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Handle and test ArrayPartition linear algebra over Arrays #106

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Jun 9, 2020
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Handle and test ArrayPartition linear algebra over Arrays #106

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2 changes: 1 addition & 1 deletion Project.toml
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Original file line number Diff line number Diff line change
@@ -1,7 +1,7 @@
name = "RecursiveArrayTools"
uuid = "731186ca-8d62-57ce-b412-fbd966d074cd"
authors = ["Chris Rackauckas <accounts@chrisrackauckas.com>"]
version = "2.4.2"
version = "2.4.3"

[deps]
ArrayInterface = "4fba245c-0d91-5ea0-9b3e-6abc04ee57a9"
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8 changes: 4 additions & 4 deletions src/array_partition.jl
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Expand Up @@ -331,11 +331,11 @@ function LinearAlgebra.ldiv!(A::T, bb::ArrayPartition) where T<:Union{UnitUpperT
lens = map(length, b)
@inbounds for j in n:-1:1
Ajj = T(getblock(A, lens, j, j))
xj = ldiv!(Ajj, b[j])
xj = ldiv!(Ajj, vec(b[j]))
for i in j-1:-1:1
Aij = getblock(A, lens, i, j)
# bi = -Aij * xj + bi
mul!(b[i], Aij, xj, -1, true)
mul!(vec(b[i]), Aij, xj, -1, true)
end
end
return bb
Expand All @@ -348,11 +348,11 @@ function LinearAlgebra.ldiv!(A::T, bb::ArrayPartition) where T<:Union{UnitLowerT
lens = map(length, b)
@inbounds for j in 1:n
Ajj = T(getblock(A, lens, j, j))
xj = ldiv!(Ajj, b[j])
xj = ldiv!(Ajj, vec(b[j]))
for i in j+1:n
Aij = getblock(A, lens, i, j)
# bi = -Aij * xj + b[i]
mul!(b[i], Aij, xj, -1, true)
mul!(vec(b[i]), Aij, xj, -1, true)
end
end
return bb
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26 changes: 12 additions & 14 deletions test/upstream.jl
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Expand Up @@ -14,19 +14,17 @@ sol = solve(prob,AutoTsit5(Rosenbrock23()))

@test all(Array(sol) .== sol)

function f!(F, vars)
x = vars.x[1]
F.x[1][1] = (x[1]+3)*(x[2]^3-7)+18
F.x[1][2] = sin(x[2]*exp(x[1])-1)
y=vars.x[2]
F.x[2][1] = (y[1]+3)*(y[2]^3-7)+18
F.x[2][2] = sin(y[2]*exp(y[1])-1)
function mymodel(F, vars)
for i in 1:2
x = vars.x[i]
F.x[i][1,1] = (x[1,1]+3)*(x[1,2]^3-7)+18.0
F.x[i][1,2] = sin(x[1,2]*exp(x[1,1])-1)
F.x[i][2,1] = (x[2,1]+3)*(x[2,2]^3-7)+19.0
F.x[i][2,2] = sin(x[2,2]*exp(x[2,1])-3)
end
end

# To show that the function works
F = ArrayPartition([0.0 0.0],[0.0, 0.0])
u0= ArrayPartition([0.1; 1.2], [0.1; 1.2])
result = f!(F, u0)

# To show the NLsolve error that results with ArrayPartitions:
nlsolve(f!, ArrayPartition([0.1; 1.2], [0.1; 1.2]))
F = ArrayPartition([0.0 0.0; 0.0 0.0],[0.0 0.0; 0.0 0.0])
u0= ArrayPartition([0.1 1.2; 0.1 1.2], [0.1 1.2; 0.1 1.2])
result = mymodel(F, u0)
nlsolve(mymodel, u0)