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    <p>The answer depends on whether you store the columns or the rows of matrix. Assuming you have a <code>n * n</code> size matrix and you store the rows, so </p> <pre><code>A = [[1,2,3,4], [2,3,4,5], [1,2,3,4], [3,2,1,4]] </code></pre> <p>and the starting point is <code>i</code>, you should go from the array no. <code>m = i div n</code> (the integer part of the division, rounding down), and inside the array, the staring element should be the no. <code>p = i mod n</code> (the modulus). And from that point, you can select every array from <code>m</code> to <code>n</code>, and in every array, the <code>p</code>th element, until the most recent element is the same as your original.</p> <p>In Java-like code:</p> <pre><code>int n = 4; int[][] A = new int[n][n]; ... // initialize A int sumValues(int i) { int original = A[i/n][i%n]; int sum = original; int p = i % n; for (m = i/n + 1, m&lt;n, ++m) { if (A[m][p] != orginal) sum += A[m][p]; else break; } return sum; } </code></pre> <p>If you are storing the columns, so</p> <pre><code>A = [[1,2,1,3], [2,3,2,2], [3,4,3,1], [4,5,4,4]] </code></pre> <p>then <code>m</code> and <code>p</code> are inversed, meaning that from <code>A</code> you should select array no. <code>m = i mod n</code> and inside that array, element no. <code>p = i div n</code>. After this, you stay in your selected array, and just increase <code>p</code> until <code>A[m][p]</code> equals to the originally selected value.</p> <p>In Java-like code:</p> <pre><code>int n = 4; int[][] A = new int[n][n]; ... // initialize A int sumValues(int i) { int original = A[i%n][i/n]; int sum = original; int p = i / n; for (p = i/n +1, p&lt;n, ++p) { if (A[m][p] != orginal) sum += A[m][p]; else break; } return sum; } </code></pre> <p>But correct me, if I'm wrong :)</p>
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