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    <p>I actually wrote an algorithm for this by hand once. Here it is:</p> <pre><code>bool incr(int z[NUM_INDICES]){ int a=NUM_INDICES-1; for(int i=NUM_INDICES-2;i&gt;=0;i--) if(z[i]&gt;z[i+1]) a--; else break; if(a==0) return false; int b=2147483647,c; for(int i=a;i&lt;=NUM_INDICES-1;i++) if(z[i]&gt;z[a-1]&amp;&amp;z[i]-z[a-1]&lt;b){ b=z[i]-z[a-1]; c=i; } int temp=z[a-1]; z[a-1]=z[c]; z[c]=temp; qsort(z+a,NUM_INDICES-a,sizeof(int),comp); return true; } </code></pre> <p>This is the increment function (i.e. you have an array like [3,2,4,1], you pass it to this, and it modifies it to [3,4,1,2]). It works off the fact that if the last <em>d</em> elements of the array are in descending order, then the next array (in "alphabetical" order) should satisfy the following conditions: 1) the last <em>d+1</em> elements are a permutation among themselves; 2) the <em>d+1</em>-th to last element is the next highest element in the last <em>d+1</em> elements; 3) the last <em>d</em> elements should be in ascending order. You can see this intuitively when you have something like [2,5,3, 8,7,6,4,1]: <em>d</em> = 5 in this case; the 3 turns into the next highest of the last <em>d+1</em> = 6 elements; and the last <em>d</em> = 5 are arranged in ascending order, so it becomes [2,5,4, 1,3,6,7,8].</p> <p>The first loop basically determines <em>d</em>. It loops over the array backwards, comparing consecutive elements, to determine the number of elements at the end that are in descending order. At the end of the loop, <code>a</code> becomes the first element that is in the descending order sequence. If <code>a==0</code>, then the whole array is in descending order and nothing more can be done. </p> <p>The next loop determines what the <em>d+1</em>-th-to-last element should be. We specified that it should be the next highest element in the last <em>d+1</em> elements, so this loop determines what that is. (Note that z[a-1] is the <em>d+1</em>-th-to-last element.) By the end of that loop, <code>b</code> contains the lowest <code>z[i]-z[a-1]</code> that is positive; that is, <code>z[i]</code> should be greater than <code>z[a-1]</code>, but as low as possible (so that <code>z[a-1]</code> becomes the next highest element). <code>c</code> contains the index of the corresponding element. We discard <code>b</code> because we only need the index.</p> <p>The next three lines swap <code>z[a-1]</code> and <code>z[c]</code>, so that the <em>d+1</em>-th-to-last element gets the element next in line, and the other element (<code>z[c]</code>) gets to keep <code>z[a-1]</code>. Finally, we sort the last <em>d</em> elements using <code>qsort</code> (<code>comp</code> must be declared elsewhere; see C++ documentation on <code>qsort</code>). </p>
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    1. COthis is good thanks, so the first part goes through them in order and breaks when they aren't in decending order. Then if its only length 0 it returns 0. I'm guessing 2147483647 is just a random number. I havent figured out what variable holds the values of the permutations but i will look at it more later.
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      1. USPatrick Kennedy
    2. CO2147483647 is the highest possible signed int (2^31 - 1). We are making `b` as small as possible by comparing it to each array element and, if `b` is higher, replacing it with the array element. Thus we choose the highest possible value for `b` to start with so that it is always replaced.
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    3. COI haven't actually created the permutations yet. Again, this is just an increment function. You'd need to create an array in main to store the final values and another array to store the current permutation, and then a loop to copy the latter over to the former, call `incr()`, and quit when it returns `false`.
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