note InfiniteSilence You know, starting out with a finite number of pieces should mean that there is a <i>finite</i> number of connecting track sequences. If you are choosing from N pieces you can produce a series of subsets which can be viewed using <a href="http://mathworld.wolfram.com/Combination.html">n-choose-k</a>. <P> N-choose-k only gives us a certain number of distinct subsets. It answers the question, "How many ways can I pick k pieces from the box?" For instance, if the set was {1,2,3} we and we want sets of two we wind up with ({1,2}, {1,3}, {2,3}) or 3!/2!(3-2)! == 6/2(1)= 3 distinct sets. Running through this process iteratively should reveal all of the possible subsets (e.g. 4, 5, 6...N - 1). <P> Any given subset is going to have N! orderings or permutations. Going back to our above example of {1,2,3} a subset of {1,2} has two permutations: {1,2} and {2,1}. A set of three elements (i.e. {a,b,c}) will have 6 because 3! == 3*2*1 == 6. <P> Armed with the permutations for each k-subset we can start the process of eliminating those that cannot form a complete circuit by graphing out the connections somehow. It is fair to say that any less than four pieces will not produce the desired result, so k < 4 cases should be ignored. <P> In order to discover whether your graph is undirected or not, you would probably need to build an <a href="http://mathworld.wolfram.com/AdjacencyMatrix.html">adjacency matrix</a> and check to see if it is symmetric. <!-- Node text goes above. Div tags should contain sig only --> <div class="pmsig"><div class="pmsig-45622"> <P><I>Celebrate Intellectual Diversity</I> </div></div> 520671 520671