Zig-zag matrix: Difference between revisions
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The sequence of turns for the first half of the matrix is east to southwest to south to northeast to east. In the second half the order of turns is reversed. |
The sequence of turns for the first half of the matrix is east to southwest to south to northeast to east. In the second half the order of turns is reversed. |
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<lang Quackery> [ stack ] is stepcount ( --> s ) |
<lang Quackery> [ stack ] is stepcount ( --> s ) |
Revision as of 22:54, 22 February 2022
You are encouraged to solve this task according to the task description, using any language you may know.
- Task
Produce a zig-zag array.
A zig-zag array is a square arrangement of the first N2 natural numbers, where the
numbers increase sequentially as you zig-zag along the array's anti-diagonals.
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
- Related tasks
- See also
- Wiktionary entry: anti-diagonals
11l
<lang 11l>F zigzag(n)
F compare(xy) V (x, y) = xy R (x + y, I (x + y) % 2 {-y} E y) V xs = 0 .< n R Dict(enumerate(sorted((multiloop(xs, xs, (x, y) -> (x, y))), key' compare)), (n, index) -> (index, n))
F printzz(myarray)
V n = Int(myarray.len ^ 0.5 + 0.5) V xs = 0 .< n print((xs.map(y -> @xs.map(x -> ‘#3’.format(@@myarray[(x, @y)])).join(‘’))).join("\n"))
printzz(zigzag(6))</lang>
- Output:
0 2 3 9 10 20 1 4 8 11 19 21 5 7 12 18 22 29 6 13 17 23 28 30 14 16 24 27 31 34 15 25 26 32 33 35
360 Assembly
<lang 360asm>* Zig-zag matrix 15/08/2015 ZIGZAGMA CSECT
USING ZIGZAGMA,R12 set base register LR R12,R15 establish addressability LA R9,N n : matrix size LA R6,1 i=1 LA R7,1 j=1 LR R11,R9 n MR R10,R9 *n BCTR R11,0 R11=n**2-1 SR R8,R8 k=0
LOOPK CR R8,R11 do k=0 to n**2-1
BH ELOOPK k>limit LR R1,R6 i BCTR R1,0 -1 MR R0,R9 *n LR R2,R7 j BCTR R2,0 -1 AR R1,R2 (i-1)*n+(j-1) SLA R1,1 index=((i-1)*n+j-1)*2 STH R8,T(R1) t(i,j)=k LR R2,R6 i AR R2,R7 i+j LA R1,2 2 SRDA R2,32 shift right r1 to r2 DR R2,R1 (i+j)/2 LTR R2,R2 if mod(i+j,2)=0 BNZ ELSEMOD CR R7,R9 if j<n BNL ELSE1 LA R7,1(R7) j=j+1 B EIF1
ELSE1 LA R6,2(R6) i=i+2 EIF1 CH R6,=H'1' if i>1
BNH NOT1 BCTR R6,0 i=i-1
NOT1 B NOT2 ELSEMOD CR R6,R9 if i<n
BNL ELSE2 LA R6,1(R6) i=i+1 B EIF2
ELSE2 LA R7,2(R7) j=j+2 EIF2 CH R7,=H'1' if j>1
BNH NOT2 BCTR R7,0 j=j-1
NOT2 LA R8,1(R8) k=k+1
B LOOPK
ELOOPK LA R6,1 end k; i=1 LOOPI CR R6,R9 do i=1 to n
BH ELOOPI i>n LA R10,0 ibuf=0 buffer index MVC BUFFER,=CL80' ' LA R7,1 j=1
LOOPJ CR R7,R9 do j=1 to n
BH ELOOPJ j>n LR R1,R6 i BCTR R1,0 -1 MR R0,R9 *n LR R2,R7 j BCTR R2,0 -1 AR R1,R2 (i-1)*n+(j-1) SLA R1,1 index=((i-1)*n+j-1)*2 LH R2,T(R1) t(i,j) LA R3,BUFFER AR R3,R10 XDECO R2,XDEC edit t(i,j) length=12 MVC 0(4,R3),XDEC+8 move in buffer length=4 LA R10,4(R10) ibuf=ibuf+1 LA R7,1(R7) j=j+1 B LOOPJ
ELOOPJ XPRNT BUFFER,80 end j
LA R6,1(R6) i=i+1 B LOOPI
ELOOPI XR R15,R15 end i; return_code=0
BR R14 return to caller
N EQU 5 matrix size BUFFER DS CL80 XDEC DS CL12 T DS (N*N)H t(n,n) matrix
YREGS END ZIGZAGMA</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
Action!
<lang Action!>DEFINE MAX_SIZE="10" DEFINE MAX_MATRIX_SIZE="100"
INT FUNC Index(BYTE size,x,y) RETURN (x+y*size)
PROC PrintMatrix(BYTE ARRAY a BYTE size)
BYTE i,j,v FOR j=0 TO size-1 DO FOR i=0 TO size-1 DO v=a(Index(size,i,j)) IF v<10 THEN Print(" ") ELSE Print(" ") FI PrintB(v) OD PutE() OD
RETURN
PROC FillMatrix(BYTE ARRAY a BYTE size)
BYTE start,end INT dir,i,j
start=0 end=size*size-1 i=0 j=0 dir=1
DO a(Index(size,i,j))=start a(Index(size,size-1-i,size-1-j))=end start==+1 end==-1 i==+dir j==-dir IF i<0 THEN i==+1 dir=-dir ELSEIF j<0 THEN j==+1 dir=-dir FI UNTIL start>=end OD
IF start=end THEN a(Index(size,i,j))=start FI
RETURN
PROC Test(BYTE size)
BYTE ARRAY mat(MAX_MATRIX_SIZE) PrintF("Matrix size: %B%E",size) FillMatrix(mat,size) PrintMatrix(mat,size) PutE()
RETURN
PROC Main()
Test(5) Test(6)
RETURN</lang>
- Output:
Screenshot from Atari 8-bit computer
Matrix 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 Matrix 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
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.
- 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
Agena
Tested with Agena 2.9.5 Win32 <lang agena># zig-zag matrix
makeZigZag := proc( n :: number ) :: table is
local move := proc( x :: number, y :: number, upRight :: boolean ) is if y = n then upRight := not upRight; x := x + 1 elif x = 1 then upRight := not upRight; y := y + 1 else x := x - 1; y := y + 1 fi; return x, y, upRight end ;
# create empty table local result := []; for i to n do result[ i ] := []; for j to n do result[ i, j ] := 0 od od;
# fill the table local x, y, upRight := 1, 1, true; for i to n * n do result[ x, y ] := i - 1; if upRight then x, y, upRight := move( x, y, upRight ) else y, x, upRight := move( y, x, upRight ) fi od;
return result
end;
scope
local m := makeZigZag( 5 ); for i to size m do for j to size m do printf( " %3d", m[ i, j ] ) od; print() od
epocs</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
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>
- 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 |
ALGOL W
Based on the Agena sample. <lang algolw>begin % zig-zag matrix %
% z is returned holding a zig-zag matrix of order n, z must be at least n x n % procedure makeZigZag ( integer value n ; integer array z( *, * ) ) ; begin procedure move ; begin if y = n then begin upRight := not upRight; x := x + 1 end else if x = 1 then begin upRight := not upRight; y := y + 1 end else begin x := x - 1; y := y + 1 end end move ; procedure swapXY ; begin integer swap; swap := x; x := y; y := swap; end swapXY ; integer x, y; logical upRight; % initialise the n x n matrix in z % for i := 1 until n do for j := 1 until n do z( i, j ) := 0; % fill in the zig-zag matrix % x := y := 1; upRight := true; for i := 1 until n * n do begin z( x, y ) := i - 1; if upRight then move else begin swapXY; move; swapXY end; end; end makeZigZap ; begin integer array zigZag( 1 :: 10, 1 :: 10 ); for n := 5 do begin makeZigZag( n, zigZag ); for i := 1 until n do begin write( i_w := 4, s_w := 1, zigZag( i, 1 ) ); for j := 2 until n do writeon( i_w := 4, s_w := 1, zigZag( i, j ) ); end end end
end.</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
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
Iterative
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
"
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
"
Recursive
By functional composition: <lang AppleScript>-- zigzagMatrix on zigzagMatrix(n)
-- diagonals :: n -> n script diagonals on |λ|(n) script mf on diags(xs, iCol, iRow) if (iCol < length of xs) then if iRow < n then set iNext to iCol + 1 else set iNext to iCol - 1 end if set {headList, tail} to splitAt(iCol, xs) {headList} & diags(tail, iNext, iRow + 1) else {xs} end if end diags end script diags(enumFromTo(0, n * n - 1), 1, 1) of mf end |λ| end script -- oddReversed :: [a] -> Int -> [a] script oddReversed on |λ|(lst, i) if i mod 2 = 0 then lst else reverse of lst end if end |λ| end script rowsFromDiagonals(n, map(oddReversed, |λ|(n) of diagonals))
end zigzagMatrix
-- Rows of given length from list of diagonals -- rowsFromDiagonals :: Int -> a -> a on rowsFromDiagonals(n, lst)
if length of lst > 0 then -- lengthOverOne :: [a] -> Bool script lengthOverOne on |λ|(lst) length of lst > 1 end |λ| end script set {edge, residue} to splitAt(n, lst) {map(my head, edge)} & ¬ rowsFromDiagonals(n, ¬ map(my tail, ¬ filter(lengthOverOne, edge)) & residue) else {} end if
end rowsFromDiagonals
-- TEST -----------------------------------------------------------------------
on run
zigzagMatrix(5)
end run
-- GENERIC FUNCTIONS ----------------------------------------------------------
-- enumFromTo :: Int -> Int -> [Int] on enumFromTo(m, n)
if n < m then set d to -1 else set d to 1 end if set lst to {} repeat with i from m to n by d set end of lst to i end repeat return lst
end enumFromTo
-- filter :: (a -> Bool) -> [a] -> [a] on filter(f, xs)
tell mReturn(f) set lst to {} set lng to length of xs repeat with i from 1 to lng set v to item i of xs if |λ|(v, i, xs) then set end of lst to v end repeat return lst end tell
end filter
-- head :: [a] -> a on head(xs)
if length of xs > 0 then item 1 of xs else missing value end if
end head
-- map :: (a -> b) -> [a] -> [b] on map(f, xs)
tell mReturn(f) set lng to length of xs set lst to {} repeat with i from 1 to lng set end of lst to |λ|(item i of xs, i, xs) end repeat return lst end tell
end map
-- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Script on mReturn(f)
if class of f is script then f else script property |λ| : f end script end if
end mReturn
-- splitAt:: n -> list -> {n items from start of list, rest of list} -- splitAt :: Int -> [a] -> ([a], [a]) on splitAt(n, xs)
if n > 0 and n < length of xs then {items 1 thru n of xs, items (n + 1) thru -1 of xs} else if n < 1 then {{}, xs} else {xs, {}} end if end if
end splitAt
-- tail :: [a] -> [a] on tail(xs)
if length of xs > 1 then items 2 thru -1 of xs else {} end if
end tail</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}}
Optimised iterative
This is an optimised version of the second iterative script above, with the rendition to text kept separate and corrected. With n = 50, it's about 7.6 times as fast as the script on which it's based.
<lang applescript>on zigzagMatrix(n)
script o property matrix : {} -- Matrix list. property row : missing value -- Row sublist. end script repeat n times set end of o's matrix to {} -- Build a foundation for the matrix out of n empty lists. end repeat set {r, d} to {1, -1} -- Row index and direction to next insertion row (negative = row above). repeat with v from 0 to (n ^ 2) - 1 -- Values to insert. set o's row to o's matrix's item r repeat while ((count o's row) = n) set r to r + 1 set d to 1 set o's row to o's matrix's item r end repeat set end of o's row to v set r to r + d if (r < 1) then set r to 1 set d to -d else if (r > n) then set r to n set d to -d else if ((r > 1) and ((count o's matrix's item (r - 1)) = 1)) then set d to -d end if end repeat return o's matrix
end zigzagMatrix
-- Demo: on matrixToText(matrix, w)
script o property matrix : missing value property row : missing value end script set o's matrix to matrix set padding to " " repeat with r from 1 to (count o's matrix) set o's row to o's matrix's item r repeat with i from 1 to (count o's row) set o's row's item i to text -w thru end of (padding & o's row's item i) end repeat set o's matrix's item r to join(o's row, "") end repeat return join(o's matrix, linefeed)
end matrixToText
on join(lst, delim)
set astid to AppleScript's text item delimiters set AppleScript's text item delimiters to delim set txt to lst as text set AppleScript's text item delimiters to astid return txt
end join
set n to 5 set matrix to zigzagMatrix(n) linefeed & matrixToText(matrix, (count (n ^ 2 - 1 as integer as text)) + 2) & linefeed</lang>
- Output:
<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>
Applesoft BASIC
<lang ApplesoftBasic>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</lang>
Arturo
<lang rebol>zigzag: function [n][
result: map 1..n 'x -> map 1..n => 0
x: 1, y: 1, v: 0, d: 1
while [v < n^2][ if? all? @[1 =< x x =< n 1 =< y y =< n][ set get result (y-1) (x-1) v x: x + d, y: y - d, v: v + 1 ] else[if? x > n [x: n, y: y + 2, d: neg d] else[if? y > n [x: x + 2, y: n, d: neg d] else[if? x < 1 [x: 1, d: neg d] else[if y < 1 [y: 1, d: neg d] ] ] ] ] ] result
]
zz: zigzag 5 loop zz 'row -> print map row 'col [pad to :string col 3]</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
ATS
<lang> (* ****** ****** *) //
- include
"share/atspre_define.hats" // defines some names
- include
"share/atspre_staload.hats" // for targeting C
- include
"share/HATS/atspre_staload_libats_ML.hats" // for ... // (* ****** ****** *) // extern fun Zig_zag_matrix(n: int): void // (* ****** ****** *)
fun max(a: int, b: int): int =
if a > b then a else b
fun movex(n: int, x: int, y: int): int =
if y < n-1 then max(0, x-1) else x+1
fun movey(n: int, x: int, y: int): int =
if y < n-1 then y+1 else y
fun zigzag(n: int, i: int, row: int, x: int, y: int): void =
if i = n*n then () else let val () = (if x = row then begin print i; print ','; end else ()) //val () = (begin print x; print ' '; print y; print ' '; print i; print ' '; end) val nextX: int = if ((x+y) % 2) = 0 then movex(n, x, y) else movey(n, y, x) val nextY: int = if ((x+y) % 2) = 0 then movey(n, x, y) else movex(n, y, x) in zigzag(n, i+1, row, nextX, nextY) end
implement Zig_zag_matrix(n) =
let fun loop(row: int): void = if row = n then () else let val () = zigzag(n, 0, row, 0, 0) val () = println!(" ") in loop(row + 1) end in loop(0) end
(* ****** ****** *)
implement main0() = () where {
val () = Zig_zag_matrix(5)
} (* end of [main0] *)
(* ****** ****** *) </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
Beads
This is a translation of the C++ example. <lang Beads>beads 1 program 'Zig-zag Matrix'
calc main_init var test : array^2 of num = create_array(5) printMatrix(test)
calc create_array( dimension:num ):array^2 of num var result : array^2 of num lastValue = dimension^2 - 1 loopFrom loopTo row col currDiag = 0 currNum = 0 loop if (currDiag < dimension) // if doing the upper-left triangular half loopFrom = 1 loopTo = currDiag + 1 else // doing the bottom-right triangular half loopFrom = currDiag - dimension + 2 loopTo = dimension loop count:c from:loopFrom to:loopTo var i = loopFrom + c - 1 if (rem(currDiag, 2) == 0) // want to fill upwards row = loopTo - i + loopFrom col = i else // want to fill downwards row = i col = loopTo - i + loopFrom result[row][col] = currNum inc currNum inc currDiag if (currNum > lastValue) exit return result
calc printMatrix( matrix:array^2 of num ) var dimension = tree_count(matrix) var maxDigits = 1 + lg((dimension^2-1), base:10) loop across:matrix ptr:rowp index:row var tempstr : str loop across:rowp index:col tempstr = tempstr & " " & to_str(matrix[row][col], min:maxDigits) log(tempstr)</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
Befunge
The size, N, is specified by the first value on the stack - 5 in the example below. The upper limit is constrained only by the range of the playfield cells used for variables, since we're using an algorithm that calculates the values on the fly rather than building them up in memory. On an 8 bit interpreter this means an upper limit of at least 127, but with an extended cell range the size of N can be almost unlimited.
<lang befunge>>> 5 >>00p0010p:1:>20p030pv >0g-:0`*:*-:00g:*1-55+/>\55+/:v v:,*84< v:++!\**2p01:+1g01:g02$$_>>#^4#00#+p#1:#+1#g0#0g#3<^/+ 55\_$:>55+/\| >55+,20g!00g10g`>#^_$$$@^!`g03g00!g04++**2p03:+1g03!\*+1*2g01:g04.$<</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
BQN
<lang bqn>Flip ← {m←2|+⌜˜↕≠𝕩 ⋄ (⍉𝕩׬m)+𝕩×m} Zz ← {Flip ⍋∘⍋⌾⥊+⌜˜↕𝕩}</lang>
Example:
<lang bqn>Zz 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 ┘
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:
% ./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
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>
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 ( 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 ) ); }</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
Ceylon
<lang ceylon>class ZigZag(Integer size) {
value data = Array { for (i in 0:size) Array.ofSize(size, 0) };
variable value i = 1; variable value j = 1;
for (element in 0 : size^2) { data[j - 1]?.set(i - 1, element); if ((i + j).even) { if (j < size) { j++; } else { i += 2; } if (i > 1) { i--; } } else { if (i < size) { i++; } else { j += 2; } if (j > 1) { j--; } } }
shared void display() { for (row in data) { for (element in row) { process.write(element.string.pad(3)); } print(""); //newline } } }
shared void run() { value zz = ZigZag(5); zz.display(); }</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:
> 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 ] ]
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
Crystal
<lang ruby>def zigzag(n)
(seq=(0...n).to_a).product(seq) .sort_by {|x,y| [x+y, (x+y).even? ? y : -y]} .map_with_index{|v, i| {v, i}}.sort.map(&.last).each_slice(n).to_a
end
def print_matrix(m)
format = "%#{m.flatten.max.to_s.size}d " * m[0].size m.each {|row| puts format % row}
end
print_matrix 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
D
<lang d>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);
}</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.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);
}</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>
Elena
ELENA 5.0: <lang elena>import extensions;
extension op : IntNumber {
zigzagMatrix() { auto result := IntMatrix.allocate(self, self); int i := 0; int j := 0; int d := -1; int start := 0; int end := self*self - 1; while (start < end) { result.setAt(i, j, start); start += 1; result.setAt(self - i - 1, self - j - 1, end); end -= 1; i := i + d; j := j - d; if (i < 0) { i:=i+1; d := d.Negative } else if (j < 0) { j := j + 1; d := d.Negative } }; if (start == end) { result.setAt(i, j, start) }; ^ result }
}
public program() {
console.printLine(5.zigzagMatrix()).readChar()
}</lang>
Elixir
<lang elixir>defmodule RC do
require Integer def zigzag(n) do fmt = "~#{to_char_list(n*n-1) |> length}w " (for x <- 1..n, y <- 1..n, do: {x,y}) |> Enum.sort_by(fn{x,y}->{x+y, if(Integer.is_even(x+y), do: y, else: x)} end) |> Enum.with_index |> Enum.sort |> Enum.each(fn {{_x,y},i} -> :io.format fmt, [i] if y==n, do: IO.puts "" end) end
end
RC.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
Erlang
<lang 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}. </lang>
- 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]]
ERRE
<lang ERRE>PROGRAM ZIG_ZAG
!$DYNAMIC
DIM ARRAY%[0,0]
BEGIN
SIZE%=5 !$DIM ARRAY%[SIZE%-1,SIZE%-1]
I%=1 J%=1 FOR E%=0 TO SIZE%^2-1 DO ARRAY%[I%-1,J%-1]=E% IF ((I%+J%) AND 1)=0 THEN IF J%<SIZE% THEN J%+=1 ELSE I%+=2 END IF IF I%>1 THEN I%-=1 END IF ELSE IF I%<SIZE% THEN I%+=1 ELSE J%+=2 END IF IF J%>1 THEN J%-=1 END IF END IF END FOR
FOR ROW%=0 TO SIZE%-1 DO FOR COL%=0 TO SIZE%-1 DO WRITE("###";ARRAY%[ROW%,COL%];) END FOR PRINT END FOR
END PROGRAM</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
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} }
F#
<lang fsharp> //Produce a zig zag matrix - Nigel Galloway: April 7th., 2015 let zz l a =
let N = Array2D.create l a 0 let rec gng (n, i, g, e) = N.[n,i] <- g match e with | _ when i=a-1 && n=l-1 -> N | 1 when n = l-1 -> gng (n, i+1, g+1, 2) | 2 when i = a-1 -> gng (n+1, i, g+1, 1) | 1 when i = 0 -> gng (n+1, 0, g+1, 2) | 2 when n = 0 -> gng (0, i+1, g+1, 1) | 1 -> gng (n+1, i-1, g+1, 1) | _ -> gng (n-1, i+1, g+1, 2) gng (0, 0, 0, 2)
</lang>
- Output:
<lang fsharp>zz 5 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]]
<lang fsharp>zz 8 8</lang>
[[0; 1; 5; 6; 14; 15; 27; 28] [2; 4; 7; 13; 16; 26; 29; 42] [3; 8; 12; 17; 25; 30; 41; 43] [9; 11; 18; 24; 31; 40; 44; 53] [10; 19; 23; 32; 39; 45; 52; 54] [20; 22; 33; 38; 46; 51; 55; 60] [21; 34; 37; 47; 50; 56; 59; 61] [35; 36; 48; 49; 57; 58; 62; 63]]
Let's try something a little less square man <lang fsharp>zz 5 8</lang>
[[0; 1; 5; 6; 14; 15; 24; 25] [2; 4; 7; 13; 16; 23; 26; 33] [3; 8; 12; 17; 22; 27; 32; 34] [9; 11; 18; 21; 28; 31; 35; 38] [10; 19; 20; 29; 30; 36; 37; 39]]
Factor
This version follows the algorithm laid out in the comments of the first JavaScript (ES5) functional example, though it is not exactly a straight translation.
<lang factor>USING: columns fry kernel make math math.ranges prettyprint sequences sequences.cords sequences.extras ; IN: rosetta-code.zig-zag-matrix
- [1,b,1] ( n -- seq )
[1,b] dup but-last-slice <reversed> cord-append ;
- <reversed-evens> ( seq -- seq' )
[ even? [ <reversed> ] when ] map-index ;
- diagonals ( n -- seq )
[ sq <iota> ] [ [1,b,1] ] bi [ [ cut [ , ] dip ] each ] { } make nip <reversed-evens> ;
- zig-zag-matrix ( n -- seq )
[ diagonals ] [ dup ] bi '[ [ dup 0 <column> _ head , [ _ < [ rest-slice ] when ] map-index harvest ] until-empty ] { } make ;
- zig-zag-demo ( -- ) 5 zig-zag-matrix simple-table. ;
MAIN: zig-zag-demo</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
The following example is an implementation of a J routine with an excellent walkthrough on the talk page. Luckily, we can mimic the "classification" step with the composition of 3 existing Factor words: zip-index expand-keys-push-at values
and the inverse-permutation
word is the same concept as J's grade, so this is fairly succinct.
<lang factor>USING: assocs assocs.extras grouping io kernel math math.combinatorics math.matrices prettyprint sequences ;
- <zig-zag-matrix> ( n -- matrix )
[ dup [ + ] <matrix-by-indices> concat zip-index expand-keys-push-at values [ even? [ reverse ] when ] map-index concat inverse-permutation ] [ group ] bi ;
5 <zig-zag-matrix> simple-table.</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
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>
FreeBASIC
<lang freebasic>' FB 1.05.0 Win64
Dim As Integer n
Do
Input "Enter size of matrix "; n
Loop Until n > 0
Dim zigzag(1 To n, 1 To n) As Integer all zero by default
' enter the numbers 0 to (n^2 - 1) in the matrix's anti-diagonals zigzag(1, 1) = 0 If n > 1 Then
Dim As Integer row = 0, col = 3 Dim As Boolean down = true, increment = true Dim As Integer i = 0, j = 2, k Do If down Then For k = 1 To j i += 1 row += 1 col -= 1 zigzag(row, col) = i Next down = false Else For k = 1 To j i += 1 row -= 1 col += 1 zigzag(row, col) = i Next down = true End If If increment Then j += 1 If j > n Then j = n - 1 increment = false End If Else j -= 1 If j = 0 Then Exit Do End If If down AndAlso increment Then col += 2 row -= 1 ElseIf Not Down AndAlso increment Then row += 2 col -= 1 ElseIf down AndAlso Not increment Then col += 1 Else Not down AndAlso NotIncrement row += 1 End If Loop
End If
' print zigzag matrix if n < 20 Print If n < 20 Then
For i As Integer = 1 To n For j As Integer = 1 To n Print Using "####"; zigzag(i, j); Next j Print Next i
Else
Print "Matrix is too big to display on 80 column console"
End If
Print Print "Press any key to quit" Sleep</lang>
- Output:
Enter size of matrix ? 8 0 1 5 6 14 15 27 28 2 4 7 13 16 26 29 42 3 8 12 17 25 30 41 43 9 11 18 24 31 40 44 53 10 19 23 32 39 45 52 54 20 22 33 38 46 51 55 60 21 34 37 47 50 56 59 61 35 36 48 49 57 58 62 63
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, array, bounds, range, (!)) import Text.Printf (printf) import Data.List (sortBy)
compZig :: (Int, Int) -> (Int, Int) -> Ordering compZig (x, y) (x_, y_) = compare (x + y) (x_ + y_) <> go x y
where go x y | even (x + y) = compare x x_ | otherwise = compare y y_
zigZag :: (Int, Int) -> Array (Int, Int) Int 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>-- 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 . ('\n' :) . show2d . zigZag) [(3, 3), (4, 4), (10, 2)]</lang>
Or, building a list of lists with mapAccumL:
<lang haskell>import Data.Text (justifyRight, pack, unpack)
import Data.List (mapAccumL)
import Data.Bool (bool)
zigZag :: Int -> Int zigZag = go <*> diagonals
where go _ [] = [] go n xss = (head <$> edge) : go n (dropWhile null (tail <$> edge) <> rst) where (edge, rst) = splitAt n xss
diagonals :: Int -> Int diagonals n =
snd $ mapAccumL go [0 .. (n * n) - 1] (slope <> [n] <> reverse slope) where slope = [1 .. n - 1] go xs h = (rst, bool id reverse (0 /= mod h 2) grp) where (grp, rst) = splitAt h xs
main :: IO () main =
putStrLn $ unlines $ concatMap unpack . fmap (justifyRight 3 ' ' . pack . show) <$> zigZag 5</lang>
- Output:
0 1 5 6 14 2 4 7 13 15 3 8 12 16 21 9 11 17 20 22 10 18 19 23 24
Icon and Unicon
This solution works for both Icon and Unicon.
<lang icon>procedure main(args)
n := integer(!args) | 5 every !(A := list(n)) := list(n) A := zigzag(A) show(A)
end
procedure show(A)
every writes(right(!A,5) | "\n")
end
procedure zigzag(A)
x := [0,0] every i := 0 to (*A^2 -1) do { x := nextIndices(*A, x) A[x[1]][x[2]] := i } return A
end
procedure nextIndices(n, x)
return if (x[1]+x[2])%2 = 0 then if x[2] = n then [x[1]+1, x[2]] else [max(1, x[1]-1), x[2]+1] else if x[1] = n then [x[1], x[2]+1] else [x[1]+1, max(1, x[2]-1)]
end</lang>
- Output:
->zz 0 1 5 6 14 2 4 7 13 15 3 8 12 16 21 9 11 17 20 22 10 18 19 23 24 ->
IS-BASIC
<lang IS-BASIC>100 PROGRAM "ZigZag.bas" 110 LET SIZE=5 120 NUMERIC A(1 TO SIZE,1 TO SIZE) 130 LET I,J=1 140 FOR E=0 TO SIZE^2-1 150 LET A(I,J)=E 160 IF ((I+J) BAND 1)=0 THEN 170 IF J<SIZE THEN 180 LET J=J+1 190 ELSE 200 LET I=I+2 210 END IF 220 IF I>1 THEN LET I=I-1 230 ELSE 240 IF I<SIZE THEN 250 LET I=I+1 260 ELSE 270 LET J=J+2 280 END IF 290 IF J>1 THEN LET J=J-1 300 END IF 310 NEXT 320 FOR ROW=1 TO SIZE 330 FOR COL=1 TO SIZE 340 PRINT USING " ##":A(ROW,COL); 350 NEXT 360 PRINT 370 NEXT</lang>
J
A succinct way: <lang j> ($ [: /:@; <@|.`</.@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
Imperative
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
Functional
ES5
<lang JavaScript>(function (n) {
// Read range of values into a series of 'diagonal rows' // for a square of given dimension, // starting at diagonal row i. // [ // [0], // [1, 2], // [3, 4, 5], // [6, 7, 8, 9], // [10, 11, 12, 13, 14], // [15, 16, 17, 18], // [19, 20, 21], // [22, 23], // [24] // ]
// diagonals :: n -> n function diagonals(n) { function diags(xs, iCol, iRow) { if (iCol < xs.length) { var xxs = splitAt(iCol, xs);
return [xxs[0]].concat(diags( xxs[1], (iCol + (iRow < n ? 1 : -1)), iRow + 1 )); } else return [xs]; }
return diags(range(0, n * n - 1), 1, 1); }
// Recursively read off n heads from the diagonals (as rows) // n -> n -> n function nHeads(n, lst) { var zipEdge = lst.slice(0, n);
return lst.length ? [zipEdge.map(function (x) { return x[0]; })].concat(nHeads(n, [].concat.apply([], zipEdge.map(function ( x) { return x.length > 1 ? [x.slice(1)] : []; })) .concat(lst.slice(n)))) : []; }
// range(intFrom, intTo, optional intStep) // Int -> Int -> Maybe Int -> [Int] function range(m, n, delta) { var d = delta || 1, blnUp = n > m, lng = Math.floor((blnUp ? n - m : m - n) / d) + 1, a = Array(lng), i = lng;
if (blnUp) while (i--) a[i] = (d * i) + m; else while (i--) a[i] = m - (d * i); return a; }
// splitAt :: Int -> [a] -> ([a],[a]) function splitAt(n, xs) { return [xs.slice(0, n), xs.slice(n)]; }
// Recursively take n heads from the alternately reversed diagonals
// [ [ // [0], -> [0, 1, 5, 6, 14] and: // [1, 2], [2], // [5, 4, 3], [4, 3], // [6, 7, 8, 9], [7, 8, 9], // [14, 13, 12, 11, 10], [13, 12, 11, 10], // [15, 16, 17, 18], [15, 16, 17, 18], // [21, 20, 19], [21, 20, 19], // [22, 23], [22, 23], // [24] [24] // ] ] // // In the next recursion with the remnant on the right, the next // 5 heads will be [2, 4, 7, 13, 15] - the second row of our zig zag matrix. // (and so forth)
return nHeads(n, diagonals(n) .map(function (x, i) { i % 2 || x.reverse(); return x; }));
})(5);</lang>
- Output:
<lang JavaScript>[[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>
ES6
<lang JavaScript>(n => {
// diagonals :: n -> n function diagonals(n) { let diags = (xs, iCol, iRow) => { if (iCol < xs.length) { let xxs = splitAt(iCol, xs);
return [xxs[0]].concat(diags( xxs[1], iCol + (iRow < n ? 1 : -1), iRow + 1 )); } else return [xs]; }
return diags(range(0, n * n - 1), 1, 1); }
// Recursively read off n heads of diagonal lists // rowsFromDiagonals :: n -> n -> n function rowsFromDiagonals(n, lst) { if (lst.length) { let [edge, rest] = splitAt(n, lst);
return [edge.map(x => x[0])] .concat(rowsFromDiagonals(n, edge.filter(x => x.length > 1) .map(x => x.slice(1)) .concat(rest) )); } else return []; }
// GENERIC FUNCTIONS
// splitAt :: Int -> [a] -> ([a],[a]) function splitAt(n, xs) { return [xs.slice(0, n), xs.slice(n)]; }
// range :: From -> To -> Maybe Step -> [Int] // range :: Int -> Int -> Maybe Int -> [Int] function range(m, n, step) { let d = (step || 1) * (n >= m ? 1 : -1);
return Array.from({ length: Math.floor((n - m) / d) + 1 }, (_, i) => m + (i * d)); }
// ZIG-ZAG MATRIX
return rowsFromDiagonals(n, diagonals(n) .map((x, i) => (i % 2 || x.reverse()) && x) );
})(5);</lang>
- Output:
<lang JavaScript>[[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>
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>
jq
Infrastructure: <lang jq># Create an m x n matrix
def matrix(m; n; init): if m == 0 then [] elif m == 1 then [range(0;n)] | map(init) elif m > 0 then matrix(1;n;init) as $row | [range(0;m)] | map( $row ) else error("matrix\(m);_;_) invalid") end ;
- Print a matrix neatly, each cell occupying n spaces
def neatly(n):
def right: tostring | ( " " * (n-length) + .); . as $in | length as $length | reduce range (0;$length) as $i (""; . + reduce range(0;$length) as $j (""; "\(.) \($in[$i][$j] | right )" ) + "\n" ) ;
</lang> Create a zigzag matrix by zigzagging: <lang jq>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)</lang>
- 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
another solution
<lang jq>#!/usr/bin/env jq -Mnrc -f
- solve zigzag matrix by constructing list of 2n+1 column "runs"
- and then shifting them into final form.
- e.g. for n=3 initial runs are [[0],[1,2],[3,4,5],[6,7],[8]]
- runs below are shown as columns:
- initial column runs 0 1 3 6 8
- 2 4 7
- 5
- reverse cols 0,2,4 0 1 5 6 8
- 2 4 7
- 3
- shift cols 3,4 down 0 1 5
- 2 4 6
- 3 7 8
- shift rows left 0 1 5
- to get final zigzag 2 4 6
- 3 7 8
def N: $n ; # size of matrix def NR: 2*N - 1; # number of runs def abs: if .<0 then -. else . end ; # absolute value def runlen: N-(N-.|abs) ; # length of run def makeruns: [ foreach range(1;NR+1) as $r ( # for each run {c:0} # state counter ; .l = ($r|runlen) # length of this run | .r = [range(.c;.c+.l)] # values in this run | .c += .l # increment counter ; .r # produce run ) ] ; # collect into array def even: .%2==0 ; # is input even? def reverseruns: # reverse alternate runs .[keys|map(select(even))[]] |= reverse ; def zeros: [range(.|N-length)|0] ; # array of padding zeros def shiftdown: def pad($r): # pad run with zeros if $r < N # determine where zeros go then . = . + zeros # at back for left runs else . = zeros + . # at front for right runs end ; reduce keys[] as $r (.;.[$r] |= pad($r)); # shift rows down with pad def shiftleft: [ range(N) as $r | [ range($r;$r+N) as $c | .[$c][$r] ] ] ; def width: [.[][]]|max|tostring|1+length; # width of largest value def justify($w): (($w-length)*" ") + . ; # leading spaces def format: width as $w # compute width | map(map(tostring | justify($w)))[] # justify values | join(" ") ; makeruns # create column runs | reverseruns # reverse alternate runs | shiftdown # shift right runs down | shiftleft # shift rows left | format # format final result
</lang>
- Output:
$ ./zigzag.jq --argjson n 8 0 1 5 6 14 15 27 28 2 4 7 13 16 26 29 42 3 8 12 17 25 30 41 43 9 11 18 24 31 40 44 53 10 19 23 32 39 45 52 54 20 22 33 38 46 51 55 60 21 34 37 47 50 56 59 61 35 36 48 49 57 58 62 63
Julia
simple solution
<lang Julia>function zigzag_matrix(n::Int)
matrix = zeros(Int, n, n) x, y = 1, 1 for i = 0:(n*n-1) matrix[y,x] = i if (x + y) % 2 == 0 # Even stripes if x < n x += 1 y -= (y > 1) else y += 1 end else # Odd stripes if y < n x -= (x > 1) y += 1 else x += 1 end end end return matrix
end</lang>
- Output:
julia> zigzag_matrix(5) 5×5 Array{Int64,2}: 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
a more generic solution
Create an iterator that steps through a matrix's indices in the zig-zag pattern and use this to create zig-zag matrices and related objects.
Zig-Zag Iterator <lang Julia> immutable ZigZag
m::Int n::Int diag::Array{Int,1} cmax::Int numd::Int lohi::(Int,Int)
end
function zigzag(m::Int, n::Int)
0<m && 0<n || error("The matrix dimensions must be positive.") ZigZag(m, n, [-1,1], m*n, m+n-1, extrema([m,n]))
end zigzag(n::Int) = zigzag(n, n)
type ZZState
cnt::Int cell::Array{Int,1} dir::Int dnum::Int dlen::Int dcnt::Int
end
Base.length(zz::ZigZag) = zz.cmax Base.start(zz::ZigZag) = ZZState(1, [1,1], 1, 1, 1, 1) Base.done(zz::ZigZag, zzs::ZZState) = zzs.cnt > zz.cmax
function Base.next(zz::ZigZag, zzs::ZZState)
s = sub2ind((zz.m, zz.n), zzs.cell[1], zzs.cell[2]) if zzs.dcnt == zzs.dlen if isodd(zzs.dnum) if zzs.cell[2] < zz.n zzs.cell[2] += 1 else zzs.cell[1] += 1 end else if zzs.cell[1] < zz.m zzs.cell[1] += 1 else zzs.cell[2] += 1 end end zzs.dcnt = 1 zzs.dnum += 1 zzs.dir = -zzs.dir if zzs.dnum <= zz.lohi[1] zzs.dlen += 1 elseif zz.lohi[2] < zzs.dnum zzs.dlen -= 1 end else zzs.cell += zzs.dir*zz.diag zzs.dcnt += 1 end zzs.cnt += 1 return (s, zzs)
end </lang>
Helper Functions <lang Julia> using Formatting
function width{T<:Integer}(n::T)
w = ndigits(n) n < 0 || return w return w + 1
end
function pretty{T<:Integer}(a::Array{T,2}, indent::Int=4)
lo, hi = extrema(a) w = max(width(lo), width(hi)) id = " "^indent fe = FormatExpr(@sprintf(" {:%dd}", w)) s = id nrow = size(a)[1] for i in 1:nrow for j in a[i,:] s *= format(fe, j) end i != nrow || continue s *= "\n"*id end return s
end </lang>
Main <lang Julia> n = 5 println("The n = ", n, " zig-zag matrix:") a = zeros(Int, (n, n)) for (i, s) in enumerate(zigzag(n))
a[s] = i-1
end println(pretty(a))
m = 3 println() println("Generalize to a non-square matrix (", m, "x", n, "):") a = zeros(Int, (m, n)) for (i, s) in enumerate(zigzag(m, n))
a[s] = i-1
end println(pretty(a))
p = primes(10^3) n = 7 println() println("An n = ", n, " prime spiral matrix:") a = zeros(Int, (n, n)) for (i, s) in enumerate(zigzag(n))
a[s] = p[i]
end println(pretty(a)) </lang>
- Output:
The n = 5 zig-zag 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 Generalize to a non-square matrix (3x5): 0 1 5 6 11 2 4 7 10 12 3 8 9 13 14 An n = 7 prime spiral matrix: 2 3 13 17 47 53 107 5 11 19 43 59 103 109 7 23 41 61 101 113 167 29 37 67 97 127 163 173 31 71 89 131 157 179 199 73 83 137 151 181 197 211 79 139 149 191 193 223 227
Klingphix
<lang Klingphix>include ..\Utilitys.tlhy
%Size 5 !Size
0 ( $Size dup ) dim
%i 1 !i %j 1 !j
$Size 2 power [
1 - ( $i $j ) set $i $j + 1 band 0 == ( [$j $Size < ( [$j 1 + !j] [$i 2 + !i] ) if $i 1 > [ $i 1 - !i] if ] [$i $Size < ( [$i 1 + !i] [$j 2 + !j] ) if $j 1 > [ $j 1 - !j] if ] ) if
] for
$Size [
%row !row $Size [ %col !col ( $row $col ) get tostr 32 32 chain chain 1 3 slice print drop ] for nl
] for
nl "End " input</lang>
- Output:
0 1 5 6 14 2 4 7 13 15 3 8 12 16 21 9 11 17 20 22 10 18 19 23 24 End
Kotlin
<lang scala>// version 1.1.3
typealias Vector = IntArray typealias Matrix = Array<Vector>
fun zigzagMatrix(n: Int): Matrix {
val result = Matrix(n) { Vector(n) } var down = false var count = 0 for (col in 0 until n) { if (down) for (row in 0..col) result[row][col - row] = count++ else for (row in col downTo 0) result[row][col - row] = count++ down = !down } for (row in 1 until n) { if (down) for (col in n - 1 downTo row) result[row + n - 1 - col][col] = count++ else for (col in row until n) result[row + n - 1 - col][col] = count++ down = !down } return result
} fun printMatrix(m: Matrix) {
for (i in 0 until m.size) { for (j in 0 until m.size) print("%2d ".format(m[i][j])) println() } println()
}
fun main(args: Array<String>) {
printMatrix(zigzagMatrix(5)) printMatrix(zigzagMatrix(10))
}</lang>
- Output:
0 1 5 6 14 2 4 7 13 15 3 8 12 16 21 9 11 17 20 22 10 18 19 23 24 0 1 5 6 14 15 27 28 44 45 2 4 7 13 16 26 29 43 46 63 3 8 12 17 25 30 42 47 62 64 9 11 18 24 31 41 48 61 65 78 10 19 23 32 40 49 60 66 77 79 20 22 33 39 50 59 67 76 80 89 21 34 38 51 58 68 75 81 88 90 35 37 52 57 69 74 82 87 91 96 36 53 56 70 73 83 86 92 95 97 54 55 71 72 84 85 93 94 98 99
Ksh
<lang ksh>
- !/bin/ksh
- Produce a zig-zag array.
- # Variables:
integer DEF_SIZE=5 # Default size = 5 arr_size=${1:-$DEF_SIZE} # $1 = size, or default
# # Externals: #
- # Functions:
######
- main #
######
integer i j n typeset -a zzarr
for (( i=n=0; i<arr_size*2; i++ )); do for (( j= (i<arr_size) ? 0 : i-arr_size+1; j<=i && j<arr_size; j++ )); do (( zzarr[(i&1) ? j*(arr_size-1)+i : (i-j)*arr_size+j] = n++ )) done done
for ((i=0; i<arr_size*arr_size; i++)); do printf "%3d " ${zzarr[i]} (( (i+1)%arr_size == 0 )) && printf "\n" done</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 23 240 1 5 6 14 15 27 28 44 2 4 7 13 16 26 29 43 45 3 8 12 17 25 30 42 46 59 9 11 18 24 31 41 47 58 60 10 19 23 32 40 48 57 61 70 20 22 33 39 49 56 62 69 71 21 34 38 50 55 63 68 72 77 35 37 51 54 64 67 73 76 7836 52 53 65 66 74 75 79 80
Lasso
<lang 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; </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
Maple
Here values are starting at 1. Replace <v+~1,v+~2>
with <v,v+~1>
to start at 0.
<lang maple>zigzag1:=proc(n)
uses ArrayTools; local i,u,v,a; u:=Replicate(<-1,1>,n): v:=Vector[row](1..n,i->i*(2*i-3)): v:=Reshape(<v+~1,v+~2>,2*n): a:=Matrix(n,n): for i to n do a[...,i]:=v[i+1..i+n]; v+=u od: a
end:
zigzag2:=proc(n)
local i,v,a; a:=zigzag1(n); v:=Vector(1..n-1,i->i^2); for i from 2 to n do a[n+2-i..n,i]-=v[1..i-1] od; a
end:</lang>
<lang maple>zigzag1(6);</lang>
- Output:
Matrix(6, 6, [[ 1, 2, 6, 7, 15, 16], [ 3, 5, 8, 14, 17, 27], [ 4, 9, 13, 18, 26, 31], [10, 12, 19, 25, 32, 42], [11, 20, 24, 33, 41, 50], [21, 23, 34, 40, 51, 61]])
<lang maple>zigzag2(6);</lang>
- Output:
Matrix(6, 6, [[ 1, 2, 6, 7, 15, 16], [ 3, 5, 8, 14, 17, 26], [ 4, 9, 13, 18, 25, 27], [10, 12, 19, 24, 28, 33], [11, 20, 23, 29, 32, 34], [21, 22, 30, 31, 35, 36]])
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. <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>
- 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
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>
MiniZinc
<lang MiniZinc> %Zigzag Matrix. Nigel Galloway, February 3rd., 2020 int: Size; array [1..Size,1..Size] of var 1..Size*Size: zigzag; constraint zigzag[1,1]=1 /\ zigzag[Size,Size]=Size*Size; constraint forall(n in {2*g | g in 1..Size div 2})(zigzag[1,n]=zigzag[1,n-1]+1 /\ forall(g in 2..n)(zigzag[g,n-g+1]=zigzag[g-1,n-g+2]+1)); constraint forall(n in {2*g + ((Size-1) mod 2) | g in 1..(Size-1) div 2})(zigzag[n,Size]=zigzag[n-1,Size]+1 /\ forall(g in 1..Size-n)(zigzag[n+g,Size-g]=zigzag[n+g-1,Size-g+1]+1)); constraint forall(n in {2*g+1 | g in 1..(Size-1) div 2})(zigzag[n,1]=zigzag[n-1,1]+1 /\ forall(g in 2..n)(zigzag[n-g+1,g]=zigzag[n-g+2,g-1]+1)); constraint forall(n in {2*g+((Size) mod 2) | g in 1..(Size-1) div 2})(zigzag[Size,n]=zigzag[Size,n-1]+1 /\ forall(g in 1..Size-n)(zigzag[Size-g,n+g]=zigzag[Size-g+1,n+g-1]+1)); output [show2d(zigzag)]; </lang> {out}
minizinc -DSize=5 zigzag.mzn [| 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 |] ---------- minizinc -DSize=6 zigzag.mzn [| 1, 2, 6, 7, 15, 16 | 3, 5, 8, 14, 17, 26 | 4, 9, 13, 18, 25, 27 | 10, 12, 19, 24, 28, 33 | 11, 20, 23, 29, 32, 34 | 21, 22, 30, 31, 35, 36 |] ----------
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>
Nim
<lang nim>from algorithm import sort from strutils import align from sequtils import newSeqWith
type Pos = tuple[x, y: int]
proc `<` (a, b: Pos): bool =
a.x + a.y < b.x + b.y or a.x + a.y == b.x + b.y and (a.x < b.x xor (a.x + a.y) mod 2 == 0)
proc zigzagMatrix(n: int): auto =
var indices = newSeqOfCap[Pos](n*n) for x in 0 ..< n: for y in 0 ..< n: indices.add((x,y)) sort(indices) result = newSeqWith(n, newSeq[int](n)) for i, p in indices: result[p.x][p.y] = i
proc `$`(m: seq[seq[int]]): string =
let Width = len($m[0][^1]) + 1 for r in m: for c in r: result.add align($c, Width) result.add "\n"
echo zigzagMatrix(6)</lang>
- Output:
0 1 5 6 14 15 2 4 7 13 16 25 3 8 12 17 24 26 9 11 18 23 27 32 10 19 22 28 31 33 20 21 29 30 34 35
Direct coord to number
This calculates the number for each coordinate directly. This allows to create very large zig-zag matrices. Generates the same output as above. <lang nim>import strutils
func sumTo(n: Natural): Natural = n * (n+1) div 2
func coord2num(row, col, N: Natural): Natural =
var start, offset: Natural let diag = col + row if diag < N: start = sumTo(diag) offset = if diag mod 2 == 0: col else: row else: # N * (2*diag+1-N) - sumTo(diag), but with smaller itermediates start = N*N - sumTo(2*N-1-diag) offset = N-1 - (if diag mod 2 == 0: row else: col) start + offset
let N = 6 let width = (N*N).`$`.len + 1 for row in 0 ..< N:
for col in 0 ..< N: stdout.write(coord2num(row, col, N).`$`.align(width)) stdout.write("\n")
</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
<lang octave>function a = zigzag1(n)
j = 1:n; u = repmat([-1; 1], n, 1); v = j.*(2*j-3); v = reshape([v; v+1], 2*n, 1); a = zeros(n, n); for i = 1:n a(:, i) = v(i+j); v += u; endfor
endfunction
function a = zigzag2(n)
a = zigzag1(n); v = (1:n-1)'.^2; for i = 2:n a(n+2-i:n, i) -= v(1:i-1); endfor
endfunction
>> zigzag2(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>
Alternate solution, filling pairs of diagonals.
<lang octave>function a = zigzag3(n)
a = zeros(n, n); for k=1:n d = (2*(j = mod(k, 2))-1)*(n-1); m = (n-1)*(k-1); a(k+(1-j)*m:d:k+j*m) = k*(k-1)/2:k*(k+1)/2-1; a(n*(n+1-k)+(1-j)*m:d:n*(n+1-k)+j*m) = n*n-k*(k+1)/2:n*n-k*(k-1)/2-1; endfor
endfunction
>> zigzag3(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>
Inspired by Rascal
<lang octave>
- {
Produce a zigzag matrix. Nigel Galloway, January 26th., 2020. At the time of writing the Rascal solution is yellow flagged for producing a striped matrix. Let me make the same faux pas.
- }
n=5; g=1; for e=1:n i=1; for l=e:-1:1 zig(i++,l)=g++; endfor endfor for e=2:n i=e; for l=n:-1:e zig(i++,l)=g++; endfor endfor
- {
I then have the following, let me call it zig. 1 2 4 7 11 3 5 8 12 16 6 9 13 17 20 10 14 18 21 23 15 19 22 24 25 To avoid being yellow flagged I must convert this striped matrix into a zigzag matrix.
- }
zag=zig'
- {
So zag is the transpose of zig. 1 3 6 10 15 2 5 9 14 19 4 8 13 18 22 7 12 17 21 24 11 16 20 23 25
- }
for e=1:n for g=1:n if(mod(e+g,2))==0 zagM(e,g)=1; endif endfor endfor; zigM=1-zagM;
- {
I now have 2 masks: zigM =
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0
zagM =
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
- }
zigzag=zag.*zagM+zig.*zigM;
- {
zigzag =
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
- }
</lang>
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 {$} {MakeTuple unit W} end} T end
in
{Inspect {CreateZigZag 5}}</lang>
PARI/GP
<lang parigp>zz(n)={ my(M=matrix(n,n),i,j,d=-1,start,end=n^2-1); while(ct--, M[i+1,j+1]=start; M[n-i,n-j]=end; start++; end--; i+=d; j-=d; if(i<0, i++; d=-d , if(j<0, j++; d=-d ) ); if(start>end,return(M)) ) };</lang>
Pascal
<lang Pascal>Program zigzag( input, output );
const
size = 5;
var
zzarray: array [1..size, 1..size] of integer; element, i, j: integer; direction: integer; width, n: integer;
begin
i := 1; j := 1; direction := 1; for element := 0 to (size*size) - 1 do begin zzarray[i,j] := element; i := i + direction; j := j - direction; if (i = 0) then begin direction := -direction; i := 1; if (j > size) then begin j := size; i := 2; end; end else if (i > size) then begin direction := -direction; i := size; j := j + 2; end else if (j = 0) then begin direction := -direction; j := 1; if (i > size) then begin j := 2; i := size; end; end else if (j > size) then begin direction := -direction; j := size; i := i + 2; end; end;
width := 2; n := size; while (n > 0) do begin width := width + 1; n := n div 10; end; for j := 1 to size do begin for i := 1 to size do write(zzarray[i,j]:width); writeln; end;
end.</lang>
- Output:
0 1 5 6 14 2 4 7 13 15 3 8 12 16 21 9 11 17 20 22 10 18 19 23 24
- Output:
with size set to 6
0 1 5 6 14 15 2 4 7 13 16 25 3 8 12 17 24 26 9 11 18 23 27 32 10 19 22 28 31 33 20 21 29 30 34 35
<lang Pascal> Program zigzag; {$APPTYPE CONSOLE}
const
size = 5;
var s: array [1..size, 1..size] of integer; i, j, d, max, n: integer;
begin
i := 1; j := 1; d := -1; max := 0; n := 0; max := size * size; for n := 1 to (max div 2)+1 do begin s[i,j] := n; s[size - i + 1,size - j + 1] := max - n + 1; i:=i+d; j:=j-d; if i < 1 then begin inc(i); d := -d; end else if j < 1 then begin inc(j); d := -d; end; end;
for j := 1 to size do begin for i := 1 to size do write(s[i,j]:4); writeln; end;
end. </lang>
- Output:
Size 5
1 3 4 10 11 2 5 9 12 19 6 8 13 18 20 7 14 17 21 24 15 16 22 23 25
- Output:
Size 8
1 3 4 10 11 21 22 36 2 5 9 12 20 23 35 37 6 8 13 19 24 34 38 49 7 14 18 25 33 39 48 50 15 17 26 32 40 47 51 58 16 27 31 41 46 52 57 59 28 30 42 45 53 56 60 63 29 43 44 54 55 61 62 64
Perl
<lang perl>use 5.010;
sub zig_zag {
my $n = shift; my $max_number = $n**2; my @matrix; my $number = 0; for my $j ( 0 .. --$n ) { for my $i ( $j % 2 ? 0 .. $j : reverse 0 .. $j ) { $matrix[$i][ $j - $i ] = $number++; #next if $j == $n; $matrix[ $n - $i ][ $n - ( $j - $i ) ] = $max_number - $number; } } return @matrix;
}
my @zig_zag_matrix = zig_zag(5); say join "\t", @{$_} foreach @zig_zag_matrix; </lang>
<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>
Phix
with javascript_semantics integer n = 9 integer zstart = 0, zend = n*n-1 --integer zstart = 1, zend = n*n string fmt = sprintf("%%%dd",length(sprintf("%d",zend))) sequence m = repeat(repeat("??",n),n) integer x = 1, y = 1, d = -1 while 1 do m[x][y] = sprintf(fmt,zstart) if zstart=zend then exit end if zstart += 1 m[n-x+1][n-y+1] = sprintf(fmt,zend) zend -= 1 x += d y -= d if x<1 then x += 1 d = -d elsif y<1 then y += 1 d = -d end if end while for i=1 to n do m[i] = join(m[i]) end for puts(1,join(m,"\n"))
Alternative:
integer n = 5 string fmt = sprintf("%%%dd",length(sprintf("%d",n*n-1))) sequence m = repeat(repeat("??",n),n) integer x = 1, y = 1 for d=0 to n*n-1 do m[y][x] = sprintf(fmt,d) if mod(x+y,2) then {x,y} = iff(y<n?{x-(x>1),y+1}:{x+1,y}) else {x,y} = iff(x<n?{x+1,y-(y>1)}:{x,y+1}) end if end for for i=1 to n do m[i] = join(m[i]) end for puts(1,join(m,"\n"))
Phixmonti
<lang Phixmonti>5 var Size 0 Size repeat Size repeat
1 var i 1 var j
Size 2 power for
swap i get rot j set i set i j + 1 bitand 0 == IF j Size < IF j 1 + var j ELSE i 2 + var i ENDIF i 1 > IF i 1 - var i ENDIF ELSE i Size < IF i 1 + var i ELSE j 2 + var j ENDIF j 1 > IF j 1 - var j ENDIF ENDIF
endfor
Size FOR
var row Size FOR var col row get col get tostr 32 32 chain chain 1 3 slice print drop drop ENDFOR nl
ENDFOR</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>
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
PL/I
<lang pli>/* 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>
- 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
Plain TeX
The code works with any etex engine. <lang tex>\long\def\antefi#1#2\fi{#2\fi#1} \def\fornum#1=#2to#3(#4){% \edef#1{\number\numexpr#2}\edef\fornumtemp{\noexpand\fornumi\expandafter\noexpand\csname fornum\string#1\endcsname {\number\numexpr#3}{\ifnum\numexpr#4<0 <\else>\fi}{\number\numexpr#4}\noexpand#1}\fornumtemp } \long\def\fornumi#1#2#3#4#5#6{\def#1{\unless\ifnum#5#3#2\relax\antefi{#6\edef#5{\number\numexpr#5+(#4)\relax}#1}\fi}#1} \def\elem(#1,#2){\numexpr(#1+#2)*(#1+#2-1)/2-(\ifodd\numexpr#1+#2\relax#1\else#2\fi)\relax} \def\zzmat#1{% \noindent% quit vertical mode \fornum\yy=1to#1(+1){% \fornum\xx=1to#1(+1){% \ifnum\numexpr\xx+\yy\relax<\numexpr#1+2\relax \hbox to 2em{\hfil\number\elem(\xx,\yy)}% \else \hbox to 2em{\hfil\number\numexpr#1*#1-1-\elem(#1+1-\xx,#1+1-\yy)\relax}% \fi }% \par\noindent% next line + quit vertical mode }\par } \zzmat{5} \bye</lang>
pdf 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
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>
An Alternate Display
Display the zig-zag matrix using the Format-Wide
cmdlet:
<lang PowerShell>
zigzag 5 | Format-Wide {"{0,2}" -f $_} -Column 5 -Force
</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
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>
- 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 by paddy3118.
<lang python>def zigzag(n):
zigzag rows def compare(xy): x, y = xy return (x + y, -y if (x + y) % 2 else y) xs = range(n) return {index: n for n, index in enumerate(sorted( ((x, y) for x in xs for y in xs), key=compare ))}
def printzz(myarray):
show zigzag rows as lines n = int(len(myarray) ** 0.5 + 0.5) xs = range(n) print('\n'.join( [.join("%3i" % myarray[(x, y)] for x in xs) for y in xs] ))
printzz(zigzag(6))</lang>
- Output:
0 2 3 9 10 20 1 4 8 11 19 21 5 7 12 18 22 29 6 13 17 23 28 30 14 16 24 27 31 34 15 25 26 32 33 35
Alternative version,
<lang python>
- pylint: disable=invalid-name
- pylint: disable=unused-argument
"ZigZag iterator." import sys
if sys.version_info[0] >= 3:
xrange = range
def move(x, y, columns, rows):
"Tells us what to do next with x and y." if y < (rows - 1): return max(0, x-1), y+1 return x+1, y
def zigzag(rows, columns):
"ZigZag iterator, yields indices." x, y = 0, 0 size = rows * columns for _ in xrange(size): yield y, x if (x + y) & 1: x, y = move(x, y, columns, rows) else: y, x = move(y, x, rows, columns)
- test code
i, rows, cols = 0, 5, 5 mat = [[0 for x in range(cols)] for y in range(rows)] for (y, x) in zigzag(rows, cols):
mat[y][x], i = i, i + 1
from pprint import pprint pprint(mat) </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> 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>
- 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
Another alternative version
<lang python> from __future__ import print_function
import math
def zigzag( dimension):
generate the zigzag indexes for a square array Exploiting the fact that an array is symmetrical around its centre NUMBER_INDEXES = dimension ** 2 HALFWAY = NUMBER_INDEXES // 2 KERNEL_ODD = dimension & 1
xy = [0 for _ in range(NUMBER_INDEXES)] # start at 0,0 ix = 0 iy = 0 # 'fake' that we are going up and right direction = 1 # the first index is always 0, so start with the second # until halfway for i in range(1, HALFWAY + KERNEL_ODD): if direction > 0: # going up and right if iy == 0: # are at top ix += 1 direction = -1 else: ix += 1 iy -= 1 else: # going down and left if ix == 0: # are at left iy += 1 direction = 1 else: ix -= 1 iy += 1 # update the index position xy[iy * dimension + ix] = i
# have first half, but they are scattered over the list # so find the zeros to replace for i in range(1, NUMBER_INDEXES): if xy[i] == 0 : xy[i] = NUMBER_INDEXES - 1 - xy[NUMBER_INDEXES - 1 - i]
return xy
def main(dim):
zz = zigzag(dim) print( 'zigzag of {}:'.format(dim)) width = int(math.ceil(math.log10(dim**2))) for j in range(dim): for i in range(dim): print('{:{width}}'.format(zz[j * dim + i], width=width), end=' ') print()
if __name__ == '__main__':
main(5)
</lang>
zigzag 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
Quackery
Sorting Indices
<lang Quackery> [ ]'[ tuck do dip do ] is with2 ( x x --> x x )
[ dup temp put [] swap dup * times [ i^ join ] sortwith [ with2 [ temp share /mod tuck + 1 & if negate ] > ] sortwith [ with2 [ temp share /mod + ] > ] dup witheach [ i^ unrot poke ] [] swap temp share times [ temp share split dip [ nested join ] ] drop temp release ] is zigzag ( n --> [ )
10 zigzag witheach [ witheach [ dup 10 < if sp echo sp ] cr ]</lang>
- Output:
0 1 5 6 14 15 27 28 44 45 2 4 7 13 16 26 29 43 46 63 3 8 12 17 25 30 42 47 62 64 9 11 18 24 31 41 48 61 65 78 10 19 23 32 40 49 60 66 77 79 20 22 33 39 50 59 67 76 80 89 21 34 38 51 58 68 75 81 88 90 35 37 52 57 69 74 82 87 91 96 36 53 56 70 73 83 86 92 95 97 54 55 71 72 84 85 93 94 98 99
Turtle style
Adapted from Spiral matrix#Quackery
The sequence of turns for the first half of the matrix is east to southwest to south to northeast to east. In the second half the order of turns is reversed.
<lang Quackery> [ stack ] is stepcount ( --> s )
[ stack ] is position ( --> s ) [ stack ] is heading ( --> s )
[ heading take behead join heading put ] is turn ( --> )
[ heading share 0 peek unrot times [ position share stepcount share unrot poke over position tally 1 stepcount tally ] nip ] is walk ( [ n --> [ )
[ dip [ temp put [] ] temp share times [ temp share split dip [ nested join ] ] drop temp release ] is matrixify ( n [ --> [ )
[ 0 stepcount put ( set up... ) 0 position put ' [ 1 ] over 1 - join over join over 1 - negate join heading put 0 over dup * of over 1 - times ( turtle draws first half of zigzag ) [ 1 walk turn i^ 1+ walk turn ] heading take ( reverse the sequence of turns ) reverse heading put over 1 - times ( turtle draws second half of zigzag ) [ turn 1 walk turn i walk ] 1 walk matrixify ( ...tidy up ) heading release position release stepcount release ] is zigzag ( n --> [ )
10 zigzag witheach [ witheach [ dup 10 < if sp echo sp ] cr ]</lang>
- Output:
0 1 5 6 14 15 27 28 44 45 2 4 7 13 16 26 29 43 46 63 3 8 12 17 25 30 42 47 62 64 9 11 18 24 31 41 48 61 65 78 10 19 23 32 40 49 60 66 77 79 20 22 33 39 50 59 67 76 80 89 21 34 38 51 58 68 75 81 88 90 35 37 52 57 69 74 82 87 91 96 36 53 56 70 73 83 86 92 95 97 54 55 71 72 84 85 93 94 98 99
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>
Raku
(formerly Perl 6)
Using the same Turtle class as in the Spiral matrix task:
<lang perl6>class Turtle {
my @dv = [0,-1], [1,-1], [1,0], [1,1], [0,1], [-1,1], [-1,0], [-1,-1]; my $points = 8; # 'compass' points of neighbors on grid: north=0, northeast=1, east=2, etc. has @.loc = 0,0; has $.dir = 0; has %.world; has $.maxegg; has $.range-x; has $.range-y; method turn-left ($angle = 90) { $!dir -= $angle / 45; $!dir %= $points; } method turn-right($angle = 90) { $!dir += $angle / 45; $!dir %= $points; } method lay-egg($egg) { %!world{~@!loc} = $egg; $!maxegg max= $egg; $!range-x minmax= @!loc[0]; $!range-y minmax= @!loc[1]; } method look($ahead = 1) { my $there = @!loc »+« @dv[$!dir] »*» $ahead; %!world{~$there}; } method forward($ahead = 1) { my $there = @!loc »+« @dv[$!dir] »*» $ahead; @!loc = @($there); } method showmap() { my $form = "%{$!maxegg.chars}s"; my $endx = $!range-x.max; for $!range-y.list X $!range-x.list -> ($y, $x) { print (%!world{"$x $y"} // ).fmt($form); print $x == $endx ?? "\n" !! ' '; } }
}
sub MAIN(Int $size = 5) {
my $t = Turtle.new(dir => 1); 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>
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>
- Output:
<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>
REXX
This REXX version allows the optional specification of the start and increment values. <lang rexx>/*REXX program produces and displays a zig─zag matrix (a square array). */ parse arg n start inc . /*obtain optional arguments from the CL*/ if n== | n=="," then n= 5 /*Not specified? Then use the default.*/ if start== | start=="," then start= 0 /* " " " " " " */ if inc== | inc=="," then inc= 1 /* " " " " " " */ row= 1; col= 1; size= n**2 /*start: 1st row & column; array size.*/
do j=start by inc for size; @.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*/ /* [↑] // is REXX ÷ remainder.*/
call show /*display a (square) matrix──►terminal.*/ exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ show: w= max(length(start), length(start+size*inc)) /*max width of any matrix elements,*/
do r=1 for n ; _= right(@.r.1, w) /*show the rows of the matrix. */ do c=2 for n-1; _= _ right(@.r.c, w) /*build a line for output of a row.*/ end /*c*/; say _ /* [↑] align the matrix elements. */ end /*r*/; return</lang>
- output when using the default inputs:
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
- output when using the inputs of: 5 1
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 when using the inputs of: 4 -1000 -1
-1000 -1001 -1005 -1006 -1002 -1004 -1007 -1012 -1003 -1008 -1011 -1013 -1009 -1010 -1014 -1015
Ring
<lang ring>
- Project Zig-zag matrix
load "guilib.ring" load "stdlib.ring" new qapp
{ win1 = new qwidget() { setwindowtitle("Zig-zag matrix") setgeometry(100,100,600,400) n = 5 a = newlist(n,n) zigzag = newlist(n,n) for j = 1 to n for i = 1 to n a[j][i] = 0 next next i = 1 j = 1 k = 1 while k < n * n a[j][i] = k k = k + 1 if i = n j = j + 1 a[j][i] = k k = k + 1 di = -1 dj = 1 ok if j = 1 i = i + 1 a[j][i] = k k = k + 1 di = -1 dj = 1 ok if j = n i = i + 1 a[j][i] = k k = k + 1 di = 1 dj = -1 ok if i = 1 j = j + 1 a[j][i] = k k = k + 1 di = 1 dj = -1 ok i = i + di j = j + dj end for p = 1 to n for m = 1 to n zigzag[p][m] = new qpushbutton(win1) { x = 150+m*40 y = 30 + p*40 setgeometry(x,y,40,40) settext(string(a[p][m])) } next next show() } exec() }
</lang> Output:
Ruby
<lang ruby>def zigzag(n)
(seq=*0...n).product(seq) .sort_by {|x,y| [x+y, (x+y).even? ? y : -y]} .each_with_index.sort.map(&:last).each_slice(n).to_a
end
def print_matrix(m)
format = "%#{m.flatten.max.to_s.size}d " * m[0].size puts m.map {|row| format % row}
end
print_matrix 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
Rust
<lang rust> use std::cmp::Ordering; use std::cmp::Ordering::{Equal, Greater, Less}; use std::iter::repeat;
- [derive(Debug, PartialEq, Eq)]
struct SortIndex {
x: usize, y: usize,
}
impl SortIndex {
fn new(x: usize, y: usize) -> SortIndex { SortIndex { x, y } }
}
impl PartialOrd for SortIndex {
fn partial_cmp(&self, other: &SortIndex) -> Option<Ordering> { Some(self.cmp(other)) }
}
impl Ord for SortIndex {
fn cmp(&self, other: &SortIndex) -> Ordering { let lower = if self.x + self.y == other.x + other.y { if (self.x + self.y) % 2 == 0 { self.x < other.x } else { self.y < other.y } } else { (self.x + self.y) < (other.x + other.y) };
if lower { Less } else if self == other { Equal } else { Greater } }
}
fn zigzag(n: usize) -> Vec<Vec<usize>> {
let mut l: Vec<SortIndex> = (0..n * n).map(|i| SortIndex::new(i % n, i / n)).collect(); l.sort();
let init_vec = vec![0; n]; let mut result: Vec<Vec<usize>> = repeat(init_vec).take(n).collect(); for (i, &SortIndex { x, y }) in l.iter().enumerate() { result[y][x] = i } result
}
fn main() {
println!("{:?}", 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]]
Scala
Uses the array indices sort solution used by others here.
<lang scala> def zigzag(n: Int): Array[Array[Int]] = {
val l = for (i <- 0 until n*n) yield (i%n, i/n) val lSorted = l.sortWith { 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 } val res = Array.ofDim[Int](n, n) lSorted.zipWithIndex foreach { case ((x,y), i) => res(y)(x) = i } res } zigzag(5).foreach{ ar => ar.foreach(x => print("%3d".format(x))) println }</lang>
Output: <lang scala> 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>
Scilab
<lang>function a = zigzag3(n)
a = zeros(n, n) for k=1:n j = modulo(k, 2) d = (2*j-1)*(n-1) m = (n-1)*(k-1) a(k+(1-j)*m:d:k+j*m) = k*(k-1)/2:k*(k+1)/2-1 a(n*(n+1-k)+(1-j)*m:d:n*(n+1-k)+j*m) = n*n-k*(k+1)/2:n*n-k*(k-1)/2-1 end
endfunction
-->zigzag3(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>
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
Sidef
<lang ruby>func zig_zag(w, h) {
var r = [] var n = 0
h.of { |e| w.of { |f| [e, f] } }.reduce('+').sort { |a, b| (a[0]+a[1] <=> b[0]+b[1]) || (a[0]+a[1] -> is_even ? a[0]<=>b[0] : a[1]<=>b[1]) }.each { |a| r[a[1]][a[0]] = n++ }
return r
}
zig_zag(5, 5).each { say .join(, {|i| "%4i" % i}) }</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
Standard ML
<lang Standard ML>fun rowprint r = (List.app (fn i => print (StringCvt.padLeft #" " 3 (Int.toString i))) r;
print "\n");
fun zig lst M = List.app rowprint (lst M);
fun sign t = if t mod 2 = 0 then ~1 else 1;
fun zag n = List.tabulate (n,
fn i=> rev ( List.tabulate (n, fn j =>
let val t = n-j+i and u = n+j-i in if i <= j
then t*t div 2 + sign t * ( t div 2 - i ) else n*n - 1 - ( u*u div 2 + sign u * ( u div 2 - n + 1 + i) )
end )));
zig zag 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 val it = () : unit
Stata
The requested zig-zag matrix can be constructed as a correction of another zig-zag matrix, which is a square "view" of the infinite zig-zag matrix. Here is the latter:
<lang stata>function zigzag1(n) { j = 0::n-1 u = J(1, n, (-1, 1)) v = (j:*(2:*j:+3)) v = rowshape((v,v:+1), 1) a = J(n, n, .) for (i=1; i<=n; i++) { a[i, .] = v[j:+i] v = v+u } return(a) }
zigzag1(5)
1 2 3 4 5 +--------------------------+ 1 | 0 1 5 6 14 | 2 | 2 4 7 13 16 | 3 | 3 8 12 17 25 | 4 | 9 11 18 24 31 | 5 | 10 19 23 32 40 | +--------------------------+</lang>
Now the corrected matrix, which solves the task:
<lang stata>function zigzag2(n) { a = zigzag1(n) v = (1..n-1):^2 for (i=1; i<n; i++) { a[n-i+1, i+1..n] = a[n-i+1, i+1..n] - v[1..n-i] } return(a) }
zigzag2(5)
1 2 3 4 5 +--------------------------+ 1 | 0 1 5 6 14 | 2 | 2 4 7 13 15 | 3 | 3 8 12 16 21 | 4 | 9 11 17 20 22 | 5 | 10 18 19 23 24 | +--------------------------+</lang>
The correction is given by the difference:
<lang stata>zigzag1(5)-zigzag2(5) [symmetric]
1 2 3 4 5 +--------------------------+ 1 | 0 | 2 | 0 0 | 3 | 0 0 0 | 4 | 0 0 1 4 | 5 | 0 1 4 9 16 | +--------------------------+</lang>
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>
- 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
uBasic/4tH
<lang>S = 5
i = 1 j = 1
For e = 0 To (S*S)-1
@((i-1) * S + (j-1)) = e
If (i + j) % 2 = 0 Then
If j < S Then j = j + 1 Else i = i + 2 EndIf
If i > 1 Then i = i - 1 EndIf Else
If i < S i = i + 1 Else j = j + 2 EndIf
If j > 1 j = j - 1 EndIf EndIf
Next
For r = 0 To S-1
For c = 0 To S-1 Print Using "___#";@(r * S + c); Next Print
Next</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 OK, 0:428
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
VBScript
<lang vb>ZigZag(Cint(WScript.Arguments(0)))
Function ZigZag(n) Dim arrZ() ReDim arrZ(n-1,n-1) i = 1 j = 1 For e = 0 To (n^2) - 1 arrZ(i-1,j-1) = e If ((i + j ) And 1) = 0 Then If j < n Then j = j + 1 Else i = i + 2 End If If i > 1 Then i = i - 1 End If Else If i < n Then i = i + 1 Else j = j + 2 End If If j > 1 Then j = j - 1 End If End If Next For k = 0 To n-1 For l = 0 To n-1 WScript.StdOut.Write Right(" " & arrZ(k,l),3) Next WScript.StdOut.WriteLine Next End Function</lang>
- Output:
C:\>cscript /nologo ZigZag.vbs 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 C:\>cscript /nologo ZigZag.vbs 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
Wren
<lang ecmascript>import "/fmt" for Conv, Fmt
var zigzag = Fn.new { |n|
var r = List.filled(n*n, 0) var i = 0 var n2 = n * 2 for (d in 1..n2) { var x = d - n if (x < 0) x = 0 var y = d - 1 if (y > n - 1) y = n - 1 var j = n2 - d if (j > d) j = d for (k in 0...j) { if (d&1 == 0) { r[(x+k)*n+y-k] = i } else { r[(y-k)*n+x+k] = i } i = i + 1 } } return r
}
var n = 5 var w = Conv.itoa(n*n - 1).count var i = 0 for (e in zigzag.call(n)) {
Fmt.write("$*d ", w, e) if (i%n == n - 1) System.print() i = i + 1
}</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
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
Yabasic
<lang Yabasic>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 and((i + j), 1) = 0 THEN IF j < Size then j = j + 1 ELSE i = i + 2 end if IF i > 1 i = i - 1 ELSE IF i < Size then i = i + 1 ELSE j = j + 2 end if IF j > 1 j = j - 1 ENDIF
NEXT e
FOR row = 0 TO Size-1
FOR col = 0 TO Size-1 PRINT array(row,col) USING "##"; NEXT col PRINT
NEXT row</lang>
zkl
<lang zkl>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())});
}</lang>
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. <lang zkl>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);
}</lang> To be pedantic, the same as above, but using the output of the list comprehension: <lang zkl>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("get",1) .pump(String,T(Void.Read,m-1), ("%3s"*m+"\n").fmt);
}</lang>
- 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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