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    <p>Here in summary is the problem I think you are trying to solve and some explanation on how i might approach solving it. <a href="http://www.cs.swan.ac.uk/~csharold/cpp/SPAEcpp.pdf" rel="nofollow">http://www.cs.swan.ac.uk/~csharold/cpp/SPAEcpp.pdf</a></p> <p>You have to do some finessing to make it fit into the chinese post man problem however... Imagine solving this problem for the binary digits, three digits strings. Assume you have the first two digits, and ask your self what are my options to move to? (In regards to the next two digit string?) You are left with a Directed Graph.</p> <pre><code> /-\ / V \- 00 ----&gt; 01 ^ / ^| \/ || /\ || V \ |V /-- 11 ---&gt; 10 \ ^ \-/ </code></pre> <p>Solve the Chinese Postman, you will have all combinations and will form one string The question is now, is the Chinese postman solvable? There are algorithms which determine weather or not a DAG is solvable for the CPP, but i don't know if this particular graph is <em>necessarily</em> solvable based on the problem alone. That would be a good thing to determine. You do however know you could find out algorithmically weather it is solvable and if it is you could solve it using algorithms available in that paper (I think) and online. </p> <p>Every vertex here has 2 incoming edges and 2 outgoing edges. There are 4 (2^2) vertexes.</p> <p>In the full sized problem there are <code>19683( 3 ^ 9 )</code> vertexs and every vertex has <code>512 ( 2 ^ 9 )</code> out going and incoming vertexes. There would be a total of </p> <pre><code>19683( 3 ^ 9 ) x 512 (2 ^ 9) = 10077696 edges in your graph. </code></pre> <p>Approach to solution:</p> <pre><code>1.) Create list of all 3 digit numbers 000 to 999. 2.) Create edges for all numbers such that last two digits of first number match first two digits of next number. ie 123 -&gt; 238 (valid edge) 123 -&gt; 128 (invalid edge) 3.) use Chinese Postman solving algorithmic approaches to discover if solvable and solve </code></pre>
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    1. POAdding one digit (0-9) to the sequence/string creates new 4 digits number
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    1. COThanks. I would check the article first.
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