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    <p><strong>Okay this is a solution that</strong> </p> <ul> <li>uses just one extra long value for working storage.</li> <li>uses no recursion.</li> <li>one pass of only N, not even N*N.</li> <li>will work for other values of N but will need new #defines.</li> </ul> <blockquote> <pre><code>#include &lt;stdio.h&gt; #define BIT30 (1&lt;&lt;24) #define COLMASK 0x108421L #define ROWMASK 0x1fL </code></pre> </blockquote> <pre><code>unsigned long long STARTGRID = ((0x10 | 0x0 | 0x4 | 0x2 | 0x0) &lt;&lt; 20) | ((0x00 | 0x8 | 0x4 | 0x2 | 0x0) &lt;&lt; 15) | ((0x10 | 0x8 | 0x4 | 0x2 | 0x1) &lt;&lt; 10) | ((0x10 | 0x0 | 0x4 | 0x2 | 0x1) &lt;&lt; 5) | ((0x10 | 0x8 | 0x4 | 0x2 | 0x1) &lt;&lt; 0); void dumpGrid (char* comment, unsigned long long theGrid) { char buffer[1000]; buffer[0]='\0'; printf ("\n\n%s\n",comment); for (int j=1;j&lt;31; j++) { if (j%5!=1) printf( "%s%s", ((theGrid &amp; BIT30)==BIT30)? "1" : "0",(((j%5)==0)?"\n" : ",") ); theGrid = theGrid &lt;&lt; 1; } } int main (int argc, const char * argv[]) { unsigned long long rowgrid = STARTGRID; unsigned long long colGrid = rowgrid; unsigned long long rowmask = ROWMASK; unsigned long long colmask = COLMASK; dumpGrid("Initial Grid", rowgrid); for (int i=0; i&lt;5; i++) { if ((rowgrid &amp; rowmask)== rowmask) rowgrid |= rowmask; else rowgrid &amp;= ~rowmask; if ((colGrid &amp; colmask) == colmask) colmask |= colmask; else colGrid &amp;= ~colmask; rowmask &lt;&lt;= 5; colmask &lt;&lt;= 1; } colGrid &amp;= rowgrid; dumpGrid("RESULT Grid", colGrid); return 0; } </code></pre>
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    1. POSet every cell in matrix to 0 if that row or column contains a 0
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    1. USAnthonyLambert
    1. USAnthonyLambert
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    1. COI just worked out how to do this in even less code....
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      1. USAnthonyLambert
    2. COIt is a nice solution to be sure. And I suppose every ones solution here neglects at least one of the requirements. So having a solution with a max allowed value for N isn't the worst thing in the world, so nice job on this one :)
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      1. USAdam
    3. CORestricting N to 8 and claiming this meets the one pass requirement is just plain dumb. This is not a general solution. No restriction of the size of N was stated in the question, so you have only solved a sub-problem.
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