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    <p>I can see a problem:</p> <p>It should be: </p> <pre><code>if(board.isLegal()){ if( addSquare(depth, outStream)) return true; } </code></pre> <p>Means if the whole board is solved then roll back.</p> <p><strong>EDIT</strong> think about it:</p> <p>you put 1 in the first square, it returns false. don't you want to try put 2?</p> <p><strong>EDIT2</strong> More problem: </p> <pre><code>if(i == NINE &amp;&amp; j == NINE &amp;&amp; board.getSquare(NINE, NINE) != ZERO){ cout &lt;&lt; i &lt;&lt; " " &lt;&lt; j &lt;&lt; endl; return true; } </code></pre> <p>Is this suppose to be a completion check? you check only 1 square. and in your sample it is filled! you need to check that there is no 0.</p> <p><strong>EDIT3:</strong></p> <pre><code>bool DoTheWork::addSquare(int&amp; depth, ostream&amp; outStream){ //use flag to let you know if all completed bool zeroFound = false; for(int i = ONE; i &lt;= NINE; ++i){ for(int j = ONE; j &lt;= NINE; ++j){ //cout &lt;&lt; "original" &lt;&lt; board.getSquare(i, j) &lt;&lt; "coord: " &lt;&lt; i &lt;&lt; ", " &lt;&lt; j &lt;&lt; endl; if(board.getSquare(i, j) == ZERO){ zeroFound = true; //cout &lt;&lt; "original: " &lt;&lt; board.getSquare(i, j) &lt;&lt; "coord: " &lt;&lt; i &lt;&lt; ", " &lt;&lt; j &lt;&lt; endl; for(int k = ONE; k &lt;= NINE; ++k){ board.setSquare(i, j, k); board.display(outStream); if(board.isLegal()){ if(addSquare(depth, outStream)){ return true; } else{ board.unsetSquare(i, j); } } } } } } board.display(outStream); return !zeroFound; //true in case it is full! } </code></pre>
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    1. POTrouble understanding recursion in this application
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    1. COI tried it, and it didn't solve the problem. I believe that is the same answer @drescherjm offered, but with different syntax.
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      1. USTrevor Hutto
    2. COin case of false, you don't want to return, you want to move to next value.
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    3. COOh, I see, however it still does not get rid of the lingering zeros in the output.
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      1. USTrevor Hutto
 

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