Spiral matrix: Difference between revisions

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{{task|Matrices}}
Produce a spiral array. A spiral array is a square arrangement of the first <tt>N<sup>2</sup></tt> natural numbers, where the numbers increase sequentially as you go around the edges of the array spiralling inwards.
 
;Task:
For example, given 5, produce this array:
Produce a spiral array.
 
 
A &nbsp; ''spiral array'' &nbsp; is a square arrangement of the first &nbsp; <big> N<sup>2</sup></big> &nbsp; natural numbers, &nbsp; where the
<br>numbers increase sequentially as you go around the edges of the array spiraling inwards.
 
 
For example, given &nbsp; '''5''', &nbsp; produce this array:
<pre>
0 1 2 3 4
Line 9 ⟶ 16:
13 22 21 20 7
12 11 10 9 8
</pre>
 
 
;Related tasks:
* &nbsp; [[Zig-zag matrix]]
* &nbsp; [[Identity_matrix]]
* &nbsp; [[Ulam_spiral_(for_primes)]]
<br><br>
 
=={{header|11l}}==
{{trans|Python}}
 
<syntaxhighlight lang="11l">F spiral_matrix(n)
V m = [[0] * n] *n
V d = [(0, 1), (1, 0), (0, -1), (-1, 0)]
V xy = (0, -1)
V c = 0
L(i) 0 .< n + n - 1
L 0 .< (n + n - i) I/ 2
xy += d[i % 4]
m[xy.x][xy.y] = c
c++
R m
 
F printspiral(myarray)
L(y) 0 .< myarray.len
L(x) 0 .< myarray.len
print(‘#2’.format(myarray[y][x]), end' ‘ ’)
print()
 
printspiral(spiral_matrix(5))</syntaxhighlight>
 
{{out}}
<pre>
0 1 2 3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8
</pre>
 
=={{header|360 Assembly}}==
For maximum compatibility, this program uses only the basic instruction set.
{{trans|BBC BASIC}}
<syntaxhighlight lang="360asm">SPIRALM CSECT
USING SPIRALM,R13
SAVEAREA B STM-SAVEAREA(R15)
DC 17F'0'
DC CL8'SPIRALM'
STM STM R14,R12,12(R13)
ST R13,4(R15)
ST R15,8(R13)
LR R13,R15
* ---- CODE
LA R0,0
LA R1,1
LH R12,N n
LR R4,R1 Row=1
LR R5,R1 Col=1
LR R6,R1 BotRow=1
LR R7,R1 BotCol=1
LR R8,R12 TopRow=n
LR R9,R12 TopCol=n
LR R10,R0 Dir=0
LR R15,R12 n
MR R14,R12 R15=n*n
LA R11,1 k=1
LOOP CR R11,R15
BH ENDLOOP
LR R1,R4
BCTR R1,0
MH R1,N
AR R1,R5
LR R2,R11 k
BCTR R2,0
BCTR R1,0
SLA R1,1
STH R2,MATRIX(R1) Matrix(Row,Col)=k-1
CH R10,=H'0'
BE DIR0
CH R10,=H'1'
BE DIR1
CH R10,=H'2'
BE DIR2
CH R10,=H'3'
BE DIR3
B DIRX
DIR0 CR R5,R9 if Col<TopCol
BNL DIR0S
LA R5,1(R5) Col=Col+1
B DIRX
DIR0S LA R10,1 Dir=1
LA R4,1(R4) Row=Row+1
LA R6,1(R6) BotRow=BotRow+1
B DIRX
DIR1 CR R4,R8 if Row<TopRow
BNL DIR1S
LA R4,1(R4) Row=Row+1
B DIRX
DIR1S LA R10,2 Dir=2
BCTR R5,0 Col=Col-1
BCTR R9,0 TopCol=TopCol-1
B DIRX
DIR2 CR R5,R7 if Col>BotCol
BNH DIR2S
BCTR R5,0 Col=Col-1
B DIRX
DIR2S LA R10,3 Dir=3
BCTR R4,0 Row=Row-1
BCTR R8,0 TopRow=TopRow-1
B DIRX
DIR3 CR R4,R6 if Row>BotRow
BNH DIR3S
BCTR R4,0 Row=Row-1
B DIRX
DIR3S LA R10,0 Dir=0
LA R5,1(R5) Col=Col+1
LA R7,1(R7) BotCol=BotCol+1
DIRX EQU *
LA R11,1(R11) k=k+1
B LOOP
ENDLOOP EQU *
LA R4,1 i
LOOPI CR R4,R12
BH ENDLOOPI
XR R10,R10
LA R5,1 j
LOOPJ CR R5,R12
BH ENDLOOPJ
LR R1,R4
BCTR R1,0
MH R1,N
AR R1,R5
BCTR R1,0
SLA R1,1
LH R2,MATRIX(R1) Matrix(i,j)
LA R3,BUF
AR R3,R10
CVD R2,P8
MVC 0(4,R3),=X'40202120'
ED 0(4,R3),P8+6
LA R10,4(R10)
LA R5,1(R5)
B LOOPJ
ENDLOOPJ EQU *
WTO MF=(E,WTOMSG)
LA R4,1(R4)
B LOOPI
ENDLOOPI EQU *
* ---- END CODE
L R13,4(0,R13)
LM R14,R12,12(R13)
XR R15,R15
BR R14
* ---- DATA
N DC H'5' max=20 (20*4=80)
LTORG
P8 DS PL8
WTOMSG DS 0F
DC H'80',XL2'0000'
BUF DC CL80' '
MATRIX DS H Matrix(n,n)
YREGS
END SPIRALM</syntaxhighlight>
{{out}}
<pre> 0 1 2 3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8</pre>
 
=={{header|ABAP}}==
 
<syntaxhighlight lang="abap">REPORT zspiral_matrix.
 
CLASS lcl_spiral_matrix DEFINITION FINAL.
PUBLIC SECTION.
 
TYPES:
BEGIN OF ty_coordinates,
dy TYPE i,
dx TYPE i,
value TYPE i,
END OF ty_coordinates,
ty_t_coordinates TYPE STANDARD TABLE OF ty_coordinates WITH EMPTY KEY.
 
DATA mv_dimention TYPE i.
DATA mv_initial_value TYPE i.
 
METHODS:
constructor IMPORTING iv_dimention TYPE i
iv_initial_value TYPE i,
 
get_coordinates RETURNING VALUE(rv_result) TYPE ty_t_coordinates,
 
print.
 
PRIVATE SECTION.
DATA lt_coordinates TYPE ty_t_coordinates.
 
METHODS create RETURNING VALUE(ro_result) TYPE REF TO lcl_spiral_matrix.
 
ENDCLASS.
 
CLASS lcl_spiral_matrix IMPLEMENTATION.
METHOD constructor.
 
mv_dimention = iv_dimention.
mv_initial_value = iv_initial_value.
 
create( ).
 
ENDMETHOD.
 
METHOD create.
 
DATA dy TYPE i.
DATA dx TYPE i.
DATA value TYPE i.
DATA seq_number TYPE i.
DATA seq_dimention TYPE i.
DATA sign_coef TYPE i VALUE -1.
 
value = mv_initial_value.
 
" Fill in the first row (index 0)
DO mv_dimention TIMES.
APPEND VALUE #( dy = dy dx = dx value = value ) TO lt_coordinates.
value = value + 1.
dx = dx + 1.
ENDDO.
 
seq_dimention = mv_dimention.
 
" Find the row and column numbers and set the values.
DO ( 2 * mv_dimention - 2 ) / 2 TIMES.
sign_coef = - sign_coef.
seq_dimention = seq_dimention - 1.
 
DO 2 TIMES.
seq_number = seq_number + 1.
 
DO seq_dimention TIMES.
 
IF seq_number MOD 2 <> 0.
dy = dy + 1 * sign_coef.
ELSE.
dx = dx - 1 * sign_coef.
ENDIF.
 
APPEND VALUE #( dy = dy dx = dx value = value ) TO lt_coordinates.
value = value + 1.
ENDDO.
 
ENDDO.
 
ENDDO.
 
ro_result = me.
 
ENDMETHOD.
 
METHOD get_coordinates.
rv_result = lt_coordinates.
ENDMETHOD.
 
METHOD print.
 
DATA cnt TYPE i.
DATA line TYPE string.
 
SORT lt_coordinates BY dy dx ASCENDING.
 
LOOP AT lt_coordinates ASSIGNING FIELD-SYMBOL(<ls_coordinates>).
 
cnt = cnt + 1.
line = |{ line } { <ls_coordinates>-value }|.
 
IF cnt MOD mv_dimention = 0.
WRITE / line.
CLEAR line.
ENDIF.
 
ENDLOOP.
 
ENDMETHOD.
 
ENDCLASS.
 
START-OF-SELECTION.
 
DATA(go_spiral_matrix) = NEW lcl_spiral_matrix( iv_dimention = 5
iv_initial_value = 0 ).
go_spiral_matrix->print( ).</syntaxhighlight>
 
{{out}}
<pre>
0 1 2 3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8
</pre>
 
[https://github.com/victorizbitskiy/abapSpiralMatrix/blob/main/docs/img/Result.png Screenshot from ABAP system]
 
=={{header|Action!}}==
<syntaxhighlight lang="action!">DEFINE MAX_SIZE="10"
DEFINE MAX_MATRIX_SIZE="100"
 
INT FUNC Index(BYTE size,x,y)
RETURN (x+y*size)
 
PROC PrintMatrix(BYTE ARRAY a BYTE size)
BYTE i,j,v
FOR j=0 TO size-1
DO
FOR i=0 TO size-1
DO
v=a(Index(size,i,j))
IF v<10 THEN
Print(" ")
ELSE
Print(" ")
FI
PrintB(v)
OD
PutE()
OD
RETURN
 
PROC FillMatrix(BYTE ARRAY a BYTE size)
INT lev,maxLev,dist,maxDist,v
 
maxLev=size/2
IF (size&1)=0 THEN
maxLev==-1
FI
maxDist=size-1
v=1
FOR lev=0 TO maxLev
DO
FOR dist=0 TO maxDist
DO
a(Index(size,lev+dist,lev))=v v==+1
OD
FOR dist=0 TO maxDist-1
DO
a(Index(size,size-1-lev,lev+dist+1))=v v==+1
OD
FOR dist=0 TO maxDist-1
DO
a(Index(size,size-2-lev-dist,size-1-lev))=v v==+1
OD
FOR dist=0 TO maxDist-2
DO
a(Index(size,lev,size-2-lev-dist))=v v==+1
OD
maxDist==-2
OD
RETURN
 
PROC Test(BYTE size)
BYTE ARRAY mat(MAX_MATRIX_SIZE)
PrintF("Matrix size: %B%E",size)
FillMatrix(mat,size)
PrintMatrix(mat,size)
PutE()
RETURN
 
PROC Main()
Test(5)
Test(6)
RETURN</syntaxhighlight>
{{out}}
[https://gitlab.com/amarok8bit/action-rosetta-code/-/raw/master/images/Spiral_matrix.png Screenshot from Atari 8-bit computer]
<pre>
Matrix size: 5
1 2 3 4 5
16 17 18 19 6
15 24 25 20 7
14 23 22 21 8
13 12 11 10 9
 
Matrix size: 6
1 2 3 4 5 6
20 21 22 23 24 7
19 32 33 34 25 8
18 31 36 35 26 9
17 30 29 28 27 10
16 15 14 13 12 11
</pre>
 
=={{header|Ada}}==
<langsyntaxhighlight lang="ada">-- Spiral Square
with Ada.Text_Io; use Ada.Text_Io;
with Ada.Integer_Text_Io; use Ada.Integer_Text_Io;
Line 78 ⟶ 478:
begin
Print(Spiral(5));
end Spiral_Square;</langsyntaxhighlight>
The following is a variant using a different algorithm (which can also be used recursively):
<langsyntaxhighlight lang="ada">function Spiral (N : Positive) return Array_Type is
Result : Array_Type (1..N, 1..N);
Left : Positive := 1;
Line 112 ⟶ 512:
Result (Top, Left) := Index;
return Result;
end Spiral;</langsyntaxhighlight>
 
=={{header|ALGOL 68}}==
Line 119 ⟶ 519:
{{works with|ALGOL 68G|Any - tested with release [http://sourceforge.net/projects/algol68/files/algol68g/algol68g-1.18.0/algol68g-1.18.0-9h.tiny.el5.centos.fc11.i386.rpm/download 1.18.0-9h.tiny]}}
{{works with|ELLA ALGOL 68|Any (with appropriate job cards) - tested with release [http://sourceforge.net/projects/algol68/files/algol68toc/algol68toc-1.8.8d/algol68toc-1.8-8d.fc9.i386.rpm/download 1.8-8d]}}
<langsyntaxhighlight lang="algol68">INT empty=0;
 
PROC spiral = (INT n)[,]INT: (
Line 152 ⟶ 552:
);
print spiral(spiral(5))</langsyntaxhighlight>
{{out}}
Output:
<pre>
1 2 3 4 5
Line 161 ⟶ 561:
13 12 11 10 9
</pre>
 
=={{header|AppleScript}}==
 
{{Trans|JavaScript}} (ES6)
<syntaxhighlight lang="applescript">---------------------- SPIRAL MATRIX ---------------------
 
-- spiral :: Int -> [[Int]]
on spiral(n)
script go
on |λ|(rows, cols, start)
if 0 < rows then
{enumFromTo(start, start + pred(cols))} & ¬
map(my |reverse|, ¬
transpose(|λ|(cols, pred(rows), start + cols)))
else
{{}}
end if
end |λ|
end script
go's |λ|(n, n, 0)
end spiral
 
 
--------------------------- TEST -------------------------
on run
wikiTable(spiral(5), ¬
false, ¬
"text-align:center;width:12em;height:12em;table-layout:fixed;")
end run
 
 
-------------------- WIKI TABLE FORMAT -------------------
 
-- wikiTable :: [Text] -> Bool -> Text -> Text
on wikiTable(lstRows, blnHdr, strStyle)
script fWikiRows
on |λ|(lstRow, iRow)
set strDelim to if_(blnHdr and (iRow = 0), "!", "|")
set strDbl to strDelim & strDelim
linefeed & "|-" & linefeed & strDelim & space & ¬
intercalateS(space & strDbl & space, lstRow)
end |λ|
end script
linefeed & "{| class=\"wikitable\" " & ¬
if_(strStyle ≠ "", "style=\"" & strStyle & "\"", "") & ¬
intercalateS("", ¬
map(fWikiRows, lstRows)) & linefeed & "|}" & linefeed
end wikiTable
 
 
------------------------- GENERIC ------------------------
 
-- comparing :: (a -> b) -> (a -> a -> Ordering)
on comparing(f)
script
on |λ|(a, b)
tell mReturn(f)
set fa to |λ|(a)
set fb to |λ|(b)
if fa < fb then
-1
else if fa > fb then
1
else
0
end if
end tell
end |λ|
end script
end comparing
 
 
-- concatMap :: (a -> [b]) -> [a] -> [b]
on concatMap(f, xs)
set lng to length of xs
set acc to {}
tell mReturn(f)
repeat with i from 1 to lng
set acc to acc & (|λ|(item i of xs, i, xs))
end repeat
end tell
if {text, string} contains class of xs then
acc as text
else
acc
end if
end concatMap
 
 
-- enumFromTo :: Int -> Int -> [Int]
on enumFromTo(m, n)
if m ≤ n then
set lst to {}
repeat with i from m to n
set end of lst to i
end repeat
return lst
else
return {}
end if
end enumFromTo
 
 
-- foldl :: (a -> b -> a) -> a -> [b] -> a
on foldl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldl
 
 
-- if_ :: Bool -> a -> a -> a
on if_(bool, x, y)
if bool then
x
else
y
end if
end if_
 
 
-- intercalateS :: String -> [String] -> String
on intercalateS(sep, xs)
set {dlm, my text item delimiters} to {my text item delimiters, sep}
set s to xs as text
set my text item delimiters to dlm
return s
end intercalateS
 
 
-- length :: [a] -> Int
on |length|(xs)
length of xs
end |length|
 
 
-- max :: Ord a => a -> a -> a
on max(x, y)
if x > y then
x
else
y
end if
end max
 
 
-- maximumBy :: (a -> a -> Ordering) -> [a] -> a
on maximumBy(f, xs)
set cmp to mReturn(f)
script max
on |λ|(a, b)
if a is missing value or cmp's |λ|(a, b) < 0 then
b
else
a
end if
end |λ|
end script
foldl(max, missing value, xs)
end maximumBy
 
 
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
 
 
-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map
 
 
-- pred :: Enum a => a -> a
on pred(x)
x - 1
end pred
 
 
-- Egyptian multiplication - progressively doubling a list, appending
-- stages of doubling to an accumulator where needed for binary
-- assembly of a target length
-- replicate :: Int -> a -> [a]
on replicate(n, a)
set out to {}
if n < 1 then return out
set dbl to {a}
repeat while (n > 1)
if (n mod 2) > 0 then set out to out & dbl
set n to (n div 2)
set dbl to (dbl & dbl)
end repeat
return out & dbl
end replicate
 
 
-- reverse :: [a] -> [a]
on |reverse|(xs)
if class of xs is text then
(reverse of characters of xs) as text
else
reverse of xs
end if
end |reverse|
 
 
-- Simplified version - assuming rows of unvarying length.
-- transpose :: [[a]] -> [[a]]
on transpose(rows)
script cols
on |λ|(_, iCol)
script cell
on |λ|(row)
item iCol of row
end |λ|
end script
concatMap(cell, rows)
end |λ|
end script
map(cols, item 1 of rows)
end transpose
 
 
-- unlines :: [String] -> String
on unlines(xs)
set {dlm, my text item delimiters} to ¬
{my text item delimiters, linefeed}
set str to xs as text
set my text item delimiters to dlm
str
end unlines
 
 
-- unwords :: [String] -> String
on unwords(xs)
intercalateS(space, xs)
end unwords</syntaxhighlight>
{{Out}}
{| class="wikitable" style="text-align:center;width:12em;height:12em;table-layout:fixed;"
|-
| 0 || 1 || 2 || 3 || 4
|-
| 15 || 16 || 17 || 18 || 5
|-
| 14 || 23 || 24 || 19 || 6
|-
| 13 || 22 || 21 || 20 || 7
|-
| 12 || 11 || 10 || 9 || 8
|}
 
=={{header|Arturo}}==
{{trans|Python}}
<syntaxhighlight lang="rebol">spiralMatrix: function [n][
m: new array.of: @[n,n] null
 
[dx, dy, x, y]: [1, 0, 0, 0]
 
loop 0..dec n^2 'i [
m\[y]\[x]: i
 
[nx,ny]: @[x+dx, y+dy]
 
if? and? [and? [in? nx 0..n-1][in? ny 0..n-1]][
null? m\[ny]\[nx]
][
[x,y]: @[nx, ny]
]
else [
bdx: dx
[dx, dy]: @[neg dy, bdx]
[x, y]: @[x+dx, y+dy]
]
]
 
return m
]
 
loop spiralMatrix 5 'row [
print map row 'x -> pad to :string x 4
]</syntaxhighlight>
 
{{out}}
 
<pre> 0 1 2 3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8</pre>
 
=={{header|AutoHotkey}}==
{{trans|Python|}}
ahk forum: [http://www.autohotkey.com/forum/post-276718.html#276718 discussion]
<langsyntaxhighlight AutoHotkeylang="autohotkey">n := 5, dx := x := y := v := 1, dy := 0
 
Loop % n*n {
Line 190 ⟶ 903:
13 12 11 10 9
---------------------------
*/</langsyntaxhighlight>
 
=={{header|AWK}}==
<syntaxhighlight lang="awk">
# syntax: GAWK -f SPIRAL_MATRIX.AWK [-v offset={0|1}] [size]
# converted from BBC BASIC
BEGIN {
# offset: "0" prints 0 to size^2-1 while "1" prints 1 to size^2
offset = (offset == "") ? 0 : offset
size = (ARGV[1] == "") ? 5 : ARGV[1]
if (offset !~ /^[01]$/) { exit(1) }
if (size !~ /^[0-9]+$/) { exit(1) }
bot_col = bot_row = 0
top_col = top_row = size - 1
direction = col = row = 0
for (i=0; i<=size*size-1; i++) { # build
arr[col,row] = i + offset
if (direction == 0) {
if (col < top_col) { col++ }
else { direction = 1 ; row++ ; bot_row++ }
}
else if (direction == 1) {
if (row < top_row) { row++ }
else { direction = 2 ; col-- ; top_col-- }
}
else if (direction == 2) {
if (col > bot_col) { col-- }
else { direction = 3 ; row-- ; top_row-- }
}
else if (direction == 3) {
if (row > bot_row) { row-- }
else { direction = 0 ; col++ ; bot_col++ }
}
}
width = length(size ^ 2 - 1 + offset) + 1 # column width
for (i=0; i<size; i++) { # print
for (j=0; j<size; j++) {
printf("%*d",width,arr[j,i])
}
printf("\n")
}
exit(0)
}
</syntaxhighlight>
{{out}}
<pre>
0 1 2 3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8
</pre>
 
=={{header|BBC BASIC}}==
<langsyntaxhighlight lang="bbcbasic"> N%=5
@%=LENSTR$(N%*N%-1)+1
BotCol%=0 : TopCol%=N%-1
Line 210 ⟶ 974:
ENDCASE
NEXT
END</langsyntaxhighlight>
 
=={{header|C}}==
Note: program produces a matrix starting from 1 instead of 0, because task says "natural numbers".
<langsyntaxhighlight lang="c">#include <stdio.h>
#include <stdlib.h>
Line 248 ⟶ 1,012:
 
return 0;
}</langsyntaxhighlight>
 
Recursive method, width and height given on command line:
<syntaxhighlight lang="c">#include <stdio.h>
#include <stdlib.h>
 
int spiral(int w, int h, int x, int y)
{
return y ? w + spiral(h - 1, w, y - 1, w - x - 1) : x;
}
 
int main(int argc, char **argv)
{
int w = atoi(argv[1]), h = atoi(argv[2]), i, j;
for (i = 0; i < h; i++) {
for (j = 0; j < w; j++)
printf("%4d", spiral(w, h, j, i));
putchar('\n');
}
return 0;
}</syntaxhighlight>
 
=={{header|C sharp|C#}}==
Solution based on the [[#J|J]] hints:
<syntaxhighlight lang="csharp">public int[,] Spiral(int n) {
int[,] result = new int[n, n];
 
int pos = 0;
int count = n;
int value = -n;
int sum = -1;
 
do {
value = -1 * value / n;
for (int i = 0; i < count; i++) {
sum += value;
result[sum / n, sum % n] = pos++;
}
value *= n;
count--;
for (int i = 0; i < count; i++) {
sum += value;
result[sum / n, sum % n] = pos++;
}
} while (count > 0);
 
return result;
}
 
 
// Method to print arrays, pads numbers so they line up in columns
public void PrintArray(int[,] array) {
int n = (array.GetLength(0) * array.GetLength(1) - 1).ToString().Length + 1;
 
for (int i = 0; i < array.GetLength(0); i++) {
for (int j = 0; j < array.GetLength(1); j++) {
Console.Write(array[i, j].ToString().PadLeft(n, ' '));
}
Console.WriteLine();
}
}</syntaxhighlight>
 
Translated proper C++ solution:
<syntaxhighlight lang="csharp">
 
//generate spiral matrix for given N
int[,] CreateMatrix(int n){
int[] dx = {0, 1, 0, -1}, dy = {1, 0, -1, 0};
int x = 0, y = -1, c = 0;
int[,] m = new int[n,n];
for (int i = 0, im = 0; i < n + n - 1; ++i, im = i % 4)
for (int j = 0, jlen = (n + n - i) / 2; j < jlen; ++j)
m[x += dx[im],y += dy[im]] = ++c;
return n;
}
 
//print aligned matrix
void Print(int[,] matrix) {
var len = (int)Math.Ceiling(Math.Log10(m.GetLength(0) * m.GetLength(1)))+1;
for(var y = 0; y<m.GetLength(1); y++){
for(var x = 0; x<m.GetLength(0); x++){
Console.Write(m[y, x].ToString().PadRight(len, ' '));
}
Console.WriteLine();
}
}
</syntaxhighlight>
 
====Spiral Matrix without using an Array====
 
<syntaxhighlight lang="csharp">
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
 
namespace spiralmat
{
class spiral
{
public static int lev;
int lev_lim, count, bk, cd, low, l, m;
spiral()
{
lev = lev_lim = count = bk = cd = low = l = m = 0;
}
void level(int n1, int r, int c)
{
lev_lim = n1 % 2 == 0 ? n1 / 2 : (n1 + 1) / 2;
if ((r <= lev_lim) && (c <= lev_lim))
lev = Math.Min(r, c);
else
{
bk = r > c ? (n1 + 1) - r : (n1 + 1) - c;
low = Math.Min(r, c);
if (low <= lev_lim)
cd = low;
lev = cd < bk ? cd : bk;
}
}
 
int func(int n2, int xo, int lo)
{
l = xo;
m = lo;
count = 0;
level(n2, l, m);
 
for (int ak = 1; ak < lev; ak++)
count += 4 * (n2 - 1 - 2 * (ak - 1));
return count;
}
 
public static void Main(string[] args)
{
spiral ob = new spiral();
Console.WriteLine("Enter Order..");
int n = int.Parse(Console.ReadLine());
Console.WriteLine("The Matrix of {0} x {1} Order is=>\n", n, n);
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= n; j++)
Console.Write("{0,3:D} ",
ob.func(n, i, j)
+ Convert.ToInt32(
((j >= i) && (i == lev))
? ((j - i) + 1)
: ((j == ((n + 1) - lev) && (i > lev) && (i <= j)))
? (n - 2 * (lev - 1) + (i - 1) - (n - j))
: ((i == ((n + 1) - lev) && (j < i)))
? ((n - 2 * (lev - 1)) + ((n - 2 * (lev - 1)) - 1) + (i - j))
: ((j == lev) && (i > lev) && (i < ((n + 1) - lev)))
? ((n - 2 * (lev - 1)) + ((n - 2 * (lev - 1)) - 1) + ((n - 2 * (lev - 1)) - 1) + (((n + 1) - lev) - i))
: 0));
Console.WriteLine();
}
Console.ReadKey();
}
}
}
</syntaxhighlight>
{{Out}}
<syntaxhighlight lang="sh">INPUT:-
 
Enter order..
5
 
OUTPUT:-
 
The Matrix of 5 x 5 Order is=>
 
1 2 3 4 5
16 17 18 19 6
15 24 25 20 7
14 23 22 21 8
13 12 11 10 9
 
INPUT:-
 
Enter order..
6
 
OUTPUT:-
 
The Matrix of 6 x 6 Order is=>
 
1 2 3 4 5 6
20 21 22 23 24 7
19 32 33 34 25 8
18 31 36 35 26 9
17 30 29 28 27 10
16 15 14 13 12 11
</syntaxhighlight>
 
=={{header|C++}}==
<langsyntaxhighlight lang="cpp">#include <vector>
#include <memory> // for auto_ptr
#include <cmath> // for the ceil and log10 and floor functions
Line 317 ⟶ 1,275:
{
printSpiralArray( getSpiralArray( 5 ) );
}</langsyntaxhighlight>
 
C++ solution done properly:
=={{header|C sharp|C#}}==
<syntaxhighlight lang="cpp">#include <vector>
Solution based on the [[#J|J]] hints:
#include <iostream>
<lang csharp>public int[,] Spiral(int n) {
using namespace std;
int[,] result = new int[n, n];
int main() {
 
const int posn = 05;
const int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
int count = n;
int x = 0, int valuey = -n1, c = 0;
vector<vector<int>> m(n, vector<int>(n));
int sum = -1;
for (int i = 0, im = 0; i < n + n - 1; ++i, im = i % 4)
 
for (int j = 0, jlen = (n + n - i) / 2; j < jlen; ++j)
do {
m[x += dx[im]][y += dy[im]] = ++c;
value = -1 * value / n;
for (int i = 0;auto i& <r count;: i++m) {
for (auto & v : r)
sum += value;
cout << v << ' ';
result[sum / n, sum % n] = pos++;
cout << endl;
}
}
value *= n;
}</syntaxhighlight>
count--;
for (int i = 0; i < count; i++) {
sum += value;
result[sum / n, sum % n] = pos++;
}
} while (count > 0);
 
return result;
}
 
 
// Method to print arrays, pads numbers so they line up in columns
public void PrintArray(int[,] array) {
int n = (array.GetLength(0) * array.GetLength(1) - 1).ToString().Length + 1;
 
for (int i = 0; i < array.GetLength(0); i++) {
for (int j = 0; j < array.GetLength(1); j++) {
Console.Write(array[i, j].ToString().PadLeft(n, ' '));
}
Console.WriteLine();
}
}</lang>
 
=={{header|Clojure}}==
Based on the [[#J|J]] hints (almost as incomprehensible, maybe)
<langsyntaxhighlight lang="clojure">(defn spiral [n]
(let [fmtcyc (strcycle "[1 ~{~<~%~,"n -1 (*- n 3) ":;~2d ~>~}~%"])]
counts (cons n (mapcat #(repeat 2 %)->> (range (dec n) 0 -1)))
ones-and-ns (mapcat #(repeat %12 %2) counts (cycle [1 n -1 (- n)]))]
(cons n)
(->> (map vector (range 0 (* n n)) (reductions + ones-and-ns))
(mapcat #(repeat %2 %) cyc)
(reductions +)
(map vector (range 0 (* n n)))
(sort-by second)
(map first)))
 
(clojure.pprint/cl-format true fmt))))</lang>
(let [n 5]
(clojure.pprint/cl-format
true
(str " ~{~<~%~," (* n 3) ":;~2d ~>~}~%")
(spiral n)))</syntaxhighlight>
Recursive generation:
{{trans|Common Lisp}}
<syntaxhighlight lang="clojure">
(defn spiral-matrix [m n & [start]]
(let [row (list (map #(+ start %) (range m)))]
(if (= 1 n) row
(concat row (map reverse
(apply map list
(spiral-matrix (dec n) m (+ start m))))))))
 
(defn spiral [n m] (spiral-matrix n m 1))
</syntaxhighlight>
 
=={{header|CoffeeScript}}==
<langsyntaxhighlight lang="coffeescript">
# Let's say you want to arrange the first N-squared natural numbers
# in a spiral, where you fill in the numbers clockwise, starting from
Line 418 ⟶ 1,375:
console.log "\n----Spiral n=#{n}"
console.log spiral_matrix n
</syntaxhighlight>
</lang>
{{out}}
output
<syntaxhighlight lang="text">
> coffee spiral.coffee
 
Line 439 ⟶ 1,396:
[ 19, 36, 35, 34, 33, 32, 11 ],
[ 18, 17, 16, 15, 14, 13, 12 ] ]
</syntaxhighlight>
</lang>
 
 
=={{header|Common Lisp}}==
{{trans|Python}}
<langsyntaxhighlight lang="lisp">(defun spiral (rows columns)
(do ((N (* rows columns))
(spiral (make-array (list rows columns) :initial-element nil))
Line 461 ⟶ 1,417:
dy dx)
(setf x (+ x dx)
y (+ y dy)))))))</langsyntaxhighlight>
<pre>> (pprint (spiral 6 6))
 
Line 478 ⟶ 1,434:
(9 14 5)
(8 7 6))</pre>
RecuriveRecursive generation:
<langsyntaxhighlight lang="lisp">(defun spiral (m n &optional (start 1))
(let ((row (list (loop for x from 0 to (1- m) collect (+ x start)))))
(if (= 1 n) row
Line 489 ⟶ 1,445:
;; test
(loop for row in (spiral 4 3) do
(format t "~{~4d~^~}~%" row))</langsyntaxhighlight>
 
=={{header|D}}==
<langsyntaxhighlight lang="d">void main() {
import std.stdio;
enum n = 5;
Line 498 ⟶ 1,454:
int pos, side = n;
 
foreach (immutable i; 0 .. (n / 2 + n % 2)) {
foreach (immutable j; 0 .. side)
M[i][i + j] = pos++;
foreach (immutable j; 1 .. side)
M[i + j][n - 1 - i] = pos++;
foreach_reverse (immutable j; 0 .. side - 1)
M[n - 1 - i][i + j] = pos++;
foreach_reverse (immutable j; 1 .. side - 1)
M[i + j][i] = pos++;
side -= 2;
Line 511 ⟶ 1,467:
 
writefln("%(%(%2d %)\n%)", M);
}</langsyntaxhighlight>
{{out}}
<pre> 0 1 2 3 4
Line 519 ⟶ 1,475:
12 11 10 9 8</pre>
Using a generator for any rectangular array:
<langsyntaxhighlight lang="d">import std.stdio;
 
/// 2D spiral generator
Line 570 ⟶ 1,526:
foreach (r; spiralMatrix(9, 4))
writefln("%(%2d %)", r);
}</langsyntaxhighlight>
{{out}}
Output:
<pre> 0 1 2 3 4 5 6 7 8
21 22 23 24 25 26 27 28 9
20 35 34 33 32 31 30 29 10
19 18 17 16 15 14 13 12 11</pre>
 
=={{header|DCL}}==
<syntaxhighlight lang="dcl">$ p1 = f$integer( p1 )
$ max = p1 * p1
$
$ i = 0
$ r = 1
$ rd = 0
$ c = 1
$ cd = 1
$ loop:
$ a'r'_'c' = i
$ nr = r + rd
$ nc = c + cd
$ if nr .eq. 0 .or. nc .eq. 0 .or. nr .gt. p1 .or. nc .gt. p1 .or. f$type( a'nr'_'nc' ) .nes. ""
$ then
$ gosub change_directions
$ endif
$ r = r + rd
$ c = c + cd
$ i = i + 1
$ if i .lt. max then $ goto loop
$ length = f$length( f$string( max - 1 ))
$ r = 1
$ loop2:
$ c = 1
$ output = ""
$ loop3:
$ output = output + f$fao( "!#UL ", length, a'r'_'c' )
$ c = c + 1
$ if c .le. p1 then $ goto loop3
$ write sys$output output
$ r = r + 1
$ if r .le. p1 then $ goto loop2
$ exit
$
$ change_directions:
$ if rd .eq. 0 .and cd .eq. 1
$ then
$ rd = 1
$ cd = 0
$ else
$ if rd .eq. 1 .and. cd .eq. 0
$ then
$ rd = 0
$ cd = -1
$ else
$ if rd .eq. 0 .and. cd .eq. -1
$ then
$ rd = -1
$ cd = 0
$ else
$ rd = 0
$ cd = 1
$ endif
$ endif
$ endif
$ return</syntaxhighlight>
{{out}}
<pre>$ @spiral_matrix 3
0 1 2
7 8 3
6 5 4
$ @spiral_matrix 5
0 1 2 3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8
 
...</pre>
 
=={{header|Delphi}}==
{{works with|Delphi|6.0}}
{{libheader|SysUtils,StdCtrls, Windows}}
This code actually creates a matrix in memory and stores values in the matrix, instead of just simulating one by drawing the pattern. It can also create matrices of any size and the matrices don't have to be square. It works by creating a rectangle of the same size as the matrix. It enters the values in the matrix along circumference of the matrix. It then uses the Windows library routine "InflateRect" to decrease the size of the rectangle until the whole matrix is filled with spiraling values. Since a rectangle can be any size and doesn't have to be square, it works with any size matrix, including non-square matrices.
 
<syntaxhighlight lang="Delphi">
 
 
type TMatrix = array of array of double;
 
 
procedure DisplayMatrix(Memo: TMemo; Mat: TMatrix);
{Display specified matrix}
var X,Y: integer;
var S: string;
begin
S:='';
for Y:=0 to High(Mat[0]) do
begin
S:=S+'[';
for X:=0 to High(Mat) do
S:=S+Format('%4.0f',[Mat[X,Y]]);
S:=S+']'+#$0D#$0A;
end;
Memo.Lines.Add(S);
end;
 
 
procedure MakeSpiralMatrix(var Mat: TMatrix; SizeX,SizeY: integer);
{Create a spiral matrix of specified size}
var Inx: integer;
var R: TRect;
 
procedure DoRect(R: TRect; var Inx: integer);
{Create on turn of the spiral base on the rectangle}
var X,Y: integer;
begin
{Do top part of rectangle}
for X:=R.Left to R.Right do
begin
Mat[X,R.Top]:=Inx;
Inc(Inx);
end;
{Do Right part of rectangle}
for Y:=R.Top+1 to R.Bottom do
begin
Mat[R.Right,Y]:=Inx;
Inc(Inx);
end;
{Do bottom part of rectangle}
for X:= R.Right-1 downto R.Left do
begin
Mat[X,R.Bottom]:=Inx;
Inc(Inx);
end;
{Do left part of rectangle}
for Y:=R.Bottom-1 downto R.Top+1 do
begin
Mat[R.Left,Y]:=Inx;
Inc(Inx);
end;
end;
 
begin
{Set matrix size}
SetLength(Mat,SizeX,SizeY);
{create matching rectangle}
R:=Rect(0,0,SizeX-1,SizeY-1);
Inx:=0;
{draw and deflate rectangle until spiral is done}
while (R.Left<=R.Right) and (R.Top<=R.Bottom) do
begin
DoRect(R,Inx);
InflateRect(R,-1,-1);
end;
end;
 
 
 
procedure SpiralMatrix(Memo: TMemo);
{Display spiral matrix}
var Mat: TMatrix;
begin
Memo.Lines.Add('5x5 Matrix');
MakeSpiralMatrix(Mat,5,5);
DisplayMatrix(Memo,Mat);
 
Memo.Lines.Add('8x8 Matrix');
MakeSpiralMatrix(Mat,8,8);
DisplayMatrix(Memo,Mat);
 
Memo.Lines.Add('14x8 Matrix');
MakeSpiralMatrix(Mat,14,8);
DisplayMatrix(Memo,Mat);
end;
</syntaxhighlight>
{{out}}
<pre>
5x5 Matrix
[ 0 1 2 3 4]
[ 15 16 17 18 5]
[ 14 23 24 19 6]
[ 13 22 21 20 7]
[ 12 11 10 9 8]
 
8x8 Matrix
[ 0 1 2 3 4 5 6 7]
[ 27 28 29 30 31 32 33 8]
[ 26 47 48 49 50 51 34 9]
[ 25 46 59 60 61 52 35 10]
[ 24 45 58 63 62 53 36 11]
[ 23 44 57 56 55 54 37 12]
[ 22 43 42 41 40 39 38 13]
[ 21 20 19 18 17 16 15 14]
 
14x8 Matrix
[ 0 1 2 3 4 5 6 7 8 9 10 11 12 13]
[ 39 40 41 42 43 44 45 46 47 48 49 50 51 14]
[ 38 71 72 73 74 75 76 77 78 79 80 81 52 15]
[ 37 70 95 96 97 98 99 100 101 102 103 82 53 16]
[ 36 69 94 111 110 109 108 107 106 105 104 83 54 17]
[ 35 68 93 92 91 90 89 88 87 86 85 84 55 18]
[ 34 67 66 65 64 63 62 61 60 59 58 57 56 19]
[ 33 32 31 30 29 28 27 26 25 24 23 22 21 20]
 
Elapsed Time: 11.242 ms.
 
</pre>
 
=={{header|E}}==
Line 581 ⟶ 1,737:
 
{{E 2D utilities}}
<langsyntaxhighlight lang="e">def spiral(size) {
def array := makeFlex2DArray(size, size)
var i := -1 # Counter of numbers to fill
Line 605 ⟶ 1,761:
return array
}</langsyntaxhighlight>
Example:
<langsyntaxhighlight lang="e">? print(spiral(5))
0 1 2 3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8</langsyntaxhighlight>
 
=={{header|EasyLang}}==
<syntaxhighlight>
proc mkspiral n . t[] .
subr side
for i to l
ind += d
t[ind] = cnt
cnt += 1
.
.
len t[] n * n
l = n
while cnt < len t[]
d = 1
side
l -= 1
d = n
side
d = -1
side
l -= 1
d = -n
side
.
.
n = 5
mkspiral n t[]
numfmt 0 3
for i to n * n
write t[i]
if i mod n = 0
print ""
.
.
</syntaxhighlight>
{{out}}
<pre>
0 1 2 3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8
</pre>
 
=={{header|Elixir}}==
{{trans|Ruby}}
<syntaxhighlight lang="elixir">defmodule RC do
def spiral_matrix(n) do
wide = length(to_char_list(n*n-1))
fmt = String.duplicate("~#{wide}w ", n) <> "~n"
runs = Enum.flat_map(n..1, &[&1,&1]) |> tl
delta = Stream.cycle([{0,1},{1,0},{0,-1},{-1,0}])
running(Enum.zip(runs,delta),0,-1,[])
|> Enum.with_index |> Enum.sort |> Enum.chunk(n)
|> Enum.each(fn row -> :io.format fmt, (for {_,i} <- row, do: i) end)
end
defp running([{run,{dx,dy}}|rest], x, y, track) do
new_track = Enum.reduce(1..run, track, fn i,acc -> [{x+i*dx, y+i*dy} | acc] end)
running(rest, x+run*dx, y+run*dy, new_track)
end
defp running([],_,_,track), do: track |> Enum.reverse
end
 
RC.spiral_matrix(5)</syntaxhighlight>
 
'''The other way'''
<syntaxhighlight lang="elixir">defmodule RC do
def spiral_matrix(n) do
wide = String.length(to_string(n*n-1))
fmt = String.duplicate("~#{wide}w ", n) <> "~n"
right(n,n-1,0,[]) |> Enum.reverse |> Enum.with_index |> Enum.sort |> Enum.chunk(n) |>
Enum.each(fn row ->
:io.format fmt, (for {_,i} <- row, do: i)
end)
end
def right(n, side, i, coordinates) do
down(n, side, i, Enum.reduce(0..side, coordinates, fn j,acc -> [{i, i+j} | acc] end))
end
def down(_, 0, _, coordinates), do: coordinates
def down(n, side, i, coordinates) do
left(n, side-1, i, Enum.reduce(1..side, coordinates, fn j,acc -> [{i+j, n-1-i} | acc] end))
end
def left(n, side, i, coordinates) do
up(n, side, i, Enum.reduce(side..0, coordinates, fn j,acc -> [{n-1-i, i+j} | acc] end))
end
def up(_, 0, _, coordinates), do: coordinates
def up(n, side, i, coordinates) do
right(n, side-1, i+1, Enum.reduce(side..1, coordinates, fn j,acc -> [{i+j, i} | acc] end))
end
end
 
RC.spiral_matrix(5)</syntaxhighlight>
 
'''Another way'''
<syntaxhighlight lang="elixir">defmodule RC do
def spiral_matrix(n) do
fmt = String.duplicate("~#{length(to_char_list(n*n-1))}w ", n) <> "~n"
Enum.flat_map(n..1, &[&1, &1])
|> tl
|> Enum.reduce({{0,-1},{0,1},[]}, fn run,{{x,y},{dx,dy},acc} ->
side = for i <- 1..run, do: {x+i*dx, y+i*dy}
{{x+run*dx, y+run*dy}, {dy, -dx}, acc++side}
end)
|> elem(2)
|> Enum.with_index
|> Enum.sort
|> Enum.map(fn {_,i} -> i end)
|> Enum.chunk(n)
|> Enum.each(fn row -> :io.format fmt, row end)
end
end
 
RC.spiral_matrix(5)</syntaxhighlight>
 
{{out}}
<pre>
0 1 2 3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8
</pre>
 
=={{header|Euphoria}}==
<langsyntaxhighlight Euphorialang="euphoria">function spiral(integer dimension)
integer side, curr, curr2
sequence s
Line 643 ⟶ 1,927:
end function
 
? spiral(5)</langsyntaxhighlight>
 
{{out}}
Output:
{
{0,1,2,3,4},
Line 653 ⟶ 1,937:
{12,11,10,9,8}
}
 
=={{header|F_Sharp|F#}}==
No fancy schmancy elegance here, just putting the numbers in the right place (though I commend the elegance)...
<syntaxhighlight lang="fsharp">let Spiral n =
let sq = Array2D.create n n 0 // Set up an output array
let nCur = ref -1 // Current value being inserted
let NextN() = nCur := (!nCur+1) ; !nCur // Inc current value and return new value
let Frame inset = // Create the "frame" at an offset from the outside
let rangeF = [inset..(n - inset - 2)] // Range we use going forward
let rangeR = [(n - inset - 1)..(-1)..(inset + 1)] // Range we use going backward
rangeF |> Seq.iter (fun i -> sq.[inset,i] <- NextN()) // Top of frame
rangeF |> Seq.iter (fun i -> sq.[i,n-inset-1] <- NextN()) // Right side of frame
rangeR |> Seq.iter (fun i -> sq.[n-inset-1,i] <- NextN()) // Bottom of frame
rangeR |> Seq.iter (fun i -> sq.[i,inset] <- NextN()) // Left side of frame
[0..(n/2 - 1)] |> Seq.iter (fun i -> Frame i) // Fill in all frames
if n &&& 1 = 1 then sq.[n/2,n/2] <- n*n - 1 // If n is odd, fill in the last single value
sq // Return our output array</syntaxhighlight>
 
=={{header|Factor}}==
This is an implementation of Joey Tuttle's method for computing a spiral directly as a list and then reshaping it into a matrix, as described in the [http://rosettacode.org/wiki/Spiral_matrix#J J entry]. To summarize, we construct a list with <code>n*n</code> elements by following some simple rules, then take its cumulative sum, and finally its inverse permutation (or grade in J parlance). This gives us a list which can be reshaped to the final matrix.
<syntaxhighlight lang="factor">USING: arrays grouping io kernel math math.combinatorics
math.ranges math.statistics prettyprint sequences
sequences.repeating ;
IN: rosetta-code.spiral-matrix
 
: counts ( n -- seq ) 1 [a,b] 2 repeat rest ;
 
: vals ( n -- seq )
[ 1 swap 2dup [ neg ] bi@ 4array ] [ 2 * 1 - cycle ] bi ;
 
: evJKT2 ( n -- seq )
[ counts ] [ vals ] bi [ <array> ] 2map concat ;
 
: spiral ( n -- matrix )
[ evJKT2 cum-sum inverse-permutation ] [ group ] bi ;
 
: spiral-demo ( -- ) 5 9 [ spiral simple-table. nl ] bi@ ;
 
MAIN: spiral-demo</syntaxhighlight>
{{out}}
<pre>
0 1 2 3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8
 
0 1 2 3 4 5 6 7 8
31 32 33 34 35 36 37 38 9
30 55 56 57 58 59 60 39 10
29 54 71 72 73 74 61 40 11
28 53 70 79 80 75 62 41 12
27 52 69 78 77 76 63 42 13
26 51 68 67 66 65 64 43 14
25 50 49 48 47 46 45 44 15
24 23 22 21 20 19 18 17 16
</pre>
 
=={{header|Fortran}}==
{{works with|Fortran|90 and later}}
<langsyntaxhighlight lang="fortran">PROGRAM SPIRAL
 
IMPLICIT NONE
Line 704 ⟶ 2,045:
END DO
 
END PROGRAM SPIRAL</langsyntaxhighlight>
 
=={{header|F_Sharp|F#FreeBASIC}}==
<syntaxhighlight lang="freebasic">' FB 1.05.0 Win64
No fancy schmancy elegance here, just putting the numbers in the right place (though I commend the elegance)...
 
<lang fsharp>let Spiral n =
Enum Direction
let sq = Array2D.create n n 0 // Set up an output array
across
let nCur = ref -1 // Current value being inserted
down
let NextN() = nCur := (!nCur+1) ; !nCur // Inc current value and return new value
back
let Frame inset = // Create the "frame" at an offset from the outside
up
let rangeF = [inset..(n - inset - 2)] // Range we use going forward
End Enum
let rangeR = [(n - inset - 1)..(-1)..(inset + 1)] // Range we use going backward
rangeF |> Seq.iter (fun i -> sq.[inset,i] <- NextN()) // Top of frame
Dim As Integer n
rangeF |> Seq.iter (fun i -> sq.[i,n-inset-1] <- NextN()) // Right side of frame
 
rangeR |> Seq.iter (fun i -> sq.[n-inset-1,i] <- NextN()) // Bottom of frame
Do
rangeR |> Seq.iter (fun i -> sq.[i,inset] <- NextN()) // Left side of frame
Input "Enter size of matrix "; n
[0..(n/2 - 1)] |> Seq.iter (fun i -> Frame i) // Fill in all frames
Loop Until n > 0
if n &&& 1 = 1 then sq.[n/2,n/2] <- n*n - 1 // If n is odd, fill in the last single value
 
sq // Return our output array</lang>
Dim spiral(1 To n, 1 To n) As Integer '' all zero by default
 
' enter the numbers 0 to (n^2 - 1) spirally in the matrix
 
Dim As Integer row = 1, col = 1, lowRow = 1, highRow = n, lowCol = 1, highCol = n
Dim d As Direction = across
 
For i As Integer = 0 To (n * n - 1)
spiral(row, col) = i
Select Case d
Case across
col += 1
If col > highCol Then
col = highCol
row += 1
d = down
End if
Case down
row += 1
If row > highRow Then
row = highRow
col -= 1
d = back
End if
Case back
col -= 1
If col < lowCol Then
col = lowCol
row -= 1
d = up
lowRow += 1
End If
Case up
row -= 1
If row < lowRow Then
row = lowRow
col += 1
d = across
highRow -= 1
lowCol += 1
highCol -= 1
End If
End Select
Next
 
' print spiral matrix if n < 20
Print
If n < 20 Then
For i As Integer = 1 To n
For j As Integer = 1 To n
Print Using "####"; spiral(i, j);
Next j
Print
Next i
Else
Print "Matrix is too big to display on 80 column console"
End If
 
Print
Print "Press any key to quit"
Sleep</syntaxhighlight>
 
{{out}}
<pre>
Enter size of matrix ? 5
 
0 1 2 3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8
</pre>
 
=={{header|GAP}}==
<langsyntaxhighlight lang="gap"># Spiral matrix with numbers 1 .. n<sup>2</sup>, more natural in GAP
SpiralMatrix := function(n)
local i, j, k, di, dj, p, vi, vj, imin, imax, jmin, jmax;
Line 770 ⟶ 2,183:
# [ 15, 24, 25, 20, 7 ],
# [ 14, 23, 22, 21, 8 ],
# [ 13, 12, 11, 10, 9 ] ]</langsyntaxhighlight>
 
=={{header|Go}}==
<langsyntaxhighlight lang="go">package main
 
import (
Line 830 ⟶ 2,243:
}
}
}</langsyntaxhighlight>
 
=={{header|Groovy}}==
Naive "path-walking" solution:
<langsyntaxhighlight lang="groovy">enum Direction {
East([0,1]), South([1,0]), West([0,-1]), North([-1,0]);
private static _n
Line 887 ⟶ 2,300:
}
M
}</langsyntaxhighlight>
Test:
<langsyntaxhighlight lang="groovy">(1..10).each { n ->
spiralMatrix(n).each { row ->
row.each { printf "%5d", it }
Line 895 ⟶ 2,308:
}
println ()
}</langsyntaxhighlight>
{{out}}
Output:
<pre style="height:30ex;overflow:scroll;"> 0
 
Line 964 ⟶ 2,377:
=={{header|Haskell}}==
Solution based on the [[#J|J]] hints:
<syntaxhighlight lang ="haskell">module Spiralimport whereData.List
 
import Data.List
import Control.Monad
import Control.Monad.Instances
 
grade xs = map snd. sort $ zip xs [0..]
values n = cycle [1,n,-1,-n]
counts n = (n:).concatMap (ap (:) return) $ [n-1,n-2..1]
reshape n = unfoldr (\xs -> if null xs then Nothing else Just (splitAt n xs))
spiral n = reshape n . grade. scanl1 (+). concat $ zipWith replicate (counts n) (values n)
displayRow = putStrLn . intercalate " " . map show
main = mapM displayRow $ spiral 5</syntaxhighlight>
 
An alternative, point-free solution based on the same J source.
 
<syntaxhighlight lang="haskell">import Data.List
import Control.Applicative
counts = tail . reverse . concat . map (replicate 2) . enumFromTo 1
values = cycle . ((++) <$> map id <*> map negate) . (1 :) . (: [])
grade = map snd . sort . flip zip [0..]
copies = grade . scanl1 (+) . concat . map (uncurry replicate) . (zip <$> counts <*> values)
parts = (<*>) take $ (.) <$> (map . take) <*> (iterate . drop) <*> copies
disp = (>> return ()) . mapM (putStrLn . intercalate " " . map show) . parts
main = disp 5</syntaxhighlight>
 
Another alternative:
<syntaxhighlight lang="haskell">import Data.List (transpose)
import Text.Printf (printf)
 
-- spiral is the first row plus a smaller spiral rotated 90 deg
spiral 0 _ _ = [[]]
spiral h w s = [s .. s+w-1] : rot90 (spiral w (h-1) (s+w))
where rot90 = (map reverse).transpose
 
-- this is sort of hideous, someone may want to fix it
main = mapM_ (\row->mapM_ ((printf "%4d").toInteger) row >> putStrLn "") (spiral 10 9 1)</syntaxhighlight>
 
 
Or less ambitiously,
{{Trans|AppleScript}}
<syntaxhighlight lang="haskell">import Data.List (intercalate, transpose)
 
---------------------- SPIRAL MATRIX ---------------------
spiral :: Int -> [[Int]]
spiral n = go n n 0
where
go rows cols x
| 0 < rows =
[x .. pred cols + x] :
fmap
reverse
(transpose $ go cols (pred rows) (x + cols))
| otherwise = [[]]
 
 
--------------------------- TEST -------------------------
main :: IO ()
main = putStrLn $ wikiTable $ spiral 5
 
 
--------------------- TABLE FORMATTING -------------------
spiral n = reshape n . grade. scanl1 (+). concat $ zipWith replicate (counts n) (values n)</lang>
wikiTable :: Show a => [[a]] -> String
wikiTable =
concat
. ("{| class=\"wikitable\" style=\"text-align: right;" :)
. ("width:12em;height:12em;table-layout:fixed;\"\n|-\n" :)
. return
. (<> "\n|}")
. intercalate "\n|-\n"
. fmap (('|' :) . (' ' :) . intercalate " || " . fmap show)</syntaxhighlight>
{{Out}}
{| class="wikitable" style="text-align: right;width:12em;height:12em;table-layout:fixed;"|-
| 0 || 1 || 2 || 3 || 4
|-
| 15 || 16 || 17 || 18 || 5
|-
| 14 || 23 || 24 || 19 || 6
|-
| 13 || 22 || 21 || 20 || 7
|-
| 12 || 11 || 10 || 9 || 8
|}
 
=={{header|Icon}} and {{header|Unicon}}==
At first I looked at keeping the filling of the matrix on track using /M[r,c] which fails when out of bounds or if the cell is null, but then I noticed the progression of the row and column increments from corner to corner reminded me of sines and cosines. I'm not sure if the use of a trigonometric function counts as elegance, perversity, or both. The generator could be easily modified to start at an arbitrary corner. Or count down to produce and evolute.
<langsyntaxhighlight Iconlang="icon">procedure main(A) # spiral matrix
N := 0 < integer(\A[1]|5) # N=1... (dfeault 5)
WriteMatrix(SpiralMatrix(N))
Line 1,006 ⟶ 2,486:
}
return M
end</langsyntaxhighlight>
 
{{out}}
 
Output:<pre> 0 1 2 3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8</pre>
 
=={{header|IS-BASIC}}==
<syntaxhighlight lang="is-basic">100 PROGRAM "SpiralMa.bas"
110 TEXT 80
120 INPUT PROMPT "Enter size of matrix (max. 10): ":N
130 NUMERIC A(1 TO N,1 TO N)
140 CALL INIT(A)
150 CALL WRITE(A)
160 DEF INIT(REF T)
170 LET BCOL,BROW,COL,ROW=1:LET TCOL,TROW=N:LET DIR=0
180 FOR I=0 TO N^2-1
190 LET T(COL,ROW)=I
200 SELECT CASE DIR
210 CASE 0
220 IF ROW<TROW THEN
230 LET ROW=ROW+1
240 ELSE
250 LET DIR=1:LET COL=COL+1:LET BCOL=BCOL+1
260 END IF
270 CASE 1
280 IF COL<TCOL THEN
290 LET COL=COL+1
300 ELSE
310 LET DIR=2:LET ROW=ROW-1:LET TROW=TROW-1
320 END IF
330 CASE 2
340 IF ROW>BROW THEN
350 LET ROW=ROW-1
360 ELSE
370 LET DIR=3:LET COL=COL-1:LET TCOL=TCOL-1
380 END IF
390 CASE 3
400 IF COL>BCOL THEN
410 LET COL=COL-1
420 ELSE
430 LET DIR=0:LET ROW=ROW+1:LET BROW=BROW+1
440 END IF
450 END SELECT
460 NEXT
470 END DEF
480 DEF WRITE(REF T)
490 FOR I=LBOUND(T,1) TO UBOUND(T,1)
500 FOR J=LBOUND(T,2) TO UBOUND(T,2)
510 PRINT USING " ##":T(I,J);
520 NEXT
530 PRINT
540 NEXT
550 END DEF</syntaxhighlight>
 
=={{header|J}}==
This function is the result of some [http://www.jsoftware.com/papers/play132.htm beautiful insights]:
some [http://www.jsoftware.com/papers/play132.htm beautiful insights]:
<lang j>spiral =. ,~ $ [: /: }.@(2 # >:@i.@-) +/\@# <:@+: $ (, -)@(1&,)
<syntaxhighlight lang="j">spiral =: ,~ $ [: /: }.@(2 # >:@i.@-) +/\@# <:@+: $ (, -)@(1&,)
 
spiral 5
Line 1,024 ⟶ 2,553:
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8</langsyntaxhighlight>
Would you like [[Talk:Spiral#J|some hints]] that will allow you to reimplement it in another language?
 
Line 1,032 ⟶ 2,561:
{{trans|C++}}
{{works with|Java|1.5+}}
<langsyntaxhighlight lang="java5">public class Blah {
 
public static void main(String[] args) {
Line 1,082 ⟶ 2,611:
}
}
}</langsyntaxhighlight>
{{out}}
Output:
<pre> 0 1 2 3 4
15 16 17 18 5
Line 1,091 ⟶ 2,620:
 
=={{header|JavaScript}}==
 
<lang javascript>spiralArray = function (edge) {
===Imperative===
 
<syntaxhighlight lang="javascript">spiralArray = function (edge) {
var arr = Array(edge),
x = 0, y = edge,
Line 1,112 ⟶ 2,644:
arr = spiralArray(edge = 5);
for (y= 0; y < edge; y++) console.log(arr[y].join(" "));
</syntaxhighlight>
</lang>
{{out}}
output
<pre>
0 1 2 3 4
Line 1,120 ⟶ 2,652:
13 22 21 20 7
12 11 10 9 8</pre>
 
===Functional===
 
====ES5====
 
Translating one of the Haskell versions:
 
<syntaxhighlight lang="javascript">(function (n) {
 
// Spiral: the first row plus a smaller spiral rotated 90 degrees clockwise
function spiral(lngRows, lngCols, nStart) {
return lngRows ? [range(nStart, (nStart + lngCols) - 1)].concat(
transpose(
spiral(lngCols, lngRows - 1, nStart + lngCols)
).map(reverse)
) : [
[]
];
}
 
// rows and columns transposed (for 90 degree rotation)
function transpose(lst) {
return lst.length > 1 ? lst[0].map(function (_, col) {
return lst.map(function (row) {
return row[col];
});
}) : lst;
}
 
// elements in reverse order (for 90 degree rotation)
function reverse(lst) {
return lst.length > 1 ? lst.reduceRight(function (acc, x) {
return acc.concat(x);
}, []) : lst;
}
 
// [m..n]
function range(m, n) {
return Array.apply(null, Array(n - m + 1)).map(function (x, i) {
return m + i;
});
}
 
// TESTING
var lstSpiral = spiral(n, n, 0);
 
 
// OUTPUT FORMATTING - JSON and wikiTable
function wikiTable(lstRows, blnHeaderRow, strStyle) {
return '{| class="wikitable" ' + (
strStyle ? 'style="' + strStyle + '"' : ''
) + lstRows.map(function (lstRow, iRow) {
var strDelim = ((blnHeaderRow && !iRow) ? '!' : '|');
 
return '\n|-\n' + strDelim + ' ' + lstRow.map(function (v) {
return typeof v === 'undefined' ? ' ' : v;
}).join(' ' + strDelim + strDelim + ' ');
}).join('') + '\n|}';
}
 
return [
wikiTable(
 
lstSpiral,
 
false,
'text-align:center;width:12em;height:12em;table-layout:fixed;'
),
JSON.stringify(lstSpiral)
].join('\n\n');
 
})(5);</syntaxhighlight>
 
Output:
 
{| class="wikitable" style="text-align:center;width:12em;height:12em;table-layout:fixed;"
|-
| 0 || 1 || 2 || 3 || 4
|-
| 15 || 16 || 17 || 18 || 5
|-
| 14 || 23 || 24 || 19 || 6
|-
| 13 || 22 || 21 || 20 || 7
|-
| 12 || 11 || 10 || 9 || 8
|}
 
<syntaxhighlight lang="javascript">[[0,1,2,3,4],[15,16,17,18,5],[14,23,24,19,6],[13,22,21,20,7],[12,11,10,9,8]]</syntaxhighlight>
 
 
====ES6====
{{Trans|Haskell}}
<syntaxhighlight lang="javascript">(() => {
"use strict";
 
// ------------------ SPIRAL MATRIX ------------------
 
// spiral :: Int -> [[Int]]
const spiral = n => {
const go = (rows, cols, start) =>
Boolean(rows) ? [
enumFromTo(start)(start + pred(cols)),
...transpose(
go(
cols,
pred(rows),
start + cols
)
).map(reverse)
] : [
[]
];
 
return go(n, n, 0);
};
 
 
// ---------------------- TEST -----------------------
// main :: () -> String
const main = () => {
const
n = 5,
cellWidth = 1 + `${pred(n ** 2)}`.length;
 
return unlines(
spiral(n).map(
row => (
row.map(x => `${x}`
.padStart(cellWidth, " "))
)
.join("")
)
);
};
 
 
// --------------------- GENERIC ---------------------
 
// enumFromTo :: Int -> Int -> [Int]
const enumFromTo = m =>
n => Array.from({
length: 1 + n - m
}, (_, i) => m + i);
 
 
// pred :: Enum a => a -> a
const pred = x => x - 1;
 
 
// reverse :: [a] -> [a]
const reverse = xs =>
"string" === typeof xs ? (
xs.split("").reverse()
.join("")
) : xs.slice(0).reverse();
 
 
// transpose :: [[a]] -> [[a]]
const transpose = rows =>
// The columns of the input transposed
// into new rows.
// Simpler version of transpose, assuming input
// rows of even length.
Boolean(rows.length) ? rows[0].map(
(_, i) => rows.flatMap(
v => v[i]
)
) : [];
 
 
// unlines :: [String] -> String
const unlines = xs =>
// A single string formed by the intercalation
// of a list of strings with the newline character.
xs.join("\n");
 
 
// MAIN ---
return main();
})();</syntaxhighlight>
{{Out}}
<pre> 0 1 2 3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8</pre>
 
=={{header|jq}}==
The strategy employed here is to start at [0,0] and move to the right ([0,1] == same row, next column)
until we reach a boundary or a populated cell; then turn right, and proceed as before.
 
Initially fill the matrix with "false" so we can easily distinguish between unvisited cells (false) and non-existent cells (null).
 
'''Infrastructure''':
<syntaxhighlight lang="jq"># Create an m x n matrix
def matrix(m; n; init):
if m == 0 then []
elif m == 1 then [range(0;n)] | map(init)
elif m > 0 then
matrix(1;n;init) as $row
| [range(0;m)] | map( $row )
else error("matrix\(m);_;_) invalid")
end ;
 
# Print a matrix neatly, each cell occupying n spaces
def neatly(n):
def right: tostring | ( " " * (n-length) + .);
. as $in
| length as $length
| reduce range (0;$length) as $i
(""; . + reduce range(0;$length) as $j
(""; "\(.)\($in[$i][$j] | right )" ) + "\n" ) ;
 
def right:
if . == [1, 0] then [ 0, -1]
elif . == [0, -1] then [-1, 0]
elif . == [-1, 0] then [ 0, 1]
elif . == [0, 1] then [ 1, 0]
else error("invalid direction: \(.)")
end;</syntaxhighlight>
'''Create a spiral n by n matrix'''
<syntaxhighlight lang="jq">def spiral(n):
# we just placed m at i,j, and we are moving in the direction d
def _next(i; j; m; d):
if m == (n*n) - 1 then .
elif .[i+d[0]][j+d[1]] == false
then .[i+d[0]][j+d[1]] = m+1 | _next(i+d[0]; j+d[1]; m+1; d)
else (d|right) as $d
| .[i+$d[0]][j+$d[1]] = m+1 | _next(i+$d[0]; j+$d[1]; m+1; $d)
end;
 
matrix(n;n;false) | .[0][0] = 0 | _next(0;0;0; [0,1]) ;
 
# Example
spiral(5) | neatly(3)</syntaxhighlight>
{{Out}}
$ jq -n -r -f spiral.jq
0 1 2 3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8
 
=={{header|Julia}}==
Define an iterator that marches through matrix indices in the spiral pattern, which makes it easy to generate spiral matrices and related objects. Note that Julia uses column major ordering of matrices and that Julia allows multi-dimensional arrays to be addressed by scalar index as well as by subscripts.
 
'''Spiral Matrix Iterator'''
<syntaxhighlight lang="julia">
immutable Spiral
m::Int
n::Int
cmax::Int
dir::Array{Array{Int,1},1}
bdelta::Array{Array{Int,1},1}
end
 
function Spiral(m::Int, n::Int)
cmax = m*n
dir = Array{Int,1}[[0,1], [1,0], [0,-1], [-1,0]]
bdelta = Array{Int,1}[[0,0,0,1], [-1,0,0,0],
[0,-1,0,0], [0,0,1,0]]
Spiral(m, n, cmax, dir, bdelta)
end
function spiral(m::Int, n::Int)
0<m&&0<n || error("The matrix dimensions must be positive.")
Spiral(m, n)
end
spiral(n::Int) = spiral(n, n)
 
type SpState
cnt::Int
dirdex::Int
cell::Array{Int,1}
bounds::Array{Int,1}
end
 
Base.length(sp::Spiral) = sp.cmax
Base.start(sp::Spiral) = SpState(1, 1, [1,1], [sp.n,sp.m,1,1])
Base.done(sp::Spiral, sps::SpState) = sps.cnt > sp.cmax
 
function Base.next(sp::Spiral, sps::SpState)
s = sub2ind((sp.m, sp.n), sps.cell[1], sps.cell[2])
if sps.cell[rem1(sps.dirdex+1, 2)] == sps.bounds[sps.dirdex]
sps.bounds += sp.bdelta[sps.dirdex]
sps.dirdex = rem1(sps.dirdex+1, 4)
end
sps.cell += sp.dir[sps.dirdex]
sps.cnt += 1
return (s, sps)
end
</syntaxhighlight>
 
'''Helper Functions'''
<syntaxhighlight lang="julia">
using Formatting
 
function width{T<:Integer}(n::T)
w = ndigits(n)
n < 0 || return w
return w + 1
end
 
function pretty{T<:Integer}(a::Array{T,2}, indent::Int=4)
lo, hi = extrema(a)
w = max(width(lo), width(hi))
id = " "^indent
fe = FormatExpr(@sprintf(" {:%dd}", w))
s = id
nrow = size(a)[1]
for i in 1:nrow
for j in a[i,:]
s *= format(fe, j)
end
i != nrow || continue
s *= "\n"*id
end
return s
end
</syntaxhighlight>
 
'''Main'''
<syntaxhighlight lang="julia">
n = 5
println("The n = ", n, " spiral matrix:")
a = zeros(Int, (n, n))
for (i, s) in enumerate(spiral(n))
a[s] = i-1
end
println(pretty(a))
 
m = 3
println()
println("Generalize to a non-square matrix (", m, "x", n, "):")
a = zeros(Int, (m, n))
for (i, s) in enumerate(spiral(m, n))
a[s] = i-1
end
println(pretty(a))
 
p = primes(10^3)
n = 7
println()
println("An n = ", n, " prime spiral matrix:")
a = zeros(Int, (n, n))
for (i, s) in enumerate(spiral(n))
a[s] = p[i]
end
println(pretty(a))
</syntaxhighlight>
 
{{out}}
<pre>
The n = 5 spiral matrix:
0 1 2 3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8
 
Generalize to a non-square matrix (3x5):
0 1 2 3 4
11 12 13 14 5
10 9 8 7 6
 
An n = 7 prime spiral matrix:
2 3 5 7 11 13 17
89 97 101 103 107 109 19
83 173 179 181 191 113 23
79 167 223 227 193 127 29
73 163 211 199 197 131 31
71 157 151 149 139 137 37
67 61 59 53 47 43 41
</pre>
 
=={{header|Kotlin}}==
{{trans|C#}}
<syntaxhighlight lang="scala">// version 1.1.3
 
typealias Vector = IntArray
typealias Matrix = Array<Vector>
 
fun spiralMatrix(n: Int): Matrix {
val result = Matrix(n) { Vector(n) }
var pos = 0
var count = n
var value = -n
var sum = -1
do {
value = -value / n
for (i in 0 until count) {
sum += value
result[sum / n][sum % n] = pos++
}
value *= n
count--
for (i in 0 until count) {
sum += value
result[sum / n][sum % n] = pos++
}
}
while (count > 0)
return result
}
 
fun printMatrix(m: Matrix) {
for (i in 0 until m.size) {
for (j in 0 until m.size) print("%2d ".format(m[i][j]))
println()
}
println()
}
 
fun main(args: Array<String>) {
printMatrix(spiralMatrix(5))
printMatrix(spiralMatrix(10))
}</syntaxhighlight>
 
{{out}}
<pre>
0 1 2 3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8
 
0 1 2 3 4 5 6 7 8 9
35 36 37 38 39 40 41 42 43 10
34 63 64 65 66 67 68 69 44 11
33 62 83 84 85 86 87 70 45 12
32 61 82 95 96 97 88 71 46 13
31 60 81 94 99 98 89 72 47 14
30 59 80 93 92 91 90 73 48 15
29 58 79 78 77 76 75 74 49 16
28 57 56 55 54 53 52 51 50 17
27 26 25 24 23 22 21 20 19 18
</pre>
 
=={{header|Liberty BASIC}}==
Extended to include automatic scaling of the display scale and font. See [http://www.diga.me.uk/spiralM5.gif spiralM5]
<langsyntaxhighlight lang="lb">nomainwin
 
UpperLeftX = 50
Line 1,198 ⟶ 3,170:
[quit]
close #w
end</langsyntaxhighlight>
 
=={{header|Lua}}==
===Original===
<lang lua>av, sn = math.abs, function(s) return s~=0 and s/av(s) or 0 end
<syntaxhighlight lang="lua">av, sn = math.abs, function(s) return s~=0 and s/av(s) or 0 end
function sindex(y, x) -- returns the value at (x, y) in a spiral that starts at 1 and goes outwards
if y == -x and y >= x then return (2*y+1)^2 end
Line 1,219 ⟶ 3,192:
end
 
for i,v in ipairs(spiralt(8)) do for j, u in ipairs(v) do io.write(u .. " ") end print() end</langsyntaxhighlight>
 
===Alternate===
If only the printed output is required, without intermediate array storage, then:
<syntaxhighlight lang="lua">local function printspiral(n)
local function z(x,y)
local m = math.min(x, y, n-1-x, n-1-y)
return x<y and (n-2*m-2)^2+(x-m)+(y-m) or (n-2*m)^2-(x-m)-(y-m)
end
for y = 1, n do
for x = 1, n do
io.write(string.format("%2d ", n^2-z(x-1,y-1)))
end
print()
end
end
printspiral(9)</syntaxhighlight>
If the intermediate array storage ''is'' required, then:
<syntaxhighlight lang="lua">local function makespiral(n)
local t, z = {}, function(x,y)
local m = math.min(x, y, n-1-x, n-1-y)
return x<y and (n-2*m-2)^2+(x-m)+(y-m) or (n-2*m)^2-(x-m)-(y-m)
end
for y = 1, n do t[y] = {}
for x = 1, n do t[y][x] = n^2-z(x-1,y-1) end
end
return t
end
local function printspiral(t)
for y = 1, #t do
for x = 1, #t[y] do
io.write(string.format("%2d ", t[y][x]))
end
print()
end
end
printspiral(makespiral(9))</syntaxhighlight>
{{out}}
(same for both)
<pre> 0 1 2 3 4 5 6 7 8
31 32 33 34 35 36 37 38 9
30 55 56 57 58 59 60 39 10
29 54 71 72 73 74 61 40 11
28 53 70 79 80 75 62 41 12
27 52 69 78 77 76 63 42 13
26 51 68 67 66 65 64 43 14
25 50 49 48 47 46 45 44 15
24 23 22 21 20 19 18 17 16</pre>
 
=={{header|Maple}}==
 
<syntaxhighlight lang="maple">
with(ArrayTools):
 
spiralArray := proc(size::integer)
local M, sideLength, count, i, j:
M := Matrix(size):
count := 0:
sideLength := size:
for i from 1 to ceil(sideLength / 2) do
for j from 1 to sideLength do
M[i,i + j - 1] := count++:
end:
for j from 1 to sideLength - 1 do
M[i + j, sideLength + i - 1] := count++:
end:
for j from 1 to sideLength - 1 do
M[i + sideLength - 1, sideLength - j + i - 1] := count++:
end:
for j from 1 to sideLength - 2 do
M[sideLength + i - j - 1, i] := count++
end:
sideLength -= 2:
end:
return M;
end proc:
 
spiralArray(5);
 
</syntaxhighlight>
{{out}}<pre>
[ 0 1 2 3 4]
[ ]
[15 16 17 18 5]
[ ]
[14 23 24 19 6]
[ ]
[13 22 21 20 7]
[ ]
[12 11 10 9 8]
 
</pre>
 
=={{header|Mathematica}} / {{header|Wolfram Language}}==
We split the task up in 2 functions, one that adds a 'ring' around a present matrix. And a function that adds rings to a 'core':
<langsyntaxhighlight Mathematicalang="mathematica">AddSquareRing[x_List/;Equal@@Dimensions[x] && Length[Dimensions[x]]==2]:=Module[{new=x,size,smallest},
size=Length[x];
smallest=x[[1,1]];
Line 1,238 ⟶ 3,302:
times=If[Mod[size,2]==0,size/2-1,(size-1)/2];
Nest[AddSquareRing,start,times]
]</langsyntaxhighlight>
Examples:
<langsyntaxhighlight Mathematicalang="mathematica">MakeSquareSpiral[2] // MatrixForm
MakeSquareSpiral[7] // MatrixForm</langsyntaxhighlight>
gives back:
<math>
Line 1,272 ⟶ 3,336:
<math>(-spiral(n))+n^2</math>
Then depending on if n is odd or even we use either an up/down or left/right mirror transformation.
<langsyntaxhighlight MATLABlang="matlab">function matrix = reverseSpiral(n)
matrix = (-spiral(n))+n^2;
Line 1,282 ⟶ 3,346:
end
end %reverseSpiral</langsyntaxhighlight>
Sample Usage:
<langsyntaxhighlight MATLABlang="matlab">>> reverseSpiral(5)
 
ans =
Line 1,292 ⟶ 3,356:
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8</langsyntaxhighlight>
 
=={{header|Maxima}}==
<langsyntaxhighlight lang="maxima">spiral(n) := block([a, i, j, k, p, di, dj, vi, vj, imin, imax, jmin, jmax],
a: zeromatrix(n, n),
vi: [1, 0, -1, 0],
Line 1,334 ⟶ 3,398:
[15, 24, 25, 20, 7],
[14, 23, 22, 21, 8],
[13, 12, 11, 10, 9]) */</langsyntaxhighlight>
 
=={{header|MiniZinc}}==
<syntaxhighlight lang="minizinc">
%Spiral Matrix. Nigel Galloway, February 3rd., 2020
int: Size;
array [1..Size,1..Size] of var 1..Size*Size: spiral;
constraint spiral[1,1..]=1..Size;
constraint forall(n in 2..(Size+1) div 2)(forall(g in n..Size+1-n)(spiral[n,g]=spiral[n,g-1]+1));
constraint forall(n in 1..(Size+1) div 2)(forall(g in n+1..Size+1-n)(spiral[g,Size-n+1]=spiral[g-1,Size-n+1]+1));
constraint forall(n in 1..Size div 2)(forall(g in n..Size-n)(spiral[Size-n+1,g]=spiral[Size-n+1,g+1]+1)) /\ forall(n in 1..Size div 2)(forall(g in n+1..Size-n)(spiral[g,n]=spiral[g+1,n]+1));
output [show2d(spiral)];
</syntaxhighlight>
{{out}}
<pre>
minizinc -DSize= spiral.mzn
 
[| 1, 2, 3, 4 |
12, 13, 14, 5 |
11, 16, 15, 6 |
10, 9, 8, 7 |]
----------
 
minizinc -DSize=5 zigzag.mzn
 
[| 1, 2, 3, 4, 5 |
16, 17, 18, 19, 6 |
15, 24, 25, 20, 7 |
14, 23, 22, 21, 8 |
13, 12, 11, 10, 9 |]
----------
 
minizinc -DSize=6 zigzag.mzn
 
[| 1, 2, 3, 4, 5, 6 |
20, 21, 22, 23, 24, 7 |
19, 32, 33, 34, 25, 8 |
18, 31, 36, 35, 26, 9 |
17, 30, 29, 28, 27, 10 |
16, 15, 14, 13, 12, 11 |]
----------
</pre>
 
=={{header|NetRexx}}==
{{Trans|ooRexx}}
<langsyntaxhighlight NetRexxlang="netrexx">/* NetRexx */
options replace format comments java crossref symbols binary
 
Line 1,430 ⟶ 3,535:
return maxNum
 
</syntaxhighlight>
</lang>
 
{{out}}
:Output
<pre>
| 0 1 2 |
Line 1,449 ⟶ 3,554:
| 12 11 10 9 8 |
</pre>
 
=={{header|Nim}}==
<syntaxhighlight lang="nim">import sequtils, strutils
 
proc `$`(m: seq[seq[int]]): string =
for r in m:
let lg = result.len
for c in r:
result.addSep(" ", lg)
result.add align($c, 2)
result.add '\n'
 
proc spiral(n: Positive): seq[seq[int]] =
result = newSeqWith(n, repeat(-1, n))
var dx = 1
var dy, x, y = 0
for i in 0 ..< (n * n):
result[y][x] = i
let (nx, ny) = (x+dx, y+dy)
if nx in 0 ..< n and ny in 0 ..< n and result[ny][nx] == -1:
x = nx
y = ny
else:
swap dx, dy
dx = -dx
x += dx
y += dy
 
echo spiral(5)</syntaxhighlight>
{{out}}
<pre> 0 1 2 3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8</pre>
 
=={{header|OCaml}}==
<langsyntaxhighlight lang="ocaml">let next_dir = function
| 1, 0 -> 0, -1
| 0, 1 -> 1, 0
Line 1,489 ⟶ 3,629:
print_newline())
 
let () = print(spiral 5)</langsyntaxhighlight>
 
Another implementation:
<langsyntaxhighlight lang="ocaml">let spiral n =
let ar = Array.make_matrix n n (-1) in
let out i = i < 0 || i >= n in
Line 1,508 ⟶ 3,648:
Array.iter (fun v -> Array.iter (Printf.printf " %2d") v; print_newline())
 
let _ = show (spiral 5)</langsyntaxhighlight>
 
=={{header|Octave}}==
{{trans|Stata}}
The function <code>make_spiral</code> (and helper functions) are modelled after the J solution.
<langsyntaxhighlight lang="octave">function rsa = runsumspiral(vn)
for ia = 1:numelones(vn*n, 1);
u = rs-(i) = sumn) * (v = ones(n, 1:i));
for k = n-1:-1:1
j = 1:k;
a(j+i) = u(j) = -u(j);
a(j+(i+k)) = v(j) = -v(j);
i += 2*k;
endfor
a(cumsum(a)) = 1:n*n;
a = reshape(a, n, n)'-1;
endfunction
 
>> spiral(5)
function g = grade(v)
ans =
for i = 1:numel(v)
g(v(i)+1) = i-1;
endfor
endfunction
 
0 1 2 3 4
function spiral = make_spiral(spirald)
15 16 17 18 5
series = ones(1,spirald^2);
14 23 24 19 6
l = spirald-1; p = spirald+1;
13 22 21 20 7
s = 1;
12 11 10 9 8</syntaxhighlight>
while(l>0)
series(p:p+l-1) *= spirald*s;
series(p+l:p+l*2-1) *= -s;
p += l*2;
l--; s *= -1;
endwhile
series(1) = 0;
spiral = reshape(grade(runsum(series)), spirald, spirald)';
endfunction
 
make_spiral(5)</lang>
 
=={{header|Opal}}==
{{incorrect|Opal|It is incomplete.}}
Recursive functional solution
<lang opal>IMPLEMENTATION Spiral
 
IMPORT Nat COMPLETELY
IMPORT Seq COMPLETELY
 
DATA matrix == node(x:nat, y:nat, val:nat)
 
FUN spiral: nat -> seq[matrix]
DEF spiral(size) == </lang>
 
=={{header|ooRexx}}==
<syntaxhighlight lang="oorexx">
<lang ooRexx>
call printArray generateArray(3)
say
Line 1,610 ⟶ 3,731:
say line
end
</syntaxhighlight>
</lang>
{{out}}
Output:
<pre>
| 0 1 2 |
Line 1,628 ⟶ 3,749:
| 12 11 10 9 8 |
</pre>
 
 
=={{header|Oz}}==
Simple, recursive solution:
<langsyntaxhighlight lang="oz">declare
fun {Spiral N}
%% create nested array
Line 1,673 ⟶ 3,793:
end
in
{Inspect {Spiral 5}}</langsyntaxhighlight>
 
=={{header|PARI/GP}}==
 
<syntaxhighlight lang="parigp">spiral(dim) = {
my (M = matrix(dim, dim), p = s = 1, q = i = 0);
for (n=1, dim,
for (b=1, dim-n+1, M[p,q+=s] = i; i++);
for (b=1, dim-n, M[p+=s,q] = i; i++);
s = -s;
);
M
}</syntaxhighlight>
 
Output:<pre>spiral(7)
 
[ 0 1 2 3 4 5 6]
 
[23 24 25 26 27 28 7]
 
[22 39 40 41 42 29 8]
 
[21 38 47 48 43 30 9]
 
[20 37 46 45 44 31 10]
 
[19 36 35 34 33 32 11]
 
[18 17 16 15 14 13 12]</pre>
 
=={{header|Pascal}}==
<syntaxhighlight lang="pascal">program Spiralmat;
type
tDir = (left,down,right,up);
tdxy = record
dx,dy: longint;
end;
tdeltaDir = array[tDir] of tdxy;
const
Nextdir : array[tDir] of tDir = (down,right,up,left);
cDir : tDeltaDir = ((dx:1;dy:0),(dx:0;dy:1),(dx:-1;dy:0),(dx:0;dy:-1));
cMaxN = 32;
type
tSpiral = array[0..cMaxN,0..cMaxN] of LongInt;
 
function FillSpiral(n:longint):tSpiral;
var
b,i,k, dn,x,y : longInt;
dir : tDir;
tmpSp : tSpiral;
BEGIN
b := 0;
x := 0;
y := 0;
//only for the first line
k := -1;
dn := n-1;
tmpSp[x,y] := b;
dir := left;
repeat
i := 0;
while i < dn do
begin
inc(b);
tmpSp[x,y] := b;
inc(x,cDir[dir].dx);
inc(y,cDir[dir].dy);
inc(i);
end;
Dir:= NextDir[dir];
inc(k);
IF k > 1 then
begin
k := 0;
//shorten the line every second direction change
dn := dn-1;
if dn <= 0 then
BREAK;
end;
until false;
//the last
tmpSp[x,y] := b+1;
FillSpiral := tmpSp;
end;
 
var
a : tSpiral;
x,y,n : LongInt;
BEGIN
For n := 1 to 5{cMaxN} do
begin
A:=FillSpiral(n);
For y := 0 to n-1 do
begin
For x := 0 to n-1 do
write(A[x,y]:4);
writeln;
end;
writeln;
end;
END.
</syntaxhighlight>
{{out}}
<pre> 1
 
1 2
4 3
....
1 2 3 4 5
16 17 18 19 6
15 24 25 20 7
14 23 22 21 8
13 12 11 10 9
</pre>
 
=={{header|Perl}}==
<langsyntaxhighlight lang="perl">sub spiral
{my ($n, $x, $y, $dx, $dy, @a) = (shift, 0, 0, 1, 0);
foreach (0 .. $n**2 - 1)
Line 1,696 ⟶ 3,929:
foreach (spiral 5)
{printf "%3d", $_ foreach @$_;
print "\n";}</langsyntaxhighlight>
 
=={{header|Perl 6Phix}}==
{{trans|Python}}
===Object-oriented Solution===
Simple is better.
Suppose we set up a Turtle class like this:
<!--<syntaxhighlight lang="phix">(phixonline)-->
<lang perl6>enum Dir < north northeast east southeast south southwest west northwest >;
<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>
my $debug = 0;
<span style="color: #004080;">integer</span> <span style="color: #000000;">n</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">6</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">x</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">1</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">y</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">0</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">counter</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">0</span><span style="color: #0000FF;">,</span>
<span style="color: #000000;">len</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">n</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">dx</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">0</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">dy</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">1</span>
<span style="color: #004080;">string</span> <span style="color: #000000;">fmt</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">sprintf</span><span style="color: #0000FF;">(</span><span style="color: #008000;">"%%%dd"</span><span style="color: #0000FF;">,</span><span style="color: #7060A8;">length</span><span style="color: #0000FF;">(</span><span style="color: #7060A8;">sprintf</span><span style="color: #0000FF;">(</span><span style="color: #008000;">"%d"</span><span style="color: #0000FF;">,</span><span style="color: #000000;">n</span><span style="color: #0000FF;">*</span><span style="color: #000000;">n</span><span style="color: #0000FF;">)))</span>
<span style="color: #004080;">sequence</span> <span style="color: #000000;">m</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">repeat</span><span style="color: #0000FF;">(</span><span style="color: #7060A8;">repeat</span><span style="color: #0000FF;">(</span><span style="color: #008000;">"??"</span><span style="color: #0000FF;">,</span><span style="color: #000000;">n</span><span style="color: #0000FF;">),</span><span style="color: #000000;">n</span><span style="color: #0000FF;">)</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">i</span><span style="color: #0000FF;">=</span><span style="color: #000000;">1</span> <span style="color: #008080;">to</span> <span style="color: #000000;">2</span><span style="color: #0000FF;">*</span><span style="color: #000000;">n</span> <span style="color: #008080;">do</span> <span style="color: #000080;font-style:italic;">-- 2n runs..</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">j</span><span style="color: #0000FF;">=</span><span style="color: #000000;">1</span> <span style="color: #008080;">to</span> <span style="color: #000000;">len</span> <span style="color: #008080;">do</span> <span style="color: #000080;font-style:italic;">-- of a length...</span>
<span style="color: #000000;">x</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">dx</span>
<span style="color: #000000;">y</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">dy</span>
<span style="color: #000000;">m</span><span style="color: #0000FF;">[</span><span style="color: #000000;">x</span><span style="color: #0000FF;">][</span><span style="color: #000000;">y</span><span style="color: #0000FF;">]</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">sprintf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">fmt</span><span style="color: #0000FF;">,</span><span style="color: #000000;">counter</span><span style="color: #0000FF;">)</span>
<span style="color: #000000;">counter</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">1</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #000000;">len</span> <span style="color: #0000FF;">-=</span> <span style="color: #7060A8;">odd</span><span style="color: #0000FF;">(</span><span style="color: #000000;">i</span><span style="color: #0000FF;">)</span> <span style="color: #000080;font-style:italic;">-- ..-1 every other </span>
<span style="color: #0000FF;">{</span><span style="color: #000000;">dx</span><span style="color: #0000FF;">,</span><span style="color: #000000;">dy</span><span style="color: #0000FF;">}</span> <span style="color: #0000FF;">=</span> <span style="color: #0000FF;">{</span><span style="color: #000000;">dy</span><span style="color: #0000FF;">,-</span><span style="color: #000000;">dx</span><span style="color: #0000FF;">}</span> <span style="color: #000080;font-style:italic;">-- in new direction</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"%s\n"</span><span style="color: #0000FF;">,{</span><span style="color: #7060A8;">join</span><span style="color: #0000FF;">(</span><span style="color: #7060A8;">apply</span><span style="color: #0000FF;">(</span><span style="color: #000000;">m</span><span style="color: #0000FF;">,</span><span style="color: #7060A8;">join</span><span style="color: #0000FF;">),</span><span style="color: #008000;">"\n"</span><span style="color: #0000FF;">)})</span>
<!--</syntaxhighlight>-->
{{out}}
<pre>
0 1 2 3 4 5
19 20 21 22 23 6
18 31 32 33 24 7
17 30 35 34 25 8
16 29 28 27 26 9
15 14 13 12 11 10
</pre>
 
=={{header|PicoLisp}}==
class Turtle {
This example uses 'grid' from "lib/simul.l", which maintains a two-dimensional structure and is normally used for simulations and board games.
has @.loc = 0,0;
<syntaxhighlight lang="picolisp">(load "@lib/simul.l")
has Dir $.dir = north;
 
(de spiral (N)
my @dv = [0,-1], [1,-1], [1,0], [1,1], [0,1], [-1,1], [-1,0], [-1,-1];
(prog1 (grid N N)
my @num-to-dir = Dir.invert.sort».value;
(let (Dir '(north east south west .) This 'a1)
my $points = +Dir;
(for Val (* N N)
(=: val Val)
(setq This
(or
(with ((car Dir) This)
(unless (: val) This) )
(with ((car (setq Dir (cdr Dir))) This)
(unless (: val) This) ) ) ) ) ) ) )
 
(mapc
my %world;
'((L)
my $maxegg;
(for This L (prin (align 3 (: val))))
my $range-x;
my $range-y; (prinl) )
(spiral 5) )</syntaxhighlight>
 
{{out}}
method turn-left ($angle = 90) { $!dir -= $angle / 45; $!dir %= $points; }
<pre> 1 2 3 4 5
method turn-right($angle = 90) { $!dir += $angle / 45; $!dir %= $points; }
16 17 18 19 6
 
15 24 25 20 7
method lay-egg($egg) {
14 23 22 21 8
%world{~@!loc} = $egg;
13 12 11 10 9</pre>
$maxegg max= $egg;
$range-x minmax= @!loc[0];
$range-y minmax= @!loc[1];
}
 
method look($ahead = 1) {
my $there = @!loc »+« (@dv[$!dir] X* $ahead);
say "looking @num-to-dir[$!dir] to $there" if $debug;
%world{~$there};
}
 
method forward($ahead = 1) {
my $there = @!loc »+« (@dv[$!dir] X* $ahead);
@!loc = @($there);
say " moving @num-to-dir[$!dir] to @!loc[]" if $debug;
}
 
method showmap() {
my $form = "%{$maxegg.chars}s";
my $endx = $range-x.max;
for $range-y.list X $range-x.list -> $y, $x {
print (%world{"$x $y"} // '').fmt($form);
print $x == $endx ?? "\n" !! ' ';
}
}
}</lang>
Now we can build the spiral in the normal way from outside-in like this:
<lang perl6>sub MAIN($size as Int) {
my $t = Turtle.new(dir => east);
my $counter = 0;
$t.forward(-1);
for 0..^ $size -> $ {
$t.forward;
$t.lay-egg($counter++);
}
for $size-1 ... 1 -> $run {
$t.turn-right;
$t.forward, $t.lay-egg($counter++) for 0..^$run;
$t.turn-right;
$t.forward, $t.lay-egg($counter++) for 0..^$run;
}
$t.showmap;
}</lang>
Or we can build the spiral from inside-out like this:
<lang perl6>sub MAIN($size as Int) {
my $t = Turtle.new(dir => ($size %% 2 ?? south !! north));
my $counter = $size * $size;
while $counter {
$t.lay-egg(--$counter);
$t.turn-left;
$t.turn-right if $t.look;
$t.forward;
}
$t.showmap;
}</lang>
Note that with these "turtle graphics" we don't actually have to care about the coordinate system, since the <code>showmap</code> method can show whatever rectangle was modified by the turtle. So unlike the standard inside-out algorithm, we don't have to find the center of the matrix first.
===Procedural Solution===
<lang perl6>sub spiral_matrix ( $n ) {
my @sm;
my $len = $n;
my $pos = 0;
 
for ^($n/2).ceiling -> $i {
my $j = $i + 1;
my $e = $n - $j;
 
@sm[$i ][$i + $_] = $pos++ for ^( $len); # Top
@sm[$j + $_][$e ] = $pos++ for ^(--$len); # Right
@sm[$e ][$i + $_] = $pos++ for reverse ^( $len); # Bottom
@sm[$j + $_][$i ] = $pos++ for reverse ^(--$len); # Left
}
 
return @sm;
}
 
say .fmt('%3d') for spiral_matrix(5);</lang>
Output:<pre> 0 1 2 3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8</pre>
 
=={{header|PL/I}}==
<langsyntaxhighlight PL/Ilang="pli">/* Generates a square matrix containing the integers from 0 to N**2-1, */
/* where N is the length of one side of the square. */
/* Written 22 February 2010. */
Line 1,859 ⟶ 4,046:
end;
 
end;</langsyntaxhighlight>
 
=={{header|PicoLispPowerShell}}==
<syntaxhighlight lang="powershell">function Spiral-Matrix ( [int]$N )
This example uses 'grid' from "lib/simul.l", which maintains a two-dimensional structure and is normally used for simulations and board games.
{
<lang PicoLisp>(load "@lib/simul.l")
# Initialize variables
$X = 0
$Y = -1
$i = 0
$Sign = 1
# Intialize array
$A = New-Object 'int[,]' $N, $N
# Set top row
1..$N | ForEach { $Y += $Sign; $A[$X,$Y] = ++$i }
# For each remaining half spiral...
ForEach ( $M in ($N-1)..1 )
{
# Set the vertical quarter spiral
1..$M | ForEach { $X += $Sign; $A[$X,$Y] = ++$i }
# Curve the spiral
$Sign = -$Sign
# Set the horizontal quarter spiral
1..$M | ForEach { $Y += $Sign; $A[$X,$Y] = ++$i }
}
# Convert the array to text output
$Spiral = ForEach ( $X in 1..$N ) { ( 1..$N | ForEach { $A[($X-1),($_-1)] } ) -join "`t" }
return $Spiral
}
Spiral-Matrix 5
""
Spiral-Matrix 7</syntaxhighlight>
{{out}}
<pre>1 2 3 4 5
16 17 18 19 6
15 24 25 20 7
14 23 22 21 8
13 12 11 10 9
 
1 2 3 4 5 6 7
(de spiral (N)
24 25 26 27 28 29 8
(prog1 (grid N N)
23 40 41 42 43 30 9
(let (Dir '(north east south west .) This 'a1)
22 39 48 49 44 31 10
(for Val (* N N)
21 38 47 46 45 32 11
(=: val Val)
20 37 36 35 34 33 12
(setq This
19 18 17 16 15 14 13</pre>
(or
(with ((car Dir) This)
(unless (: val) This) )
(with ((car (setq Dir (cdr Dir))) This)
(unless (: val) This) ) ) ) ) ) ) )
 
=={{header|Prolog}}==
(mapc
<syntaxhighlight lang="prolog">
'((L)
% Prolog implementation: SWI-Prolog 7.2.3
(for This L (prin (align 3 (: val))))
 
(prinl) )
replace([_|T], 0, E, [E|T]) :- !.
(spiral 5) )</lang>
replace([H|T], N, E, Xs) :-
Output:
succ(N1, N), replace(T, N1, E, Xs1), Xs = [H|Xs1].
<pre> 1 2 3 4 5
 
16 17 18 19 6
% True if Xs is the Original grid with the element at (X, Y) replaces by E.
15 24 25 20 7
replace_in([H|T], (0, Y), E, Xs) :- replace(H, Y, E, NH), Xs = [NH|T], !.
14 23 22 21 8
replace_in([H|T], (X, Y), E, Xs) :-
13 12 11 10 9</pre>
succ(X1, X), replace_in(T, (X1, Y), E, Xs1), Xs = [H|Xs1].
 
% True, if E is the value at (X, Y) in Xs
get_in(Xs, (X, Y), E) :- nth0(X, Xs, L), nth0(Y, L, E).
 
create(N, Mx) :- % NxN grid full of nils
numlist(1, N, Ns),
findall(X, (member(_, Ns), X = nil), Ls),
findall(X, (member(_, Ns), X = Ls), Mx).
 
% Depending of the direction, returns two possible coordinates and directions
% (C,D) that will be used in case of a turn, and (A,B) otherwise.
ops(right, (X,Y), (A,B), (C,D), D1, D2) :-
A is X, B is Y+1, D1 = right, C is X+1, D is Y, D2 = down.
 
ops(left, (X,Y), (A,B), (C,D), D1, D2) :-
A is X, B is Y-1, D1 = left, C is X-1, D is Y, D2 = up.
 
ops(up, (X,Y), (A,B), (C,D), D1, D2) :-
A is X-1, B is Y, D1 = up, C is X, D is Y+1, D2 = right.
 
ops(down, (X,Y), (A,B), (C,D), D1, D2) :-
A is X+1, B is Y, D1 = down, C is X, D is Y-1, D2 = left.
 
% True if NCoor is the right coor in spiral shape. Returns a new direction also.
next(Dir, Mx, Coor, NCoor, NDir) :-
ops(Dir, Coor, C1, C2, D1, D2),
(get_in(Mx, C1, nil) -> NCoor = C1, NDir = D1
; NCoor = C2, NDir = D2).
 
% Returns an spiral with [H|Vs] elements called R, only work if the length of
% [H|Vs], is the square of the size of the grid.
spiralH(Dir, Mx, Coor, [H|Vs], R) :-
replace_in(Mx, Coor, H, NMx),
(Vs = [] -> R = NMx
; next(Dir, Mx, Coor, NCoor, NDir),
spiralH(NDir, NMx, NCoor, Vs, R)).
 
% True if Mx is the grid in spiral shape of the numbers from 0 to N*N-1.
spiral(N, Mx) :-
Sq is N*N-1, numlist(0, Sq, Ns),
create(N, EMx), spiralH(right, EMx, (0,0), Ns, Mx).
</syntaxhighlight>
{{out}}
<pre>
?- spiral(6,Mx), forall(member(X,Mx), writeln(X)).
[0,1,2,3,4,5]
[19,20,21,22,23,6]
[18,31,32,33,24,7]
[17,30,35,34,25,8]
[16,29,28,27,26,9]
[15,14,13,12,11,10]
</pre>
 
=={{header|PureBasic}}==
{{trans|Fortran}}
<langsyntaxhighlight PureBasiclang="purebasic">Procedure spiralMatrix(size = 1)
Protected i, x = -1, y, count = size, n
Dim a(size - 1,size - 1)
Line 1,947 ⟶ 4,223:
Input()
CloseConsole()
EndIf</langsyntaxhighlight>
{{out}}
Sample output:
<pre>Spiral: 2
 
Line 1,964 ⟶ 4,240:
 
=={{header|Python}}==
<langsyntaxhighlight lang="python">def spiral(n):
dx,dy = 1,0 # Starting increments
x,y = 0,0 # Starting location
Line 1,985 ⟶ 4,261:
print
 
printspiral(spiral(5))</langsyntaxhighlight>
{{out}}
Sample output:
<pre>
0 1 2 3 4
Line 1,994 ⟶ 4,270:
12 11 10 9 8
</pre>
 
===Recursive Solution===
<langsyntaxhighlight lang="python">def spiral(n):
def spiral_part(x, y, n):
if x == -1 and y == 0:
Line 2,017 ⟶ 4,294:
 
for row in spiral(5):
print " ".join("%2s" % x for x in row)</langsyntaxhighlight>
Adding a cache for the ''spiral_part'' function it could be quite efficient.
 
 
Recursion by rotating the solution for rest of the square except the first row,
<syntaxhighlight lang="python">def rot_right(a):
return zip(*a[::-1])
 
def sp(m, n, start = 0):
""" Generate number range spiral of dimensions m x n
"""
if n == 0:
yield ()
else:
yield tuple(range(start, m + start))
for row in rot_right(list(sp(n - 1, m, m + start))):
yield row
 
def spiral(m):
return sp(m, m)
 
for row in spiral(5):
print(''.join('%3i' % i for i in row))</syntaxhighlight>
 
 
Another way, based on preparing lists ahead
<langsyntaxhighlight lang="python">def spiral(n):
dat = [[None] * n for i in range(n)]
le = [[i + 1, i + 1] for i in reversed(range(n))]
Line 2,036 ⟶ 4,335:
 
for row in spiral(5): # calc spiral and print it
print ' '.join('%3s' % x for x in row)</langsyntaxhighlight>
 
===Functional Solution===
===Functional Solutions===
{{works with|Python|2.6, 3.0}}
<langsyntaxhighlight lang="python">import itertools
 
concat = itertools.chain.from_iterable
Line 2,057 ⟶ 4,357:
 
for row in spiral(5):
print(' '.join('%3s' % x for x in row))</langsyntaxhighlight>
 
 
Or, as an alternative to generative mutation:
{{Works with|Python|3.7}}
{{Trans|Haskell}}
<syntaxhighlight lang="python">'''Spiral Matrix'''
 
 
# spiral :: Int -> [[Int]]
def spiral(n):
'''The rows of a spiral matrix of order N.
'''
def go(rows, cols, x):
return [range(x, x + cols)] + [
reversed(x) for x
in zip(*go(cols, rows - 1, x + cols))
] if 0 < rows else [[]]
return go(n, n, 0)
 
 
# ------------------------- TEST -------------------------
# main :: IO ()
def main():
'''Spiral matrix of order 5, in wiki table markup.
'''
print(
wikiTable(spiral(5))
)
 
 
# ---------------------- FORMATTING ----------------------
 
# wikiTable :: [[a]] -> String
def wikiTable(rows):
'''Wiki markup for a no-frills tabulation of rows.'''
return '{| class="wikitable" style="' + (
'width:12em;height:12em;table-layout:fixed;"|-\n'
) + '\n|-\n'.join(
'| ' + ' || '.join(
str(cell) for cell in row
)
for row in rows
) + '\n|}'
 
 
# MAIN ---
if __name__ == '__main__':
main()
</syntaxhighlight>
{| class="wikitable" style="width:12em;height:12em;table-layout:fixed;"|-
| 0 || 1 || 2 || 3 || 4
|-
| 15 || 16 || 17 || 18 || 5
|-
| 14 || 23 || 24 || 19 || 6
|-
| 13 || 22 || 21 || 20 || 7
|-
| 12 || 11 || 10 || 9 || 8
|}
 
=== Simple solution ===
<syntaxhighlight lang="python">def spiral_matrix(n):
m = [[0] * n for i in range(n)]
dx, dy = [0, 1, 0, -1], [1, 0, -1, 0]
x, y, c = 0, -1, 1
for i in range(n + n - 1):
for j in range((n + n - i) // 2):
x += dx[i % 4]
y += dy[i % 4]
m[x][y] = c
c += 1
return m
for i in spiral_matrix(5): print(*i)</syntaxhighlight>
{{out}}
<syntaxhighlight lang="text">1 2 3 4 5
16 17 18 19 6
15 24 25 20 7
14 23 22 21 8
13 12 11 10 9</syntaxhighlight>
 
=={{header|Quackery}}==
 
This task really lends itself to a turtle graphics metaphor.
 
<syntaxhighlight lang="quackery"> [ stack ] is stepcount ( --> s )
[ stack ] is position ( --> s )
[ stack ] is heading ( --> s )
[ heading take
behead join
heading put ] is right ( --> )
[ heading share 0 peek
unrot times
[ position share
stepcount share
unrot poke
over position tally
1 stepcount tally ]
nip ] is walk ( [ n --> [ )
 
[ dip [ temp put [] ]
temp share times
[ temp share split
dip
[ nested join ] ]
drop temp release ] is matrixify ( n [ --> [ )
 
[ 0 stepcount put ( set up... )
0 position put
' [ 1 ] over join
-1 join over negate join
heading put
0 over dup * of
over 1 - walk right ( turtle draws spiral )
over 1 - times
[ i 1+ walk right
i 1+ walk right ]
1 walk
matrixify ( ...tidy up )
heading release
position release
stepcount release ] is spiral ( n --> [ )
 
9 spiral
witheach
[ witheach
[ dup 10 < if sp echo sp ]
cr ]</syntaxhighlight>
 
{{out}}
 
<pre> 0 1 2 3 4 5 6 7 8
31 32 33 34 35 36 37 38 9
30 55 56 57 58 59 60 39 10
29 54 71 72 73 74 61 40 11
28 53 70 79 80 75 62 41 12
27 52 69 78 77 76 63 42 13
26 51 68 67 66 65 64 43 14
25 50 49 48 47 46 45 44 15
24 23 22 21 20 19 18 17 16</pre>
 
=={{header|R}}==
===Sequence Solution===
{{trans|Octave}}
<syntaxhighlight lang="rsplus">spiral <- function(n) matrix(order(cumsum(rep(rep_len(c(1, n, -1, -n), 2 * n - 1), n - seq(2 * n - 1) %/% 2))), n, byrow = T) - 1
<lang R>runsum <- function(v) {
rs <- c()
for(i in 1:length(v)) {
rs <- c(rs, sum(v[1:i]))
}
rs
}
 
spiral(5)</syntaxhighlight>
grade <- function(v) {
{{out}}
g <- vector("numeric", length(v))
<pre> [,1] [,2] [,3] [,4] [,5]
for(i in 1:length(v)) {
[1,] 0 1 2 3 4
g[v[i]] <- i-1
[2,] 15 16 17 18 5
[3,] 14 23 24 19 6
[4,] 13 22 21 20 7
[5,] 12 11 10 9 8</pre>
 
===Recursive Solution===
<syntaxhighlight lang="rsplus">spiral_matrix <- function(n) {
spiralv <- function(v) {
n <- sqrt(length(v))
if (n != floor(n))
stop("length of v should be a square of an integer")
if (n == 0)
stop("v should be of positive length")
if (n == 1)
m <- matrix(v, 1, 1)
else
m <- rbind(v[1:n], cbind(spiralv(v[(2 * n):(n^2)])[(n - 1):1, (n - 1):1], v[(n + 1):(2 * n - 1)]))
m
}
spiralv(1:(n^2))
}</syntaxhighlight>
 
===Iterative Solution===
Not the most elegant, but certainly distinct from the other R solutions. The key is the observation that we need to produce n elements from left to right, then n-1 elements down, then n-1 left, then n-2 right, then n-2 down, ... . This gives us two patterns. One in the direction that we need to write and another in the number of elements to write. After this, all that is left is battling R's indexing system.
<syntaxhighlight lang="rsplus">spiralMatrix <- function(n)
{
spiral <- matrix(0, nrow = n, ncol = n)
firstNumToWrite <- 0
neededLength <- n
startPt <- cbind(1, 0)#(1, 0) is needed for the first call to writeRight to work. We need to start in row 1.
writingDirectionIndex <- 0
#These two functions select a collection of adjacent elements and replaces them with the needed sequence.
#This heavily uses R's vector recycling rules.
writeDown <- function(seq) spiral[startPt[1] + seq, startPt[2]] <<- seq_len(neededLength) - 1 + firstNumToWrite
writeRight <- function(seq) spiral[startPt[1], startPt[2] + seq] <<- seq_len(neededLength) - 1 + firstNumToWrite
while(firstNumToWrite != n^2)
{
writingDirectionIndex <- writingDirectionIndex %% 4 + 1
seq <- seq_len(neededLength)
switch(writingDirectionIndex,
writeRight(seq),
writeDown(seq),
writeRight(-seq),
writeDown(-seq))
if(writingDirectionIndex %% 2) neededLength <- neededLength - 1
max <- max(spiral)
firstNumToWrite <- max + 1
startPt <- which(max == spiral, arr.ind = TRUE)
}
gspiral
}</syntaxhighlight>
 
=={{header|Racket}}==
<syntaxhighlight lang="racket">
#lang racket
(require math)
 
(define (spiral rows columns)
(define (index x y) (+ (* x columns) y))
(do ((N (* rows columns))
(spiral (make-vector (* rows columns) #f))
(dx 1) (dy 0) (x 0) (y 0)
(i 0 (+ i 1)))
((= i N) spiral)
(vector-set! spiral (index y x) i)
(let ((nx (+ x dx)) (ny (+ y dy)))
(cond
((and (< -1 nx columns)
(< -1 ny rows)
(not (vector-ref spiral (index ny nx))))
(set! x nx)
(set! y ny))
(else
(set!-values (dx dy) (values (- dy) dx))
(set! x (+ x dx))
(set! y (+ y dy)))))))
 
(vector->matrix 4 4 (spiral 4 4))
</syntaxhighlight>
{{out}}
<syntaxhighlight lang="racket">
(mutable-array #[#[0 1 2 3] #[11 12 13 4] #[10 15 14 5] #[9 8 7 6]])
</syntaxhighlight>
 
=={{header|Raku}}==
(formerly Perl 6)
===Object-oriented Solution===
Suppose we set up a Turtle class like this:
<syntaxhighlight lang="raku" line>class Turtle {
my @dv = [0,-1], [1,-1], [1,0], [1,1], [0,1], [-1,1], [-1,0], [-1,-1];
my $points = 8; # 'compass' points of neighbors on grid: north=0, northeast=1, east=2, etc.
 
has @.loc = 0,0;
has $.dir = 0;
has %.world;
has $.maxegg;
has $.range-x;
has $.range-y;
 
method turn-left ($angle = 90) { $!dir -= $angle / 45; $!dir %= $points; }
method turn-right($angle = 90) { $!dir += $angle / 45; $!dir %= $points; }
 
method lay-egg($egg) {
%!world{~@!loc} = $egg;
$!maxegg max= $egg;
$!range-x minmax= @!loc[0];
$!range-y minmax= @!loc[1];
}
 
method look($ahead = 1) {
my $there = @!loc »+« @dv[$!dir] »*» $ahead;
%!world{~$there};
}
 
method forward($ahead = 1) {
my $there = @!loc »+« @dv[$!dir] »*» $ahead;
@!loc = @($there);
}
 
method showmap() {
my $form = "%{$!maxegg.chars}s";
my $endx = $!range-x.max;
for $!range-y.list X $!range-x.list -> ($y, $x) {
print (%!world{"$x $y"} // '').fmt($form);
print $x == $endx ?? "\n" !! ' ';
}
}
}
 
# Now we can build the spiral in the normal way from outside-in like this:
makespiral <- function(spirald) {
 
series <- vector("numeric", spirald^2)
sub MAIN(Int $size = 5) {
series[] <- 1
my $t = Turtle.new(dir => 2);
l <- spirald-1; p <- spirald+1
my $counter = 0;
s <- 1
$t.forward(-1);
while(l > 0) {
for 0..^ $size -> $ {
series[p:(p+l-1)] <- series[p:(p+l-1)] * spirald*s
$t.forward;
series[(p+l):(p+l*2-1)] <- -s*series[(p+l):(p+l*2-1)]
p <$t.lay- p egg($counter+ l*2+);
}
l <- l - 1; s <- -s
for $size-1 ... 1 -> $run {
}
$t.turn-right;
matrix(grade(runsum(series)), spirald, spirald, byrow=TRUE)
$t.forward, $t.lay-egg($counter++) for 0..^$run;
$t.turn-right;
$t.forward, $t.lay-egg($counter++) for 0..^$run;
}
$t.showmap;
}</syntaxhighlight>
 
Or we can build the spiral from inside-out like this:
print(makespiral(5))</lang>
 
===Recursive Solution===
<syntaxhighlight lang="raku" line>sub MAIN(Int $size = 5) {
<lang R>#more general function, v is assumed to be a vector
my $t = Turtle.new(dir => ($size %% 2 ?? 4 !! 0));
spiralv<-function(v){
my $counter = $size * $size;
n<-sqrt(length(v))
while $counter {
if(n!=floor(n)) stop(simpleError("length of v should be a square of an integer"))
$t.lay-egg(--$counter);
if(n==0) stop(simpleError("v should be of positive length"))
$t.turn-left;
if(n==1) M<-matrix(v,1,1)
$t.turn-right if $t.look;
else M<-rbind(v[1:n],cbind(spiralv(v[(2*n):(n^2)])[(n-1):1,(n-1):1],v[(n+1):(2*n-1)]))
$t.forward;
M
}
$t.showmap;
#wrapper
}</syntaxhighlight>
spiral<-function(n){spiralv(0:(n^2-1))}
Note that with these "turtle graphics" we don't actually have to care about the coordinate system, since the <code>showmap</code> method can show whatever rectangle was modified by the turtle. So unlike the standard inside-out algorithm, we don't have to find the center of the matrix first.
#check:
spiral(5)</lang>
 
===Procedural Solution===
=={{header|REXX}}==
<syntaxhighlight lang="raku" line>sub spiral_matrix ( $n ) {
Logic stolen (mostly) from the [[#Fortran|Fortran]] example.
my @sm;
<lang rexx>/*REXX program to show a spiral in a square array (of any size). */
my $len = $n;
my $pos = 0;
 
for ^($n/2).ceiling -> $i {
arg size . /*get the array size from arg. */
my $j = $i + 1;
if size=='' then size=5 /*if no argument, use the default*/
my $e = $n - $j;
tot=size**2 /*total # of elements in spiral. */
k=size /*K is the counter for the sprial*/
row=1 /*start with row one. */
col=0 /*start with col zero. */
n=0 /*start the sprial at 0 (zero).*/
 
@sm[$i ][$i + $_] = $pos++ for ^( $len); # Top
/*─────────────────────────────────────build the spiral─────────────────*/
do n=0 for@sm[$j k;+ $_][$e ] col=col $pos++1; for @.col.row=n ^(--$len); # endRight
@sm[$e ][$i + $_] = $pos++ for reverse ^( $len); # Bottom
do until n>=tot
@sm[$j + $_][$i ] = $pos++ for reverse ^(--$len); # Left
k=k-1
}
do n=n for k; row=row+1; @.col.row=n; end
 
do n=n for k; col=col-1; @.col.row=n; end
if n>=tot thenreturn leave@sm;
}
k=k-1
 
do n=n for k; row=row-1; @.col.row=n; end
say .fmt('%3d') for spiral_matrix(5);</syntaxhighlight>
do n=n for k; col=col+1; @.col.row=n; end
{{out}}
end
<pre> 0 1 2 3 4
/*─────────────────────────────────────display the spiral───────────────*/
15 do16 col=1 for17 size; 18 _='' 5
14 23 do24 row=1 for19 size 6
13 22 21 20 7
_=_ right(@.row.col,length(tot))
12 11 10 9 8</pre>
end
 
say substr(_,2)
=={{header|REXX}}==
end</lang>
Original logic borrowed (mostly) from the [[#Fortran|Fortran]] example.
Output (using the default array size of 5):
===static column width===
<syntaxhighlight lang="rexx">/*REXX program displays a spiral in a square array (of any size) starting at START. */
parse arg size start . /*obtain optional arguments from the CL*/
if size =='' | size =="," then size =5 /*Not specified? Then use the default.*/
if start=='' | start=="," then start=0 /*Not specified? Then use the default.*/
tot=size**2; L=length(tot + start) /*total number of elements in spiral. */
k=size /*K: is the counter for the spiral. */
row=1; col=0 /*start spiral at row 1, column 0. */
/* [↓] construct the numbered spiral. */
do n=0 for k; col=col + 1; @.col.row=n + start; end; if k==0 then exit
/* [↑] build the first row of spiral. */
do until n>=tot /*spiral matrix.*/
do one=1 to -1 by -2 until n>=tot; k=k-1 /*perform twice.*/
do n=n for k; row=row + one; @.col.row=n + start; end /*for the row···*/
do n=n for k; col=col - one; @.col.row=n + start; end /* " " col···*/
end /*one*/ /* ↑↓ direction.*/
end /*until n≥tot*/ /* [↑] done with the matrix spiral. */
/* [↓] display spiral to the screen. */
do r=1 for size; _= right(@.1.r, L) /*construct display row by row. */
do c=2 for size -1; _=_ right(@.c.r, L) /*construct a line for the display. */
end /*col*/ /* [↑] line has an extra leading blank*/
say _ /*display a line (row) of the spiral. */
end /*row*/ /*stick a fork in it, we're all done. */</syntaxhighlight>
{{out|output|text=&nbsp; using the default array size of: &nbsp; '''5'''}}
<pre>
0 1 2 3 4
Line 2,146 ⟶ 4,734:
12 11 10 9 8
</pre>
{{out|output|text=&nbsp; using an array size &nbsp; and &nbsp; start value of: &nbsp; '''10''' &nbsp; '''-70000'''}}
Output (using an array size of 36):
<pre>
<pre style="height:30ex;overflow:scroll">
-70000 -69999 -69998 -69997 -69996 -69995 -69994 -69993 -69992 -69991
-69965 -69964 -69963 -69962 -69961 -69960 -69959 -69958 -69957 -69990
-69966 -69937 -69936 -69935 -69934 -69933 -69932 -69931 -69956 -69989
-69967 -69938 -69917 -69916 -69915 -69914 -69913 -69930 -69955 -69988
-69968 -69939 -69918 -69905 -69904 -69903 -69912 -69929 -69954 -69987
-69969 -69940 -69919 -69906 -69901 -69902 -69911 -69928 -69953 -69986
-69970 -69941 -69920 -69907 -69908 -69909 -69910 -69927 -69952 -69985
-69971 -69942 -69921 -69922 -69923 -69924 -69925 -69926 -69951 -69984
-69972 -69943 -69944 -69945 -69946 -69947 -69948 -69949 -69950 -69983
-69973 -69974 -69975 -69976 -69977 -69978 -69979 -69980 -69981 -69982
</pre>
{{out|output|text=&nbsp; (shown at <sup>3</sup>/<sub>4</sub> size) &nbsp; using an array size of: &nbsp; '''36'''
<b>
<pre style="font-size:75%">
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35
139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 36
Line 2,184 ⟶ 4,786:
106 239 238 237 236 235 234 233 232 231 230 229 228 227 226 225 224 223 222 221 220 219 218 217 216 215 214 213 212 211 210 209 208 207 206 69
105 104 103 102 101 100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70
</pre>
</b>
 
===minimum column width===
This REXX version automatically adjusts the width of the spiral matrix columns to minimize the area of the matrix display (so more elements may be shown on a display screen).
<syntaxhighlight lang="rexx">/*REXX program displays a spiral in a square array (of any size) starting at START. */
parse arg size start . /*obtain optional arguments from the CL*/
if size =='' | size =="," then size =5 /*Not specified? Then use the default.*/
if start=='' | start=="," then start=0 /*Not specified? Then use the default.*/
tot=size**2; L=length(tot + start) /*total number of elements in spiral. */
k=size /*K: is the counter for the spiral. */
row=1; col=0 /*start spiral at row 1, column 0. */
/* [↓] construct the numbered spiral. */
do n=0 for k; col=col + 1; @.col.row=n + start; end; if k==0 then exit
/* [↑] build the first row of spiral. */
do until n>=tot /*spiral matrix.*/
do one=1 to -1 by -2 until n>=tot; k=k - 1 /*perform twice.*/
do n=n for k; row=row + one; @.col.row=n + start; end /*for the row···*/
do n=n for k; col=col - one; @.col.row=n + start; end /* " " col···*/
end /*one*/ /* ↑↓ direction.*/
end /*until n≥tot*/ /* [↑] done with the matrix spiral. */
!.=0 /* [↓] display spiral to the screen. */
do two=0 for 2 /*1st time? Find max column and width.*/
do r=1 for size; _= /*construct display row by row. */
do c=1 for size; x=@.c.r /*construct a line column by column. */
if two then _=_ right(x, !.c) /*construct a line for the display. */
else !.c=max(!.c, length(x)) /*find the maximum width of the column.*/
end /*c*/ /* [↓] line has an extra leading blank*/
if two then say substr(_, 2) /*this SUBSTR ignores the first blank. */
end /*r*/
end /*two*/ /*stick a fork in it, we're all done. */</syntaxhighlight>
{{out|output|text=&nbsp; (shown at <sup>3</sup>/<sub>4</sub> size) &nbsp; using an array size of: &nbsp; '''36'''
<b>
<pre style="font-size:75%">
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35
139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 36
138 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 174 37
137 270 395 396 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 304 175 38
136 269 394 511 512 513 514 515 516 517 518 519 520 521 522 523 524 525 526 527 528 529 530 531 532 533 534 535 536 537 538 539 426 305 176 39
135 268 393 510 619 620 621 622 623 624 625 626 627 628 629 630 631 632 633 634 635 636 637 638 639 640 641 642 643 644 645 540 427 306 177 40
134 267 392 509 618 719 720 721 722 723 724 725 726 727 728 729 730 731 732 733 734 735 736 737 738 739 740 741 742 743 646 541 428 307 178 41
133 266 391 508 617 718 811 812 813 814 815 816 817 818 819 820 821 822 823 824 825 826 827 828 829 830 831 832 833 744 647 542 429 308 179 42
132 265 390 507 616 717 810 895 896 897 898 899 900 901 902 903 904 905 906 907 908 909 910 911 912 913 914 915 834 745 648 543 430 309 180 43
131 264 389 506 615 716 809 894 971 972 973 974 975 976 977 978 979 980 981 982 983 984 985 986 987 988 989 916 835 746 649 544 431 310 181 44
130 263 388 505 614 715 808 893 970 1039 1040 1041 1042 1043 1044 1045 1046 1047 1048 1049 1050 1051 1052 1053 1054 1055 990 917 836 747 650 545 432 311 182 45
129 262 387 504 613 714 807 892 969 1038 1099 1100 1101 1102 1103 1104 1105 1106 1107 1108 1109 1110 1111 1112 1113 1056 991 918 837 748 651 546 433 312 183 46
128 261 386 503 612 713 806 891 968 1037 1098 1151 1152 1153 1154 1155 1156 1157 1158 1159 1160 1161 1162 1163 1114 1057 992 919 838 749 652 547 434 313 184 47
127 260 385 502 611 712 805 890 967 1036 1097 1150 1195 1196 1197 1198 1199 1200 1201 1202 1203 1204 1205 1164 1115 1058 993 920 839 750 653 548 435 314 185 48
126 259 384 501 610 711 804 889 966 1035 1096 1149 1194 1231 1232 1233 1234 1235 1236 1237 1238 1239 1206 1165 1116 1059 994 921 840 751 654 549 436 315 186 49
125 258 383 500 609 710 803 888 965 1034 1095 1148 1193 1230 1259 1260 1261 1262 1263 1264 1265 1240 1207 1166 1117 1060 995 922 841 752 655 550 437 316 187 50
124 257 382 499 608 709 802 887 964 1033 1094 1147 1192 1229 1258 1279 1280 1281 1282 1283 1266 1241 1208 1167 1118 1061 996 923 842 753 656 551 438 317 188 51
123 256 381 498 607 708 801 886 963 1032 1093 1146 1191 1228 1257 1278 1291 1292 1293 1284 1267 1242 1209 1168 1119 1062 997 924 843 754 657 552 439 318 189 52
122 255 380 497 606 707 800 885 962 1031 1092 1145 1190 1227 1256 1277 1290 1295 1294 1285 1268 1243 1210 1169 1120 1063 998 925 844 755 658 553 440 319 190 53
121 254 379 496 605 706 799 884 961 1030 1091 1144 1189 1226 1255 1276 1289 1288 1287 1286 1269 1244 1211 1170 1121 1064 999 926 845 756 659 554 441 320 191 54
120 253 378 495 604 705 798 883 960 1029 1090 1143 1188 1225 1254 1275 1274 1273 1272 1271 1270 1245 1212 1171 1122 1065 1000 927 846 757 660 555 442 321 192 55
119 252 377 494 603 704 797 882 959 1028 1089 1142 1187 1224 1253 1252 1251 1250 1249 1248 1247 1246 1213 1172 1123 1066 1001 928 847 758 661 556 443 322 193 56
118 251 376 493 602 703 796 881 958 1027 1088 1141 1186 1223 1222 1221 1220 1219 1218 1217 1216 1215 1214 1173 1124 1067 1002 929 848 759 662 557 444 323 194 57
117 250 375 492 601 702 795 880 957 1026 1087 1140 1185 1184 1183 1182 1181 1180 1179 1178 1177 1176 1175 1174 1125 1068 1003 930 849 760 663 558 445 324 195 58
116 249 374 491 600 701 794 879 956 1025 1086 1139 1138 1137 1136 1135 1134 1133 1132 1131 1130 1129 1128 1127 1126 1069 1004 931 850 761 664 559 446 325 196 59
115 248 373 490 599 700 793 878 955 1024 1085 1084 1083 1082 1081 1080 1079 1078 1077 1076 1075 1074 1073 1072 1071 1070 1005 932 851 762 665 560 447 326 197 60
114 247 372 489 598 699 792 877 954 1023 1022 1021 1020 1019 1018 1017 1016 1015 1014 1013 1012 1011 1010 1009 1008 1007 1006 933 852 763 666 561 448 327 198 61
113 246 371 488 597 698 791 876 953 952 951 950 949 948 947 946 945 944 943 942 941 940 939 938 937 936 935 934 853 764 667 562 449 328 199 62
112 245 370 487 596 697 790 875 874 873 872 871 870 869 868 867 866 865 864 863 862 861 860 859 858 857 856 855 854 765 668 563 450 329 200 63
111 244 369 486 595 696 789 788 787 786 785 784 783 782 781 780 779 778 777 776 775 774 773 772 771 770 769 768 767 766 669 564 451 330 201 64
110 243 368 485 594 695 694 693 692 691 690 689 688 687 686 685 684 683 682 681 680 679 678 677 676 675 674 673 672 671 670 565 452 331 202 65
109 242 367 484 593 592 591 590 589 588 587 586 585 584 583 582 581 580 579 578 577 576 575 574 573 572 571 570 569 568 567 566 453 332 203 66
108 241 366 483 482 481 480 479 478 477 476 475 474 473 472 471 470 469 468 467 466 465 464 463 462 461 460 459 458 457 456 455 454 333 204 67
107 240 365 364 363 362 361 360 359 358 357 356 355 354 353 352 351 350 349 348 347 346 345 344 343 342 341 340 339 338 337 336 335 334 205 68
106 239 238 237 236 235 234 233 232 231 230 229 228 227 226 225 224 223 222 221 220 219 218 217 216 215 214 213 212 211 210 209 208 207 206 69
105 104 103 102 101 100 99 98 97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70
</pre>
</b>
 
=={{header|Ring}}==
<syntaxhighlight lang="ring">
# Project : Spiral matrix
 
load "guilib.ring"
load "stdlib.ring"
new qapp
{
win1 = new qwidget() {
setwindowtitle("Spiral matrix")
setgeometry(100,100,600,400)
n = 5
result = newlist(n,n)
spiral = newlist(n,n)
k = 1
top = 1
bottom = n
left = 1
right = n
while (k <= n*n)
for i= left to right
result[top][i] = k
k = k + 1
next
top = top + 1
for i = top to bottom
result[i][right] = k
k = k + 1
next
right = right - 1
for i = right to left step -1
result[bottom][i] = k
k = k + 1
next
bottom = bottom - 1
for i = bottom to top step -1
result[i][left] = k
k = k + 1
next
left = left + 1
end
for m = 1 to n
for p = 1 to n
spiral[p][m] = new qpushbutton(win1) {
x = 150+m*40
y = 30 + p*40
setgeometry(x,y,40,40)
settext(string(result[m][p]))
}
next
next
show()
}
exec()
}
</syntaxhighlight>
Output:
 
[http://keptarhely.eu/view.php?file=20220218v00xuh6r2.jpeg Spiral matrix]
 
=={{header|RPL}}==
{{works with|RPL|HP48-R}}
« { 0 1 } → n step
« { 1 1 } n DUP 2 →LIST -1 CON <span style="color:grey">@ empty cell = -1</span>
1 n SQ '''FOR''' j
OVER j PUT
DUP2 SWAP step ADD
'''IF IFERR''' GET '''THEN''' DROP2 1 '''ELSE''' -1 ≠ '''END''' <span style="color:grey">@ if next step is out of border or an already written cell</span>
'''THEN''' step REVLIST { 1 -1 } * 'step' STO '''END''' <span style="color:grey">@ then turn right</span>
SWAP step ADD SWAP
'''NEXT'''
» » '<span style="color:blue">SPIRAL</span>' STO
 
5 <span style="color:blue">SPIRAL</span>
{{out}}
<pre>
1: [[1 2 3 4 5]
[16 17 18 19 6]
[15 24 25 20 7]
[14 23 22 21 8]
[13 12 11 10 9]]
</pre>
 
=={{header|Ruby}}==
Using the print_matrix method from [[Reduced row echelon form#Ruby]]
{{trans|Python}}
<langsyntaxhighlight lang="ruby">def spiral(n)
spiral = Array.new(n) {Array.new(n, nil)} # n x n array of nils
runs = n.downto(0).each_cons(2).to_a.flatten # n==5; [5,4,4,3,3,2,2,1,1,0]
delta = [[1,0], [0,1], [-1,0], [0,-1]].eachcycle
x, y, value = -1, 0, -1
for run in runs
dx, dy = begindelta.next
run.times { spiral[y+=dy][x+=dx] = (value+=1) }
delta.next
rescue StopIteration
delta.rewind
retry
end
run.times do |i|
x += dx
y += dy
value += 1
spiral[y][x] = value
end
end
spiral
end
 
def print_matrix(m)
print_matrix spiral(5)</lang>
width = m.flatten.map{|x| x.to_s.size}.max
<pre> 0 1 2 3 4
m.each {|row| puts row.map {|x| "%#{width}s " % x}.join}
end
 
print_matrix spiral(5)</syntaxhighlight>
{{out}}
<pre>
0 1 2 3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8</pre>
</pre>
 
The other way
{{trans|D}}
<syntaxhighlight lang="ruby">n = 5
m = Array.new(n){Array.new(n)}
pos, side = -1, n
for i in 0 .. (n-1)/2
(0...side).each{|j| m[i][i+j] = (pos+=1) }
(1...side).each{|j| m[i+j][n-1-i] = (pos+=1) }
side -= 2
side.downto(0) {|j| m[n-1-i][i+j] = (pos+=1) }
side.downto(1) {|j| m[i+j][i] = (pos+=1) }
end
 
fmt = "%#{(n*n-1).to_s.size}d " * n
puts m.map{|row| fmt % row}</syntaxhighlight>
 
Output as above.
 
 
It processes the Array which is for work without creating it.
<syntaxhighlight lang="ruby">def spiral_matrix(n)
x, y, dx, dy = -1, 0, 0, -1
fmt = "%#{(n*n-1).to_s.size}d " * n
n.downto(1).flat_map{|x| [x, x-1]}.flat_map{|run|
dx, dy = -dy, dx # turn 90
run.times.map { [y+=dy, x+=dx] }
}.each_with_index.sort.map(&:last).each_slice(n){|row| puts fmt % row}
end
 
spiral_matrix(5)</syntaxhighlight>
 
=={{header|Rust}}==
<syntaxhighlight lang="rust">const VECTORS: [(isize, isize); 4] = [(1, 0), (0, 1), (-1, 0), (0, -1)];
 
pub fn spiral_matrix(size: usize) -> Vec<Vec<u32>> {
let mut matrix = vec![vec![0; size]; size];
let mut movement = VECTORS.iter().cycle();
let (mut x, mut y, mut n) = (-1, 0, 1..);
 
for (move_x, move_y) in std::iter::once(size)
.chain((1..size).rev().flat_map(|n| std::iter::repeat(n).take(2)))
.flat_map(|steps| std::iter::repeat(movement.next().unwrap()).take(steps))
{
x += move_x;
y += move_y;
matrix[y as usize][x as usize] = n.next().unwrap();
}
 
matrix
}
 
fn main() {
for i in spiral_matrix(4).iter() {
for j in i.iter() {
print!("{:>2} ", j);
}
println!();
}
}</syntaxhighlight>
{{out}}
<pre>
0 1 2 3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8
</pre>
 
=={{header|Scala}}==
<langsyntaxhighlight lang="scala">class Folder(){
var dir = (1,0)
var pos = (-1,0)
Line 2,240 ⟶ 5,061:
folds.foreach {len => fold(seq.take(len),array); seq = seq.drop(len)}
array
}</langsyntaxhighlight>
Explanation: if you see the sequence of numbers to spiral around as a tape to fold around, you can see this pattern on the lenght of tape segment to fold in each step:
:<math>N,\ N-1,\ N-1,\ \ldots,\ 1,\ 1</math>
Line 2,249 ⟶ 5,070:
It's simple to make this generic, changing start position, initial direction, etc.
The code could be more compact, but I'm leaving it like this for clarity.
 
=={{header|Scilab}}==
{{trans|Octave}}
<syntaxhighlight lang="text">function a = spiral(n)
a = ones(n*n, 1)
v = ones(n, 1)
u = -n*v;
i = n
for k = n-1:-1:1
j = 1:k
u(j) = -u(j)
a(j+i) = u(j)
v(j) = -v(j)
a(j+(i+k)) = v(j)
i = i+2*k
end
a(cumsum(a)) = (1:n*n)'
a = matrix(a, n, n)'-1
endfunction
 
-->spiral(5)
ans =
0. 1. 2. 3. 4.
15. 16. 17. 18. 5.
14. 23. 24. 19. 6.
13. 22. 21. 20. 7.
12. 11. 10. 9. 8.</syntaxhighlight>
 
=={{header|Seed7}}==
<langsyntaxhighlight lang="seed7">$ include "seed7_05.s7i";
 
const type: matrix is array array integer;
Line 2,302 ⟶ 5,151:
begin
writeMatrix(spiral(5));
end func;</langsyntaxhighlight>
{{out}}
Output:
<pre>
1 2 3 4 5
Line 2,311 ⟶ 5,160:
13 12 11 10 9
</pre>
 
=={{header|Sidef}}==
{{trans|Perl}}
<syntaxhighlight lang="ruby">func spiral(n) {
var (x, y, dx, dy, a) = (0, 0, 1, 0, [])
{ |i|
a[y][x] = i
var (nx, ny) = (x+dx, y+dy)
( if (dx == 1 && (nx == n || a[ny][nx]!=nil)) { [ 0, 1] }
elsif (dy == 1 && (ny == n || a[ny][nx]!=nil)) { [-1, 0] }
elsif (dx == -1 && (nx < 0 || a[ny][nx]!=nil)) { [ 0, -1] }
elsif (dy == -1 && (ny < 0 || a[ny][nx]!=nil)) { [ 1, 0] }
else { [dx, dy] }
) » (\dx, \dy)
x = x+dx
y = y+dy
} << (1 .. n**2)
return a
}
 
spiral(5).each { |row|
row.map {"%3d" % _}.join(' ').say
}</syntaxhighlight>
{{out}}
<pre>
1 2 3 4 5
16 17 18 19 6
15 24 25 20 7
14 23 22 21 8
13 12 11 10 9
</pre>
 
=={{header|Stata}}==
<syntaxhighlight lang="stata">function spiral_mat(n) {
a = J(n*n, 1, 1)
u = J(n, 1, -n)
v = J(n, 1, 1)
for (k=(i=n)-1; k>=1; i=i+2*k--) {
j = 1..k
a[j:+i] = u[j] = -u[j]
a[j:+(i+k)] = v[j] = -v[j]
}
return(rowshape(invorder(runningsum(a)),n):-1)
}
 
spiral_mat(5)
1 2 3 4 5
+--------------------------+
1 | 0 1 2 3 4 |
2 | 15 16 17 18 5 |
3 | 14 23 24 19 6 |
4 | 13 22 21 20 7 |
5 | 12 11 10 9 8 |
+--------------------------+</syntaxhighlight>
 
=={{header|Tcl}}==
Using <code>print_matrix</code> from [[Matrix Transpose#Tcl]]
<langsyntaxhighlight lang="tcl">package require Tcl 8.5
namespace path {::tcl::mathop}
proc spiral size {
Line 2,347 ⟶ 5,250:
}
 
print_matrix [spiral 5]</langsyntaxhighlight>
<pre> 0 1 2 3 4
15 16 17 18 5
Line 2,353 ⟶ 5,256:
13 22 21 20 7
12 11 10 9 8</pre>
 
=={{header|TI-83 BASIC}}==
{{trans|BBC Basic}}
<syntaxhighlight lang="ti83b">5->N
DelVar [F]
{N,N}→dim([F])
1→A: N→B
1→C: N→D
0→E: E→G
1→I: 1→J
For(K,1,N*N)
K-1→[F](I,J)
If E=0: Then
If J<D: Then
J+1→J
Else: 1→G
I+1→I: A+1→A
End
End
If E=1: Then
If I<B: Then
I+1→I
Else: 2→G
J-1→J: D-1→D
End
End
If E=2: Then
If J>C: Then
J-1→J
Else: 3→G
I-1→I: B-1→B
End
End
If E=3: Then
If I>A: Then
I-1→I
Else: 0→G
J+1→J: C+1→C
End
End
G→E
End
[F]</syntaxhighlight>
{{out}}
<pre>[[0 1 2 3 4]
[15 16 17 18 5]
[14 23 24 19 6]
[13 22 21 20 7]
[12 11 10 9 8]]</pre>
 
=={{header|TSE SAL}}==
<syntaxhighlight lang="tse sal">
 
// library: math: create: array: spiral: inwards <description></description> <version control></version control> <version>1.0.0.0.15</version> (filenamemacro=creamasi.s) [<Program>] [<Research>] [kn, ri, mo, 31-12-2012 01:15:43]
PROC PROCMathCreateArraySpiralInwards( INTEGER nI )
// e.g. PROC Main()
// e.g. STRING s1[255] = "5"
// e.g. IF ( NOT ( Ask( "math: create: array: spiral: inwards: nI = ", s1, _EDIT_HISTORY_ ) ) AND ( Length( s1 ) > 0 ) ) RETURN() ENDIF
// e.g. PROCMathCreateArraySpiralInwards( Val( s1 ) )
// e.g. END
// e.g.
// e.g. <F12> Main()
//
INTEGER columnEndI = 0
//
INTEGER columnBeginI = nI - 1
//
INTEGER rowEndI = 0
//
INTEGER rowBeginI = nI - 1
//
INTEGER columnI = 0
//
INTEGER rowI = 0
//
INTEGER minI = 0
INTEGER maxI = nI * nI - 1
INTEGER I = 0
//
INTEGER columnWidthI = Length( Str( nI * nI - 1 ) ) + 1
//
INTEGER directionRightI = 0
INTEGER directionLeftI = 1
INTEGER directionDownI = 2
INTEGER directionUpI = 3
//
INTEGER directionI = directionRightI
//
FOR I = minI TO maxI
//
SetGlobalInt( Format( "MatrixS", columnI, ",", rowI ), I )
// SetGlobalInt( Format( "MatrixS", columnI, ",", rowI ), I )
//
PutStrXY( ( Query( ScreenCols ) / 8 ) + columnI * columnWidthI, ( Query( ScreenRows ) / 8 ) + rowI, Str( I ), Color( BRIGHT RED ON WHITE ) )
// PutStrXY( ( Query( ScreenCols ) / 8 ) + columnI * columnWidthI, ( Query( ScreenRows ) / 8 ) + rowI, Str( I + 1 ), Color( BRIGHT RED ON WHITE ) )
//
CASE directionI
//
WHEN directionRightI
//
IF ( columnI < columnBeginI )
//
columnI = columnI + 1
//
ELSE
//
directionI = directionDownI
//
rowI = rowI + 1
//
rowEndI = rowEndI + 1
//
ENDIF
//
WHEN directionDownI
//
IF ( rowI < rowBeginI )
//
rowI = rowI + 1
//
ELSE
//
directionI = directionLeftI
//
columnI = columnI - 1
//
columnBeginI = columnBeginI - 1
//
ENDIF
//
WHEN directionLeftI
//
IF ( columnI > columnEndI )
//
columnI = columnI - 1
//
ELSE
//
directionI = directionUpI
//
rowI = rowI - 1
//
rowBeginI = rowBeginI - 1
//
ENDIF
//
WHEN directionUpI
//
IF ( rowI > rowEndI )
//
rowI = rowI - 1
//
ELSE
//
directionI = directionRightI
//
columnI = columnI + 1
//
columnEndI = columnEndI + 1
//
ENDIF
//
OTHERWISE
//
Warn( Format( "PROCMathCreateArraySpiralInwards(", " ", "case", " ", ":", " ", Str( directionI ), ": not known" ) )
//
RETURN()
//
ENDCASE
//
ENDFOR
//
END
 
PROC Main()
STRING s1[255] = "5"
IF ( NOT ( Ask( "math: create: array: spiral: inwards: nI = ", s1, _EDIT_HISTORY_ ) ) AND ( Length( s1 ) > 0 ) ) RETURN() ENDIF
PROCMathCreateArraySpiralInwards( Val( s1 ) )
END
 
</syntaxhighlight>
 
=={{header|uBasic/4tH}}==
{{trans|C}}
This recursive version is quite compact.
<syntaxhighlight lang="text">Input "Width: ";w
Input "Height: ";h
Print
 
For i = 0 To h-1
For j = 0 To w-1
Print Using "__#"; FUNC(_Spiral(w,h,j,i));
Next
Print
Next
End
 
 
_Spiral Param(4)
If d@ Then
Return (a@ + FUNC(_Spiral(b@-1, a@, d@ - 1, a@ - c@ - 1)))
Else
Return (c@)
EndIf</syntaxhighlight>
 
=={{header|Ursala}}==
Helpful hints from the [[#J|J]] example are gratefully acknowledged. The spiral function works for any n, and results are shown for n equal to 5, 6, and 7. The results are represented as lists of lists rather than arrays.
<langsyntaxhighlight Ursalalang="ursala">#import std
#import nat
#import int
Line 2,368 ⟶ 5,475:
#cast %nLLL
 
examples = spiral* <5,6,7></langsyntaxhighlight>
{{out}}
output:
<pre>
<
Line 2,393 ⟶ 5,500:
<19,36,35,34,33,32,11>,
<18,17,16,15,14,13,12>>></pre>
 
=={{header|VBScript}}==
{{trans|BBC BASIC}}
<syntaxhighlight lang="vb">
Function build_spiral(n)
botcol = 0 : topcol = n - 1
botrow = 0 : toprow = n - 1
'declare a two dimensional array
Dim matrix()
ReDim matrix(topcol,toprow)
dir = 0 : col = 0 : row = 0
'populate the array
For i = 0 To n*n-1
matrix(col,row) = i
Select Case dir
Case 0
If col < topcol Then
col = col + 1
Else
dir = 1 : row = row + 1 : botrow = botrow + 1
End If
Case 1
If row < toprow Then
row = row + 1
Else
dir = 2 : col = col - 1 : topcol = topcol - 1
End If
Case 2
If col > botcol Then
col = col - 1
Else
dir = 3 : row = row - 1 : toprow = toprow - 1
End If
Case 3
If row > botrow Then
row = row - 1
Else
dir = 0 : col = col + 1 : botcol = botcol + 1
End If
End Select
Next
'print the array
For y = 0 To n-1
For x = 0 To n-1
WScript.StdOut.Write matrix(x,y) & vbTab
Next
WScript.StdOut.WriteLine
Next
End Function
 
build_spiral(CInt(WScript.Arguments(0)))
</syntaxhighlight>
 
{{Out}}
<pre>
F:\>cscript /nologo build_spiral.vbs 5
0 1 2 3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8
 
F:\>cscript /nologo build_spiral.vbs 7
0 1 2 3 4 5 6
23 24 25 26 27 28 7
22 39 40 41 42 29 8
21 38 47 48 43 30 9
20 37 46 45 44 31 10
19 36 35 34 33 32 11
18 17 16 15 14 13 12
</pre>
 
=={{header|Visual Basic}}==
==={{header|VB6}}===
{{trans|Java}}
This requires VB6.
<langsyntaxhighlight lang="vb">Option Explicit
 
Sub Main()
Line 2,454 ⟶ 5,633:
Debug.Print arr(row, UBound(arr, 2))
Next
End Sub</langsyntaxhighlight>
 
==={{header|VBA}}===
====Solution 1====
{{trans|Java}}
{{works with|VBA/Excel}}
<langsyntaxhighlight lang="vb">Sub spiral()
Dim n As Integer, a As Integer, b As Integer
Dim numCsquares As Integer, sideLen As Integer, currNum As Integer
Line 2,510 ⟶ 5,689:
Next b
Next a
End Sub</langsyntaxhighlight>
 
===Solution 2===
====Solution 2====
<lang vb>Sub spiral(n As Integer)
<syntaxhighlight lang="vb">Sub spiral(n As Integer)
Const FREE = -9 'negative number indicates unoccupied cell
Dim A() As Integer
Line 2,579 ⟶ 5,759:
Next
End Sub</langsyntaxhighlight>
{{out}}
Sample output:
<pre>
spiral 5
Line 2,596 ⟶ 5,776:
16 29 28 27 26 9
15 14 13 12 11 10
</pre>
 
==={{header|Visual Basic .NET}}===
'''Platform:''' [[.NET]]<br>
From VB6. This requires Visual Basic .Net.
<syntaxhighlight lang="vbnet">Module modSpiralArray
Sub Main()
print2dArray(getSpiralArray(5))
End Sub
 
Function getSpiralArray(dimension As Integer) As Object
Dim spiralArray(,) As Integer
Dim numConcentricSquares As Integer
 
ReDim spiralArray(dimension - 1, dimension - 1)
numConcentricSquares = dimension \ 2
If (dimension Mod 2) Then numConcentricSquares = numConcentricSquares + 1
 
Dim j As Integer, sideLen As Integer, currNum As Integer
sideLen = dimension
 
Dim i As Integer
For i = 0 To numConcentricSquares - 1
' do top side
For j = 0 To sideLen - 1
spiralArray(i, i + j) = currNum
currNum = currNum + 1
Next
' do right side
For j = 1 To sideLen - 1
spiralArray(i + j, dimension - 1 - i) = currNum
currNum = currNum + 1
Next
' do bottom side
For j = sideLen - 2 To 0 Step -1
spiralArray(dimension - 1 - i, i + j) = currNum
currNum = currNum + 1
Next
' do left side
For j = sideLen - 2 To 1 Step -1
spiralArray(i + j, i) = currNum
currNum = currNum + 1
Next
sideLen = sideLen - 2
Next
getSpiralArray = spiralArray
End Function
 
Sub print2dArray(arr)
Dim row As Integer, col As Integer, s As String
For row = 0 To UBound(arr, 1)
s = ""
For col = 0 To UBound(arr, 2)
s = s & " " & Right(" " & arr(row, col), 3)
Next
Debug.Print(s)
Next
End Sub
 
End Module
</syntaxhighlight>
 
=={{header|Wren}}==
{{trans|Go}}
{{libheader|Wren-fmt}}
<syntaxhighlight lang="wren">import "./fmt" for Conv, Fmt
 
var n = 5
var top = 0
var left = 0
var bottom = n - 1
var right = n - 1
var sz = n * n
var a = List.filled(sz, 0)
var i = 0
while (left < right) {
// work right, along top
var c = left
while (c <= right) {
a[top*n+c] = i
i = i + 1
c = c + 1
}
top = top + 1
// work down right side
var r = top
while (r <= bottom) {
a[r*n+right] = i
i = i + 1
r = r + 1
}
right = right - 1
if (top == bottom) break
// work left, along bottom
c = right
while (c >= left) {
a[bottom*n+c] = i
i = i + 1
c = c - 1
}
bottom = bottom - 1
r = bottom
// work up left side
while (r >= top) {
a[r*n+left] = i
i = i + 1
r = r - 1
}
left = left + 1
}
// center (last) element
a[top*n+left] = i
 
// print
var w = Conv.itoa(n*n - 1).count
i = 0
for (e in a) {
Fmt.write("$*d ", w, e)
if (i%n == n - 1) System.print()
i = i + 1
}</syntaxhighlight>
 
{{out}}
<pre>
0 1 2 3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8
</pre>
 
=={{header|XPL0}}==
<langsyntaxhighlight XPL0lang="xpl0">def N=5;
int A(N,N);
int I, J, X, Y, Steps, Dir;
Line 2,620 ⟶ 5,929:
until Steps = 0;
Cursor(0,N);
]</langsyntaxhighlight>
 
{{out}}
Output:
<pre>
0 1 2 3 4
Line 2,629 ⟶ 5,938:
13 22 21 20 7
12 11 10 9 8
</pre>
 
=={{header|Z80 Assembly}}==
N is set at beginning of code (valid range 1..150-ish, then you soon run out of memory), sjasmplus syntax, CP/M executable:
<syntaxhighlight lang="z80">; Spiral matrix in Z80 assembly (for CP/M OS - you can use `tnylpo` or `z88dk-ticks` on PC)
OPT --syntax=abf : OUTPUT "spiralmt.com" ; asm syntax for z00m's variant of sjasmplus
ORG $100
spiral_matrix:
ld a,5 ; N matrix size (argument for the code) (valid range: 1..150)
; setup phase
push af
ld l,a
ld h,0
add hl,hl
ld (delta_d),hl ; down-direction address delta = +N*2
neg
ld l,a
ld h,$FF
add hl,hl
ld (delta_u),hl ; up-direction address delta = -N*2
neg
ld hl,matrix
ld de,2 ; delta_r value to move right in matrix
ld bc,0 ; starting value
dec a ; first sequences will be N-1 long
jr z,.finish ; 1x1 doesn't need any sequence, just set last element
call set_sequence ; initial entry sequence has N-1 elements (same as two more)
; main loop - do twice same length sequence, then decrement length, until zero
.loop:
call set_sequence_twice
dec a
jr nz,.loop
.finish: ; whole spiral is set except last element, set it now
ld (hl),c
inc hl
ld (hl),b
; print matrix - reading it by POP HL (destructive, plus some memory ahead of matrix too)
pop de ; d = N
ld (.oldsp+1),sp
ld sp,matrix ; set stack to beginning of matrix (call/push does damage memory ahead)
ld c,d ; c = N (lines counter)
.print_rows:
ld b,d ; b = N (value per row counter)
.print_row:
pop hl
push de
push bc
call print_hl
pop bc
pop de
djnz .print_row
push de
call print_crlf
pop de
dec c
jr nz,.print_rows
.oldsp:
ld sp,0
rst 0 ; return to CP/M
 
print_crlf:
ld e,10
call print_char
ld e,13
jr print_char
print_hl:
ld b,' '
ld e,b
call print_char
ld de,-10000
call extract_digit
ld de,-1000
call extract_digit
ld de,-100
call extract_digit
ld de,-10
call extract_digit
ld a,l
print_digit:
ld b,'0'
add a,b
ld e,a
print_char:
push bc
push hl
ld c,2
call 5
pop hl
pop bc
ret
extract_digit:
ld a,-1
.digit_loop:
inc a
add hl,de
jr c,.digit_loop
sbc hl,de
or a
jr nz,print_digit
ld e,b
jr print_char
 
set_sequence_twice:
call set_sequence
set_sequence:
; A = length, HL = next_to_matrix, DE = delta to advance hl, BC = next_value
push af
.set_loop:
ld (hl),c
inc hl
ld (hl),b
dec hl ; [HL] = BC
add hl,de ; HL += DE
inc bc ; ++BC
dec a
jr nz,.set_loop
push hl ; change DE for next direction (right->down->left->up->right->...)
.d: ld hl,delta_d ; self-modify-code: pointer to next delta
ld e,(hl)
inc hl
ld d,(hl) ; de = address delta for next sequence
inc hl
ld a,low delta_u+2 ; if hl == delta_u+2 => reset it back to delta_r
cp l
jr nz,.next_delta
ld hl,delta_r
.next_delta:
ld (.d+1),hl ; self modify code pointer for next delta value
pop hl
pop af
ret
 
delta_r: dw +2 ; value to add to move right in matrix
delta_d: dw 0 ; value to add to move down in matrix (set to +N*2)
delta_l: dw -2 ; value to add to move left in matrix
delta_u: dw 0 ; value to add to move up in matrix (set to -N*2)
 
matrix:
; following memory is used for NxN matrix of uint16_t values (150x150 needs 45000 bytes)</syntaxhighlight>
 
{{out}}
<pre>
$ tnylpo spiralmt.com
0 1 2 3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8
</pre>
 
=={{header|zkl}}==
{{trans|C}} which turns out to be the same as Ruby
<syntaxhighlight lang="zkl">fcn spiralMatrix(n){
sm:=(0).pump(n,List,(0).pump(n,List,False).copy); //L(L(False,False..), L(F,F,..) ...)
drc:=Walker.cycle(T(0,1,0), T(1,0,1), T(0,-1,0), T(-1,0,1)); // deltas
len:=n; r:=0; c:=-1; z:=-1; while(len>0){ //or do(2*n-1){
dr,dc,dl:=drc.next();
do(len-=dl){ sm[r+=dr][c+=dc]=(z+=1); }
}
sm
}</syntaxhighlight>
<syntaxhighlight lang="zkl">foreach n in (T(5,-1,0,1,2)){
spiralMatrix(n).pump(Console.println,fcn(r){ r.apply("%4d".fmt).concat() });
println("---");
}</syntaxhighlight>
{{out}}
<pre>
0 1 2 3 4
15 16 17 18 5
14 23 24 19 6
13 22 21 20 7
12 11 10 9 8
---
---
---
0
---
0 1
3 2
---
</pre>
1,150

edits