Re: Complexity explosion in linear solve

*To*: mathgroup at smc.vnet.net*Subject*: [mg80229] Re: Complexity explosion in linear solve*From*: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>*Date*: Wed, 15 Aug 2007 04:18:19 -0400 (EDT)*Organization*: Uni Leipzig*References*: <f9s3f8$a8r$1@smc.vnet.net>*Reply-to*: kuska at informatik.uni-leipzig.de

Hi, what do you expect ? matrix inversion is an N^3 process when N is the dimension of the matrix. This mean for numerical data, that you have to do N^3 operations to get the inverse. For symbolic data a single operation gives not a single number it simply add another leaf to your expression because 1+2 is 3 (leaf count 1) but a+b is a+b (leaf count 2) so every of your N^3 operations add a new leaf to your expression and you end up (in the worst case) with N^3 more leafs in your expression .. Regards Jens carlos at colorado.edu wrote: > LinearSolve (and cousins Inverse, LUDecomposition etc) works > efficiently with > numerical float data. But it is prone to complexity explosion in > solving symbolic > systems of moderate size. Here is an example from an actual > application. > > Task: solve A x = b for x where A is 16 x 16 and b is 16 x 1: > > A={{7/12,1/24,-1/2,0,-1/4,-5/24,-1/2,0, > 0,-1/24,-1/4,0,-1/3,5/24,-1/4,0},{1/24,91/144,-2/3,0, > 5/24,-3/16,-1/3,0,-1/24,0,-1/3,0,-5/24,-4/9,-2/3, > 0},{-1/2,-2/3,(-9+(1378*C11*hh)/225)/2,(-4*C13*hh)/15, > 1/2,-1/3,(-3-2*C11*hh)/2,4*C13*hh,1/4, > 1/3,(-3-(32*C11*hh)/25)/2,0,-1/4, > 2/3,(-3-(128*C11*hh)/45)/2,(-64*C13*hh)/15},{0, > 0,(-4*C13*hh)/15,(224*C22*hh)/5,0,0,-4*C13*hh,16*C23*hh,0,0,0, > 16*C23*hh,0,0,(64*C13*hh)/15,(64*C23*hh)/5},{-1/4,5/24,1/2,0, > 7/12,-1/24,1/2,0,-1/3,-5/24,1/4,0,0,1/24,1/4, > 0},{-5/24,-3/16,-1/3,0,-1/24,91/144,-2/3,0,5/24,-4/9,-2/3, > 0,1/24,0,-1/3,0},{-1/2,-1/3,(-3-2*C11*hh)/2,-4*C13*hh, > 1/2,-2/3,(-9+(314*C11*hh)/45)/2,(4*C13*hh)/15,1/4, > 2/3,(-3-(128*C11*hh)/45)/2,(64*C13*hh)/15,-1/4, > 1/3,(-3-(32*C11*hh)/15)/2,0},{0,0,4*C13*hh,16*C23*hh,0, > 0,(4*C13*hh)/15,(832*C22*hh)/15,0, > 0,(-64*C13*hh)/15,(64*C23*hh)/5,0,0,0,(80*C23*hh)/3},{0,-1/24, > 1/4,0,-1/3,5/24,1/4,0,7/12,1/24,1/2,0,-1/4,-5/24,1/2, > 0},{-1/24,0,1/3,0,-5/24,-4/9,2/3,0,1/24,91/144,2/3,0, > 5/24,-3/16,1/3,0},{-1/4,-1/3,(-3-(32*C11*hh)/25)/2,0, > 1/4,-2/3,(-3-(128*C11*hh)/45)/2,(-64*C13*hh)/15,1/2, > 2/3,(-9+(1378*C11*hh)/225)/2,(-4*C13*hh)/15,-1/2, > 1/3,(-3-2*C11*hh)/2,4*C13*hh},{0,0,0,16*C23*hh,0, > 0,(64*C13*hh)/15,(64*C23*hh)/5,0,0,(-4*C13*hh)/15,(224*C22*hh)/5, > 0,0,-4*C13*hh,16*C23*hh},{-1/3,-5/24,-1/4,0,0,1/24,-1/4, > 0,-1/4,5/24,-1/2,0,7/12,-1/24,-1/2,0},{5/24,-4/9,2/3,0, > 1/24,0,1/3,0,-5/24,-3/16,1/3,0,-1/24,91/144,2/3, > 0},{-1/4,-2/3,(-3-(128*C11*hh)/45)/2,(64*C13*hh)/15, > 1/4,-1/3,(-3-(32*C11*hh)/15)/2,0,1/2, > 1/3,(-3-2*C11*hh)/2,-4*C13*hh,-1/2, > 2/3,(-9+(314*C11*hh)/45)/2,(4*C13*hh)/15},{0, > 0,(-64*C13*hh)/15,(64*C23*hh)/5,0,0,0,(80*C23*hh)/3,0,0, > 4*C13*hh,16*C23*hh,0,0,(4*C13*hh)/15,(832*C22*hh)/15}} > > b={0,0,0,0,q*b/2,0,0,0,q*b/2,0,0,0,0,0,0,0} > > in which all symbolic variables are atoms. Matrix A has rank 13. > Three BCs: x1=x2=x13=0 are imposed by removing equations 1, 2 and > 13, > which gives the reduced 13-system Ahat xhat = bhat. Matrix Ahat is > symmetric but indefinite, so Cholesky is not an option. Instead > LinearSolve is used > > xhat=LinearSolve[Ahat,bhat] > > Solving takes about 3 mins on a dual G5 running 5.2. LeafCounts of > the xhat entries reach hundreds of millions: > > {325197436,235675446,292306655,256982512,146153454,73076907,35324210, > 18877804,9441784,4745440,2429139,1354800,903023} > > Obviously Simplify would take a long, long time so I didnt attempt it. > Another solution method, however, gave this solution in about 10 sec: > > xhat={q/3, 0, 4*q, 0, q/3, 0, 4*q, 0, q/3, 0, 0, q/3, 0} > > My question is: is there a way to tell LinearSolve to Simplify as it > goes > along? That would preclude or at least alleviate the leafcount > explosion. > >