Tensor decomposition and homotopy continuation

Using homotopy continuation for computing the CPD of a tensor

In this blog I want to repeat the computations from the article “Homotopy Techniques for Tensor Decomposition and Perfect Identifiability” by Hauenstein, Oeding, Ottaviani and Sommese. In that article they compute the number of canonical polyadic decompositions (CPD) of a random $2\times 2\times 2\times 3$-tensor by using the monodromy method. It is known that with probability one a random complex $2\times 2\times 2\times 3$-tensor has rank equal to 4:

$$ T = \sum_{i=1}^4 a_i \otimes b_i \otimes c_i \otimes d_i,$$


$$a_i = \begin{bmatrix} a_i^{(1)} \\ a_i^{(2)} \end{bmatrix}\in \mathbb{C}^2, b_i = \begin{bmatrix} 1 \\ b_i^{(1)} \end{bmatrix} \in \mathbb{C}^2, c_i = \begin{bmatrix} 1 \\ c_i^{(1)} \end{bmatrix}\in \mathbb{C}^2, d_i = \begin{bmatrix} 1 \\ d_i^{(1)} \\ d_i^{(2)}\end{bmatrix} \in \mathbb{C}^3$$

(the $1$s compensate for the multilinear property of the Kronecker product $\otimes$ to get a unique representation of the rank one tensors $a_i\otimes b_i\otimes c_i \otimes d_i$ via the factors $a_i,b_i,c_i,d_i$). Let us decompose a random $T$ in Julia. First, we define the rank.

r = 4

Now, I initialize the variables for the factors $a_i,b_i,c_i,d_i, 1\leq i\leq 4$. The variables for the factors $a_i$, $1\leq i\leq 4$, are being accumulated in one matrix $a$. Similar for the other factors:

using HomotopyContinuation
N = 2 * 2 * 2 * 3

#define variables for the tensor T
@polyvar T[1:N]
#define variables for the factors
@polyvar a[1:2, 1:r] b[1:1, 1:r] c[1:1, 1:r] d[1:2, 1:r]

A = a; B = [ones(1,r); b]; C = [ones(1,r); c]; D = [ones(1,r); d];

Then, I define the right-hand side $\sum_{i=1}^4 a_i \otimes b_i \otimes c_i \otimes d_i$:

S = sum(kron(A[:,i], B[:,i], C[:,i], D[:,i]) for i in 1:r)

To get an initial solution for the monodromy method, I simply assign random values to the factors $a,b,c,d$ and compute the corresponding value of S, which is called T₀.

a₀ = randn(ComplexF64, 2,r);
b₀ = randn(ComplexF64, 1,r);
c₀ = randn(ComplexF64, 1,r);
d₀ = randn(ComplexF64, 2,r);

initial_decomposition = [vec(a₀); vec(b₀); vec(c₀); vec(d₀)];
T₀ = [s([vec(a); vec(b); vec(c); vec(d)] => initial_decomposition) for s in S]

Finally, I use initial_decomposition for monodromy loops with starting parameter T₀.

F = T - S
result = monodromy_solve(F, [initial_decomposition], T₀, parameters = T)

Here is the result:

• 24 solutions (0 real)
• return code → heuristic_stop
• 564 tracked paths

There are 24 solutions. However, since 1 solution gives $4!=24$ solutions, corresponding to all orderings of the summands, I actually found one solution. This confirms the computations from the paper: a random complex $2\times 2\times 2\times 3$-tensor has a unique CPD.

Cite this example:
@Misc{ CPD2022 ,
    author =  { Paul Breiding },
    title = { Tensor decomposition and homotopy continuation },
    howpublished = { \url{ https://www.JuliaHomotopyContinuation.org/examples/cpd/ } },
    note = { Accessed: June 27, 2022 }

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