mismatches between mathematical essences and data structures

Oct 13, 2009 20:15

Lately, I've been faced with problems of the following type ( Read more... )

computer_science

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Comments 34

aleffert October 14 2009, 04:23:24 UTC
Graph isomorphism is NP-complete so if you had such a representation, creating an instance for a given graph would have to be Hard.

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gustavolacerda October 14 2009, 04:32:27 UTC
I was just thinking that.

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gustavolacerda October 14 2009, 04:33:15 UTC
OR you'd have a proof of P=NP.

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gustavolacerda October 14 2009, 04:40:46 UTC
... hard in the worst case. One can still hope that there exists something that works fast most of the time.
See http://en.wikipedia.org/wiki/Smoothed_analysis

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Graphs anonymous October 14 2009, 08:19:21 UTC
Forming a canonicalization algorithm for representing graphs modulo isomorphism is clearly at least as hard as checking whether two graphs are, in fact, isomorphic (proof: just run the canonicalizer on both graphs and check if the results are equal).

And the graph isomorphism problem currently has no known polynomial-time solution.

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Re: Graphs anonymous October 14 2009, 08:20:51 UTC
...... aaaaand I didn't notice that there were 8 comments before mine, because they weren't on the same page :)

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Re: Graphs gustavolacerda October 14 2009, 08:29:22 UTC
yes, that's what aleffert and I were saying.

btw, who is this?

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Re: Graphs anonymous October 14 2009, 08:32:02 UTC
Chris

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mdinitz October 14 2009, 11:13:23 UTC
As a theoretician it pains me to do this, but there are some decent algorithms out there for testing graph isomorphism and reducing to a canonical graph. In a previous life, when I was doing combinatorial design theory, I used Brendan McKay's nauty, which does exactly what you asked for -- given a graph, it returns a representation that is canonical up to isomorphism. I have no idea what algorithm it uses, but it was reasonably fast when I used it 7 or 8 years ago.

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simrob October 14 2009, 14:49:14 UTC
Wait why are you in pain?

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mdinitz October 14 2009, 14:59:26 UTC
Because I just suggested that someone use an algorithm that is exponential time in the worst case, and moreover suggested that it was a pretty decent algorithm. Considering that I firmly believe in the fundamental importance of the worst case polytime vs. worst case exptime distinction, this was a painful suggestion to make.

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neelk October 15 2009, 21:41:29 UTC
I take it you don't like ML type inference? Worst case exptime, but in practice it is never anything other than linear-times-inverse-Ackermann unless you very, very carefully engineer your program to hit the bad case.

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anonymous October 15 2009, 08:01:50 UTC
is (1) just an example or you actually have some problem which requires some sort of optimization over the space of "k-flats"? if you have such a problem, maybe you can exploit better the structure of it in order to simplify matters. for example, you can characterize the k-th largest eigenvalue of a hermitian matrix as the solution of a certain non-linear min-max problem in which the min is taken over the space of "k-flats" and the max is taken in that space (removing the origin, actually) that's the result of the Courant-Fischer Minimax Theorem, you can take a look at its statement here http://www.cs.tau.ac.il/~haima/gen_cf.pdf

my point is, exploiting the structure of your problem may lead to other more useful characterizations of the solution-set which wouldn't require you to actually try to find a representation for the set of k-flats... there are many different methods to compute eigenvalues...

a propósito, sou eu, Ives...

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gustavolacerda October 15 2009, 17:59:21 UTC
Thanks, Ives!

Both of these are actual problems motivated by research ideas. #1 is like a regression problem: given points on a unit hypersphere centered at the origin, which k-dim "great circle" minimizes the residual sum of squares? The great circles correspond to the set of k-flats (aka "k-planes") through the origin (the circle is just the intersection of the flat and the sphere).

I don't see how eigenvalues might help here. Do you? I don't have a square matrix anywhere.

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anonymous October 15 2009, 20:55:17 UTC
what do you mean by "residual sum of squares"? is it the sum of the squared distances to the k-flat? could it be the maximum of the squared distances? I will think about it...

Ives

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gustavolacerda October 15 2009, 20:57:28 UTC
<< is it the sum of the squared distances to the k-flat? >>

yup. You could use either geodesic distance (ideal) or orthogonal distance. It probably doesn't matter which.

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