Zig-zag matrix

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Task
Zig-zag matrix
You are encouraged to solve this task according to the task description, using any language you may know.

Produce a zig-zag array. A zig-zag array is a square arrangement of the first N2 integers, where the numbers increase sequentially as you zig-zag along the anti-diagonals of the array.
For a graphical representation, see JPG zigzag (JPG uses such arrays to encode images).

For example, given 5, produce this array:

 0  1  5  6 14
 2  4  7 13 15
 3  8 12 16 21
 9 11 17 20 22
10 18 19 23 24
See also Spiral matrix

Contents

[edit] ActionScript

 
package
{
public class ZigZagMatrix extends Array
{
 
private var height:uint;
private var width:uint;
public var mtx:Array = [];
 
public function ZigZagMatrix(size:uint)
{
this.height = size;
this.width = size;
 
this.mtx = [];
for (var i:uint = 0; i < size; i++) {
this.mtx[i] = [];
}
i = 1;
var j:uint = 1;
for (var e:uint = 0; e < size*size; e++) {
this.mtx[i-1][j-1] = e;
if ((i + j) % 2 == 0) {
// Even stripes
if (j < size) j ++;
else i += 2;
if (i > 1) i --;
} else {
// Odd stripes
if (i < size) i ++;
else j += 2;
if (j > 1) j --;
}
}
}
}
}
 

[edit] Ada

with Ada.Text_IO;  use Ada.Text_IO;
 
procedure Test_Zig_Zag is
 
type Matrix is array (Positive range <>, Positive range <>) of Natural;
function Zig_Zag (Size : Positive) return Matrix is
Data : Matrix (1..Size, 1..Size);
I, J : Integer := 1;
begin
Data (1, 1) := 0;
for Element in 1..Size**2 - 1 loop
if (I + J) mod 2 = 0 then
-- Even stripes
if J < Size then
J := J + 1;
else
I := I + 2;
end if;
if I > 1 then
I := I - 1;
end if;
else
-- Odd stripes
if I < Size then
I := I + 1;
else
J := J + 2;
end if;
if J > 1 then
J := J - 1;
end if;
end if;
Data (I, J) := Element;
end loop;
return Data;
end Zig_Zag;
 
procedure Put (Data : Matrix) is
begin
for I in Data'Range (1) loop
for J in Data'Range (2) loop
Put (Integer'Image (Data (I, J)));
end loop;
New_Line;
end loop;
end Put;
 
begin
Put (Zig_Zag (5));
end Test_Zig_Zag;

The function Zig_Zag generates a square matrix filled as requested by the task.

Output:
 0 1 5 6 14
 2 4 7 13 15
 3 8 12 16 21
 9 11 17 20 22
 10 18 19 23 24

[edit] ALGOL 68

Translation of: D
Works with: ALGOL 68 version Standard - no extensions to language used
Works with: ALGOL 68G version Any - tested with release mk15-0.8b.fc9.i386
Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - tested with release 1.8.8d.fc9.i386
PROC zig zag = (INT n)[,]INT: (
PROC move = (REF INT i, j)VOID: (
IF j < n THEN
i := ( i <= 1 | 1 | i-1 );
j +:= 1
ELSE
i +:= 1
FI
);
 
[n, n]INT a;
INT x:=LWB a, y:=LWB a;
 
FOR v FROM 0 TO n**2-1 DO
a[y, x] := v;
IF ODD (x + y) THEN
move(x, y)
ELSE
move(y, x)
FI
OD;
a
);
 
INT dim = 5;
#IF formatted transput possible THEN
FORMAT d = $z-d$;
FORMAT row = $"("n(dim-1)(f(d)",")f(d)")"$;
FORMAT block = $"("n(dim-1)(f(row)","lx)f(row)")"l$;
 
printf((block, zig zag(dim)))
ELSE#

[,]INT result = zig zag(dim);
FOR i TO dim DO
print((result[i,], new line))
OD
#FI#
Output:
With formatted transput possible, e.g. ALGOL 68G not formatted transput possible, e.g. ELLA ALGOL 68
((  0,  1,  5,  6, 14),
 (  2,  4,  7, 13, 15),
 (  3,  8, 12, 16, 21),
 (  9, 11, 17, 20, 22),
 ( 10, 18, 19, 23, 24))
         +0          +1          +5          +6         +14
         +2          +4          +7         +13         +15
         +3          +8         +12         +16         +21
         +9         +11         +17         +20         +22
        +10         +18         +19         +23         +24

[edit] APL

Works with: Dyalog APL
Translation of: J
      zz   ←  {⍵⍴⎕IO-⍨⍋⊃,/{(2|⍴⍵):⌽⍵⋄⍵}¨(⊂w)/¨⍨w{↓⍵∘.=⍨∪⍵}+/[1]⍵⊤w←⎕IO-⍨⍳×/⍵}   ⍝  General zigzag (any rectangle)
zzSq ← {zz,⍨⍵} ⍝ Square zigzag
zzSq 5
0 1 5 6 14
2 4 7 13 15
3 8 12 16 21
9 11 17 20 22
10 18 19 23 24

[edit] AppleScript

Here's a vector & matrix boundary detection approach to the Zig-zap matrix:

set n to 5 -- Size of zig-zag matrix (n^2 cells).
 
-- Create an empty matrix.
set m to {}
repeat with i from 1 to n
set R to {}
repeat with j from 1 to n
set end of R to 0
end repeat
set end of m to R
end repeat
 
-- Populate the matrix in a zig-zag manner.
set {x, y, v, d} to {1, 1, 0, 1}
repeat while v < (n ^ 2)
if 1 ≤ x and x ≤ n and 1 ≤ y and y ≤ n then
set {m's item y's item x, x, y, v} to {v, x + d, y - d, v + 1}
else if x > n then
set {x, y, d} to {n, y + 2, -d}
else if y > n then
set {x, y, d} to {x + 2, n, -d}
else if x < 1 then
set {x, y, d} to {1, y, -d}
else if y < 1 then
set {x, y, d} to {x, 1, -d}
end if
end repeat
--> R = {{0, 1, 5, 6, 14}, {2, 4, 7, 13, 15}, {3, 8, 12, 16, 21}, {9, 11, 17, 20, 22}, {10, 18, 19, 23, 24}}
 
-- Reformat the matrix into a table for viewing.
repeat with i in m
repeat with j in i
set j's contents to (characters -(length of (n ^ 2 as string)) thru -1 of (" " & j)) as string
end repeat
set end of i to return
end repeat
return return & m as string
But this can be improved upon by building the matrix by populating empty AppleScript lists (it's about 50% faster when n=50):
set n to 5
 
set m to {}
repeat with i from 1 to n
set end of m to {} -- Built a foundation for the matrix out of n empty lists.
end repeat
 
set {v, d, i} to {0, -1, 1}
repeat while v < n ^ 2
if length of m's item i < n then
set {end of m's item i, i, v} to {f(v, n), i + d, v + 1}
if i < 1 then
set {i, d} to {1, -d}
else if i > n then
set {i, d} to {n, -d}
else if i > 1 and (count of m's item (i - 1)) = 1 then
set d to -d
end if
else
set {i, d} to {i + 1, 1}
end if
end repeat
 
-- Handler/function to format the cells on the fly.
on f(v, n)
return (characters -(length of (n ^ 2 as string)) thru -1 of (" " & v)) as string
end f
 
-- Reformat the matrix into a table for viewing.
set text item delimiters to ""
repeat with i in m
set i's contents to (i as string) & return
end repeat
return return & m as string
 
Output:
for both scripts is:
"

  0   1   5   6  14
  2   4   7  13  15
  3   8  12  16  21
  9  11  17  20  22
 10  18  19  23  24
"

[edit] Applesoft BASIC

100 S = 5
110 S2 = S ^ 2 : REM SQUARED
120 H = S2 / 2 : REM HALFWAY
130 S2 = S2 - 1
140 DX = 1 : REM INITIAL
150 DY = 0 : REM DIRECTION
160 N = S - 1
170 DIM A%(N, N)
 
200 FOR I = 0 TO H
210 A%(X, Y) = I
220 A%(N - X, N - Y) = S2 - I
230 X = X + DX
240 Y = Y + DY
250 IF Y = 0 THEN DY = DY + 1 : IF DY THEN DX = -DX
260 IF X = 0 THEN DX = DX + 1 : IF DX THEN DY = -DY
270 NEXT I
 
300 FOR Y = 0 TO N
310 FOR X = 0 TO N
320 IF X THEN PRINT TAB(X * (LEN(STR$(S2)) + 1) + 1);
330 PRINT A%(X, Y);
340 NEXT X
350 PRINT
360 NEXT Y

[edit] AutoHotkey

Translation of: lisp

contributed by Laszlo on the ahk forum.

n = 5                           ; size
v := x := y := 1 ; initial values
Loop % n*n { ; for every array element
a_%x%_%y% := v++ ; assign the next index
If ((x+y)&1) ; odd diagonal
If (x < n) ; while inside the square
y -= y<2 ? 0 : 1, x++ ; move right-up
Else y++ ; on the edge increment y, but not x: to even diagonal
Else ; even diagonal
If (y < n) ; while inside the square
x -= x<2 ? 0 : 1, y++ ; move left-down
Else x++ ; on the edge increment x, but not y: to odd diagonal
}
 
Loop %n% { ; generate printout
x := A_Index ; for each row
Loop %n% ; and for each column
t .= a_%x%_%A_Index% "`t" ; attach stored index
t .= "`n" ; row is complete
}
MsgBox %t% ; show output

[edit] AutoIt

 
#include <Array.au3>
$Array = ZigZag(5)
_ArrayDisplay($Array)
 
Func ZigZag($int)
Local $av_array[$int][$int]
Local $x = 1, $y = 1
For $I = 0 To $int ^ 2 -1
$av_array[$x-1][$y-1] = $I
If Mod(($x + $y), 2) = 0 Then ;Even
if ($y < $int) Then
$y += 1
Else
$x += 2
EndIf
if ($x > 1) Then $x -= 1
Else ; ODD
if ($x < $int) Then
$x += 1
Else
$y += 2
EndIf
If $y > 1 Then $y -= 1
EndIf
Next
Return $av_array
EndFunc ;==>ZigZag
 

[edit] AWK

 
# syntax: GAWK -f ZIG-ZAG_MATRIX.AWK [-v offset={0|1}] [size]
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) }
width = length(size ^ 2 - 1 + offset) + 1
i = j = 1
for (n=0; n<=size^2-1; n++) { # build array
arr[i-1,j-1] = n + offset
if ((i+j) % 2 == 0) {
if (j < size) { j++ } else { i+=2 }
if (i > 1) { i-- }
}
else {
if (i < size) { i++ } else { j+=2 }
if (j > 1) { j-- }
}
}
for (row=0; row<size; row++) { # show array
for (col=0; col<size; col++) {
printf("%*d",width,arr[row,col])
}
printf("\n")
}
exit(0)
}
 
Output:
  0  1  5  6 14
  2  4  7 13 15
  3  8 12 16 21
  9 11 17 20 22
 10 18 19 23 24

[edit] BBC BASIC

      Size% = 5
DIM array%(Size%-1,Size%-1)
 
i% = 1
j% = 1
FOR e% = 0 TO Size%^2-1
array%(i%-1,j%-1) = e%
IF ((i% + j%) AND 1) = 0 THEN
IF j% < Size% j% += 1 ELSE i% += 2
IF i% > 1 i% -= 1
ELSE
IF i% < Size% i% += 1 ELSE j% += 2
IF j% > 1 j% -= 1
ENDIF
NEXT
 
@% = &904
FOR row% = 0 TO Size%-1
FOR col% = 0 TO Size%-1
PRINT array%(row%,col%);
NEXT
PRINT
NEXT row%
Output:
   0   1   5   6  14
   2   4   7  13  15
   3   8  12  16  21
   9  11  17  20  22
  10  18  19  23  24

[edit] C

#include <stdio.h>
#include <stdlib.h>
 
int main(int c, char **v)
{
int i, j, m, n, *s;
 
/* default size: 5 */
if (c < 2 || ((m = atoi(v[1]))) <= 0) m = 5;
 
/* alloc array*/
s = malloc(sizeof(int) * m * m);
 
for (i = n = 0; i < m * 2; i++)
for (j = (i < m) ? 0 : i-m+1; j <= i && j < m; j++)
s[(i&1)? j*(m-1)+i : (i-j)*m+j ] = n++;
 
for (i = 0; i < m * m; putchar((++i % m) ? ' ':'\n'))
printf("%3d", s[i]);
 
/* free(s) */
return 0;
}
Output:
% ./a.out 7
  0  1  5  6 14 15 27
  2  4  7 13 16 26 28
  3  8 12 17 25 29 38
  9 11 18 24 30 37 39
 10 19 23 31 36 40 45
 20 22 32 35 41 44 46
 21 33 34 42 43 47 48

[edit] C++

#include <vector>
#include <memory> // for auto_ptr
#include <cmath> // for the log10 and floor functions
#include <iostream>
#include <iomanip> // for the setw function
 
using namespace std;
 
typedef vector< int > IntRow;
typedef vector< IntRow > IntTable;
 
auto_ptr< IntTable > getZigZagArray( int dimension )
{
auto_ptr< IntTable > zigZagArrayPtr( new IntTable(
dimension, IntRow( dimension ) ) );
 
// fill along diagonal stripes (oriented as "/")
int lastValue = dimension * dimension - 1;
int currDiag = 0;
int loopFrom;
int loopTo;
int i;
int row;
int col;
do
{
if ( currDiag < dimension ) // if doing the upper-left triangular half
{
loopFrom = 0;
loopTo = currDiag;
}
else // doing the bottom-right triangular half
{
loopFrom = currDiag - dimension + 1;
loopTo = dimension - 1;
}
 
for ( i = loopFrom; i <= loopTo; i++ )
{
if ( currDiag % 2 == 0 ) // want to fill upwards
{
row = loopTo - i + loopFrom;
col = i;
}
else // want to fill downwards
{
row = i;
col = loopTo - i + loopFrom;
}
 
( *zigZagArrayPtr )[ row ][ col ] = currNum++;
}
 
currDiag++;
}
while ( currDiag <= lastValue );
 
return zigZagArrayPtr;
}
 
void printZigZagArray( const auto_ptr< IntTable >& zigZagArrayPtr )
{
size_t dimension = zigZagArrayPtr->size();
 
int fieldWidth = static_cast< int >( floor( log10(
static_cast< double >( dimension * dimension - 1 ) ) ) ) + 2;
 
size_t col;
for ( size_t row = 0; row < dimension; row++ )
{
for ( col = 0; col < dimension; col++ )
cout << setw( fieldWidth ) << ( *zigZagArrayPtr )[ row ][ col ];
cout << endl;
}
}
 
int main()
{
printZigZagArray( getZigZagArray( 5 ) );
}

[edit] C#

public static int[,] ZigZag(int n)
{
int[,] result = new int[n, n];
int i = 0, j = 0;
int d = -1; // -1 for top-right move, +1 for bottom-left move
int start = 0, end = n * n - 1;
do
{
result[i, j] = start++;
result[n - i - 1, n - j - 1] = end--;
 
i += d; j -= d;
if (i < 0)
{
i++; d = -d; // top reached, reverse
}
else if (j < 0)
{
j++; d = -d; // left reached, reverse
}
} while (start < end);
if (start == end)
result[i, j] = start;
return result;
}

[edit] Clojure

Purely functional approach.

(defn partitions [sizes coll]
(lazy-seq
(when-let [n (first sizes)]
(when-let [s (seq coll)]
(cons (take n coll)
(partitions (next sizes) (drop n coll)))))))
 
(defn take-from [n colls]
(lazy-seq
(when-let [s (seq colls)]
(let [[first-n rest-n] (split-at n s)]
(cons (map first first-n)
(take-from n (concat (filter seq (map rest first-n)) rest-n)))))))
 
(defn zig-zag [n]
(->> (partitions (concat (range 1 (inc n)) (range (dec n) 0 -1)) (range (* n n)))
(map #(%1 %2) (cycle [reverse identity]) ,)
(take-from n ,)))
 
user> (zig-zag 5)
(( 0 1 5 6 14)
( 2 4 7 13 15)
( 3 8 12 16 21)
( 9 11 17 20 22)
(10 18 19 23 24))
 
user> (zig-zag 6)
(( 0 1 5 6 14 15)
( 2 4 7 13 16 25)
( 3 8 12 17 24 26)
( 9 11 18 23 27 32)
(10 19 22 28 31 33)
(20 21 29 30 34 35))

[edit] CoffeeScript

 
# Calculate a zig-zag pattern of numbers like so:
# 0 1 5
# 2 4 6
# 3 7 8
#
# There are many interesting ways to solve this; we
# try for an algebraic approach, calculating triangle
# areas, so that me minimize space requirements.
 
zig_zag_value = (x, y, n) ->
 
upper_triangle_zig_zag = (x, y) ->
# calculate the area of the triangle from the prior
# diagonals
diag = x + y
triangle_area = diag * (diag+1) / 2
# then add the offset along the diagonal
if diag % 2 == 0
triangle_area + y
else
triangle_area + x
 
if x + y < n
upper_triangle_zig_zag x, y
else
# For the bottom right part of the matrix, we essentially
# use reflection to count backward.
bottom_right_cell = n * n - 1
n -= 1
v = upper_triangle_zig_zag(n-x, n-y)
bottom_right_cell - v
 
zig_zag_matrix = (n) ->
row = (i) -> (zig_zag_value i, j, n for j in [0...n])
(row i for i in [0...n])
 
do ->
for n in [4..6]
console.log "---- n=#{n}"
console.log zig_zag_matrix(n)
console.log "\n"
 
Output:
> coffee zigzag.coffee 
---- n=4
[ [ 0, 1, 5, 6 ],
  [ 2, 4, 7, 12 ],
  [ 3, 8, 11, 13 ],
  [ 9, 10, 14, 15 ] ]


---- n=5
[ [ 0, 1, 5, 6, 14 ],
  [ 2, 4, 7, 13, 15 ],
  [ 3, 8, 12, 16, 21 ],
  [ 9, 11, 17, 20, 22 ],
  [ 10, 18, 19, 23, 24 ] ]


---- n=6
[ [ 0, 1, 5, 6, 14, 15 ],
  [ 2, 4, 7, 13, 16, 25 ],
  [ 3, 8, 12, 17, 24, 26 ],
  [ 9, 11, 18, 23, 27, 32 ],
  [ 10, 19, 22, 28, 31, 33 ],
  [ 20, 21, 29, 30, 34, 35 ] ]


[edit] Common Lisp

[edit]
Translation of: Java
(but with zero-based indexes and combining the even and odd cases)

(defun zigzag (n)
(flet ((move (i j)
(if (< j (1- n))
(values (max 0 (1- i)) (1+ j))
(values (1+ i) j))))
(loop with a = (make-array (list n n) :element-type 'integer)
with x = 0
with y = 0
for v from 0 below (* n n)
do (setf (aref a x y) v)
(if (evenp (+ x y))
(setf (values x y) (move x y))
(setf (values y x) (move y x)))
finally (return a))))

[edit] An alternative approach

 
; ZigZag
;
; Nigel Galloway.
; June 4th., 2012
;
(defun ZigZag (COLS)
(let ((cs 2) (st '(1 2)) (dx '(-1 1)))
(defun new_cx (i)
(setq st (append st (list (setq cs (+ cs (* 2 i))) (setq cs (+ 1 cs))))
dx (append dx '(-1 1))))
(do ((i 2 (+ 2 i))) ((>= i COLS)) (new_cx i))
(do ((i (- COLS 1 (mod COLS 2)) (+ -2 i))) ((<= i 0)) (new_cx i))
(do ((i 0 (+ 1 i))) ((>= i COLS))
(format t "~%")
(do ((j i (+ 1 j))) ((>= j (+ COLS i)))
(format t "~3d" (nth j st))
(setf (nth j st) (+ (nth j st) (nth j dx)))))))
 

(ZigZag 5) Produces:

  1  2  6  7 15
  3  5  8 14 16
  4  9 13 17 22
 10 12 18 21 23
 11 19 20 24 25

(ZigZag 8) Produces:

  1  2  6  7 15 16 28 29
  3  5  8 14 17 27 30 43
  4  9 13 18 26 31 42 44
 10 12 19 25 32 41 45 54
 11 20 24 33 40 46 53 55
 21 23 34 39 47 52 56 61
 22 35 38 48 51 57 60 62
 36 37 49 50 58 59 63 64

(ZigZag 9) Produces:

  1  2  6  7 15 16 28 29 45
  3  5  8 14 17 27 30 44 46
  4  9 13 18 26 31 43 47 60
 10 12 19 25 32 42 48 59 61
 11 20 24 33 41 49 58 62 71
 21 23 34 40 50 57 63 70 72
 22 35 39 51 56 64 69 73 78
 36 38 52 55 65 68 74 77 79
 37 53 54 66 67 75 76 80 81

[edit] D

Translation of: Common Lisp
int[][] zigZag(in int n) pure nothrow @safe {
static void move(in int n, ref int i, ref int j)
pure nothrow @safe @nogc {
if (j < n - 1) {
if (i > 0) i--;
j++;
} else
i++;
}
 
auto a = new int[][](n, n);
int x, y;
foreach (v; 0 .. n ^^ 2) {
a[y][x] = v;
(x + y) % 2 ? move(n, x, y) : move(n, y, x);
}
return a;
}
 
void main() {
import std.stdio;
 
writefln("%(%(%2d %)\n%)", 5.zigZag);
}
Output:
 0  1  5  6 14
 2  4  7 13 15
 3  8 12 16 21
 9 11 17 20 22
10 18 19 23 24

[edit] Alternative Version

Translation of: Scala

Same output.

import std.stdio, std.algorithm, std.range, std.array;
 
int[][] zigZag(in int n) pure nothrow {
static struct P2 { int x, y; }
const L = iota(n ^^ 2).map!(i => P2(i % n, i / n)).array
.sort!q{ (a.x + a.y == b.x + b.y) ?
((a.x + a.y) % 2 ? a.y < b.y : a.x < b.x) :
(a.x + a.y) < (b.x + b.y) }.release;
 
auto result = new typeof(return)(n, n);
foreach (immutable i, immutable p; L)
result[p.y][p.x] = i;
return result;
}
 
void main() {
writefln("%(%(%2d %)\n%)", 5.zigZag);
}

[edit] E

First, some tools originally written for Spiral (only the array is used):

/** Missing scalar multiplication, but we don't need it. */
def makeVector2(x, y) {
return def vector {
to x() { return x }
to y() { return y }
to add(other) { return makeVector2(x + other.x(), y + other.y()) }
to clockwise() { return makeVector2(-y, x) }
}
}
 
/** Bugs: (1) The printing is specialized. (2) No bounds check on the column. */
def makeFlex2DArray(rows, cols) {
def storage := ([null] * (rows * cols)).diverge()
return def flex2DArray {
to __printOn(out) {
for y in 0..!rows {
for x in 0..!cols {
out.print(<import:java.lang.makeString>.format("%3d", [flex2DArray[y, x]]))
}
out.println()
}
}
to get(r, c) { return storage[r * cols + c] }
to put(r, c, v) { storage[r * cols + c] := v }
}
}
Then the code.
Translation of: D
def zigZag(n) {
def move(&i, &j) {
if (j < (n - 1)) {
i := 0.max(i - 1)
j += 1
} else {
i += 1
}
}
 
def array := makeFlex2DArray(n, n)
var x := 0
var y := 0
 
for i in 1..n**2 {
array[y, x] := i
if ((x + y) % 2 == 0) {
move(&x, &y)
} else {
move(&y, &x)
}
}
return array
}

[edit] Erlang

 
-module( zigzag ).
 
-export( [matrix/1, task/0] ).
 
matrix( N ) ->
{{_X_Y, N}, Proplist} = lists:foldl( fun matrix_as_proplist/2, {{{0, 0}, N}, []}, lists:seq(0, (N * N) - 1) ),
[columns( X, Proplist ) || X <- lists:seq(0, N - 1)].
 
task() -> matrix( 5 ).
 
 
 
columns( Column, Proplist ) -> lists:sort( [Value || {{_X, Y}, Value} <- Proplist, Y =:= Column] ).
 
matrix_as_proplist( N, {{X_Y, Max}, Acc} ) ->
Next = next_indexes( X_Y, Max ),
{{Next, Max}, [{X_Y, N} | Acc]}.
 
next_indexes( {X, Y}, Max ) when Y + 1 =:= Max, (X + Y) rem 2 =:= 0 -> {X + 1, Y - 1};
next_indexes( {X, Y}, Max ) when Y + 1 =:= Max, (X + Y) rem 2 =:= 1 -> {X + 1, Y};
next_indexes( {X, Y}, Max ) when X + 1 =:= Max, (X + Y) rem 2 =:= 0 -> {X, Y + 1};
next_indexes( {X, Y}, Max ) when X + 1 =:= Max, (X + Y) rem 2 =:= 1 -> {X - 1, Y + 1};
next_indexes( {X, 0}, _Max ) when X rem 2 =:= 0 -> {X + 1, 0};
next_indexes( {X, 0}, _Max ) when X rem 2 =:= 1 -> {X - 1, 1};
next_indexes( {0, Y}, _Max ) when Y rem 2 =:= 0 -> {1, Y - 1};
next_indexes( {0, Y}, _Max ) when Y rem 2 =:= 1 -> {0, Y + 1};
next_indexes( {X, Y}, _Max ) when (X + Y) rem 2 =:= 0 -> {X + 1, Y - 1};
next_indexes( {X, Y}, _Max ) when (X + Y) rem 2 =:= 1 -> {X - 1, Y + 1}.
 
Output:
71> zigzag:task().
[[0,1,5,6,14],
 [2,4,7,13,15],
 [3,8,12,16,21],
 [9,11,17,20,22],
 [10,18,19,23,24]]

[edit] Euphoria

Translation of: C#
function zigzag(integer size)
sequence s
integer i, j, d, max
s = repeat(repeat(0,size),size)
i = 1 j = 1 d = -1
max = size*size
for n = 1 to floor(max/2)+1 do
s[i][j] = n
s[size-i+1][size-j+1] = max-n+1
i += d j-= d
if i < 1 then
i += 1 d = -d
elsif j < 1 then
j += 1 d = -d
end if
end for
return s
end function
 
? zigzag(5)
Output:
{
  {1,2,6,7,15},
  {3,5,8,14,16},
  {4,9,13,17,22},
  {10,12,18,21,23},
  {11,19,20,24,25}
}

[edit] Fan

using gfx  // for Point; convenient x/y wrapper
 
**
** A couple methods for generating a 'zigzag' array like
**
** 0 1 5 6
** 2 4 7 12
** 3 8 11 13
** 9 10 14 15
**
class ZigZag
{
** return an n x n array of uninitialized Int
static Int[][] makeSquareArray(Int n)
{
Int[][] grid := Int[][,] {it.size=n}
n.times |i| { grid[i] = Int[,] {it.size=n} }
return grid
}
 
 
Int[][] zig(Int n)
{
grid := makeSquareArray(n)
 
move := |Int i, Int j->Point|
{ return j < n - 1 ? Point(i <= 0 ? 0 : i-1, j+1) : Point(i+1, j) }
pt := Point(0,0)
(n*n).times |i| {
grid[pt.y][pt.x] = i
if ((pt.x+pt.y)%2 != 0) pt = move(pt.x,pt.y)
else {tmp:= move(pt.y,pt.x); pt = Point(tmp.y, tmp.x) }
}
return grid
}
 
public static Int[][] zag(Int size)
{
data := makeSquareArray(size)
 
Int i := 1
Int j := 1
for (element:=0; element < size * size; element++)
{
data[i - 1][j - 1] = element
if((i + j) % 2 == 0) {
// Even stripes
if (j < size) {
j++
} else {
i += 2
}
if (i > 1) {
i--
}
} else {
// Odd stripes
if (i < size) {
i++;
} else {
j += 2
}
if (j > 1) {
j--
}
}
}
return data;
}
 
Void print(Int[][] data)
{
data.each |row|
{
buf := StrBuf()
row.each |num|
{
buf.add(num.toStr.justr(3))
}
echo(buf)
}
}
 
Void main()
{
echo("zig method:")
print(zig(8))
echo("\nzag method:")
print(zag(8))
}
}

[edit] Forth

0 value diag
 
: south diag abs + cell+ ;
 
' cell+ value zig
' south value zag
 
: init ( n -- )
1- cells negate to diag
['] cell+ to zig
['] south to zag ;
 
: swap-diag zig zag to zig to zag ;
 
: put ( n addr -- n+1 addr )
2dup ! swap 1+ swap ;
 
: turn ( addr -- addr+E/S )
zig execute swap-diag
diag negate to diag ;
 
: zigzag ( matrix n -- )
{ n } n init
0 swap
n 1 ?do
put turn
i 0 do put diag + loop
loop
swap-diag
n 1 ?do
put turn
n i 1+ ?do put diag + loop
loop
 ! ;
 
: .matrix ( n matrix -- )
over 0 do
cr
over 0 do
dup @ 3 .r cell+
loop
loop 2drop ;
 
: test ( n -- ) here over zigzag here .matrix ;
5 test
0 1 5 6 14
2 4 7 13 15
3 8 12 16 21
9 11 17 20 22
10 18 19 23 24 ok


[edit] Fortran

Works with: Fortran version 90 and later
PROGRAM ZIGZAG
 
IMPLICIT NONE
INTEGER, PARAMETER :: size = 5
INTEGER :: zzarray(size,size), x(size*size), y(size*size), i, j
 
! index arrays
x = (/ ((j, i = 1, size), j = 1, size) /)
y = (/ ((i, i = 1, size), j = 1, size) /)
 
! Sort indices
DO i = 2, size*size
j = i - 1
DO WHILE (j>=1 .AND. (x(j)+y(j)) > (x(i)+y(i)))
j = j - 1
END DO
x(j+1:i) = cshift(x(j+1:i),-1)
y(j+1:i) = cshift(y(j+1:i),-1)
END DO
 
! Create zig zag array
DO i = 1, size*size
IF (MOD(x(i)+y(i), 2) == 0) THEN
zzarray(x(i),y(i)) = i - 1
ELSE
zzarray(y(i),x(i)) = i - 1
END IF
END DO
 
! Print zig zag array
DO j = 1, size
DO i = 1, size
WRITE(*, "(I5)", ADVANCE="NO") zzarray(i,j)
END DO
WRITE(*,*)
END DO
 
END PROGRAM ZIGZAG

[edit] GAP

ZigZag := function(n)
local a, i, j, k;
a := NullMat(n, n);
i := 1;
j := 1;
for k in [0 .. n*n - 1] do
a[i][j] := k;
if (i + j) mod 2 = 0 then
if j < n then
j := j + 1;
else
i := i + 2;
fi;
if i > 1 then
i := i - 1;
fi;
else
if i < n then
i := i + 1;
else
j := j + 2;
fi;
if j > 1 then
j := j - 1;
fi;
fi;
od;
return a;
end;
 
PrintArray(ZigZag(5));
# [ [ 0, 1, 5, 6, 14 ],
# [ 2, 4, 7, 13, 15 ],
# [ 3, 8, 12, 16, 21 ],
# [ 9, 11, 17, 20, 22 ],
# [ 10, 18, 19, 23, 24 ] ]

[edit] Go

Translation of: Groovy
Edge direct algorithm
package main
 
import (
"fmt"
"strconv"
)
 
func zz(n int) []int {
r := make([]int, n*n)
i := 0
n2 := n * 2
for d := 1; d <= n2; d++ {
x := d - n
if x < 0 {
x = 0
}
y := d - 1
if y > n-1 {
y = n - 1
}
j := n2 - d
if j > d {
j = d
}
for k := 0; k < j; k++ {
if d&1 == 0 {
r[(x+k)*n+y-k] = i
} else {
r[(y-k)*n+x+k] = i
}
i++
}
}
 
return r
}
 
func main() {
const n = 5
w := len(strconv.Itoa(n*n - 1))
for i, e := range zz(n) {
fmt.Printf("%*d ", w, e)
if i%n == n-1 {
fmt.Println("")
}
}
}
Output:
 0  1  5  6 14
 2  4  7 13 15
 3  8 12 16 21
 9 11 17 20 22
10 18 19 23 24

[edit] Groovy

[edit] Edge

An odd technique that traverses the grid edges directly and calculates the transform onto the grid.

def zz = { n ->
grid = new int[n][n]
i = 0
for (d in 1..n*2) {
(x, y) = [Math.max(0, d - n), Math.min(n - 1, d - 1)]
Math.min(d, n*2 - d).times {
grid[d%2?y-it:x+it][d%2?x+it:y-it] = i++;
}
}
grid
}
Output:
 > zz(5).each { it.each { print("${it}".padLeft(3)) }; println() }
  0  1  5  6 14
  2  4  7 13 15
  3  8 12 16 21
  9 11 17 20 22
 10 18 19 23 24

[edit] Cursor

Ported from the Java example

def zz = { n->
move = { i, j -> j < n - 1 ? [i <= 0 ? 0 : i-1, j+1] : [i+1, j] }
grid = new int[n][n]
(x, y) = [0, 0]
(n**2).times {
grid[y][x] = it
if ((x+y)%2) (x,y) = move(x,y)
else (y,x) = move(y,x)
}
grid
}
Output:
 > zz(5).each { it.each { print("${it}".padLeft(3)) }; println() }
  0  1  5  6 14
  2  4  7 13 15
  3  8 12 16 21
  9 11 17 20 22
 10 18 19 23 24

[edit] Sorting

Ported from the Python example with some input from J

def zz = { n ->
(0..<n*n).collect { [x:it%n,y:(int)(it/n)] }.sort { c->
[c.x+c.y, (((c.x+c.y)%2) ? c.y : -c.y)]
}.with { l -> l.inject(new int[n][n]) { a, c -> a[c.y][c.x] = l.indexOf(c); a } }
}
Output:
 > zz(5).each { it.each { print("${it}".padLeft(3)) }; println() }
  0  1  5  6 14
  2  4  7 13 15
  3  8 12 16 21
  9 11 17 20 22
 10 18 19 23 24

[edit] Haskell

Computing the array:

import Data.Array (array, bounds, range, (!))
import Data.Monoid (mappend)
import Data.List (sortBy)
 
compZig (x,y) (x',y') = compare (x+y) (x'+y')
`mappend` if even (x+y) then compare x x'
else compare y y'

 
zigZag upper = array b $ zip (sortBy compZig (range b))
[0..]
where b = ((0,0),upper)

compZig compares coordinates using the order of a zigzag walk: primarily, the antidiagonals; secondarily, alternating directions along them.

In zigZag, array takes the bounds and a list of indexes paired with values. We take the list of all indexes, range b, and sort it in the zigzag order, then zip that with the integers starting from 0. (This algorithm was inspired by the explanation of the J example.)

Displaying the array (not part of the task):

import Text.Printf (printf)
 
-- format a 2d array of integers neatly
show2d a = unlines [unwords [printf "%3d" (a ! (x,y) :: Integer) | x <- axis fst] | y <- axis snd]
where (l, h) = bounds a
axis f = [f l .. f h]
 
main = mapM_ (putStr . show2d . zigZag) [(3,3), (4,4), (10,2)]

[edit] Icon and Unicon

This solution works for both Icon and Unicon.

procedure main(args)
n := integer(!args) | 5
every !(A := list(n)) := list(n)
A := zigzag(A)
show(A)
end
 
procedure show(A)
every writes(right(!A,5) | "\n")
end
 
procedure zigzag(A)
x := [0,0]
every i := 0 to (*A^2 -1) do {
x := nextIndices(*A, x)
A[x[1]][x[2]] := i
}
return A
end
 
procedure nextIndices(n, x)
return if (x[1]+x[2])%2 = 0
then if x[2] = n then [x[1]+1, x[2]] else [max(1, x[1]-1), x[2]+1]
else if x[1] = n then [x[1], x[2]+1] else [x[1]+1, max(1, x[2]-1)]
end
Output:
->zz
    0    1    5    6   14
    2    4    7   13   15
    3    8   12   16   21
    9   11   17   20   22
   10   18   19   23   24
->

[edit] J

A succinct way:

   ($ [: /:@; <@|.`</.@i.)@,~ 5
0 1 5 6 14
2 4 7 13 15
3 8 12 16 21
9 11 17 20 22
10 18 19 23 24

This version is longer, but more "mathematical" and less "procedural":

   ($ [: /:@; [: <@(A.~_2|#)/. i.)@,~ 5
0 1 5 6 14
2 4 7 13 15
3 8 12 16 21
9 11 17 20 22
10 18 19 23 24

Leveraging a useful relationship among the indices:

   ($ ([: /:@;@(+/"1 <@|.`</. ]) (#: i.@(*/))))@,~ 5 
0 1 5 6 14
2 4 7 13 15
3 8 12 16 21
9 11 17 20 22
10 18 19 23 24

By the way, all the edge cases are handled transparently, without any special checks. Furthermore, by simply removing the trailing @,~ from the solutions, they automatically generalize to rectangular (non-square) matrices:

   ($ [: /:@; [: <@|.`</. i.) 5 3
0 1 5
2 4 6
3 7 11
8 10 12
9 13 14

[edit] Java

Translation of: Ada
public static int[][] Zig_Zag(final int size)
{
int[][] data = new int[size][size];
int i = 1;
int j = 1;
for (int element = 0; element < size * size; element++)
{
data[i - 1][j - 1] = element;
if ((i + j) % 2 == 0)
{
// Even stripes
if (j < size)
j++;
else
i+= 2;
if (i > 1)
i--;
}
else
{
// Odd stripes
if (i < size)
i++;
else
j+= 2;
if (j > 1)
j--;
}
}
return data;
}

[edit] JavaScript

Works with: SpiderMonkey
for the print() function.
Translation of: Java

Subclasses the Matrix class defined at Matrix Transpose#JavaScript

function ZigZagMatrix(n) {
this.height = n;
this.width = n;
 
this.mtx = [];
for (var i = 0; i < n; i++)
this.mtx[i] = [];
 
var i=1, j=1;
for (var e = 0; e < n*n; e++) {
this.mtx[i-1][j-1] = e;
if ((i + j) % 2 == 0) {
// Even stripes
if (j < n) j ++;
else i += 2;
if (i > 1) i --;
} else {
// Odd stripes
if (i < n) i ++;
else j += 2;
if (j > 1) j --;
}
}
}
ZigZagMatrix.prototype = Matrix.prototype;
 
var z = new ZigZagMatrix(5);
print(z);
print();
 
z = new ZigZagMatrix(4);
print(z);
Output:
0,1,5,6,14
2,4,7,13,15
3,8,12,16,21
9,11,17,20,22
10,18,19,23,24

0,1,5,6
2,4,7,12
3,8,11,13
9,10,14,15

[edit] Joy

 
(*
From the library.
*)
DEFINE reverse == [] swap shunt;
shunt == [swons] step.
 
(*
Split according to the parameter given.
*)
DEFINE take-drop == [dup] swap dup [[] cons [take swap] concat concat] dip []
cons concat [drop] concat.
 
(*
Take the first of a list of lists.
*)
DEFINE take-first == [] cons 3 [dup] times [dup] swap concat [take [first] map
swap dup] concat swap concat [drop swap] concat swap
concat [take [rest] step []] concat swap concat [[cons]
times swap concat 1 drop] concat.
 
DEFINE zigzag ==
 
(*
Use take-drop to generate a list of lists.
*)
4 [dup] times 1 swap from-to-list swap pred 1 swap from-to-list reverse concat
swap dup * pred 0 swap from-to-list swap [take-drop i] step [pop list] [cons]
while
 
(*
The odd numbers must be modified with reverse.
*)
[dup size 2 div popd [1 =] [pop reverse] [pop] ifte] map
 
(*
Take the first of the first of n lists.
*)
swap dup take-first [i] cons times pop
 
(*
Merge the n separate lists.
*)
[] [pop list] [cons] while
 
(*
And print them.
*)
swap dup * pred 'd 1 1 format size succ [] cons 'd swons [1 format putchars]
concat [step '\n putch] cons step.
 
11 zigzag.

[edit] jq

Infrastructure:

# 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" ) ;
 

Create a zigzag matrix by zigzagging:

def zigzag(n):
 
# unless m == n*n, place m at (i,j), pointing
# in the direction d, where d = [drow, dcolumn]:
def _next(i; j; m; d):
if m == (n*n) then . else .[i][j] = m end
| if m == (n*n) - 1 then .
elif i == n-1 then if j+1 < n then .[i][j+1] = m+1 | _next(i-1; j+2; m+2; [-1, 1]) else . end
elif i == 0 then if j+1 < n then .[i][j+1] = m+1 | _next(i+1; j  ; m+2; [ 1,-1])
else .[i+1][j] = m+1 | _next(i+2; j-1; m+2; [ 1,-1]) end
elif j == n-1 then if i+1 < n then .[i+1][j] = m+1 | _next(i+2; j-1; m+2; [ 1,-1]) else . end
elif j == 0 then if i+1 < n then .[i+1][j] = m+1 | _next(i; j+1; m+2; [-1, 1])
else .[i][j+1] = m+1 | _next(i-1; j+1; m+2; [-1, 1]) end
else _next(i+ d[0]; j+ d[1]; m+1; d)
end ;
matrix(n;n;-1) | _next(0;0; 0; [0,1]) ;
 
# Example
zigzag(5) | neatly(4)
Output:
$ jq -n -r -f zigzag.jq
    0    1    5    6   14
    2    4    7   13   15
    3    8   12   16   21
    9   11   17   20   22
   10   18   19   23   24

[edit] Lasso

 
var(
'square' = array
,'size' = integer( 5 )// for a 5 X 5 square
,'row' = array
,'x' = integer( 1 )
,'y' = integer( 1 )
,'counter' = integer( 1 )
);
 
// create place-holder matrix
loop( $size );
$row = array;
 
loop( $size );
$row->insert( 0 );
 
/loop;
 
$square->insert( $row );
 
/loop;
 
while( $counter < $size * $size );
// check downward diagonal
if(
$x > 1
&&
$y < $square->size
&&
$square->get( $y + 1 )->get( $x - 1 ) == 0
);
 
$x -= 1;
$y += 1;
 
// check upward diagonal
else(
$x < $square->size
&&
$y > 1
&&
$square->get( $y - 1 )->get( $x + 1 ) == 0
);
 
$x += 1;
$y -= 1;
 
// check right
else(
(
$y == 1
||
$y == $square->size
)
&&
$x < $square->size
&&
$square->get( $y )->get( $x + 1 ) == 0
);
 
$x += 1;
 
// down
else;
$y += 1;
 
/if;
 
$square->get( $y )->get( $x ) = loop_count;
 
$counter += 1;
 
/while;
 
$square;
 

[edit] Lua

 
local zigzag = {}
 
function zigzag.new(n)
local a = {}
local i -- cols
local j -- rows
 
a.n = n
a.val = {}
 
for j = 1, n do
a.val[j] = {}
for i = 1, n do
a.val[j][i] = 0
end
end
 
i = 1
j = 1
 
local di
local dj
local k = 0
 
while k < n * n do
a.val[j][i] = k
k = k + 1
if i == n then
j = j + 1
a.val[j][i] = k
k = k + 1
di = -1
dj = 1
end
if j == 1 then
i = i + 1
a.val[j][i] = k
k = k + 1
di = -1
dj = 1
end
if j == n then
i = i + 1
a.val[j][i] = k
k = k + 1
di = 1
dj = -1
end
if i == 1 then
j = j + 1
a.val[j][i] = k
k = k + 1
di = 1
dj = -1
end
i = i + di
j = j + dj
end
 
setmetatable(a, {__index = zigzag, __tostring = zigzag.__tostring})
return a
end
 
function zigzag:__tostring()
local s = {}
for j = 1, self.n do
local row = {}
for i = 1, self.n do
row[i] = string.format('%d', self.val[j][i])
end
s[j] = table.concat(row, ' ')
end
return table.concat(s, '\n')
end
 
print(zigzag.new(5))
 

[edit] M4

divert(-1)
 
define(`set2d',`define(`$1[$2][$3]',`$4')')
define(`get2d',`defn(`$1[$2][$3]')')
define(`for',
`ifelse($#,0,``$0'',
`ifelse(eval($2<=$3),1,
`pushdef(`$1',$2)$4`'popdef(`$1')$0(`$1',incr($2),$3,`$4')')')')
define(`show2d',
`for(`x',0,decr($2),
`for(`y',0,decr($3),`format(`%2d',get2d($1,x,y)) ')
')')
 
dnl <name>,<size>
define(`zigzag',
`define(`j',1)`'define(`k',1)`'for(`e',0,eval($2*$2-1),
`set2d($1,decr(j),decr(k),e)`'ifelse(eval((j+k)%2),0,
`ifelse(eval(k<$2),1,
`define(`k',incr(k))',
`define(`j',eval(j+2))')`'ifelse(eval(j>1),1,
`define(`j',decr(j))')',
`ifelse(eval(j<$2),1,
`define(`j',incr(j))',
`define(`k',eval(k+2))')`'ifelse(eval(k>1),1,
`define(`k',decr(k))')')')')
 
divert
 
zigzag(`a',5)
show2d(`a',5,5)
Output:
 0  1  5  6 14
 2  4  7 13 15
 3  8 12 16 21
 9 11 17 20 22
10 18 19 23 24

[edit] Mathematica / Wolfram Language

Rule-based implementation, the upper-left half is correctly calculated using a direct formula. The lower-right half is then 'mirrored' from the upper-left half.

ZigZag[size_Integer/;size>0]:=Module[{empty=ConstantArray[0,{size,size}]},
empty=ReplacePart[empty,{i_,j_}:>1/2 (i+j)^2-(i+j)/2-i (1-Mod[i+j,2])-j Mod[i+j,2]];
ReplacePart[empty,{i_,j_}/;i+j>size+1:> size^2-tmp[[size-i+1,size-j+1]]-1]
]

Ported from the java-example:

ZigZag2[size_] := Module[{data, i, j, elem},
data = ConstantArray[0, {size, size}];
i = j = 1;
For[elem = 0, elem < size^2, elem++,
data[[i, j]] = elem;
If[Mod[i + j, 2] == 0,
If[j < size, j++, i += 2];
If[i > 1, i--]
,
If[i < size, i++, j += 2];
If[j > 1, j--];
];
];
data
]

Examples:

ZigZag[5] // MatrixForm
ZigZag2[6] // MatrixForm

gives back:


\left(
\begin{array}{ccccc}
 0 & 1 & 5 & 6 & 14 \\
 2 & 4 & 7 & 13 & 15 \\
 3 & 8 & 12 & 16 & 21 \\
 9 & 11 & 17 & 20 & 22 \\
 10 & 18 & 19 & 23 & 24
\end{array}
\right)


\left(
\begin{array}{cccccc}
 0 & 1 & 5 & 6 & 14 & 15 \\
 2 & 4 & 7 & 13 & 16 & 25 \\
 3 & 8 & 12 & 17 & 24 & 26 \\
 9 & 11 & 18 & 23 & 27 & 32 \\
 10 & 19 & 22 & 28 & 31 & 33 \\
 20 & 21 & 29 & 30 & 34 & 35
\end{array}
\right)

[edit] MATLAB

This isn't the best way to solve this task and the algorithm is completely unintuitive without some major exploration of the code. But! It is pretty fast for n < 10000.

function matrix = zigZag(n)
 
%This is very unintiutive. This algorithm parameterizes the
%zig-zagging movement along the matrix indicies. The easiest way to see
%what this algorithm does is to go through line-by-line and write out
%what the algorithm does on a peace of paper.
 
matrix = zeros(n);
counter = 1;
flipCol = true;
flipRow = false;
 
%This for loop does the top-diagonal of the matrix
for i = (2:n)
row = (1:i);
column = (1:i);
 
%Causes the zig-zagging. Without these conditionals,
%you would end up with a diagonal matrix.
%To see what happens, comment these conditionals out.
if flipCol
column = fliplr(column);
flipRow = true;
flipCol = false;
elseif flipRow
row = fliplr(row);
flipRow = false;
flipCol = true;
end
 
%Selects a diagonal of the zig-zag matrix and places the
%correct integer value in each index along that diagonal
for j = (1:numel(row))
matrix(row(j),column(j)) = counter;
counter = counter + 1;
end
end
 
%This for loop does the bottom-diagonal of the matrix
for i = (2:n)
row = (i:n);
column = (i:n);
 
%Causes the zig-zagging. Without these conditionals,
%you would end up with a diagonal matrix.
%To see what happens comment these conditionals out.
if flipCol
column = fliplr(column);
flipRow = true;
flipCol = false;
elseif flipRow
row = fliplr(row);
flipRow = false;
flipCol = true;
end
 
%Selects a diagonal of the zig-zag matrix and places the
%correct integer value in each index along that diagonal
for j = (1:numel(row))
matrix(row(j),column(j)) = counter;
counter = counter + 1;
end
end
 
 
end
Output:
>> zigZag(5)

ans =

     0     1     5     6    14
     2     4     7    13    15
     3     8    12    16    21
     9    11    17    20    22
    10    18    19    23    24


[edit] Maxima

zigzag(n) := block([a, i, j],
a: zeromatrix(n, n),
i: 1,
j: 1,
for k from 0 thru n*n - 1 do (
a[i, j]: k,
if evenp(i + j) then (
if j < n then j: j + 1 else i: i + 2,
if i > 1 then i: i - 1
) else (
if i < n then i: i + 1 else j: j + 2,
if j > 1 then j: j - 1
)
),
a)$
 
zigzag(5);
/* matrix([ 0, 1, 5, 6, 14],
[ 2, 4, 7, 13, 15],
[ 3, 8, 12, 16, 21],
[ 9, 11, 17, 20, 22],
[10, 18, 19, 23, 24]) */

[edit] Modula-3

MODULE ZigZag EXPORTS Main;
 
IMPORT IO, Fmt;
 
TYPE Matrix = REF ARRAY OF ARRAY OF CARDINAL;
 
PROCEDURE Create(size: CARDINAL): Matrix =
PROCEDURE move(VAR i, j: INTEGER) =
BEGIN
IF j < (size - 1) THEN
IF (i - 1) < 0 THEN
i := 0;
ELSE
i := i - 1;
END;
INC(j);
ELSE
INC(i);
END;
END move;
 
VAR data := NEW(Matrix, size, size);
x, y: INTEGER := 0;
BEGIN
FOR v := 0 TO size * size - 1 DO
data[y, x] := v;
IF (x + y) MOD 2 = 0 THEN
move(y, x);
ELSE
move(x, y);
END;
END;
RETURN data;
END Create;
 
PROCEDURE Print(data: Matrix) =
BEGIN
FOR i := FIRST(data^) TO LAST(data^) DO
FOR j := FIRST(data[0]) TO LAST(data[0]) DO
IO.Put(Fmt.F("%3s", Fmt.Int(data[i, j])));
END;
IO.Put("\n");
END;
END Print;
 
BEGIN
Print(Create(5));
END ZigZag.
Output:
  0  1  5  6 14
  2  4  7 13 15
  3  8 12 16 21
  9 11 17 20 22
 10 18 19 23 24

[edit] NetRexx

Translation of: REXX
/* NetRexx */
options replace format comments java crossref savelog symbols binary
 
zigzag(5)
 
return
 
method zigzag(msize) public static
 
row = 1
col = 1
 
ziggy = Rexx(0)
loop j_ = 0 for msize * msize
ziggy[row, col] = j_
if (row + col) // 2 == 0 then do
if col < msize then -
col = col + 1
else row = row + 2
if row \== 1 then -
row = row - 1
end
else do
if row < msize then -
row = row + 1
else col = col + 2
if col \== 1 then -
col = col - 1
end
end j_
 
L = (msize * msize - 1).length /*for a constant element width. */
loop row = 1 for msize /*show all the matrix's rows. */
rowOut = ''
loop col = 1 for msize
rowOut = rowOut ziggy[row, col].right(L)
end col
say rowOut
end row
 
return
 

[edit] Nimrod

Translation of: Python
import algorithm, strutils
 
type Pos = tuple[x, y: int]
 
template newSeqWith(len: int, init: expr): expr =
var result {.gensym.} = newSeq[type(init)](len)
for i in 0 .. <len:
result[i] = init
result
 
proc `$`(m: seq[seq[int]]): string =
result = ""
for r in m:
for c in r:
result.add align($c, 2) & " "
result.add "\n"
 
proc zigzagMatrix(n): auto =
result = newSeqWith(n, newSeq[int](n))
 
var indices = newSeq[Pos]()
 
for x in 0 .. <n:
for y in 0 .. <n:
indices.add((x,y))
 
sort(indices, proc(a, b: Pos): int =
result = a.x + a.y - b.x - b.y
if result == 0: result =
(if (a.x + a.y) mod 2 == 0: a.y else: -a.y) -
(if (b.x + b.y) mod 2 == 0: b.y else: -b.y))
 
for i, p in indices:
result[p.x][p.y] = i
 
echo zigzagMatrix(6)
Output:
 0  1  5  6 14 15 
 2  4  7 13 16 25 
 3  8 12 17 24 26 
 9 11 18 23 27 32 
10 19 22 28 31 33 
20 21 29 30 34 35

[edit] Objeck

Translation of: Java
 
function : native : ZigZag(size : Int) ~ Int[,] {
data := Int->New[size, size];
i := 1;
j := 1;
 
max := size * size;
for(element := 0; element < max ; element += 1;) {
data[i - 1, j - 1] := element;
 
if((i + j) % 2 = 0) {
# even stripes
if(j < size){
j += 1;
}
else{
i+= 2;
};
 
if(i > 1) {
i -= 1;
};
}
else{
# ddd stripes
if(i < size){
i += 1;
}
else{
j+= 2;
};
 
if(j > 1){
j -= 1;
};
};
};
 
return data;
}
 


[edit] OCaml

Translation of: Common Lisp
let zigzag n =
(* move takes references and modifies them directly *)
let move i j =
if !j < n - 1 then begin
i := max 0 (!i - 1);
incr j
end else
incr i
in
let a = Array.make_matrix n n 0
and x = ref 0 and y = ref 0 in
for v = 0 to n * n - 1 do
a.(!x).(!y) <- v;
if (!x + !y) mod 2 = 0 then
move x y
else
move y x
done;
a

[edit] Octave

See the MATLAB solution, which works perfectly in Octave too.

[edit] ooRexx

Translation of: Java
 
call printArray zigzag(3)
say
call printArray zigzag(4)
say
call printArray zigzag(5)
 
::routine zigzag
use strict arg size
 
data = .array~new(size, size)
row = 1
col = 1
 
loop element = 0 to (size * size) - 1
data[row, col] = element
-- even stripes
if (row + col) // 2 = 0 then do
if col < size then col += 1
else row += 2
if row > 1 then row -= 1
end
-- odd rows
else do
if row < size then row += 1
else col += 2
if col > 1 then col -= 1
end
end
 
return data
 
::routine printArray
use arg array
dimension = array~dimension(1)
loop i = 1 to dimension
line = "|"
loop j = 1 to dimension
line = line array[i, j]~right(2)
end
line = line "|"
say line
end
 
Output:
|  0  1  5 |
|  2  4  6 |
|  3  7  8 |

|  0  1  5  6 |
|  2  4  7 12 |
|  3  8 11 13 |
|  9 10 14 15 |

|  0  1  5  6 14 |
|  2  4  7 13 15 |
|  3  8 12 16 21 |
|  9 11 17 20 22 |
| 10 18 19 23 24 |

[edit] Oz

Implemented as a state machine:

declare
%% state move success failure
States = unit(right: [ 1# 0 downLeft downInstead]
downInstead: [ 0# 1 downLeft terminate]
downLeft: [~1# 1 downLeft down]
down: [ 0# 1 topRight rightInstead]
rightInstead: [ 1# 0 topRight terminate]
topRight: [ 1#~1 topRight right])
 
fun {CreateZigZag N}
ZZ = {Create2DTuple N N}
 
%% recursively walk through 2D tuple and set values
proc {Walk Pos=X#Y Count State}
[Dir Success Failure] = States.State
NextPos = {Record.zip Pos Dir Number.'+'}
Valid = {Record.all NextPos fun {$ C} C > 0 andthen C =< N end}
NewPos = if Valid then NextPos else Pos end
NewCount = if Valid then Count + 1 else Count end
NewState = if Valid then Success else Failure end
in
ZZ.Y.X = Count
if NewState \= terminate then
{Walk NewPos NewCount NewState}
end
end
in
{Walk 1#1 0 right}
ZZ
end
 
fun {Create2DTuple W H}
T = {MakeTuple unit H}
in
{Record.forAll T fun {$} {MakeTuple unit W} end}
T
end
in
{Inspect {CreateZigZag 5}}

[edit] PARI/GP

Translation of: C.23
zz(n)={
my(M=matrix(n,n),i,j,d=-1,start,end=n^2-1);
while(ct--,
M[i+1,j+1]=start;
M[n-i,n-j]=end;
start++;
end--;
i+=d;
j-=d;
if(i<0,
i++;
d=-d
,
if(j<0,
j++;
d=-d
)
);
if(start>end,return(M))
)
};

[edit] Pascal

Program zigzag;
 
const
size = 5;
var
zzarray: array [1..size, 1..size] of integer;
element, i, j: integer;
direction: integer;
 
begin
i := 1;
j := 1;
direction := 1;
for element := 1 to size*size do
begin
zzarray[i,j] := element;
i := i + direction;
j := j - direction;
if (i = 0) then
begin
direction := -direction;
i := i + 1;
end;
if (i = size +1) then
begin
direction := -direction;
i := i - 1;
j := j + 2;
end;
if (j = 0) then
begin
direction := -direction;
j := j + 1;
end;
if (j = size + 1) then
begin
direction := -direction;
j := j - 1;
i := i + 2;
end;
end;
 
for j := 1 to size do
begin
for i := 1 to size do
write(zzarray[i,j]:3);
writeln;
end;
end.
Output:
  1  2  6  7 15
  3  5  8 14 16
  4  9 13 17 22
 10 12 18 21 23
 11 19 20 24 25

[edit] Perl

Translation of: Haskell
sub zig_zag {
my ($w, $h, @r, $n) = @_;
 
$r[ $_->[1] ][ $_->[0] ] = $n++ for
sort { $a->[0] + $a->[1] <=> $b->[0] + $b->[1] or
($a->[0] + $a->[1]) % 2
? $a->[1] <=> $b->[1]
: $a->[0] <=> $b->[0]
}
map { my $e = $_;
map{ [$e, $_] } 0 .. $w-1
} 0 .. $h - 1;
@r
}
 
print map{ "@$_\n" } zig_zag(3, 5);

[edit] Perl 6

Assuming the same Turtle class that is used in Spiral_matrix:

sub MAIN($size as Int) {
my $t = Turtle.new(dir => northeast);
my $counter = 0;
for 1 ..^ $size -> $run {
for ^$run {
$t.lay-egg($counter++);
$t.forward;
}
my $yaw = $run %% 2 ?? -1 !! 1;
$t.turn-right($yaw * 135); $t.forward; $t.turn-right($yaw * 45);
}
for $size ... 1 -> $run {
for ^$run -> $ {
$t.lay-egg($counter++);
$t.forward;
}
$t.turn-left(180); $t.forward;
my $yaw = $run %% 2 ?? 1 !! -1;
$t.turn-right($yaw * 45); $t.forward; $t.turn-left($yaw * 45);
}
$t.showmap;
}

[edit] PHP

function ZigZagMatrix($num) {
$matrix = array();
for ($i = 0; $i < $num; $i++){
$matrix[$i] = array();
}
 
$i=1;
$j=1;
for ($e = 0; $e < $num*$num; $e++) {
$matrix[$i-1][$j-1] = $e;
if (($i + $j) % 2 == 0) {
if ($j < $num){
$j++;
}else{
$i += 2;
}
if ($i > 1){
$i --;
}
} else {
if ($i < $num){
$i++;
}else{
$j += 2;
}
if ($j > 1){
$j --;
}
}
}
return $matrix;
}

[edit] PicoLisp

This example uses 'grid' from "lib/simul.l", which maintains a two-dimensional structure and is normally used for simulations and board games.

(load "@lib/simul.l")
 
(de zigzag (N)
(prog1 (grid N N)
(let (D '(north west south east .) E '(north east .) This 'a1)
(for Val (* N N)
(=: val Val)
(setq This
(or
((cadr D) ((car D) This))
(prog
(setq D (cddr D))
((pop 'E) This) )
((pop 'E) This) ) ) ) ) ) )
 
(mapc
'((L)
(for This L (prin (align 3 (: val))))
(prinl) )
(zigzag 5) )
Output:
  1  2  6  7 15
  3  5  8 14 16
  4  9 13 17 22
 10 12 18 21 23
 11 19 20 24 25

[edit] PL/I

/* Fill a square matrix with the values 0 to N**2-1,     */
/* in a zig-zag fashion. */
/* N is the length of one side of the square. */
/* Written 22 February 2010. */
 
declare n fixed binary;
 
put skip list ('Please type the size of the matrix:');
get list (n);
 
begin;
declare A(n,n) fixed binary;
declare (i, j, inc, q) fixed binary;
 
on subrg snap begin;
declare i fixed binary;
do i = 1 to n;
put skip edit (a(i,*)) (f(4));
end;
stop;
end;
 
A = -1;
inc = -1;
i, j = 1;
 
loop:
do q = 0 to n**2-1;
a(i,j) = q;
if q = n**2-1 then leave;
if i = 1 & j = n then
if iand(j,1) = 1 then /* odd-sided matrix */
do; i = i + 1; inc = -inc; iterate loop; end;
else /* an even-sided matrix */
do; i = i + inc; j = j - inc; iterate loop; end;
if inc = -1 then if i+inc < 1 then
do; inc = -inc; j = j + 1; a(i,j) = q; iterate loop; end;
if inc = 1 then if i+inc > n then
do; inc = -inc; j = j + 1; a(i,j) = q; iterate loop; end;
if inc = 1 then if j-inc < 1 then
do; inc = -inc; i = i + 1; a(i,j) = q; iterate loop; end;
if inc = -1 then if j - inc > n then
do; inc = -inc; i = i + 1; a(i,j) = q; iterate loop; end;
i = i + inc; j = j - inc;
end;
 
/* Display the square. */
do i = 1 to n;
put skip edit (a(i,*)) (f(4));
end;
end;
Output:
   0   1   5   6  14
   2   4   7  13  15
   3   8  12  16  21
   9  11  17  20  22
  10  18  19  23  24 

[edit] PostScript

This implementation is far from being elegant or smart, but it builds the zigzag how a human being could do, and also draws lines to show the path.

%!PS
%%BoundingBox: 0 0 300 200
/size 9 def % defines row * column (9*9 -> 81 numbers,
 % from 0 to 80)
/itoa { 2 string cvs } bind def
% visual bounding box...
% 0 0 moveto 300 0 lineto 300 200 lineto 0 200 lineto
% closepath stroke
20 150 translate
% it can be easily enhanced to support more columns and
% rows. This limit is put here just to avoid more than 2
% digits, mainly because of formatting
size size mul 99 le {
/Helvetica findfont 14 scalefont setfont
/ulimit size size mul def
/sizem1 size 1 sub def
 % prepare the number list
0 ulimit 1 sub { dup 1 add } repeat
ulimit array astore
/di -1 def /dj 1 def
/ri 1 def /rj 0 def /pus true def
0 0 moveto
/i 0 def /j 0 def
{  % can be rewritten a lot better :)
0.8 setgray i 30 mul j 15 mul neg lineto stroke
0 setgray i 30 mul j 15 mul neg moveto itoa show
i 30 mul j 15 mul neg moveto
pus {
i ri add size ge {
/ri 0 def /rj 1 def
} if
j rj add size ge {
/ri 1 def /rj 0 def
} if
/pus false def
/i i ri add def
/j j rj add def
/ri rj /rj ri def def
} {
i di add dup 0 le
exch sizem1 ge or
j dj add dup 0 le
exch sizem1 ge or
or {
/pus true def
/i i di add def /j j dj add def
/di di neg def /dj dj neg def
} {
/i i di add def /j j dj add def
} ifelse
} ifelse
} forall
stroke showpage
} if
%%EOF

[edit] PowerShell

function zigzag( [int] $n ) {
$zigzag=New-Object 'Object[,]' $n,$n
$nodd = $n -band 1
$nm1 = $n - 1
$i=0;
$j=0;
foreach( $k in 0..( $n * $n - 1 ) ) {
$zigzag[$i,$j] = $k
$iodd = $i -band 1
$jodd = $j -band 1
if( ( $j -eq $nm1 ) -and ( $iodd -ne $nodd ) ) {
$i++
} elseif( ( $i -eq $nm1 ) -and ( $jodd -eq $nodd ) ) {
$j++
} elseif( ( $i -eq 0 ) -and ( -not $jodd ) ) {
$j++
} elseif( ( $j -eq 0 ) -and $iodd ) {
$i++
} elseif( $iodd -eq $jodd ) {
$i--
$j++
} else {
$i++
$j--
}
}
,$zigzag
}
 
function displayZigZag( [int] $n ) {
$a = zigzag $n
0..$n | ForEach-Object {
$b=$_
$pad=($n*$n-1).ToString().Length
"$(0..$n | ForEach-Object {
"
{0,$pad}" -f $a[$b,$_]
} )"

}
}

[edit] Prolog

Works with: SWi-Prolog
zig_zag(N) :-
zig_zag(N, N).
 
% compute zig_zag for a matrix of Lig lines of Col columns
zig_zag(Lig, Col) :-
length(M, Lig),
maplist(init(Col), M),
fill(M, 0, 0, 0, Lig, Col, up),
% display the matrix
maplist(print_line, M).
 
 
fill(M, Cur, L, C, NL, NC, _) :-
L is NL - 1,
C is NC - 1,
nth0(L, M, Line),
nth0(C, Line, Cur).
 
fill(M, Cur, L, C, NL, NC, Sens) :-
nth0(L, M, Line),
nth0(C, Line, Cur),
Cur1 is Cur + 1,
compute_next(NL, NC, L, C, Sens, L1, C1, Sens1),
fill(M, Cur1, L1, C1, NL, NC, Sens1).
 
 
init(N, L) :-
length(L, N).
 
% compute_next
% arg1 : Number of lnes of the matrix
% arg2 : number of columns of the matrix
% arg3 : current line
% arg4 : current column
% arg5 : current direction of movement
% arg6 : nect line
% arg7 : next column
% arg8 : next direction of movement
compute_next(_NL, NC, 0, Col, up, 0, Col1, down) :-
Col < NC - 1,
Col1 is Col+1.
 
compute_next(_NL, NC, 0, Col, up, 1, Col, down) :-
Col is NC - 1.
 
compute_next(NL, _NC, Lig, 0, down, Lig1, 0, up) :-
Lig < NL - 1,
Lig1 is Lig+1.
 
compute_next(NL, _NC, Lig, 0, down, Lig, 1, up) :-
Lig is NL - 1.
 
compute_next(NL, _NC, Lig, Col, down, Lig1, Col1, down) :-
Lig < NL - 1,
Lig1 is Lig + 1,
Col1 is Col-1.
 
compute_next(NL, _NC, Lig, Col, down, Lig, Col1, up) :-
Lig is NL - 1,
Col1 is Col+1.
 
compute_next(_NL, NC, Lig, Col, up, Lig1, Col1, up) :-
Col < NC - 1,
Lig1 is Lig - 1,
Col1 is Col+1.
 
compute_next(_NL, NC, Lig, Col, up, Lig1, Col, down) :-
Col is NC - 1,
Lig1 is Lig + 1.
 
 
print_line(L) :-
maplist(print_val, L),
nl.
 
print_val(V) :-
writef('%3r ', [V]).
 
Output:
?- zig_zag(5).
  0   1   5   6  14 
  2   4   7  13  15 
  3   8  12  16  21 
  9  11  17  20  22 
 10  18  19  23  24 
true .

?- zig_zag(5, 7).
  0   1   5   6  14  15  24 
  2   4   7  13  16  23  25 
  3   8  12  17  22  26  31 
  9  11  18  21  27  30  32 
 10  19  20  28  29  33  34 
true .

?- zig_zag(7,5).
  0   1   5   6  14 
  2   4   7  13  15 
  3   8  12  16  24 
  9  11  17  23  25 
 10  18  22  26  31 
 19  21  27  30  32 
 20  28  29  33  34 
true .

[edit] PureBasic

Translation of: AutoHotkey

Procedure zigZag(size)
Protected i, v, x, y
 
Dim a(size - 1, size - 1)
 
x = 1
y = 1
For i = 1 To size * size ;loop once for each element
a(x - 1, y - 1) = v ;assign the next index
 
If (x + y) & 1 = 0 ;even diagonal (zero based count)
If x < size ;while inside the square
If y > 1 ;move right-up
y - 1
EndIf
x + 1
Else
y + 1 ;on the edge increment y, but not x until diagonal is odd
EndIf
Else ;odd diagonal (zero based count)
If y < size ;while inside the square
If x > 1 ;move left-down
x - 1
EndIf
y + 1
Else
x + 1 ;on the edge increment x, but not y until diagonal is even
EndIf
EndIf
v + 1
Next
 
 
;generate and show printout
PrintN("Zig-zag matrix of size " + Str(size) + #CRLF$)
maxDigitCount = Len(Str(size * size)) + 1
For y = 0 To size - 1
For x = 0 To size - 1
Print(RSet(Str(a(x, y)), maxDigitCount, " "))
Next
PrintN("")
Next
PrintN("")
EndProcedure
 
If OpenConsole()
zigZag(5)
zigZag(6)
 
Print(#CRLF$ + #CRLF$ + "Press ENTER to exit")
Input()
CloseConsole()
EndIf
Output:
Zig-zag matrix of size 5

  0  1  5  6 14
  2  4  7 13 15
  3  8 12 16 21
  9 11 17 20 22
 10 18 19 23 24

Zig-zag matrix of size 6

  0  1  5  6 14 15
  2  4  7 13 16 25
  3  8 12 17 24 26
  9 11 18 23 27 32
 10 19 22 28 31 33
 20 21 29 30 34 35

[edit] Python

[edit] Python: By sorting indices

There is a full explanation of the algorithm used by paddy3118.

def zigzag(n):
indexorder = sorted(((x,y) for x in range(n) for y in range(n)),
key = lambda (x,y): (x+y, -y if (x+y) % 2 else y) )
return {index: n for n,index in enumerate(indexorder)}
 
def printzz(myarray):
n = int(len(myarray)** 0.5 +0.5)
for x in range(n):
for y in range(n):
print "%2i" % myarray[(x,y)],
print
 
printzz(zigzag(6))
Output:
     0  1  5  6 14 15
     2  4  7 13 16 25
     3  8 12 17 24 26
     9 11 18 23 27 32
    10 19 22 28 31 33
    20 21 29 30 34 35

[edit] Alternative version,
Translation of: Common Lisp

def zigzag(n):
def move(i, j):
if j < (n - 1):
return max(0, i-1), j+1
else:
return i+1, j
a = [[0] * n for _ in xrange(n)]
x, y = 0, 0
for v in xrange(n * n):
a[y][x] = v
if (x + y) & 1:
x, y = move(x, y)
else:
y, x = move(y, x)
return a
 
from pprint import pprint
pprint(zigzag(5))
Output:
[[0, 1, 5, 6, 14],
[2, 4, 7, 13, 15],
[3, 8, 12, 16, 21],
[9, 11, 17, 20, 22],
[10, 18, 19, 23, 24]]

[edit] Alternative version, inspired by the Common Lisp Alternative Approach

 
COLS = 9
def CX(x, ran):
while True:
x += 2 * next(ran)
yield x
x += 1
yield x
ran = []
d = -1
for V in CX(1,iter(list(range(0,COLS,2)) + list(range(COLS-1-COLS%2,0,-2)))):
ran.append(iter(range(V, V+COLS*d, d)))
d *= -1
for x in range(0,COLS):
for y in range(x, x+COLS):
print(repr(next(ran[y])).rjust(3), end = ' ')
print()
 
Output:
COLS = 5 Produces:
  1   2   6   7  15 
  3   5   8  14  16 
  4   9  13  17  22 
 10  12  18  21  23 
 11  19  20  24  25 
Output:
COLS = 8 Produces:
  1   2   6   7  15  16  28  29 
  3   5   8  14  17  27  30  43 
  4   9  13  18  26  31  42  44 
 10  12  19  25  32  41  45  54 
 11  20  24  33  40  46  53  55 
 21  23  34  39  47  52  56  61 
 22  35  38  48  51  57  60  62 
 36  37  49  50  58  59  63  64 
Output:
COLS = 9 Produces:
  1   2   6   7  15  16  28  29  45 
  3   5   8  14  17  27  30  44  46 
  4   9  13  18  26  31  43  47  60 
 10  12  19  25  32  42  48  59  61 
 11  20  24  33  41  49  58  62  71 
 21  23  34  40  50  57  63  70  72 
 22  35  39  51  56  64  69  73  78 
 36  38  52  55  65  68  74  77  79 
 37  53  54  66  67  75  76  80  81 

[edit] Rascal

This example is incorrect. Output is striped rather than zig-zag

i.e. your numbers always increase going diagonally down and to the left when it should alternativly increase/decrease. Please fix the code and remove this message.

This is a translation of the Python example. As explained on the Talk page, the key way to understand a zig-zag matrix is to write down an example with coordinates:

0 (0,0), 1 (0,1), 3 (0,2)
2 (1,0), 4 (1,1), 6 (1,2)
5 (2,0), 7 (2,1), 8 (2,2)

If you order these coordinates on the number, you create the order:

 0 (0,0), 1 (0,1), 2 (1,0), 3 (0,2), 4 (1,1), 5 (2,0), 6 (1,2), 7 (2,1), 8 (2,2)

One can observe that this increases with the sum of the coordinates, and secondly with the the first number of the coordinates. The Rascal example uses this phenomenon:

import util::Math;
import List;
import Set;
import IO;
 
alias cd = tuple[int,int];
 
public rel[cd, int] zz(int n){
indexorder = sort([<x,y>| x <- [0..n], y <- [0..n]],
bool (cd a, cd b){
if (a[0]+a[1] > b[0]+b[1])
return false;
elseif(a[0] < b[0])
return false;
else
return true;
;
});
return {<indexorder[z] , z> | z <- index(indexorder)};
}
 
public void printzz(rel[cd, int] myarray){
n = floor(sqrt(size(myarray)));
for (x <- [0..n-1]){
for (y <- [0..n-1]){
print("<myarray[<y,x>]>\t");}
println();}
}
Output:
rascal>printzz(zz(4))
{0} {1} {3} {6} {10}
{2} {4} {7} {11} {15}
{5} {8} {12} {16} {19}
{9} {13} {17} {20} {22}
{14} {18} {21} {23} {24}
ok

[edit] Qi

This is a purely functional, very inefficient, and straight forward solution. The code can probably be simplified somewhat.

 
(define odd? A -> (= 1 (MOD A 2)))
(define even? A -> (= 0 (MOD A 2)))
 
(define zigzag-val
0 0 N -> 0
 
X 0 N -> (1+ (zigzag-val (1- X) 0 N)) where (odd? X)
X 0 N -> (1+ (zigzag-val (1- X) 1 N))
 
0 Y N -> (1+ (zigzag-val 1 (1- Y) N)) where (odd? Y)
0 Y N -> (1+ (zigzag-val 0 (1- Y) N))
 
X Y N -> (1+ (zigzag-val (MAX 0 (1- X)) (MIN (1- N) (1+ Y)) N)) where (even? (+ X Y))
X Y N -> (1+ (zigzag-val (MIN (1- N) (1+ X)) (MAX 0 (1- Y)) N)))
 
(define range
E E -> []
S E -> [S|(range (1+ S) E)])
 
(define zigzag
N -> (map (/. Y
(map (/. X
(zigzag-val X Y N))
(range 0 N)))
(range 0 N)))
 

[edit] R

Translation of: Java
zigzag <- function(size)
{
digits <- seq_len(size^2) - 1
mat <- matrix(0, nrow = size, ncol=size)
i <- 1
j <- 1
for(element in digits)
{
mat[i,j] <- element
if((i + j) %% 2 == 0)
{
# Even stripes
if(j < size) j <- j + 1 else i <- i + 2
if(i > 1) i <- i - 1
} else
{
# Odd stripes
if(i < size) i <- i + 1 else j <- j + 2
if(j > 1) j <- j - 1
}
}
mat
}
 
zigzag(5)

[edit] Racket

 
#lang racket
 
(define/match (compare i j)
[((list x y) (list a b)) (or (< x a) (and (= x a) (< y b)))])
 
(define/match (key i)
[((list x y)) (list (+ x y) (if (even? (+ x y)) (- y) y))])
 
(define (zigzag-ht n)
(define indexorder
(sort (for*/list ([x n] [y n]) (list x y))
compare #:key key))
(for/hash ([(n i) (in-indexed indexorder)]) (values n i)))
 
(define (zigzag n)
(define ht (zigzag-ht n))
(for/list ([x n])
(for/list ([y n])
(hash-ref ht (list x y)))))
 
(zigzag 4)
 
Output:
 
'((0 2 3 9)
(1 4 8 10)
(5 7 11 14)
(6 12 13 15))
 

[edit] REXX

/*REXX program to produce a  zig-zag  matrix  (array)  and display it.  */
 
parse arg n start inc . /*get any and/or all arguments. */
if n=='' then n=5 /*if not specified, use default. */
if start=='' then start=0 /* " " " " " */
if inc=='' then inc=1 /* " " " " " */
row=1; col=1 /*start with 1st row, 1st column.*/
 
do j=start by inc for n*n; @.row.col=j
if (row+col)//2==0 then do
if col<n then col=col+1
else row=row+2
if row\==1 then row=row-1
end
else do
if row<n then row=row+1
else col=col+2
if col\==1 then col=col-1
end
end /*j*/
 
L=max(length(start),length(start+n*n*inc)) /*max length of any element.*/
 
do row=1 for n; _= /*show all the matrix's rows. */
do col=1 for n; _=_ right(@.row.col,L); end /*col*/
say _ /*show the row just constructed. */
end /*row*/
/*stick a fork in it, we're done.*/
Output:
when using the default of: 5
  0  1  5  6 14
  2  4  7 13 15
  3  8 12 16 21
  9 11 17 20 22
 10 18 19 23 24

[edit] Ruby

Translation of: Python
def zigzag(n)
indices = []
n.times {|x| n.times {|y| indices << [x,y] }}
zigzag = Array.new(n) {Array.new(n)} # n x n array of nils
indices.sort_by {|x,y| [x+y, (x+y).even? ? y : -y]} \
.each_with_index {|(x,y),i| zigzag[x][y] = i}
zigzag
end
 
def print_matrix(m)
format = "%#{m.flatten.max.to_s.size}d "
puts m.map {|row| row.map {|val| format % val}.join}
end
 
print_matrix zigzag(5)
Output:
 0  1  5  6 14 
 2  4  7 13 15 
 3  8 12 16 21 
 9 11 17 20 22 
10 18 19 23 24

[edit] Scala

Uses the array indices sort solution used by others here.

def zigzag(n:int) = {
var l = List[Tuple2[int,int]]()
(0 until n*n) foreach {i=>l = l + (i%n,i/n)}
l = l.sort{case ((x,y),(u,v)) => if (x+y == u+v)
if ((x+y) % 2 == 0) x<u else y<v
else (x+y) < (u+v) }
var a = new Array[Array[int]](n,n)
l.zipWithIndex foreach {case ((x,y),i) => a(y)(x) = i}
a
}

Or, compressed into just one statement

def zigzag(n:int) = {
var indices = List[Tuple2[Int,Int]]()
var array = new Array[Array[Int]](n,n)
 
(0 until n*n).foldLeft(indices)((l,i) => l + (i%n,i/n)).
sort{case ((x,y),(u,v)) => if (x+y == u+v)
if ((x+y) % 2 == 0) x<u else y<v
else (x+y) < (u+v) }.
zipWithIndex.foldLeft(array) {case (a,((x,y),i)) => a(y)(x) = i; a}
}
 

[edit] Seed7

$ include "seed7_05.s7i";
 
const type: matrix is array array integer;
 
const func matrix: zigzag (in integer: size) is func
result
var matrix: s is matrix.value;
local
var integer: i is 1;
var integer: j is 1;
var integer: d is -1;
var integer: max is 0;
var integer: n is 0;
begin
s := size times size times 0;
max := size ** 2;
for n range 1 to max div 2 + 1 do
s[i][j] := n;
s[size - i + 1][size - j + 1] := max - n + 1;
i +:= d;
j -:= d;
if i < 1 then
incr(i);
d := -d;
elsif j < 1 then
incr(j);
d := -d;
end if;
end for;
end func;
 
const proc: main is func
local
var matrix: s is matrix.value;
var integer: i is 0;
var integer: num is 0;
begin
s := zigzag(7);
for i range 1 to length(s) do
for num range s[i] do
write(num lpad 4);
end for;
writeln;
end for;
end func;
Output:
   1   2   6   7  15  16  28
   3   5   8  14  17  27  29
   4   9  13  18  26  30  39
  10  12  19  25  31  38  40
  11  20  24  32  37  41  46
  21  23  33  36  42  45  47
  22  34  35  43  44  48  49

[edit] Tcl

Using print_matrix from Matrix Transpose

proc zigzag {size} {
set m [lrepeat $size [lrepeat $size .]]
set x 0; set dx -1
set y 0; set dy 1
 
for {set i 0} {$i < $size ** 2} {incr i} {
if {$x >= $size} {
incr x -1
incr y 2
negate dx dy
} elseif {$y >= $size} {
incr x 2
incr y -1
negate dx dy
} elseif {$x < 0 && $y >= 0} {
incr x
negate dx dy
} elseif {$x >= 0 && $y < 0} {
incr y
negate dx dy
}
lset m $x $y $i
incr x $dx
incr y $dy
}
return $m
}
 
proc negate {args} {
foreach varname $args {
upvar 1 $varname var
set var [expr {-1 * $var}]
}
}
 
print_matrix [zigzag 5]
Output:
 0  1  5  6 14 
 2  4  7 13 15 
 3  8 12 16 21 
 9 11 17 20 22 
10 18 19 23 24

[edit] Ursala

adapted from the J solution

#import std
#import nat
 
zigzag = ~&mlPK2xnSS+ num+ ==+sum~~|=xK9xSL@iiK0+ iota

test program (three examples):

#cast %nLLL
 
tests = zigzag* <4,5,6>
Output:
<
   <
      <0,1,5,6>,
      <2,4,7,12>,
      <3,8,11,13>,
      <9,10,14,15>>,
   <
      <0,1,5,6,14>,
      <2,4,7,13,15>,
      <3,8,12,16,21>,
      <9,11,17,20,22>,
      <10,18,19,23,24>>,
   <
      <0,1,5,6,14,15>,
      <2,4,7,13,16,25>,
      <3,8,12,17,24,26>,
      <9,11,18,23,27,32>,
      <10,19,22,28,31,33>,
      <20,21,29,30,34,35>>>

[edit] VBA

 
Public Sub zigzag(n)
Dim a() As Integer
'populate a (1,1) to a(n,n) in zigzag pattern
 
'check if n too small
If n < 1 Then
Debug.Print "zigzag: enter a number greater than 1"
Exit Sub
End If
 
'initialize
ReDim a(1 To n, 1 To n)
i = 1 'i is the row
j = 1 'j is the column
P = 0 'P is the next number
a(i, j) = P 'fill in initial value
 
'now zigzag through the matrix and fill it in
Do While (i <= n) And (j <= n)
'move one position to the right or down the rightmost column, if possible
If j < n Then
j = j + 1
ElseIf i < n Then
i = i + 1
Else
Exit Do
End If
'fill in
P = P + 1: a(i, j) = P
'move down to the left
While (j > 1) And (i < n)
i = i + 1: j = j - 1
P = P + 1: a(i, j) = P
Wend
'move one position down or to the right in the bottom row, if possible
If i < n Then
i = i + 1
ElseIf j < n Then
j = j + 1
Else
Exit Do
End If
P = P + 1: a(i, j) = P
'move back up to the right
While (i > 1) And (j < n)
i = i - 1: j = j + 1
P = P + 1: a(i, j) = P
Wend
Loop
 
'print result
Debug.Print "Result for n="; n; ":"
For i = 1 To n
For j = 1 To n
Debug.Print a(i, j),
Next
Debug.Print
Next
End Sub
 
Output:
zigzag 5
Result for n= 5 :
 0             1             5             6             14           
 2             4             7             13            15           
 3             8             12            16            21           
 9             11            17            20            22           
 10            18            19            23            24           

zigzag 6
Result for n= 6 :
 0             1             5             6             14            15           
 2             4             7             13            16            25           
 3             8             12            17            24            26           
 9             11            18            23            27            32           
 10            19            22            28            31            33           
 20            21            29            30            34            35           

[edit] XPL0

include c:\cxpl\codes;
def N=6;
int A(N,N), X, Y, I, D;
[I:=0; X:=0; Y:=0; D:=1;
repeat A(X,Y):=I;
case of
X+D>=N: [D:=-D; Y:=Y+1];
Y-D>=N: [D:=-D; X:=X+1];
X+D<0: [D:=-D; Y:=Y+1];
Y-D<0: [D:=-D; X:=X+1]
other [X:=X+D; Y:=Y-D];
I:=I+1;
until I>=N*N;
for Y:=0 to N-1 do
[for X:=0 to N-1 do
[I:=A(X,Y);
ChOut(0,^ );
if I<10 then ChOut(0,^ );
IntOut(0, I);
];
CrLf(0);
];
]
Output:
  0  1  5  6 14 15
  2  4  7 13 16 25
  3  8 12 17 24 26
  9 11 18 23 27 32
 10 19 22 28 31 33
 20 21 29 30 34 35

[edit] zkl

Translation of: Groovy
fcn zz(n){
grid := (0).pump(n,List, (0).pump(n,List).copy).copy();
ri := Ref(0);
foreach d in ([1..n*2]){
x:=(0).max(d - n); y:=(n - 1).min(d - 1);
(0).pump(d.min(n*2 - d),Void,'wrap(it){
grid[if(d%2)y-it else x+it][if(d%2)x+it else y-it] = ri.inc();
});
}
grid.pump(String,'wrap(r){("%3s"*n+"\n").fmt(r.xplode())});
}
Translation of: C

Using list comprehension (and side effects) for the double loop, the resulting list is just thrown away, which is easier than creating an enumerated list and sorting.

fcn ceg(m){
s  := (0).pump(m*m,List).copy(); // copy to make writable
rn := Ref(0);
[[(i,j); [0..m*2-1]; '{[(0).max(i-m+1) .. i.min(m-1)]};
'{ s[ if(i.isOdd) j*(m-1)+i else (i-j)*m+j ] = rn.inc(); }]];
s.pump(String,T(Void.Read,m-1), ("%3s"*m+"\n").fmt);
}

To be pedantic, the same as above, but using the output of the list comprehension:

fcn ceg2(m){
rn := Ref(0);
[[(i,j); [0..m*2-1]; '{[(0).max(i-m+1) .. i.min(m-1)]};
'{ T( if(i.isOdd) j*(m-1)+i else (i-j)*m+j;, rn.inc() ) }]]
.sort(fcn([(a,_)], [(b,_)]){ a<b }).apply("__sGet",1)
.pump(String,T(Void.Read,m-1), ("%3s"*m+"\n").fmt);
}
Output:
The results are the same:
zz(5).println();
  0  1  5  6 14
  2  4  7 13 15
  3  8 12 16 21
  9 11 17 20 22
 10 18 19 23 24
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