Re: PentadiagonalSolve?

*To*: mathgroup at smc.vnet.net*Subject*: [mg22362] Re: PentadiagonalSolve?*From*: Daniel Lichtblau <danl at wolfram.com>*Date*: Fri, 25 Feb 2000 21:14:16 -0500 (EST)*Organization*: Wolfram Research, Inc.*References*: <88lbad$bso@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Kurt Taretto wrote: > > Solving a system of linear equations with a matrix with 5 diagonals is > done pretty well by LinearSolve. But since I work with large matrices, > and all of them have only 5 nonzero diagonals, I don't have to use > memory for the nonzero entries in order to give LinearSolve a "good and > square" matrix. > Does anybody know how could I, for example, do some transformation from > 5 to 3 diagonals and then use TridiagonalSolve? I would > appreciate any help. Thanks! If you require only machine arithmetic this can be done effectively using the sparse linear solver in the Developer context (version 4 only). Below is an example. First we create a diagonally-dominant random pentadiagonal sparse matrix. It has the form { {row,col}->value} ... }. random5Mat[n_] := Module[{mat}, mat = Table[{j+If[k>0,k,0],j+If[k<0,-k,0]}->Random[], {k,-2,2}, {j,n-Abs[k]}]; mat[[3]] = Map[#[[1]]->#[[2]]+1&, mat[[3]]]; Flatten[mat] ] Now we'll run an example of dimension 1000. SeedRandom[1111]; m = 1000; sp5mat = random5Mat[m]; rhs = Table[Random[],{m}]; On my machine (300 mhz, Linux OS): In[198]:= Timing[sol1 = Developer`SparseLinearSolve[sp5mat, rhs];] Out[198]= {0.06 Second, Null} We can compare to LinearSolve by converting to an ordinary matrix, as below. denseMat[smat_] := Module[{dim, res}, dim = Max[Map[#[[1]]&,smat]]; res = Table[{j,k}, {j,dim}, {k,dim}] /. smat; Map[If[ListQ[#],0,#]&, res, {2}] ] In[203]:= Timing[dens5mat = denseMat[sp5mat];] Out[203]= {56. Second, Null} In[204]:= Timing[sol2 = LinearSolve[dens5mat, rhs];] Out[204]= {3.36 Second, Null} Finally we check that the results are in fairly close agreement. In[205]:= Max[Abs[sol2-sol1]] -12 Out[205]= 4.1922 10 When I do the sparse solving at dimension 10000, I get Out[210]= {1.87 Second, Null} Daniel Lichtblau Wolfram Research