Zach Langley
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 Aug 10 comment Generating Linear Extensions of a Partial Order Although it's not as concise, I'm accepting this answer because its performance trumps the others'. Aug 8 comment How to perform a breadth-first traversal of an expression? @Mr.Wizard Yes, that is a breadth-first traversal. Aug 7 comment Word Squares and Beyond @R.M Thanks, definitely looks interesting! Aug 6 comment Generating Linear Extensions of a Partial Order Ohhh, I see. Nice! I didn't think to use integer programming for this. This is certainly the most efficient solution so far. Aug 6 comment Generating Linear Extensions of a Partial Order Can you explain the fourth constraint? Aug 6 comment Generating Linear Extensions of a Partial Order @alancalvitti $\Omega$ is an asymptotic lower bound. Think of $O$ as $\le$, $o$ as $<$, and $\Omega$ as $\ge$. (The analogy falls apart for some pairs of functions which are neither $O$ nor $\Omega$ of each other.) Aug 6 comment Word Squares and Beyond @Verbeia I haven't tried anything yet. I'm not as interested in having a solution to the problem as I am in seeing how expert Mathematica users would approach it. As you pointed out, this problem becomes quickly intractable, so I'm not looking for solutions whose asymptotic running time remarkable, but rather clever ways to utilize Mathematica's toolbox to solve the problem with not too many lines of code. Aug 5 comment Generating Linear Extensions of a Partial Order Interestingly, the pattern matching seems to be much faster than this. Try, for example, with linearExtensions[{a, b, c, d, e, f, g, h, i}, {{a, c}, {b, c}, {f, g}, {g, e}, {d, a}, {h, i}, {i, d}, {g, h}}]. Aug 4 comment Generating Linear Extensions of a Partial Order Nice! I didn't realize Subsets preserved order. Is there no nicer way to determine if a list is a subset of another than what you're doing here?