Wed, 22 Dec 2021 11:13:38 +0200
Planar finite elements, simple linear solvers for fixed dimensions
""" Linear solvers for small problems. """ module LinSolve using ..Metaprogramming export linsolve, TupleMatrix const TupleMatrix{M,N} = NTuple{M, NTuple{N, Float64}} """ `linsolve(AB :: TupleMatrix{M,N}, :: Type{TupleMatrix{M, K}}) :: TupleMatrix{M, K}` where `TupleMatrix{M, N} = NTuple{M, NTuple{N, Float64}}` “Immutable” Gaussian elimination on tuples: solve AX=B for X, Both A and B are stored in AB. The second type parameter indicates the size of B. """ @polly function linsolve₀(AB :: TupleMatrix{M,N}, :: Val{K}) :: TupleMatrix{M, K} where {N,M,K} @assert(M == N - K) k = 0 # Convert to row-echelon form for h = 1:(M-1) # Find pivotable column (has some non-zero entries in rows ≥ h) v = 0.0 î = h while k ≤ N-1 && v == 0 k = k + 1 v = abs(AB[h][k]) # Find row ≥ h of maximum absolute value in this column for i=(h+1):M local v′ = abs(AB[i][k]) if v′ > v î = i v = v′ end end end if v > 0 AB = ( AB[1:(h-1)]..., AB[î], (let ĩ = (i==î ? h : i), h̃ = î, f = AB[ĩ][k] / AB[h̃][k] ((0.0 for _ = 1:k)..., (AB[ĩ][j]-AB[h̃][j]*f for j = (k+1):N)...,) end for i = (h+1):M)... ) end end # Solve UAX=UB for X where UA with U presenting the transformations above an # upper triangular matrix. X = () for i=M:-1:1 r = .+(AB[i][M+1:end], (-AB[i][j].*X[j-i] for j=i+1:M)...)./AB[i][i] X = (r, X...) end return X end @inline function linsolve₀(AB :: TupleMatrix{M,N}) :: NTuple{M,Float64} where {N,M} X = linsolve₀(AB, Val(1)) return ((x for (x,) ∈ X)...,) end #@generated function linsolve₁(AB :: TupleMatrix{M,N}, :: Type{TupleMatrix{M, K}}) :: TupleMatrix{M, K} where {N,M,K} function generate_linsolve(AB :: Symbol, M :: Int, N :: Int, K :: Int) @assert(M == N - K) # The variables of ABN collect the stepwise stages of the transformed matrix. # Initial i-1 entries of ABN[i] are never used, as the previous steps have already # finalised the corresponding rows, but are included as “missing” (at compile time) # for the sake of indexing clarity. The M-1:th row-wise step finalises the row-echelon # form, so ABN has M-1 rows itself. step_ABN(step) = ((missing for i=1:step-1)..., (gensym("ABN_$(step)_$(i)") for i ∈ step:M)...,) ABN = ((step_ABN(step) for step=1:M-1)...,) # UAB “diagonally” refers to ABN to collate the final rows of the transformed matrix UAB. # Since the M-1:th row-wise step finalises the row-echelon form, the last row comes already # from the M-1:th step. UAB = ((ABN[i][i] for i=1:M-1)..., ABN[M-1][M]) # The variables of X collect the rows of the solution to AX=B. X = ((gensym("X_$(i)") for i ∈ 1:M)...,) # Convert to row-echelon form. On each step we strip leading zeroes from ABN. # In the end UAB should be upper-triangular. convert_row(ABNout, h, ABNin) = quote # Find pivotable column (has some non-zero entries in rows ≥ h) $(ABNout[h]) = $(ABNin[h]) local v = abs($(ABNout[h])[1]) # Find row ≥ h of maximum absolute value in this column $(sequence_exprs(:( let v′ = abs($(ABNin[i])[1]) if v′ > v $(ABNout[h]), $(ABNin[i]) = $(ABNin[i]), $(ABNout[h]) v = v′ end end ) for i=(h+1):M)) $(lift_exprs(ABNout[h+1:M])) = v > 0 ? ( $(lift_exprs( :( # Transform $(ABNin[i])[2:$N-$h+1] .- $(ABNout[h])[2:$N-$h+1].*( $(ABNin[i])[1] / $(ABNout[h])[1]) ) for i=h+1:M)) ) : ( $(lift_exprs( :( # Strip leading zeroes $(ABNin[i])[2:$N-$h+1] ) for i=h+1:M)) ) end # Solve UAX=UB for X where UA with U presenting the transformations above an # upper triangular matrix. solve_row(UAB, i) = :( $(X[i]) = $(lift_exprs( :( +($(UAB[i])[$M-$i+1+$k], $(( :( -$(UAB[i])[$j-$i+1]*$(X[j])[$k] ) for j=i+1:M)...)) ) for k=1:K )) ./ $(UAB[i])[1] ) return X, quote $(lift_exprs(ABN[1][i] for i=1:M)) = $(lift_exprs( :( $AB[$i] ) for i=1:M)) $(convert_row(ABN[1], 1, ABN[1])) $((convert_row(ABN[h], h, ABN[h-1]) for h = 2:(M-1))...) $((solve_row(UAB, i) for i=M:-1:1)...) end end @inline @generated function linsolve₁(AB :: TupleMatrix{M,N}, :: Val{K}) :: TupleMatrix{M, K} where {N,M,K} X, solver = generate_linsolve(:AB, M, N, K) return quote $solver return $(lift_exprs( X[i] for i=1:M )) end end @inline @generated function linsolve₁(AB :: TupleMatrix{M,N}) :: NTuple{M,Float64} where {N,M} X, solver = generate_linsolve(:AB, M, N, 1) return quote $solver return $(lift_exprs( :( $(X[i])[1] ) for i=1:M )) end end const linsolve = linsolve₁ function tuplify(M, N, A, b) ((((A[i][j] for j=1:N)..., b[j]) for i=1:M)...,) end function compare(; dim=5, n_matrices=10000, n_testvectors=100) testmatrices=[] testvectors=[] while length(testmatrices)<n_matrices A=randn(dim, dim) push!(testmatrices, A) end test_vectors=[randn(dim) for _ = 1:n_testvectors] function evaluate(fn) for A ∈ testmatrices for b ∈ testvectors fn(A, b) end end end function evaluate_and_report(fn, name) printstyled("Evaluating $name…\n", color=:cyan) @time evaluate(fn) end evaluate_and_report("tuple-linsolve, ungenerated") do A, b linsolve₀(tuplify(A, b)) end evaluate_and_report("tuple-linsolve, generated") do A, b linsolve₁(tuplify(A, b)) end evaluate_and_report("backslash") do A, b A \ b end end end # module