Re: Speed of dot product in Mathematica

*To*: mathgroup at smc.vnet.net*Subject*: [mg5317] Re: [mg5230] Speed of dot product in Mathematica*From*: carlos at mars.Colorado.EDU (Carlos A. Felippa)*Date*: Wed, 27 Nov 1996 01:47:35 -0500*Organization*: University of Colorado, Boulder*Sender*: owner-wri-mathgroup at wolfram.com

In article <56p225$m3c at dragonfly.wolfram.com> BobHanlon at aol.com writes: >You can also use > >Plus @@ (a b) > >If you want the dot product of portions of the factors, use only the >portions. For example, > >Take[a, {100, 1000}].Take[b, {200, 1100}] > > Of course I use Take where possible. The following skyline factorization module is an example. SymmSkyMatrixFactor[S_,tol_]:= Module[ {p,a,fail,i,j,k,l,m,n,ii,ij,jj,jk,jmj,d,s,row,v}, row=SymmSkyMatrixRowLengths[S]; s=Max[row]; {p,a}=S; n=Length[p]-1; v=Table[0,{n}]; fail=0; Do [jj=p[[j+1]]; If [jj<0|row[[j]]==0, Continue[]]; d=a[[jj]]; jmj=Abs[p[[j]]]; jk=jj-jmj; Do [i=j-jk+k; v[[k]]=0; ii=p[[i+1]]; If [ii<0, Continue[]]; m=Min[ii-Abs[p[[i]]],k]-1; ij=jmj+k; v[[k]]=a[[ij]]; v[[k]]-=Take[a,{ii-m,ii-1}].Take[v,{k-m,k-1}]; a[[ij]]=v[[k]]*a[[ii]], {k,1,jk-1}]; d-=Take[a,{jmj+1,jmj+jk-1}].Take[v,{1,jk-1}]; If [Abs[d]<tol*row[[j]], fail=j; a[[jj]]=Infinity; Break[] ]; a[[jj]]=1/d, {j,1,n}]; Return[{{p,a},fail}] ]; However, the dot product cannot always be done using unit strides in both arrays, and use of Sum, Do or For exert big speed penalties. To overcome that problem I once tried the form but that was rejected as illegal. It seems an unreasonable restriction for the Take function.