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|---|---|---|---|---|---|
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 | #REXX | REXX | /*REXX program that implements various List Manager functions. */
/*┌────────────────────────────────────────────────────────────────────┐
┌─┘ Functions of the List Manager └─┐
│ │
│ @init ─── initializes the List. │
│ │
│ @size ─── returns the size of the List [could be 0 (zero)]. │
│ │
│ @show ─── shows (displays) the complete List. │
│ @show k,1 ─── shows (displays) the Kth item. │
│ @show k,m ─── shows (displays) M items, starting with Kth item. │
│ @show ,,─1 ─── shows (displays) the complete List backwards. │
│ │
│ @get k ─── returns the Kth item. │
│ @get k,m ─── returns the M items starting with the Kth item. │
│ │
│ @put x ─── adds the X items to the end (tail) of the List. │
│ @put x,0 ─── adds the X items to the start (head) of the List. │
│ @put x,k ─── adds the X items to before of the Kth item. │
│ │
│ @del k ─── deletes the item K. │
│ @del k,m ─── deletes the M items starting with item K. │
└─┐ ┌─┘
└────────────────────────────────────────────────────────────────────┘*/
call sy 'initializing the list.' ; call @init
call sy 'building list: Was it a cat I saw'; call @put 'Was it a cat I saw'
call sy 'displaying list size.' ; say 'list size='@size()
call sy 'forward list' ; call @show
call sy 'backward list' ; call @show ,,-1
call sy 'showing 4th item' ; call @show 4,1
call sy 'showing 6th & 6th items' ; call @show 5,2
call sy 'adding item before item 4: black' ; call @put 'black',4
call sy 'showing list' ; call @show
call sy 'adding to tail: there, in the ...'; call @put 'there, in the shadows, stalking its prey (and next meal).'
call sy 'showing list' ; call @show
call sy 'adding to head: Oy!' ; call @put 'Oy!',0
call sy 'showing list' ; call @show
exit /*stick a fork in it, we're done.*/
/*──────────────────────────────────subroutines─────────────────────────*/
p: return word(arg(1),1)
sy: say; say left('',30) "───" arg(1) '───'; return
@hasopt: arg o; return pos(o,opt)\==0
@size: return $.#
@init: $.@=''; $.#=0; return 0
@adjust: $.@=space($.@); $.#=words($.@); return 0
@parms: arg opt
if @hasopt('k') then k=min($.#+1,max(1,p(k 1)))
if @hasopt('m') then m=p(m 1)
if @hasopt('d') then dir=p(dir 1)
return
@show: procedure expose $.; parse arg k,m,dir
if dir==-1 & k=='' then k=$.#
m=p(m $.#);
call @parms 'kmd'
say @get(k,m,dir);
return 0
@get: procedure expose $.; arg k,m,dir,_
call @parms 'kmd'
do j=k for m by dir while j>0 & j<=$.#
_=_ subword($.@,j,1)
end /*j*/
return strip(_)
@put: procedure expose $.; parse arg x,k
k=p(k $.#+1)
call @parms 'k'
$.@=subword($.@,1,max(0,k-1)) x subword($.@,k)
call @adjust
return 0
@del: procedure expose $.; arg k,m
call @parms 'km'
_=subword($.@,k,k-1) subword($.@,k+m)
$.@=_
call @adjust
return |
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
| #Delphi | Delphi |
unit main;
interface
uses
Winapi.Windows, System.SysUtils, System.Classes, Vcl.Graphics, Vcl.Forms,
Vcl.ExtCtrls;
type
TClock = class(TForm)
tmrTimer: TTimer;
procedure FormResize(Sender: TObject);
procedure tmrTimerTimer(Sender: TObject);
private
{ Private declarations }
const
degrees06 = PI / 30;
degrees30 = degrees06 * 5;
degrees90 = degrees30 * 3;
margin = 20;
var
p0: TPoint;
MinP0XY: Integer;
class function IfThen(Condition: Boolean; TrueValue, FalseValue: Integer):
Integer; overload; static;
class function IfThen(Condition: Boolean; TrueValue, FalseValue: Double):
Double; overload; static;
procedure Paint; override;
procedure DrawHand(Color: TColor; Angle, Size: Double; aWidth: Integer = 2);
procedure DrawFace;
procedure DrawCenter;
procedure DrawNumbers(Angle: Double; Value: Integer);
public
{ Public declarations }
end;
var
Clock: TClock;
implementation
{$R *.dfm}
{ TClock }
procedure TClock.DrawCenter;
var
radius: Integer;
begin
radius := 6;
with Canvas do
begin
pen.Color := clNone;
Brush.Color := clBlack;
Ellipse(p0.x - radius, p0.y - radius, p0.x + radius, p0.y + radius);
end;
end;
procedure TClock.DrawFace;
var
radius, h, m: Integer;
begin
radius := MinP0XY - margin;
with Canvas do
begin
Pen.Color := clBlack;
Pen.Width := 2;
Brush.Color := clWhite;
Ellipse(p0.x - radius, p0.y - radius, p0.x + radius, p0.y + radius);
for m := 0 to 59 do
DrawHand(clGray, m * degrees06, -0.08, 2);
for h := 0 to 11 do
begin
DrawHand(clBlack, h * degrees30, -0.09, 3);
DrawNumbers((h + 3) * degrees30, 12 - h);
end;
end;
end;
procedure TClock.DrawHand(Color: TColor; Angle, Size: Double; aWidth: Integer = 2);
var
radius, x0, y0, x1, y1: Integer;
begin
radius := MinP0XY - margin;
x0 := p0.X + (IfThen(Size > 0, 0, Round(radius * (Size + 1) * cos(Angle))));
y0 := p0.Y + (IfThen(Size > 0, 0, Round(radius * (Size + 1) * sin(-Angle))));
x1 := p0.X + round(radius * IfThen(Size > 0, Size, 1) * cos(Angle));
y1 := p0.y + round(radius * IfThen(Size > 0, Size, 1) * sin(-Angle));
with Canvas do
begin
Pen.Color := Color;
pen.Width := aWidth;
MoveTo(x0, y0);
LineTo(x1, y1);
end;
end;
procedure TClock.DrawNumbers(Angle: Double; Value: Integer);
var
radius, x0, y0, x1, y1, h, w: Integer;
Size: Double;
s: string;
begin
radius := MinP0XY - margin;
Size := 0.85;
s := (Value).ToString;
x1 := p0.X + round(radius * Size * cos(Angle));
y1 := p0.y + round(radius * Size * sin(-Angle));
with Canvas do
begin
radius := 5;
Font.Size := 12;
w := TextWidth(s);
h := TextHeight(s);
x0 := x1 - (w div 2);
y0 := y1 - (h div 2);
TextOut(x0, y0, s);
end;
end;
procedure TClock.FormResize(Sender: TObject);
begin
p0 := Tpoint.create(ClientRect.CenterPoint);
MinP0XY := p0.x;
if MinP0XY > p0.Y then
MinP0XY := p0.y;
Refresh();
end;
class function TClock.IfThen(Condition: Boolean; TrueValue, FalseValue: Double): Double;
begin
if Condition then
exit(TrueValue);
exit(FalseValue);
end;
class function TClock.IfThen(Condition: Boolean; TrueValue, FalseValue: Integer): Integer;
begin
if Condition then
exit(TrueValue);
exit(FalseValue);
end;
procedure TClock.Paint;
var
t: TDateTime;
second, minute, hour: Integer;
angle, minsecs, hourmins: Double;
begin
inherited;
t := time;
second := trunc(Frac(t * 24 * 60) * 60);
minute := trunc(Frac(t * 24) * 60);
hour := trunc(t * 24);
DrawFace;
angle := degrees90 - (degrees06 * second);
DrawHand(clred, angle, 0.95, 3);
minsecs := (minute + second / 60.0);
angle := degrees90 - (degrees06 * minsecs);
DrawHand(clGreen, angle, 0.8, 4);
hourmins := (hour + minsecs / 60.0);
angle := degrees90 - (degrees30 * hourmins);
DrawHand(clBlue, angle, 0.6, 5);
DrawCenter;
Caption := Format('%.2d:%.2d:%.2d', [hour, minute, second]);
end;
procedure TClock.tmrTimerTimer(Sender: TObject);
begin
Refresh;
end;
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
| #Phix | Phix | enum NEXT,PREV,DATA
constant empty_dll = {{1,1}}
sequence dll = deep_copy(empty_dll)
procedure insert_after(object data, integer pos=1)
integer prv = dll[pos][PREV]
dll = append(dll,{pos,prv,data})
if prv!=0 then
dll[prv][NEXT] = length(dll)
end if
dll[pos][PREV] = length(dll)
end procedure
insert_after("ONE")
insert_after("TWO")
insert_after("THREE")
?dll
procedure show(integer d)
integer idx = dll[1][d]
while idx!=1 do
?dll[idx][DATA]
idx = dll[idx][d]
end while
end procedure
show(NEXT)
?"=="
show(PREV)
|
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
| #PicoLisp | PicoLisp | # Print the elements a doubly-linked list
(de 2print (DLst)
(for (L (car DLst) L (cddr L))
(printsp (car L)) )
(prinl) )
# Print the elements a doubly-linked list in reverse order
(de 2printReversed (DLst)
(for (L (cdr DLst) L (cadr L))
(printsp (car L)) )
(prinl) ) |
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
| #Plain_English | Plain English | An element is a thing with a number. |
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
| #Pop11 | Pop11 | uses objectclass;
define :class Link;
slot next = [];
slot prev = [];
slot data = [];
enddefine; |
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
| #PureBasic | PureBasic | Structure node
*prev.node
*next.node
value.i
EndStructure |
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
| #Python | Python | class Node(object):
def __init__(self, data = None, prev = None, next = None):
self.prev = prev
self.next = next
self.data = data
def __str__(self):
return str(self.data)
def __repr__(self):
return repr(self.data)
def iter_forward(self):
c = self
while c != None:
yield c
c = c.next
def iter_backward(self):
c = self
while c != None:
yield c
c = c.prev |
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)
| #Julia | Julia |
const COLORS = ["red", "white", "blue"]
function dutchsort!(a::Array{ASCIIString,1}, lo=COLORS[1], hi=COLORS[end])
i = 1
j = 1
n = length(a)
while j <= n
if a[j] == lo
a[i], a[j] = a[j], a[i]
i += 1
j += 1
elseif a[j] == hi
a[j], a[n] = a[n], a[j]
n -= 1
else
j += 1
end
end
return a
end
function dutchsort(a::Array{ASCIIString,1}, lo=COLORS[1], hi=COLORS[end])
dutchsort!(copy(a), lo, hi)
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)
| #Kotlin | Kotlin | // version 1.1.4
import java.util.Random
enum class DutchColors { RED, WHITE, BLUE }
fun Array<DutchColors>.swap(i: Int, j: Int) {
val temp = this[i]
this[i] = this[j]
this[j] = temp
}
fun Array<DutchColors>.sort() {
var lo = 0
var mid = 0
var hi = this.lastIndex
while (mid <= hi) {
when (this[mid]) {
DutchColors.RED -> this.swap(lo++, mid++)
DutchColors.WHITE -> mid++
DutchColors.BLUE -> this.swap(mid, hi--)
}
}
}
fun Array<DutchColors>.isSorted(): Boolean {
return (1 until this.size)
.none { this[it].ordinal < this[it - 1].ordinal }
}
const val NUM_BALLS = 9
fun main(args: Array<String>) {
val r = Random()
val balls = Array(NUM_BALLS) { DutchColors.RED }
val colors = DutchColors.values()
// give balls random colors whilst ensuring they're not already sorted
do {
for (i in 0 until NUM_BALLS) balls[i] = colors[r.nextInt(3)]
}
while (balls.isSorted())
// print the colors of the balls before sorting
println("Before sorting : ${balls.contentToString()}")
// sort the balls in DutchColors order
balls.sort()
// print the colors of the balls after sorting
println("After sorting : ${balls.contentToString()}")
} |
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
| #J | J | vectors =. ((% +/&.:*:"1) _1 1 0,:_1 _1 3) +/@:*"1/~ 2 3 4*=i.3
' .*o' {~ +/ 1 2 3* (|:"2 -."_ 1~ vectors) ([:*./ 1 = 0 1 I. %.~)"_ 1"_1 _ ]4j21 ,~"0/&:i: 4j41
oooo
ooooooooooooo
oooooooooooooo....
*****oooo.........
*******...........
*******...........
*******...........
*******...........
*******...........
*******...........
*******...........
*******.........
*****.....
|
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 | #True_BASIC | True BASIC | SET WINDOW 0, 320, 0, 240
SET COLOR 4
PLOT 100,100
END |
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 | #VBA | VBA | Sub draw()
Dim sh As Shape, sl As Shape
Set sh = ActiveDocument.Shapes.AddCanvas(100, 100, 320, 240)
Set sl = sh.CanvasItems.AddLine(100, 100, 101, 100)
sl.Line.ForeColor.RGB = RGB(Red:=255, Green:=0, Blue:=0)
End Sub |
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
| #TI-83_BASIC | TI-83 BASIC | :-1→Xmin:1→Xmax
:-1→Ymin:1→Ymax
:AxesOff
:Degrees
:While 1
:For(X,0,359,5
:sin(X-120→I%
:sin(X→PV
:sin(X+120→FV
:Line(0,1,I%,.3
:Line(0,1,PV,.3
:Line(0,1,FV,.3
:Line(0,-1,-I%,-.3
:Line(0,-1,-PV,-.3
:Line(0,-1,-FV,-.3
:Line(.3,I%,-.3,-PV
:Line(.3,I%,-.3,-FV
:Line(.3,PV,-.3,-I%
:Line(.3,PV,-.3,-FV
:Line(.3,FV,-.3,-I%
:Line(.3,FV,-.3,-PV
:End
:End |
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
| #Wren | Wren | import "graphics" for Canvas, Color
import "dome" for Window
import "math" for Math
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]
]
class RotatingCube {
construct new(width, height) {
Window.title = "Rotating cube"
Window.resize(width, height)
Canvas.resize(width, height)
_width = width
_height = height
_fore = Color.blue
}
init() {
scale(100)
rotateCube(Num.pi / 4, Math.atan(2.sqrt))
drawCube()
}
update() {
rotateCube(Num.pi / 180, 0)
}
draw(alpha) {
drawCube()
}
scale(s) {
for (node in Nodes) {
node[0] = node[0] * s
node[1] = node[1] * s
node[2] = node[2] * s
}
}
drawCube() {
Canvas.cls(Color.white)
Canvas.offset(_width / 2, _height / 2)
for (edge in Edges) {
var xy1 = Nodes[edge[0]]
var xy2 = Nodes[edge[1]]
Canvas.line(Math.round(xy1[0]), Math.round(xy1[1]),
Math.round(xy2[0]), Math.round(xy2[1]), _fore)
}
for (node in Nodes) {
Canvas.rectfill(Math.round(node[0]) - 4, Math.round(node[1]) - 4, 8, 8, _fore)
}
}
rotateCube(angleX, angleY) {
var sinX = Math.sin(angleX)
var cosX = Math.cos(angleX)
var sinY = Math.sin(angleY)
var cosY = Math.cos(angleY)
for (node in Nodes) {
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
}
}
}
var Game = RotatingCube.new(640, 640) |
http://rosettacode.org/wiki/Dragon_curve | Dragon curve |
Create and display a dragon curve fractal.
(You may either display the curve directly or write it to an image file.)
Algorithms
Here are some brief notes the algorithms used and how they might suit various languages.
Recursively a right curling dragon is a right dragon followed by a left dragon, at 90-degree angle. And a left dragon is a left followed by a right.
*---R----* expands to * *
\ /
R L
\ /
*
*
/ \
L R
/ \
*---L---* expands to * *
The co-routines dcl and dcr in various examples do this recursively to a desired expansion level.
The curl direction right or left can be a parameter instead of two separate routines.
Recursively, a curl direction can be eliminated by noting the dragon consists of two copies of itself drawn towards a central point at 45-degrees.
*------->* becomes * * Recursive copies drawn
\ / from the ends towards
\ / the centre.
v v
*
This can be seen in the SVG example. This is best suited to off-line drawing since the reversal in the second half means the drawing jumps backward and forward (in binary reflected Gray code order) which is not very good for a plotter or for drawing progressively on screen.
Successive approximation repeatedly re-writes each straight line as two new segments at a right angle,
*
*-----* becomes / \ bend to left
/ \ if N odd
* *
* *
*-----* becomes \ / bend to right
\ / if N even
*
Numbering from the start of the curve built so far, if the segment is at an odd position then the bend introduced is on the right side. If the segment is an even position then on the left. The process is then repeated on the new doubled list of segments. This constructs a full set of line segments before any drawing.
The effect of the splitting is a kind of bottom-up version of the recursions. See the Asymptote example for code doing this.
Iteratively the curve always turns 90-degrees left or right at each point. The direction of the turn is given by the bit above the lowest 1-bit of n. Some bit-twiddling can extract that efficiently.
n = 1010110000
^
bit above lowest 1-bit, turn left or right as 0 or 1
LowMask = n BITXOR (n-1) # eg. giving 0000011111
AboveMask = LowMask + 1 # eg. giving 0000100000
BitAboveLowestOne = n BITAND AboveMask
The first turn is at n=1, so reckon the curve starting at the origin as n=0 then a straight line segment to position n=1 and turn there.
If you prefer to reckon the first turn as n=0 then take the bit above the lowest 0-bit instead. This works because "...10000" minus 1 is "...01111" so the lowest 0 in n-1 is where the lowest 1 in n is.
Going by turns suits turtle graphics such as Logo or a plotter drawing with a pen and current direction.
If a language doesn't maintain a "current direction" for drawing then you can always keep that separately and apply turns by bit-above-lowest-1.
Absolute direction to move at point n can be calculated by the number of bit-transitions in n.
n = 11 00 1111 0 1
^ ^ ^ ^ 4 places where change bit value
so direction=4*90degrees=East
This can be calculated by counting the number of 1 bits in "n XOR (n RIGHTSHIFT 1)" since such a shift and xor leaves a single 1 bit at each position where two adjacent bits differ.
Absolute X,Y coordinates of a point n can be calculated in complex numbers by some powers (i+1)^k and add/subtract/rotate. This is done in the gnuplot code. This might suit things similar to Gnuplot which want to calculate each point independently.
Predicate test for whether a given X,Y point or segment is on the curve can be done. This might suit line-by-line output rather than building an entire image before printing. See M4 for an example of this.
A predicate works by dividing out complex number i+1 until reaching the origin, so it takes roughly a bit at a time from X and Y is thus quite efficient. Why it works is slightly subtle but the calculation is not difficult. (Check segment by applying an offset to move X,Y to an "even" position before dividing i+1. Check vertex by whether the segment either East or West is on the curve.)
The number of steps in the predicate corresponds to doublings of the curve, so stopping the check at say 8 steps can limit the curve drawn to 2^8=256 points. The offsets arising in the predicate are bits of n the segment number, so can note those bits to calculate n and limit to an arbitrary desired length or sub-section.
As a Lindenmayer system of expansions. The simplest is two symbols F and S both straight lines, as used by the PGF code.
Axiom F, angle 90 degrees
F -> F+S
S -> F-S
This always has F at even positions and S at odd. Eg. after 3 levels F_S_F_S_F_S_F_S. The +/- turns in between bend to the left or right the same as the "successive approximation" method above. Read more at for instance Joel Castellanos' L-system page.
Variations are possible if you have only a single symbol for line draw, for example the Icon and Unicon and Xfractint code. The angles can also be broken into 45-degree parts to keep the expansion in a single direction rather than the endpoint rotating around.
The string rewrites can be done recursively without building the whole string, just follow its instructions at the target level. See for example C by IFS Drawing code. The effect is the same as "recursive with parameter" above but can draw other curves defined by L-systems.
| #Befunge | Befunge | " :htpeD">:#,_&>:00p:2%10p:2/:1+1>\#<1#*-#2:#\_$:1-20p510g2*-*1+610g4vv<v<
| v%2\/3-1$_\#!4#:*#-\#1<\1+1:/4+1g00:\_\#$1<%2/2+1\g02\-1+%-g012\/-*<v"*/
_ >!>0$#0\#$\_-10p20p::00g4/:1+1>\#<1#*-#4:#\_$1-2*3/\2%!>0$#0\#$\_--vv|+2
v:\p06!*-1::p05<g00+1--g01g03\+-g01-p04+1:<0p03:-1_>>$$$1-\1->>:v:+1\<v~::
>:1+*!60g*!#v_!\!*50g0`*!40gg,::30g40g:2-#^_>>$>>:^:+1g02::\,+55_55+,@v":*
v%2/2+*">~":< ^\-1g05-*">~"/2+*"|~"-%*"|~"\/*"|~":\-*">~"/2+%*"|~"\/*<^<:
>60p\:"~>"*+2/2%60g+2%70p:"kI"*+2/2%60p\:"kI"*+2/2%60g+2%-\70g-"~|"**+"}"^ |
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
| #Emacs_Lisp | Emacs Lisp | ; Draw a sphere
(defun normalize (v)
"Normalize a vector."
(setq invlen (/ 1.0 (sqrt (dot v v))))
(mapcar (lambda (x) (* invlen x)) v))
(defun dot (v1 v2)
"Dot product of two vectors."
(+ (* (car v1) (car v2))
(* (cadr v1) (cadr v2))
(* (caddr v1) (caddr v2))))
(defun make-array (size)
"Create an empty array with size*size elements."
(setq m-array (make-vector size nil))
(dotimes (i size)
(setf (aref m-array i) (make-vector size 0)))
m-array)
(defun pic-lines (arr size)
"Turn array into a string."
(setq all "")
(dotimes (y size)
(setq line "")
(dotimes (x size)
(setq line (concat line (format "%i \n" (elt (elt arr y) x)))))
(setq all (concat all line "\n")))
all)
(defun pic-show (arr size)
"Convert size*size array to grayscale PBM image and show it."
(insert-image (create-image (concat (format "P2
%i %i 255\n" size size) (pic-lines arr size)) 'pbm t)))
(defun sphere (size k amb dir)
"Draw a sphere."
(let ((arr (make-array size))
(ndir (normalize dir))
(r (/ size 2)))
(dotimes (yp size)
(dotimes (xp size)
(setq x (- xp r))
(setq y (- yp r))
(setq z (- (* r r) (* x x) (* y y)))
(if (>= z 0)
(let* ((vec (normalize (list x y (sqrt z))))
(s (max 0 (dot vec ndir)))
(lum (max 0 (min 255 (* 255 (+ amb (expt s k))
(/ (1+ amb)))))))
(setf (elt (elt arr yp) xp) lum)))))
(pic-show arr size)))
(sphere 200 1.5 0.2 '(-30 -30 50)) |
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 | #Ruby | Ruby | class DListNode
def insert_after(search_value, new_value)
if search_value == value
new_node = self.class.new(new_value, nil, nil)
next_node = self.succ
self.succ = new_node
new_node.prev = self
new_node.succ = next_node
next_node.prev = new_node
elsif self.succ.nil?
raise StandardError, "value #{search_value} not found in list"
else
self.succ.insert_after(search_value, new_value)
end
end
end
head = DListNode.from_array([:a, :b])
head.insert_after(:a, :c) |
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 | #Rust | Rust | use std::collections::LinkedList;
fn main() {
let mut list = LinkedList::new();
list.push_front(8);
} |
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 | #Swift | Swift | typealias NodePtr<T> = UnsafeMutablePointer<Node<T>>
class Node<T> {
var value: T
fileprivate var prev: NodePtr<T>?
fileprivate var next: NodePtr<T>?
init(value: T, prev: NodePtr<T>? = nil, next: NodePtr<T>? = nil) {
self.value = value
self.prev = prev
self.next = next
}
}
@discardableResult
func insert<T>(_ val: T, after: Node<T>? = nil, list: NodePtr<T>? = nil) -> NodePtr<T> {
let node = NodePtr<T>.allocate(capacity: 1)
node.initialize(to: Node(value: val))
var n = list
while n != nil {
if n?.pointee !== after {
n = n?.pointee.next
continue
}
node.pointee.prev = n
node.pointee.next = n?.pointee.next
n?.pointee.next?.pointee.prev = node
n?.pointee.next = node
break
}
return node
}
// [1]
let list = insert(1)
// [1, 2]
insert(2, after: list.pointee, list: list)
// [1, 3, 2]
insert(3, after: list.pointee, list: list) |
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 | #Tcl | Tcl | oo::define List {
method insertBefore {elem} {
$elem next [self]
$elem previous $prev
if {$prev ne ""} {
$prev next $elem
}
set prev $elem
}
method insertAfter {elem} {
$elem previous [self]
$elem next $next
if {$next ne ""} {
$next previous $elem
}
set next $elem
}
} |
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
| #EasyLang | EasyLang | # Clock
#
func draw hour min sec . .
# dial
color 333
move 50 50
circle 45
color 797
circle 44
color 333
for i range 60
a = i * 6
move 50 + sin a * 40 50 - cos a * 40
circle 0.25
.
for i range 12
a = i * 30
move 50 + sin a * 40 50 - cos a * 40
circle 1
.
# hour
linewidth 2
color 000
a = (hour * 60 + min) / 2
move 50 50
line 50 + sin a * 32 50 - cos a * 32
# min
linewidth 1.5
a = (sec + min * 60) / 10
move 50 50
line 50 + sin a * 40 50 - cos a * 40
# sec
linewidth 1
color 700
a = sec * 6
move 50 50
line 50 + sin a * 40 50 - cos a * 40
.
on timer
if t <> floor systime
t = floor systime
h$ = timestr t
sec = number substr h$ 17 2
min = number substr h$ 14 2
hour = number substr h$ 11 2
if hour > 12
hour -= 12
.
call draw hour min sec
.
timer 0.1
.
timer 0 |
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
| #PL.2FI | PL/I | /* To implement a doubly-linked list -- i.e., a 2-way linked list. */
doubly_linked_list: proc options (main);
define structure
1 node,
2 value fixed,
2 fwd_pointer handle(node),
2 back_pointer handle(node);
declare (head, tail, t) handle (node);
declare null builtin;
declare i fixed binary;
head, tail = bind(:node, null:);
do i = 1 to 10; /* Add ten items to the tail of the queue. */
if head = bind(:node, null:) then
do;
head,tail = new(:node:);
get list (head => value);
put skip list (head => value);
head => back_pointer,
head => fwd_pointer = bind(:node, null:); /* A NULL link */
end;
else
do;
t = new(:node:);
t => back_pointer = tail; /* Point the new tail back to the old */
/* tail. */
tail => fwd_pointer = t; /* Point the tail to the new node. */
t => back_pointer = tail; /* Point the new tail back to the old */
/* tail. */
tail = t; /* Point at the new tail. */
tail => fwd_pointer = bind(:node, null:);
/* Set the tail link to NULL */
get list (tail => value) copy;
put skip list (tail => value);
end;
end;
if head = bind(:node, null:) then return; /* Empty list. */
/* Traverse the list from the head. */
put skip list ('In a forwards direction, the list has:');
t = head;
do while (t ^= bind(:node, null:));
put skip list (t => value);
t = t => fwd_pointer;
end;
/* Traverse the list from the tail to the head. */
put skip list ('In the reverse direction, the list has:');
t = tail;
do while (t ^= bind(:node, null:));
put skip list (t => value);
t = t => back_pointer;
end;
end doubly_linked_list; |
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
| #PureBasic | PureBasic | NewList MyData.i() ; Create a double linked list holding a single value (integer)
;Set up a randomly long linked list in the range 25-125 elements
For i=0 To (Random(100)+25)
AddElement(MyData()) ; Create a new tailing element
MyData()=Random(314) ; Inert a vale into it
Next
;
;Traverse from the beginning of a doubly-linked list to the end.
FirstElement(MyData())
Repeat
Debug MyData() ; Present the value in the current cell
Until Not NextElement(MyData())
;
;Traverse from the end to the beginning.
LastElement(MyData())
Repeat
Debug MyData() ; Present the value in the current cell
Until Not PreviousElement(MyData()) |
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
| #Racket | Racket |
(define-struct dlist (head tail) #:mutable)
(define-struct dlink (content prev next) #:mutable)
|
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
| #Raku | Raku | role DLElem[::T] {
has DLElem[T] $.prev is rw;
has DLElem[T] $.next is rw;
has T $.payload = T;
method pre-insert(T $payload) {
die "Can't insert before beginning" unless $!prev;
my $elem = ::?CLASS.new(:$payload);
$!prev.next = $elem;
$elem.prev = $!prev;
$elem.next = self;
$!prev = $elem;
$elem;
}
method post-insert(T $payload) {
die "Can't insert after end" unless $!next;
my $elem = ::?CLASS.new(:$payload);
$!next.prev = $elem;
$elem.next = $!next;
$elem.prev = self;
$!next = $elem;
$elem;
}
method delete {
die "Can't delete a sentinel" unless $!prev and $!next;
$!next.prev = $!prev;
$!prev.next = $!next; # conveniently returns next element
}
} |
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)
| #Lasso | Lasso | define orderdutchflag(a) => {
local(r = array, w = array, b = array)
with i in #a do => {
match(#i) => {
case('Red')
#r->insert(#i)
case('White')
#w->insert(#i)
case('Blue')
#b->insert(#i)
}
}
return #r + #w + #b
}
orderdutchflag(array('Red', 'Red', 'Blue', 'Blue', 'Blue', 'Red', 'Red', 'Red', 'White', 'Blue')) |
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)
| #Logo | Logo | ; We'll just use words for the balls
make "colors {red white blue}
; to get a mapping from colors back to a numeric value,
; we make variables out of the color names (e.g. the variable
; "red" has value "1").
foreach arraytolist :colors [
make ? #
]
; Make a random list of a given size
to random_balls :n
local "balls
make "balls array n
repeat n [
setitem # :balls pick :colors
]
output :balls
end
; Test for Dutchness
to dutch? :array
output dutchlist? arraytolist :array
end
; List is easier than array to test
to dutchlist? :list
output cond [
[(less? count :list 2) "true]
[(greater? thing first :list thing item 2 :list) "false ]
[else dutchlist? butfirst :list]
]
end
; But array is better for sorting algorithm
to dutch :array
local "lo
make "lo 0
local "hi
make "hi sum 1 count :array
local "i
make "i 1
while [:i < :hi] [
case (item :i :array) [
[[red]
make "lo sum :lo 1
swap :array :lo :i
make "i sum :i 1
]
[[white]
make "i sum :i 1
]
[[blue]
make "hi difference :hi 1
swap :array :hi :i
]
]
]
output :array
end
; utility routine to swap array elements
to swap :array :a :b
local "temp
make "temp item :a :array
setitem :a :array item :b :array
setitem :b :array :temp
end |
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
| #Java | Java | import java.awt.*;
import java.awt.event.*;
import static java.lang.Math.*;
import javax.swing.*;
public class Cuboid extends JPanel {
double[][] 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}};
int[][] 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}};
int mouseX, prevMouseX, mouseY, prevMouseY;
public Cuboid() {
setPreferredSize(new Dimension(640, 640));
setBackground(Color.white);
scale(80, 120, 160);
rotateCube(PI / 5, PI / 9);
addMouseListener(new MouseAdapter() {
@Override
public void mousePressed(MouseEvent e) {
mouseX = e.getX();
mouseY = e.getY();
}
});
addMouseMotionListener(new MouseAdapter() {
@Override
public void mouseDragged(MouseEvent e) {
prevMouseX = mouseX;
prevMouseY = mouseY;
mouseX = e.getX();
mouseY = e.getY();
double incrX = (mouseX - prevMouseX) * 0.01;
double incrY = (mouseY - prevMouseY) * 0.01;
rotateCube(incrX, incrY);
repaint();
}
});
}
private void scale(double sx, double sy, double sz) {
for (double[] node : nodes) {
node[0] *= sx;
node[1] *= sy;
node[2] *= sz;
}
}
private void rotateCube(double angleX, double angleY) {
double sinX = sin(angleX);
double cosX = cos(angleX);
double sinY = sin(angleY);
double cosY = cos(angleY);
for (double[] node : 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;
}
}
void drawCube(Graphics2D g) {
g.translate(getWidth() / 2, getHeight() / 2);
for (int[] edge : edges) {
double[] xy1 = nodes[edge[0]];
double[] xy2 = nodes[edge[1]];
g.drawLine((int) round(xy1[0]), (int) round(xy1[1]),
(int) round(xy2[0]), (int) round(xy2[1]));
}
for (double[] node : nodes) {
g.fillOval((int) round(node[0]) - 4, (int) round(node[1]) - 4, 8, 8);
}
}
@Override
public void paintComponent(Graphics gg) {
super.paintComponent(gg);
Graphics2D g = (Graphics2D) gg;
g.setRenderingHint(RenderingHints.KEY_ANTIALIASING,
RenderingHints.VALUE_ANTIALIAS_ON);
drawCube(g);
}
public static void main(String[] args) {
SwingUtilities.invokeLater(() -> {
JFrame f = new JFrame();
f.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
f.setTitle("Cuboid");
f.setResizable(false);
f.add(new Cuboid(), BorderLayout.CENTER);
f.pack();
f.setLocationRelativeTo(null);
f.setVisible(true);
});
}
} |
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 | #Vlang | Vlang | import gg
import gx
fn main() {
mut context := gg.new_context(
width: 320
height: 240
frame_fn: frame
)
context.run()
}
fn frame(mut ctx gg.Context) {
ctx.begin()
ctx.draw_pixel(100, 100, gx.red)
ctx.end()
} |
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 | #Wee_Basic | Wee Basic | keyhide
plot 0 100,100,5
end |
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
| #XPL0 | XPL0 | def Size=100., Speed=0.05; \drawing size and rotation speed
real X, Y, Z, Farthest; \arrays: 3D coordinates of vertices
int I, J, K, ZI, Edge;
def R2=sqrt(2.), R3=sqrt(3.), R13=sqrt(1./3.), R23=sqrt(2./3.), R232=R23*2.;
\vertex:0 1 2 3 4 5 6 7
[X:= [ 0., R2, 0., -R2, 0., R2, 0., -R2];
Y:= [ -R3, -R13, R13, -R13, -R13, R13, R3, R13];
Z:= [ 0., -R23, -R232, -R23, R232, R23, 0., R23];
Edge:= [0,1, 1,2, 2,3, 3,0, 4,5, 5,6, 6,7, 7,4, 0,4, 1,5, 2,6, 3,7];
SetVid($101); \set 640x480x8 graphics
repeat Farthest:= 0.0; \find the farthest vertex
for I:= 0 to 8-1 do
if Z(I) > Farthest then [Farthest:= Z(I); ZI:= I];
Clear; \erase screen
for I:= 0 to 2*12-1 do \for all the vertices...
[J:= Edge(I); I:= I+1; \get vertex numbers for edge
Move(Fix(X(J)*Size)+640/2, Fix(Y(J)*Size)+480/2);
K:= Edge(I);
Line(Fix(X(K)*Size)+640/2, Fix(Y(K)*Size)+480/2,
if J=ZI ! K=ZI then $F001 \dashed blue\ else $0C \red\);
];
DelayUS(55000);
for I:= 0 to 8-1 do
[X(I):= X(I) + Z(I)*Speed; \rotate vertices about Y axis
Z(I):= Z(I) - X(I)*Speed; \ (which rotates in X-Z plane)
];
until KeyHit; \run until a key is struck
SetVid(3); \restore normal text mode
] |
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
| #Yabasic | Yabasic | // Rosetta Code problem: http://rosettacode.org/wiki/Draw_a_rotating_cube
// adapted to Yabasic by Galileo, 05/2022
// GFA Punch (code from tigen.ti-fr.com/)
// Carré 3D en rotation
open window 50, 70
backcolor 0,0,0
clear window
color 255,255,255
do
clear window
x = COS(T) * 20
y = SIN(T) * 18
r = SIN(T + T)
line (x + 40), (y + 40 - r), (-y + 40), (x + 40 - r)
line (-y + 40), (x + 40 - r), (-x + 40), (-y + 40 - r)
line (-x + 40), (-y + 40 - r), (y + 40), (-x + 40 - r)
line (y + 40), (-x + 40 - r), (x + 40), (y + 40 - r)
line (x + 40), (y + 20 + r), (-y + 40), (x + 20 + r)
line (-y + 40), (x + 20 + r), (-x + 40), (-y + 20 + r)
line (-x + 40), (-y + 20 + r), (y + 40), (-x + 20 + r)
line (y + 40), (-x + 20 + r), (x + 40), (y + 20 + r)
line (x + 40), (y + 40 - r), (x + 40), (y + 20 + r)
line (-y + 40), (x + 40 - r), (-y + 40), (x + 20 + r)
line (-x + 40), (-y + 40 - r), (-x + 40), (-y + 20 + r)
line (y + 40), (-x + 40 - r), (y + 40), (-x + 20 + r)
pause 0.02
T = T + 0.15
loop |
http://rosettacode.org/wiki/Dragon_curve | Dragon curve |
Create and display a dragon curve fractal.
(You may either display the curve directly or write it to an image file.)
Algorithms
Here are some brief notes the algorithms used and how they might suit various languages.
Recursively a right curling dragon is a right dragon followed by a left dragon, at 90-degree angle. And a left dragon is a left followed by a right.
*---R----* expands to * *
\ /
R L
\ /
*
*
/ \
L R
/ \
*---L---* expands to * *
The co-routines dcl and dcr in various examples do this recursively to a desired expansion level.
The curl direction right or left can be a parameter instead of two separate routines.
Recursively, a curl direction can be eliminated by noting the dragon consists of two copies of itself drawn towards a central point at 45-degrees.
*------->* becomes * * Recursive copies drawn
\ / from the ends towards
\ / the centre.
v v
*
This can be seen in the SVG example. This is best suited to off-line drawing since the reversal in the second half means the drawing jumps backward and forward (in binary reflected Gray code order) which is not very good for a plotter or for drawing progressively on screen.
Successive approximation repeatedly re-writes each straight line as two new segments at a right angle,
*
*-----* becomes / \ bend to left
/ \ if N odd
* *
* *
*-----* becomes \ / bend to right
\ / if N even
*
Numbering from the start of the curve built so far, if the segment is at an odd position then the bend introduced is on the right side. If the segment is an even position then on the left. The process is then repeated on the new doubled list of segments. This constructs a full set of line segments before any drawing.
The effect of the splitting is a kind of bottom-up version of the recursions. See the Asymptote example for code doing this.
Iteratively the curve always turns 90-degrees left or right at each point. The direction of the turn is given by the bit above the lowest 1-bit of n. Some bit-twiddling can extract that efficiently.
n = 1010110000
^
bit above lowest 1-bit, turn left or right as 0 or 1
LowMask = n BITXOR (n-1) # eg. giving 0000011111
AboveMask = LowMask + 1 # eg. giving 0000100000
BitAboveLowestOne = n BITAND AboveMask
The first turn is at n=1, so reckon the curve starting at the origin as n=0 then a straight line segment to position n=1 and turn there.
If you prefer to reckon the first turn as n=0 then take the bit above the lowest 0-bit instead. This works because "...10000" minus 1 is "...01111" so the lowest 0 in n-1 is where the lowest 1 in n is.
Going by turns suits turtle graphics such as Logo or a plotter drawing with a pen and current direction.
If a language doesn't maintain a "current direction" for drawing then you can always keep that separately and apply turns by bit-above-lowest-1.
Absolute direction to move at point n can be calculated by the number of bit-transitions in n.
n = 11 00 1111 0 1
^ ^ ^ ^ 4 places where change bit value
so direction=4*90degrees=East
This can be calculated by counting the number of 1 bits in "n XOR (n RIGHTSHIFT 1)" since such a shift and xor leaves a single 1 bit at each position where two adjacent bits differ.
Absolute X,Y coordinates of a point n can be calculated in complex numbers by some powers (i+1)^k and add/subtract/rotate. This is done in the gnuplot code. This might suit things similar to Gnuplot which want to calculate each point independently.
Predicate test for whether a given X,Y point or segment is on the curve can be done. This might suit line-by-line output rather than building an entire image before printing. See M4 for an example of this.
A predicate works by dividing out complex number i+1 until reaching the origin, so it takes roughly a bit at a time from X and Y is thus quite efficient. Why it works is slightly subtle but the calculation is not difficult. (Check segment by applying an offset to move X,Y to an "even" position before dividing i+1. Check vertex by whether the segment either East or West is on the curve.)
The number of steps in the predicate corresponds to doublings of the curve, so stopping the check at say 8 steps can limit the curve drawn to 2^8=256 points. The offsets arising in the predicate are bits of n the segment number, so can note those bits to calculate n and limit to an arbitrary desired length or sub-section.
As a Lindenmayer system of expansions. The simplest is two symbols F and S both straight lines, as used by the PGF code.
Axiom F, angle 90 degrees
F -> F+S
S -> F-S
This always has F at even positions and S at odd. Eg. after 3 levels F_S_F_S_F_S_F_S. The +/- turns in between bend to the left or right the same as the "successive approximation" method above. Read more at for instance Joel Castellanos' L-system page.
Variations are possible if you have only a single symbol for line draw, for example the Icon and Unicon and Xfractint code. The angles can also be broken into 45-degree parts to keep the expansion in a single direction rather than the endpoint rotating around.
The string rewrites can be done recursively without building the whole string, just follow its instructions at the target level. See for example C by IFS Drawing code. The effect is the same as "recursive with parameter" above but can draw other curves defined by L-systems.
| #BQN | BQN | Rot ← ¬⊸{-⌾(𝕨⊸⊑)⌽𝕩}
Turns ← {{𝕩∾1∾¬⌾((⌊2÷˜≠)⊸⊑)𝕩}⍟𝕩 ⥊1}
•Plot´ <˘⍉>+` (<0‿1)(Rot˜)`Turns 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
| #ERRE | ERRE | PROGRAM SPHERE
CONST SHADES$=".:!*oe&#%@"
DIM LIGHT[2],X[2],Y[2],V[2],VEC[2]
PROCEDURE DOT(X[],Y[]->D)
D=X[0]*Y[0]+X[1]*Y[1]+X[2]*Y[2]
IF D<0 THEN D=-D ELSE D=0 END IF
END PROCEDURE
PROCEDURE NORMALIZE(V[]->V[])
LUNG=SQR(V[0]*V[0]+V[1]*V[1]+V[2]*V[2])
V[0]=V[0]/LUNG
V[1]=V[1]/LUNG
V[2]=V[2]/LUNG
END PROCEDURE
PROCEDURE PDRAW(R,K,AMBIENT)
FOR I=INT(-R) TO INT(R) DO
X=I+0.5
FOR J=INT(-2*R) TO INT(2*R) DO
Y=J/2+0.5
IF (X*X+Y*Y<=R*R) THEN
VEC[0]=X
VEC[1]=Y
VEC[2]=SQR(R*R-X*X-Y*Y)
NORMALIZE(VEC[]->VEC[])
DOT(LIGHT[],VEC[]->D)
B=D^K+AMBIENT
INTENSITY%=(1-B)*(LEN(SHADES$)-1)
IF (INTENSITY%<0) THEN INTENSITY%=0 END IF
IF (INTENSITY%>=LEN(SHADES$)-1) THEN
INTENSITY%=LEN(SHADES$)-2
END IF
PRINT(#1,MID$(SHADES$,INTENSITY%+1,1);)
ELSE
PRINT(#1,(" ");)
END IF
END FOR
PRINT(#1,)
END FOR
END PROCEDURE
BEGIN
LIGHT[]=(30,30,-50)
OPEN("O",1,"SPHERE.PRN")
NORMALIZE(LIGHT[]->LIGHT[])
PDRAW(10,2,0.4)
PRINT(#1,STRING$(79,"="))
PDRAW(20,4,0.1)
CLOSE(1)
END PROGRAM
|
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 | #Visual_Basic_.NET | Visual Basic .NET | Public Sub Insert(ByVal a As Node(Of T), ByVal b As Node(Of T), ByVal c As T)
Dim node As New Node(Of T)(value)
a.Next = node
node.Previous = a
b.Previous = node
node.Next = b
End Sub |
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 | #Wren | Wren | import "/llist" for DLinkedList
var dll = DLinkedList.new(["A", "B"])
dll.insertAfter("A", "C")
System.print(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 | #Yabasic | Yabasic | // Rosetta Code problem: http://rosettacode.org/wiki/Doubly-linked_list/Element_insertion & removal & traverse
// by Galileo, 02/2022
FIL = 1 : DATO = 2 : LPREV = 3 : LNEXT = 4
countNodes = 0 : Nodes = 10
dim list(Nodes, 4)
list(0, LNEXT) = 1
sub searchNode(node)
local i, Node
for i = 0 to node-1
Node = list(Node, LNEXT)
next
return Node
end sub
sub insertNode(node, newNode)
local Node, i
if not countNodes node = 2
for i = 1 to Nodes
if not list(i, FIL) break
next
list(i, FIL) = true
list(i, DATO) = newNode
Node = searchNode(node)
list(i, LPREV) = list(Node, LPREV) : list(list(i, LPREV), LNEXT) = i
if i <> Node list(i, LNEXT) = Node : list(Node, LPREV) = i
countNodes = countNodes + 1
if countNodes = Nodes then Nodes = Nodes + 10 : redim list(Nodes, 4) : end if
end sub
sub removeNode(n)
local Node
Node = searchNode(n)
list(list(Node, LPREV), LNEXT) = list(Node, LNEXT)
list(list(Node, LNEXT), LPREV) = list(Node, LPREV)
list(Node, LNEXT) = 0 : list(Node, LPREV) = 0
list(Node, FIL) = false
countNodes = countNodes - 1
end sub
sub printNode(node)
local Node
Node = searchNode(node)
print list(Node, DATO);
print
end sub
sub traverseList()
local i
for i = 1 to countNodes
printNode(i)
next
end sub
insertNode(1, 1000)
insertNode(2, 2000)
insertNode(2, 3000)
traverseList()
removeNode(2)
print
traverseList() |
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
| #F.23 | F# | open System.Text.RegularExpressions
let numberTemplate = """
_ _ _ _ __ _ _
/ \ /| ) _)|_||_ / /(_)(_) *
\_/ | /_ _) | _)(_) / (_) / *
"""
let g =
numberTemplate.Split([|'\n';'\r'|], System.StringSplitOptions.RemoveEmptyEntries)
|> Array.map (fun s ->
Regex.Matches(s, "...")
|> Seq.cast<Match>
|> Seq.map (fun m -> m.ToString())
|> Seq.toArray)
let idx c =
let v c = ((int) c) - ((int) '0')
let i = v c
if 0 <= i && i <= 9 then i
elif c = ':' then 10
else failwith ("Cannot draw character " + c.ToString())
let draw (s :string) =
System.Console.Clear()
g
|> Array.iter (fun a ->
s.ToCharArray() |> Array.iter (fun c ->
let i = idx c
printf "%s" (a.[i]))
printfn ""
)
[<EntryPoint>]
let main argv =
let showTime _ = draw (System.String.Format("{0:HH:mm:ss}", (System.DateTime.Now)))
let timer = new System.Timers.Timer(500.)
timer.AutoReset <- true // The timer triggers cyclically
timer.Elapsed // An event stream
|> Observable.subscribe showTime |> ignore // Subscribe to the event stream
timer.Start() // Now it counts
System.Console.ReadLine() |> ignore // Until return is hit
showTime ()
0 |
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
| #Python | Python | class List:
def __init__(self, data, next=None, prev=None):
self.data = data
self.next = next
self.prev = prev
def append(self, data):
if self.next == None:
self.next = List(data, None, self)
return self.next
else:
return self.next.append(data)
# Build the list
tail = head = List(10)
for i in [ 20, 30, 40 ]:
tail = tail.append(i)
# Traverse forwards
node = head
while node != None:
print(node.data)
node = node.next
# Traverse Backwards
node = tail
while node != None:
print(node.data)
node = node.prev |
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
| #Racket | Racket | (define (dlist-elements dlist)
(let loop ([elements '()] [dlink (dlist-tail dlist)])
(if dlink
(loop (cons (dlink-content dlink) elements) (dlink-prev dlink))
elements)))
(define (dlist-elements/reverse dlist)
(let loop ([elements '()] [dlink (dlist-head dlist)])
(if dlink
(loop (cons (dlink-content dlink) elements) (dlink-next dlink))
elements))) |
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
| #REXX | REXX | ╔═════════════════════════════════════════════════════════════════════════╗
║ ☼☼☼☼☼☼☼☼☼☼☼ Functions of the List Manager ☼☼☼☼☼☼☼☼☼☼☼ ║
║ @init ─── initializes the List. ║
║ ║
║ @size ─── returns the size of the List [could be a 0 (zero)]. ║
║ ║
║ @show ─── shows (displays) the complete List. ║
║ @show k,1 ─── shows (displays) the Kth item. ║
║ @show k,m ─── shows (displays) M items, starting with Kth item. ║
║ @show ,,─1 ─── shows (displays) the complete List backwards. ║
║ ║
║ @get k ─── returns the Kth item. ║
║ @get k,m ─── returns the M items starting with the Kth item. ║
║ ║
║ @put x ─── adds the X items to the end (tail) of the List. ║
║ @put x,0 ─── adds the X items to the start (head) of the List. ║
║ @put x,k ─── adds the X items to before of the Kth item. ║
║ ║
║ @del k ─── deletes the item K. ║
║ @del k,m ─── deletes the M items starting with item K. ║
╚═════════════════════════════════════════════════════════════════════════╝
|
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
| #Ruby | Ruby | class DListNode < ListNode
attr_accessor :prev
# accessors :succ and :value are inherited
def initialize(value, prev=nil, succ=nil)
@value = value
@prev = prev
@prev.succ = self if prev
@succ = succ
@succ.prev = self if succ
end
def self.from_values(*ary)
ary << (f = ary.pop)
ary.map! {|i| new i }
ary.inject(f) {|p, c| p.succ = c; c.prev = p; c }
end
end
list = DListNode.from_values 1,2,3,4 |
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
| #Rust | Rust | use std::collections::LinkedList;
fn main() {
// Doubly linked list containing 32-bit integers
let list = LinkedList::<i32>::new();
} |
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)
| #Lua | Lua | -- "1. Generate a randomized order of balls.."
math.randomseed(os.time())
N, balls, colors = 10, {}, { "red", "white", "blue" }
for i = 1, N do balls[i] = colors[math.random(#colors)] end
-- "..ensuring that they are not in the order of the Dutch national flag."
order = { red=1, white=2, blue=3 }
function issorted(t)
for i = 2, #t do
if order[t[i]] < order[t[i-1]] then return false end
end
return true
end
local function shuffle(t)
for i = #t, 2, -1 do
local j = math.random(i)
t[i], t[j] = t[j], t[i]
end
end
while issorted(balls) do shuffle(balls) end
print("RANDOM: "..table.concat(balls,","))
-- "2. Sort the balls in a way idiomatic to your language."
table.sort(balls, function(a, b) return order[a] < order[b] end)
-- "3. Check the sorted balls are in the order of the Dutch national flag."
print("SORTED: "..table.concat(balls,","))
print(issorted(balls) and "Properly sorted." or "IMPROPERLY 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
| #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");
canvas.addEventListener("mousemove", function (event) {
prevMouseX = mouseX;
prevMouseY = mouseY;
mouseX = event.x;
mouseY = event.y;
var incrX = (mouseX - prevMouseX) * 0.01;
var incrY = (mouseY - prevMouseY) * 0.01;
rotateCuboid(incrX, incrY);
drawCuboid();
});
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]];
var mouseX = 0, prevMouseX, mouseY = 0, prevMouseY;
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(80, 120, 160);
rotateCuboid(Math.PI / 5, Math.PI / 9);
</script>
</body>
</html> |
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 | #Wren | Wren | import "dome" for Window
import "graphics" for Canvas, Color
class Game {
static init() {
Window.title = "Draw a pixel"
Window.resize(320, 240)
Canvas.resize(320, 240)
var red = Color.rgb(255, 0, 0)
Canvas.pset(100, 100, red)
// check it worked
var col = Canvas.pget(100, 100)
System.print("The color of the pixel at (100, 100) is %(getRGB(col))")
}
static update() {}
static draw(dt) {}
static getRGB(col) { "{%(col.r), %(col.g), %(col.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 | #X86_Assembly | X86 Assembly |
.model tiny
.code
org 100h
start: mov ax, 0013h ;set 320x200x8 graphic screen
int 10h
push 0A000h ;point to graphic memory segment
pop es
mov byte ptr es:[320*100+100], 28h ;draw bright red pixel
mov ah, 0 ;wait for keystroke
int 16h
mov ax, 0003h ;restore normal text mode screen
int 10h
ret ;return to OS
end start
|
http://rosettacode.org/wiki/Dragon_curve | Dragon curve |
Create and display a dragon curve fractal.
(You may either display the curve directly or write it to an image file.)
Algorithms
Here are some brief notes the algorithms used and how they might suit various languages.
Recursively a right curling dragon is a right dragon followed by a left dragon, at 90-degree angle. And a left dragon is a left followed by a right.
*---R----* expands to * *
\ /
R L
\ /
*
*
/ \
L R
/ \
*---L---* expands to * *
The co-routines dcl and dcr in various examples do this recursively to a desired expansion level.
The curl direction right or left can be a parameter instead of two separate routines.
Recursively, a curl direction can be eliminated by noting the dragon consists of two copies of itself drawn towards a central point at 45-degrees.
*------->* becomes * * Recursive copies drawn
\ / from the ends towards
\ / the centre.
v v
*
This can be seen in the SVG example. This is best suited to off-line drawing since the reversal in the second half means the drawing jumps backward and forward (in binary reflected Gray code order) which is not very good for a plotter or for drawing progressively on screen.
Successive approximation repeatedly re-writes each straight line as two new segments at a right angle,
*
*-----* becomes / \ bend to left
/ \ if N odd
* *
* *
*-----* becomes \ / bend to right
\ / if N even
*
Numbering from the start of the curve built so far, if the segment is at an odd position then the bend introduced is on the right side. If the segment is an even position then on the left. The process is then repeated on the new doubled list of segments. This constructs a full set of line segments before any drawing.
The effect of the splitting is a kind of bottom-up version of the recursions. See the Asymptote example for code doing this.
Iteratively the curve always turns 90-degrees left or right at each point. The direction of the turn is given by the bit above the lowest 1-bit of n. Some bit-twiddling can extract that efficiently.
n = 1010110000
^
bit above lowest 1-bit, turn left or right as 0 or 1
LowMask = n BITXOR (n-1) # eg. giving 0000011111
AboveMask = LowMask + 1 # eg. giving 0000100000
BitAboveLowestOne = n BITAND AboveMask
The first turn is at n=1, so reckon the curve starting at the origin as n=0 then a straight line segment to position n=1 and turn there.
If you prefer to reckon the first turn as n=0 then take the bit above the lowest 0-bit instead. This works because "...10000" minus 1 is "...01111" so the lowest 0 in n-1 is where the lowest 1 in n is.
Going by turns suits turtle graphics such as Logo or a plotter drawing with a pen and current direction.
If a language doesn't maintain a "current direction" for drawing then you can always keep that separately and apply turns by bit-above-lowest-1.
Absolute direction to move at point n can be calculated by the number of bit-transitions in n.
n = 11 00 1111 0 1
^ ^ ^ ^ 4 places where change bit value
so direction=4*90degrees=East
This can be calculated by counting the number of 1 bits in "n XOR (n RIGHTSHIFT 1)" since such a shift and xor leaves a single 1 bit at each position where two adjacent bits differ.
Absolute X,Y coordinates of a point n can be calculated in complex numbers by some powers (i+1)^k and add/subtract/rotate. This is done in the gnuplot code. This might suit things similar to Gnuplot which want to calculate each point independently.
Predicate test for whether a given X,Y point or segment is on the curve can be done. This might suit line-by-line output rather than building an entire image before printing. See M4 for an example of this.
A predicate works by dividing out complex number i+1 until reaching the origin, so it takes roughly a bit at a time from X and Y is thus quite efficient. Why it works is slightly subtle but the calculation is not difficult. (Check segment by applying an offset to move X,Y to an "even" position before dividing i+1. Check vertex by whether the segment either East or West is on the curve.)
The number of steps in the predicate corresponds to doublings of the curve, so stopping the check at say 8 steps can limit the curve drawn to 2^8=256 points. The offsets arising in the predicate are bits of n the segment number, so can note those bits to calculate n and limit to an arbitrary desired length or sub-section.
As a Lindenmayer system of expansions. The simplest is two symbols F and S both straight lines, as used by the PGF code.
Axiom F, angle 90 degrees
F -> F+S
S -> F-S
This always has F at even positions and S at odd. Eg. after 3 levels F_S_F_S_F_S_F_S. The +/- turns in between bend to the left or right the same as the "successive approximation" method above. Read more at for instance Joel Castellanos' L-system page.
Variations are possible if you have only a single symbol for line draw, for example the Icon and Unicon and Xfractint code. The angles can also be broken into 45-degree parts to keep the expansion in a single direction rather than the endpoint rotating around.
The string rewrites can be done recursively without building the whole string, just follow its instructions at the target level. See for example C by IFS Drawing code. The effect is the same as "recursive with parameter" above but can draw other curves defined by L-systems.
| #C | C | #include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
/* x, y: coordinates of current point; dx, dy: direction of movement.
* Think turtle graphics. They are divided by scale, so as to keep
* very small coords/increments without losing precission. clen is
* the path length travelled, which should equal to scale at the end
* of the curve.
*/
long long x, y, dx, dy, scale, clen;
typedef struct { double r, g, b; } rgb;
rgb ** pix;
/* for every depth increase, rotate 45 degrees and scale up by sqrt(2)
* Note how coords can still be represented by integers.
*/
void sc_up()
{
long long tmp = dx - dy; dy = dx + dy; dx = tmp;
scale *= 2; x *= 2; y *= 2;
}
/* Hue changes from 0 to 360 degrees over entire length of path; Value
* oscillates along the path to give some contrast between segments
* close to each other spatially. RGB derived from HSV gets *added*
* to each pixel reached; they'll be dealt with later.
*/
void h_rgb(long long x, long long y)
{
rgb *p = &pix[y][x];
# define SAT 1
double h = 6.0 * clen / scale;
double VAL = 1 - (cos(3.141592653579 * 64 * clen / scale) - 1) / 4;
double c = SAT * VAL;
double X = c * (1 - fabs(fmod(h, 2) - 1));
switch((int)h) {
case 0: p->r += c; p->g += X; return;
case 1: p->r += X; p->g += c; return;
case 2: p->g += c; p->b += X; return;
case 3: p->g += X; p->b += c; return;
case 4: p->r += X; p->b += c; return;
default:
p->r += c; p->b += X;
}
}
/* string rewriting. No need to keep the string itself, just execute
* its instruction recursively.
*/
void iter_string(const char * str, int d)
{
long tmp;
# define LEFT tmp = -dy; dy = dx; dx = tmp
# define RIGHT tmp = dy; dy = -dx; dx = tmp
while (*str != '\0') {
switch(*(str++)) {
case 'X': if (d) iter_string("X+YF+", d - 1); continue;
case 'Y': if (d) iter_string("-FX-Y", d - 1); continue;
case '+': RIGHT; continue;
case '-': LEFT; continue;
case 'F':
/* draw: increment path length; add color; move. Here
* is why the code does not allow user to choose arbitrary
* image size: if it's not a power of two, aliasing will
* occur and grid-like bright or dark lines will result
* when normalized later. It can be gotten rid of, but that
* involves computing multiplicative order and would be a huge
* bore.
*/
clen ++;
h_rgb(x/scale, y/scale);
x += dx; y += dy;
continue;
}
}
}
void dragon(long leng, int depth)
{
long i, d = leng / 3 + 1;
long h = leng + 3, w = leng + d * 3 / 2 + 2;
/* allocate pixel buffer */
rgb *buf = malloc(sizeof(rgb) * w * h);
pix = malloc(sizeof(rgb *) * h);
for (i = 0; i < h; i++)
pix[i] = buf + w * i;
memset(buf, 0, sizeof(rgb) * w * h);
/* init coords; scale up to desired; exec string */
x = y = d; dx = leng; dy = 0; scale = 1; clen = 0;
for (i = 0; i < depth; i++) sc_up();
iter_string("FX", depth);
/* write color PNM file */
unsigned char *fpix = malloc(w * h * 3);
double maxv = 0, *dbuf = (double*)buf;
/* find highest value among pixels; normalize image according
* to it. Highest value would be at points most travelled, so
* this ends up giving curve edge a nice fade -- it's more apparaent
* if we increase iteration depth by one or two.
*/
for (i = 3 * w * h - 1; i >= 0; i--)
if (dbuf[i] > maxv) maxv = dbuf[i];
for (i = 3 * h * w - 1; i >= 0; i--)
fpix[i] = 255 * dbuf[i] / maxv;
printf("P6\n%ld %ld\n255\n", w, h);
fflush(stdout); /* printf and fwrite may treat buffer differently */
fwrite(fpix, h * w * 3, 1, stdout);
}
int main(int c, char ** v)
{
int size, depth;
depth = (c > 1) ? atoi(v[1]) : 10;
size = 1 << depth;
fprintf(stderr, "size: %d depth: %d\n", size, depth);
dragon(size, depth * 2);
return 0;
} |
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
| #Factor | Factor | USING: classes.struct kernel raylib.ffi ;
640 480 "sphere" init-window
S{ Camera3D
{ position S{ Vector3 f 4.5 4.5 4.5 } }
{ target S{ Vector3 f 0 0 0 } }
{ up S{ Vector3 f 0 1 0 } }
{ fovy 45.0 }
{ type 0 }
}
60 set-target-fps
[ window-should-close ] [
begin-drawing
BLACK clear-background dup
begin-mode-3d
S{ Vector3 f 0 0 0 } 2 20 20 LIME draw-sphere-wires
end-mode-3d
end-drawing
] until drop close-window |
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 | #zkl | zkl | class Node{
fcn init(_value,_prev=Void,_next=Void)
{ var value=_value, prev=_prev, next=_next; }
fcn toString{ value.toString() }
fcn append(value){
b,c := Node(value,self,next),next;
next=b;
if(c) c.prev=b;
b
}
} |
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
| #Forth | Forth |
HEX
8379 CONSTANT TICKER \ address of 1/60 second counter
CREATE PDT ( -- addr) \ bit pattern descriptors for 0..9 and colon
0038 , 444C , 5464 , 4438 , ( 0)
0010 , 3010 , 1010 , 1038 , ( 1)
0038 , 4404 , 1820 , 407C , ( 2)
007C , 0810 , 0804 , 4438 , ( 3)
0008 , 1828 , 487C , 0808 , ( 4)
007C , 4078 , 0404 , 4438 , ( 5)
0038 , 4040 , 7844 , 4438 , ( 6)
007C , 0408 , 1020 , 2020 , ( 7)
0038 , 4444 , 3844 , 4438 , ( 8)
0038 , 4444 , 3C04 , 0438 , ( 9)
0000 , 3030 , 0030 , 3000 , ( :)
: ]PDT ( 0..9 -- addr) [CHAR] 0 - 8 * PDT + ;
: BIG.TYPE ( caddr len -- )
8 0
DO
CR
2DUP BOUNDS
?DO
I C@ ]PDT J + C@ \ PDT char, byte# J
2 7 DO \ from bit# 7 to 2
DUP 1 I LSHIFT AND \ mask out each bit
IF [char] * EMIT \ if true emit a character
ELSE SPACE \ else print space
THEN
-1 +LOOP DROP
LOOP
LOOP
2DROP ;
DECIMAL
CREATE SECONDS 0 , 0 , \ 2 CELLS, holds a double integer
: SECONDS++ ( -- ) SECONDS 2@ 1 M+ SECONDS 2! ;
\ subtract old value from new value until ticker changes.
: 1/60 ( -- )
TICKER DUP @ ( -- addr value)
BEGIN
PAUSE \ *Gives time to other Forth processes while we wait
OVER @ \ read ticker addr
OVER - \ subtract from old value
UNTIL
2DROP ;
: SEXTAL ( -- ) 6 BASE ! ;
: 1SEC ( -- ) 60 0 DO 1/60 LOOP SECONDS++ ;
: ##: ( -- ) # SEXTAL # DECIMAL [CHAR] : HOLD ;
: .TIME ( d --) <# ##: ##: # # #> BIG.TYPE ;
: CLOCK ( -- )
DECIMAL \ set task's local radix
BEGIN
1SEC
0 0 AT-XY SECONDS 2@ .TIME
?TERMINAL
UNTIL
2DROP ; |
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
| #Raku | Raku | say $dll.list;
say $dll.reverse; |
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
| #REXX | REXX | /*REXX program that implements various List Manager functions. */
/*┌────────────────────────────────────────────────────────────────────┐
┌─┘ Functions of the List Manager └─┐
│ │
│ @init ─── initializes the List. │
│ │
│ @size ─── returns the size of the List [could be 0 (zero)]. │
│ │
│ @show ─── shows (displays) the complete List. │
│ @show k,1 ─── shows (displays) the Kth item. │
│ @show k,m ─── shows (displays) M items, starting with Kth item. │
│ @show ,,─1 ─── shows (displays) the complete List backwards. │
│ │
│ @get k ─── returns the Kth item. │
│ @get k,m ─── returns the M items starting with the Kth item. │
│ │
│ @put x ─── adds the X items to the end (tail) of the List. │
│ @put x,0 ─── adds the X items to the start (head) of the List. │
│ @put x,k ─── adds the X items to before of the Kth item. │
│ │
│ @del k ─── deletes the item K. │
│ @del k,m ─── deletes the M items starting with item K. │
└─┐ ┌─┘
└────────────────────────────────────────────────────────────────────┘*/
call sy 'initializing the list.' ; call @init
call sy 'building list: Was it a cat I saw'; call @put 'Was it a cat I saw'
call sy 'displaying list size.' ; say 'list size='@size()
call sy 'forward list' ; call @show
call sy 'backward list' ; call @show ,,-1
call sy 'showing 4th item' ; call @show 4,1
call sy 'showing 6th & 6th items' ; call @show 5,2
call sy 'adding item before item 4: black' ; call @put 'black',4
call sy 'showing list' ; call @show
call sy 'adding to tail: there, in the ...'; call @put 'there, in the shadows, stalking its prey (and next meal).'
call sy 'showing list' ; call @show
call sy 'adding to head: Oy!' ; call @put 'Oy!',0
call sy 'showing list' ; call @show
exit /*stick a fork in it, we're done.*/
/*──────────────────────────────────subroutines─────────────────────────*/
p: return word(arg(1),1)
sy: say; say left('',30) "───" arg(1) '───'; return
@hasopt: arg o; return pos(o,opt)\==0
@size: return $.#
@init: $.@=''; $.#=0; return 0
@adjust: $.@=space($.@); $.#=words($.@); return 0
@parms: arg opt
if @hasopt('k') then k=min($.#+1,max(1,p(k 1)))
if @hasopt('m') then m=p(m 1)
if @hasopt('d') then dir=p(dir 1)
return
@show: procedure expose $.; parse arg k,m,dir
if dir==-1 & k=='' then k=$.#
m=p(m $.#);
call @parms 'kmd'
say @get(k,m,dir);
return 0
@get: procedure expose $.; arg k,m,dir,_
call @parms 'kmd'
do j=k for m by dir while j>0 & j<=$.#
_=_ subword($.@,j,1)
end /*j*/
return strip(_)
@put: procedure expose $.; parse arg x,k
k=p(k $.#+1)
call @parms 'k'
$.@=subword($.@,1,max(0,k-1)) x subword($.@,k)
call @adjust
return 0
@del: procedure expose $.; arg k,m
call @parms 'km'
_=subword($.@,k,k-1) subword($.@,k+m)
$.@=_
call @adjust
return |
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
| #Sidef | Sidef | var node = Hash.new(
data => 'say what',
next => foo_node,
prev => bar_node,
);
node{:next} = quux_node; # mutable |
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
| #Swift | Swift | typealias NodePtr<T> = UnsafeMutablePointer<Node<T>>
class Node<T> {
var value: T
fileprivate var prev: NodePtr<T>?
fileprivate var next: NodePtr<T>?
init(value: T, prev: NodePtr<T>? = nil, next: NodePtr<T>? = nil) {
self.value = value
self.prev = prev
self.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
| #Tcl | Tcl | oo::class create List {
variable content next prev
constructor {value {list ""}} {
set content $value
set next $list
set prev ""
if {$next ne ""} {
$next previous [self]
}
}
method value args {
set content {*}$args
}
method next args {
set next {*}$args
}
method previous args {
set prev {*}$args
}
} |
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)
| #M2000_Interpreter | M2000 Interpreter |
Report "Dutch Flag from Dijkstra"
const center=2
enum balls {Red, White, Blue}
fillarray=lambda a=(Red, White, Blue) (size as long=10)-> {
if size<1 then size=1
randomitem=lambda a->a#val(random(0,2))
dim a(size)<<randomitem()
=a()
}
Display$=lambda$ (s as array) ->{
Document r$=eval$(array(s))
if len(s)>1 then
For i=1 to len(s)-1 {
r$=", "+eval$(array(s,i))
}
end if
=r$
}
TestSort$=lambda$ (s as array)-> {
="unsorted: "
x=array(s)
for i=1 to len(s)-1 {
k=array(s,i)
if x>k then break
swap x, k
}
="sorted: "
}
Positions=lambda mid=White (a as array) ->{
m=len(a)
dim Base 0, b(m)=-1
low=-1
high=m
m--
i=0
medpos=stack
link a to a()
for i=m to 0 {
if a(i)<=mid then exit
high--
b(high)=high
}
for i=0 to m {
if a(i)>=mid then exit
low++
b(low)=low
}
if high-low>1 then
for i=low+1 to high-1 {
select case a(i)<=>Mid
case -1
low++ : b(low)=i
case 1
{
high-- :b(high)=i
if High<i then swap b(high), b(i)
}
else case
stack medpos {data i}
end select
}
end if
if Len(medpos)>0 then
dim c()
c()=array(medpos)
stock c(0) keep len(c()), b(low+1)
for i=low+1 to high-1
if b(i)>low and b(i)<high and b(i)<>i then swap b(b(i)), b(i)
next i
end if
if low>0 then
for i=0 to low
if b(i)<=low and b(i)<>i then swap b(b(i)), b(i)
next
end if
if High<m then
for i=m to High
if b(i)>=High and b(i)<>i then swap b(b(i)), b(i)
next
end if
=b()
}
InPlace=Lambda (&p(), &Final()) ->{
def i=0, j=-1, k=-1, many=0
for i=0 to len(p())-1
if p(i)<>i then
j=i
z=final(j)
do
final(j)=final(p(j))
k=j
j=p(j)
p(k)=k
many++
until j=i
final(k)=z
end if
next
=many
}
Dim final(), p(), second(), p1()
Rem final()=(White,Red,Blue,White,Red, Red, Blue)
Rem final()=(white, blue, red, blue, white)
final()=fillarray(30)
Print "Items: ";len(final())
Report TestSort$(final())+Display$(final())
\\ backup for final() for second example
second()=final()
p()=positions(final())
\\ backup p() to p1() for second example
p1()=p()
Report Center, "InPlace"
rem Print p() ' show array items
many=InPlace(&p(), &final())
rem print p() ' show array items
Report TestSort$(final())+Display$(final())
print "changes: "; many
Report Center, "Using another array to make the changes"
final()=second()
\\ using a second array to place only the changes
item=each(p1())
many=0
While item {
if item^=array(item) else final(item^)=second(array(item)) : many++
}
Report TestSort$(final())+Display$(final())
print "changes: "; many
Module three_way_partition (A as array, mid as balls, &swaps) {
Def i, j, k
k=Len(A)
Link A to A()
While j < k
if A(j) < mid Then
Swap A(i), A(j)
swaps++
i++
j++
Else.if A(j) > mid Then
k--
Swap A(j), A(k)
swaps++
Else
j++
End if
End While
}
Many=0
Z=second()
Print
Report center, {Three Way Partition
}
Report TestSort$(Z)+Display$(Z)
three_way_partition Z, White, &many
Print
Report TestSort$(Z)+Display$(Z)
Print "changes: "; many
|
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)
| #Mathematica_.2F_Wolfram_Language | Mathematica / Wolfram Language | flagSort[data_List] := Sort[data, (#1 === RED || #2 === BLUE) &] |
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
| #Julia | Julia | _pr(t::Dict, x::Int, y::Int, z::Int) = join((rstrip(join(t[(n, m)] for n in range(0, 3+x+z))) for m in reverse(range(0, 3+y+z))), "\n")
function cuboid(x::Int, y::Int, z::Int)
t = Dict((n, m) => " " for n in range(0, 3 + x + z), m in range(0, 3 + y + z))
xrow = vcat("+", collect("$(i % 10)" for i in range(0, x)), "+")
for (i, ch) in enumerate(xrow) t[(i, 0)] = t[(i, 1+y)] = t[(1+z+i, 2+y+z)] = ch end
yrow = vcat("+", collect("$(j % 10)" for j in range(0, y)), "+")
for (j, ch) in enumerate(yrow) t[(0, j)] = t[(x+1, j)] = t[(2+x+z, 1+z+j)] = ch end
zdep = vcat("+", collect("$(k % 10)" for k in range(0, y)), "+")
for (k, ch) in enumerate(xrow) t[(k, 1+y+k)] = t[(1+x+k, 1+y+k)] = t[(1+x+k, k)] = ch end
return _pr(t, x, y, z)
end
for (x, y, z) in [(2, 3, 4), (3, 4, 2), (4, 2, 3), (5, 5, 6)]
println("\nCUBOID($x, $y, $z)\n")
println(cuboid(x, y, z))
end |
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 | #XPL0 | XPL0 | [SetFB(320, 240, 24); Point(100, 100, $FF0000)] |
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 | #Yabasic | Yabasic | open window 320, 240
color 255, 0, 0
dot 100, 100 |
http://rosettacode.org/wiki/Dragon_curve | Dragon curve |
Create and display a dragon curve fractal.
(You may either display the curve directly or write it to an image file.)
Algorithms
Here are some brief notes the algorithms used and how they might suit various languages.
Recursively a right curling dragon is a right dragon followed by a left dragon, at 90-degree angle. And a left dragon is a left followed by a right.
*---R----* expands to * *
\ /
R L
\ /
*
*
/ \
L R
/ \
*---L---* expands to * *
The co-routines dcl and dcr in various examples do this recursively to a desired expansion level.
The curl direction right or left can be a parameter instead of two separate routines.
Recursively, a curl direction can be eliminated by noting the dragon consists of two copies of itself drawn towards a central point at 45-degrees.
*------->* becomes * * Recursive copies drawn
\ / from the ends towards
\ / the centre.
v v
*
This can be seen in the SVG example. This is best suited to off-line drawing since the reversal in the second half means the drawing jumps backward and forward (in binary reflected Gray code order) which is not very good for a plotter or for drawing progressively on screen.
Successive approximation repeatedly re-writes each straight line as two new segments at a right angle,
*
*-----* becomes / \ bend to left
/ \ if N odd
* *
* *
*-----* becomes \ / bend to right
\ / if N even
*
Numbering from the start of the curve built so far, if the segment is at an odd position then the bend introduced is on the right side. If the segment is an even position then on the left. The process is then repeated on the new doubled list of segments. This constructs a full set of line segments before any drawing.
The effect of the splitting is a kind of bottom-up version of the recursions. See the Asymptote example for code doing this.
Iteratively the curve always turns 90-degrees left or right at each point. The direction of the turn is given by the bit above the lowest 1-bit of n. Some bit-twiddling can extract that efficiently.
n = 1010110000
^
bit above lowest 1-bit, turn left or right as 0 or 1
LowMask = n BITXOR (n-1) # eg. giving 0000011111
AboveMask = LowMask + 1 # eg. giving 0000100000
BitAboveLowestOne = n BITAND AboveMask
The first turn is at n=1, so reckon the curve starting at the origin as n=0 then a straight line segment to position n=1 and turn there.
If you prefer to reckon the first turn as n=0 then take the bit above the lowest 0-bit instead. This works because "...10000" minus 1 is "...01111" so the lowest 0 in n-1 is where the lowest 1 in n is.
Going by turns suits turtle graphics such as Logo or a plotter drawing with a pen and current direction.
If a language doesn't maintain a "current direction" for drawing then you can always keep that separately and apply turns by bit-above-lowest-1.
Absolute direction to move at point n can be calculated by the number of bit-transitions in n.
n = 11 00 1111 0 1
^ ^ ^ ^ 4 places where change bit value
so direction=4*90degrees=East
This can be calculated by counting the number of 1 bits in "n XOR (n RIGHTSHIFT 1)" since such a shift and xor leaves a single 1 bit at each position where two adjacent bits differ.
Absolute X,Y coordinates of a point n can be calculated in complex numbers by some powers (i+1)^k and add/subtract/rotate. This is done in the gnuplot code. This might suit things similar to Gnuplot which want to calculate each point independently.
Predicate test for whether a given X,Y point or segment is on the curve can be done. This might suit line-by-line output rather than building an entire image before printing. See M4 for an example of this.
A predicate works by dividing out complex number i+1 until reaching the origin, so it takes roughly a bit at a time from X and Y is thus quite efficient. Why it works is slightly subtle but the calculation is not difficult. (Check segment by applying an offset to move X,Y to an "even" position before dividing i+1. Check vertex by whether the segment either East or West is on the curve.)
The number of steps in the predicate corresponds to doublings of the curve, so stopping the check at say 8 steps can limit the curve drawn to 2^8=256 points. The offsets arising in the predicate are bits of n the segment number, so can note those bits to calculate n and limit to an arbitrary desired length or sub-section.
As a Lindenmayer system of expansions. The simplest is two symbols F and S both straight lines, as used by the PGF code.
Axiom F, angle 90 degrees
F -> F+S
S -> F-S
This always has F at even positions and S at odd. Eg. after 3 levels F_S_F_S_F_S_F_S. The +/- turns in between bend to the left or right the same as the "successive approximation" method above. Read more at for instance Joel Castellanos' L-system page.
Variations are possible if you have only a single symbol for line draw, for example the Icon and Unicon and Xfractint code. The angles can also be broken into 45-degree parts to keep the expansion in a single direction rather than the endpoint rotating around.
The string rewrites can be done recursively without building the whole string, just follow its instructions at the target level. See for example C by IFS Drawing code. The effect is the same as "recursive with parameter" above but can draw other curves defined by L-systems.
| #C.23 | C# | using System;
using System.Collections.Generic;
using System.Drawing;
using System.Drawing.Drawing2D;
using System.Windows.Forms;
public class DragonCurve : Form
{
private List<int> turns;
private double startingAngle, side;
public DragonCurve(int iter)
{
Size = new Size(800, 600);
StartPosition = FormStartPosition.CenterScreen;
DoubleBuffered = true;
BackColor = Color.White;
startingAngle = -iter * (Math.PI / 4);
side = 400 / Math.Pow(2, iter / 2.0);
turns = getSequence(iter);
}
private List<int> getSequence(int iter)
{
var turnSequence = new List<int>();
for (int i = 0; i < iter; i++)
{
var copy = new List<int>(turnSequence);
copy.Reverse();
turnSequence.Add(1);
foreach (int turn in copy)
{
turnSequence.Add(-turn);
}
}
return turnSequence;
}
protected override void OnPaint(PaintEventArgs e)
{
base.OnPaint(e);
e.Graphics.SmoothingMode = SmoothingMode.AntiAlias;
double angle = startingAngle;
int x1 = 230, y1 = 350;
int x2 = x1 + (int)(Math.Cos(angle) * side);
int y2 = y1 + (int)(Math.Sin(angle) * side);
e.Graphics.DrawLine(Pens.Black, x1, y1, x2, y2);
x1 = x2;
y1 = y2;
foreach (int turn in turns)
{
angle += turn * (Math.PI / 2);
x2 = x1 + (int)(Math.Cos(angle) * side);
y2 = y1 + (int)(Math.Sin(angle) * side);
e.Graphics.DrawLine(Pens.Black, x1, y1, x2, y2);
x1 = x2;
y1 = y2;
}
}
[STAThread]
static void Main()
{
Application.Run(new DragonCurve(14));
}
} |
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
| #Forth | Forth | : 3dup 2 pick 2 pick 2 pick ;
: sqrt ( u -- sqrt ) ( Babylonian method )
dup 2/ ( first square root guess is half )
dup 0= if drop exit then ( sqrt[0]=0, sqrt[1]=1 )
begin dup >r 2dup / r> + 2/ ( stack: square old-guess new-guess )
2dup > while ( as long as guess is decreasing )
nip repeat ( forget old-guess and repeat )
drop nip ;
: normalize ( x1 y1 z1 -- x1' y1' z1' ) ( normalise down to 1000 )
3dup dup * rot dup * rot dup * + + sqrt 1000 / >r ( length )
r@ / rot r@ / rot r> / rot ;
: r2-y2-x2 ( x y r -- z2 ) dup * swap dup * - swap dup * - ;
: shade ( u -- c ) C" @#&eo%*!:. " + c@ ;
: map-to-shade ( u -- u ) 0 shade * 1000 / 1 max 0 shade min ;
: dot-light ( x y z -- i ) ( hard coded light vector z, y, x )
-770 * rot 461 * rot 461 * + +
0 min 1000 / ;
: intensity ( x y z -- u ) dot-light dup * 1000 / map-to-shade ;
: pixel ( x y r -- c )
3dup r2-y2-x2 dup 0> if ( if in disk )
sqrt nip normalize intensity shade ( z=sqrt[r2-x2-y2] )
else 2drop 2drop bl ( else blank )
then ;
: draw ( r -- ) ( r x1000 )
1000 * dup dup negate do
cr
dup dup negate do
dup I 500 + J 500 + rot pixel emit
500 +loop
1000 +loop drop ;
20 draw
10 draw |
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
| #Fortran | Fortran |
!Digital Text implemented as in C version - Anant Dixit (Oct, 2014)
program clock
implicit none
integer :: t(8)
do
call date_and_time(values=t)
call sleep(1)
call system('clear')
call digital_display(t(5),t(6),t(7))
end do
end program
subroutine digital_display(H,M,S)
!arguments
integer :: H, M, S
!local
character(len=*), parameter :: nfmt='(A8)', cfmt='(A6)'
character(len=88), parameter :: d1 = ' 00000 1 22222 33333 4 5555555 66666 7777777 88888 99999 '
character(len=88), parameter :: d2 = '0 0 11 2 2 3 3 44 5 6 6 7 7 8 8 9 9 :: '
character(len=88), parameter :: d3 = '0 00 1 1 2 3 4 4 5 6 7 8 8 9 9 :: '
character(len=88), parameter :: d4 = '0 0 0 1 2 3 4 4 5 6 7 8 8 9 9 :: '
character(len=88), parameter :: d5 = '0 0 0 1 2 333 4444444 555555 666666 7 88888 999999 '
character(len=88), parameter :: d6 = '0 0 0 1 2 3 4 5 6 6 7 8 8 9 :: '
character(len=88), parameter :: d7 = '00 0 1 2 3 4 5 6 6 7 8 8 9 :: '
character(len=88), parameter :: d8 = '0 0 1 2 3 3 4 5 5 6 6 7 8 8 9 9 :: '
character(len=88), parameter :: d9 = ' 00000 1111111 2222222 33333 4 55555 66666 7 88888 99999 '
integer :: h1, h2, m1, m2, s1, s2
h1 = 1+8*floor(dble(H)/10.D0)
h2 = 1+8*modulo(H,10)
m1 = 1+8*floor(dble(M)/10.D0)
m2 = 1+8*modulo(M,10)
s1 = 1+8*floor(dble(S)/10.D0)
s2 = 1+8*modulo(S,10)
write(*,nfmt,advance='no') d1(h1:h1+8)
write(*,nfmt,advance='no') d1(h2:h2+8)
write(*,cfmt,advance='no') d1(81:88)
write(*,nfmt,advance='no') d1(m1:m1+8)
write(*,nfmt,advance='no') d1(m2:m2+8)
write(*,cfmt,advance='no') d1(81:88)
write(*,nfmt,advance='no') d1(s1:s1+8)
write(*,nfmt) d1(s2:s2+8)
write(*,nfmt,advance='no') d2(h1:h1+8)
write(*,nfmt,advance='no') d2(h2:h2+8)
write(*,cfmt,advance='no') d2(81:88)
write(*,nfmt,advance='no') d2(m1:m1+8)
write(*,nfmt,advance='no') d2(m2:m2+8)
write(*,cfmt,advance='no') d2(81:88)
write(*,nfmt,advance='no') d2(s1:s1+8)
write(*,nfmt) d2(s2:s2+8)
write(*,nfmt,advance='no') d3(h1:h1+8)
write(*,nfmt,advance='no') d3(h2:h2+8)
write(*,cfmt,advance='no') d3(81:88)
write(*,nfmt,advance='no') d3(m1:m1+8)
write(*,nfmt,advance='no') d3(m2:m2+8)
write(*,cfmt,advance='no') d3(81:88)
write(*,nfmt,advance='no') d3(s1:s1+8)
write(*,nfmt) d3(s2:s2+8)
write(*,nfmt,advance='no') d4(h1:h1+8)
write(*,nfmt,advance='no') d4(h2:h2+8)
write(*,cfmt,advance='no') d4(81:88)
write(*,nfmt,advance='no') d4(m1:m1+8)
write(*,nfmt,advance='no') d4(m2:m2+8)
write(*,cfmt,advance='no') d4(81:88)
write(*,nfmt,advance='no') d4(s1:s1+8)
write(*,nfmt) d4(s2:s2+8)
write(*,nfmt,advance='no') d5(h1:h1+8)
write(*,nfmt,advance='no') d5(h2:h2+8)
write(*,cfmt,advance='no') d5(81:88)
write(*,nfmt,advance='no') d5(m1:m1+8)
write(*,nfmt,advance='no') d5(m2:m2+8)
write(*,cfmt,advance='no') d5(81:88)
write(*,nfmt,advance='no') d5(s1:s1+8)
write(*,nfmt) d5(s2:s2+8)
write(*,nfmt,advance='no') d6(h1:h1+8)
write(*,nfmt,advance='no') d6(h2:h2+8)
write(*,cfmt,advance='no') d6(81:88)
write(*,nfmt,advance='no') d6(m1:m1+8)
write(*,nfmt,advance='no') d6(m2:m2+8)
write(*,cfmt,advance='no') d6(81:88)
write(*,nfmt,advance='no') d6(s1:s1+8)
write(*,nfmt) d6(s2:s2+8)
write(*,nfmt,advance='no') d7(h1:h1+8)
write(*,nfmt,advance='no') d7(h2:h2+8)
write(*,cfmt,advance='no') d7(81:88)
write(*,nfmt,advance='no') d7(m1:m1+8)
write(*,nfmt,advance='no') d7(m2:m2+8)
write(*,cfmt,advance='no') d7(81:88)
write(*,nfmt,advance='no') d7(s1:s1+8)
write(*,nfmt) d7(s2:s2+8)
write(*,nfmt,advance='no') d8(h1:h1+8)
write(*,nfmt,advance='no') d8(h2:h2+8)
write(*,cfmt,advance='no') d8(81:88)
write(*,nfmt,advance='no') d8(m1:m1+8)
write(*,nfmt,advance='no') d8(m2:m2+8)
write(*,cfmt,advance='no') d8(81:88)
write(*,nfmt,advance='no') d8(s1:s1+8)
write(*,nfmt) d8(s2:s2+8)
write(*,nfmt,advance='no') d9(h1:h1+8)
write(*,nfmt,advance='no') d9(h2:h2+8)
write(*,cfmt,advance='no') d9(81:88)
write(*,nfmt,advance='no') d9(m1:m1+8)
write(*,nfmt,advance='no') d9(m2:m2+8)
write(*,cfmt,advance='no') d9(81:88)
write(*,nfmt,advance='no') d9(s1:s1+8)
write(*,nfmt) d9(s2:s2+8)
end subroutine
|
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
| #Ring | Ring |
# Project : Doubly-linked list/Traversal
trav = [123, 789, 456]
travfor = sort(trav)
see "Traverse forwards:" + nl
see travfor
see nl
travback = reverse(travfor)
see "Traverse backwards:" + nl
see travback
see nl
|
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
| #Ruby | Ruby | class DListNode
def get_tail
# parent class (ListNode) includes Enumerable, so the find method is available to us
self.find {|node| node.succ.nil?}
end
def each_previous(&b)
yield self
self.prev.each_previous(&b) if self.prev
end
end
head = DListNode.from_array([:a, :b, :c])
head.each {|node| p node.value}
head.get_tail.each_previous {|node| p node.value} |
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
| #Visual_Basic_.NET | Visual Basic .NET | Public Class Node(Of T)
Public Value As T
Public [Next] As Node(Of T)
Public Previous As Node(Of T)
End Class |
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
| #Wren | Wren | import "/llist" for DNode
var dn1 = DNode.new(1)
var dn2 = DNode.new(2)
dn1.next = dn2
dn1.prev = null
dn2.prev = dn1
dn2.next = null
System.print(["node 1", "data = %(dn1.data)", "prev = %(dn1.prev)", "next = %(dn1.next)"])
System.print(["node 2", "data = %(dn2.data)", "prev = %(dn2.prev)", "next = %(dn2.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
| #zkl | zkl | class Node{
fcn init(_value,_prev=Void,_next=Void)
{ var value=_value, prev=_prev, next=_next; }
fcn toString{ value.toString() }
} |
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)
| #Nim | Nim | import os, random, strutils
type Color {.pure.} = enum Red = "R", White = "W", Blue = "B"
#---------------------------------------------------------------------------------------------------
proc isSorted(a: openArray[Color]): bool =
# Check if an array of colors is in the order of the dutch national flag.
var prevColor = Red
for color in a:
if color < prevColor:
return false
prevColor = color
return true
#---------------------------------------------------------------------------------------------------
proc threeWayPartition(a: var openArray[Color]; mid: Color) =
## Dijkstra way to sort the colors.
var i, j = 0
var k = a.high
while j <= k:
if a[j] < mid:
swap a[i], a[j]
inc i
inc j
elif a[j] > mid:
swap a[j], a[k]
dec k
else:
inc j
#———————————————————————————————————————————————————————————————————————————————————————————————————
var n: Positive = 10
# Get the number of colors.
if paramCount() > 0:
try:
n = paramStr(1).parseInt()
if n <= 1:
raise newException(ValueError, "")
except ValueError:
echo "Wrong number of colors"
quit(QuitFailure)
# Create the colors.
randomize()
var colors = newSeqOfCap[Color](n)
while true:
for i in 0..<n:
colors.add(Color(rand(ord(Color.high))))
if not colors.isSorted():
break
colors.setLen(0) # Reset for next try.
echo "Original: ", colors.join("")
# Sort the colors.
var sortedColors = colors
threeWayPartition(sortedColors, White)
doAssert sortedColors.isSorted()
echo "Sorted: ", sortedColors.join("") |
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)
| #PARI.2FGP | PARI/GP | compare(a,b)={
if (a==b,
0
,
if(a=="red" || b=="blue", -1, 1)
)
};
r(n)=vector(n,i,if(random(3),if(random(2),"red","white"),"blue"));
inorder(v)=for(i=2,#v,if(compare(v[i-1],v[i])>0,return(0)));1;
v=r(10);
while(inorder(v), v=r(10));
v=vecsort(v,compare);
inorder(v) |
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
| #Kotlin | Kotlin | // version 1.1
import java.awt.*
import java.awt.event.MouseAdapter
import java.awt.event.MouseEvent
import javax.swing.*
class Cuboid: 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)
)
private var mouseX: Int = 0
private var prevMouseX: Int = 0
private var mouseY: Int = 0
private var prevMouseY: Int = 0
init {
preferredSize = Dimension(640, 640)
background = Color.white
scale(80.0, 120.0, 160.0)
rotateCube(Math.PI / 5.0, Math.PI / 9.0)
addMouseListener(object: MouseAdapter() {
override fun mousePressed(e: MouseEvent) {
mouseX = e.x
mouseY = e.y
}
})
addMouseMotionListener(object: MouseAdapter() {
override fun mouseDragged(e: MouseEvent) {
prevMouseX = mouseX
prevMouseY = mouseY
mouseX = e.x
mouseY = e.y
val incrX = (mouseX - prevMouseX) * 0.01
val incrY = (mouseY - prevMouseY) * 0.01
rotateCube(incrX, incrY)
repaint()
}
})
}
private fun scale(sx: Double, sy: Double, sz: Double) {
for (node in nodes) {
node[0] *= sx
node[1] *= sy
node[2] *= sz
}
}
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 = "Cuboid"
f.isResizable = false
f.add(Cuboid(), BorderLayout.CENTER)
f.pack()
f.setLocationRelativeTo(null)
f.isVisible = true
}
} |
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 | #ZX_Spectrum_Basic | ZX Spectrum Basic | PLOT INK 2;100,100 |
http://rosettacode.org/wiki/Dragon_curve | Dragon curve |
Create and display a dragon curve fractal.
(You may either display the curve directly or write it to an image file.)
Algorithms
Here are some brief notes the algorithms used and how they might suit various languages.
Recursively a right curling dragon is a right dragon followed by a left dragon, at 90-degree angle. And a left dragon is a left followed by a right.
*---R----* expands to * *
\ /
R L
\ /
*
*
/ \
L R
/ \
*---L---* expands to * *
The co-routines dcl and dcr in various examples do this recursively to a desired expansion level.
The curl direction right or left can be a parameter instead of two separate routines.
Recursively, a curl direction can be eliminated by noting the dragon consists of two copies of itself drawn towards a central point at 45-degrees.
*------->* becomes * * Recursive copies drawn
\ / from the ends towards
\ / the centre.
v v
*
This can be seen in the SVG example. This is best suited to off-line drawing since the reversal in the second half means the drawing jumps backward and forward (in binary reflected Gray code order) which is not very good for a plotter or for drawing progressively on screen.
Successive approximation repeatedly re-writes each straight line as two new segments at a right angle,
*
*-----* becomes / \ bend to left
/ \ if N odd
* *
* *
*-----* becomes \ / bend to right
\ / if N even
*
Numbering from the start of the curve built so far, if the segment is at an odd position then the bend introduced is on the right side. If the segment is an even position then on the left. The process is then repeated on the new doubled list of segments. This constructs a full set of line segments before any drawing.
The effect of the splitting is a kind of bottom-up version of the recursions. See the Asymptote example for code doing this.
Iteratively the curve always turns 90-degrees left or right at each point. The direction of the turn is given by the bit above the lowest 1-bit of n. Some bit-twiddling can extract that efficiently.
n = 1010110000
^
bit above lowest 1-bit, turn left or right as 0 or 1
LowMask = n BITXOR (n-1) # eg. giving 0000011111
AboveMask = LowMask + 1 # eg. giving 0000100000
BitAboveLowestOne = n BITAND AboveMask
The first turn is at n=1, so reckon the curve starting at the origin as n=0 then a straight line segment to position n=1 and turn there.
If you prefer to reckon the first turn as n=0 then take the bit above the lowest 0-bit instead. This works because "...10000" minus 1 is "...01111" so the lowest 0 in n-1 is where the lowest 1 in n is.
Going by turns suits turtle graphics such as Logo or a plotter drawing with a pen and current direction.
If a language doesn't maintain a "current direction" for drawing then you can always keep that separately and apply turns by bit-above-lowest-1.
Absolute direction to move at point n can be calculated by the number of bit-transitions in n.
n = 11 00 1111 0 1
^ ^ ^ ^ 4 places where change bit value
so direction=4*90degrees=East
This can be calculated by counting the number of 1 bits in "n XOR (n RIGHTSHIFT 1)" since such a shift and xor leaves a single 1 bit at each position where two adjacent bits differ.
Absolute X,Y coordinates of a point n can be calculated in complex numbers by some powers (i+1)^k and add/subtract/rotate. This is done in the gnuplot code. This might suit things similar to Gnuplot which want to calculate each point independently.
Predicate test for whether a given X,Y point or segment is on the curve can be done. This might suit line-by-line output rather than building an entire image before printing. See M4 for an example of this.
A predicate works by dividing out complex number i+1 until reaching the origin, so it takes roughly a bit at a time from X and Y is thus quite efficient. Why it works is slightly subtle but the calculation is not difficult. (Check segment by applying an offset to move X,Y to an "even" position before dividing i+1. Check vertex by whether the segment either East or West is on the curve.)
The number of steps in the predicate corresponds to doublings of the curve, so stopping the check at say 8 steps can limit the curve drawn to 2^8=256 points. The offsets arising in the predicate are bits of n the segment number, so can note those bits to calculate n and limit to an arbitrary desired length or sub-section.
As a Lindenmayer system of expansions. The simplest is two symbols F and S both straight lines, as used by the PGF code.
Axiom F, angle 90 degrees
F -> F+S
S -> F-S
This always has F at even positions and S at odd. Eg. after 3 levels F_S_F_S_F_S_F_S. The +/- turns in between bend to the left or right the same as the "successive approximation" method above. Read more at for instance Joel Castellanos' L-system page.
Variations are possible if you have only a single symbol for line draw, for example the Icon and Unicon and Xfractint code. The angles can also be broken into 45-degree parts to keep the expansion in a single direction rather than the endpoint rotating around.
The string rewrites can be done recursively without building the whole string, just follow its instructions at the target level. See for example C by IFS Drawing code. The effect is the same as "recursive with parameter" above but can draw other curves defined by L-systems.
| #C.2B.2B | C++ |
#include <windows.h>
#include <iostream>
//-----------------------------------------------------------------------------------------
using namespace std;
//-----------------------------------------------------------------------------------------
const int BMP_SIZE = 800, NORTH = 1, EAST = 2, SOUTH = 4, WEST = 8, LEN = 1;
//-----------------------------------------------------------------------------------------
class myBitmap
{
public:
myBitmap() : pen( NULL ), brush( NULL ), clr( 0 ), wid( 1 ) {}
~myBitmap()
{
DeleteObject( pen ); DeleteObject( brush );
DeleteDC( hdc ); DeleteObject( bmp );
}
bool create( int w, int h )
{
BITMAPINFO bi;
ZeroMemory( &bi, sizeof( bi ) );
bi.bmiHeader.biSize = sizeof( bi.bmiHeader );
bi.bmiHeader.biBitCount = sizeof( DWORD ) * 8;
bi.bmiHeader.biCompression = BI_RGB;
bi.bmiHeader.biPlanes = 1;
bi.bmiHeader.biWidth = w;
bi.bmiHeader.biHeight = -h;
HDC dc = GetDC( GetConsoleWindow() );
bmp = CreateDIBSection( dc, &bi, DIB_RGB_COLORS, &pBits, NULL, 0 );
if( !bmp ) return false;
hdc = CreateCompatibleDC( dc );
SelectObject( hdc, bmp );
ReleaseDC( GetConsoleWindow(), dc );
width = w; height = h;
return true;
}
void clear( BYTE clr = 0 )
{
memset( pBits, clr, width * height * sizeof( DWORD ) );
}
void setBrushColor( DWORD bClr )
{
if( brush ) DeleteObject( brush );
brush = CreateSolidBrush( bClr );
SelectObject( hdc, brush );
}
void setPenColor( DWORD c )
{
clr = c; createPen();
}
void setPenWidth( int w )
{
wid = w; createPen();
}
void saveBitmap( string path )
{
BITMAPFILEHEADER fileheader;
BITMAPINFO infoheader;
BITMAP bitmap;
DWORD wb;
GetObject( bmp, sizeof( bitmap ), &bitmap );
DWORD* dwpBits = new DWORD[bitmap.bmWidth * bitmap.bmHeight];
ZeroMemory( dwpBits, bitmap.bmWidth * bitmap.bmHeight * sizeof( DWORD ) );
ZeroMemory( &infoheader, sizeof( BITMAPINFO ) );
ZeroMemory( &fileheader, sizeof( BITMAPFILEHEADER ) );
infoheader.bmiHeader.biBitCount = sizeof( DWORD ) * 8;
infoheader.bmiHeader.biCompression = BI_RGB;
infoheader.bmiHeader.biPlanes = 1;
infoheader.bmiHeader.biSize = sizeof( infoheader.bmiHeader );
infoheader.bmiHeader.biHeight = bitmap.bmHeight;
infoheader.bmiHeader.biWidth = bitmap.bmWidth;
infoheader.bmiHeader.biSizeImage = bitmap.bmWidth * bitmap.bmHeight * sizeof( DWORD );
fileheader.bfType = 0x4D42;
fileheader.bfOffBits = sizeof( infoheader.bmiHeader ) + sizeof( BITMAPFILEHEADER );
fileheader.bfSize = fileheader.bfOffBits + infoheader.bmiHeader.biSizeImage;
GetDIBits( hdc, bmp, 0, height, ( LPVOID )dwpBits, &infoheader, DIB_RGB_COLORS );
HANDLE file = CreateFile( path.c_str(), GENERIC_WRITE, 0, NULL, CREATE_ALWAYS, FILE_ATTRIBUTE_NORMAL, NULL );
WriteFile( file, &fileheader, sizeof( BITMAPFILEHEADER ), &wb, NULL );
WriteFile( file, &infoheader.bmiHeader, sizeof( infoheader.bmiHeader ), &wb, NULL );
WriteFile( file, dwpBits, bitmap.bmWidth * bitmap.bmHeight * 4, &wb, NULL );
CloseHandle( file );
delete [] dwpBits;
}
HDC getDC() const { return hdc; }
int getWidth() const { return width; }
int getHeight() const { return height; }
private:
void createPen()
{
if( pen ) DeleteObject( pen );
pen = CreatePen( PS_SOLID, wid, clr );
SelectObject( hdc, pen );
}
HBITMAP bmp;
HDC hdc;
HPEN pen;
HBRUSH brush;
void *pBits;
int width, height, wid;
DWORD clr;
};
//-----------------------------------------------------------------------------------------
class dragonC
{
public:
dragonC() { bmp.create( BMP_SIZE, BMP_SIZE ); dir = WEST; }
void draw( int iterations ) { generate( iterations ); draw(); }
private:
void generate( int it )
{
generator.push_back( 1 );
string temp;
for( int y = 0; y < it - 1; y++ )
{
temp = generator; temp.push_back( 1 );
for( string::reverse_iterator x = generator.rbegin(); x != generator.rend(); x++ )
temp.push_back( !( *x ) );
generator = temp;
}
}
void draw()
{
HDC dc = bmp.getDC();
unsigned int clr[] = { 0xff, 0xff00, 0xff0000, 0x00ffff };
int mov[] = { 0, 0, 1, -1, 1, -1, 1, 0 }; int i = 0;
for( int t = 0; t < 4; t++ )
{
int a = BMP_SIZE / 2, b = a; a += mov[i++]; b += mov[i++];
MoveToEx( dc, a, b, NULL );
bmp.setPenColor( clr[t] );
for( string::iterator x = generator.begin(); x < generator.end(); x++ )
{
switch( dir )
{
case NORTH:
if( *x ) { a += LEN; dir = EAST; }
else { a -= LEN; dir = WEST; }
break;
case EAST:
if( *x ) { b += LEN; dir = SOUTH; }
else { b -= LEN; dir = NORTH; }
break;
case SOUTH:
if( *x ) { a -= LEN; dir = WEST; }
else { a += LEN; dir = EAST; }
break;
case WEST:
if( *x ) { b -= LEN; dir = NORTH; }
else { b += LEN; dir = SOUTH; }
}
LineTo( dc, a, b );
}
}
// !!! change this path !!!
bmp.saveBitmap( "f:/rc/dragonCpp.bmp" );
}
int dir;
myBitmap bmp;
string generator;
};
//-----------------------------------------------------------------------------------------
int main( int argc, char* argv[] )
{
dragonC d; d.draw( 17 );
return system( "pause" );
}
//-----------------------------------------------------------------------------------------
|
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
| #Frink | Frink | res = 254 / in
v = callJava["frink.graphics.VoxelArray", "makeSphere", [1/2 inch res]]
v.projectX[undef].show["X"]
v.projectY[undef].show["Y"]
v.projectZ[undef].show["Z"]
filename = "sphere.stl"
print["Writing $filename..."]
w = new Writer[filename]
w.println[v.toSTLFormat["sphere", 1/(res mm)]]
w.close[]
println["done."] |
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
| #FreeBASIC | FreeBASIC | ' version 05-04-2017
' compile with: fbc -s gui
Const As Double deg2rad = Atn(1) / 45
Const As UInteger w = 199, h = 199
Const As UInteger x0 = w \ 2, y0 = h \ 2 ' center
Dim As UInteger x, x1, x2, x3, y, y1, y2, y3
Dim As String sys_time, press
Dim As Integer hours, minutes, seconds
Dim As Double angle, a_sin, a_cos
ScreenRes w, h, 8 ' 8bit color depth (palette)
WindowTitle "Simple Clock"
' create image 8bit (palette) and set pixels to 15 (white)
Dim clockdial As Any Ptr = ImageCreate(w, h, 15, 8)
If clockdial = 0 Then
Print "Failed to create image."
Sleep
End -1
End If
' draw clockdial in memory
Circle clockdial, (x0, y0), 94 ,0
Circle clockdial, (x0, y0), 90 ,0
For x = 0 To 174 Step 6
a_sin = Sin(x * deg2rad)
a_cos = Cos(x * deg2rad)
x1 = 94 * a_sin : y1 = 94 * a_cos
If x Mod 30 = 0 Then
x2 = 85 * a_sin : y2 = 85 * a_cos
Else
x2 = 90 * a_sin : y2 = 90 * a_cos
End If
Line clockdial, (x0 + x1, y0 + y1) - (x0 + x2, y0 + y2), 0
Line clockdial, (x0 - x1, y0 - y1) - (x0 - x2, y0 - y2), 0
Next
'draw clock
Do
sys_time = Time
hours = (sys_time[0] - Asc("0")) * 10 + sys_time[1] - Asc("0")
minutes = (sys_time[3] - Asc("0")) * 10 + sys_time[4] - Asc("0")
seconds = (sys_time[6] - Asc("0")) * 10 + sys_time[7] - Asc("0")
If hours > 12 Then hours -= 12
angle = (180 - (hours * 30 + minutes / 2)) * deg2rad
x1 = 65 * Sin(angle)
y1 = 65 * Cos(angle)
angle = (180 - (minutes * 6 + seconds / 10)) * deg2rad
x2 = 80 * Sin(angle)
y2 = 80 * Cos(angle)
angle = (180 - seconds * 6) * deg2rad
x3 = 90 * Sin(angle)
y3 = 90 * Cos(angle)
ScreenLock
' load image, setting pixels
Put (0, 0), clockdial, PSet
Line (x0, y0) - (x0 + x1, y0 + y1), 1 ' hour hand blue
Line (x0, y0) - (x0 + x2, y0 + y2), 2 ' minute hand green
Line (x0, y0) - (x0 + x3, y0 + y3), 12 ' second hand red
ScreenUnLock
Sleep 300, 1 ' wait 300 ms, don't respond to keys pressed
' press esc or mouse click on close window to stop program
press = InKey
If press = Chr(27) Or press = Chr(255) + "k" Then Exit Do
Loop
ImageDestroy(clockdial)
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
| #Salmon | Salmon | class item(init_data)
{
variable next,
previous,
data := init_data;
};
function prepend(tail, new_data)
{
immutable result := item(new_data);
result.next := tail;
result.previous := null;
if (tail != null)
tail.previous := result;;
return result;
};
variable my_list := null;
my_list := prepend(my_list, "R");
my_list := prepend(my_list, "o");
my_list := prepend(my_list, "s");
my_list := prepend(my_list, "e");
my_list := prepend(my_list, "t");
my_list := prepend(my_list, "t");
my_list := prepend(my_list, "a");
"Items in the list going forward:"!
variable follow := my_list;
while (true)
{
follow.data!
if (follow.next == null)
break;;
}
step
follow := follow.next;;
"Items in the list going backward:"!
while (follow != null)
follow.data!
step
follow := follow.previous;; |
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
| #Scala | Scala | import java.util
object DoublyLinkedListTraversal extends App {
private val ll = new util.LinkedList[String]
private def traverse(iter: util.Iterator[String]) =
while (iter.hasNext) iter.next
traverse(ll.iterator)
traverse(ll.descendingIterator)
} |
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)
| #Perl | Perl | use warnings;
use strict;
use 5.010; # //
use List::Util qw( shuffle );
my @colours = qw( blue white red );
sub are_ordered {
my $balls = shift;
my $last = 0;
for my $ball (@$balls) {
return if $ball < $last;
$last = $ball;
}
return 1;
}
sub show {
my $balls = shift;
print join(' ', map $colours[$_], @$balls), "\n";
}
sub debug {
return unless $ENV{DEBUG};
my ($pos, $top, $bottom, $balls) = @_;
for my $i (0 .. $#$balls) {
my ($prefix, $suffix) = (q()) x 2;
($prefix, $suffix) = qw/( )/ if $i == $pos;
$prefix .= '>' if $i == $top;
$suffix .= '<' if $i == $bottom;
print STDERR " $prefix$colours[$balls->[$i]]$suffix";
}
print STDERR "\n";
}
my $count = shift // 10;
die "$count: Not enough balls\n" if $count < 3;
my $balls = [qw( 2 1 0 )];
push @$balls, int rand 3 until @$balls == $count;
do { @$balls = shuffle @$balls } while are_ordered($balls);
show($balls);
my $top = 0;
my $bottom = $#$balls;
my $i = 0;
while ($i <= $bottom) {
debug($i, $top, $bottom, $balls);
my $col = $colours[ $balls->[$i] ];
if ('red' eq $col and $i < $bottom) {
@{$balls}[$bottom, $i] = @{$balls}[$i, $bottom];
$bottom--;
} elsif ('blue' eq $col and $i > $top) {
@{$balls}[$top, $i] = @{$balls}[$i, $top];
$top++;
} else {
$i++;
}
}
debug($i, $top, $bottom, $balls);
show($balls);
are_ordered($balls) or die "Incorrect\n"; |
http://rosettacode.org/wiki/Draw_a_cuboid | Draw a cuboid | Task
Draw a cuboid with relative dimensions of 2 × 3 × 4.
The cuboid can be represented graphically, or in ASCII art, depending on the language capabilities.
To fulfill the criteria of being a cuboid, three faces must be visible.
Either static or rotational projection is acceptable for this task.
Related tasks
draw a sphere
draw a rotating cube
write language name in 3D ASCII
draw a Deathstar
| #Liberty_BASIC | Liberty BASIC |
Call cuboid 1,3,4
End
Sub cuboid width, height, depth
wd=width*7+2: hi=height*3: dp=depth
For i=1 To wd-2
w$=w$+"-":h$=h$+" "
Next
w$="+"+w$+"+":d$="/"+h$+"/":h$="|"+h$+"|"
px=dp+2:py=1:Locate dp+2,py:Print w$;
For i=2 To hi+1
Locate wd+dp+1,i:Print"|";
Next
Locate wd+dp+1, i: Print "+";
For i=dp+1 To 1 Step -1
py=py+1:Locate i,py:Print d$;
Next
For i=1 To dp
Locate wd+(dp+1)-i,hi+d+2+i:Print "/";
Next
Locate 1, dp+2: Print w$;
For i=dp+3 To hi+dp+2
Locate 1,i:Print h$;
Next
Locate 1, dp+hi+3: Print w$
End Sub
|
http://rosettacode.org/wiki/Dragon_curve | Dragon curve |
Create and display a dragon curve fractal.
(You may either display the curve directly or write it to an image file.)
Algorithms
Here are some brief notes the algorithms used and how they might suit various languages.
Recursively a right curling dragon is a right dragon followed by a left dragon, at 90-degree angle. And a left dragon is a left followed by a right.
*---R----* expands to * *
\ /
R L
\ /
*
*
/ \
L R
/ \
*---L---* expands to * *
The co-routines dcl and dcr in various examples do this recursively to a desired expansion level.
The curl direction right or left can be a parameter instead of two separate routines.
Recursively, a curl direction can be eliminated by noting the dragon consists of two copies of itself drawn towards a central point at 45-degrees.
*------->* becomes * * Recursive copies drawn
\ / from the ends towards
\ / the centre.
v v
*
This can be seen in the SVG example. This is best suited to off-line drawing since the reversal in the second half means the drawing jumps backward and forward (in binary reflected Gray code order) which is not very good for a plotter or for drawing progressively on screen.
Successive approximation repeatedly re-writes each straight line as two new segments at a right angle,
*
*-----* becomes / \ bend to left
/ \ if N odd
* *
* *
*-----* becomes \ / bend to right
\ / if N even
*
Numbering from the start of the curve built so far, if the segment is at an odd position then the bend introduced is on the right side. If the segment is an even position then on the left. The process is then repeated on the new doubled list of segments. This constructs a full set of line segments before any drawing.
The effect of the splitting is a kind of bottom-up version of the recursions. See the Asymptote example for code doing this.
Iteratively the curve always turns 90-degrees left or right at each point. The direction of the turn is given by the bit above the lowest 1-bit of n. Some bit-twiddling can extract that efficiently.
n = 1010110000
^
bit above lowest 1-bit, turn left or right as 0 or 1
LowMask = n BITXOR (n-1) # eg. giving 0000011111
AboveMask = LowMask + 1 # eg. giving 0000100000
BitAboveLowestOne = n BITAND AboveMask
The first turn is at n=1, so reckon the curve starting at the origin as n=0 then a straight line segment to position n=1 and turn there.
If you prefer to reckon the first turn as n=0 then take the bit above the lowest 0-bit instead. This works because "...10000" minus 1 is "...01111" so the lowest 0 in n-1 is where the lowest 1 in n is.
Going by turns suits turtle graphics such as Logo or a plotter drawing with a pen and current direction.
If a language doesn't maintain a "current direction" for drawing then you can always keep that separately and apply turns by bit-above-lowest-1.
Absolute direction to move at point n can be calculated by the number of bit-transitions in n.
n = 11 00 1111 0 1
^ ^ ^ ^ 4 places where change bit value
so direction=4*90degrees=East
This can be calculated by counting the number of 1 bits in "n XOR (n RIGHTSHIFT 1)" since such a shift and xor leaves a single 1 bit at each position where two adjacent bits differ.
Absolute X,Y coordinates of a point n can be calculated in complex numbers by some powers (i+1)^k and add/subtract/rotate. This is done in the gnuplot code. This might suit things similar to Gnuplot which want to calculate each point independently.
Predicate test for whether a given X,Y point or segment is on the curve can be done. This might suit line-by-line output rather than building an entire image before printing. See M4 for an example of this.
A predicate works by dividing out complex number i+1 until reaching the origin, so it takes roughly a bit at a time from X and Y is thus quite efficient. Why it works is slightly subtle but the calculation is not difficult. (Check segment by applying an offset to move X,Y to an "even" position before dividing i+1. Check vertex by whether the segment either East or West is on the curve.)
The number of steps in the predicate corresponds to doublings of the curve, so stopping the check at say 8 steps can limit the curve drawn to 2^8=256 points. The offsets arising in the predicate are bits of n the segment number, so can note those bits to calculate n and limit to an arbitrary desired length or sub-section.
As a Lindenmayer system of expansions. The simplest is two symbols F and S both straight lines, as used by the PGF code.
Axiom F, angle 90 degrees
F -> F+S
S -> F-S
This always has F at even positions and S at odd. Eg. after 3 levels F_S_F_S_F_S_F_S. The +/- turns in between bend to the left or right the same as the "successive approximation" method above. Read more at for instance Joel Castellanos' L-system page.
Variations are possible if you have only a single symbol for line draw, for example the Icon and Unicon and Xfractint code. The angles can also be broken into 45-degree parts to keep the expansion in a single direction rather than the endpoint rotating around.
The string rewrites can be done recursively without building the whole string, just follow its instructions at the target level. See for example C by IFS Drawing code. The effect is the same as "recursive with parameter" above but can draw other curves defined by L-systems.
| #COBOL | COBOL | >>SOURCE FORMAT FREE
*> This code is dedicated to the public domain
identification division.
program-id. dragon.
environment division.
configuration section.
repository. function all intrinsic.
data division.
working-storage section.
01 segment-length pic 9 value 2.
01 mark pic x value '.'.
01 segment-count pic 9999 value 513.
01 segment pic 9999.
01 point pic 9999 value 1.
01 point-max pic 9999.
01 point-lim pic 9999 value 8192.
01 dragon-curve.
03 filler occurs 8192.
05 ydragon pic s9999.
05 xdragon pic s9999.
01 x pic s9999 value 1.
01 y pic S9999 value 1.
01 xdelta pic s9 value 1. *> start pointing east
01 ydelta pic s9 value 0.
01 x-max pic s9999 value -9999.
01 x-min pic s9999 value 9999.
01 y-max pic s9999 value -9999.
01 y-min pic s9999 value 9999.
01 n pic 9999.
01 r pic 9.
01 xupper pic s9999.
01 yupper pic s9999.
01 window-line-number pic 99.
01 window-width pic 99 value 64.
01 window-height pic 99 value 22.
01 window.
03 window-line occurs 22.
05 window-point occurs 64 pic x.
01 direction pic x.
procedure division.
start-dragon.
if segment-count * segment-length > point-lim
*> too many segments for the point-table
compute segment-count = point-lim / segment-length
end-if
perform varying segment from 1 by 1
until segment > segment-count
*>===========================================
*> segment = n * 2 ** b
*> if mod(n,4) = 3, turn left else turn right
*>===========================================
*> calculate the turn
divide 2 into segment giving n remainder r
perform until r <> 0
divide 2 into n giving n remainder r
end-perform
divide 2 into n giving n remainder r
*> perform the turn
evaluate r also xdelta also ydelta
when 0 also 1 also 0 *> turn right from east
when 1 also -1 also 0 *> turn left from west
*> turn to south
move 0 to xdelta
move 1 to ydelta
when 1 also 1 also 0 *> turn left from east
when 0 also -1 also 0 *> turn right from west
*> turn to north
move 0 to xdelta
move -1 to ydelta
when 0 also 0 also 1 *> turn right from south
when 1 also 0 also -1 *> turn left from north
*> turn to west
move 0 to ydelta
move -1 to xdelta
when 1 also 0 also 1 *> turn left from south
when 0 also 0 also -1 *> turn right from north
*> turn to east
move 0 to ydelta
move 1 to xdelta
end-evaluate
*> plot the segment points
perform segment-length times
add xdelta to x
add ydelta to y
move x to xdragon(point)
move y to ydragon(point)
add 1 to point
end-perform
*> update the limits for the display
compute x-max = max(x, x-max)
compute x-min = min(x, x-min)
compute y-max = max(y, y-max)
compute y-min = min(y, y-min)
move point to point-max
end-perform
*>==========================================
*> display the curve
*> hjkl corresponds to left, up, down, right
*> anything else ends the program
*>==========================================
move 1 to yupper xupper
perform with test after
until direction <> 'h' and 'j' and 'k' and 'l'
*>==========================================
*> (yupper,xupper) maps to window-point(1,1)
*>==========================================
*> move the window
evaluate true
when direction = 'h' *> move window left
and xupper > x-min + window-width
subtract 1 from xupper
when direction = 'j' *> move window up
and yupper < y-max - window-height
add 1 to yupper
when direction = 'k' *> move window down
and yupper > y-min + window-height
subtract 1 from yupper
when direction = 'l' *> move window right
and xupper < x-max - window-width
add 1 to xupper
end-evaluate
*> plot the dragon points in the window
move spaces to window
perform varying point from 1 by 1
until point > point-max
if ydragon(point) >= yupper and < yupper + window-height
and xdragon(point) >= xupper and < xupper + window-width
*> we're in the window
compute y = ydragon(point) - yupper + 1
compute x = xdragon(point) - xupper + 1
move mark to window-point(y, x)
end-if
end-perform
*> display the window
perform varying window-line-number from 1 by 1
until window-line-number > window-height
display window-line(window-line-number)
end-perform
*> get the next window move or terminate
display 'hjkl?' with no advancing
accept direction
end-perform
stop run
.
end program dragon. |
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
| #FutureBasic | FutureBasic |
include "Tlbx agl.incl"
include "Tlbx glut.incl"
output file "Rotating Sphere"
local fn SphereDraw
'~'1
begin globals
dim as double sRotation // 'static' var
end globals
// Speed of rotation
sRotation += 2.9
glMatrixMode( _GLMODELVIEW )
glLoadIdentity()
// Position parameters: x axis, y axis, z axis
// Set to center of screen:
glTranslated( 0.0, 0.0, 0.0 )
// Rotation (wobble) parameters: angle, x, y
glRotated( sRotation, -0.45, -0.8, -0.6 )
// Set color of sphere's wireframe
glColor3d( 1.0, 0.0, 0.3 )
// Set width of sphere's wireframe lines
glLineWidth( 1.5 )
// Apply above to GLUT's built-in sphere wireframe
// Size & frame parameters: radius, slices, stack
fn glutWireSphere( 0.8, 25, 25 )
end fn
// main program
dim as GLint attrib(2)
dim as CGrafPtr port
dim as AGLPixelFormat fmt
dim as AGLContext glContext
dim as EventRecord ev
dim as GLboolean yesOK
// Make a window
window 1, @"Rotating Sphere", (0,0) - (500,500)
// Minimal setup of OpenGL
attrib(0) = _AGLRGBA
attrib(1) = _AGLDOUBLEBUFFER
attrib(2) = _AGLNONE
fmt = fn aglChoosePixelFormat( 0, 0, attrib(0) )
glContext = fn aglCreateContext( fmt, 0 )
aglDestroyPixelFormat( fmt )
// Set the FB window as port for drawing
port = window( _wndPort )
yesOK = fn aglSetDrawable( glContext, port )
yesOK = fn aglSetCurrentContext( glContext )
// Background color: red, green, blue, alpha
glClearColor( 0.0, 0.0, 0.0, 0.0 )
// 60/s HandleEvents Trigger
poke long event - 8, 1
do
// Clear window
glClear( _GLCOLORBUFFERBIT )
// Run animation
fn SphereDraw
aglSwapBuffers( glContext )
HandleEvents
until gFBquit
|
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
| #FunL | FunL | import concurrent.{scheduleAtFixedRate, scheduler}
val ROW = 10
val COL = 20
val digits = array( [
" __",
" / /",
"/__/ ",
" ",
" /",
" / ",
" __",
" __/",
"/__ ",
" __",
" __/",
" __/ ",
" ",
" /__/",
" / ",
" __",
" /__ ",
" __/ ",
" __",
" /__ ",
"/__/ ",
" __",
" /",
" / ",
" __",
" /__/",
"/__/ ",
" __",
" /__/",
" __/ "
] )
val colon = array( [
" ",
" .",
". "
] )
def displayTime =
def pad( n ) = if n < 10 then '0' + n else n
t = $time
s = (t + $timeZoneOffset)\1000%86400
time = pad( s\3600 ) + ':' + pad( s%3600\60 ) + ':' + pad( s%60 )
for row <- 0:3
print( if $os.startsWith('Windows') then '\n' else '\u001B[' + (ROW + row) + ';' + COL + 'H' )
for ch <- time
print( if ch == ':' then colon(row) else digits(int(ch)*3 + row) )
println()
t
if not $os.startsWith( 'Windows' )
print( '\u001B[2J\u001B[?25l' )
scheduleAtFixedRate( displayTime, 1000 - displayTime()%1000, 1000 )
readLine()
scheduler().shutdown()
if not $os.startsWith( 'Windows' )
print( '\u001B[?25h' ) |
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
| #Tcl | Tcl | # Modify the List class to add the iterator methods
oo::define List {
method foreach {varName script} {
upvar 1 $varName v
for {set node [self]} {$node ne ""} {set node [$node next]} {
set v [$node value]
uplevel 1 $script
}
}
method revforeach {varName script} {
upvar 1 $varName v
for {set node [self]} {$node ne ""} {set node [$node previous]} {
set v [$node value]
uplevel 1 $script
}
}
}
# Demonstrating...
set first [List new a [List new b [List new c [set last [List new d]]]]]
puts "Forward..."
$first foreach char { puts $char }
puts "Backward..."
$last revforeach char { puts $char } |
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
| #Wren | Wren | import "/llist" for DLinkedList
import "/fmt" for Fmt
// create a new doubly-linked list and add the first 50 positive integers to it
var dll = DLinkedList.new(1..50)
// traverse the doubly-linked list from head to tail
for (i in dll) {
Fmt.write("$4d ", i)
if (i % 10 == 0) System.print()
}
System.print()
// traverse the doubly-linked list from tail to head
for (i in dll.reversed) {
Fmt.write("$4d ", i)
if (i % 10 == 1) System.print()
} |
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)
| #Phix | Phix | with javascript_semantics
function three_way_partition(sequence s, integer mid)
integer i=1, j=1, n = length(s)
while j <= n do
if s[j] < mid then
{s[i],s[j]} = {s[j],s[i]}
i += 1
j += 1
elsif s[j] > mid then
{s[j],s[n]} = {s[n],s[j]}
n -= 1
else
j += 1
end if
end while
return s
end function
constant colours = {"red","white","blue"}
enum /*red,*/ white = 2, blue, maxc = blue
procedure show(string msg, sequence s)
sequence t = repeat(0,length(s))
for i=1 to length(s) do
t[i] = colours[s[i]]
end for
printf(1,"%s: %s\n",{msg,join(t)})
end procedure
sequence unsorted, sorted
while 1 do
unsorted = sq_rand(repeat(maxc,12))
-- sorted = sort(deep_copy(unsorted)) -- (works just as well)
sorted = three_way_partition(deep_copy(unsorted), white)
if unsorted!=sorted then exit end if
?"oops"
end while
show("Unsorted",unsorted)
show("Sorted",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
| #Logo | Logo | to cuboid :l1 :l2 :l3
cs perspective ;making the room ready to use
setxyz :l1 0 0
setxyz :l1 :l2 0
setxyz 0 :l2 0
setxyz 0 0 0
setxyz :l1 0 0
setxyz :l1 0 -:l3
setxyz :l1 :l2 -:l3
setxyz :l1 :l2 0
setxyz 0 :l2 0
setxyz 0 :l2 -:l3
setxyz :l1 :l2 -:l3
end |
http://rosettacode.org/wiki/Dragon_curve | Dragon curve |
Create and display a dragon curve fractal.
(You may either display the curve directly or write it to an image file.)
Algorithms
Here are some brief notes the algorithms used and how they might suit various languages.
Recursively a right curling dragon is a right dragon followed by a left dragon, at 90-degree angle. And a left dragon is a left followed by a right.
*---R----* expands to * *
\ /
R L
\ /
*
*
/ \
L R
/ \
*---L---* expands to * *
The co-routines dcl and dcr in various examples do this recursively to a desired expansion level.
The curl direction right or left can be a parameter instead of two separate routines.
Recursively, a curl direction can be eliminated by noting the dragon consists of two copies of itself drawn towards a central point at 45-degrees.
*------->* becomes * * Recursive copies drawn
\ / from the ends towards
\ / the centre.
v v
*
This can be seen in the SVG example. This is best suited to off-line drawing since the reversal in the second half means the drawing jumps backward and forward (in binary reflected Gray code order) which is not very good for a plotter or for drawing progressively on screen.
Successive approximation repeatedly re-writes each straight line as two new segments at a right angle,
*
*-----* becomes / \ bend to left
/ \ if N odd
* *
* *
*-----* becomes \ / bend to right
\ / if N even
*
Numbering from the start of the curve built so far, if the segment is at an odd position then the bend introduced is on the right side. If the segment is an even position then on the left. The process is then repeated on the new doubled list of segments. This constructs a full set of line segments before any drawing.
The effect of the splitting is a kind of bottom-up version of the recursions. See the Asymptote example for code doing this.
Iteratively the curve always turns 90-degrees left or right at each point. The direction of the turn is given by the bit above the lowest 1-bit of n. Some bit-twiddling can extract that efficiently.
n = 1010110000
^
bit above lowest 1-bit, turn left or right as 0 or 1
LowMask = n BITXOR (n-1) # eg. giving 0000011111
AboveMask = LowMask + 1 # eg. giving 0000100000
BitAboveLowestOne = n BITAND AboveMask
The first turn is at n=1, so reckon the curve starting at the origin as n=0 then a straight line segment to position n=1 and turn there.
If you prefer to reckon the first turn as n=0 then take the bit above the lowest 0-bit instead. This works because "...10000" minus 1 is "...01111" so the lowest 0 in n-1 is where the lowest 1 in n is.
Going by turns suits turtle graphics such as Logo or a plotter drawing with a pen and current direction.
If a language doesn't maintain a "current direction" for drawing then you can always keep that separately and apply turns by bit-above-lowest-1.
Absolute direction to move at point n can be calculated by the number of bit-transitions in n.
n = 11 00 1111 0 1
^ ^ ^ ^ 4 places where change bit value
so direction=4*90degrees=East
This can be calculated by counting the number of 1 bits in "n XOR (n RIGHTSHIFT 1)" since such a shift and xor leaves a single 1 bit at each position where two adjacent bits differ.
Absolute X,Y coordinates of a point n can be calculated in complex numbers by some powers (i+1)^k and add/subtract/rotate. This is done in the gnuplot code. This might suit things similar to Gnuplot which want to calculate each point independently.
Predicate test for whether a given X,Y point or segment is on the curve can be done. This might suit line-by-line output rather than building an entire image before printing. See M4 for an example of this.
A predicate works by dividing out complex number i+1 until reaching the origin, so it takes roughly a bit at a time from X and Y is thus quite efficient. Why it works is slightly subtle but the calculation is not difficult. (Check segment by applying an offset to move X,Y to an "even" position before dividing i+1. Check vertex by whether the segment either East or West is on the curve.)
The number of steps in the predicate corresponds to doublings of the curve, so stopping the check at say 8 steps can limit the curve drawn to 2^8=256 points. The offsets arising in the predicate are bits of n the segment number, so can note those bits to calculate n and limit to an arbitrary desired length or sub-section.
As a Lindenmayer system of expansions. The simplest is two symbols F and S both straight lines, as used by the PGF code.
Axiom F, angle 90 degrees
F -> F+S
S -> F-S
This always has F at even positions and S at odd. Eg. after 3 levels F_S_F_S_F_S_F_S. The +/- turns in between bend to the left or right the same as the "successive approximation" method above. Read more at for instance Joel Castellanos' L-system page.
Variations are possible if you have only a single symbol for line draw, for example the Icon and Unicon and Xfractint code. The angles can also be broken into 45-degree parts to keep the expansion in a single direction rather than the endpoint rotating around.
The string rewrites can be done recursively without building the whole string, just follow its instructions at the target level. See for example C by IFS Drawing code. The effect is the same as "recursive with parameter" above but can draw other curves defined by L-systems.
| #Commodore_BASIC | Commodore BASIC |
10 REM DRAGON CURVE
20 REM SIN, COS IN ARRAYS FOR PI/4 MULTIPL.
30 DIM S(7),C(7)
40 QPI=ATN(1)
50 FOR I=0 TO 7
60 S(I)=SIN(I*QPI)
70 C(I)=COS(I*QPI)
80 NEXT I
90 LEVEL=15
100 INSIZE=128:REM 2^WHOLE_NUM (LOOKS BETTER)
110 X=112
120 Y=70
130 SQ=SQR(2)
140 ROTQPI=0:ITER=0:RQ=1
150 DIM R(LEVEL)
160 GRAPHIC 2,1
170 GOSUB 190
180 END
190 REM DRAGON
200 IF ROTQPI<0 THEN ROTQPI=ROTQPI+8:GOTO 220
210 IF ROTQPI>7 THEN ROTQPI=ROTQPI-8
220 IF LEVEL>1 THEN GO TO 290
230 YN=S(ROTQPI)*INSIZE+Y
240 XN=C(ROTQPI)*INSIZE+X
250 DRAW ,X,Y TO XN,YN
260 ITER=ITER+1
270 X=XN:Y=YN
280 RETURN
290 INSIZE=INSIZE*SQ/2
300 ROTQPI=ROTQPI+RQ
310 IF ROTQPI<0 THEN ROTQPI=ROTQPI+8:GOTO 330
320 IF ROTQPI>7 THEN ROTQPI=ROTQPI-8
330 LEVEL=LEVEL-1
340 R(LEVEL)=RQ:RQ=1
350 GOSUB 190
360 ROTQPI=ROTQPI-R(LEVEL)*2
370 IF ROTQPI<0 THEN ROTQPI=ROTQPI+8:GOTO 390
380 IF ROTQPI>7 THEN ROTQPI=ROTQPI-8
390 RQ=-1
400 GOSUB 190
410 RQ=R(LEVEL)
420 ROTQPI=ROTQPI+RQ
430 IF ROTQPI<0 THEN ROTQPI=ROTQPI+8:GOTO 450
440 IF ROTQPI>7 THEN ROTQPI=ROTQPI-8
450 LEVEL=LEVEL+1
460 INSIZE=INSIZE*SQ
470 RETURN
|
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
| #Go | Go | package main
import (
"fmt"
"image"
"image/color"
"image/png"
"math"
"os"
)
type vector [3]float64
func normalize(v *vector) {
invLen := 1 / math.Sqrt(dot(v, v))
v[0] *= invLen
v[1] *= invLen
v[2] *= invLen
}
func dot(x, y *vector) float64 {
return x[0]*y[0] + x[1]*y[1] + x[2]*y[2]
}
func drawSphere(r int, k, amb float64, dir *vector) *image.Gray {
w, h := r*4, r*3
img := image.NewGray(image.Rect(-w/2, -h/2, w/2, h/2))
vec := new(vector)
for x := -r; x < r; x++ {
for y := -r; y < r; y++ {
if z := r*r - x*x - y*y; z >= 0 {
vec[0] = float64(x)
vec[1] = float64(y)
vec[2] = math.Sqrt(float64(z))
normalize(vec)
s := dot(dir, vec)
if s < 0 {
s = 0
}
lum := 255 * (math.Pow(s, k) + amb) / (1 + amb)
if lum < 0 {
lum = 0
} else if lum > 255 {
lum = 255
}
img.SetGray(x, y, color.Gray{uint8(lum)})
}
}
}
return img
}
func main() {
dir := &vector{-30, -30, 50}
normalize(dir)
img := drawSphere(200, 1.5, .2, dir)
f, err := os.Create("sphere.png")
if err != nil {
fmt.Println(err)
return
}
if err = png.Encode(f, img); err != nil {
fmt.Println(err)
}
if err = f.Close(); err != nil {
fmt.Println(err)
}
} |
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
| #Go | Go | package main
import (
"golang.org/x/net/websocket"
"flag"
"fmt"
"html/template"
"io"
"math"
"net/http"
"time"
)
var (
Portnum string
Hostsite string
)
type PageSettings struct {
Host string
Port string
}
const (
Canvaswidth = 512
Canvasheight = 512
//color constants
HourColor = "#ff7373" // pinkish
MinuteColor = "#00b7e4" //light blue
SecondColor = "#b58900" //gold
)
func main() {
flag.StringVar(&Portnum, "Port", "1234", "Port to host server.")
flag.StringVar(&Hostsite, "Site", "localhost", "Site hosting server")
flag.Parse()
http.HandleFunc("/", webhandler)
http.Handle("/ws", websocket.Handler(wshandle))
err := http.ListenAndServe(Hostsite+":"+Portnum, nil)
if err != nil {
fmt.Println(err)
}
fmt.Println("server running")
}
func webhandler(w http.ResponseWriter, r *http.Request) {
wsurl := PageSettings{Host: Hostsite, Port: Portnum}
template, _ := template.ParseFiles("clock.html")
template.Execute(w, wsurl)
}
//Given a websocket connection,
//serves updating time function
func wshandle(ws *websocket.Conn) {
for {
hour, min, sec := time.Now().Clock()
hourx, houry := HourCords(hour, Canvasheight/2)
minx, miny := MinSecCords(min, Canvasheight/2)
secx, secy := MinSecCords(sec, Canvasheight/2)
msg := "CLEAR\n"
msg += fmt.Sprintf("HOUR %d %d %s\n", hourx, houry, HourColor)
msg += fmt.Sprintf("MIN %d %d %s\n", minx, miny, MinuteColor)
msg += fmt.Sprintf("SEC %d %d %s", secx, secy, SecondColor)
io.WriteString(ws, msg)
time.Sleep(time.Second / 60.0)
}
}
//Given current minute or second time(i.e 30 min, 60 minutes)
//and the radius, returns pair of cords to draw line to
func MinSecCords(ctime int, radius int) (int, int) {
//converts min/sec to angle and then to radians
theta := ((float64(ctime)*6 - 90) * (math.Pi / 180))
x := float64(radius) * math.Cos(theta)
y := float64(radius) * math.Sin(theta)
return int(x) + 256, int(y) + 256
}
//Given current hour time(i.e. 12, 8) and the radius,
//returns pair of cords to draw line to
func HourCords(ctime int, radius int) (int, int) {
//converts hours to angle and then to radians
theta := ((float64(ctime)*30 - 90) * (math.Pi / 180))
x := float64(radius) * math.Cos(theta)
y := float64(radius) * math.Sin(theta)
return int(x) + 256, int(y) + 256
} |
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
| #zkl | zkl | class Node{
fcn init(_value,_prev=Void,_next=Void)
{ var value=_value, prev=_prev, next=_next; }
fcn toString{ value.toString() }
fcn append(value){
b,c := Node(value,self,next),next;
next=b;
if(c) c.prev=b;
b
}
} |
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)
| #Picat | Picat | go =>
_ = random2(), % random seed
N = 21,
Map = new_map([1=red,2=white,3=blue]),
[Rand,Sorted] = dutch_random_sort(N,Map,Map.inverse()),
println('rand '=Rand),
println(sorted=Sorted),
nl.
% generate a random order and ensure it's not already dutch sorted
dutch_random_sort(N,Map,MapInv) = [Rand,Sorted] =>
Rand = dutch_random1(N,Map),
Sorted = dutch_sort(Rand,MapInv),
while (Rand == Sorted)
println("Randomize again"),
Rand := dutch_random1(N,Map),
Sorted := dutch_sort(Rand,MapInv)
end.
dutch_random1(N,Map) = [Map.get(1+(random() mod Map.map_to_list().length)) : _I in 1..N].
dutch_sort(L,MapInv) = [R : _=R in [MapInv.get(R)=R : R in L].sort()].
inverse(Map) = new_map([V=K : K=V in Map]). |
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)
| #PicoLisp | PicoLisp | (def 'Colors
(list
(def 'RED 1)
(def 'WHITE 2)
(def 'BLUE 3) ) )
(let (L (make (do 9 (link (get Colors (rand 1 3))))) S (by val sort L))
(prin "Original balls ")
(print L)
(prinl (unless (= L S) " not sorted"))
(prin "Sorted balls ")
(print S)
(prinl " are 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
| #LSL | LSL | vector vSCALE = <2.0, 3.0, 4.0>;
default {
state_entry() {
llSetScale(vSCALE);
}
} |
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
| #Lua | Lua | -- needed for actual task
cube.scale = function(self, sx, sy, sz)
for i,v in ipairs(self.verts) do
v[1], v[2], v[3] = v[1]*sx, v[2]*sy, v[3]*sz
end
end
-- only needed for output
-- (to size it for screen, given a limited camera)
cube.translate = function(self, tx, ty, tz)
for i,v in ipairs(self.verts) do
v[1], v[2], v[3] = v[1]+tx, v[2]+ty, v[3]+tz
end
end |
http://rosettacode.org/wiki/Dragon_curve | Dragon curve |
Create and display a dragon curve fractal.
(You may either display the curve directly or write it to an image file.)
Algorithms
Here are some brief notes the algorithms used and how they might suit various languages.
Recursively a right curling dragon is a right dragon followed by a left dragon, at 90-degree angle. And a left dragon is a left followed by a right.
*---R----* expands to * *
\ /
R L
\ /
*
*
/ \
L R
/ \
*---L---* expands to * *
The co-routines dcl and dcr in various examples do this recursively to a desired expansion level.
The curl direction right or left can be a parameter instead of two separate routines.
Recursively, a curl direction can be eliminated by noting the dragon consists of two copies of itself drawn towards a central point at 45-degrees.
*------->* becomes * * Recursive copies drawn
\ / from the ends towards
\ / the centre.
v v
*
This can be seen in the SVG example. This is best suited to off-line drawing since the reversal in the second half means the drawing jumps backward and forward (in binary reflected Gray code order) which is not very good for a plotter or for drawing progressively on screen.
Successive approximation repeatedly re-writes each straight line as two new segments at a right angle,
*
*-----* becomes / \ bend to left
/ \ if N odd
* *
* *
*-----* becomes \ / bend to right
\ / if N even
*
Numbering from the start of the curve built so far, if the segment is at an odd position then the bend introduced is on the right side. If the segment is an even position then on the left. The process is then repeated on the new doubled list of segments. This constructs a full set of line segments before any drawing.
The effect of the splitting is a kind of bottom-up version of the recursions. See the Asymptote example for code doing this.
Iteratively the curve always turns 90-degrees left or right at each point. The direction of the turn is given by the bit above the lowest 1-bit of n. Some bit-twiddling can extract that efficiently.
n = 1010110000
^
bit above lowest 1-bit, turn left or right as 0 or 1
LowMask = n BITXOR (n-1) # eg. giving 0000011111
AboveMask = LowMask + 1 # eg. giving 0000100000
BitAboveLowestOne = n BITAND AboveMask
The first turn is at n=1, so reckon the curve starting at the origin as n=0 then a straight line segment to position n=1 and turn there.
If you prefer to reckon the first turn as n=0 then take the bit above the lowest 0-bit instead. This works because "...10000" minus 1 is "...01111" so the lowest 0 in n-1 is where the lowest 1 in n is.
Going by turns suits turtle graphics such as Logo or a plotter drawing with a pen and current direction.
If a language doesn't maintain a "current direction" for drawing then you can always keep that separately and apply turns by bit-above-lowest-1.
Absolute direction to move at point n can be calculated by the number of bit-transitions in n.
n = 11 00 1111 0 1
^ ^ ^ ^ 4 places where change bit value
so direction=4*90degrees=East
This can be calculated by counting the number of 1 bits in "n XOR (n RIGHTSHIFT 1)" since such a shift and xor leaves a single 1 bit at each position where two adjacent bits differ.
Absolute X,Y coordinates of a point n can be calculated in complex numbers by some powers (i+1)^k and add/subtract/rotate. This is done in the gnuplot code. This might suit things similar to Gnuplot which want to calculate each point independently.
Predicate test for whether a given X,Y point or segment is on the curve can be done. This might suit line-by-line output rather than building an entire image before printing. See M4 for an example of this.
A predicate works by dividing out complex number i+1 until reaching the origin, so it takes roughly a bit at a time from X and Y is thus quite efficient. Why it works is slightly subtle but the calculation is not difficult. (Check segment by applying an offset to move X,Y to an "even" position before dividing i+1. Check vertex by whether the segment either East or West is on the curve.)
The number of steps in the predicate corresponds to doublings of the curve, so stopping the check at say 8 steps can limit the curve drawn to 2^8=256 points. The offsets arising in the predicate are bits of n the segment number, so can note those bits to calculate n and limit to an arbitrary desired length or sub-section.
As a Lindenmayer system of expansions. The simplest is two symbols F and S both straight lines, as used by the PGF code.
Axiom F, angle 90 degrees
F -> F+S
S -> F-S
This always has F at even positions and S at odd. Eg. after 3 levels F_S_F_S_F_S_F_S. The +/- turns in between bend to the left or right the same as the "successive approximation" method above. Read more at for instance Joel Castellanos' L-system page.
Variations are possible if you have only a single symbol for line draw, for example the Icon and Unicon and Xfractint code. The angles can also be broken into 45-degree parts to keep the expansion in a single direction rather than the endpoint rotating around.
The string rewrites can be done recursively without building the whole string, just follow its instructions at the target level. See for example C by IFS Drawing code. The effect is the same as "recursive with parameter" above but can draw other curves defined by L-systems.
| #Common_Lisp | Common Lisp | (defpackage #:dragon
(:use #:clim-lisp #:clim)
(:export #:dragon #:dragon-part))
(in-package #:dragon)
(defun dragon-part (depth bend-direction)
(if (zerop depth)
(draw-line* *standard-output* 0 0 1 0)
(with-scaling (t (/ (sqrt 2)))
(with-rotation (t (* pi -1/4 bend-direction))
(dragon-part (1- depth) 1)
(with-translation (t 1 0)
(with-rotation (t (* pi 1/2 bend-direction))
(dragon-part (1- depth) -1)))))))
(defun dragon (&optional (depth 7) (size 100))
(with-room-for-graphics ()
(with-scaling (t size)
(dragon-part depth 1)))) |
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
| #Haskell | Haskell | import Graphics.Rendering.OpenGL.GL
import Graphics.UI.GLUT.Objects
import Graphics.UI.GLUT
setProjection :: IO ()
setProjection = do
matrixMode $= Projection
ortho (-1) 1 (-1) 1 0 (-1)
grey1,grey9,red,white :: Color4 GLfloat
grey1 = Color4 0.1 0.1 0.1 1
grey9 = Color4 0.9 0.9 0.9 1
red = Color4 1 0 0 1
white = Color4 1 1 1 1
setLights :: IO ()
setLights = do
let l = Light 0
ambient l $= grey1
diffuse l $= white
specular l $= white
position l $= Vertex4 (-4) 4 3 (0 :: GLfloat)
light l $= Enabled
lighting $= Enabled
setMaterial :: IO ()
setMaterial = do
materialAmbient Front $= grey1
materialDiffuse Front $= red
materialSpecular Front $= grey9
materialShininess Front $= (32 :: GLfloat)
display :: IO()
display = do
clear [ColorBuffer]
renderObject Solid $ Sphere' 0.8 64 64
swapBuffers
main :: IO()
main = do
_ <- getArgsAndInitialize
_ <- createWindow "Sphere"
clearColor $= Color4 0.0 0.0 0.0 0.0
setProjection
setLights
setMaterial
displayCallback $= display
mainLoop |
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
| #GUISS | GUISS | Start,Programs,Accessories,Analogue Clock |
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)
| #PowerShell | PowerShell |
$Colors = 'red', 'white','blue'
# Select 10 random colors
$RandomBalls = 1..10 | ForEach { $Colors | Get-Random }
# Ensure we aren't finished before we start. For some reason. It's in the task requirements.
While ( $RandomBalls -eq $RandomBalls | Sort { $Colors.IndexOf( $_ ) } )
{ $RandomBalls = 1..10 | ForEach { $Colors | Get-Random } }
# Sort the colors
$SortedBalls = $RandomBalls | Sort { $Colors.IndexOf( $_ ) }
# Display the results
$RandomBalls
''
$SortedBalls
|
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