Spiral matrix

From Rosetta Code
Task
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>

C

Translation of: Python

(First code)

<lang c>#include <stdio.h>

  1. include <stdlib.h>
  1. define SPIADDR(baseptr,x,y,dim) (&((baseptr)[(y)*(dim)+(x)]))
  2. 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

Works with: Fortran version 90 and later
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>