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  1. PO
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    <p>Here is an implementation in Mathematica, from the package <a href="http://www.combinatorica.com/" rel="nofollow"><em>Combinatorica</em></a>. The semantics are fairly generic, so I think it is helpful. Please leave a comment if you need anything explained.</p> <pre><code>UnrankKSubset::usage = "UnrankKSubset[m, k, l] gives the mth k-subset of set l, listed in lexicographic order." UnrankKSubset[m_Integer, 1, s_List] := {s[[m + 1]]} UnrankKSubset[0, k_Integer, s_List] := Take[s, k] UnrankKSubset[m_Integer, k_Integer, s_List] := Block[{i = 1, n = Length[s], x1, u, $RecursionLimit = Infinity}, u = Binomial[n, k]; While[Binomial[i, k] &lt; u - m, i++]; x1 = n - (i - 1); Prepend[UnrankKSubset[m - u + Binomial[i, k], k-1, Drop[s, x1]], s[[x1]]] ] </code></pre> <p>Usage is like:</p> <pre><code>UnrankKSubset[1, 4, {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}] </code></pre> <pre><b> {1, 2, 3, 5}</b></pre> <p>As you can see this function operates on sets.</p> <hr> <p>Below is my attempt to explain the code above.</p> <h3>UnrankKSubset is a recursive function with three arguments, called (m, k, s):</h3> <ol> <li><code>m</code> an Integer, the "rank" of the combination in lexigraphical order, starting from zero.</li> <li><code>k</code> an Integer, the number of elements in each combination</li> <li><code>s</code> a List, the elements from which to assemble combinations</li> </ol> <h3>There are two boundary conditions on the recursion:</h3> <ol> <li><p>for any rank <code>m</code>, and any list <code>s</code>, if the number of elements in each combination <code>k</code> is <code>1</code>, then:</p> <p>return the <code>m + 1</code> element of the list <code>s</code>, itself in a list.</p> <p>(<code>+ 1</code> is needed because Mathematica indexes from one, rather than zero. I believe in C++ this would be s[m] )</p></li> <li><p>if rank <code>m</code> is <code>0</code> then for any <code>k</code> and any <code>s</code>:</p> <p>return the first <code>k</code> elements of <code>s</code></p></li> </ol> <h3>The main recursive function, for any other arguments than ones specified above:</h3> <p>local variables: (<code>i</code>, <code>n</code>, <code>x1</code>, <code>u</code>)</p> <p><code>Binomial</code> is binomial coefficient: <code>Binomial[7, 5]</code> = 21</p> <p>Do:</p> <pre><code>i = 1 n = Length[s] u = Binomial[n, k] While[Binomial[i, k] &lt; u - m, i++]; x1 = n - (i - 1); </code></pre> <p>Then return:</p> <pre><code>Prepend[ UnrankKSubset[m - u + Binomial[i, k], k - 1, Drop[s, x1]], s[[x1]] ] </code></pre> <p>That is, "prepend" the <code>x1</code> element of list <code>s</code> (remember Mathematica indexes from one, so I believe this would be the <code>x1 - 1</code> index in C++) to the list returned by the recursively called UnrankKSubset function with the arguments:</p> <ul> <li><code>m - u + Binomial[i, k]</code></li> <li><code>k - 1</code></li> <li><code>Drop[s, x1]</code></li> </ul> <p><code>Drop[s, x1]</code> is the rest of list <code>s</code> with the first <code>x1</code> elements removed.</p> <hr> <p>If anything above is not understandable, or if what you wanted was an explanation of the algorithm, rather than an explanation of the code, please leave a comment and I will try again.</p>
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    1. POCalculate Combination based on position
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    1. COI don't use mathematica i need write this as c++ function can you know how this gonna be in c++ ?
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      1. USjosh
    2. CO@josh, I am sorry, I only know Mathematica. I was hoping that since this Mathematica code is primarily procedural, rather than functional, it would translate to C/C++ well enough. Would you like me to try explaining the code above, step by step?
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    3. COYes if you can, thank you very much!
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