Matrix arrangement issues in php

Question

I would like to know some solutions to such a problem.

It is given a number lets say 16 and you have to arrange a matrix this way

1  2  3  4
12 13 14 5
         


        

Answer 1

OK, I'm just posting this answer for fun.

The other solutions use variables to accumulate information iteratively. I wanted to try a functional solution, where the number of any table cell (or alternatively, the table cell for any number) could be known without iterating through any others.

Here it is in javascript. I know, it's not pure functional programming, nor is it extremely elegant, but the computation of the number for each table cell is done without reference to prior iterations. So it's multi-core friendly.

Now we just need somebody to do it in haskell. ;-)

BTW this was written before the comment that the 1 should end up in a certain location that is not necessarily the northwest corner (still unspecified as of now).

Since somebody mentioned the Ulam spiral, just for kicks I added code to put a red border around the primes (even though the spiral is inside out). Interestingly enough, there do seem to be diagonal streaks of primes, though I don't know if it's significantly different from the streaks you'd get with random odd numbers.

The code:

// http://stackoverflow.com/questions/3584557/logical-problem

/* Return a square array initialized to the numbers 1...n2, arranged in a spiral */
function spiralArray(n2) {
   var n = Math.round(Math.sqrt(n2));
   if (n * n != n2) {
      alert('' + n2 + ' is not a square.');
      return 0;
   }
   var h = n / 2;
   var arr = new Array(n);
   var i, j;

   for (i = 0; i < n; i++) {
      arr[i] = new Array(n);
      for (j = 0; j < n; j++) {
         // upper rows and lower rows of spiral already completed
         var ur = Math.min(i, n - i, j + 1, n - j, h),
            lr = Math.min(i, n - i - 1, j + 1, n - j - 1, h);
         // count of cells in completed rows
         // n + n-2 + n-4 ... n - 2*(ur-1) = ur*n - (ur*2*(ur - 1)/2) = ur * (n - ur + 1)
         // n-1 + n-3 + ... n-1 - 2*(lr-1) = lr*(n-1) - (lr*2*(lr - 1)/2) = lr * (n - 1 - lr + 1)
         var compr = ur * (n - ur + 1) + lr * (n - lr);
         // e.g. ur = 2, cr = 2*(5 - 2 + 1) = 2*4 = 8
         // left columns and right columns of spiral already completed
         var rc = Math.min(n - j - 1, i, n - i, j + 1, h),
            lc = Math.min(n - j - 1, i, n - i - 1, j, h);
         // count of cells in completed columns
         var compc = rc * (n - rc) + lc * (n - lc - 1);
         // offset along current row/column
         var offset;
         // Which direction are we going?
         if (ur > rc) {
            // going south
            offset = i - (n - j) + 1;
         } else if (rc > lr) {
            // going west
            offset = i - j;
         } else if (lr > lc) {
            // going north
            offset = n - i - 1 - j;
         } else {
            // going east
            offset = j - i + 1;
         }

         arr[i][j] = compr + compc + offset;
      }
   }
   return arr;
}

function isPrime(n) {
    // no fancy sieve... we're not going to be testing large primes.
    var lim = Math.floor(Math.sqrt(n));
    var i;
    if (n == 2) return true;
    else if (n == 1 || n % 2 == 0) return false;
    for (i = 3; i <= lim; i += 2) {
        if (n % i == 0) return false;
    }
    return true;
}

// display the given array as a table, with fancy background shading
function writeArray(arr, tableId, m, n) {
   var tableElt = document.getElementById(tableId);
   var s = '';
   var scale = 1 / (m * n);
   var i, j;
   for (i = 0; i < m; i++) {
      s += '';
      for (j = 0; j < n; j++) {
         var border = isPrime(arr[i][j]) ? "border: solid red 1px;" : "";
         s += '';
      }
      s += '';
   }
   s += '
' + arr[i][j] + '
';

   tableElt.innerHTML = s;
}

function tryIt(tableId) {
   var sizeElt = document.getElementById('size');
   var size = parseInt(sizeElt.value);
   writeArray(spiralArray(size * size), 'spiral', size, size);
}

The HTML page to exercise it:

    
        
        
    
    
        

            Size of spiral: