Earlier today I had a discussion with a representative at Premier Support about the 2 questions I've asked here over the past couple of days:
Neither the conversation nor the answers to the above questions have produced a solution as good as compiling my function to C code. This leaves me with some problems as the function I need to deploy uses 4 high-level Mathematica functions which don't (or maybe don't readily) compile:
CholeskyDecomposition[]
IdentityMatrix[]
LinearSolve[]
Simplify[]
Yesterday I made some progress developing/adapting solutions for procedural versions of CholeskyDecomposition[]
and IdentityMatrix[]
, so these at least should compile.
Given the anticipated use of my function, maybe I can get away without using Simplify[]
.
This would leave me still needing a procedural equivalent to LinearSolve[]
or some means of compiling it.
Maybe simpler because, I don't need a myLinearSolve[]
to do everything Mathematica's version does.
Interestingly, in my conversation with the guy from support, he suggested that given the limited nature of what I wanted to do, he thought that setting options on LinearSolve[]
should give me a reduced scope version of LinearSolve[]
that Mathematica could compile. He couldn't specify which options.
So let's explore this.
A typical use of LinearSolve[]
in my current function looks like this:
c = {0.516344, 0.287671, 0.216344, 0.176796, 1};
A = {{0, 1, 1/2, 1/2, 1/2}, {0, 0, Sqrt[3]/2, 1/(2 Sqrt[3]), 1/(2 Sqrt[3])}, {0, 0, 0, Sqrt[2/3], 1/(2 Sqrt[6])}, {0, 0, 0, 0, Sqrt[5/2]/2}, {1, 1, 1, 1, 1}};
LinearSolve[A, c]
{0.173339, 0.206027, 0.187944, 0.209058, 0.223631}
No symbolic processing. No imaginary numbers. No SparseArray
objects. A
always a square matrix; c
always a vector; and its usage always outputs a real vector.
Question 1 -- Does anyone have any insight into how to set options for the above use of LinearSolve[] so it will compile or if this is even possible?
Note: Oleksandr's answer in Why is MainEvaluate being used when LinearSolve can be compiled? may have some bearing on this.
...
If no way forward comes from the above, I may still have a chance to implement a limited procedurally based proceduralLinearSolve[]
.
If very distant memory serves me, one can use the inverse of a square matrix for some problems addressable with LinearSolve[]
. This might give me a more specific way forward except Mathematica's implementation of it, Inverse[]
, doesn't fall on the list of compilable functions either.
I have found some code for calculating the inverse of a square matrix. Not even certain of the language, but I can probably follow its logic and port it to a proceduralInverse[] function in Mathematica.
This background takes me to my second question...
Question 2 -- Does creating a proceduralLinearSolve[]
or proceduralInverse[]
seem like a viable way to go (has anyone tried this kind of thing and succeeded) and/or can anyone point me in the direction of resources or solutions that could help me do this?
Answer
In many cases you don't really need to have LinearSolve compiles, since you are dealing for instance with only a specific type of matrices, or maybe just want to solve the same system with different right hand sides. In such cases you can evaluate the definiions once and compile the resulting expressions. Here is a rough example where I just injected two unkowns into your matrix.
c = {0.516344, 0.287671, 0.216344, 0.176796, 1};
A = {{0, b, 1/2, 1/2, 1/2}, {0, 0, Sqrt[3]/2, a/(2 Sqrt[3]),
1/(2 Sqrt[3])}, {0, 0, 0, Sqrt[2/3], 1/(2 Sqrt[6])}, {0, 0, 0, 0,
Sqrt[5/2]/2}, {1, 1, 1, 1, 1}};
fun[b_, a_] = LinearSolve[A, c]
funC = Compile[{{a, _Real}, {b, _Real}}, Evaluate[fun[a, b]]]
CompiledFunction[{a,b},
Block[{Compile`$2,Compile`$9},
Compile`$2=1/a;
Compile`$9=1. b;
{0.0696861 Compile`$2 (-2.4565+4.44393 a-0.5 b+1. a b),
0.0348431 Compile`$2 (4.913 +Compile`$9),
-0.0696861 (-3.69701+Compile`$9),
0.209058,
0.223631}] ,-CompiledCode-]
Now however when doing this it's important to remember that you need to check your inputs yourself, for instance in the case of A={{p1,0},{0,p2}}
, you would need to check that neither p1
or p2
are zero.
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