Skip to main content

algebraic manipulation - Generating an efficient way to compute einsum?


Given an einsum like below, how could I generate an efficient computation graph for it?


$$X_{ik} M_{ij}M_{kl} X_{jl}$$


The indices range from $1$ to $d$ and the goal is to minimize computation time assuming $d$ is large. IE, prefer $O(d^{k})$ to $O(d^{k+1})$. For the sum above, it can be computed as follows:


$$A_{kj}=X_{ik} M_{ij}\\B_{kj} = M_{kl} X_{jl}\\c=A_{kj}B_{kj}$$


You could specify this solution in terms of indices occurring in the expression


A={ik,ij}
B={kl,jl}
c={A,B}


More compactly, the problem and solution can be encoded as follows:


input: {ik, ij, kl, jl}
output: {{ik, ij}, {kl, jl}}

This is likely to be an NP-complete problem, but there are probably heuristics to find near-optimal solution most of the time.


Edit: the most important case for practical applications was when result can be expressed in terms matrices, which can be done using Carl Woll's package in the answer. Specifically, it seems to work to get efficient matrix expression for the following einsum


$$X_{ik} (M_{ij}^{(1)} M_{kl}^{(2)} + M_{ik}^{(3)} M_{jl}^{(4)} + M_{il}^{(5)} M_{jk}^{(6)}) X_{jl}$$


as


$$\text{tr}(M_2' X' M_1 X)+\text{tr}(M_3' X)\text{tr}(M_4' X)+\text{tr}(M_6' X M_5' X)$$



This was computed using the answer below as


PacletInstall[
"TensorSimplify",
"Site" -> "http://raw.githubusercontent.com/carlwoll/TensorSimplify/master"
]

<< TensorSimplify`
einsum[in_List -> out_, arrays__] :=
Module[{res = isum[in -> out, {arrays}]}, res /; res =!= $Failed];


isum[in_List -> out_, arrays_List] :=
Catch@Module[{indices, contracted, uncontracted, contractions,
transpose},
If[Length[in] != Length[arrays],
Message[einsum::length, Length[in], Length[arrays]];
Throw[$Failed]];
MapThread[
If[IntegerQ@TensorRank[#1] && Length[#1] != TensorRank[#2],
Message[einsum::shape, #1, #2];
Throw[$
Failed]] &, {in, arrays}];

indices = Tally[Flatten[in, 1]];
If[DeleteCases[indices, {_, 1 | 2}] =!= {},
Message[einsum::repeat,
Cases[indices, {x_, Except[1 | 2]} :> x]];
Throw[$Failed]];
uncontracted = Cases[indices, {x_, 1} :> x];
If[Sort[uncontracted] =!= Sort[out],
Message[einsum::output, uncontracted, out];
Throw[$
Failed]];
contracted = Cases[indices, {x_, 2} :> x];

contractions = Flatten[Position[Flatten[in, 1], #]] & /@ contracted;
transpose = FindPermutation[uncontracted, out];
Activate@
TensorTranspose[
TensorContract[Inactive[TensorProduct] @@ arrays, contractions],
transpose]]

einsum::length =
"Number of index specifications (`1`) does not match the number of \
arrays (`2`)";

einsum::shape =
"Index specification `1` does not match the array depth of `2`";
einsum::repeat =
"Index specifications `1` are repeated more than twice";
einsum::output =
"The uncontracted indices don't match the desired output";

$Assumptions = (X | M | M1 | M2 | M3 | M4 | M5 | M6) \[Element]
Matrices[{d, d}];
FromTensor@einsum[{{1, 3}, {1, 2}, {3, 4}, {2, 4}} -> {}, X, M1, M2, X]

FromTensor@
TensorReduce@
einsum[{{1, 3}, {2, 4}, {1, 3}, {2, 4}} -> {}, M3, M4, X, X]
FromTensor@
TensorReduce@
einsum[{{1, 4}, {2, 3}, {1, 3}, {2, 4}} -> {}, M5, M6, X, X]

Answer



Maybe the following will be useful for you.


You can combine my FromTensor function (part of my TensorSimplify paclet) with my einsum function to convert your einsum representation into Tr + Dot.


$Assumptions = (X|M) ∈ Matrices[{d,d}];


FromTensor @ einsum[{{1,3}, {1,2}, {3,4}, {2,4}}->{}, X, M, M, X]


Tr[Transpose[M].Transpose[X].M.X]



Hopefully the loading instructions for these functions is clear from the above links. If not, I can add them here again.


Addendum


If your tensor has disconnected pieces, then FromTensor doesn't currently work. A simple fix is to include TensorReduce. From the comments in the examples (I think I fixed a typo in the second example):


$Assumptions = (X | M) ∈ Matrices[{d,d}];


FromTensor @ TensorReduce @ einsum[{{1, 3}, {2, 4}, {1, 3}} -> {2, 4}, M, M, X]
FromTensor @ TensorReduce @ einsum[{{1, 3}, {2, 4}, {1, 3}, {2, 4}} -> {}, M, M, X, X]


M Tr[Transpose[M].X]


Tr[Transpose[M].X]^2



Comments

Popular posts from this blog

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.