I want to make a table in which each element is a matrix product:
<< Developer`
n = 2;
m = 6;
matrix = Map[ToPackedArray[#] &, Table[Sin[1.0*i*j], {i, 1, m}, {j, 1, n}, {k, 1, n}]];
vector = {IdentityMatrix[n][[1]]}\[Transpose];
Do[Flatten[Table[ #1\[ConjugateTranspose].(matrix[[#2]].#1) &[vector, i], {i, 1, m}]], {ite, 1, 2*10^5}] // AbsoluteTiming
(*{7.430010, Null}*)
In the code above there are two important parts:
First part = Choosing one matrix from the lists of matrices matrix
and multiply it by vector
.
Second part = Multiplying vector\[ConjugateTranspose]
by the result of the first part.
I thought if I convert matrix[[#2]].#1
, the first part, into one matrix multiplication, it would be more efficient as instead of many multiplications I calculate all of them in one go:
matrix2 = ToPackedArray[Flatten[Table[matrix[[i]]\[Transpose], {i, 1, m}], 1]];
matrix2
is the desired matrix to do the next step. If you use matrix//MatrixForm
, then both the matrices look like each other.
Do[x = matrix2.vector; Flatten[Table[ #1\[ConjugateTranspose].(x[[#2 ;; #2 + n - 1,All]]) &[vector, i], {i, 1, n*m, n}]], {ite, 1, 2*10^5}] // AbsoluteTiming
(*{9.080013, Null}*)
I had to still choose different parts of (matrix2.#1)
,first part, to multiply that part with #1\[ConjugateTranspose]
,second part.
As you see although I calculate all the matrix multiplications in the first part at once, it is slower. I thought it might be because of this part of code [[#2 ;; #2 + n - 1, All]]
in which I had to choose different parts of the result. I tried:
Do[x = matrix2.vector; Flatten[Table[ (x[[#1 ;; #1 + n - 1, All]]) &[i], {i, 1, n*m, n}]], {ite, 1, 2*10^5}] // AbsoluteTiming
(*{5.290007, Null}*)
Do[Flatten[Table[ (matrix[[#2]].#1) &[vector, i], {i, 1, m}]], {ite, 1, 2*10^5}] // AbsoluteTiming
(*{4.630007, Null}*)
At each iteration for matrix2
I just do the calculation once but for that of matrix
at each iteration there are 6 multiplications. In the below I remove the table for matrix2
and replace it with Do
for matrix
:
Do[x = matrix2.vector, {ite, 1, 2*10^5}] // AbsoluteTiming
(*{0.240000, Null}*)
Do[ Do[(matrix[[#2]].#1) &[vector, i], {i, 1, m}], {ite, 1, 2*10^5}] // AbsoluteTiming
(*{2.620004, Null}*)
Now the calculations for matrix2
are faster. Thus, I believe that choosing parts of matrix2.vector
causes the calculation gets slower.
How can I overcome this bottleneck? How can I choose parts of a matrix faster?
Answer
You can increase the performance by transposing your matrix and using Dot
without explicit [[...]]
<< Developer`
n = 2;
m = 6;
matrix = Table[Sin[1.0*i*j], {i, m}, {j, n}, {k, n}];
matrix3 = ToPackedArray@Transpose@matrix;
vector = N@Transpose@{IdentityMatrix[n]};
vectorHC = ConjugateTranspose@vector;
Flatten[vectorHC.matrix3.vector, {{1, 3, 2, 4, 5}}] ==
Flatten@Table[#1\[ConjugateTranspose].(matrix[[#2]].#1) &[vector, i], {i, 1, m}]
(* True *)
Do[Flatten[vectorHC.matrix3.vector, {{1, 3, 2, 4, 5}}], {ite, 10^5}] // AbsoluteTiming
(* {0.894130, Null} *)
Do[Flatten[Table[#1\[ConjugateTranspose].(matrix[[#2]].#1) &[vector, i], {i, m}]],
{ite, 10^5}] // AbsoluteTiming
(* {8.979821, Null} *)
Converting vector
to a floating-point array (N[...]
) also increase the performance a bit.
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