 0, Real $divisor where * > 0) {
my $accumulator = 0;
([1, $divisor], { [.[0] + .[0], .[1] + .[1]] } … ^ *.[1] > $dividend)
.reverse.map: { $dividend -= .[1], $accumulator += .[0] if $dividend >= .[1] }
$accumulator, $dividend;
}
#TESTING
for 580,34, 578,34, 7532795332300578,235117 -> $n, $d {
printf "%s divmod %s = %s remainder %s\n",
$n, $d, |egyptian-divmod( $n, $d )
} |
http://rosettacode.org/wiki/Egyptian_fractions | Egyptian fractions | An Egyptian fraction is the sum of distinct unit fractions such as:
1
2
+
1
3
+
1
16
(
=
43
48
)
{\displaystyle {\tfrac {1}{2}}+{\tfrac {1}{3}}+{\tfrac {1}{16}}\,(={\tfrac {43}{48}})}
Each fraction in the expression has a numerator equal to 1 (unity) and a denominator that is a positive integer, and all the denominators are distinct (i.e., no repetitions).
Fibonacci's Greedy algorithm for Egyptian fractions expands the fraction
x
y
{\displaystyle {\tfrac {x}{y}}}
to be represented by repeatedly performing the replacement
x
y
=
1
⌈
y
/
x
⌉
+
(
−
y
)
mod
x
y
⌈
y
/
x
⌉
{\displaystyle {\frac {x}{y}}={\frac {1}{\lceil y/x\rceil }}+{\frac {(-y)\!\!\!\!\mod x}{y\lceil y/x\rceil }}}
(simplifying the 2nd term in this replacement as necessary, and where
⌈
x
⌉
{\displaystyle \lceil x\rceil }
is the ceiling function).
For this task, Proper and improper fractions must be able to be expressed.
Proper fractions are of the form
a
b
{\displaystyle {\tfrac {a}{b}}}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are positive integers, such that
a
<
b
{\displaystyle a<b}
, and
improper fractions are of the form
a
b
{\displaystyle {\tfrac {a}{b}}}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are positive integers, such that a ≥ b.
(See the REXX programming example to view one method of expressing the whole number part of an improper fraction.)
For improper fractions, the integer part of any improper fraction should be first isolated and shown preceding the Egyptian unit fractions, and be surrounded by square brackets [n].
Task requirements
show the Egyptian fractions for:
43
48
{\displaystyle {\tfrac {43}{48}}}
and
5
121
{\displaystyle {\tfrac {5}{121}}}
and
2014
59
{\displaystyle {\tfrac {2014}{59}}}
for all proper fractions,
a
b
{\displaystyle {\tfrac {a}{b}}}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are positive one-or two-digit (decimal) integers, find and show an Egyptian fraction that has:
the largest number of terms,
the largest denominator.
for all one-, two-, and three-digit integers, find and show (as above). {extra credit}
Also see
Wolfram MathWorld™ entry: Egyptian fraction
| #Raku | Raku | role Egyptian {
method gist {
join ' + ',
("[{self.floor}]" if self.abs >= 1),
map {"1/$_"}, self.denominators;
}
method denominators {
my ($x, $y) = self.nude;
$x %= $y;
my @denom = gather ($x, $y) = -$y % $x, $y * take ($y / $x).ceiling
while $x;
}
}
say .nude.join('/'), " = ", $_ but Egyptian for 43/48, 5/121, 2014/59;
my @sample = map { $_ => .denominators },
grep * < 1,
map {$_ but Egyptian},
(2 .. 99 X/ 2 .. 99);
say .key.nude.join("/"),
" has max denominator, namely ",
.value.max
given max :by(*.value.max), @sample;
say .key.nude.join("/"),
" has max number of denominators, namely ",
.value.elems
given max :by(*.value.elems), @sample; |
http://rosettacode.org/wiki/Ethiopian_multiplication | Ethiopian multiplication | Ethiopian multiplication is a method of multiplying integers using only addition, doubling, and halving.
Method:
Take two numbers to be multiplied and write them down at the top of two columns.
In the left-hand column repeatedly halve the last number, discarding any remainders, and write the result below the last in the same column, until you write a value of 1.
In the right-hand column repeatedly double the last number and write the result below. stop when you add a result in the same row as where the left hand column shows 1.
Examine the table produced and discard any row where the value in the left column is even.
Sum the values in the right-hand column that remain to produce the result of multiplying the original two numbers together
For example: 17 × 34
17 34
Halving the first column:
17 34
8
4
2
1
Doubling the second column:
17 34
8 68
4 136
2 272
1 544
Strike-out rows whose first cell is even:
17 34
8 68
4 136
2 272
1 544
Sum the remaining numbers in the right-hand column:
17 34
8 --
4 ---
2 ---
1 544
====
578
So 17 multiplied by 34, by the Ethiopian method is 578.
Task
The task is to define three named functions/methods/procedures/subroutines:
one to halve an integer,
one to double an integer, and
one to state if an integer is even.
Use these functions to create a function that does Ethiopian multiplication.
References
Ethiopian multiplication explained (BBC Video clip)
A Night Of Numbers - Go Forth And Multiply (Video)
Russian Peasant Multiplication
Programming Praxis: Russian Peasant Multiplication
| #UNIX_Shell | UNIX Shell | halve()
{
expr "$1" / 2
}
double()
{
expr "$1" \* 2
}
is_even()
{
expr "$1" % 2 = 0 >/dev/null
}
ethiopicmult()
{
plier=$1
plicand=$2
r=0
while [ "$plier" -ge 1 ]; do
is_even "$plier" || r=`expr $r + "$plicand"`
plier=`halve "$plier"`
plicand=`double "$plicand"`
done
echo $r
}
ethiopicmult 17 34
# => 578 |
http://rosettacode.org/wiki/Elementary_cellular_automaton | Elementary cellular automaton | An elementary cellular automaton is a one-dimensional cellular automaton where there are two possible states (labeled 0 and 1) and the rule to determine the state of a cell in the next generation depends only on the current state of the cell and its two immediate neighbors. Those three values can be encoded with three bits.
The rules of evolution are then encoded with eight bits indicating the outcome of each of the eight possibilities 111, 110, 101, 100, 011, 010, 001 and 000 in this order. Thus for instance the rule 13 means that a state is updated to 1 only in the cases 011, 010 and 000, since 13 in binary is 0b00001101.
Task
Create a subroutine, program or function that allows to create and visualize the evolution of any of the 256 possible elementary cellular automaton of arbitrary space length and for any given initial state. You can demonstrate your solution with any automaton of your choice.
The space state should wrap: this means that the left-most cell should be considered as the right neighbor of the right-most cell, and reciprocally.
This task is basically a generalization of one-dimensional cellular automata.
See also
Cellular automata (natureofcode.com)
| #Wren | Wren | import "/fmt" for Conv
var SIZE = 32
var LINES = SIZE / 2
var RULE = 90
var ruleTest = Fn.new { |x| (RULE & (1 << (7 & x))) != 0 }
var bshl = Fn.new { |b, bitCount| Conv.btoi(b) << bitCount }
var evolve = Fn.new { |s|
var t = List.filled(SIZE, false)
t[SIZE-1] = ruleTest.call(bshl.call(s[0], 2) | bshl.call(s[SIZE-1], 1) | Conv.btoi(s[SIZE-2]))
t[0] = ruleTest.call(bshl.call(s[1], 2) | bshl.call(s[0], 1) | Conv.btoi(s[SIZE-1]))
for (i in 1...SIZE - 1) {
t[i] = ruleTest.call(bshl.call(s[i+1], 2) | bshl.call(s[i], 1) | Conv.btoi(s[i-1]))
}
for (i in 0...SIZE) s[i] = t[i]
}
var show = Fn.new { |s|
for (i in SIZE - 1..0) System.write(s[i] ? "*" : " ")
System.print()
}
var state = List.filled(SIZE, false)
state[LINES] = true
System.print("Rule %(RULE):")
for (i in 0...LINES) {
show.call(state)
evolve.call(state)
} |
http://rosettacode.org/wiki/Factorial | Factorial | Definitions
The factorial of 0 (zero) is defined as being 1 (unity).
The Factorial Function of a positive integer, n, is defined as the product of the sequence:
n, n-1, n-2, ... 1
Task
Write a function to return the factorial of a number.
Solutions can be iterative or recursive.
Support for trapping negative n errors is optional.
Related task
Primorial numbers
| #Wrapl | Wrapl | DEF fac(n) n <= 1 | PROD 1:to(n); |
http://rosettacode.org/wiki/Echo_server | Echo server | Create a network service that sits on TCP port 12321, which accepts connections on that port, and which echoes complete lines (using a carriage-return/line-feed sequence as line separator) back to clients. No error handling is required. For the purposes of testing, it is only necessary to support connections from localhost (127.0.0.1 or perhaps ::1). Logging of connection information to standard output is recommended.
The implementation must be able to handle simultaneous connections from multiple clients. A multi-threaded or multi-process solution may be used. Each connection must be able to echo more than a single line.
The implementation must not stop responding to other clients if one client sends a partial line or stops reading responses.
| #REALbasic | REALbasic |
Class EchoSocket
Inherits TCPSocket
Sub DataAvailable()
If Instr(Me.LookAhead, EndofLine.Windows) > 0 Then
Dim data As String = Me.ReadAll
Dim lines() As String = Split(data, EndofLine.Windows)
For i As Integer = 0 To Ubound(lines)
Me.Write(lines(i) + EndOfLine.Windows)
Print(lines(i))
Next
End If
End Sub
End Class
Class EchoServer
Inherits ServerSocket
Function AddSocket() As TCPSocket
Return New EchoSocket
End Function
End Class
Class App
Inherits ConsoleApplication
Function Run(args() As String) As Integer
Listener = New EchoServer
Listener.Port = 12321
Listener.Listen()
While True
DoEvents() 'pump the event loop
Wend
End Function
Private Listener As EchoServer
End Class
|
http://rosettacode.org/wiki/Eban_numbers | Eban numbers |
Definition
An eban number is a number that has no letter e in it when the number is spelled in English.
Or more literally, spelled numbers that contain the letter e are banned.
The American version of spelling numbers will be used here (as opposed to the British).
2,000,000,000 is two billion, not two milliard.
Only numbers less than one sextillion (1021) will be considered in/for this task.
This will allow optimizations to be used.
Task
show all eban numbers ≤ 1,000 (in a horizontal format), and a count
show all eban numbers between 1,000 and 4,000 (inclusive), and a count
show a count of all eban numbers up and including 10,000
show a count of all eban numbers up and including 100,000
show a count of all eban numbers up and including 1,000,000
show a count of all eban numbers up and including 10,000,000
show all output here.
See also
The MathWorld entry: eban numbers.
The OEIS entry: A6933, eban numbers.
| #Ruby | Ruby | def main
intervals = [
[2, 1000, true],
[1000, 4000, true],
[2, 10000, false],
[2, 100000, false],
[2, 1000000, false],
[2, 10000000, false],
[2, 100000000, false],
[2, 1000000000, false]
]
for intv in intervals
(start, ending, display) = intv
if start == 2 then
print "eban numbers up to and including %d:\n" % [ending]
else
print "eban numbers between %d and %d (inclusive):\n" % [start, ending]
end
count = 0
for i in (start .. ending).step(2)
b = (i / 1000000000).floor
r = (i % 1000000000)
m = (r / 1000000).floor
r = (r % 1000000)
t = (r / 1000).floor
r = (r % 1000)
if m >= 30 and m <= 66 then
m = m % 10
end
if t >= 30 and t <= 66 then
t = t % 10
end
if r >= 30 and r <= 66 then
r = r % 10
end
if b == 0 or b == 2 or b == 4 or b == 6 then
if m == 0 or m == 2 or m == 4 or m == 6 then
if t == 0 or t == 2 or t == 4 or t == 6 then
if r == 0 or r == 2 or r == 4 or r == 6 then
if display then
print ' ', i
end
count = count + 1
end
end
end
end
end
if display then
print "\n"
end
print "count = %d\n\n" % [count]
end
end
main() |
http://rosettacode.org/wiki/Draw_a_rotating_cube | Draw a rotating cube | Task
Draw a rotating cube.
It should be oriented with one vertex pointing straight up, and its opposite vertex on the main diagonal (the one farthest away) straight down. It can be solid or wire-frame, and you can use ASCII art if your language doesn't have graphical capabilities. Perspective is optional.
Related tasks
Draw a cuboid
write language name in 3D ASCII
| #Julia | Julia | using Makie, LinearAlgebra
N = 40
interval = 0.10
scene = mesh(FRect3D(Vec3f0(-0.5), Vec3f0(1)), color = :skyblue2)
rect = scene[end]
for rad in 0.5:1/N:8.5
arr = normalize([cospi(rad/2), 0, sinpi(rad/2), -sinpi(rad/2)])
Makie.rotate!(rect, Quaternionf0(arr[1], arr[2], arr[3], arr[4]))
sleep(interval)
end |
http://rosettacode.org/wiki/Element-wise_operations | Element-wise operations | This task is similar to:
Matrix multiplication
Matrix transposition
Task
Implement basic element-wise matrix-matrix and scalar-matrix operations, which can be referred to in other, higher-order tasks.
Implement:
addition
subtraction
multiplication
division
exponentiation
Extend the task if necessary to include additional basic operations, which should not require their own specialised task.
| #PicoLisp | PicoLisp | (de elementWiseMatrix (Fun Mat1 Mat2)
(mapcar '((L1 L2) (mapcar Fun L1 L2)) Mat1 Mat2) )
(de elementWiseScalar (Fun Mat Scalar)
(elementWiseMatrix Fun Mat (circ (circ Scalar))) ) |
http://rosettacode.org/wiki/Element-wise_operations | Element-wise operations | This task is similar to:
Matrix multiplication
Matrix transposition
Task
Implement basic element-wise matrix-matrix and scalar-matrix operations, which can be referred to in other, higher-order tasks.
Implement:
addition
subtraction
multiplication
division
exponentiation
Extend the task if necessary to include additional basic operations, which should not require their own specialised task.
| #PL.2FI | PL/I | declare (matrix(3,3), vector(3), scalar) fixed;
declare (m(3,3), v(3) fixed;
m = scalar * matrix;
m = vector * matrix;
m = matrix * matrix;
v = scalar * vector;
v = vector * vector; |
http://rosettacode.org/wiki/Draw_a_sphere | Draw a sphere | Task
Draw a sphere.
The sphere can be represented graphically, or in ASCII art, depending on the language capabilities.
Either static or rotational projection is acceptable for this task.
Related tasks
draw a cuboid
draw a rotating cube
write language name in 3D ASCII
draw a Deathstar
| #Batch_File | Batch File | :: Draw a Sphere Task from Rosetta Code
:: Batch File Implementation
@echo off
rem -------------- define arithmetic "functions"
rem more info: https://www.dostips.com/forum/viewtopic.php?f=3&t=6744
rem integer sqrt arithmetic function by Aacini and penpen
rem source: https://www.dostips.com/forum/viewtopic.php?f=3&t=5819&start=30#p44016
set "sqrt(N)=( M=(N),j=M/(11*1024)+40, j=(M/j+j)>>1, j=(M/j+j)>>1, j=(M/j+j)>>1, j=(M/j+j)>>1, j=(M/j+j)>>1, j+=(M-j*j)>>31 )"
rem -------------- define batch file macros with parameters appended
rem more info: https://www.dostips.com/forum/viewtopic.php?f=3&t=2518
setlocal disabledelayedexpansion % == required for macro ==%
(set \n=^^^
%== this creates escaped line feed for macro ==%
)
rem normalize macro
rem argument: v
set normalize=for %%# in (1 2) do if %%#==2 ( %\n%
for /f "tokens=1" %%a in ("!args!") do set "v=%%a" %\n%
set /a "length_sqrd=!v![0]*!v![0]+!v![1]*!v![1]+!v![2]*!v![2]" %\n%
set /a "length=%sqrt(N):N=!length_sqrd!%" %== sqrt(N) applied ==% %\n%
set /a "!v![0]*=100", "!v![1]*=100", "!v![2]*=100" %== normalized elements mult. by 100 ==% %\n%
set /a "!v![0]/=length", "!v![1]/=length", "!v![2]/=length" %\n%
) else set args=
rem dot macro
rem arguments: s t outputvar
set dot=for %%# in (1 2) do if %%#==2 ( %\n%
for /f "tokens=1,2,3" %%a in ("!args!") do ( %\n%
set "s=%%a" %\n%
set "t=%%b" %\n%
set "outputvar=%%c" %\n%
) %\n%
set /a "d=!s![0]*!t![0]+!s![1]*!t![1]+!s![2]*!t![2]" %\n%
if !d! lss 0 (set /a "!outputvar!=-d") else (set "!outputvar!=0") %\n%
) else set args=
rem -------------- define pseudo-arrays
set "shades[0]=."
set "shades[1]=:"
set "shades[2]=!"
set "shades[3]=*"
set "shades[4]=o"
set "shades[5]=e"
set "shades[6]=&"
set "shades[7]=#"
set "shades[8]=%%"
set "shades[9]=@"
set "num_shades=9" %== start at 0 ==%
set "light[0]=30" & set "light[1]=30" & set "light[2]=-50"
rem -------------- main thing: execute drawSphere
setlocal enabledelayedexpansion
%normalize% light %== normalize macro applied ==%
call :drawSphere 20 1
exit /b 0
rem -------------- the function to draw the sphere
rem arguments: R ambient
:drawSphere
rem initialize variables from arguments
set /a "R=%1", "negR=-R", "twiceR=R*2", "twiceNegR=negR*2"
set /a "sqrdR=R*R*100*100" %== R*R is mult. by 100*100 ==%
set "k=2" %== k is hardcoded to 2 ==%
set "ambient=%3"
rem start draw line-by-line
for /l %%i in (%negR%, 1, %R%) do (
set /a "x=100*%%i+(100/2)" %== x is mult. by 100 ==%
set "line="
for /l %%j in (%twiceNegR%, 1, %twiceR%) do (
set /a "y=(100/2)*%%j+(100/2)" %== y is mult. by 100 ==%
set /a "pythag = x*x + y*y"
if !pythag! lss !sqrdR! (
set /a "vec[0]=x"
set /a "vec[1]=y"
set /a "vec[2]=sqrdR-pythag"
set /a "vec[2]=%sqrt(N):N=!vec[2]!%" %== sqrt(N) applied ==%
%normalize% vec %== normalize macro applied ==%
%dot% light vec dot_out %== dot macro applied ==%
rem since both light and vec are normalized to 100,
rem then dot_out is scaled up by 100*100 now.
set /a "dot_out/=100" %== scale-down to 100*[actual] to prevent overflow before exponentiation ==%
set "scaleup=1" %== after exponentiation, b would be scaleup*[actual] ==%
set "b=1"
for /l %%? in (1,1,%k%) do set /a "b*=dot_out","scaleup*=100" %== exponentiation ==%
set /a "b+=ambient*scaleup/10" %== add ambient/10 to b ==%
set /a "b/=scaleup/100" %== scale-down to 100*[actual] ==%
set /a "intensity=(100-b)*num_shades"
set /a "intensity/=100" %== final scale-down ==%
if !intensity! lss 0 set "intensity=0"
if !intensity! gtr %num_shades% set "intensity=%num_shades%"
for %%c in (!intensity!) do set "line=!line!!shades[%%c]!"
) else (
set "line=!line! "
)
)
echo(!line!
)
goto :EOF |
http://rosettacode.org/wiki/Doubly-linked_list/Element_insertion | Doubly-linked list/Element insertion | Doubly-Linked List (element)
This is much like inserting into a Singly-Linked List, but with added assignments so that the backwards-pointing links remain correct.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack | #Go | Go | package main
import "fmt"
type dlNode struct {
string
next, prev *dlNode
}
type dlList struct {
head, tail *dlNode
}
func (list *dlList) String() string {
if list.head == nil {
return fmt.Sprint(list.head)
}
r := "[" + list.head.string
for p := list.head.next; p != nil; p = p.next {
r += " " + p.string
}
return r + "]"
}
func (list *dlList) insertTail(node *dlNode) {
if list.tail == nil {
list.head = node
} else {
list.tail.next = node
}
node.next = nil
node.prev = list.tail
list.tail = node
}
func (list *dlList) insertAfter(existing, insert *dlNode) {
insert.prev = existing
insert.next = existing.next
existing.next.prev = insert
existing.next = insert
if existing == list.tail {
list.tail = insert
}
}
func main() {
dll := &dlList{}
fmt.Println(dll)
a := &dlNode{string: "A"}
dll.insertTail(a)
dll.insertTail(&dlNode{string: "B"})
fmt.Println(dll)
dll.insertAfter(a, &dlNode{string: "C"})
fmt.Println(dll)
} |
http://rosettacode.org/wiki/Doubly-linked_list/Element_insertion | Doubly-linked list/Element insertion | Doubly-Linked List (element)
This is much like inserting into a Singly-Linked List, but with added assignments so that the backwards-pointing links remain correct.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack | #Haskell | Haskell |
insert _ Leaf = Leaf
insert nv l@(Node pl v r) = (\(Node c _ _) -> c) new
where new = updateLeft left . updateRight right $ Node l nv r
left = Node pl v new
right = case r of
Leaf -> Leaf
Node _ v r -> Node new v r
|
http://rosettacode.org/wiki/Doubly-linked_list/Traversal | Doubly-linked list/Traversal | Traverse from the beginning of a doubly-linked list to the end, and from the end to the beginning.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Erlang | Erlang |
open System.Collections.Generic
let first (l: LinkedList<char>) = l.First
let last (l: LinkedList<char>) = l.Last
let next (l: LinkedListNode<char>) = l.Next
let prev (l: LinkedListNode<char>) = l.Previous
let traverse g f (ls: LinkedList<char>) =
let rec traverse (l: LinkedListNode<char>) =
match l with
| null -> ()
| _ ->
printf "%A" l.Value
traverse (f l)
traverse (g ls)
let traverseForward = traverse first next
let traverseBackward = traverse last prev
let cs = LinkedList(['a'..'z'])
traverseForward cs
printfn ""
traverseBackward cs
|
http://rosettacode.org/wiki/Doubly-linked_list/Traversal | Doubly-linked list/Traversal | Traverse from the beginning of a doubly-linked list to the end, and from the end to the beginning.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #F.23 | F# |
open System.Collections.Generic
let first (l: LinkedList<char>) = l.First
let last (l: LinkedList<char>) = l.Last
let next (l: LinkedListNode<char>) = l.Next
let prev (l: LinkedListNode<char>) = l.Previous
let traverse g f (ls: LinkedList<char>) =
let rec traverse (l: LinkedListNode<char>) =
match l with
| null -> ()
| _ ->
printf "%A" l.Value
traverse (f l)
traverse (g ls)
let traverseForward = traverse first next
let traverseBackward = traverse last prev
let cs = LinkedList(['a'..'z'])
traverseForward cs
printfn ""
traverseBackward cs
|
http://rosettacode.org/wiki/Doubly-linked_list/Element_definition | Doubly-linked list/Element definition | Task
Define the data structure for a doubly-linked list element.
The element should include a data member to hold its value and pointers to both the next element in the list and the previous element in the list.
The pointers should be mutable.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Clojure | Clojure | (defrecord Node [prev next data])
(defn new-node [prev next data]
(Node. (ref prev) (ref next) data)) |
http://rosettacode.org/wiki/Doubly-linked_list/Element_definition | Doubly-linked list/Element definition | Task
Define the data structure for a doubly-linked list element.
The element should include a data member to hold its value and pointers to both the next element in the list and the previous element in the list.
The pointers should be mutable.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Common_Lisp | Common Lisp | (defstruct dlist head tail)
(defstruct dlink content prev next) |
http://rosettacode.org/wiki/Doubly-linked_list/Element_definition | Doubly-linked list/Element definition | Task
Define the data structure for a doubly-linked list element.
The element should include a data member to hold its value and pointers to both the next element in the list and the previous element in the list.
The pointers should be mutable.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #D | D | struct Node(T) {
T data;
typeof(this)* prev, next;
}
void main() {
alias N = Node!int;
N* n = new N(10);
} |
http://rosettacode.org/wiki/Dutch_national_flag_problem | Dutch national flag problem |
The Dutch national flag is composed of three coloured bands in the order:
red (top)
then white, and
lastly blue (at the bottom).
The problem posed by Edsger Dijkstra is:
Given a number of red, blue and white balls in random order, arrange them in the order of the colours in the Dutch national flag.
When the problem was first posed, Dijkstra then went on to successively refine a solution, minimising the number of swaps and the number of times the colour of a ball needed to determined and restricting the balls to end in an array, ...
Task
Generate a randomized order of balls ensuring that they are not in the order of the Dutch national flag.
Sort the balls in a way idiomatic to your language.
Check the sorted balls are in the order of the Dutch national flag.
C.f.
Dutch national flag problem
Probabilistic analysis of algorithms for the Dutch national flag problem by Wei-Mei Chen. (pdf)
| #D | D | import std.stdio, std.random, std.algorithm, std.traits, std.array;
enum DutchColors { red, white, blue }
void dutchNationalFlagSort(DutchColors[] items) pure nothrow @nogc {
int lo, mid, hi = items.length - 1;
while (mid <= hi)
final switch (items[mid]) {
case DutchColors.red:
swap(items[lo++], items[mid++]);
break;
case DutchColors.white:
mid++;
break;
case DutchColors.blue:
swap(items[mid], items[hi--]);
break;
}
}
void main() {
DutchColors[12] balls;
foreach (ref ball; balls)
ball = uniform!DutchColors;
writeln("Original Ball order:\n", balls);
balls.dutchNationalFlagSort;
writeln("\nSorted Ball Order:\n", balls);
assert(balls[].isSorted, "Balls not sorted.");
} |
http://rosettacode.org/wiki/Draw_a_cuboid | Draw a cuboid | Task
Draw a cuboid with relative dimensions of 2 × 3 × 4.
The cuboid can be represented graphically, or in ASCII art, depending on the language capabilities.
To fulfill the criteria of being a cuboid, three faces must be visible.
Either static or rotational projection is acceptable for this task.
Related tasks
draw a sphere
draw a rotating cube
write language name in 3D ASCII
draw a Deathstar
| #C.23 | C# | using System;
using System.Drawing;
using System.Drawing.Drawing2D;
using System.Windows.Forms;
namespace Cuboid
{
public partial class Form1 : Form
{
double[][] nodes = {
new double[] {-1, -1, -1}, new double[] {-1, -1, 1}, new double[] {-1, 1, -1},
new double[] {-1, 1, 1}, new double[] {1, -1, -1}, new double[] {1, -1, 1},
new double[] {1, 1, -1}, new double[] {1, 1, 1} };
int[][] edges = {
new int[] {0, 1}, new int[] {1, 3}, new int[] {3, 2}, new int[] {2, 0}, new int[] {4, 5},
new int[] {5, 7}, new int[] {7, 6}, new int[] {6, 4}, new int[] {0, 4}, new int[] {1, 5},
new int[] {2, 6}, new int[] {3, 7}};
private int mouseX;
private int prevMouseX;
private int prevMouseY;
private int mouseY;
public Form1()
{
Width = Height = 640;
StartPosition = FormStartPosition.CenterScreen;
SetStyle(
ControlStyles.AllPaintingInWmPaint |
ControlStyles.UserPaint |
ControlStyles.DoubleBuffer,
true);
MouseMove += (s, e) =>
{
prevMouseX = mouseX;
prevMouseY = mouseY;
mouseX = e.X;
mouseY = e.Y;
double incrX = (mouseX - prevMouseX) * 0.01;
double incrY = (mouseY - prevMouseY) * 0.01;
RotateCuboid(incrX, incrY);
Refresh();
};
MouseDown += (s, e) =>
{
mouseX = e.X;
mouseY = e.Y;
};
Scale(80, 120, 160);
RotateCuboid(Math.PI / 5, Math.PI / 9);
}
private void RotateCuboid(double angleX, double angleY)
{
double sinX = Math.Sin(angleX);
double cosX = Math.Cos(angleX);
double sinY = Math.Sin(angleY);
double cosY = Math.Cos(angleY);
foreach (var node in nodes)
{
double x = node[0];
double y = node[1];
double z = node[2];
node[0] = x * cosX - z * sinX;
node[2] = z * cosX + x * sinX;
z = node[2];
node[1] = y * cosY - z * sinY;
node[2] = z * cosY + y * sinY;
}
}
private void Scale(int v1, int v2, int v3)
{
foreach (var item in nodes)
{
item[0] *= v1;
item[1] *= v2;
item[2] *= v3;
}
}
protected override void OnPaint(PaintEventArgs args)
{
var g = args.Graphics;
g.SmoothingMode = SmoothingMode.HighQuality;
g.Clear(Color.White);
g.TranslateTransform(Width / 2, Height / 2);
foreach (var edge in edges)
{
double[] xy1 = nodes[edge[0]];
double[] xy2 = nodes[edge[1]];
g.DrawLine(Pens.Black, (int)Math.Round(xy1[0]), (int)Math.Round(xy1[1]),
(int)Math.Round(xy2[0]), (int)Math.Round(xy2[1]));
}
foreach (var node in nodes)
{
g.FillEllipse(Brushes.Black, (int)Math.Round(node[0]) - 4,
(int)Math.Round(node[1]) - 4, 8, 8);
}
}
}
} |
http://rosettacode.org/wiki/Dynamic_variable_names | Dynamic variable names | Task
Create a variable with a user-defined name.
The variable name should not be written in the program text, but should be taken from the user dynamically.
See also
Eval in environment is a similar task.
| #Prolog | Prolog | test :- read(Name), atomics_to_string([Name, "= 50, writeln('", Name, "' = " , Name, ")"], String), term_string(Term, String), Term. |
http://rosettacode.org/wiki/Dynamic_variable_names | Dynamic variable names | Task
Create a variable with a user-defined name.
The variable name should not be written in the program text, but should be taken from the user dynamically.
See also
Eval in environment is a similar task.
| #Python | Python | >>> name = raw_input("Enter a variable name: ")
Enter a variable name: X
>>> globals()[name] = 42
>>> X
42 |
http://rosettacode.org/wiki/Draw_a_pixel | Draw a pixel | Task
Create a window and draw a pixel in it, subject to the following:
the window is 320 x 240
the color of the pixel must be red (255,0,0)
the position of the pixel is x = 100, y = 100 | #OCaml | OCaml | module G = Graphics
let () =
G.open_graph "";
G.resize_window 320 240;
G.set_color G.red;
G.plot 100 100;
ignore (G.read_key ()) |
http://rosettacode.org/wiki/Draw_a_pixel | Draw a pixel | Task
Create a window and draw a pixel in it, subject to the following:
the window is 320 x 240
the color of the pixel must be red (255,0,0)
the position of the pixel is x = 100, y = 100 | #Ol | Ol |
(import (lib gl))
(import (OpenGL version-1-0))
; init
(glClearColor 0.3 0.3 0.3 1)
(glOrtho 0 320 0 240 0 1)
; draw loop
(gl:set-renderer (lambda (mouse)
(glClear GL_COLOR_BUFFER_BIT)
(glBegin GL_POINTS)
(glColor3f 255 0 0)
(glVertex2f 100 100)
(glEnd)))
|
http://rosettacode.org/wiki/Egyptian_division | Egyptian division | Egyptian division is a method of dividing integers using addition and
doubling that is similar to the algorithm of Ethiopian multiplication
Algorithm:
Given two numbers where the dividend is to be divided by the divisor:
Start the construction of a table of two columns: powers_of_2, and doublings; by a first row of a 1 (i.e. 2^0) in the first column and 1 times the divisor in the first row second column.
Create the second row with columns of 2 (i.e 2^1), and 2 * divisor in order.
Continue with successive i’th rows of 2^i and 2^i * divisor.
Stop adding rows, and keep only those rows, where 2^i * divisor is less than or equal to the dividend.
We now assemble two separate sums that both start as zero, called here answer and accumulator
Consider each row of the table, in the reverse order of its construction.
If the current value of the accumulator added to the doublings cell would be less than or equal to the dividend then add it to the accumulator, as well as adding the powers_of_2 cell value to the answer.
When the first row has been considered as above, then the integer division of dividend by divisor is given by answer.
(And the remainder is given by the absolute value of accumulator - dividend).
Example: 580 / 34
Table creation:
powers_of_2
doublings
1
34
2
68
4
136
8
272
16
544
Initialization of sums:
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
8
272
16
544
0
0
Considering table rows, bottom-up:
When a row is considered it is shown crossed out if it is not accumulated, or bold if the row causes summations.
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
8
272
16
544
16
544
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
8
272
16
544
16
544
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
16
544
8
272
16
544
powers_of_2
doublings
answer
accumulator
1
34
2
68
16
544
4
136
8
272
16
544
powers_of_2
doublings
answer
accumulator
1
34
17
578
2
68
4
136
8
272
16
544
Answer
So 580 divided by 34 using the Egyptian method is 17 remainder (578 - 580) or 2.
Task
The task is to create a function that does Egyptian division. The function should
closely follow the description above in using a list/array of powers of two, and
another of doublings.
Functions should be clear interpretations of the algorithm.
Use the function to divide 580 by 34 and show the answer here, on this page.
Related tasks
Egyptian fractions
References
Egyptian Number System
| #REXX | REXX | /*REXX program performs division on positive integers using the Egyptian division method*/
numeric digits 1000 /*support gihugic numbers & be gung-ho.*/
parse arg n d . /*obtain optional arguments from the CL*/
if d=='' | d=="," then do; n= 580; d= 34 /*Not specified? Then use the defaults*/
end
call EgyptDiv n, d /*invoke the Egyptian Division function*/
parse var result q r /*extract the quotient & the remainder.*/
say n ' divided by ' d " is " q ' with a remainder of ' r
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
EgyptDiv: procedure; parse arg num,dem /*obtain the numerator and denominator.*/
p= 1; t= dem /*initialize the double & power values.*/
do #=1 until t>num /*construct the power & doubling lists.*/
pow.#= p; p= p + p /*build power entry; bump power value.*/
dbl.#= t; t= t + t /* " doubling " ; " doubling val.*/
end /*#*/
acc=0; ans=0 /*initialize accumulator & answer to 0 */
do s=# by -1 for # /* [↓] process the table "backwards". */
sum= acc + dbl.s /*compute the sum (to be used for test)*/
if sum>num then iterate /*Is sum to big? Then ignore this step*/
acc= sum /*use the "new" sum for the accumulator*/
ans= ans + pow.s /*calculate the (newer) running answer.*/
end /*s*/
return ans num-acc /*return the answer and the remainder. */ |
http://rosettacode.org/wiki/Egyptian_division | Egyptian division | Egyptian division is a method of dividing integers using addition and
doubling that is similar to the algorithm of Ethiopian multiplication
Algorithm:
Given two numbers where the dividend is to be divided by the divisor:
Start the construction of a table of two columns: powers_of_2, and doublings; by a first row of a 1 (i.e. 2^0) in the first column and 1 times the divisor in the first row second column.
Create the second row with columns of 2 (i.e 2^1), and 2 * divisor in order.
Continue with successive i’th rows of 2^i and 2^i * divisor.
Stop adding rows, and keep only those rows, where 2^i * divisor is less than or equal to the dividend.
We now assemble two separate sums that both start as zero, called here answer and accumulator
Consider each row of the table, in the reverse order of its construction.
If the current value of the accumulator added to the doublings cell would be less than or equal to the dividend then add it to the accumulator, as well as adding the powers_of_2 cell value to the answer.
When the first row has been considered as above, then the integer division of dividend by divisor is given by answer.
(And the remainder is given by the absolute value of accumulator - dividend).
Example: 580 / 34
Table creation:
powers_of_2
doublings
1
34
2
68
4
136
8
272
16
544
Initialization of sums:
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
8
272
16
544
0
0
Considering table rows, bottom-up:
When a row is considered it is shown crossed out if it is not accumulated, or bold if the row causes summations.
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
8
272
16
544
16
544
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
8
272
16
544
16
544
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
16
544
8
272
16
544
powers_of_2
doublings
answer
accumulator
1
34
2
68
16
544
4
136
8
272
16
544
powers_of_2
doublings
answer
accumulator
1
34
17
578
2
68
4
136
8
272
16
544
Answer
So 580 divided by 34 using the Egyptian method is 17 remainder (578 - 580) or 2.
Task
The task is to create a function that does Egyptian division. The function should
closely follow the description above in using a list/array of powers of two, and
another of doublings.
Functions should be clear interpretations of the algorithm.
Use the function to divide 580 by 34 and show the answer here, on this page.
Related tasks
Egyptian fractions
References
Egyptian Number System
| #Ring | Ring |
load "stdlib.ring"
table = newlist(32, 2)
dividend = 580
divisor = 34
i = 1
table[i][1] = 1
table[i][2] = divisor
while table[i] [2] < dividend
i = i + 1
table[i][1] = table[i -1] [1] * 2
table[i][2] = table[i -1] [2] * 2
end
i = i - 1
answer = table[i][1]
accumulator = table[i][2]
while i > 1
i = i - 1
if table[i][2]+ accumulator <= dividend
answer = answer + table[i][1]
accumulator = accumulator + table[i][2]
ok
end
see string(dividend) + " divided by " + string(divisor) + " using egytian division" + nl
see " returns " + string(answer) + " mod(ulus) " + string(dividend-accumulator)
|
http://rosettacode.org/wiki/Egyptian_fractions | Egyptian fractions | An Egyptian fraction is the sum of distinct unit fractions such as:
1
2
+
1
3
+
1
16
(
=
43
48
)
{\displaystyle {\tfrac {1}{2}}+{\tfrac {1}{3}}+{\tfrac {1}{16}}\,(={\tfrac {43}{48}})}
Each fraction in the expression has a numerator equal to 1 (unity) and a denominator that is a positive integer, and all the denominators are distinct (i.e., no repetitions).
Fibonacci's Greedy algorithm for Egyptian fractions expands the fraction
x
y
{\displaystyle {\tfrac {x}{y}}}
to be represented by repeatedly performing the replacement
x
y
=
1
⌈
y
/
x
⌉
+
(
−
y
)
mod
x
y
⌈
y
/
x
⌉
{\displaystyle {\frac {x}{y}}={\frac {1}{\lceil y/x\rceil }}+{\frac {(-y)\!\!\!\!\mod x}{y\lceil y/x\rceil }}}
(simplifying the 2nd term in this replacement as necessary, and where
⌈
x
⌉
{\displaystyle \lceil x\rceil }
is the ceiling function).
For this task, Proper and improper fractions must be able to be expressed.
Proper fractions are of the form
a
b
{\displaystyle {\tfrac {a}{b}}}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are positive integers, such that
a
<
b
{\displaystyle a<b}
, and
improper fractions are of the form
a
b
{\displaystyle {\tfrac {a}{b}}}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are positive integers, such that a ≥ b.
(See the REXX programming example to view one method of expressing the whole number part of an improper fraction.)
For improper fractions, the integer part of any improper fraction should be first isolated and shown preceding the Egyptian unit fractions, and be surrounded by square brackets [n].
Task requirements
show the Egyptian fractions for:
43
48
{\displaystyle {\tfrac {43}{48}}}
and
5
121
{\displaystyle {\tfrac {5}{121}}}
and
2014
59
{\displaystyle {\tfrac {2014}{59}}}
for all proper fractions,
a
b
{\displaystyle {\tfrac {a}{b}}}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are positive one-or two-digit (decimal) integers, find and show an Egyptian fraction that has:
the largest number of terms,
the largest denominator.
for all one-, two-, and three-digit integers, find and show (as above). {extra credit}
Also see
Wolfram MathWorld™ entry: Egyptian fraction
| #REXX | REXX | /*REXX program converts a fraction (can be improper) to an Egyptian fraction. */
parse arg fract '' -1 t; z=$egyptF(fract) /*compute the Egyptian fraction. */
if t\==. then say fract ' ───► ' z /*show Egyptian fraction from C.L.*/
return z /*stick a fork in it, we're done.*/
/*────────────────────────────────$EGYPTF subroutine──────────────────────────*/
$egyptF: parse arg z 1 zn '/' zd,,$; if zd=='' then zd=1 /*whole number ?*/
if z='' then call erx "no fraction was specified."
if zd==0 then call erx "denominator can't be zero:" zd
if zn==0 then call erx "numerator can't be zero:" zn
if zd<0 | zn<0 then call erx "fraction can't be negative" z
if \datatype(zn,'W') then call erx "numerator must be an integer:" zn
if \datatype(zd,'W') then call erx "denominator must be an integer:" zd
_=zn%zd /*check if it's an improper fraction. */
if _>=1 then do /*if improper fraction, then append it.*/
$='['_"]" /*append the whole # part of fraction. */
zn=zn-_*zd /*now, just use the proper fraction. */
if zn==0 then return $ /*Is there no fraction? Then we're done*/
end
if zd//zn==0 then do; zd=zd%zn; zn=1; end
do forever
if zn==1 & datatype(zd,'W') then return $ "1/"zd /*append Egyptian fract.*/
nd=zd%zn+1; $=$ '1/'nd /*add unity to integer fraction, append*/
z=$fractSub(zn'/'zd, "-", 1'/'nd) /*go and subtract the two fractions. */
parse var z zn '/' zd /*extract the numerator and denominator*/
L=2*max(length(zn),length(zd)) /*calculate if need more decimal digits*/
if L>=digits() then numeric digits L+L /*yes, then bump the decimal digits*/
end /*forever*/ /* [↑] the DO forever ends when zn==1.*/
/*────────────────────────────────$FRACTSUB subroutine────────────────────────*/
$fractSub: procedure; parse arg z.1,,z.2 1 zz.2; arg ,op
do j=1 for 2; z.j=translate(z.j,'/',"_"); end
if z.1=='' then z.1=(op\=="+" & op\=='-') /*unary +,- first fraction.*/
if z.2=='' then z.2=(op\=="+" & op\=='-') /*unary +.- second fraction.*/
do j=1 for 2 /*process both of the fractions*/
if pos('/',z.j)==0 then z.j=z.j"/1"; parse var z.j n.j '/' d.j
if \datatype(n.j,'N') then call erx "numerator isn't an integer:" n.j
if \datatype(d.j,'N') then call erx "denominator isn't an integer:" d.j
n.j=n.j/1; d.j=d.j/1 /*normalize numerator/denominator.*/
do while \datatype(n.j,'W'); n.j=n.j*10/1; d.j=d.j*10/1; end /*while*/
/* [↑] normalize both numbers. */
if d.j=0 then call erx "denominator can't be zero:" z.j
g=gcd(n.j,d.j); if g=0 then iterate; n.j=n.j/g; d.j=d.j/g
end /*j*/
l=lcm(d.1 d.2); do j=1 for 2; n.j=l*n.j/d.j; d.j=l; end /*j*/
if op=='-' then n.2=-n.2
t=n.1+n.2; u=l; if t==0 then return 0
g=gcd(t,u); t=t/g; u=u/g; if u==1 then return t
return t'/'u
/*─────────────────────────────general 1─line subs────────────────────────────*/
erx: say; say '***error!***' arg(1); say; exit 13
gcd:procedure;$=;do i=1 for arg();$=$ arg(i);end;parse var $ x z .;if x=0 then x=z;x=abs(x);do j=2 to words($);y=abs(word($,j));if y=0 then iterate;do until _==0;_=x//y;x=y;y=_;end;end;return x
lcm:procedure;y=;do j=1 for arg();y=y arg(j);end;x=word(y,1);do k=2 to words(y);!=abs(word(y,k));if !=0 then return 0;x=x*!/gcd(x,!);end;return x
p: return word(arg(1),1) |
http://rosettacode.org/wiki/Ethiopian_multiplication | Ethiopian multiplication | Ethiopian multiplication is a method of multiplying integers using only addition, doubling, and halving.
Method:
Take two numbers to be multiplied and write them down at the top of two columns.
In the left-hand column repeatedly halve the last number, discarding any remainders, and write the result below the last in the same column, until you write a value of 1.
In the right-hand column repeatedly double the last number and write the result below. stop when you add a result in the same row as where the left hand column shows 1.
Examine the table produced and discard any row where the value in the left column is even.
Sum the values in the right-hand column that remain to produce the result of multiplying the original two numbers together
For example: 17 × 34
17 34
Halving the first column:
17 34
8
4
2
1
Doubling the second column:
17 34
8 68
4 136
2 272
1 544
Strike-out rows whose first cell is even:
17 34
8 68
4 136
2 272
1 544
Sum the remaining numbers in the right-hand column:
17 34
8 --
4 ---
2 ---
1 544
====
578
So 17 multiplied by 34, by the Ethiopian method is 578.
Task
The task is to define three named functions/methods/procedures/subroutines:
one to halve an integer,
one to double an integer, and
one to state if an integer is even.
Use these functions to create a function that does Ethiopian multiplication.
References
Ethiopian multiplication explained (BBC Video clip)
A Night Of Numbers - Go Forth And Multiply (Video)
Russian Peasant Multiplication
Programming Praxis: Russian Peasant Multiplication
| #Ursala | Ursala | odd = ~&ihB
double = ~&iNiCB
half = ~&itB |
http://rosettacode.org/wiki/Elementary_cellular_automaton | Elementary cellular automaton | An elementary cellular automaton is a one-dimensional cellular automaton where there are two possible states (labeled 0 and 1) and the rule to determine the state of a cell in the next generation depends only on the current state of the cell and its two immediate neighbors. Those three values can be encoded with three bits.
The rules of evolution are then encoded with eight bits indicating the outcome of each of the eight possibilities 111, 110, 101, 100, 011, 010, 001 and 000 in this order. Thus for instance the rule 13 means that a state is updated to 1 only in the cases 011, 010 and 000, since 13 in binary is 0b00001101.
Task
Create a subroutine, program or function that allows to create and visualize the evolution of any of the 256 possible elementary cellular automaton of arbitrary space length and for any given initial state. You can demonstrate your solution with any automaton of your choice.
The space state should wrap: this means that the left-most cell should be considered as the right neighbor of the right-most cell, and reciprocally.
This task is basically a generalization of one-dimensional cellular automata.
See also
Cellular automata (natureofcode.com)
| #zkl | zkl | fcn rule(n){ n=n.toString(2); "00000000"[n.len() - 8,*] + n }
fcn applyRule(rule,cells){
cells=String(cells[-1],cells,cells[0]); // wrap cell ends
(cells.len() - 2).pump(String,'wrap(n){ rule[7 - cells[n,3].toInt(2)] })
} |
http://rosettacode.org/wiki/Factorial | Factorial | Definitions
The factorial of 0 (zero) is defined as being 1 (unity).
The Factorial Function of a positive integer, n, is defined as the product of the sequence:
n, n-1, n-2, ... 1
Task
Write a function to return the factorial of a number.
Solutions can be iterative or recursive.
Support for trapping negative n errors is optional.
Related task
Primorial numbers
| #Wren | Wren | import "/fmt" for Fmt
import "/big" for BigInt
class Factorial {
static iterative(n) {
if (n < 2) return BigInt.one
var fact = BigInt.one
for (i in 2..n.toSmall) fact = fact * i
return fact
}
static recursive(n) {
if (n < 2) return BigInt.one
return n * recursive(n-1)
}
}
var n = BigInt.new(24)
Fmt.print("Factorial(%(n)) iterative -> $,s", Factorial.iterative(n))
Fmt.print("Factorial(%(n)) recursive -> $,s", Factorial.recursive(n)) |
http://rosettacode.org/wiki/Echo_server | Echo server | Create a network service that sits on TCP port 12321, which accepts connections on that port, and which echoes complete lines (using a carriage-return/line-feed sequence as line separator) back to clients. No error handling is required. For the purposes of testing, it is only necessary to support connections from localhost (127.0.0.1 or perhaps ::1). Logging of connection information to standard output is recommended.
The implementation must be able to handle simultaneous connections from multiple clients. A multi-threaded or multi-process solution may be used. Each connection must be able to echo more than a single line.
The implementation must not stop responding to other clients if one client sends a partial line or stops reading responses.
| #REBOL | REBOL | server-port: open/lines tcp://:12321
forever [
connection-port: first server-port
until [
wait connection-port
error? try [insert connection-port first connection-port]
]
close connection-port
]
close server-port |
http://rosettacode.org/wiki/Eban_numbers | Eban numbers |
Definition
An eban number is a number that has no letter e in it when the number is spelled in English.
Or more literally, spelled numbers that contain the letter e are banned.
The American version of spelling numbers will be used here (as opposed to the British).
2,000,000,000 is two billion, not two milliard.
Only numbers less than one sextillion (1021) will be considered in/for this task.
This will allow optimizations to be used.
Task
show all eban numbers ≤ 1,000 (in a horizontal format), and a count
show all eban numbers between 1,000 and 4,000 (inclusive), and a count
show a count of all eban numbers up and including 10,000
show a count of all eban numbers up and including 100,000
show a count of all eban numbers up and including 1,000,000
show a count of all eban numbers up and including 10,000,000
show all output here.
See also
The MathWorld entry: eban numbers.
The OEIS entry: A6933, eban numbers.
| #Scala | Scala | object EbanNumbers {
class ERange(s: Int, e: Int, p: Boolean) {
val start: Int = s
val end: Int = e
val print: Boolean = p
}
def main(args: Array[String]): Unit = {
val rgs = List(
new ERange(2, 1000, true),
new ERange(1000, 4000, true),
new ERange(2, 10000, false),
new ERange(2, 100000, false),
new ERange(2, 1000000, false),
new ERange(2, 10000000, false),
new ERange(2, 100000000, false),
new ERange(2, 1000000000, false)
)
for (rg <- rgs) {
if (rg.start == 2) {
println(s"eban numbers up to an including ${rg.end}")
} else {
println(s"eban numbers between ${rg.start} and ${rg.end}")
}
var count = 0
for (i <- rg.start to rg.end) {
val b = i / 1000000000
var r = i % 1000000000
var m = r / 1000000
r = i % 1000000
var t = r / 1000
r %= 1000
if (m >= 30 && m <= 66) {
m %= 10
}
if (t >= 30 && t <= 66) {
t %= 10
}
if (r >= 30 && r <= 66) {
r %= 10
}
if (b == 0 || b == 2 || b == 4 || b == 6) {
if (m == 0 || m == 2 || m == 4 || m == 6) {
if (t == 0 || t == 2 || t == 4 || t == 6) {
if (r == 0 || r == 2 || r == 4 || r == 6) {
if (rg.print) {
print(s"$i ")
}
count += 1
}
}
}
}
}
if (rg.print) {
println()
}
println(s"count = $count")
println()
}
}
} |
http://rosettacode.org/wiki/Draw_a_rotating_cube | Draw a rotating cube | Task
Draw a rotating cube.
It should be oriented with one vertex pointing straight up, and its opposite vertex on the main diagonal (the one farthest away) straight down. It can be solid or wire-frame, and you can use ASCII art if your language doesn't have graphical capabilities. Perspective is optional.
Related tasks
Draw a cuboid
write language name in 3D ASCII
| #Kotlin | Kotlin | // version 1.1
import java.awt.*
import javax.swing.*
class RotatingCube : JPanel() {
private val nodes = arrayOf(
doubleArrayOf(-1.0, -1.0, -1.0),
doubleArrayOf(-1.0, -1.0, 1.0),
doubleArrayOf(-1.0, 1.0, -1.0),
doubleArrayOf(-1.0, 1.0, 1.0),
doubleArrayOf( 1.0, -1.0, -1.0),
doubleArrayOf( 1.0, -1.0, 1.0),
doubleArrayOf( 1.0, 1.0, -1.0),
doubleArrayOf( 1.0, 1.0, 1.0)
)
private val edges = arrayOf(
intArrayOf(0, 1),
intArrayOf(1, 3),
intArrayOf(3, 2),
intArrayOf(2, 0),
intArrayOf(4, 5),
intArrayOf(5, 7),
intArrayOf(7, 6),
intArrayOf(6, 4),
intArrayOf(0, 4),
intArrayOf(1, 5),
intArrayOf(2, 6),
intArrayOf(3, 7)
)
init {
preferredSize = Dimension(640, 640)
background = Color.white
scale(100.0)
rotateCube(Math.PI / 4.0, Math.atan(Math.sqrt(2.0)))
Timer(17) {
rotateCube(Math.PI / 180.0, 0.0)
repaint()
}.start()
}
private fun scale(s: Double) {
for (node in nodes) {
node[0] *= s
node[1] *= s
node[2] *= s
}
}
private fun rotateCube(angleX: Double, angleY: Double) {
val sinX = Math.sin(angleX)
val cosX = Math.cos(angleX)
val sinY = Math.sin(angleY)
val cosY = Math.cos(angleY)
for (node in nodes) {
val x = node[0]
val y = node[1]
var z = node[2]
node[0] = x * cosX - z * sinX
node[2] = z * cosX + x * sinX
z = node[2]
node[1] = y * cosY - z * sinY
node[2] = z * cosY + y * sinY
}
}
private fun drawCube(g: Graphics2D) {
g.translate(width / 2, height / 2)
for (edge in edges) {
val xy1 = nodes[edge[0]]
val xy2 = nodes[edge[1]]
g.drawLine(Math.round(xy1[0]).toInt(), Math.round(xy1[1]).toInt(),
Math.round(xy2[0]).toInt(), Math.round(xy2[1]).toInt())
}
for (node in nodes) {
g.fillOval(Math.round(node[0]).toInt() - 4, Math.round(node[1]).toInt() - 4, 8, 8)
}
}
override public fun paintComponent(gg: Graphics) {
super.paintComponent(gg)
val g = gg as Graphics2D
g.setRenderingHint(RenderingHints.KEY_ANTIALIASING, RenderingHints.VALUE_ANTIALIAS_ON)
g.color = Color.blue
drawCube(g)
}
}
fun main(args: Array<String>) {
SwingUtilities.invokeLater {
val f = JFrame()
f.defaultCloseOperation = JFrame.EXIT_ON_CLOSE
f.title = "Rotating cube"
f.isResizable = false
f.add(RotatingCube(), BorderLayout.CENTER)
f.pack()
f.setLocationRelativeTo(null)
f.isVisible = true
}
} |
http://rosettacode.org/wiki/Element-wise_operations | Element-wise operations | This task is similar to:
Matrix multiplication
Matrix transposition
Task
Implement basic element-wise matrix-matrix and scalar-matrix operations, which can be referred to in other, higher-order tasks.
Implement:
addition
subtraction
multiplication
division
exponentiation
Extend the task if necessary to include additional basic operations, which should not require their own specialised task.
| #Python | Python | >>> import random
>>> from operator import add, sub, mul, floordiv
>>> from pprint import pprint as pp
>>>
>>> def ewise(matrix1, matrix2, op):
return [[op(e1,e2) for e1,e2 in zip(row1, row2)] for row1,row2 in zip(matrix1, matrix2)]
>>> m,n = 3,4 # array dimensions
>>> a0 = [[random.randint(1,9) for y in range(n)] for x in range(m)]
>>> a1 = [[random.randint(1,9) for y in range(n)] for x in range(m)]
>>> pp(a0); pp(a1)
[[7, 8, 7, 4], [4, 9, 4, 1], [2, 3, 6, 4]]
[[4, 5, 1, 6], [6, 8, 3, 4], [2, 2, 6, 3]]
>>> pp(ewise(a0, a1, add))
[[11, 13, 8, 10], [10, 17, 7, 5], [4, 5, 12, 7]]
>>> pp(ewise(a0, a1, sub))
[[3, 3, 6, -2], [-2, 1, 1, -3], [0, 1, 0, 1]]
>>> pp(ewise(a0, a1, mul))
[[28, 40, 7, 24], [24, 72, 12, 4], [4, 6, 36, 12]]
>>> pp(ewise(a0, a1, floordiv))
[[1, 1, 7, 0], [0, 1, 1, 0], [1, 1, 1, 1]]
>>> pp(ewise(a0, a1, pow))
[[2401, 32768, 7, 4096], [4096, 43046721, 64, 1], [4, 9, 46656, 64]]
>>> pp(ewise(a0, a1, lambda x, y:2*x - y))
[[10, 11, 13, 2], [2, 10, 5, -2], [2, 4, 6, 5]]
>>>
>>> def s_ewise(scalar1, matrix1, op):
return [[op(scalar1, e1) for e1 in row1] for row1 in matrix1]
>>> scalar = 10
>>> a0
[[7, 8, 7, 4], [4, 9, 4, 1], [2, 3, 6, 4]]
>>> for op in ( add, sub, mul, floordiv, pow, lambda x, y:2*x - y ):
print("%10s :" % op.__name__, s_ewise(scalar, a0, op))
add : [[17, 18, 17, 14], [14, 19, 14, 11], [12, 13, 16, 14]]
sub : [[3, 2, 3, 6], [6, 1, 6, 9], [8, 7, 4, 6]]
mul : [[70, 80, 70, 40], [40, 90, 40, 10], [20, 30, 60, 40]]
floordiv : [[1, 1, 1, 2], [2, 1, 2, 10], [5, 3, 1, 2]]
pow : [[10000000, 100000000, 10000000, 10000], [10000, 1000000000, 10000, 10], [100, 1000, 1000000, 10000]]
<lambda> : [[13, 12, 13, 16], [16, 11, 16, 19], [18, 17, 14, 16]]
>>> |
http://rosettacode.org/wiki/Draw_a_sphere | Draw a sphere | Task
Draw a sphere.
The sphere can be represented graphically, or in ASCII art, depending on the language capabilities.
Either static or rotational projection is acceptable for this task.
Related tasks
draw a cuboid
draw a rotating cube
write language name in 3D ASCII
draw a Deathstar
| #Befunge | Befunge | 45*65*65*"2"30p20p10p::00p2*40p4*5vv<
>60p140g->:::*00g50g*60g40g-:*-\-v0>1
^_@#`\g0<|`\g04:+1, <*84$$_v#`\0:<>p^
>v>g2+:5^$>g:*++*/7g^>*:9$#<"~"/:"~"v
g:^06,+55<^03*<v09p07%"~"p09/"~"p08%<
^>#0 *#12#<0g:^>+::"~~"90g*80g+*70gv|
g-10g*+:9**00gv|!*`\2\`-20::/2-\/\+<>
%#&eo*!:..^g05<>$030g-*9/\20g*+60g40^ |
http://rosettacode.org/wiki/Doubly-linked_list/Element_insertion | Doubly-linked list/Element insertion | Doubly-Linked List (element)
This is much like inserting into a Singly-Linked List, but with added assignments so that the backwards-pointing links remain correct.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack | #Icon_and_Unicon | Icon and Unicon |
class DoubleLink (value, prev_link, next_link)
# insert given node after this one, losing its existing connections
method insert_after (node)
node.prev_link := self
if (\next_link) then next_link.prev_link := node
node.next_link := next_link
self.next_link := node
end
initially (value, prev_link, next_link)
self.value := value
self.prev_link := prev_link # links are 'null' if not given
self.next_link := next_link
end
|
http://rosettacode.org/wiki/Doubly-linked_list/Element_insertion | Doubly-linked list/Element insertion | Doubly-Linked List (element)
This is much like inserting into a Singly-Linked List, but with added assignments so that the backwards-pointing links remain correct.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack | #J | J | (pred;succ;<data) conew 'DoublyLinkedListElement'
|
http://rosettacode.org/wiki/Doubly-linked_list/Element_insertion | Doubly-linked list/Element insertion | Doubly-Linked List (element)
This is much like inserting into a Singly-Linked List, but with added assignments so that the backwards-pointing links remain correct.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack | #Java | Java |
import java.util.LinkedList;
@SuppressWarnings("serial")
public class DoublyLinkedListInsertion<T> extends LinkedList<T> {
public static void main(String[] args) {
DoublyLinkedListInsertion<String> list = new DoublyLinkedListInsertion<String>();
list.addFirst("Add First 1");
list.addFirst("Add First 2");
list.addFirst("Add First 3");
list.addFirst("Add First 4");
list.addFirst("Add First 5");
traverseList(list);
list.addAfter("Add First 3", "Add New");
traverseList(list);
}
/*
* Add after indicated node. If not in the list, added as the last node.
*/
public void addAfter(T after, T element) {
int index = indexOf(after);
if ( index >= 0 ) {
add(index + 1, element);
}
else {
addLast(element);
}
}
private static void traverseList(LinkedList<String> list) {
System.out.println("Traverse List:");
for ( int i = 0 ; i < list.size() ; i++ ) {
System.out.printf("Element number %d - Element value = '%s'%n", i, list.get(i));
}
System.out.println();
}
}
|
http://rosettacode.org/wiki/Draw_a_clock | Draw a clock | Task
Draw a clock.
More specific:
Draw a time keeping device. It can be a stopwatch, hourglass, sundial, a mouth counting "one thousand and one", anything. Only showing the seconds is required, e.g.: a watch with just a second hand will suffice. However, it must clearly change every second, and the change must cycle every so often (one minute, 30 seconds, etc.) It must be drawn; printing a string of numbers to your terminal doesn't qualify. Both text-based and graphical drawing are OK.
The clock is unlikely to be used to control space flights, so it needs not be hyper-accurate, but it should be usable, meaning if one can read the seconds off the clock, it must agree with the system clock.
A clock is rarely (never?) a major application: don't be a CPU hog and poll the system timer every microsecond, use a proper timer/signal/event from your system or language instead. For a bad example, many OpenGL programs update the frame-buffer in a busy loop even if no redraw is needed, which is very undesirable for this task.
A clock is rarely (never?) a major application: try to keep your code simple and to the point. Don't write something too elaborate or convoluted, instead do whatever is natural, concise and clear in your language.
Key points
animate simple object
timed event
polling system resources
code clarity
| #ActionScript | ActionScript |
package {
import flash.display.Graphics;
import flash.display.Shape;
import flash.display.Sprite;
import flash.events.Event;
import flash.events.TimerEvent;
import flash.utils.Timer;
public class Clock extends Sprite {
// Changes of hands (in degrees) per second
private static const HOUR_HAND_CHANGE:Number = 1 / 120; // 360 / (60 * 60 * 12)
private static const MINUTE_HAND_CHANGE:Number = 0.1; // 360 / (60 * 60)
private static const SECOND_HAND_CHANGE:Number = 6; // 360 / 60
private var _timer:Timer;
private var _hHand:Shape;
private var _mHand:Shape;
private var _sHand:Shape;
public function Clock() {
if ( stage ) _init();
else addEventListener(Event.ADDED_TO_STAGE, _init);
}
private function _init(e:Event = null):void {
var i:uint;
var base:Shape = new Shape(), hHand:Shape = new Shape(), mHand:Shape = new Shape();
var sHand:Shape = new Shape(), hub:Shape = new Shape();
var size:Number = 500;
var c:Number = size / 2;
x = 30;
y = 30;
var baseGraphics:Graphics = base.graphics;
baseGraphics.lineStyle(5, 0xEE0000);
baseGraphics.beginFill(0xFFDDDD);
baseGraphics.drawCircle(c, c, c);
var uAngle:Number = Math.PI / 30;
var markerStart:Number = c - 30;
var markerEnd:Number = c - 15;
var markerX1:Number, markerY1:Number, markerX2:Number, markerY2:Number;
var angle:Number, angleSin:Number, angleCos:Number;
baseGraphics.endFill();
var isMajorMarker:Boolean = true;
for ( i = 0; i < 60; i++ ) {
// Draw the markers
angle = uAngle * i;
angleSin = Math.sin(angle);
angleCos = Math.cos(angle);
markerX1 = c + markerStart * angleCos;
markerY1 = c + markerStart * angleSin;
markerX2 = c + markerEnd * angleCos;
markerY2 = c + markerEnd * angleSin;
if ( i % 5 == 0 ) {
baseGraphics.lineStyle(3, 0x000080);
isMajorMarker = true;
}
else if ( isMajorMarker ) {
baseGraphics.lineStyle(1, 0x000080);
isMajorMarker = false;
}
baseGraphics.moveTo(markerX1, markerY1);
baseGraphics.lineTo(markerX2, markerY2);
}
addChild(base);
sHand.graphics.lineStyle(2, 0x00BB00);
sHand.graphics.moveTo(0, 0);
sHand.graphics.lineTo(0, 40 - c);
sHand.x = sHand.y = c;
mHand.graphics.lineStyle(8, 0x444444);
mHand.graphics.moveTo(0, 0);
mHand.graphics.lineTo(0, 50 - c);
mHand.x = mHand.y = c;
hHand.graphics.lineStyle(8, 0x777777);
hHand.graphics.moveTo(0, 0);
hHand.graphics.lineTo(0, 120 - c);
hHand.x = hHand.y = c;
hub.graphics.lineStyle(4, 0x664444);
hub.graphics.beginFill(0xCC9999);
hub.graphics.drawCircle(c, c, 5);
_hHand = hHand;
_mHand = mHand;
_sHand = sHand;
addChild(mHand);
addChild(hHand);
addChild(sHand);
addChild(hub);
var date:Date = new Date();
// Since millisecond precision is not needed, round it up to the nearest second.
var seconds:Number = date.seconds + ((date.milliseconds > 500) ? 1 : 0);
var minutes:Number = date.minutes + seconds / 60;
var hours:Number = (date.hours + minutes / 60) % 12;
sHand.rotation = seconds * 6;
mHand.rotation = minutes * 6;
hHand.rotation = hours * 30;
_timer = new Timer(1000); // 1 second = 1000 ms
_timer.addEventListener(TimerEvent.TIMER, _onTimerTick);
_timer.start();
}
private function _onTimerTick(e:TimerEvent):void {
_hHand.rotation += HOUR_HAND_CHANGE;
_mHand.rotation += MINUTE_HAND_CHANGE;
_sHand.rotation += SECOND_HAND_CHANGE;
}
}
}
|
http://rosettacode.org/wiki/Doubly-linked_list/Traversal | Doubly-linked list/Traversal | Traverse from the beginning of a doubly-linked list to the end, and from the end to the beginning.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Fortran | Fortran | Dim As Integer i, MiLista()
For i = 0 To Int(Rnd * 100)+25
Redim Preserve MiLista(i)
MiLista(i) = Rnd * 314
Next
'Tour from the beginning
For i = Lbound(MiLista) To Ubound(MiLista)
Print MiLista(i)
Next i
Print
'Travel from the end
For i = Ubound(MiLista) To Lbound(MiLista) Step -1
Print MiLista(i)
Next i
Sleep |
http://rosettacode.org/wiki/Doubly-linked_list/Traversal | Doubly-linked list/Traversal | Traverse from the beginning of a doubly-linked list to the end, and from the end to the beginning.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #FreeBASIC | FreeBASIC | Dim As Integer i, MiLista()
For i = 0 To Int(Rnd * 100)+25
Redim Preserve MiLista(i)
MiLista(i) = Rnd * 314
Next
'Tour from the beginning
For i = Lbound(MiLista) To Ubound(MiLista)
Print MiLista(i)
Next i
Print
'Travel from the end
For i = Ubound(MiLista) To Lbound(MiLista) Step -1
Print MiLista(i)
Next i
Sleep |
http://rosettacode.org/wiki/Doubly-linked_list/Element_definition | Doubly-linked list/Element definition | Task
Define the data structure for a doubly-linked list element.
The element should include a data member to hold its value and pointers to both the next element in the list and the previous element in the list.
The pointers should be mutable.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Delphi | Delphi | struct Node(T) {
type
pList = ^List ;
list = record
data : pointer ;
prev : pList ;
next : pList ;
end;
} |
http://rosettacode.org/wiki/Doubly-linked_list/Element_definition | Doubly-linked list/Element definition | Task
Define the data structure for a doubly-linked list element.
The element should include a data member to hold its value and pointers to both the next element in the list and the previous element in the list.
The pointers should be mutable.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #E | E | def makeElement(var value, var next, var prev) {
def element {
to setValue(v) { value := v }
to getValue() { return value }
to setNext(n) { next := n }
to getNext() { return next }
to setPrev(p) { prev := p }
to getPrev() { return prev }
}
return element
} |
http://rosettacode.org/wiki/Doubly-linked_list/Element_definition | Doubly-linked list/Element definition | Task
Define the data structure for a doubly-linked list element.
The element should include a data member to hold its value and pointers to both the next element in the list and the previous element in the list.
The pointers should be mutable.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Erlang | Erlang |
new( Data ) -> erlang:spawn( fun() -> loop( Data, noprevious, nonext ) end ).
|
http://rosettacode.org/wiki/Dutch_national_flag_problem | Dutch national flag problem |
The Dutch national flag is composed of three coloured bands in the order:
red (top)
then white, and
lastly blue (at the bottom).
The problem posed by Edsger Dijkstra is:
Given a number of red, blue and white balls in random order, arrange them in the order of the colours in the Dutch national flag.
When the problem was first posed, Dijkstra then went on to successively refine a solution, minimising the number of swaps and the number of times the colour of a ball needed to determined and restricting the balls to end in an array, ...
Task
Generate a randomized order of balls ensuring that they are not in the order of the Dutch national flag.
Sort the balls in a way idiomatic to your language.
Check the sorted balls are in the order of the Dutch national flag.
C.f.
Dutch national flag problem
Probabilistic analysis of algorithms for the Dutch national flag problem by Wei-Mei Chen. (pdf)
| #Elixir | Elixir | defmodule Dutch_national_flag do
defp ball(:red), do: 1
defp ball(:white), do: 2
defp ball(:blue), do: 3
defp random_ball, do: Enum.random([:red, :white, :blue])
defp random_ball(n), do: (for _ <- 1..n, do: random_ball())
defp is_dutch([]), do: true
defp is_dutch([_]), do: true
defp is_dutch([b,h|l]), do: ball(b) < ball(h) and is_dutch([h|l])
defp is_dutch(_), do: false
def dutch(list), do: dutch([], [], [], list)
defp dutch(r, w, b, []), do: r ++ w ++ b
defp dutch(r, w, b, [:red | list]), do: dutch([:red | r], w, b, list)
defp dutch(r, w, b, [:white | list]), do: dutch(r, [:white | w], b, list)
defp dutch(r, w, b, [:blue | list]), do: dutch(r, w, [:blue | b], list)
def problem(n \\ 10) do
list = random_ball(n)
if is_dutch(list) do
IO.puts "The random sequence #{inspect list} is already in the order of the Dutch flag!"
else
IO.puts "The starting random sequence is #{inspect list};"
IO.puts "The ordered sequence is #{inspect dutch(list)}."
end
end
end
Dutch_national_flag.problem |
http://rosettacode.org/wiki/Dutch_national_flag_problem | Dutch national flag problem |
The Dutch national flag is composed of three coloured bands in the order:
red (top)
then white, and
lastly blue (at the bottom).
The problem posed by Edsger Dijkstra is:
Given a number of red, blue and white balls in random order, arrange them in the order of the colours in the Dutch national flag.
When the problem was first posed, Dijkstra then went on to successively refine a solution, minimising the number of swaps and the number of times the colour of a ball needed to determined and restricting the balls to end in an array, ...
Task
Generate a randomized order of balls ensuring that they are not in the order of the Dutch national flag.
Sort the balls in a way idiomatic to your language.
Check the sorted balls are in the order of the Dutch national flag.
C.f.
Dutch national flag problem
Probabilistic analysis of algorithms for the Dutch national flag problem by Wei-Mei Chen. (pdf)
| #Erlang | Erlang | -module(dutch).
-export([random_balls/1, is_dutch/1, dutch/1]).
ball(red) -> 1;
ball(white) -> 2;
ball(blue) -> 3.
random_ball() -> lists:nth(random:uniform(3), [red, white, blue]).
random_balls(N) -> random_balls(N,[]).
random_balls(0,L) -> L;
random_balls(N,L) when N > 0 ->
B = random_ball(),
random_balls(N-1, [B|L]).
is_dutch([]) -> true;
is_dutch([_]) -> true;
is_dutch([B|[H|L]]) -> (ball(B) < ball(H)) and is_dutch([H|L]);
is_dutch(_) -> false.
dutch(L) -> dutch([],[],[],L).
dutch(R, W, B, []) -> R ++ W ++ B;
dutch(R, W, B, [red | L]) -> dutch([red|R], W, B, L);
dutch(R, W, B, [white | L]) -> dutch(R, [white|W], B, L);
dutch(R, W, B, [blue | L]) -> dutch(R, W, [blue|B], L). |
http://rosettacode.org/wiki/Draw_a_cuboid | Draw a cuboid | Task
Draw a cuboid with relative dimensions of 2 × 3 × 4.
The cuboid can be represented graphically, or in ASCII art, depending on the language capabilities.
To fulfill the criteria of being a cuboid, three faces must be visible.
Either static or rotational projection is acceptable for this task.
Related tasks
draw a sphere
draw a rotating cube
write language name in 3D ASCII
draw a Deathstar
| #C.2B.2B | C++ |
#include<graphics.h>
#include<iostream>
int main()
{
int k;
initwindow(1500,810,"Rosetta Cuboid");
do{
std::cout<<"Enter ratio of sides ( 0 or -ve to exit) : ";
std::cin>>k;
if(k>0){
bar3d(100, 100, 100 + 2*k, 100 + 4*k, 3*k, 1);
}
}while(k>0);
return 0;
}
|
http://rosettacode.org/wiki/Dynamic_variable_names | Dynamic variable names | Task
Create a variable with a user-defined name.
The variable name should not be written in the program text, but should be taken from the user dynamically.
See also
Eval in environment is a similar task.
| #Quackery | Quackery | [ say "The word "
dup echo$
names find names found iff
[ say " exists." ]
else
[ say " does not exist." ] ] is exists? ( $ --> )
[ $ "Please enter a name: " input
cr
dup exists?
cr cr
dup say "Creating " echo$
say "..."
$ "[ stack ] is " over join quackery
cr cr
exists? cr ] is task ( --> ) |
http://rosettacode.org/wiki/Dynamic_variable_names | Dynamic variable names | Task
Create a variable with a user-defined name.
The variable name should not be written in the program text, but should be taken from the user dynamically.
See also
Eval in environment is a similar task.
| #R | R | # Read the name in from a command prompt
varname <- readline("Please name your variable >")
# Make sure the name is valid for a variable
varname <- make.names(varname)
message(paste("The variable being assigned is '", varname, "'"))
# Assign the variable (with value 42) into the user workspace (global environment)
assign(varname, 42)
#Check that the value has been assigned ok
ls(pattern=varname)
get(varname) |
http://rosettacode.org/wiki/Dynamic_variable_names | Dynamic variable names | Task
Create a variable with a user-defined name.
The variable name should not be written in the program text, but should be taken from the user dynamically.
See also
Eval in environment is a similar task.
| #Racket | Racket |
-> (begin (printf "Enter some name: ")
(namespace-set-variable-value! (read) 123))
Enter some name: bleh
-> bleh
123
|
http://rosettacode.org/wiki/Draw_a_pixel | Draw a pixel | Task
Create a window and draw a pixel in it, subject to the following:
the window is 320 x 240
the color of the pixel must be red (255,0,0)
the position of the pixel is x = 100, y = 100 | #Perl | Perl | use Gtk3 '-init';
my $window = Gtk3::Window->new();
$window->set_default_size(320, 240);
$window->set_border_width(10);
$window->set_title("Draw a Pixel");
$window->set_app_paintable(TRUE);
my $da = Gtk3::DrawingArea->new();
$da->signal_connect('draw' => \&draw_in_drawingarea);
$window->add($da);
$window->show_all();
Gtk3->main;
sub draw_in_drawingarea
{
my ($widget, $cr, $data) = @_;
$cr->set_source_rgb(1, 0, 0);
$cr->set_line_width(1);
$cr->rectangle( 100, 100, 1, 1);
$cr->stroke;
} |
http://rosettacode.org/wiki/Draw_a_pixel | Draw a pixel | Task
Create a window and draw a pixel in it, subject to the following:
the window is 320 x 240
the color of the pixel must be red (255,0,0)
the position of the pixel is x = 100, y = 100 | #Phix | Phix | with javascript_semantics
include pGUI.e
cdCanvas cdcanvas
function redraw_cb(Ihandle /*ih*/, integer /*posx*/, integer /*posy*/)
cdCanvasActivate(cdcanvas)
cdCanvasPixel(cdcanvas, 100, 100, CD_RED)
cdCanvasFlush(cdcanvas)
return IUP_DEFAULT
end function
function map_cb(Ihandle ih)
cdcanvas = cdCreateCanvas(CD_IUP, ih)
return IUP_DEFAULT
end function
procedure main()
IupOpen()
Ihandle canvas = IupCanvas(NULL)
IupSetAttribute(canvas, "RASTERSIZE", "320x240")
IupSetCallback(canvas, "MAP_CB", Icallback("map_cb"))
Ihandle dlg = IupDialog(canvas)
IupSetAttribute(dlg, "TITLE", "Draw a pixel")
IupSetCallback(canvas, "ACTION", Icallback("redraw_cb"))
IupShow(dlg)
if platform()!=JS then
IupMainLoop()
IupClose()
end if
end procedure
main()
|
http://rosettacode.org/wiki/Egyptian_division | Egyptian division | Egyptian division is a method of dividing integers using addition and
doubling that is similar to the algorithm of Ethiopian multiplication
Algorithm:
Given two numbers where the dividend is to be divided by the divisor:
Start the construction of a table of two columns: powers_of_2, and doublings; by a first row of a 1 (i.e. 2^0) in the first column and 1 times the divisor in the first row second column.
Create the second row with columns of 2 (i.e 2^1), and 2 * divisor in order.
Continue with successive i’th rows of 2^i and 2^i * divisor.
Stop adding rows, and keep only those rows, where 2^i * divisor is less than or equal to the dividend.
We now assemble two separate sums that both start as zero, called here answer and accumulator
Consider each row of the table, in the reverse order of its construction.
If the current value of the accumulator added to the doublings cell would be less than or equal to the dividend then add it to the accumulator, as well as adding the powers_of_2 cell value to the answer.
When the first row has been considered as above, then the integer division of dividend by divisor is given by answer.
(And the remainder is given by the absolute value of accumulator - dividend).
Example: 580 / 34
Table creation:
powers_of_2
doublings
1
34
2
68
4
136
8
272
16
544
Initialization of sums:
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
8
272
16
544
0
0
Considering table rows, bottom-up:
When a row is considered it is shown crossed out if it is not accumulated, or bold if the row causes summations.
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
8
272
16
544
16
544
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
8
272
16
544
16
544
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
16
544
8
272
16
544
powers_of_2
doublings
answer
accumulator
1
34
2
68
16
544
4
136
8
272
16
544
powers_of_2
doublings
answer
accumulator
1
34
17
578
2
68
4
136
8
272
16
544
Answer
So 580 divided by 34 using the Egyptian method is 17 remainder (578 - 580) or 2.
Task
The task is to create a function that does Egyptian division. The function should
closely follow the description above in using a list/array of powers of two, and
another of doublings.
Functions should be clear interpretations of the algorithm.
Use the function to divide 580 by 34 and show the answer here, on this page.
Related tasks
Egyptian fractions
References
Egyptian Number System
| #Ruby | Ruby | def egyptian_divmod(dividend, divisor)
table = [[1, divisor]]
table << table.last.map{|e| e*2} while table.last.first * 2 <= dividend
answer, accumulator = 0, 0
table.reverse_each do |pow, double|
if accumulator + double <= dividend
accumulator += double
answer += pow
end
end
[answer, dividend - accumulator]
end
puts "Quotient = %s Remainder = %s" % egyptian_divmod(580, 34)
|
http://rosettacode.org/wiki/Egyptian_division | Egyptian division | Egyptian division is a method of dividing integers using addition and
doubling that is similar to the algorithm of Ethiopian multiplication
Algorithm:
Given two numbers where the dividend is to be divided by the divisor:
Start the construction of a table of two columns: powers_of_2, and doublings; by a first row of a 1 (i.e. 2^0) in the first column and 1 times the divisor in the first row second column.
Create the second row with columns of 2 (i.e 2^1), and 2 * divisor in order.
Continue with successive i’th rows of 2^i and 2^i * divisor.
Stop adding rows, and keep only those rows, where 2^i * divisor is less than or equal to the dividend.
We now assemble two separate sums that both start as zero, called here answer and accumulator
Consider each row of the table, in the reverse order of its construction.
If the current value of the accumulator added to the doublings cell would be less than or equal to the dividend then add it to the accumulator, as well as adding the powers_of_2 cell value to the answer.
When the first row has been considered as above, then the integer division of dividend by divisor is given by answer.
(And the remainder is given by the absolute value of accumulator - dividend).
Example: 580 / 34
Table creation:
powers_of_2
doublings
1
34
2
68
4
136
8
272
16
544
Initialization of sums:
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
8
272
16
544
0
0
Considering table rows, bottom-up:
When a row is considered it is shown crossed out if it is not accumulated, or bold if the row causes summations.
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
8
272
16
544
16
544
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
8
272
16
544
16
544
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
16
544
8
272
16
544
powers_of_2
doublings
answer
accumulator
1
34
2
68
16
544
4
136
8
272
16
544
powers_of_2
doublings
answer
accumulator
1
34
17
578
2
68
4
136
8
272
16
544
Answer
So 580 divided by 34 using the Egyptian method is 17 remainder (578 - 580) or 2.
Task
The task is to create a function that does Egyptian division. The function should
closely follow the description above in using a list/array of powers of two, and
another of doublings.
Functions should be clear interpretations of the algorithm.
Use the function to divide 580 by 34 and show the answer here, on this page.
Related tasks
Egyptian fractions
References
Egyptian Number System
| #Rust | Rust | fn egyptian_divide(dividend: u32, divisor: u32) -> (u32, u32) {
let dividend = dividend as u64;
let divisor = divisor as u64;
let pows = (0..32).map(|p| 1 << p);
let doublings = (0..32).map(|p| divisor << p);
let (answer, sum) = doublings
.zip(pows)
.rev()
.skip_while(|(i, _)| i > ÷nd )
.fold((0, 0), |(answer, sum), (double, power)| {
if sum + double < dividend {
(answer + power, sum + double)
} else {
(answer, sum)
}
});
(answer as u32, (dividend - sum) as u32)
}
fn main() {
let (div, rem) = egyptian_divide(580, 34);
println!("580 divided by 34 is {} remainder {}", div, rem);
} |
http://rosettacode.org/wiki/Egyptian_fractions | Egyptian fractions | An Egyptian fraction is the sum of distinct unit fractions such as:
1
2
+
1
3
+
1
16
(
=
43
48
)
{\displaystyle {\tfrac {1}{2}}+{\tfrac {1}{3}}+{\tfrac {1}{16}}\,(={\tfrac {43}{48}})}
Each fraction in the expression has a numerator equal to 1 (unity) and a denominator that is a positive integer, and all the denominators are distinct (i.e., no repetitions).
Fibonacci's Greedy algorithm for Egyptian fractions expands the fraction
x
y
{\displaystyle {\tfrac {x}{y}}}
to be represented by repeatedly performing the replacement
x
y
=
1
⌈
y
/
x
⌉
+
(
−
y
)
mod
x
y
⌈
y
/
x
⌉
{\displaystyle {\frac {x}{y}}={\frac {1}{\lceil y/x\rceil }}+{\frac {(-y)\!\!\!\!\mod x}{y\lceil y/x\rceil }}}
(simplifying the 2nd term in this replacement as necessary, and where
⌈
x
⌉
{\displaystyle \lceil x\rceil }
is the ceiling function).
For this task, Proper and improper fractions must be able to be expressed.
Proper fractions are of the form
a
b
{\displaystyle {\tfrac {a}{b}}}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are positive integers, such that
a
<
b
{\displaystyle a<b}
, and
improper fractions are of the form
a
b
{\displaystyle {\tfrac {a}{b}}}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are positive integers, such that a ≥ b.
(See the REXX programming example to view one method of expressing the whole number part of an improper fraction.)
For improper fractions, the integer part of any improper fraction should be first isolated and shown preceding the Egyptian unit fractions, and be surrounded by square brackets [n].
Task requirements
show the Egyptian fractions for:
43
48
{\displaystyle {\tfrac {43}{48}}}
and
5
121
{\displaystyle {\tfrac {5}{121}}}
and
2014
59
{\displaystyle {\tfrac {2014}{59}}}
for all proper fractions,
a
b
{\displaystyle {\tfrac {a}{b}}}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are positive one-or two-digit (decimal) integers, find and show an Egyptian fraction that has:
the largest number of terms,
the largest denominator.
for all one-, two-, and three-digit integers, find and show (as above). {extra credit}
Also see
Wolfram MathWorld™ entry: Egyptian fraction
| #Ruby | Ruby | def ef(fr)
ans = []
if fr >= 1
return [[fr.to_i], Rational(0, 1)] if fr.denominator == 1
intfr = fr.to_i
ans, fr = [intfr], fr - intfr
end
x, y = fr.numerator, fr.denominator
while x != 1
ans << Rational(1, (1/fr).ceil)
fr = Rational(-y % x, y * (1/fr).ceil)
x, y = fr.numerator, fr.denominator
end
ans << fr
end
for fr in [Rational(43, 48), Rational(5, 121), Rational(2014, 59)]
puts '%s => %s' % [fr, ef(fr).join(' + ')]
end
lenmax = denommax = [0]
for b in 2..99
for a in 1...b
fr = Rational(a,b)
e = ef(fr)
elen, edenom = e.length, e[-1].denominator
lenmax = [elen, fr] if elen > lenmax[0]
denommax = [edenom, fr] if edenom > denommax[0]
end
end
puts 'Term max is %s with %i terms' % [lenmax[1], lenmax[0]]
dstr = denommax[0].to_s
puts 'Denominator max is %s with %i digits' % [denommax[1], dstr.size], dstr |
http://rosettacode.org/wiki/Ethiopian_multiplication | Ethiopian multiplication | Ethiopian multiplication is a method of multiplying integers using only addition, doubling, and halving.
Method:
Take two numbers to be multiplied and write them down at the top of two columns.
In the left-hand column repeatedly halve the last number, discarding any remainders, and write the result below the last in the same column, until you write a value of 1.
In the right-hand column repeatedly double the last number and write the result below. stop when you add a result in the same row as where the left hand column shows 1.
Examine the table produced and discard any row where the value in the left column is even.
Sum the values in the right-hand column that remain to produce the result of multiplying the original two numbers together
For example: 17 × 34
17 34
Halving the first column:
17 34
8
4
2
1
Doubling the second column:
17 34
8 68
4 136
2 272
1 544
Strike-out rows whose first cell is even:
17 34
8 68
4 136
2 272
1 544
Sum the remaining numbers in the right-hand column:
17 34
8 --
4 ---
2 ---
1 544
====
578
So 17 multiplied by 34, by the Ethiopian method is 578.
Task
The task is to define three named functions/methods/procedures/subroutines:
one to halve an integer,
one to double an integer, and
one to state if an integer is even.
Use these functions to create a function that does Ethiopian multiplication.
References
Ethiopian multiplication explained (BBC Video clip)
A Night Of Numbers - Go Forth And Multiply (Video)
Russian Peasant Multiplication
Programming Praxis: Russian Peasant Multiplication
| #VBA | VBA | Private Function lngHalve(Nb As Long) As Long
lngHalve = Nb / 2
End Function
Private Function lngDouble(Nb As Long) As Long
lngDouble = Nb * 2
End Function
Private Function IsEven(Nb As Long) As Boolean
IsEven = (Nb Mod 2 = 0)
End Function |
http://rosettacode.org/wiki/Factorial | Factorial | Definitions
The factorial of 0 (zero) is defined as being 1 (unity).
The Factorial Function of a positive integer, n, is defined as the product of the sequence:
n, n-1, n-2, ... 1
Task
Write a function to return the factorial of a number.
Solutions can be iterative or recursive.
Support for trapping negative n errors is optional.
Related task
Primorial numbers
| #x86_Assembly | x86 Assembly | global factorial
section .text
; Input in ECX register (greater than 0!)
; Output in EAX register
factorial:
mov eax, 1
.factor:
mul ecx
loop .factor
ret |
http://rosettacode.org/wiki/Echo_server | Echo server | Create a network service that sits on TCP port 12321, which accepts connections on that port, and which echoes complete lines (using a carriage-return/line-feed sequence as line separator) back to clients. No error handling is required. For the purposes of testing, it is only necessary to support connections from localhost (127.0.0.1 or perhaps ::1). Logging of connection information to standard output is recommended.
The implementation must be able to handle simultaneous connections from multiple clients. A multi-threaded or multi-process solution may be used. Each connection must be able to echo more than a single line.
The implementation must not stop responding to other clients if one client sends a partial line or stops reading responses.
| #Ruby | Ruby | require 'socket'
server = TCPServer.new(12321)
while (connection = server.accept)
Thread.new(connection) do |conn|
port, host = conn.peeraddr[1,2]
client = "#{host}:#{port}"
puts "#{client} is connected"
begin
loop do
line = conn.readline
puts "#{client} says: #{line}"
conn.puts(line)
end
rescue EOFError
conn.close
puts "#{client} has disconnected"
end
end
end |
http://rosettacode.org/wiki/Eban_numbers | Eban numbers |
Definition
An eban number is a number that has no letter e in it when the number is spelled in English.
Or more literally, spelled numbers that contain the letter e are banned.
The American version of spelling numbers will be used here (as opposed to the British).
2,000,000,000 is two billion, not two milliard.
Only numbers less than one sextillion (1021) will be considered in/for this task.
This will allow optimizations to be used.
Task
show all eban numbers ≤ 1,000 (in a horizontal format), and a count
show all eban numbers between 1,000 and 4,000 (inclusive), and a count
show a count of all eban numbers up and including 10,000
show a count of all eban numbers up and including 100,000
show a count of all eban numbers up and including 1,000,000
show a count of all eban numbers up and including 10,000,000
show all output here.
See also
The MathWorld entry: eban numbers.
The OEIS entry: A6933, eban numbers.
| #Tailspin | Tailspin |
templates isEban
def number: $;
'$;' -> \(<'([246]|[3456][0246])(0[03456][0246])*'> $ !\) -> $number !
end isEban
|
http://rosettacode.org/wiki/Eban_numbers | Eban numbers |
Definition
An eban number is a number that has no letter e in it when the number is spelled in English.
Or more literally, spelled numbers that contain the letter e are banned.
The American version of spelling numbers will be used here (as opposed to the British).
2,000,000,000 is two billion, not two milliard.
Only numbers less than one sextillion (1021) will be considered in/for this task.
This will allow optimizations to be used.
Task
show all eban numbers ≤ 1,000 (in a horizontal format), and a count
show all eban numbers between 1,000 and 4,000 (inclusive), and a count
show a count of all eban numbers up and including 10,000
show a count of all eban numbers up and including 100,000
show a count of all eban numbers up and including 1,000,000
show a count of all eban numbers up and including 10,000,000
show all output here.
See also
The MathWorld entry: eban numbers.
The OEIS entry: A6933, eban numbers.
| #Visual_Basic_.NET | Visual Basic .NET | Module Module1
Structure Interval
Dim start As Integer
Dim last As Integer
Dim print As Boolean
Sub New(s As Integer, l As Integer, p As Boolean)
start = s
last = l
print = p
End Sub
End Structure
Sub Main()
Dim intervals As Interval() = {
New Interval(2, 1_000, True),
New Interval(1_000, 4_000, True),
New Interval(2, 10_000, False),
New Interval(2, 100_000, False),
New Interval(2, 1_000_000, False),
New Interval(2, 10_000_000, False),
New Interval(2, 100_000_000, False),
New Interval(2, 1_000_000_000, False)
}
For Each intv In intervals
If intv.start = 2 Then
Console.WriteLine("eban numbers up to and including {0}:", intv.last)
Else
Console.WriteLine("eban numbers between {0} and {1} (inclusive):", intv.start, intv.last)
End If
Dim count = 0
For i = intv.start To intv.last Step 2
Dim b = i \ 1_000_000_000
Dim r = i Mod 1_000_000_000
Dim m = r \ 1_000_000
r = i Mod 1_000_000
Dim t = r \ 1_000
r = r Mod 1_000
If m >= 30 AndAlso m <= 66 Then
m = m Mod 10
End If
If t >= 30 AndAlso t <= 66 Then
t = t Mod 10
End If
If r >= 30 AndAlso r <= 66 Then
r = r Mod 10
End If
If b = 0 OrElse b = 2 OrElse b = 4 OrElse b = 6 Then
If m = 0 OrElse m = 2 OrElse m = 4 OrElse m = 6 Then
If t = 0 OrElse t = 2 OrElse t = 4 OrElse t = 6 Then
If r = 0 OrElse r = 2 OrElse r = 4 OrElse r = 6 Then
If intv.print Then
Console.Write("{0} ", i)
End If
count += 1
End If
End If
End If
End If
Next
If intv.print Then
Console.WriteLine()
End If
Console.WriteLine("count = {0}", count)
Console.WriteLine()
Next
End Sub
End Module |
http://rosettacode.org/wiki/Draw_a_rotating_cube | Draw a rotating cube | Task
Draw a rotating cube.
It should be oriented with one vertex pointing straight up, and its opposite vertex on the main diagonal (the one farthest away) straight down. It can be solid or wire-frame, and you can use ASCII art if your language doesn't have graphical capabilities. Perspective is optional.
Related tasks
Draw a cuboid
write language name in 3D ASCII
| #Lua | Lua | local abs,atan,cos,floor,pi,sin,sqrt = math.abs,math.atan,math.cos,math.floor,math.pi,math.sin,math.sqrt
local bitmap = {
init = function(self, w, h, value)
self.w, self.h, self.pixels = w, h, {}
for y=1,h do self.pixels[y]={} end
self:clear(value)
end,
clear = function(self, value)
for y=1,self.h do
for x=1,self.w do
self.pixels[y][x] = value or " "
end
end
end,
set = function(self, x, y, value)
x,y = floor(x),floor(y)
if x>0 and y>0 and x<=self.w and y<=self.h then
self.pixels[y][x] = value or "#"
end
end,
line = function(self, x1, y1, x2, y2, c)
x1,y1,x2,y2 = floor(x1),floor(y1),floor(x2),floor(y2)
local dx, sx = abs(x2-x1), x1<x2 and 1 or -1
local dy, sy = abs(y2-y1), y1<y2 and 1 or -1
local err = floor((dx>dy and dx or -dy)/2)
while(true) do
self:set(x1, y1, c)
if (x1==x2 and y1==y2) then break end
if (err > -dx) then
err, x1 = err-dy, x1+sx
if (x1==x2 and y1==y2) then
self:set(x1, y1, c)
break
end
end
if (err < dy) then
err, y1 = err+dx, y1+sy
end
end
end,
render = function(self)
for y=1,self.h do
print(table.concat(self.pixels[y]))
end
end,
}
screen = {
clear = function()
os.execute("cls") -- or? os.execute("clear"), or? io.write("\027[2J\027[H"), or etc?
end,
}
local camera = { fl = 2.5 }
local L = 0.5
local cube = {
verts = { {L,L,L}, {L,-L,L}, {-L,-L,L}, {-L,L,L}, {L,L,-L}, {L,-L,-L}, {-L,-L,-L}, {-L,L,-L} },
edges = { {1,2}, {2,3}, {3,4}, {4,1}, {5,6}, {6,7}, {7,8}, {8,5}, {1,5}, {2,6}, {3,7}, {4,8} },
rotate = function(self, rx, ry)
local cx,sx = cos(rx),sin(rx)
local cy,sy = cos(ry),sin(ry)
for i,v in ipairs(self.verts) do
local x,y,z = v[1],v[2],v[3]
v[1], v[2], v[3] = x*cx-z*sx, y*cy-x*sx*sy-z*cx*sy, x*sx*cy+y*sy+z*cx*cy
end
end,
}
local renderer = {
render = function(self, shape, camera, bitmap)
local fl = camera.fl
local ox, oy = bitmap.w/2, bitmap.h/2
local mx, my = bitmap.w/2, bitmap.h/2
local rpverts = {}
for i,v in ipairs(shape.verts) do
local x,y,z = v[1],v[2],v[3]
local px = ox + mx * (fl*x)/(fl-z)
local py = oy + my * (fl*y)/(fl-z)
rpverts[i] = { px,py }
end
for i,e in ipairs(shape.edges) do
local v1, v2 = rpverts[e[1]], rpverts[e[2]]
bitmap:line( v1[1], v1[2], v2[1], v2[2], "██" )
end
end
}
--
bitmap:init(40,40)
cube:rotate(pi/4, atan(sqrt(2)))
for i=1,60 do
cube:rotate(pi/60,0)
bitmap:clear("··")
renderer:render(cube, camera, bitmap)
screen:clear()
bitmap:render()
end |
http://rosettacode.org/wiki/Element-wise_operations | Element-wise operations | This task is similar to:
Matrix multiplication
Matrix transposition
Task
Implement basic element-wise matrix-matrix and scalar-matrix operations, which can be referred to in other, higher-order tasks.
Implement:
addition
subtraction
multiplication
division
exponentiation
Extend the task if necessary to include additional basic operations, which should not require their own specialised task.
| #R | R | # create a 2-times-2 matrix
mat <- matrix(1:4, 2, 2)
# matrix with scalar
mat + 2
mat * 2
mat ^ 2
# matrix with matrix
mat + mat
mat * mat
mat ^ mat |
http://rosettacode.org/wiki/Draw_a_sphere | Draw a sphere | Task
Draw a sphere.
The sphere can be represented graphically, or in ASCII art, depending on the language capabilities.
Either static or rotational projection is acceptable for this task.
Related tasks
draw a cuboid
draw a rotating cube
write language name in 3D ASCII
draw a Deathstar
| #Brlcad | Brlcad | opendb balls.g y # Create a database to hold our shapes
units cm # Set the unit of measure
in ball.s sph 0 0 0 3 # Create a sphere of radius 3 cm named ball.s with its centre at 0,0,0 |
http://rosettacode.org/wiki/Doubly-linked_list/Element_insertion | Doubly-linked list/Element insertion | Doubly-Linked List (element)
This is much like inserting into a Singly-Linked List, but with added assignments so that the backwards-pointing links remain correct.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack | #JavaScript | JavaScript | DoublyLinkedList.prototype.insertAfter = function(searchValue, nodeToInsert) {
if (this._value == searchValue) {
var after = this.next();
this.next(nodeToInsert);
nodeToInsert.prev(this);
nodeToInsert.next(after);
after.prev(nodeToInsert);
}
else if (this.next() == null)
throw new Error(0, "value '" + searchValue + "' not found in linked list.")
else
this.next().insertAfter(searchValue, nodeToInsert);
}
var list = createDoublyLinkedListFromArray(['A','B']);
list.insertAfter('A', new DoublyLinkedList('C', null, null)); |
http://rosettacode.org/wiki/Doubly-linked_list/Element_insertion | Doubly-linked list/Element insertion | Doubly-Linked List (element)
This is much like inserting into a Singly-Linked List, but with added assignments so that the backwards-pointing links remain correct.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack | #Julia | Julia | mutable struct DLNode{T}
value::T
pred::Union{DLNode{T}, Nothing}
succ::Union{DLNode{T}, Nothing}
DLNode(v) = new{typeof(v)}(v, nothing, nothing)
end
function insertpost(prevnode, node)
succ = prevnode.succ
prevnode.succ = node
node.pred = prevnode
node.succ = succ
if succ != nothing
succ.pred = node
end
node
end
function printfromroot(root)
print(root.value)
while root.succ != nothing
root = root.succ
print(" -> $(root.value)")
end
println()
end
node1 = DLNode(1)
printfromroot(node1)
node2 = DLNode(2)
node3 = DLNode(3)
insertpost(node1, node2)
printfromroot(node1)
insertpost(node1, node3)
printfromroot(node1)
|
http://rosettacode.org/wiki/Draw_a_clock | Draw a clock | Task
Draw a clock.
More specific:
Draw a time keeping device. It can be a stopwatch, hourglass, sundial, a mouth counting "one thousand and one", anything. Only showing the seconds is required, e.g.: a watch with just a second hand will suffice. However, it must clearly change every second, and the change must cycle every so often (one minute, 30 seconds, etc.) It must be drawn; printing a string of numbers to your terminal doesn't qualify. Both text-based and graphical drawing are OK.
The clock is unlikely to be used to control space flights, so it needs not be hyper-accurate, but it should be usable, meaning if one can read the seconds off the clock, it must agree with the system clock.
A clock is rarely (never?) a major application: don't be a CPU hog and poll the system timer every microsecond, use a proper timer/signal/event from your system or language instead. For a bad example, many OpenGL programs update the frame-buffer in a busy loop even if no redraw is needed, which is very undesirable for this task.
A clock is rarely (never?) a major application: try to keep your code simple and to the point. Don't write something too elaborate or convoluted, instead do whatever is natural, concise and clear in your language.
Key points
animate simple object
timed event
polling system resources
code clarity
| #Ada | Ada | with Ada.Numerics.Elementary_Functions;
with Ada.Calendar.Formatting;
with Ada.Calendar.Time_Zones;
with SDL.Video.Windows.Makers;
with SDL.Video.Renderers.Makers;
with SDL.Events.Events;
procedure Draw_A_Clock is
use Ada.Calendar;
use Ada.Calendar.Formatting;
Window : SDL.Video.Windows.Window;
Renderer : SDL.Video.Renderers.Renderer;
Event : SDL.Events.Events.Events;
Offset : Time_Zones.Time_Offset;
procedure Draw_Clock (Stamp : Time)
is
use SDL.C;
use Ada.Numerics.Elementary_Functions;
Radi : constant array (0 .. 59) of int := (0 | 15 | 30 | 45 => 2,
5 | 10 | 20 | 25 | 35 | 40 | 50 | 55 => 1,
others => 0);
Diam : constant array (0 .. 59) of int := (0 | 15 | 30 | 45 => 5,
5 | 10 | 20 | 25 | 35 | 40 | 50 | 55 => 3,
others => 1);
Width : constant int := Window.Get_Surface.Size.Width;
Height : constant int := Window.Get_Surface.Size.Height;
Radius : constant Float := Float (int'Min (Width, Height));
R_1 : constant Float := 0.48 * Radius;
R_2 : constant Float := 0.35 * Radius;
R_3 : constant Float := 0.45 * Radius;
R_4 : constant Float := 0.47 * Radius;
Hour : constant Hour_Number := Formatting.Hour (Stamp, Offset);
Minute : constant Minute_Number := Formatting.Minute (Stamp, Offset);
Second : constant Second_Number := Formatting.Second (Stamp);
function To_X (A : Float; R : Float) return int is
(Width / 2 + int (R * Sin (A, 60.0)));
function To_Y (A : Float; R : Float) return int is
(Height / 2 - int (R * Cos (A, 60.0)));
begin
SDL.Video.Renderers.Makers.Create (Renderer, Window.Get_Surface);
Renderer.Set_Draw_Colour ((0, 0, 150, 255));
Renderer.Fill (Rectangle => (0, 0, Width, Height));
Renderer.Set_Draw_Colour ((200, 200, 200, 255));
for A in 0 .. 59 loop
Renderer.Fill (Rectangle => (To_X (Float (A), R_1) - Radi (A),
To_Y (Float (A), R_1) - Radi (A), Diam (A), Diam (A)));
end loop;
Renderer.Set_Draw_Colour ((200, 200, 0, 255));
Renderer.Draw (Line => ((Width / 2, Height / 2),
(To_X (5.0 * (Float (Hour) + Float (Minute) / 60.0), R_2),
To_Y (5.0 * (Float (Hour) + Float (Minute) / 60.0), R_2))));
Renderer.Draw (Line => ((Width / 2, Height / 2),
(To_X (Float (Minute) + Float (Second) / 60.0, R_3),
To_Y (Float (Minute) + Float (Second) / 60.0, R_3))));
Renderer.Set_Draw_Colour ((220, 0, 0, 255));
Renderer.Draw (Line => ((Width / 2, Height / 2),
(To_X (Float (Second), R_4),
To_Y (Float (Second), R_4))));
Renderer.Fill (Rectangle => (Width / 2 - 3, Height / 2 - 3, 7, 7));
end Draw_Clock;
function Poll_Quit return Boolean is
use type SDL.Events.Event_Types;
begin
while SDL.Events.Events.Poll (Event) loop
if Event.Common.Event_Type = SDL.Events.Quit then
return True;
end if;
end loop;
return False;
end Poll_Quit;
begin
Offset := Time_Zones.UTC_Time_Offset;
if not SDL.Initialise (Flags => SDL.Enable_Screen) then
return;
end if;
SDL.Video.Windows.Makers.Create (Win => Window,
Title => "Draw a clock",
Position => SDL.Natural_Coordinates'(X => 10, Y => 10),
Size => SDL.Positive_Sizes'(300, 300),
Flags => SDL.Video.Windows.Resizable);
loop
Draw_Clock (Clock);
Window.Update_Surface;
delay 0.200;
exit when Poll_Quit;
end loop;
Window.Finalize;
SDL.Finalise;
end Draw_A_Clock; |
http://rosettacode.org/wiki/Doubly-linked_list/Traversal | Doubly-linked list/Traversal | Traverse from the beginning of a doubly-linked list to the end, and from the end to the beginning.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Go | Go | package main
import "fmt"
type dlNode struct {
string
next, prev *dlNode
}
type dlList struct {
head, tail *dlNode
}
func (list *dlList) String() string {
if list.head == nil {
return fmt.Sprint(list.head)
}
r := "[" + list.head.string
for p := list.head.next; p != nil; p = p.next {
r += " " + p.string
}
return r + "]"
}
func (list *dlList) insertTail(node *dlNode) {
if list.tail == nil {
list.head = node
} else {
list.tail.next = node
}
node.next = nil
node.prev = list.tail
list.tail = node
}
func (list *dlList) insertAfter(existing, insert *dlNode) {
insert.prev = existing
insert.next = existing.next
existing.next.prev = insert
existing.next = insert
if existing == list.tail {
list.tail = insert
}
}
func main() {
dll := &dlList{}
fmt.Println(dll)
a := &dlNode{string: "A"}
dll.insertTail(a)
dll.insertTail(&dlNode{string: "B"})
fmt.Println(dll)
dll.insertAfter(a, &dlNode{string: "C"})
fmt.Println(dll)
// traverse from end to beginning
fmt.Print("From tail:")
for p := dll.tail; p != nil; p = p.prev {
fmt.Print(" ", p.string)
}
fmt.Println("")
} |
http://rosettacode.org/wiki/Doubly-linked_list/Element_definition | Doubly-linked list/Element definition | Task
Define the data structure for a doubly-linked list element.
The element should include a data member to hold its value and pointers to both the next element in the list and the previous element in the list.
The pointers should be mutable.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #F.23 | F# |
type 'a DLElm = {
mutable prev: 'a DLElm option
data: 'a
mutable next: 'a DLElm option
}
|
http://rosettacode.org/wiki/Doubly-linked_list/Element_definition | Doubly-linked list/Element definition | Task
Define the data structure for a doubly-linked list element.
The element should include a data member to hold its value and pointers to both the next element in the list and the previous element in the list.
The pointers should be mutable.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Factor | Factor | TUPLE: node data next prev ; |
http://rosettacode.org/wiki/Doubly-linked_list/Element_definition | Doubly-linked list/Element definition | Task
Define the data structure for a doubly-linked list element.
The element should include a data member to hold its value and pointers to both the next element in the list and the previous element in the list.
The pointers should be mutable.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #Fortran | Fortran | type node
real :: data
type(node), pointer :: next => null(), previous => null()
end type node
!
! . . . .
!
type( node ), target :: head |
http://rosettacode.org/wiki/Doubly-linked_list/Element_definition | Doubly-linked list/Element definition | Task
Define the data structure for a doubly-linked list element.
The element should include a data member to hold its value and pointers to both the next element in the list and the previous element in the list.
The pointers should be mutable.
See also
Array
Associative array: Creation, Iteration
Collections
Compound data type
Doubly-linked list: Definition, Element definition, Element insertion, List Traversal, Element Removal
Linked list
Queue: Definition, Usage
Set
Singly-linked list: Element definition, Element insertion, List Traversal, Element Removal
Stack
| #FreeBASIC | FreeBASIC | type node
nxt as node ptr
prv as node ptr
dat as any ptr 'points to any kind of data; user's responsibility
'to keep track of what's actually in it
end type |
http://rosettacode.org/wiki/Dutch_national_flag_problem | Dutch national flag problem |
The Dutch national flag is composed of three coloured bands in the order:
red (top)
then white, and
lastly blue (at the bottom).
The problem posed by Edsger Dijkstra is:
Given a number of red, blue and white balls in random order, arrange them in the order of the colours in the Dutch national flag.
When the problem was first posed, Dijkstra then went on to successively refine a solution, minimising the number of swaps and the number of times the colour of a ball needed to determined and restricting the balls to end in an array, ...
Task
Generate a randomized order of balls ensuring that they are not in the order of the Dutch national flag.
Sort the balls in a way idiomatic to your language.
Check the sorted balls are in the order of the Dutch national flag.
C.f.
Dutch national flag problem
Probabilistic analysis of algorithms for the Dutch national flag problem by Wei-Mei Chen. (pdf)
| #F.23 | F# | (* Since the task description here does not impose Dijsktra's original restrictions
* Changing the order is only allowed by swapping 2 elements
* Every element must only be inspected once
we have several options ...
One way -- especially when we work with immutable data structures --
is to scan the unordered list, collect the different
colours on our way and append the 3 sub-lists in the correct order.
*)
let rnd = System.Random()
type color = | Red | White | Blue
let isDutch s =
Seq.forall2 (fun last this ->
match (last, this) with
| (Red, Red) | (Red, White) | (White, White) | (White, Blue) | (Blue, Blue) -> true | _ -> false
) s (Seq.skip 1 s)
[<EntryPoint>]
let main argv =
let n = 10
let rec getBallsToSort n s =
let sn = Seq.take n s
if (isDutch sn) then (getBallsToSort n (Seq.skip 1 s)) else sn
let balls = getBallsToSort n (Seq.initInfinite (fun _ -> match (rnd.Next(3)) with | 0 -> Red | 1 -> White | _ -> Blue))
printfn "Sort the sequence of %i balls: %A" n (Seq.toList balls)
let (rs,ws,bs) =
balls
|> Seq.fold (fun (rs,ws,bs) b ->
match b with | Red -> (b::rs,ws,bs) | White -> (rs,b::ws,bs) | Blue -> (rs,ws,b::bs))
([],[],[])
let sorted = rs @ ws @ bs
printfn "The sequence %A is sorted: %b" sorted (isDutch sorted)
0 |
http://rosettacode.org/wiki/Draw_a_cuboid | Draw a cuboid | Task
Draw a cuboid with relative dimensions of 2 × 3 × 4.
The cuboid can be represented graphically, or in ASCII art, depending on the language capabilities.
To fulfill the criteria of being a cuboid, three faces must be visible.
Either static or rotational projection is acceptable for this task.
Related tasks
draw a sphere
draw a rotating cube
write language name in 3D ASCII
draw a Deathstar
| #Clojure | Clojure |
(use 'quil.core)
(def w 500)
(def h 400)
(defn setup []
(background 0))
(defn draw []
(push-matrix)
(translate (/ w 2) (/ h 2) 0)
(rotate-x 0.7)
(rotate-z 0.5)
(box 100 150 200) ; 2x3x4 relative dimensions
(pop-matrix))
(defsketch main
:title "cuboid"
:setup setup
:size [w h]
:draw draw
:renderer :opengl)
|
http://rosettacode.org/wiki/Dynamic_variable_names | Dynamic variable names | Task
Create a variable with a user-defined name.
The variable name should not be written in the program text, but should be taken from the user dynamically.
See also
Eval in environment is a similar task.
| #Raku | Raku | our $our-var = 'The our var';
my $my-var = 'The my var';
my $name = prompt 'Variable name: ';
my $value = $::('name'); # use the right sigil, etc
put qq/Var ($name) starts with value ｢$value｣/;
$::('name') = 137;
put qq/Var ($name) ends with value ｢{$::('name')}｣/;
|
http://rosettacode.org/wiki/Dynamic_variable_names | Dynamic variable names | Task
Create a variable with a user-defined name.
The variable name should not be written in the program text, but should be taken from the user dynamically.
See also
Eval in environment is a similar task.
| #REBOL | REBOL | rebol [
Title: "Dynamic Variable Name"
URL: http://rosettacode.org/wiki/Dynamic_variable_names
]
; Here, I ask the user for a name, then convert it to a word and
; assign the value "Hello!" to it. To read this phrase, realize that
; REBOL collects terms from right to left, so "Hello!" is stored for
; future use, then the prompt string "Variable name? " is used as the
; argument to ask (prompts user for input). The result of ask is
; converted to a word so it can be an identifier, then the 'set' word
; accepts the new word and the string ("Hello!") to be assigned.
set to-word ask "Variable name? " "Hello!" |
http://rosettacode.org/wiki/Draw_a_pixel | Draw a pixel | Task
Create a window and draw a pixel in it, subject to the following:
the window is 320 x 240
the color of the pixel must be red (255,0,0)
the position of the pixel is x = 100, y = 100 | #Processing | Processing | size(320, 240);
set(100, 100, color(255,0,0)); |
http://rosettacode.org/wiki/Draw_a_pixel | Draw a pixel | Task
Create a window and draw a pixel in it, subject to the following:
the window is 320 x 240
the color of the pixel must be red (255,0,0)
the position of the pixel is x = 100, y = 100 | #PureBasic | PureBasic |
If OpenWindow(0, 0, 0, 320, 240, "Rosetta Code Draw A Pixel in PureBasic")
If CreateImage(0, 320, 240) And StartDrawing(ImageOutput(0))
Plot(100, 100,RGB(255,0,0))
StopDrawing()
ImageGadget(0, 0, 0, 320, 240, ImageID(0))
EndIf
Repeat
Event = WaitWindowEvent()
Until Event = #PB_Event_CloseWindow
EndIf
|
http://rosettacode.org/wiki/Egyptian_division | Egyptian division | Egyptian division is a method of dividing integers using addition and
doubling that is similar to the algorithm of Ethiopian multiplication
Algorithm:
Given two numbers where the dividend is to be divided by the divisor:
Start the construction of a table of two columns: powers_of_2, and doublings; by a first row of a 1 (i.e. 2^0) in the first column and 1 times the divisor in the first row second column.
Create the second row with columns of 2 (i.e 2^1), and 2 * divisor in order.
Continue with successive i’th rows of 2^i and 2^i * divisor.
Stop adding rows, and keep only those rows, where 2^i * divisor is less than or equal to the dividend.
We now assemble two separate sums that both start as zero, called here answer and accumulator
Consider each row of the table, in the reverse order of its construction.
If the current value of the accumulator added to the doublings cell would be less than or equal to the dividend then add it to the accumulator, as well as adding the powers_of_2 cell value to the answer.
When the first row has been considered as above, then the integer division of dividend by divisor is given by answer.
(And the remainder is given by the absolute value of accumulator - dividend).
Example: 580 / 34
Table creation:
powers_of_2
doublings
1
34
2
68
4
136
8
272
16
544
Initialization of sums:
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
8
272
16
544
0
0
Considering table rows, bottom-up:
When a row is considered it is shown crossed out if it is not accumulated, or bold if the row causes summations.
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
8
272
16
544
16
544
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
8
272
16
544
16
544
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
16
544
8
272
16
544
powers_of_2
doublings
answer
accumulator
1
34
2
68
16
544
4
136
8
272
16
544
powers_of_2
doublings
answer
accumulator
1
34
17
578
2
68
4
136
8
272
16
544
Answer
So 580 divided by 34 using the Egyptian method is 17 remainder (578 - 580) or 2.
Task
The task is to create a function that does Egyptian division. The function should
closely follow the description above in using a list/array of powers of two, and
another of doublings.
Functions should be clear interpretations of the algorithm.
Use the function to divide 580 by 34 and show the answer here, on this page.
Related tasks
Egyptian fractions
References
Egyptian Number System
| #Scala | Scala | object EgyptianDivision extends App {
private def divide(dividend: Int, divisor: Int): Unit = {
val powersOf2, doublings = new collection.mutable.ListBuffer[Integer]
//populate the powersof2- and doublings-columns
var line = 0
while ((math.pow(2, line) * divisor) <= dividend) {
val powerOf2 = math.pow(2, line).toInt
powersOf2 += powerOf2
doublings += (powerOf2 * divisor)
line += 1
}
var answer, accumulator = 0
//Consider the rows in reverse order of their construction (from back to front of the List)
var i = powersOf2.size - 1
for (i <- powersOf2.size - 1 to 0 by -1)
if (accumulator + doublings(i) <= dividend) {
accumulator += doublings(i)
answer += powersOf2(i)
}
println(f"$answer%d, remainder ${dividend - accumulator}%d")
}
divide(580, 34)
} |
http://rosettacode.org/wiki/Egyptian_division | Egyptian division | Egyptian division is a method of dividing integers using addition and
doubling that is similar to the algorithm of Ethiopian multiplication
Algorithm:
Given two numbers where the dividend is to be divided by the divisor:
Start the construction of a table of two columns: powers_of_2, and doublings; by a first row of a 1 (i.e. 2^0) in the first column and 1 times the divisor in the first row second column.
Create the second row with columns of 2 (i.e 2^1), and 2 * divisor in order.
Continue with successive i’th rows of 2^i and 2^i * divisor.
Stop adding rows, and keep only those rows, where 2^i * divisor is less than or equal to the dividend.
We now assemble two separate sums that both start as zero, called here answer and accumulator
Consider each row of the table, in the reverse order of its construction.
If the current value of the accumulator added to the doublings cell would be less than or equal to the dividend then add it to the accumulator, as well as adding the powers_of_2 cell value to the answer.
When the first row has been considered as above, then the integer division of dividend by divisor is given by answer.
(And the remainder is given by the absolute value of accumulator - dividend).
Example: 580 / 34
Table creation:
powers_of_2
doublings
1
34
2
68
4
136
8
272
16
544
Initialization of sums:
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
8
272
16
544
0
0
Considering table rows, bottom-up:
When a row is considered it is shown crossed out if it is not accumulated, or bold if the row causes summations.
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
8
272
16
544
16
544
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
8
272
16
544
16
544
powers_of_2
doublings
answer
accumulator
1
34
2
68
4
136
16
544
8
272
16
544
powers_of_2
doublings
answer
accumulator
1
34
2
68
16
544
4
136
8
272
16
544
powers_of_2
doublings
answer
accumulator
1
34
17
578
2
68
4
136
8
272
16
544
Answer
So 580 divided by 34 using the Egyptian method is 17 remainder (578 - 580) or 2.
Task
The task is to create a function that does Egyptian division. The function should
closely follow the description above in using a list/array of powers of two, and
another of doublings.
Functions should be clear interpretations of the algorithm.
Use the function to divide 580 by 34 and show the answer here, on this page.
Related tasks
Egyptian fractions
References
Egyptian Number System
| #Sidef | Sidef | func egyptian_divmod(dividend, divisor) {
var table = [[1, divisor]]
table << table[-1].map{|e| 2*e } while (2*table[-1][0] <= dividend)
var (answer, accumulator) = (0, 0)
table.reverse.each { |pair|
var (pow, double) = pair...
if (accumulator + double <= dividend) {
accumulator += double
answer += pow
}
}
return (answer, dividend - accumulator)
}
say ("Quotient = %s Remainder = %s" % egyptian_divmod(580, 34)) |
http://rosettacode.org/wiki/Egyptian_fractions | Egyptian fractions | An Egyptian fraction is the sum of distinct unit fractions such as:
1
2
+
1
3
+
1
16
(
=
43
48
)
{\displaystyle {\tfrac {1}{2}}+{\tfrac {1}{3}}+{\tfrac {1}{16}}\,(={\tfrac {43}{48}})}
Each fraction in the expression has a numerator equal to 1 (unity) and a denominator that is a positive integer, and all the denominators are distinct (i.e., no repetitions).
Fibonacci's Greedy algorithm for Egyptian fractions expands the fraction
x
y
{\displaystyle {\tfrac {x}{y}}}
to be represented by repeatedly performing the replacement
x
y
=
1
⌈
y
/
x
⌉
+
(
−
y
)
mod
x
y
⌈
y
/
x
⌉
{\displaystyle {\frac {x}{y}}={\frac {1}{\lceil y/x\rceil }}+{\frac {(-y)\!\!\!\!\mod x}{y\lceil y/x\rceil }}}
(simplifying the 2nd term in this replacement as necessary, and where
⌈
x
⌉
{\displaystyle \lceil x\rceil }
is the ceiling function).
For this task, Proper and improper fractions must be able to be expressed.
Proper fractions are of the form
a
b
{\displaystyle {\tfrac {a}{b}}}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are positive integers, such that
a
<
b
{\displaystyle a<b}
, and
improper fractions are of the form
a
b
{\displaystyle {\tfrac {a}{b}}}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are positive integers, such that a ≥ b.
(See the REXX programming example to view one method of expressing the whole number part of an improper fraction.)
For improper fractions, the integer part of any improper fraction should be first isolated and shown preceding the Egyptian unit fractions, and be surrounded by square brackets [n].
Task requirements
show the Egyptian fractions for:
43
48
{\displaystyle {\tfrac {43}{48}}}
and
5
121
{\displaystyle {\tfrac {5}{121}}}
and
2014
59
{\displaystyle {\tfrac {2014}{59}}}
for all proper fractions,
a
b
{\displaystyle {\tfrac {a}{b}}}
where
a
{\displaystyle a}
and
b
{\displaystyle b}
are positive one-or two-digit (decimal) integers, find and show an Egyptian fraction that has:
the largest number of terms,
the largest denominator.
for all one-, two-, and three-digit integers, find and show (as above). {extra credit}
Also see
Wolfram MathWorld™ entry: Egyptian fraction
| #Rust | Rust |
use num_bigint::BigInt;
use num_integer::Integer;
use num_traits::{One, Zero};
use std::fmt;
#[derive(Debug, Clone, PartialEq, PartialOrd)]
struct Rational {
nominator: BigInt,
denominator: BigInt,
}
impl Rational {
fn new(n: &BigInt, d: &BigInt) -> Rational {
assert!(!d.is_zero(), "denominator cannot be 0");
// simplify if possible
let c = n.gcd(d);
Rational {
nominator: n / &c,
denominator: d / &c,
}
}
fn is_proper(&self) -> bool {
self.nominator < self.denominator
}
fn to_egyptian(&self) -> Vec<Rational> {
let mut frac: Vec<Rational> = Vec::new();
let mut current: Rational;
if !self.is_proper() {
// input is grater than 1
// store the integer part
frac.push(Rational::new(
&self.nominator.div_floor(&self.denominator),
&One::one(),
));
// calculate the remainder
current = Rational::new(
&self.nominator.mod_floor(&self.denominator),
&self.denominator,
);
} else {
current = self.clone();
}
while !current.nominator.is_one() {
let div = current.denominator.div_ceil(¤t.nominator);
// store the term
frac.push(Rational::new(&One::one(), &div));
current = Rational::new(
&(-¤t.denominator).mod_floor(¤t.nominator),
match current.denominator.checked_mul(&div).as_ref() {
Some(r) => r,
_ => break,
},
);
}
frac.push(current);
frac
}
}
impl fmt::Display for Rational {
fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
if self.denominator.is_one() {
// for integers only display the integer part
write!(f, "{}", self.nominator)
} else {
write!(f, "{}/{}", self.nominator, self.denominator)
}
}
}
fn rational_vec_to_string(vec: Vec<Rational>) -> String {
let mut p = vec
.iter()
.fold(String::new(), |acc, num| (acc + &num.to_string() + ", "));
if p.len() > 1 {
p.truncate(p.len() - 2);
}
format!("[{}]", p)
}
fn run_max_searches(x: usize) {
// generate all proper fractions with 2 digits
let pairs = (1..x).flat_map(move |i| (i + 1..x).map(move |j| (i, j)));
let mut max_length = (0, Rational::new(&BigInt::from(1), &BigInt::from(1)));
let mut max_denom = (
Zero::zero(),
Rational::new(&BigInt::from(1), &BigInt::from(1)),
);
for (i, j) in pairs {
let e = Rational::new(&BigInt::from(i), &BigInt::from(j)).to_egyptian();
if e.len() > max_length.0 {
max_length = (e.len(), Rational::new(&BigInt::from(i), &BigInt::from(j)));
}
if e.last().unwrap().denominator > max_denom.0 {
max_denom = (
e.last().unwrap().denominator.clone(),
Rational::new(&BigInt::from(i), &BigInt::from(j)),
);
}
}
println!(
"Maximum length of terms is for {} with {} terms",
max_length.1, max_length.0
);
println!("{}", rational_vec_to_string(max_length.1.to_egyptian()));
println!(
"Maximum denominator is for {} with {} terms",
max_denom.1, max_denom.0
);
println!("{}", rational_vec_to_string(max_denom.1.to_egyptian()));
}
fn main() {
let tests = [
Rational::new(&BigInt::from(43), &BigInt::from(48)),
Rational::new(&BigInt::from(5), &BigInt::from(121)),
Rational::new(&BigInt::from(2014), &BigInt::from(59)),
];
for test in tests.iter() {
println!("{} -> {}", test, rational_vec_to_string(test.to_egyptian()));
}
run_max_searches(100);
run_max_searches(1000);
}
|
http://rosettacode.org/wiki/Ethiopian_multiplication | Ethiopian multiplication | Ethiopian multiplication is a method of multiplying integers using only addition, doubling, and halving.
Method:
Take two numbers to be multiplied and write them down at the top of two columns.
In the left-hand column repeatedly halve the last number, discarding any remainders, and write the result below the last in the same column, until you write a value of 1.
In the right-hand column repeatedly double the last number and write the result below. stop when you add a result in the same row as where the left hand column shows 1.
Examine the table produced and discard any row where the value in the left column is even.
Sum the values in the right-hand column that remain to produce the result of multiplying the original two numbers together
For example: 17 × 34
17 34
Halving the first column:
17 34
8
4
2
1
Doubling the second column:
17 34
8 68
4 136
2 272
1 544
Strike-out rows whose first cell is even:
17 34
8 68
4 136
2 272
1 544
Sum the remaining numbers in the right-hand column:
17 34
8 --
4 ---
2 ---
1 544
====
578
So 17 multiplied by 34, by the Ethiopian method is 578.
Task
The task is to define three named functions/methods/procedures/subroutines:
one to halve an integer,
one to double an integer, and
one to state if an integer is even.
Use these functions to create a function that does Ethiopian multiplication.
References
Ethiopian multiplication explained (BBC Video clip)
A Night Of Numbers - Go Forth And Multiply (Video)
Russian Peasant Multiplication
Programming Praxis: Russian Peasant Multiplication
| #VBScript | VBScript | option explicit
class List
private theList
private nOccupiable
private nTop
sub class_initialize
nTop = 0
nOccupiable = 100
redim theList( nOccupiable )
end sub
public sub store( x )
if nTop >= nOccupiable then
nOccupiable = nOccupiable + 100
redim preserve theList( nOccupiable )
end if
theList( nTop ) = x
nTop = nTop + 1
end sub
public function recall( n )
if n >= 0 and n <= nOccupiable then
recall = theList( n )
else
err.raise vbObjectError + 1000,,"Recall bounds error"
end if
end function
public sub replace( n, x )
if n >= 0 and n <= nOccupiable then
theList( n ) = x
else
err.raise vbObjectError + 1001,,"Replace bounds error"
end if
end sub
public property get listCount
listCount = nTop
end property
end class
function halve( n )
halve = int( n / 2 )
end function
function twice( n )
twice = int( n * 2 )
end function
function iseven( n )
iseven = ( ( n mod 2 ) = 0 )
end function
function multiply( n1, n2 )
dim LL
set LL = new List
dim RR
set RR = new List
LL.store n1
RR.store n2
do while n1 <> 1
n1 = halve( n1 )
LL.store n1
n2 = twice( n2 )
RR.store n2
loop
dim i
for i = 0 to LL.listCount
if iseven( LL.recall( i ) ) then
RR.replace i, 0
end if
next
dim total
total = 0
for i = 0 to RR.listCount
total = total + RR.recall( i )
next
multiply = total
end function
|
http://rosettacode.org/wiki/Factorial | Factorial | Definitions
The factorial of 0 (zero) is defined as being 1 (unity).
The Factorial Function of a positive integer, n, is defined as the product of the sequence:
n, n-1, n-2, ... 1
Task
Write a function to return the factorial of a number.
Solutions can be iterative or recursive.
Support for trapping negative n errors is optional.
Related task
Primorial numbers
| #XL | XL | 0! -> 1
N! -> N * (N-1)! |
http://rosettacode.org/wiki/Echo_server | Echo server | Create a network service that sits on TCP port 12321, which accepts connections on that port, and which echoes complete lines (using a carriage-return/line-feed sequence as line separator) back to clients. No error handling is required. For the purposes of testing, it is only necessary to support connections from localhost (127.0.0.1 or perhaps ::1). Logging of connection information to standard output is recommended.
The implementation must be able to handle simultaneous connections from multiple clients. A multi-threaded or multi-process solution may be used. Each connection must be able to echo more than a single line.
The implementation must not stop responding to other clients if one client sends a partial line or stops reading responses.
| #Rust | Rust |
use std::net::{TcpListener, TcpStream};
use std::io::{BufReader, BufRead, Write};
use std::thread;
fn main() {
let listener = TcpListener::bind("127.0.0.1:12321").unwrap();
println!("server is running on 127.0.0.1:12321 ...");
for stream in listener.incoming() {
let stream = stream.unwrap();
thread::spawn(move || handle_client(stream));
}
}
fn handle_client(stream: TcpStream) {
let mut stream = BufReader::new(stream);
loop {
let mut buf = String::new();
if stream.read_line(&mut buf).is_err() {
break;
}
stream
.get_ref()
.write(buf.as_bytes())
.unwrap();
}
}
|
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