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  1. PO
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    <p>Define P(n) as the number of ways to partition n, restricting n to be an integer, n >= 1.</p> <p>Define p(n, k) as the number of ways to partition n using numbers no larger than k, restricting k to be an integer, k >= 1, k &lt;= n.</p> <p>It follows that P(n) = sum (i: 1..n) p(n, i).</p> <p>To find p(n, k), first note that we neatly avoid double-counting by simply keeping the partition sorted: take the largest chunk first. So the first chunk may have any size from 1 up to k, and then the rest of the chunks must account for the rest of the total n, and be no larger than the first chunk.</p> <p>Thus p(n, k) = sum (i: 1..k) p(n - i, i).</p> <p>As a base case, p(1, 1) = 1.</p> <hr> <p>A sample implementation, very much <strong>not</strong> guaranteed to be at all efficient, or even to compile (but it should give you the basic idea) - spoilers!</p> <pre><code>// A utility function to represent the result of appending to a vector, // as a new vector (without affecting the previous one). template &lt;typename T&gt; vector&lt;T&gt; operator&lt;&lt;(vector&lt;T&gt; copy, T to_append) { // We passed by value to get a copy implicitly. copy.push_back(to_append); return copy; } // A utility function to append one vector to another. template &lt;typename T&gt; vector&lt;T&gt;&amp; operator+=(vector&lt;T&gt;&amp; original, const vector&lt;T&gt;&amp; to_append) { // 'copy' is in namespace std:: and defined in &lt;algorithm&gt;. // 'back_inserter' is in namespace std:: and defined in &lt;iterator&gt;. copy(to_append.begin(), to_append.end(), back_inserter(original)); return original; } vector&lt;vector&lt;int&gt; &gt; partitions(int remaining, int limit, vector&lt;int&gt; prefix) { // Finds the partitions of some larger number which start with the // numbers in 'prefix', such that the rest of the "chunks" sum to // 'remaining' and are all no larger than 'limit'. // 'max' is in namespace std:: and defined in &lt;algorithm&gt;. We restrict // the 'limit' to be no more than 'remaining'. limit = max(remaining, limit); vector&lt;vector&lt;int&gt; &gt; result; // Base case. if (remaining == 1) { return result &lt;&lt; (prefix &lt;&lt; 1); // note the parentheses are required! } for (int i = limit; i &gt; 0; --i) { result += partitions(remaining - i, i, prefix &lt;&lt; i); } return result; } </code></pre>
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    1. POtrying to write a recursive function that counts the number of sequences that sum up to that number C++
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    1. USKarl Knechtel
    1. USKarl Knechtel
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    1. COWould you mind taking a look at my code and seeing if you can tell whats wrong.
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    2. COThe most obvious problem: you are expecting `counter` to be passed by reference and thus to accumulate values across all the recursive calls. You could just change the parameter to be passed by reference, but you'd miss a valuable lesson about how recursion works. What you really want to do is have `counter` as a local variable, and then **add in** the values that are returned by the recursive calls. In general, ignoring the return value of a function that returns a value is a sign that you might be doing something wrong; you might be able to turn up compiler warnings high enough to flag this.
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      1. USKarl Knechtel
    3. COThe next most obvious problem: trying to print the partitions as you go is much harder than you expect. Your output line only allows for outputting two numbers, and a partition may consist of more than that. In general, you would have to "carry forward" the "chunks" of the partition found so far into the next recursive step. I'm working on a solution that I'll edit with, which assembles the partitions in a vector of vectors of ints, which you can then cycle through to print. No guarantees about efficiency :)
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      1. USKarl Knechtel
 

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