To convert a two dimensional matrix to spiral order

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Hi CPs,
Its been a long time. And I'm stuck at an idea and I need your opinions or solutions to it. So here I go.

Lets consider a 4x4 matrix

|1  2  3  4  |
|5  6  7  8  |
|9  10 11 12 |
|13 14 15 16 |

In a spiral order it will be re-written as 1 2 3 4 8 12 16 15 14 13 9 5 6 7 11 10.

So how would my conditions go? I wrote a program for a 4x4 matrix but the IF-conditions has to be increased each time the matrix grows in size. And my input won't always be a square matrix. So how to make this dynamic?

Posted 4-Aug-14 8:44am

Ashwin2013

Updated 4-Aug-14 10:08am

PIEBALDconsult

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PIEBALDconsult 4-Aug-14 15:26pm

Oh, come on, how am I supposed to remember something I wrote thirty years ago? Well, OK, I have it written down in my BASICplus book from high school. :D
Except mine was to display the content of a matrix on a VT100 screen and I wanted to do something more interesting than left-to-right, top-to-bottom.

As I recall I had a number of for/next loops -- one to go across the top, one to go down the right, one to go across the bottom, one to go up the left, and repeat.

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Solution 4

Wow thank you everyone for replying.

I am using C to do the task. So I cant use the class and other functions :)

But I get the idea. Thank you very much.

Posted 13-Aug-14 4:06am

Ashwin2013

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PIEBALDconsult · Accepted Answer · 2014-08-07T11:00:00

Sorry I didn't get back to you earlier. Wondering what came of this.

Here's what I devised:

C#

public static partial class Array
{
  public static System.Collections.Generic.IEnumerable<T>
  Spiral<T>
  (
    this T[,] Matrix
  )
  {
    int top = Matrix.GetLowerBound ( 0 )
      , bot = Matrix.GetUpperBound ( 0 )
      , lft = Matrix.GetLowerBound ( 1 )
      , rgt = Matrix.GetUpperBound ( 1 ) ;

    while ( top <= bot )
    {
      for ( int i = lft ; i <= rgt ; i++ ) yield return ( Matrix [ top ,   i ] ) ;
      top++ ;

      for ( int i = top ; i <= bot ; i++ ) yield return ( Matrix [   i , rgt ] ) ;
      rgt-- ;

      for ( int i = rgt ; i >= lft ; i-- ) yield return ( Matrix [ bot ,   i ] ) ;
      bot-- ;

      for ( int i = bot ; i >= top ; i-- ) yield return ( Matrix [   i , lft ] ) ;
      lft++ ;
    }

    yield break ;
  }
}

nv3 · Accepted Answer · 2014-08-07T12:05:00

How about doing it this way: Your spiral is nothing but nested rectangles. The outer rectangle starts at (0,0), the next inner one at (1,1), and so forth until you reach (width+1)/2; (note the +1 to cater for odd widths).

So I would create an outer loop for that, say

C++

int i;
int cntRects = (width + 1) / 2;

for (i = 0; i < cntRects; ++i)
{
    ...

Inside that outer loop you do four inner loops for the four sides of the rectangle, i.e. top, right, bottom, left. You can construct the indices easily from i, width and height. Here I am going to do the first one for you:

C++

int x, y;

// top side
y = i;
for (x = i; x < width - i; ++i)
    AddPoint (x, y);

...

CPallini · Accepted Answer · 2014-08-04T09:12:00

You could create a bookkeeping matrix, say 'visited', with visited[j][k]=1 if the same item in the original matrix has been visited. Now suppose you are proceeding on the positive x-axis, then if visited[y][x+1] == 1 (or x == N-1, that is you are on a boundary) then you have to change direction (in this example, you have to go down).

To convert a two dimensional matrix to spiral order

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