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    <p>To begin, ask yourself: <strong>"How would I create a box?"</strong> To create a box, you need <strong>4 sides</strong>. <em>However</em>, <strong>the simplest box that one can make is involves two lines</strong>, as we can take <strong>the height of the box to be zero - or <em>infinitesimal</em></strong>. In code, you'll need at least 2 rows to make a box.</p> <p>Like so:</p> <pre><code>000000000000000 000000000000000 </code></pre> <p>However, it has <strong>no height.</strong> In other words, <strong>this is a zero-height bordered box.</strong></p> <p>Thus to create a box like this:</p> <pre><code>000000000000000 011111111111110 000000000000000 </code></pre> <p>What do you notice? <strong>The first and the last rows are all zeroes. <em>And</em>, the first and last element of the middle row is a zero, everything else on that row is a 1.</strong></p> <p>Extending further:</p> <pre><code>000000000000000 011111111111110 011111111111110 000000000000000 </code></pre> <p><strong>We see the same pattern - so this can be extended for the nth row case.</strong> Thus, the algorithm is: </p> <ol> <li>First and last rows are all zeros. </li> <li>For all other rows, the first and last columns of those rows are all 0. </li> <li>Everything else is set to 1.</li> </ol> <p>Hence, in your case:</p> <pre><code>for(r=0; r&lt; ROWS; r++) { for(c=0; c &lt; COLS; c++) { if (r == 0 || r == ROWS - 1) { M[r][c]=0; } else if(c == 0 || c == COLS -1) { M[r][c]=0; } else { M[r][c]=1; } cout&lt;&lt; M[r][c]; } cout &lt;&lt; endl; } </code></pre>
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    1. POhow to create a boundry around a grid
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    1. COWouldn't the assignments be to M[r][c]?
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    2. COThe only thing i forgot to ask is how the spot of M[1][1] becomes 2 because thats teh whole point of the program for it to start changing the values from that point
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    3. COYou're correct, @ArunSaha. Thank you! I copied the OP's code verbatim. Edited.
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