Zig-zag matrix: Difference between revisions
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===Alternative Version=== |
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{{trans|Scala}} |
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Same output. |
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<lang d>import std.stdio, std.algorithm, std.typecons, std.range, std.array; |
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int[][] zigZag(int n) { |
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alias P2 = Tuple!(int,"x", int,"y"); |
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auto L = iota(n ^^ 2).map!(i => P2(i % n, i / n)).array; |
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L.sort!q{ (a.x + a.y == b.x + b.y) ? |
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((a.x + a.y) % 2 ? a.y < b.y : a.x < b.x) : |
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(a.x + a.y) < (b.x + b.y) }; |
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auto result = new typeof(return)(n, n); |
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foreach (i, p; L) |
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result[p.y][p.x] = i; |
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return result; |
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} |
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void main() { |
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writefln("%(%(%2d %)\n%)", 5.zigZag); |
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}</lang> |
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=={{header|E}}== |
=={{header|E}}== |
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Revision as of 23:05, 7 May 2013
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
ActionScript
<lang as> 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 --;
}
}
}
}
} </lang>
Ada
<lang 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;</lang> The function Zig_Zag generates a square matrix filled as requested by the task.
Sample 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
ALGOL 68
<lang algol68>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#</lang>
Sample 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
|
APL
<lang apl> 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</lang>
AppleScript
Here's a vector & matrix boundary detection approach to the Zig-zap matrix: <lang AppleScript>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</lang> But this can be improved upon by building the matrix by populating empty AppleScript lists (it's about 50% faster when n=50):<lang AppleScript>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 </lang>Output for both scripts is:<lang AppleScript>"
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
"</lang>
AutoHotkey
contributed by Laszlo on the ahk forum. <lang AutoHotkey>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</lang>
AutoIt
<lang 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 $av_array EndFunc ;==>ZigZag </lang>
AWK
<lang AWK>
- syntax: GAWK -f ZIG-ZAG_MATRIX.AWK [-v offset={0|1}] [size]
BEGIN {
- offset: "0" prints 0 to size^2-1 while "1" prints 1 to size^2
offset = (offset == "") ? 0 : offset
size = (ARGV[1] == "") ? 5 : ARGV[1]
if (offset !~ /^[01]$/) { 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)
} </lang>
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
BBC BASIC
<lang bbcbasic> 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%</lang>
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
C
<lang 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; }</lang>output<lang>% ./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</lang>
C++
<lang cpp>#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 currNum = 0; 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 ( currNum <= 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 ) ); }</lang>
C#
<lang csharp>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;
}</lang>
Clojure
Purely functional approach. <lang Clojure>(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))</lang>
CoffeeScript
<lang 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"
</lang>
output <lang> > 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 ] ]
</lang>
Common Lisp
(but with zero-based indexes and combining the even and odd cases)
<lang lisp>(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))))</lang>
An alternative approach
<lang lisp>
- 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)))))))
</lang> (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
D
<lang d>int[][] zigZag(in int n) pure nothrow {
static void move(in int n, ref int i, ref int j) pure nothrow {
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%)", zigZag(5));
}</lang>
- 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
Alternative Version
Same output. <lang d>import std.stdio, std.algorithm, std.typecons, std.range, std.array;
int[][] zigZag(int n) {
alias P2 = Tuple!(int,"x", int,"y"); auto L = iota(n ^^ 2).map!(i => P2(i % n, i / n)).array;
L.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) };
auto result = new typeof(return)(n, n);
foreach (i, p; L)
result[p.y][p.x] = i;
return result;
}
void main() {
writefln("%(%(%2d %)\n%)", 5.zigZag);
}</lang>
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.
<lang e>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
}</lang>
Euphoria
<lang Euphoria>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)</lang>
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}
}
Fan
<lang 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))
}
}</lang>
Forth
<lang 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</lang>
Fortran
<lang fortran>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</lang>
GAP
<lang 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 ] ]</lang>
Go
Edge direct algorithm
<lang go>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("")
}
}
}</lang> 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
Groovy
Edge
An odd technique that traverses the grid edges directly and calculates the transform onto the grid.
<lang groovy>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
}</lang>
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
Cursor
Ported from the Java example
<lang groovy>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
}</lang>
Output
> zz(5).each { it.each { print("${it}".padLeft(3)) }; println() }
0 1 5 6 14
2 4 7 13 15
3 8 12 16 21
9 11 17 20 22
10 18 19 23 24
Sorting
Ported from the Python example with some input from J
<lang groovy>def zz = { n ->
(0..<n*n).collect { [x:it%n,y:(int)(it/n)] }.sort { c->
[c.x+c.y, (((c.x+c.y)%2) ? c.y : -c.y)]
}.with { l -> l.inject(new int[n][n]) { a, c -> a[c.y][c.x] = l.indexOf(c); a } }
}</lang>
Output
> zz(5).each { it.each { print("${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
Haskell
Computing the array:
<lang haskell>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)</lang>
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):
<lang haskell>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)]</lang>
Icon and Unicon
This solution works for both Icon and Unicon.
<lang icon>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</lang>
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
->
J
A succinct way: <lang j> ($ [: /:@; <@|.`</.@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</lang>
This version is longer, but more "mathematical" and less "procedural": <lang j> ($ [: /:@; [: <@(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</lang>
Leveraging a useful relationship among the indices: <lang j> ($ ([: /:@;@(+/"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</lang>
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: <lang j> ($ [: /:@; [: <@|.`</. i.) 5 3 0 1 5 2 4 6 3 7 11 8 10 12 9 13 14</lang>
Java
<lang java>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;
}</lang>
JavaScript
for the print() function.
Subclasses the Matrix class defined at Matrix Transpose#JavaScript <lang 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);</lang> 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
Joy
<lang 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. </lang>
Lua
<lang 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)) </lang>
M4
<lang 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)</lang>
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
Mathematica
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. <lang Mathematica>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-tmpsize-i+1,size-j+1-1]
]</lang> Ported from the java-example: <lang Mathematica>ZigZag2[size_] := Module[{data, i, j, elem},
data = ConstantArray[0, {size, size}];
i = j = 1;
For[elem = 0, elem < size^2, elem++,
datai, 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
]</lang>
Examples: <lang Mathematica>ZigZag[5] // MatrixForm ZigZag2[6] // MatrixForm</lang> gives back:
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.
<lang MATLAB>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</lang>
Sample Output: <lang MATLAB>>> 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</lang>
Maxima
<lang 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]) */</lang>
Modula-3
<lang modula3>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.</lang> 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
NetRexx
<lang NetRexx>/* 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
</lang>
Objeck
<lang ocaml> 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;
} </lang>
OCaml
<lang ocaml>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</lang>
Octave
See the MATLAB solution, which works perfectly in Octave too.
ooRexx
<lang ooRexx> 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
</lang> 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 |
Oz
Implemented as a state machine: <lang oz>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 {$} {MakeTuple unit W} end}
T
end
in
{Inspect {CreateZigZag 5}}</lang>
PARI/GP
<lang parigp>zz(n)={ my(M=matrix(n,n),i,j,d=-1,start,end=n^2-1); while(ct--, M[i+1,j+1]=start; M[n-i,n-j]=end; start++; end--; i+=d; j-=d; if(i<0, i++; d=-d , if(j<0, j++; d=-d ) ); if(start>end,return(M)) ) };</lang>
Pascal
<lang Pascal>Program zigzag;
const
size = 5;
var
zzarray: array [1..size, 1..size] of integer; element, i, j: integer; direction: integer;
begin
i := 1;
j := 1;
direction := 1;
for element := 1 to size*size do
begin
zzarray[i,j] := element;
i := i + direction;
j := j - direction;
if (i = 0) then
begin
direction := -direction;
i := i + 1;
end;
if (i = size +1) then
begin
direction := -direction;
i := i - 1;
j := j + 2;
end;
if (j = 0) then
begin
direction := -direction;
j := j + 1;
end;
if (j = size + 1) then
begin
direction := -direction;
j := j - 1;
i := i + 2;
end;
end;
for j := 1 to size do
begin
for i := 1 to size do
write(zzarray[i,j]:3);
writeln;
end;
end.</lang> output
1 2 6 7 15 3 5 8 14 16 4 9 13 17 22 10 12 18 21 23 11 19 20 24 25
Perl
<lang perl>sub zig_zag {
my ($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);</lang>
Perl 6
Assuming the same Turtle class that is used in Spiral_matrix: <lang perl6>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);
} $t.showmap;
}</lang>
PHP
<lang php>function ZigZagMatrix($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; }</lang>
PL/I
<lang 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; </lang>
PicoLisp
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")
(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) )</lang>
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
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.
<lang postscript>%!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</lang>
PowerShell
<lang 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,$_]
} )"
}
}</lang>
Prolog
Works with SWi-Prolog. <lang 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]). </lang> 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 .
PureBasic
<lang purebasic>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</lang> Sample 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
Python
Python: By sorting indices
There is a full explanation of the algorithm used here. <lang python>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))</lang> Program 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
Alternative version,
<lang python>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))</lang> Output: <lang python>[[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]]</lang>
Alternative version, inspired by the Common Lisp Alternative Approach
<lang python>
- ZigZag
- Nigel Galloway: June 7th., 2012,
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()
</lang> 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
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
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
Rascal
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: <lang rascal>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)</lang> If you order these coordinates on the number, you create the order: <lang rascal> 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)</lang> 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: <lang rascal>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();}
}</lang> This has as result: <lang rascal>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</lang>
Qi
This is a purely functional, very inefficient, and straight forward solution. The code can probably be simplified somewhat.
<lang qi> (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)))
</lang>
R
<lang R>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)</lang>
Racket
<lang 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) </lang> Output: <lang racket> '((0 2 3 9)
(1 4 8 10) (5 7 11 14) (6 12 13 15))
</lang>
REXX
<lang 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.*/</lang>
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
Ruby
<lang ruby>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)
width = m.flatten.max.to_s.size
m.each {|row| row.each {|val| print "%#{width}d " % val}; puts}
end
print_matrix zigzag(5)</lang>
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
Scala
Uses the array indices sort solution used by others here.
<lang scala>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
}</lang>
Or, compressed into just one statement
<lang scala>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}
} </lang>
Seed7
<lang 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;</lang>
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
Tcl
Using print_matrix from Matrix Transpose…
<lang tcl>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]</lang>
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
Ursala
adapted from the J solution <lang Ursala>#import std
- import nat
zigzag = ~&mlPK2xnSS+ num+ ==+sum~~|=xK9xSL@iiK0+ iota</lang> test program (three examples): <lang Ursala>#cast %nLLL
tests = zigzag* <4,5,6></lang> 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>>>
VBA
<lang 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 </lang>
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
XPL0
<lang 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);
];
]</lang>
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
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