Spiral matrix
You are encouraged to solve this task according to the task description, using any language you may know.
Produce a spiral array. A spiral array is a square arrangement of the first N2 natural numbers, where the numbers increase sequentially as you go around the edges of the array spiralling inwards.
For example, given 5, produce this array:
0 1 2 3 4 15 16 17 18 5 14 23 24 19 6 13 22 21 20 7 12 11 10 9 8
Ada
<lang ada>-- Spiral Square with Ada.Text_Io; use Ada.Text_Io; with Ada.Integer_Text_Io; use Ada.Integer_Text_Io; with Ada.Numerics.Elementary_Functions; use Ada.Numerics.Elementary_Functions;
procedure Spiral_Square is
type Array_Type is array(Positive range <>, Positive range <>) of Natural; function Spiral (N : Positive) return Array_Type is Result : Array_Type(1..N, 1..N); Row : Natural := 1; Col : Natural := 1; Max_Row : Natural := N; Max_Col : Natural := N; Min_Row : Natural := 1; Min_Col : Natural := 1; begin for I in 0..N**2 - 1 loop Result(Row, Col) := I; if Row = Min_Row then Col := Col + 1; if Col > Max_Col then Col := Max_Col; Row := Row + 1; end if; elsif Col = Max_Col then Row := Row + 1; if Row > Max_Row then Row := Max_Row; Col := Col - 1; end if; elsif Row = Max_Row then Col := Col - 1; if Col < Min_Col then Col := Min_Col; Row := Row - 1; end if; elsif Col = Min_Col then Row := Row - 1; if Row = Min_Row then -- Reduce spiral Min_Row := Min_Row + 1; Max_Row := Max_Row - 1; Row := Min_Row; Min_Col := Min_Col + 1; Max_Col := Max_Col - 1; Col := Min_Col; end if; end if; end loop; return Result; end Spiral; procedure Print(Item : Array_Type) is Num_Digits : constant Float := Log(X => Float(Item'Length(1)**2), Base => 10.0); Spacing : constant Positive := Integer(Num_Digits) + 2; begin for I in Item'range(1) loop for J in Item'range(2) loop Put(Item => Item(I,J), Width => Spacing); end loop; New_Line; end loop; end Print;
begin
Print(Spiral(5));
end Spiral_Square; </lang> The following is a variant using a different algorithm (which can also be used recursively): <lang ada>
function Spiral (N : Positive) return Array_Type is Result : Array_Type (1..N, 1..N); Left : Positive := 1; Right : Positive := N; Top : Positive := 1; Bottom : Positive := N; Index : Natural := 0; begin while Left < Right loop for I in Left..Right - 1 loop Result (Top, I) := Index; Index := Index + 1; end loop; for J in Top..Bottom - 1 loop Result (J, Right) := Index; Index := Index + 1; end loop; for I in reverse Left + 1..Right loop Result (Bottom, I) := Index; Index := Index + 1; end loop; for J in reverse Top + 1..Bottom loop Result (J, Left) := Index; Index := Index + 1; end loop; Left := Left + 1; Right := Right - 1; Top := Top + 1; Bottom := Bottom - 1; end loop; Result (Top, Left) := Index; return Result; end Spiral;
</lang>
ALGOL 68
INT empty=0; PROC spiral = (INT n)[,]INT: ( INT dx:=1, dy:=0; # Starting increments # INT x:=0, y:=0; # Starting location # [0:n-1,0:n-1]INT my array; FOR y FROM LWB my array TO UPB my array DO FOR x FROM LWB my array TO UPB my array DO my array[x,y]:=empty OD OD; FOR i TO n**2 DO my array[x,y] := i; INT nx:=x+dx, ny:=y+dy; IF ( 0<=nx AND nx<n AND 0<=ny AND ny<n | my array[nx,ny] = empty | FALSE ) THEN x:=nx; y:=ny ELSE INT swap:=dx; dx:=-dy; dy:=swap; x+:=dx; y+:=dy FI OD; my array ); PROC print spiral = ([,]INT my array)VOID:( FOR y FROM LWB my array TO UPB my array DO FOR x FROM LWB my array TO UPB my array DO print(whole(my array[x,y],-3)) OD; print(new line) OD ); print spiral(spiral(5))
Output:
1 2 3 4 5 16 17 18 19 6 15 24 25 20 7 14 23 22 21 8 13 12 11 10 9
C
(First code)
<lang c>#include <stdio.h>
- include <stdlib.h>
- define SPIADDR(baseptr,x,y,dim) (&((baseptr)[(y)*(dim)+(x)]))
- define None -1
int *spiral(int n) {
int dx,dy,x,y,i,nx,ny; x=y=0; dx=1; dy=0; int *buf = malloc(n*n*sizeof(int)); if (buf==NULL) return NULL; for(i=0;i<(n*n);i++) buf[i] = None; for(i=0;i<(n*n);i++) { *SPIADDR(buf,x,y,n) = i; nx = x + dx; ny = y + dy; if ( (nx<n) && (0<=nx) && (ny<n) && (0<=ny) && (*SPIADDR(buf,nx,ny,n) == None) ) { x=nx; y=ny; } else { int t = dx; dx=-dy; dy=t; x+=dx; y+=dy; } } return buf;
}
void printspiral(int *spi, int n) {
int x,y; if ( spi==NULL ) return; for(y=0; y<n; y++) { for(x=0; x<n; x++) { printf("%2d ", *SPIADDR(spi,x,y,n)); } printf("\n"); }
}
int main() {
int *the_spiral = NULL; the_spiral = spiral(5); printspiral(the_spiral, 5); if (the_spiral!=NULL) free(the_spiral); return 0;
}</lang>
Fortran
PROGRAM SPIRAL IMPLICIT NONE INTEGER, PARAMETER :: size = 5 INTEGER :: i, x = 0, y = 1, count = size, n = 0 INTEGER :: array(size,size) DO i = 1, count x = x + 1 array(x,y) = n n = n + 1 END DO DO count = count - 1 DO i = 1, count y = y + 1 array(x,y) = n n = n + 1 END DO DO i = 1, count x = x - 1 array(x,y) = n n = n + 1 END DO IF (n > size*size-1) EXIT count = count - 1 DO i = 1, count y = y - 1 array(x,y) = n n = n + 1 END DO DO i = 1, count x = x + 1 array(x,y) = n n = n + 1 END DO IF (n > size*size-1) EXIT END DO DO y = 1, size DO x = 1, size WRITE (*, "(I4)", ADVANCE="NO") array (x, y) END DO WRITE (*,*) END DO END PROGRAM SPIRAL
Haskell
Solution based on the J hints:
grade xs = map snd. sort $ zip xs [0..] values n = cycle [1,n,-1,-n] counts n = (n:).concatMap (ap (:) return) $ [n-1,n-2..1] reshape n = unfoldr (\xs -> if null xs then Nothing else Just (splitAt n xs)) spiral n = reshape n . grade. scanl1 (+). concat $ zipWith replicate (counts n) (values n)
J
This function is the result of some beautiful insights:
spiral =. ,~ $ [: /: }.@(2 # >:@i.@-) +/\@# <:@+: $ (, -)@(1&,) spiral 5 0 1 2 3 4 15 16 17 18 5 14 23 24 19 6 13 22 21 20 7 12 11 10 9 8
Would you like some hints that will allow you to reimplement it in another language?
These inward spiralling arrays are known as "involutes"; we can also generate outward-spiraling "evolutes", and we can start or end the spiral at any corner, and go in either direction (clockwise or counterclockwise). See the first link (to JSoftware.com).
Python
<lang python> def spiral(n):
dx,dy = 1,0 # Starting increments x,y = 0,0 # Starting location myarray = [[None]* n for j in range(n)] for i in xrange(n**2): myarray[x][y] = i nx,ny = x+dx, y+dy if 0<=nx<n and 0<=ny<n and myarray[nx][ny] == None: x,y = nx,ny else: dx,dy = -dy,dx x,y = x+dx, y+dy return myarray
def printspiral(myarray):
n = range(len(myarray)) for y in n: for x in n: print "%2i" % myarray[x][y], print
printspiral(spiral(5)) </lang> Sample output:
0 1 2 3 4 15 16 17 18 5 14 23 24 19 6 13 22 21 20 7 12 11 10 9 8
Recursive Solution
<lang python> def spiral_part(x,y,n):
if x==-1 and y==0: return -1 if y==(x+1) and x<(n/2): return spiral_part(x-1, y-1, n-1) + 4*(n-y)
if x<(n-y) and y<=x: return spiral_part(y-1, y, n) + (x-y) + 1 if x>=(n-y) and y<=x: return spiral_part(x, y-1, n) + 1 if x>=(n-y) and y>x: return spiral_part(x+1, y, n) + 1 if x<(n-y) and y>x: return spiral_part(x, y-1, n) - 1
def spiral(n):
array = [[None]*n for j in range(n)] for x in range(n): for y in range(n): array[x][y] = spiral_part(x,y,n) return array
</lang>
Adding a cache for the spiral_part function it could be quite efficient.
Another way, based on preparing lists ahead
<lang python> def spiral(n):
dat = [[None] * n for i in range(n)] le = [[i + 1, i + 1] for i in reversed(range(n))] le = sum(le, [])[1:] # for n = 5 le will be [5, 4, 4, 3, 3, 2, 2, 1, 1] dxdy = [[1, 0], [0, 1], [-1, 0], [0, -1]] * ((len(le) + 4) / 4) # long enough x, y, val = -1, 0, -1 for steps, (dx, dy) in zip(le, dxdy): x, y, val = x + dx, y + dy, val + 1 for j in range(steps): dat[y][x] = val if j != steps-1: x, y, val = x + dx, y + dy, val + 1 return dat
for row in spiral(5): # calc spiral and print it
print ' '.join('%3s' % x for x in row)
</lang>