Improved method for finding optimal formulae for bilinear maps in a finite field

TL;DR

This paper introduces a new pruning criterion that enhances the search for optimal bilinear map formulae over finite fields, leading to new optimal solutions and insights into matrix product decompositions.

Contribution

It presents a novel pruning criterion for exhaustive search, enabling discovery of new optimal formulae and proving uniqueness of certain matrix product decompositions.

Findings

New optimal formulae for short product modulo X^5 and circulant product modulo (X^5 - 1)

Proof of uniqueness of the optimal decomposition for 3x2 by 2x3 matrix products

Enhanced search efficiency for bilinear map formulae

Abstract

In 2012, Barbulescu, Detrey, Estibals and Zimmermann proposed a new framework to exhaustively search for optimal formulae for evaluating bilinear maps, such as Strassen or Karatsuba formulae. The main contribution of this work is a new criterion to aggressively prune useless branches in the exhaustive search, thus leading to the computation of new optimal formulae, in particular for the short product modulo X 5 and the circulant product modulo (X 5 -- 1). Moreover , we are able to prove that there is essentially only one optimal decomposition of the product of 3 x 2 by 2 x 3 matrices up to the action of some group of automorphisms.

Tables

Table 1. Table 1: Comparison of the cardinality for ℓ = 3 ℓ 3 \ell=3 of three coverings of T 𝑇 {T} for K = 𝔽 2 𝐾 subscript 𝔽 2 K=\mathbb{F}_{2} .

set	cardinality
${𝒮}_2\left(\mathrm{Span}\left(\varnothing \right)\right)={𝒮}_2$	$980$
${𝒮}_3\left(\mathrm{Span}\left({\Phi }_0\right)\right)$	$28$
${𝒮}_4\left(\mathrm{Span}({\Phi }_0,{\Phi }_1)\right)$	$6$

Table 2. Table 2: Timings for our approach to compute the sets Ω d subscript Ω 𝑑 \Omega_{d} over K = 𝔽 2 𝐾 subscript 𝔽 2 K=\mathbb{F}_{2} on a single core of a 3.3 GHz Intel Core i5-4590.

set	${\Omega }_1$	${\Omega }_2$	${\Omega }_3$	${\Omega }_4$	${\Omega }_5$	${\Omega }_6$	${\Omega }_7$	${\Omega }_8$
cardinality	$1$	$3$	$9$	$31$	$141$	$969$	$11,289$	$265,577$
upper bound	$1$	$9$	$4.4\cdot {10}^2$	$9.9\cdot {10}^4$	$9.5\cdot {10}^7$	$3.8\cdot {10}^{11}$	$6.1\cdot {10}^{15}$	$4.0\cdot {10}^{20}$
time (s)	$0$	$4.0\cdot {10}^{-2}$	$6.0\cdot {10}^{-2}$	$1.8\cdot {10}^{-1}$	$1.5$	$1.8\cdot 10$	$4.7\cdot {10}^2$	$1.8\cdot {10}^4$

Table 3. Table 3: Timings obtained with Algorithm  BDEZ and  BDEZStab for various bilinear maps over K = 𝔽 2 𝐾 subscript 𝔽 2 K=\mathbb{F}_{2} .

bilinear map	rank	algorithm	nb. of tests	time (s)
${\text{MatProduct}}_{(2,2,2)}$	$7$	BDEZ	$1.05\cdot {10}^6$	$8.5\cdot 10$
		BDEZStab	$6.8\cdot {10}^3$	$5.0\cdot {10}^{-1}$
${\text{MatProduct}}_{(3,2,3)}$	$15$	BDEZ	$9.2\cdot {10}^{19}$ (est.)	$1.1\cdot {10}^{17}$ (est.)
		BDEZStab	$2.6\cdot {10}^{13}$ (est.)	$3.4\cdot {10}^{10}$ (est.)
		CoveringSetsMethod	$1.6\cdot {\mathrm{𝟏𝟎}}^{\mathrm{𝟗}}$	$8.5\cdot {\mathrm{𝟏𝟎}}^{\mathrm{𝟓}}$
${\text{MatProduct}}_{(2,3,2)}$	$11$	BDEZ	$2.3\cdot {10}^{23}$ (est.)	$2.7\cdot {10}^{20}$ (est.)
		BDEZStab	$4.6\cdot {10}^{18}$ (est.)	$5.4\cdot {10}^{15}$ (est.)
		CoveringSetsMethod	$6.3\cdot {\mathrm{𝟏𝟎}}^{\mathrm{𝟏𝟎}}$	$4.1\cdot {\mathrm{𝟏𝟎}}^{\mathrm{𝟔}}$
${\text{ShortProduct}}_3$	$5$	BDEZ	$5.9\cdot {10}^2$	$1.4\cdot {10}^{-1}$
		BDEZStab	$3.4\cdot 10$	$0.0$
${\text{ShortProduct}}_4$	$8$	BDEZ	$5.2\cdot {10}^7$	$4.3\cdot {10}^3$
		BDEZStab	$3.1\cdot {10}^5$	$2.7\cdot 10$
		CoveringSetsMethod	$2.8\cdot {\mathrm{𝟏𝟎}}^{\mathrm{𝟐}}$	$3.0$
${\text{ShortProduct}}_5$	$11$	BDEZ	$1.8\cdot {10}^{16}$ (est.)	$5.7\cdot {10}^{12}$ (est.)
		BDEZStab	$6.9\cdot {10}^{11}$ (est.)	$2.2\cdot {10}^8$ (est.)
		CoveringSetsMethod	$6.3\cdot {\mathrm{𝟏𝟎}}^{\mathrm{𝟔}}$	$2.4\cdot {\mathrm{𝟏𝟎}}^{\mathrm{𝟑}}$
${\text{ShortProduct}}_6$	$14$	BDEZ	$3.9\cdot {10}^{26}$ (est.)	$4.7\cdot {10}^{23}$ (est.)
		BDEZStab	$2.0\cdot {10}^{19}$ (est.)	$2.7\cdot {10}^{16}$ (est.)
${\text{CirculantProduct}}_3$	$4$	BDEZ	$36$	$0.0$
		BDEZStab	$6$	$0.1\cdot {10}^{-2}$
${\text{CirculantProduct}}_4$	$8$	BDEZ	$5.2\cdot {10}^7$	$4.3\cdot {10}^3$
		BDEZStab	$3.1\cdot {10}^5$	$2.7\cdot 10$
${\text{CirculantProduct}}_5$	$10$	BDEZ	$4.0\cdot {10}^{13}$ (est.)	$1.2\cdot {10}^{10}$ (est.)
		BDEZStab	$1.0\cdot {10}^{10}$ (est.)	$3.5\cdot {10}^6$ (est.)
		CoveringSetsMethod	$8.8\cdot {\mathrm{𝟏𝟎}}^{\mathrm{𝟖}}$	$5.4\cdot {\mathrm{𝟏𝟎}}^{\mathrm{𝟑}}$
${\text{CirculantProduct}}_6$	$12$	BDEZ	$1.0\cdot {10}^{20}$ (est.)	$1.3\cdot {10}^{17}$ (est.)
		BDEZStab	$1.1\cdot {10}^{15}$ (est.)	$1.5\cdot {10}^{12}$ (est.)

Table 4. Table 4: Computation of elements of 𝒮 15 ​ ( T 3 , 2 , 3 ) subscript 𝒮 15 subscript 𝑇 3 2 3 \mathscr{S}_{{15}}(T_{3,2,3}) .

set	cardinality	nb. tests	time (s)	nb. of solutions found
${\widetilde{ℰ}}_0$	$8.8\cdot 10$	$1.2\cdot {10}^8$	$2.0\cdot {10}^5$	$5$
${\widetilde{ℰ}}_1$	$7.5\cdot {10}^5$	$2.2\cdot {10}^7$	$3.3\cdot {10}^5$	$13$
${\widetilde{ℰ}}_2$	$1.0\cdot {10}^4$	$2.8\cdot {10}^5$	$4.1\cdot {10}^2$	$1$
${\widetilde{ℰ}}_3$	$2.7\cdot {10}^5$	$5.9\cdot {10}^8$	$9.1\cdot {10}^5$	$46$
${\widetilde{ℰ}}_4^{\prime }$	$2.5\cdot {10}^7$	$9.1\cdot {10}^8$	$1.3\cdot {10}^6$	$2$

Table 5. Table 5: Computation of 𝒮 11 ​ ( T 2 , 3 , 2 ) subscript 𝒮 11 subscript 𝑇 2 3 2 \mathscr{S}_{{11}}(T_{2,3,2}) .

set	cardinality	nb. tests	time (s)	nb. of solutions found
${\widetilde{ℰ}}_0$	$139$	$5.0\cdot {10}^4$	$6.2\cdot {10}^4$	$44$
${\widetilde{ℰ}}_1$	$3.8\cdot {10}^8$	$6.3\cdot {10}^{10}$	$4.1\cdot {10}^6$	$5,614$

Table 6. Table 6: Computation of 𝒮 r ​ ( T ) subscript 𝒮 𝑟 𝑇 \mathscr{S}_{{r}}(T) .

bilinear map	nb. of tests	time (s)	nb. of solutions	equivalence classes
${\text{ShortProduct}}_4$	$2.8\cdot {10}^2$	$3.0$	$1,440$	$220$
${\text{ShortProduct}}_5$	$6.3\cdot {10}^6$	$2.4\cdot {10}^3$	$146,944$	$11,424$

Table 7. Table 7: Computation of 𝒮 10 ​ ( T ) subscript 𝒮 10 𝑇 \mathscr{S}_{{10}}(T) .

set	cardinality	nb. tests	time (s)	nb. of solutions found
${\widetilde{ℰ}}_0$	$5.2\cdot 10$	$8.7\cdot {10}^7$	$3.1\cdot {10}^3$	$0$
${\widetilde{ℰ}}_1$	$2.0\cdot {10}^3$	$6.7\cdot {10}^5$	$2.4\cdot {10}^2$	$264$

Equations

$$A=a_0+a_{1}X\text{ and }B=b_0+b_{1}X.$$

$$\mathbf{\Phi }=\left(\begin{array}{c}{\Phi }_0\\ {\Phi }_1\\ {\Phi }_2\end{array}\right):(\mathbf{a},\mathbf{b})\mapsto \left(\begin{array}{c}{a}_0{b}_0\\ a_0{b}_1+a_1{b}_0\\ a_1{b}_1\end{array}\right).$$

$$\mathbf{\Phi }={\varphi }_0\cdot \left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)+{\varphi }_1\cdot \left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)+{\varphi }_2\cdot \left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)+{\varphi }_3\cdot \left(\begin{array}{c}0\\ 0\\ 1\end{array}\right),$$

$$\mathbf{\Phi }={\varphi }_0\cdot \left(\begin{array}{c}1\\ -1\\ 0\end{array}\right)+\psi \cdot \left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)+{\varphi }_2\cdot \left(\begin{array}{c}0\\ -1\\ 1\end{array}\right).$$

$$\mathbf{\Phi }=\sum _{0\le t<r}{\varphi }_t\cdot {\mathbf{c}}_t.$$

$$\forall h\in \left\{0,\dots ,\ell -1\right\},M_h\in \mathrm{Span}(\{N_0,\dots ,N_{r-1}\}).$$

$$C=a_0{b}_0+(a_0{b}_1+a_1{b}_0)X+(a_0{b}_2+a_1{b}_1+a_2{b}_0)X^2.$$

$$\begin{array}{rl}{\Phi }_0:& (\mathbf{a},\mathbf{b})\mapsto a_0{b}_0,\\ {\Phi }_1:& (\mathbf{a},\mathbf{b})\mapsto a_0{b}_1+a_1{b}_0,\\ {\Phi }_2:& (\mathbf{a},\mathbf{b})\mapsto a_0{b}_2+a_1{b}_1+a_2{b}_0.\end{array}$$

$$\begin{array}{r}{M}_0=\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right),M_1=\left(\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 0\end{array}\right),M_2=\left(\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ 1& 0& 0\end{array}\right).\end{array}$$

$$\Phi \circ \sigma :(\mathbf{a},\mathbf{b})\mapsto \Phi (\mu (\mathbf{a}),\nu (\mathbf{b})).$$

$$\forall \mathbf{a},\mathbf{b},((\Phi \circ \sigma )\circ {\sigma }^{\prime })(\mathbf{a},\mathbf{b})=(\Phi \circ \sigma )({\mu }^{\prime }(\mathbf{a}),{\nu }^{\prime }(\mathbf{b}))=\Phi (\mu ({\mu }^{\prime }(\mathbf{a})),\nu ({\nu }^{\prime }(\mathbf{b})))=(\Phi \circ (\sigma \circ {\sigma }^{\prime }))(\mathbf{a},\mathbf{b}).$$

$$\mathrm{rk}(T\circ \sigma )=\mathrm{rk}(T).$$

$$\begin{array}{r}{M}_1=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right),M_2=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right).\end{array}$$

$$\begin{array}{r}{M}_1^{\prime }=M_1\cdot \sigma =X^{\mathrm{T}}\cdot M_1\cdot Y=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right),M_2^{\prime }=M_2\cdot \sigma =X^{\mathrm{T}}\cdot M_2\cdot Y=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right).\end{array}$$

$$\mathrm{Stab}(\mathcal{T})=\left\{\sigma \in \mathrm{GL}(K^m)\times \mathrm{GL}(K^n)|\mathcal{T}\circ \sigma =\mathcal{T}\right\}.$$

$$\forall \sigma \in \mathrm{Stab}(T),{\mathcal{S}}_r(T)\circ \sigma ={\mathcal{S}}_r(T),$$

$$\mathbf{\Phi }:\left(\begin{array}{c}{a}_0\\ ⋮\\ a_{\ell -1}\end{array}\right),\left(\begin{array}{c}{b}_0\\ ⋮\\ b_{\ell -1}\end{array}\right)\mapsto \left(\begin{array}{c}{c}_0\\ ⋮\\ c_{\ell -1}\end{array}\right)$$

$$\forall j\in \left\{0,\dots ,\ell -1\right\},M(0,\dots ,0,1,\underset{j\text{zeros}}{\underbrace{0,\dots ,0}})=N^{\ell -1-j},$$

$$(\Psi \circ \sigma ,{\Psi }^{\prime }\circ \sigma )=(I,N);$$

$$\begin{array}{cccc}{\mathbf{\Phi }}_{p,q,r}:& {\mathcal{M}}_{p,q}(K)\times {\mathcal{M}}_{q,r}(K)& ⟶& {\mathcal{M}}_{p,r}(K)\\ & (A,B)& ⟼& A\cdot B\end{array}.$$

$$T_{p,q,r}=\mathrm{Span}(\{{\Phi }_{i,j}{\}}_{\begin{array}{c}i,j\end{array}}).$$

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right),\left(\begin{array}{cccc}0& 0& 1& 0\\ 0& 0& 0& 1\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right),\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 1& 0& 0& 0\\ 0& 1& 0& 0\end{array}\right),\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right),$$

$$\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)\otimes I_2,\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\otimes I_2,\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)\otimes I_2,\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)\otimes I_2,$$

$${\Phi }_{i,j}=\sum _{0\le h<q}{e}_i\otimes f_h\otimes f_h\otimes g_j.$$

$$\left(\left(\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 0& 1\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right),\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\right),\left(\left(\begin{array}{cccc}0& 0& 1& 0\\ 0& 0& 0& 1\\ 1& 0& 0& 0\\ 0& 1& 0& 0\end{array}\right),\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\right),$$

$$\left(\left(\begin{array}{cccc}1& 1& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 1\\ 0& 0& 0& 1\end{array}\right),\left(\begin{array}{cccc}1& 1& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 1\\ 0& 0& 0& 1\end{array}\right)\right),\left(\left(\begin{array}{cccc}0& 1& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 1& 0\end{array}\right),\left(\begin{array}{cccc}0& 1& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 1& 0\end{array}\right)\right),$$

$$\left(\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right),\left(\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 0& 1\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\right),\left(\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right),\left(\begin{array}{cccc}0& 0& 1& 0\\ 0& 0& 0& 1\\ 1& 0& 0& 0\\ 0& 1& 0& 0\end{array}\right)\right).$$

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right).$$

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right).$$

$${\mathcal{S}}_r(T)\subset \{T+V|\exists i\in \left\{0,\dots ,g-1\right\},V\in {\mathcal{E}}_{i,r}\circ \mathrm{Stab}(T)\}.$$

Full text

1 Introduction

Finding optimal formulae for computing bilinear maps is a problem of algebraic complexity theory [7, 6, 26, 15], initiated by the discoveries of Karatsuba and Ofman [16] and Strassen [26]. It consists in determining almost optimal algorithms for important problems of complexity theory, among which the well studied complexity of matrix multiplication [26, 21, 9, 18] and the complexity of polynomial multiplication [16, 27, 24, 11].

As far as polynomial multiplication is concerned, the first improvement over the schoolbook method came from Karatsuba and Ofman [16] in 1962, who proposed a decomposition of the bilinear map associated to the product of two polynomials of degree $1$

[TABLE]

Using the schoolbook algorithm, computing the product $A\cdot B$ requires ${4}$ multiplications over the coefficient ring: $a_{0}b_{0}$, $a_{1}b_{0}$, $a_{0}b_{1}$, $a_{1}b_{1}$. With the algorithm proposed by Karatsuba, the coefficients of the product $A\cdot B$ can be retrieved from the computation of the ${3}$ following multiplications: $a_{0}b_{0}$, $(a_{0}+a_{1})(b_{0}+b_{1})$, $a_{1}b_{1}$. In particular, Karatsuba’s algorithm can be applied recursively to improve the binary complexity of the multiplication of two $n$-bit integers: instead of $O(n^{2})$ with the naive schoolbook algorithm, we obtain $O(n^{\log_{2}3})$.

In 1969, Strassen [26] proposed formulae improving on the cost of the product of two $2\times 2$ matrices. When applied recursively on large matrices, this leads to a binary complexity of $O(n^{\log_{2}7})$ instead of $O(n^{\log_{2}8})=O(n^{3})$. Smirnov describes in [25] practical algorithms for matrices of higher dimensions. One can notice that, most of the time, optimal algorithms for matrix multiplication are unknown. For example, it is possible compute the product of $3\times 3$ matrices over $\mathbb{C}$ with $23$ multiplications [17], but the best known lower bound is still $19$ [4].

State of the art

An obstacle to finding optimal formulae is the fact that the decomposition of bilinear maps is known to be NP-hard [14]. In terms of method, the least-squares method seems to be one of the most popular [25]. Another way to decompose a bilinear map consists in using ingredients from geometry [3] and to find a generalization of the decomposition of singular value decomposition for matrices to general tensors. However, these methods are essentially used over an algebraically closed field $K$ (e.g. $K=\mathbb{C}$) and are not meant to produce all the possible decompositions for a bilinear map. In our context, we are looking for a method computing optimal formulae (for the bilinear rank) over a finite field $K$. These formulae can be used for the same bilinear map over any extension of $K$. Thus, they can be used in the context of the asymptotic multiplication of polynomials over a finite field for example. Furthermore, for a set of formulae of $\mathbb{Q}$, we can deduce formulae over a finite field $K$. It should be possible, given all the optimal formulae over $K$ to obtain formulae over $\mathbb{Q}$. In other terms, finding optimal formulae over finite fields can be used to improve on the multiplication algorithms over larger fields.

Montgomery proposed in [19] an algorithm to compute such a decomposition for the particular case of polynomials of small degree over a finite field. The author takes advantage of the fact that the number of possible formulae is always finite on a finite field. He obtains new formulae for the multiplication of polynomials of degree $4$, $5$ and $6$ over $\mathbb{F}_{2}$. In [20], Oseledets proposes a heuristic approach to solve the bilinear rank problem for the polynomial product over $\mathbb{F}_{2}$. Later, Barbulescu et al. proposed in [1] a unified framework, extending the idea proposed by Oseledets. This allows the authors to compute the bilinear rank of different applications, such as the short product or the middle product over a finite field. Their algorithm allows one to generate all the possible rank decompositions of any bilinear map over a finite field. We extend this work in the current article.

Contributions

The work presented is an improvement to the algorithm introduced in [1], allowing one to increase the family of bilinear maps over a finite field for which we are able to compute all the optimal formulae. Our algorithm relies on the automorphism group stabilizing a bilinear map, and on the notion of “stem” of a vector space associated to such a bilinear map. The main theorem of this work is Theorem 27 and it states that Algorithm 4 is able to find all decompositions of a bilinear map over a finite field. It can be used for proving lower bounds on the rank of a bilinear map and it has applications for improving upper bounds on the Chudnovsky-Chudnovsky algorithms [8, 23, 22]. Specifically, we compute all the decompositions for the short product of polynomials $P$ and $Q$ modulo $X^{5}$ and the product of $3\times 2$ by $2\times 3$ matrices. The latter problem was out of reach with the method used in [1]. We prove, in particular, that the set of possible decompositions for this matrix product is essentially unique, up to the action of the automorphism group. It is difficult to propose a complexity analysis showing the impact of our method, since it takes into account intrinsic properties of the bilinear maps that are considered.

Roadmap

This article is organized as follows. In Section 2, we present the theoretical tools and the framework for this article, corresponding to the framework introduced in [1]. In Section 3, we present, with kind permission of the authors, unpublished improvements [2] taking into account the symmetries of bilinear maps. In Section 4, we describe the algebraic structure of specific bilinear maps. This section can be skipped on a first read, because it is only required in proofs of the following section. In Section 5, we describe the theoretical aspect of our main contribution, which relies on the construction of coverings, and illustrate it with the examples of the short product and the matrix product. We discuss specific algorithmic aspects in Section 6: this part is quite technical and can be skipped on a first read. Finally, experimental timings are given in Section 7.

2 Preliminaries

We present in this section the definition of the mathematical objects that we manipulate in this work and we define the bilinear rank. We choose the characterization given by de Groote [10] or Bürgisser et al. [7, Ch. 14]. In particular, we introduce here the framework of [1] and the underlying linear algebra problem.

2.1 Problem statement

Let ${K}$ be a field. Given a bilinear map $\mathbf{\Phi}:{K}^{m}\times{K}^{n}\rightarrow{K}^{\ell}$, the bilinear rank problem consists in finding the minimal number of multiplications between scalars used for evaluating $\mathbf{\Phi}$. The set $\mathcal{L}(K^{m},K^{n};K^{\ell})$ denotes the set of bilinear maps from $K^{m}\times K^{n}$ to $K^{\ell}$. Any bilinear map $\mathbf{\Phi}$ from $K^{m}\times K^{n}$ to $K^{\ell}$ can be seen as an element of $\mathcal{L}(K^{m},K^{n};K)^{\ell}$, whose coordinates are the bilinear forms $(\Phi_{h})_{0\leq h<\ell}$.

Example 1** (Multiplication of linear polynomials).**

Let $A=a_{0}+a_{1}X$ and $B=b_{0}+b_{1}X$ be two polynomials over $K$. The product $A\cdot B$ is associated to the bilinear map $\mathbf{\Phi}$ taking as input the vectors $\mathbf{a}=(a_{0},a_{1})$ and $\mathbf{b}=(b_{0},b_{1})$ such that

[TABLE]

Denoting by $\phi_{0}$, $\phi_{1}$, $\phi_{2}$ and $\phi_{3}$ the bilinear forms $(\mathbf{a},\mathbf{b})\mapsto a_{0}b_{0}$, $(\mathbf{a},\mathbf{b})\mapsto a_{0}b_{1}$, $(\mathbf{a},\mathbf{b})\mapsto a_{1}b_{0}$ and $(\mathbf{a},\mathbf{b})\mapsto a_{1}b_{1}$, respectively, we have

[TABLE]

which corresponds to the schoolbook algorithm.

Let $\psi$ be an element of $\mathcal{L}(K^{2},K^{2};K)$ such that $\psi:(\mathbf{a},\mathbf{b})\mapsto(a_{0}+a_{1})(b_{0}+b_{1})$. Then, since $\phi_{1}+\phi_{2}=\psi-\phi_{0}-\phi_{3}$, we can rewrite $\mathbf{\Phi}$ as

[TABLE]

The bilinear forms $\phi_{0}$, $\psi$ and $\phi_{2}$ each correspond to exactly one multiplication over $K$. This decomposition corresponds to the Karatsuba algorithm. Thus, we can deduce that the bilinear rank of $\mathbf{\Phi}$ is at most $3$. Actually, one can show that the bilinear rank of $\mathbf{\Phi}$ is equal to $3$.

Formally, a bilinear form $\phi\in\mathcal{L}(K^{m},K^{n};K)$ is said to have rank one if there exist two linear forms $\alpha\in\mathcal{L}(K^{m};K)$ and $\beta\in\mathcal{L}(K^{n};K)$ such that $\phi(\mathbf{a},\mathbf{b})=\alpha(\mathbf{a})\cdot\beta(\mathbf{b})$. For $i\in\{0,\ldots,m-1\}$ and $j\in\{0,\ldots,n-1\}$, we denote by $e_{i,j}$ the bilinear forms $e_{i,j}:(\mathbf{a},\mathbf{b})\mapsto a_{i}b_{j}$. The $e_{i,j}$’s have rank one and form the canonical basis of $\mathcal{L}(K^{m},K^{n};K)$. This implies that any bilinear form can be expressed as a linear combination of bilinear forms of rank one.

Definition 2** (Bilinear rank).**

The rank of a bilinear form $\Phi$, denoted by $\operatorname{rk}(\Phi)$, is defined as the minimal number of bilinear forms $\phi_{t}$ of rank one such that $\Phi$ is a linear combination of the $\phi_{t}$’s. Then, a family $(\phi_{t})_{t}$ of cardinality $\operatorname{rk}(\Phi)$ is said to be an optimal decomposition of $\Phi$.

We extend this definition to bilinear maps $\mathbf{\Phi}\in\mathcal{L}(K^{m},K^{n};K)^{\ell}$: the rank $r$ of $\mathbf{\Phi}$ is the cardinality of a minimal set of bilinear forms $(\phi_{t})_{0\leq t<r}$ of rank one for which there exist vectors $\mathbf{c}_{t}\in K^{\ell}$ such that

[TABLE]

We have a matrix equivalent of Definition 2. Indeed, for $\Phi\in\mathcal{L}(K^{m},K^{n};K)$, there exists a matrix $M\in\mathcal{M}_{m,n}(K)$ such that $\Phi(\mathbf{a},\mathbf{b})={\mathbf{a}}^{\mathrm{T}}\cdot M\cdot\mathbf{b}$ for $\mathbf{a}\in K^{m}$ and $\mathbf{b}\in K^{n}$. In this situation, the usual matrix rank of $M$ is equal to the rank of $\Phi$ defined as above. Let $\mathbf{\Phi}=(\Phi_{0},\ldots,\Phi_{\ell-1})$ be a bilinear map of rank $r$, for which each $\Phi_{h}$ for $0\leq h<\ell$ is represented by $M_{h}\in\mathcal{M}_{m,n}(K)$. Consequently, there exists a set of $r$ matrices $N_{t}\in\mathcal{M}_{m,n}(K)$ of rank one such that

[TABLE]

Example 3** (Short product of polynomials of degree $2$).**

We describe in this example the matrices associated to the short product of two polynomials of degree $2$.

Let $A$ and $B$ be the polynomials $A=a_{0}+a_{1}X+a_{2}X^{2}$ and $B=b_{0}+b_{1}X+b_{2}X^{2}$. We denote by $C$ the polynomial $A\cdot B\bmod X^{3}$:

[TABLE]

We consider $A$ and $B$ as vectors of $K^{3}$ denoted by $\mathbf{a}$ and $\mathbf{b}$, respectively. Let $\Phi_{0}$, $\Phi_{1}$ and $\Phi_{2}$ be bilinear forms defined as

[TABLE]

In order to represent the corresponding matrices, we use the canonical basis for $\mathcal{L}(K^{3},K^{3};K)$, i.e. the bilinear forms $e_{i,j}$ satisfying $e_{i,j}:(\mathbf{a},\mathbf{b})\mapsto a_{i}b_{j}$, for $0\leq i,j<3$. Then, the matrices $M_{h}$ associated to $\Phi_{h}$ are

[TABLE]

2.2 A linear algebra problem

The approach of [1] consists in computing the rank of a bilinear map $\mathbf{\Phi}=(\Phi_{0},\ldots,\Phi_{\ell-1})$ by considering $T=\operatorname{Span}(\{\Phi_{0},\ldots,\Phi_{\ell-1}\})$, which is a subspace of $\mathcal{L}(K^{m},K^{n};K)$. Indeed, finding formulas for computing the $\Phi_{t}$’s is equivalent to finding a family of rank-one bilinear forms generating $T$. Thus, we need to extend the definition of the rank to subspaces of $\mathcal{L}(K^{m},K^{n};K)$.

Notation 4**.**

For $T$ a subspace of $\mathcal{L}(K^{m},K^{n};K)$, we denote by $\mathscr{S}_{{m,n,r}}(T)$ the set of subspaces $V\subset\mathcal{L}(K^{m},K^{n};K)$ spanned by a free family of rank-one bilinear forms of size $r$ such that $T\subset V$.

When $T=\operatorname{Span}(\emptyset)$, $\mathscr{S}_{{m,n,r}}(T)$ is the set of subspaces $V\in\mathcal{L}(K^{m},K^{n};K)$ spanned by a free family of rank-one bilinear forms of size $r$ and we denote it simply by $\mathscr{S}_{{m,n,r}}$.

When $m$ and $n$ are clear from the context, these sets are simply denoted by $\mathscr{S}_{{r}}(T)$ and $\mathscr{S}_{{r}}$.

We use Notation 4 to define the rank of a subspace $T\in\mathcal{L}(K^{m},K^{n};K)$ in Definition 5.

Definition 5** (Rank of a subspace of $\mathcal{L}(K^{m},K^{n};K)$).**

Let $T$ be a subspace of $\mathcal{L}(K^{m},K^{n};K)$. We denote by $\operatorname{rk}(T)$ the smallest $r$ such that $\mathscr{S}_{{r}}(T)\neq\emptyset$. The set $\mathscr{S}_{{\operatorname{rk}(T)}}(T)$ is the said to be the set of optimal decompositions of $T$.

We observe that $\operatorname{rk}(T)\geq\dim(T)$.

Let $\mathbf{\Phi}=(\Phi_{0},\ldots,\Phi_{\ell-1})\in\mathcal{L}(K^{m},K^{n};K)^{\ell}$ and $T=\operatorname{Span}(\{\Phi_{0},\ldots,\Phi_{\ell-1}\})\subset\mathcal{L}(K^{m},K^{n};K)$. Decomposing a bilinear map $\mathbf{\Phi}\in\mathcal{L}(K^{m},K^{n};K)^{\ell}$ into linear combination of $r$ rank-one bilinear forms is equivalent to computing $\mathscr{S}_{{r}}(T)$. Our approach focuses on the latter point of view, which is also the point of view taken by Algorithm [1, Alg. 1].

General strategy for computing the bilinear rank

Taking into account the formalism proposed in Section 2.2, the algorithmic strategy we use to compute the bilinear rank of a bilinear map is stated as follows.

Let $T=\operatorname{Span}(\{\Phi_{0},\ldots,\Phi_{\ell-1}\})\subset\mathcal{L}(K^{m},K^{n};K)$ of dimension $\ell$;

• 2.

start with the known lower bound $r=\ell$ on the bilinear rank;

• 3.

compute $\mathscr{S}_{{r}}(T)$;

• 4.

if $\mathscr{S}_{{r}}(T)=\emptyset$, increment $r$ and return to the previous step;

• 5.

if $\mathscr{S}_{{r}}(T)\neq\emptyset$, $r$ is the bilinear rank and $\mathscr{S}_{{r}}(T)$ the set of optimal decompositions.

2.3 The BDEZ Algorithm (Barbulescu, Detrey, Estibals, Zimmermann)

We describe in this section Algorithm [1, Alg. 1], which is a recursive method to solve the bilinear rank problem for a bilinear map $\mathbf{\Phi}=(\Phi_{0},\ldots,\Phi_{\ell-1})$ over a finite field. As described above, this is essentially equivalent to computing $\mathscr{S}_{{r}}(T)$ for $T=\operatorname{Span}(\{\Phi_{0},\ldots,\Phi_{\ell-1}\})$ of dimension $\ell$.

In order to get all the vector spaces $V\in\mathscr{S}_{{r}}$ such that $T\subset V$, we compute the vector spaces $W\in\mathscr{S}_{{r-\ell}}$ such that $T\oplus W\in\mathscr{S}_{{r}}$. In other terms, instead of enumerating all the elements of $\mathscr{S}_{{r}}$, we rather enumerate complementary subspaces of $T$ in $\mathscr{S}_{{r-\ell}}$. This restriction can be done thanks to Proposition [1, Prop. 1], reformulated as Proposition 6 using the formalism of Section 2.2.

Proposition 6**.**

Let $T$ be a subspace of dimension $\ell$ of $\mathcal{L}(K^{m},K^{n};K)$, let $r\geq\ell$ be an integer. For any $V\in\mathscr{S}_{{r}}(T)$, there exists $W\in\mathscr{S}_{{r-\ell}}$ such that $T\oplus W=V$.

Proof.

Let $\mathcal{B}$ be a basis of $V$ composed of rank-one matrices. We define inductively a sequence of subspaces $(W_{t})_{0\leq t\leq r-\ell}$, such that for any $t$ we have $W_{t}\in\mathscr{S}_{{t}}$, as follows.

The set $W_{0}$ is the null subspace and satisfies $T\oplus W_{0}\subset V$ and $\dim{T\oplus W_{0}}=\ell$.

• 2.

For $t\in\left\{1,\ldots,r-\ell\right\}$, assuming that $T\oplus W_{t-1}\subset V$ and $\dim{(T\oplus W_{t-1})}=\ell+t-1$, there exists $\Phi\in\mathcal{B}$ such that $\Phi\not\in T\oplus W_{t-1}$ (otherwise $T\oplus W_{t-1}=V$ and $\dim{V}\leq r-1$, which is a contradiction). Then, we define $W_{t}$ as $W_{t}=W_{t-1}\oplus\operatorname{Span}(\{\Phi\})$. The subspace $W_{t}$ satisfies $T\oplus W_{t}\subset V$, $\dim{(T\oplus W_{t})}=\ell+t$ and $W_{t}\in\mathscr{S}_{{t}}$.

Taking $W=W_{r-\ell}$, Proposition 6 is proved. ∎

We denote by $\mathcal{G}$ the set of rank-one bilinear forms up to a multiplicative factor, isomorphic to $\mathscr{S}_{{m,n,1}}$. In a finite field, $\mathcal{G}$ is a finite set of cardinality ${(\#K^{m}-1)(\#K^{n}-1)}/{(\#K-1)^{2}}$. Algorithm BDEZ requires a test to determine whether, for $V\in\mathcal{L}(K^{m},K^{n};K)$ of dimension $r$, we have $V\in\mathscr{S}_{{r}}$: we denote by HasRankOneBasis this test. A naive method to perform this test is described in Algorithm 1. We could think of other methods based on solving bilinear systems, but it does not seem efficient in our applications. However, an optimized version of this algorithm is used for particular bilinear maps (such as product of $2\times 3$ by $3\times 2$ matrices, for example).

Algorithm BDEZ can be described as a recursive optimized version of the backtracking method constructing all the sets of cardinality $r-\ell$ of independent bilinear forms of rank one. The input of the first call to BDEZ is: a target subspace $T$ of dimension $\ell$ and an integer $r$ ($r$ is a lower bound on the rank of $T$, as explained at the end of Section 2.2).

Algorithm BDEZ takes into account, on Line 9, the equivalence relation “modulo $V$”: two distinct elements $\phi$ and $\phi^{\prime}$ of $\mathcal{H}$ may be such that $V+\operatorname{Span}(\{\phi\})=V+\operatorname{Span}(\{\phi^{\prime}\})$. Reducing each element of $\mathcal{H}$ against $V$ (via Gauss reduction) allows us to consider a single representative for each such equivalence class modulo $V$. A similar reduction is performed on Line 15 to compute $\mathcal{G}\bmod T$.

The recursive calls of this algorithm can be represented by a tree in which each node at depth $r-\ell$ corresponds to a vector space $T\oplus W_{u_{1},u_{2},\ldots,u_{r-\ell}}$ of dimension $r$ generated by a basis of $T$ and rank-one matrices $\phi_{u_{1}},\phi_{u_{2}},\ldots,\phi_{u_{r-\ell}}$. For example, assuming that the initial set of rank-one bilinear forms is $\mathcal{G}=\{\phi_{0},\phi_{1},\phi_{2},\phi_{3}\}$ and ignoring the reductions computed on Line 9, we would obtain generically, for $r-\ell=3$, the tree given in Figure 1.

3 Improving on BDEZ using symmetries

We present in this section, with kind permission from the authors, an unpublished improvement [2] to Algorithm BDEZ. This improvement takes into account the fact that we can define rank-preserving automorphisms of $\mathcal{L}(K^{m},K^{n};K)$. Their action is defined in Section 3.1.

3.1 Action of automorphisms on $\mathcal{L}(K^{m},K^{n};K)$

We work with subspaces of $\mathcal{L}(K^{m},K^{n};K)$ rather than with bilinear maps, as in Section 2.2. We describe in this section the rank-preserving group of automorphisms $\sigma$ acting on subspaces $T\subset\mathcal{L}(K^{m},K^{n};K)$, also referred to as the $\operatorname{RP}$-automorphisms group.

Definition 7**.**

An element $\sigma=(\mu,\nu)\in\operatorname{GL}({K^{m}})\times\operatorname{GL}({K^{n}})$ acts on $\mathcal{L}(K^{m},K^{n};K)$ via

[TABLE]

Such an element is called $\operatorname{RP}$-automorphism.

Proposition 8**.**

The action of $\operatorname{GL}({K^{m}})\times\operatorname{GL}({K^{n}})$ is a group action and its elements are all invertible.

Proof.

For $\sigma=(\mu,\nu),\sigma^{\prime}=(\mu^{\prime},\nu^{\prime})\in\operatorname{GL}({K^{m}})\times\operatorname{GL}({K^{n}})$ and $\Phi\in\mathcal{L}(K^{m},K^{n};K)$, we have

[TABLE]

Thus, the action that we defined is indeed a group action. Since all the elements of $\operatorname{GL}({K^{m}})\times\operatorname{GL}({K^{n}})$ are invertible, we have automorphisms. ∎

Proposition 9** ($\operatorname{RP}$-automorphisms preserve the rank).**

Let $\sigma\in\operatorname{GL}({K^{m}})\times\operatorname{GL}({K^{n}})$.

For any $\Phi\in\mathcal{L}(K^{m},K^{n};K)$, we have $\operatorname{rk}(\Phi\operatorname{\mathop{\circ}}\sigma)=\operatorname{rk}(\Phi)$.

• 2.

For any subspace $T\subset\mathcal{L}(K^{m},K^{n};K)$, we also have $\operatorname{rk}(T\operatorname{\mathop{\circ}}\sigma)=\operatorname{rk}(T)$.

Proof.

First, let $\phi\in\mathcal{L}(K^{m},K^{n};K)$ of rank one. There exist $\alpha\in\mathcal{L}(K^{m};K)$ and $\beta\in\mathcal{L}(K^{n};K)$ such that $\phi:(\mathbf{a},\mathbf{b})\mapsto\alpha(\mathbf{a})\cdot\beta(\mathbf{b})$. There exist $\mu\in\operatorname{GL}(K^{m})$ and $\nu\in\operatorname{GL}(K^{n})$ such that $\phi\operatorname{\mathop{\circ}}\sigma:(\mathbf{a},\mathbf{b})\mapsto\alpha(\mu(\mathbf{a}))\cdot\beta(\nu(\mathbf{b}))$. Since $\alpha\operatorname{\mathop{\circ}}\mu\in\mathcal{L}(K^{m};K)$ and $\beta\operatorname{\mathop{\circ}}\nu\in\mathcal{L}(K^{n};K)$, $\phi\operatorname{\mathop{\circ}}\sigma$ is a rank-one bilinear form.

Since the $\operatorname{RP}$-automorphisms in Definition 7 preserve the rank of rank-one bilinear forms, by linearity and by definition of the rank of a bilinear form, it preserves the rank of any bilinear form. For any subspace $T\subset\mathcal{L}(K^{m},K^{n};K)$ and any $\sigma\in\operatorname{GL}({K^{m}})\times\operatorname{GL}({K^{n}})$, we have

[TABLE]

∎

Remark 10**.**

Note that, when $m=n$, Proposition 7 is not the most general notion of $\operatorname{RP}$-automorphisms that we may have: for simplicity, we do not take into account the possible transposition $\tau$ acting on any $\Phi\in\mathcal{L}(K^{m},K^{m};K)$, via $\Phi\circ\tau:(\mathbf{a},\mathbf{b})\mapsto\Phi(\mathbf{b},\mathbf{a})$.

Notation 11** (Group action on matrices).**

The group $\operatorname{GL}({K^{m}})\times\operatorname{GL}({K^{n}})$ is isomorphic to the group $\operatorname{GL}_{m}(K)\times\operatorname{GL}_{n}(K)$, acting on matrices $M$ via $M\cdot(X,Y)={X}^{\mathrm{T}}\cdot M\cdot Y.$ Thus, we often consider elements of $\operatorname{GL}({K^{m}})\times\operatorname{GL}({K^{n}})$ as elements of $\operatorname{GL}_{m}(K)\times\operatorname{GL}_{n}(K)$ and vice versa.

Example 12** (Action of $\operatorname{GL}({K^{2}})\times\operatorname{GL}({K^{2}})$).**

Let us consider the subspace $V$ of $\mathcal{L}(K^{2},K^{2};K)$ generated by the bilinear forms represented by the matrices $M_{1}$ and $M_{2}$ defined as

[TABLE]

We take $\sigma=(X,Y)$ such that $X=Y=\begin{pmatrix}0&1\\ 1&0\end{pmatrix}$.

The subspace $V^{\prime}=V\circ\sigma$ is generated by $M^{\prime}_{1}$ and $M^{\prime}_{2}$, defined as

[TABLE]

Since we will often refer to subgroups of $\operatorname{GL}({K^{m}})\times\operatorname{GL}({K^{n}})$ stabilizing elements of $\mathcal{L}(K^{m},K^{n};K)$ in the following, we define the notion of setwise stabilizer.

Definition 13** (Setwise stabilizer).**

For a subset $\mathcal{T}\subset\mathcal{L}(K^{m},K^{n};K)$, we denote by $\operatorname{Stab}(\mathcal{T})$ the subgroup of $\operatorname{GL}({K^{m}})\times\operatorname{GL}({K^{n}})$ stabilizing $\mathcal{T}$:

[TABLE]

We use the same notation for a subspace $T\subset\mathcal{L}(K^{m},K^{n};K)$.

In the rest of this work, we often refer to the “stabilizer” of a given set $\mathcal{T}$. Each time, we exclusively mean the setwise stabilizer of $\mathcal{T}$, which is, in general, different from the pointwise stabilizer of $\mathcal{T}$. Indeed, the pointwise stabilizer of $\mathcal{T}$ is defined as $\{\sigma\in\operatorname{GL}({K^{m}})\times\operatorname{GL}({K^{n}})\ |\ \forall\Phi\in\mathcal{T},\ \Phi\circ\sigma=\Phi\}$.

The algorithmic improvement originally presented in [2] comes from the fact that, for any target space $T\subset\mathcal{L}(K^{m},K^{n};K)$ of dimension $\ell$ and any integer $r\geq\ell$, we have

[TABLE]

because $\sigma$ preserves the rank. Thus, we can restrict our interest to the computation of the quotient \hbox{\mathscr{S}_{{r}}(T)}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{Stab}(T)} instead of $\mathscr{S}_{{r}}(T)$.

3.2 BDEZ with stabilizer

In order to find all the elements of $\mathscr{S}_{{r}}(T)$, it is sufficient to obtain one representative per equivalence class of \hbox{\mathscr{S}{{r}}(\operatorname{Span}(T))}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{Stab}(T)}, from which one can recover the whole orbits through the group action of $\operatorname{Stab}(T)$. Moreover, we can compute \hbox{\mathscr{S}{{r}}(T)}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{Stab}(T)} faster than $\mathscr{S}_{{r}}(T)$. Thus, we adapt our general strategy to this idea.

General strategy for computing the bilinear rank using $\operatorname{RP}$-automorphisms

The new algorithmic strategy we are considering is stated as follows, for a target subspace $T\subset\mathcal{L}(K^{m},K^{n};K)$ of dimension $\ell$ and the associated subgroup $\operatorname{Stab}(T)$ of $\operatorname{RP}$-automorphisms stabilizing $T$:

start with an initial guess $r=\ell$;

• 2.

compute \hbox{\mathscr{S}_{{r}}(T)}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{Stab}(T)} (the set $\mathscr{S}_{{r}}(T)$ up to the action of $\operatorname{Stab}(T)$);

• 3.

if \hbox{\mathscr{S}_{{r}}(T)}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{Stab}(T)}=\emptyset, increment $r$ and return to the previous step;

• 4.

enumerate $\mathscr{S}_{{r}}(T)$ using the action of $\operatorname{Stab}(T)$;

• 5.

at the end, $r$ is the rank and $\mathscr{S}_{{r}}(T)$ the set of optimal decompositions.

Algorithm BDEZStab is a recursive approach for the computation of one representative per equivalence class. The input of the first call to BDEZStab is: a target subspace $T$ of dimension $\ell$, the group $\operatorname{Stab}(T)$ and an integer $r\geq\ell$.

Figure 2 describes this recursive approach using a tree and illustrates how some branches are pruned, relying on Proposition 14. We assume that the initial set of rank-one bilinear forms is $\{\phi_{0},\phi_{1},\phi_{2},\phi_{3}\}$ and that we have $\sigma\in\operatorname{Stab}(T)$ such that $\sigma(\phi_{0})=\phi_{1}$, $\sigma(\phi_{1})=\phi_{0}$, $\sigma(\phi_{2})=\phi_{3}$ and $\sigma(\phi_{3})=\phi_{2}$.

Proposition 14**.**

Let $T$ and $V$ be subspaces of $\mathcal{L}(K^{m},K^{n};K)$ such that $V\in\mathscr{S}_{{r}}(T)$. Then, given the orbit $\phi\circ\operatorname{Stab}(T)$ of a bilinear form $\phi$ of rank one, if $V$ satisfies $V\cap\left(\phi\circ\operatorname{Stab}(T)\right)\neq\emptyset$, then there exists an element $V^{\prime}$ in the equivalence class of $V$ for the action of $\operatorname{Stab}(T)$ and such that $\phi\in V^{\prime}$.

Proof.

There exists $\sigma\in\operatorname{Stab}(T)$ such that $\phi\circ\sigma\in V$. We can then take $V^{\prime}=V\circ(\sigma^{-1})$, which meets all the conditions. ∎

The particularity of BDEZStab is that, instead of enumerating all the elements of $\mathcal{H}$ as in BDEZ, we restrict the enumeration to one element per equivalence class for the action of $U\subset\operatorname{Stab}(V)$. We use in particular the fact that the additional computations such as stabilizers on Line 11 are negligible, compared to the speed-up obtained by pruning branches in BDEZ. Heuristically, BDEZStab is faster than BDEZ by a factor $\#\operatorname{Stab}{(T)}$. This method constitutes the state of the art for the current work: our contribution is compared to the performance of this algorithm.

4 Algebraic structure of some bilinear maps

In this section, we describe the structure of some vector spaces corresponding to bilinear maps that are considered in our applications. This section can be skipped on a first read. It is needed to prove properties of specific bilinear maps that are stated in Section 5. In particular, we need to know the structure of the stabilizer of a vector space in order to be able to improve on the exhaustive search.

4.1 Short product

For the purpose of this section, we restrict our discussion to the specific case defined as follows. Let $\ell$ be a positive integer, let $\boldsymbol{\Phi}$ be the bilinear map $\boldsymbol{\Phi}\in\mathcal{L}(K^{\ell},K^{\ell};K^{\ell})$ defined by the short product

[TABLE]

such that $\sum_{0\leq i<\ell}c_{i}X^{\ell-1-i}=(\sum_{0\leq i<\ell}a_{i}X^{i})(\sum_{0\leq i<\ell}b_{i}X^{\ell-1-i})\mod X^{\ell}$. Let $T$ be the subspace of $\mathcal{L}(K^{\ell},K^{\ell};K)$ spanned by the $\ell$ bilinear forms that are the coordinates of $\boldsymbol{\Phi}$, denoted by $\Phi_{0},\ldots,\Phi_{\ell-1}$.

The matrix representing the element $\sum_{0\leq i<\ell}m_{i}\Phi_{i}\in T$, where $m_{i}\in K$, is

$M(m_{0},\ldots,m_{\ell-1})=$

{m_{0}}$${m_{1}}$${m_{\ell-1}}$${0}$${m_{1}}$${0}$${0}$${m_{0}}$$\left.\vbox{\hrule height=30.22662pt,depth=30.22662pt,width=0.0pt}\right]$$\left[\vbox{\hrule height=30.22662pt,depth=30.22662pt,width=0.0pt}\right. ,

in the canonical basis. This matrix is an upper triangular Toeplitz matrix.

Let $N$ be the matrix $M(0,1,\ldots,0)$. The matrix $N$ is a nilpotent matrix such that

[TABLE]

and $N^{\ell}=0$. The elements of the algebra $K[N]$ are the upper triangular Toeplitz matrices and $K[N]\cong K[X]/(X^{\ell})$.

We provide in Theorem 15 a useful property describing the action of $\operatorname{Stab}(T)$ on $T$.

Theorem 15**.**

Let any integer $\ell\geq 2$:

the orbit of the identity matrix $I=N^{0}$ for the action of $\operatorname{Stab}(T)$ is the set of invertible matrices of $T$; 2. 2.

the orbit of $N$ for the action of $\operatorname{Stab}(T)\cap\operatorname{Stab}(I)$ is the set of nilpotent matrices of $T$; 3. 3.

for any pair $(\Psi,\Psi^{\prime})$ of elements of $T$ such that $\operatorname{rk}(\Psi)=\ell$ and $\operatorname{rk}(\Psi^{\prime})=\ell-1$, there exists $\sigma\in\operatorname{Stab}(T)$ such that

[TABLE] 4. 4.

we have $\operatorname{Stab}(I)\cap\operatorname{Stab}(N)\subset\operatorname{Stab}(T)$ and the cardinality of $\operatorname{Stab}(T)$ is $(\#K)^{3\ell-4}(\#K-1)^{3}$.

Proof.

See A.1. ∎

4.2 Matrix product

We denote by $\boldsymbol{\Phi}_{p,q,r}$ the bilinear map corresponding to the $p\times q$ by $q\times r$ matrix product:

[TABLE]

We denote by $\Phi_{i,j}$ the bilinear forms such that $\Phi_{i,j}(A,B)$ is the coefficient $(i,j)$ of $\boldsymbol{\Phi}_{p,q,r}(A,B)$ for $i\in\left\{0,\ldots,p-1\right\},j\in\left\{0,\ldots,r-1\right\}$. The elements $\Phi_{i,j}$ satisfy $\Phi_{i,j}(A,B)=\sum_{0\leq h<q}a_{i,h}b_{h,j}$.

The bilinear map $\boldsymbol{\Phi}_{p,q,r}$ is represented by a subspace of $\mathcal{L}(K^{pq},K^{qr};K)$ denoted by

[TABLE]

In order to represent the elements of $T_{p,q,r}$ in terms of matrices of $\mathcal{M}_{pq,qr}$, we need an order on the $a_{i,h}$’s and $b_{h,j}$’s.

For the $a_{i,h}$’s, we fix the following order: $a_{i,h}\leq a_{i^{\prime},h^{\prime}}$ if $i\leq i^{\prime}$ or $i=i^{\prime}$ and $h\leq h^{\prime}$, which is the row-major order.

• 2.

For the $b_{h,j}$’s, we fix the following order: $b_{h,j}\leq b_{h^{\prime},j^{\prime}}$ if $j\leq j^{\prime}$ or $j=j^{\prime}$ and $h\leq h^{\prime}$, which is the column-major order.

Then, in the bases of $\mathcal{M}_{p,q}$ and $\mathcal{M}_{q,r}$ given by the $a_{i,h}$’s and $b_{h,j}$’s ordered as above, the elements of $T_{p,q,r}$ can be represented as matrices of $\mathcal{M}_{pq,qr}$ divided in blocks of size $q\times q$ equal to $I_{q}$ the identity matrix of $\mathcal{M}_{q,q}$. Consequently, this space is isomorphic to $\mathcal{M}_{p,r}\otimes I_{q}$ and all the elements of $T_{p,q,r}$ have a rank which is multiple of $q$.

Example 16** (Matrix representation of elements of $T_{2,2,2}$).**

The elements of $T_{2,2,2}$ are represented by matrices of $\mathcal{M}_{4,4}$ spanned by

[TABLE]

corresponding to the coefficients $a_{0,0}b_{0,0}+a_{0,1}b_{1,0}$, $a_{0,0}b_{0,1}+a_{0,1}b_{1,1}$, $a_{1,0}b_{0,0}+a_{1,1}b_{1,0}$ and $a_{1,0}b_{0,1}+a_{1,1}b_{1,1}$, respectively. The previous matrices can also be expressed as

[TABLE]

respectively.

Let $(e_{i})$, $(f_{h})$ and $(g_{j})$ be the canonical bases of $K^{p}$, $K^{q}$ and $K^{r}$. The subspace $T_{p,q,r}$ can be easily characterized with the tensor notation: it is generated by the vectors, for $i\in\left\{0,\ldots,p-1\right\},j\in\left\{0,\ldots,r-1\right\}$,

[TABLE]

Theorem 17**.**

For the group action $M\cdot(X,Y)\mapsto{X}^{\mathrm{T}}MY$, the subgroup stabilizing the vector space $T_{p,q,r}$ can be described as the group given by the pairs $({P\otimes{R}^{\mathrm{T}}},Q\otimes(R^{-1}))$ for $P\in\operatorname{GL}_{p}$, $R\in\operatorname{GL}_{q}$, and $Q\in\operatorname{GL}_{r}$.

Proof.

See A.2. ∎

Corollary 18**.**

The elements of $T_{p,q,r}$ of a given rank lie in the same orbit under the action of $\operatorname{Stab}(T_{p,q,r})$.

Example 19** (Action of the stabilizer of $T_{2,2,2}$).**

The stabilizer of $T_{2,2,2}$ is generated by the following elements of $\operatorname{GL}({K^{4}})\times\operatorname{GL}({K^{4}})$:

[TABLE]

[TABLE]

[TABLE]

The vector space of $T_{2,2,2}$ is isomorphic to $\mathcal{M}_{2,2}\otimes I_{2}$. Thus the elements of $T_{2,2,2}$ have rank [math], $2$ or $4$.

Via the action of $\operatorname{Stab}(T_{2,2,2})$, all the elements of rank $2$ can all be mapped to the element

[TABLE]

Similarly, via the action of $\operatorname{Stab}(T_{2,2,2})$, all the elements of rank $4$ can all be mapped to the element

[TABLE]

5 Coverings of subspaces of bilinear forms

Our contribution consists in reducing the number of vector spaces $W$ that we need to enumerate in order to get those that satisfy $T\oplus W\in\mathscr{S}_{{r}}$, where $T$ is the vector space representing a given bilinear map. To this effect, we restrict the enumeration to vector spaces $W$ satisfying some properties which are intrinsic to $T$. In this section, the definition and theoretical aspects of the set of vector spaces satisfying these properties are treated, illustrated via the example of the short product and the matrix product. In Section 6, we deal with practical and computational aspects.

5.1 Theoretical aspect

Our strategy consists, first, for any $r\geq\ell$, in constructing $g$ sets $\mathcal{E}_{i,r}$ for $i\in\left\{0,\ldots,g-1\right\}$, that are all subsets of $\mathscr{S}_{{r-\ell+k_{i}}}$, where $k_{i}$ is a nonnegative integer, and that satisfy some property described in Definition 20.

Definition 20** (Covering of a vector space).**

Let $r$ be a nonnegative integer, and $\{k_{i}\}$ a set of nonnegative integers such that $k_{i}\leq\ell$. Let $T$ be a subspace of $\mathcal{L}(K^{m},K^{n};K)$ of dimension $\ell$. Let $\{\mathcal{E}_{i,r}\}_{0\leq i<g}$ be a set of subsets where $\mathcal{E}_{i,r}\subset\mathscr{S}_{{r-\ell+k_{i}}}$, for all $i\in\left\{0,\ldots,g-1\right\}$. Then, $(\mathcal{E}_{i,r})_{0\leq i<g}$ is said to be a covering of $T$ if and only if, for any vector space $W\in\mathscr{S}_{{r-\ell}}$ such that $T\oplus W\in\mathscr{S}_{{r}}$, there exist an index $i\in\left\{0,\ldots,g-1\right\}$, a subspace $V\in\mathcal{E}_{i,r}$, and an $\operatorname{RP}$-automorphism $\sigma\in\operatorname{Stab}(T)$ such that $T+(V\circ\sigma)=T\oplus W$.

Proposition 21**.**

Given $T\subset\mathcal{L}(K^{m},K^{n};K)$ as above and a covering $(\mathcal{E}_{i,r})_{0\leq i<g}$ of ${T}$, then, for any $r\geq\ell$, we have

[TABLE]

Proof.

Let $V\in\mathscr{S}_{{r}}(T)$. By Proposition 6, there exists $W\in\mathscr{S}_{{r-\ell}}$ such that $T\oplus W=U$. Then, by Definition 20, there exist an index $i\in\left\{0,\ldots,g-1\right\}$, a subspace $V\in\mathcal{E}_{i,r}$, and an $\operatorname{RP}$-automorphism $\sigma\in\operatorname{Stab}(T)$ such that $T+(V\circ\sigma)=T\oplus W$. Taking $V^{\prime}=V\circ\sigma$, we thus have $U=T+V^{\prime}$ and $V^{\prime}\in\mathcal{E}_{i,r}\circ\operatorname{Stab}(T)$, which proves the inclusion. ∎

Thus, assuming that we have a method for computing the $\mathcal{E}_{i,r}$’s, we are able to cover the whole set $\mathscr{S}_{{r}}(T)$. For example, the set composed of the single set \mathcal{E}_{0,r}=\hbox{\mathscr{S}_{{r-\ell}}}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{Stab}(T)} is a covering of ${T}$ and can be enumerated using BDEZStab. We describe below how we construct the $\mathcal{E}_{i,r}$’s that we use in practice.

Definition 22** (Stem of a vector space).**

For a vector space $T$, a set $\{F_{i}\}_{0\leq i<g}$ of $g$ subspaces $F_{i}\subset T$ of dimension $k_{i}$ is said to be a stem of $T$ if and only if, for any basis $\mathcal{B}$ of $T$, there exist $i\in\left\{0,\ldots,g-1\right\}$, an $\operatorname{RP}$-automorphism $\sigma\in\operatorname{Stab}(T)$ and a free family $\mathcal{F}\subset\mathcal{B}$ of size $k_{i}$ such that

[TABLE]

Proposition 23**.**

For a vector space $T\subset\mathcal{L}(K^{m},K^{n};K)$, a stem of $T$ given by $g$ subspaces $F_{i}\subset T$, and $g$ subgroups $U_{i}\subset\operatorname{Stab}(T)\cap\operatorname{Stab}(F_{i})$, the set $\{\mathcal{E}_{i,r}\}_{0\leq i<g}$, where each $\mathcal{E}_{i,r}$ is a set of representatives of the quotient \hbox{\mathscr{S}{{r-\ell+k{i}}}(F_{i})}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{U_{i}}, is a covering of $T$.

Proof.

Let $W\in\mathscr{S}_{{r-\ell}}$ be such that $T\oplus W\in\mathscr{S}_{{r}}$. Take a basis $\mathcal{W}$ of $W$, and complete it into a basis of $T\oplus W$ using $\ell$ rank-one bilinear forms, denoted by $\{\psi_{i}\}_{0\leq i<\ell}$. For all $i\in\left\{0,\ldots,\ell-1\right\}$, write $\psi_{i}=t_{i}+w_{i}$, with $t_{i}\in T$ and $w_{i}\in W$.

The $t_{i}$’s are linearly independent. Otherwise, there would exist coefficients $(\lambda_{i})_{0\leq i<\ell}$ such that $\sum_{i=0}^{\ell-1}\lambda_{i}t_{i}=0$, whence $\sum_{i=0}^{\ell-1}\lambda_{i}\psi_{i}=\sum_{i=0}^{\ell-1}\lambda_{i}w_{i}$, which would then contradict the fact that $\{\psi_{i}\}_{0\leq i<\ell}$ completes $\mathcal{W}$ into a basis of $T\oplus W$.

Consequently, $\mathcal{B}=\{t_{i}\}_{0\leq i<\ell}$ is a free family of $\ell$ vectors of $T$ and, as $\dim(T)=\ell$, $\mathcal{B}$ is a basis of $T$. Then, by Definition 22, there exist an index $i\in\left\{0,\ldots,g-1\right\}$, a subset $\mathcal{F}\subset\mathcal{B}$ of size $k_{i}=\dim(F_{i})$, and an $\operatorname{RP}$-automorphism $\sigma\in\operatorname{Stab}(T)$ such that $\operatorname{Span}(\mathcal{F})\circ\sigma=F_{i}$.

Let $V=W\oplus\operatorname{Span}(\mathcal{F})$. Writing $\mathcal{F}=\{t_{i}\}_{i\in I}$, with $I\subset\left\{0,\ldots,\ell-1\right\}$, we define $\mathcal{F}^{\prime}=\{\psi_{i}\}_{i\in I}$. Since $\psi_{i}=t_{i}+w_{i}$ and $\operatorname{Span}(\mathcal{F}^{\prime})\in\mathscr{S}_{{k_{i}}}$, we have $V=W\oplus\operatorname{Span}(\mathcal{F})=W\oplus\operatorname{Span}(\mathcal{F}^{\prime})\in\mathscr{S}_{{r-\ell+k_{i}}}$.

Now, consider $V^{\prime}=V\circ\sigma=(W\oplus\operatorname{Span}(\mathcal{F}))\circ\sigma$: we also have $V^{\prime}\in\mathscr{S}_{{r-\ell+k_{i}}}$, as $\operatorname{RP}$-automorphisms preserve the bilinear rank, and $F_{i}=\operatorname{Span}(\mathcal{F})\circ\sigma\subset V^{\prime}$, whence $V^{\prime}\in\mathscr{S}_{{r-\ell+k_{i}}}(F_{i})$.

Finally, let $V^{\prime\prime}\in\mathcal{E}_{i,r}$ be a representative of the equivalence class of $V^{\prime}$ in the quotient set \hbox{\mathscr{S}{{r-\ell+k{i}}}(F_{i})}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{U_{i}}: there exists an $\operatorname{RP}$-automorphism $\gamma\in U_{i}$ such that $V^{\prime\prime}=V^{\prime}\circ\gamma$. We then have

[TABLE]

where the last equality comes from the fact that $\operatorname{Span}(\mathcal{F})\subset T$. Finally, as $\gamma^{-1}\circ\sigma^{-1}\in\operatorname{Stab}(T)$, this proves the result. ∎

Given $T$ and a stem of $T$, we can derive a new algorithm that computes $\mathscr{S}_{{r}}(T)$ via the computation of some intermediate sets \mathcal{E}_{i,r}=\hbox{\mathscr{S}{{r-\ell+k{i}}}(F_{i})}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{U_{i}} for $i\in\left\{0,\ldots,g-1\right\}$.

Example 24** (Two examples of stems).**

For any vector space $T$, let $\mathcal{B}$ be a basis of $T$. There exists a subset of $\mathcal{B}$ generating $T$ (namely, $\mathcal{B}$): $\{T\}$ is a stem of $T$. There exists also a subset of $\mathcal{B}$ generating $\operatorname{Span}(\emptyset)$ (namely, $\emptyset$): $\{\operatorname{Span}(\emptyset)\}$ is a stem of $T$.

An enumeration algorithm that uses $\{T\}$ as a stem amounts to computing \hbox{\mathscr{S}_{{r}}(T)}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{Stab}(T)}. In this case, we did not decompose the original problem into simpler problems.

• 2.

If the stem chosen is the set $\{\operatorname{Span}(\emptyset)\}$, this is equivalent to enumerate a set of representatives of the quotient \hbox{\mathscr{S}_{{r-\ell}}(\operatorname{Span}(\emptyset))}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{Stab}(T)}. For this purpose, no better methods than BDEZStab is known.

Thus, BDEZStab can be seen as an approach derived from the stem $\{\operatorname{Span}(\emptyset)\}$. We propose here other strategies that are derived from stems, given by sets of subspaces $F_{i}\subset T$ of dimension $k_{i}$. The enumeration of a set $\mathscr{S}_{{r-\ell+k_{i}}}(F_{i})$ is interesting in practice if its cardinality is less than $\#\mathscr{S}_{{r-\ell}}$. However, its cost depends also on the algorithms used for the computation of quotients and stabilizers and on how large $k_{i}$ is, which is detailed below.

No automatic method is known to determine, how to choose a stem for a given vector space $T$: we have to provide a stem for each $T$. This task has to be done by hand specifically for each bilinear map. We will actually do so in Section 5.2 and 5.3 for the examples of the short product and the matrix. To this end, the determination of the stabilizer, as done in Section 4, plays a key role.

In order to compute a set of the form \hbox{\mathscr{S}{{r-\ell+k{i}}}(F_{i})}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{U_{i}}, we proceed in two steps. Let $\mathcal{F}_{i}$ be a basis of $F_{i}$. Our strategy assumes that we have a finite representation of a group $U_{i}$ such that $U_{i}\subset\operatorname{Stab}(T)\cap\operatorname{Stab}(F_{i})$. In Proposition 23, the larger the groups $U_{i}$ are, the smaller the $\mathcal{E}_{i,r}$’s are. And we prefer to keep the $\mathcal{E}_{i,r}$’s as small as possible, since it gives smaller sets to enumerate. Thus, this should lead us to choose $U_{i}=\operatorname{Stab}(T)\cap\operatorname{Stab}(F_{i})$. However, in practice, the method used in our implementation is specialized to the choice $U_{i}=\operatorname{Stab}(T)\cap\operatorname{Stab}(\mathcal{F}_{i})\subset\operatorname{Stab}(T)\cap\operatorname{Stab}(F_{i})$ (we have $\operatorname{Stab}(\mathcal{F}_{i})\subset\operatorname{Stab}(F_{i})$) because only in this case do we have a practical algorithm to enumerate a set of representatives for the quotient \hbox{\mathscr{S}{{r-\ell+k{i}}}(F_{i})}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{U_{i}}.

Notation 25**.**

For a free family $\mathcal{F}$ of $k$ bilinear forms and a positive integer $d$, we let

[TABLE]

In order to enumerate sets of the form \hbox{\mathscr{S}{{r-\ell+k{i}}}(F_{i})}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{Stab}(T)\cap\operatorname{Stab}(\mathcal{F}_{i})}, we adopt a two-step strategy.

Remark 26**.**

This strategy requires the precomputation of a set of representatives of the quotient

[TABLE]

Section 6.3 describes how to compute such a set.

However, there is a pratical limit on their dimension $k_{i}$, due to the precomputations that are used in our method and that constitute a bottleneck. Assuming that

[TABLE]

behaves as $(d!)^{1.1}$ over $\mathbb{F}_{2}$ (which is an empirical estimate), storing a set of representatives of

[TABLE]

for $d=13$ would require $15$ terabytes for instance. Consequently, given the largest “$d$” for which we are able to compute in practice

[TABLE]

we have a practical constraint on how large the $r-\ell+k_{i}$’s may be: we should have $r-\ell+k_{i}\leq d$ for all $i$.

Thus, we precompute the quotient \hbox{\mathscr{S}{{r-\ell+k{i}}}}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{GL}({K^{m}})\times\operatorname{GL}({K^{n}})}. The first step consists in computing $\tilde{\mathscr{S}}_{r-\ell+k_{i}}(\mathcal{F}_{i})$ and is detailed in Section 6.1. The second step applies the action of the left transversal

[TABLE]

which can be computed using the algorithms proposed in [12] for example.

We describe in Algorithm CoveringSetsMethod the global strategy to find optimal formulae for $T$ in the sense of the bilinear rank, that is, to enumerate $\mathscr{S}_{{r}}(T)$ given a stem. We assume that we are given a subspace $T$ and a set of $g$ free families $\mathcal{F}_{0},\ldots,\mathcal{F}_{g-1}$ of $T$ such that $\{\operatorname{Span}(\mathcal{F}_{i})\}_{i}$ forms a stem of $T$.

Theorem 27**.**

Let $R$ be the rank of $T$. For any positive integer $r$, Algorithm CoveringSetsMethod proves either that

$r<R$** 2. 2.

or $R\leq r$.

In the case where $R\leq r$, any element of $\mathscr{S}_{{r}}(T)$ is included in the set returned by Algorithm CoveringSetsMethod.

The computation of the quotient $\mathcal{Q}$ on Line 5 is detailed in Section 6.1.

5.2 A stem for the short product

We use the same notations as in Section 4.1: we denote by $\Phi_{0},\ldots,\Phi_{\ell-1}$ the bilinear forms such that

[TABLE]

and by $T$ the subspace $\operatorname{Span}(\{\Phi_{0},\ldots,\Phi_{\ell-1}\})$.

In order to produce a covering of the vector spaces $W$ satisfying $T\oplus W\in\mathscr{S}_{{r}}(T)$ that we compute with CoveringSetsMethod, we need a stem of $T$. This stem is given in Proposition 28.

Proposition 28** (Stem for the short product).**

For any $\ell\geq 2$ the singleton $\{\operatorname{Span}(\{\Phi_{0},\Phi_{1}\})\}$is a stem of ${T}$: for any basis $\mathcal{B}$ of $T$, there exists $\sigma\in\operatorname{Stab}(T)$ and $\mathcal{F}\subset\mathcal{B}$ of cardinality $2$ such that

[TABLE]

Proof.

We first observe that for any $\Phi\in\operatorname{Span}(\{\Phi_{\ell-1-i},\ldots,\Phi_{\ell-1}\})$, $\operatorname{rk}(\Phi)\leq i+1$. Therefore, any element of rank $\ell$ in $T$ has a nonzero coordinate over $\Phi_{0}$ in its decomposition over the basis $(\Phi_{0},\ldots,\Phi_{\ell-1})$ and, reciprocally, any element having a nonzero coordinate over $\Phi_{0}$ has rank $\ell$. Thus, a basis $\mathcal{B}$ of $T$ necessarily contains an element of rank $\ell$ denoted by $\Psi$. The element $\Psi$ has a nonzero coordinate over $\Phi_{0}$, when we decompose it over $\left\{\Phi_{0},\ldots,\Phi_{\ell-1}\right\}$. Similarly, there exist $\Psi^{\prime}\in\mathcal{B}$ and $\lambda\in K$ for which $\Psi^{\prime}-\lambda\Psi$ has rank $\ell-1$.

We then use Theorem 15 to find an element $\sigma\in\operatorname{Stab}(T)$ such that

[TABLE]

which concludes. ∎

We give in Table 1 the cardinality of coverings of $\mathscr{S}_{{r}}(T)$ given by Proposition 28.

In conclusion, we need to compute the following set: $\tilde{\mathscr{S}}_{r-\ell+2}(\{\Phi_{0},\Phi_{1}\})$. We describe in Section 6 how we perform Line 5 of Algorithm CoveringSetsMethod. The set $\mathcal{L}$ on Line 6 of CoveringSetsMethod is, for the short product, a set containing one element, which is the identity element of $\operatorname{GL}({K^{\ell}})\times\operatorname{GL}({K^{\ell}})$.

5.3 A stem for the matrix product $3\times 2$ by $2\times 3$ over $\mathbb{F}_{2}$

We focus here on the special case given by the bilinear map

[TABLE]

over $K=\mathbb{F}_{2}$. The rank of this bilinear map is known to be $15$ [13]. However, all the optimal formulae are not known. We denote by $\Phi_{i,j}$ the bilinear forms such that $\Phi_{i,j}(A,B)$ is the coefficient $(i,j)$ of $\boldsymbol{\Phi}_{3,2,3}(A,B)$ for $i,j\in\left\{0,1,2\right\}$. The elements $\Phi_{i,j}$ satisfy $\Phi_{i,j}(A,B)=a_{i,0}b_{0,j}+a_{i,1}b_{1,j}$.

The target subspace of $\mathcal{L}(K^{6},K^{6};K)$ considered is denoted by

[TABLE]

The approach proposed in this section can be generalized to any matrix product (albeit at the expense of combinatorial blowup).

We use the stem of $T_{3,2,3}$ given by Proposition 29.

Proposition 29** (Covering of the matrix product).**

The set

[TABLE]

[TABLE]

is a covering of $T_{3,2,3}$: for any basis $\mathcal{B}$ of $T_{3,2,3}$, there exists $\mathcal{F}\subset\mathcal{B}$ and $\sigma\in\operatorname{Stab}(T_{3,2,3})$ such that

[TABLE]

Proof.

Let $\mathcal{B}$ be a basis of $T_{3,2,3}$.

If there exists an element $\Phi$ of rank $6$ in $\mathcal{B}$, then, according to Corollary 18, there exists $\sigma\in\operatorname{Stab}(T_{3,2,3})$ such that $\Phi_{0,0}+\Phi_{1,1}+\Phi_{2,2}\in\mathcal{B}\circ\sigma$. Otherwise, any element $\Phi$ of $\mathcal{B}$ has rank smaller or equal to $4$ and we have to distinguish two cases.

• 2.

If there exists an element $\Phi$ of rank $4$, there exists $\sigma$ such that $\Phi_{0,0}+\Phi_{1,1}\in\mathcal{B}\circ\sigma$ and, consequently, there exists another element $\Phi^{\prime}\in\mathcal{B}$ of rank $2$ or $4$ whose coordinate over $\Phi_{2,2}$ in the basis $(\Phi_{i,j})_{i,j}$ is nonzero: we need to look at the possible orbits in which $\Phi^{\prime}$ is included under the action of the subgroup of $\operatorname{Stab}(T_{3,2,3})$ preserving the fact that $\Phi$ is in the orbit of $\Phi_{0,0}+\Phi_{1,1}$. We can prove that there exist $3$ such orbits and that there exists $\sigma\in\operatorname{Stab}(T_{3,2,3})$ and $\mathcal{F}\subset\mathcal{B}$ of cardinality $2$ such that

[TABLE]

• 3.

Otherwise, all the elements of $\mathcal{B}$ have rank $2$ and there exists $\mathcal{F}\subset\mathcal{B}$ and $\sigma\in\operatorname{Stab}(T_{3,2,3})$ such that

[TABLE]

∎

6 How to compute subspaces containing specific bilinear forms

We propose in this section a method for computing a covering of ${\mathscr{S}_{{r}}(T)}$, where $T$ is a target space of dimension $\ell$. The covering is a set of subspaces containing a specific set of bilinear forms described as in Section 5.2 or 5.3. More specifically, we are interested in computing sets defined as $\tilde{\mathscr{S}}_{r-\ell+k}(\{\Psi_{0},\ldots,\Psi_{k-1}\})$, for $\Psi_{0},\ldots,\Psi_{k-1}$ bilinear forms of $\mathcal{L}(K^{m},K^{n};K)$. Those can be described as sets of subspaces of rank $r-\ell+k$ containing a prescribed set $\left\{\Psi_{0},\ldots,\Psi_{k-1}\right\}$ of bilinear forms, up to the action of $\operatorname{Stab}(\{\Psi_{0},\ldots,\Psi_{k-1}\})$.

6.1 General approach

First, our strategy consists in precomputing the quotient \hbox{\mathscr{S}_{{m,n,r-\ell+k}}}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{GL}({K^{m}})\times\operatorname{GL}({K^{n}})}. This quotient is smaller than $\mathscr{S}_{{m,n,r-\ell+k}}$ by construction. We explain how to compute it in Section 6.3.

Algorithm 5 explains how we compute the quotient $\mathcal{Q}$ in Algorithm CoveringSetsMethod.

Correctness of Algorithm 5.

By construction, according to Line 6, any element of $\mathcal{Q}$ is an element of

[TABLE]

First, we prove that any orbit of $\tilde{\mathscr{S}}_{r-\ell+k}(\{\Psi_{0},\ldots,\Psi_{k-1}\})$ has a representative in $\mathcal{Q}$.

Let $W^{\prime}$ be a representative of an orbit in $\tilde{\mathscr{S}}_{r-\ell+k}(\{\Psi_{0},\ldots,\Psi_{k-1}\}).$ There exist $\sigma\in\operatorname{GL}({K^{m}})\times\operatorname{GL}({K^{n}})$ and $W$ a representative of an element of \hbox{\mathscr{S}_{{m,n,r-\ell+k}}}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{GL}({K^{m}})\times\operatorname{GL}({K^{n}})} such that $W\circ\sigma=W^{\prime}$. Thus, we have $\{\Psi_{0},\ldots,\Psi_{k-1}\}\circ\sigma^{-1}\subset W$ and the set

[TABLE]

satisfies the predicate on Line 6. Any $\sigma^{\prime}$ such that $\{\Psi_{0},\ldots,\Psi_{k-1}\}\circ\sigma^{-1}\circ\sigma^{\prime}=\{\Psi_{0},\ldots,\Psi_{k-1}\}$ satisfies

[TABLE]

which means that an element of $W\circ\sigma\circ\operatorname{Stab}(\{\Psi_{0},\ldots,\Psi_{k-1}\})=W^{\prime}\circ\operatorname{Stab}(\{\Psi_{0},\ldots,\Psi_{k-1}\})$ is included in the list returned by Algorithm 5. Thus, the list returned contains at least one representative per orbit of $\tilde{\mathscr{S}}_{r-\ell+k}(\{\Psi_{0},\ldots,\Psi_{k-1}\})$.

• 2.

In the following, we prove that each orbit of $\tilde{\mathscr{S}}_{r-\ell+k}(\{\Psi_{0},\ldots,\Psi_{k-1}\})$ has a unique representative in $\mathcal{Q}$.

Assume that there exist $W,W^{\prime}\in\mathcal{Q}$ and $\gamma\in\operatorname{Stab}(\{\Psi_{0},\ldots,\Psi_{k-1}\})$ such that $W=W^{\prime}\circ\gamma$. By construction, there exists $W_{0},W^{\prime}_{0}\in\mathscr{S}_{{r-\ell+k}}$ and $\sigma,\sigma^{\prime}\in\operatorname{GL}({K^{m}})\times\operatorname{GL}({K^{n}})$ such that $W=W_{0}\circ\sigma$ and $W^{\prime}=W^{\prime}_{0}\circ\sigma^{\prime}$. Then $W^{\prime}_{0}=W_{0}\circ\sigma\circ\gamma^{-1}\circ\sigma^{\prime-1}$, whence $W^{\prime}_{0}=W_{0}$ as on Line 4 of Algorithm 5 we enumerate only one representative of each orbit of \hbox{\mathscr{S}_{{r-\ell+k}}}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{GL}({K^{m}})\times\operatorname{GL}({K^{n}})}. Thus, $\sigma\circ\gamma^{-1}\circ\sigma^{\prime-1}\in\operatorname{Stab}(W_{0})$.

Still by construction, there exists $\{\Phi_{0},\ldots,\Phi_{k-1}\}$ and $\{\Phi^{\prime}_{0},\ldots,\Phi^{\prime}_{k-1}\}\subset W_{0}$ such that

[TABLE]

and

[TABLE]

Then,

[TABLE]

and $\{\Phi_{0},\ldots,\Phi_{k-1}\}$ is in the same orbit as $\{\Phi^{\prime}_{0},\ldots,\Phi^{\prime}_{k-1}\}$ under the action of $\operatorname{Stab}(W_{0})$, which is contradictory with the definition of the quotient on Line 5.

∎

Testing the predicate on Line 6 is a problem generalizing the problem of [7, Ch. 19] and [15]: given two pairs $(M_{0},M_{1})$ and $(N_{0},N_{1})$ of $(\mathcal{M}_{m,n})^{2}$, determine whether there exists two invertible matrices $X$ and $Y$ such that $({X}^{\mathrm{T}}M_{0}Y,{X}^{\mathrm{T}}M_{1}Y)=(N_{0},N_{1})$, which is done by computing a Weierstrass–Kronecker canonical form for $(M_{0},M_{1})$. When we consider more than two matrices, for example three matrices $(M_{0},M_{1},M_{2})$ mapped to $(N_{0},N_{1},N_{2})$, we compute $(X,Y)$ such that $(M_{0},M_{1})$ is mapped to $(N_{0},N_{1})$ and we compose it with elements of \hbox{\operatorname{Stab}{M_{0}}\cap\operatorname{Stab}{M_{1}}}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{Stab}{M_{2}}}, computed with the algorithms proposed in [12] for example. The complexity for finding all the $\operatorname{RP}$-automorphisms $\sigma$ in IntermediateSetViaQuotientComputation is bounded by the cardinality of $\mathscr{S}_{{r-\ell+k}}$ (which is comparable to BDEZ) by construction, and is hard to estimate more precisely. In our applications, it appears to be negligible compared to BDEZ.

6.2 Application to the short product

We come back to the example given in Section 5.2 corresponding to the short product. We recall that $T$ is the subspace obtained from the bilinear map given by the short product modulo $\ell$ and that we need to compute the set $\mathcal{Q}=\tilde{\mathscr{S}}_{r-\ell+2}(\{\Phi_{0},\Phi_{1}\})$ for a given integer $r$.

If we take $\ell=3$, we can represent $\Phi_{0}$ and $\Phi_{1}$ by the matrices

[TABLE]

Thus, for a given couple $(M_{0},M_{1})$ of matrices representing bilinear forms of a subspace W\in\hbox{\mathscr{S}_{{r-\ell+2}}}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{GL}({K^{\ell}})\times\operatorname{GL}({K^{\ell}})} , we are looking for invertible matrices $X$ and $Y$ such that

[TABLE]

which is done in Algorithm 6. As it is precised on Line 6 of Algorithm 6, we find $X$ and $Y$ such that ${X}^{\mathrm{T}}M_{0}Y=I$ via Gauss reduction. Then, we need to check whether ${X}^{\mathrm{T}}M_{1}Y$ and $N$ are similar or not ($({X}^{\mathrm{T}}M_{1}Y)^{\ell}$ should be the null matrix for this purpose), as done on Line 9 of Algorithm 6.

Once we have computed $\mathcal{Q}$, it remains to compute the left transversal

[TABLE]

and to compute $\mathcal{Q}\circ\mathcal{L}$. According to Theorem 15, we have $\#\mathcal{L}=1$, which means that Algorithm 6 actually returns $\tilde{\mathscr{S}}_{r-\ell+2}(\{I,N\})\circ\mathcal{L}$.

In terms of complexity, we do not have explicit bounds. However, we can state that the complexity depends linearly on \#\hbox{\mathscr{S}{{r-\ell+2}}}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{GL}({K^{\ell}})\times\operatorname{GL}({K^{\ell}})} and on the number of pairs of bilinear forms $(\Phi,\Psi)$ per element of \hbox{\mathscr{S}{{r-\ell+2}}}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{GL}({K^{\ell}})\times\operatorname{GL}({K^{\ell}})} such that $\operatorname{rk}(\Phi)=\ell$ and $\operatorname{rk}(\Psi)=\ell-1$.

6.3 Computing the orbits of vector spaces of bilinear forms

In this section, we propose an approach for computing the set \hbox{\mathscr{S}_{{m,n,d}}}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{GL}({K^{m}})\times\operatorname{GL}({K^{n}})}, required by the algorithm described in Section 6.1. Its cost is at least exponential in $d$, $m$ and $n$ and difficult to estimate.

Notation 30**.**

We denote by ${\Omega}_{d}$ the quotient \hbox{\mathscr{S}_{{d,d,d}}}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{GL}({K^{d}})\times\operatorname{GL}({K^{d}})} for any $d\geq 1$.

First, we describe how we represent elements of $\mathscr{S}_{{m,n,d}}$ and we prove that given the knowledge of $\Omega_{d}$ we can deduce the elements of \hbox{\mathscr{S}_{{m,n,d}}}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{GL}({K^{m}})\times\operatorname{GL}({K^{n}})} for any $m$ and $n$ from this precomputation.

Let $W$ be an element of $\mathscr{S}_{{m,n,d}}$. There exist $d$ rank-one bilinear forms $\phi_{t}:(\mathbf{a},\mathbf{b})\mapsto\alpha_{t}(\mathbf{a})\cdot\beta_{t}(\mathbf{b})$ such that $W=\operatorname{Span}\left(\{\phi_{i}\}_{i\in\left\{0,\ldots,d-1\right\}}\right)$. In the canonical basis of $K^{m}$ and $K^{n}$, we represent $\alpha_{t}$ and $\beta_{t}$ as matrices of $\mathcal{M}_{1,m}$ and $\mathcal{M}_{1,n}$. Thus, there exist two matrices $U\in\mathcal{M}_{d,m}$ and $V\in\mathcal{M}_{d,n}$, whose rows are given by the linear forms $\alpha_{t}$ and $\beta_{t}$ respectively, and $W$ can be represented by the pair $(U,V)$. Such a representation is not unique (for example, any permutation of the rows of $(U,V)$ gives a valid representation). In particular, for a pair of matrices $(U,V)$ representing some vector space $W$, there exists $\sigma=\mu\times\nu$ in $\operatorname{GL}(K^{m})\times\operatorname{GL}(K^{n})$ such that the pair of matrices $U^{\prime},V^{\prime}$, such that $(U^{\prime},V^{\prime})=(U\circ\mu,V\circ\nu)$ represents $W\circ\sigma$, are the reduced column echelon form of the matrices $U$ and $V$, respectively.

Example 31**.**

Let us consider the vector space $W$ of $\mathscr{S}_{{3,4,6}}$ generated by the rank-one bilinear forms represented by

[TABLE]

The pair of matrices $(U,V)$ associated to $W$ is

[TABLE]

Assuming that we have a representation of the elements of \hbox{\mathscr{S}_{{d,d,d}}}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{GL}({K^{d}})\times\operatorname{GL}({K^{d}})} in terms of pairs of matrices $(U,V)\in\mathcal{M}_{d,d}\times\mathcal{M}_{d,d}$ in reduced column echelon form, we obtain all the elements of

[TABLE]

by considering the subset $\Omega^{\prime}_{d}$ of \Omega_{d}=\hbox{\mathscr{S}_{{d,d,d}}}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{GL}({K^{d}})\times\operatorname{GL}({K^{d}})} of elements represented by matrices $(U,V)$ in reduced column echelon form such that $\operatorname{rk}(U)\leq\min(m,d)$ and $\operatorname{rk}(V)\leq\min(n,d)$. Given $m$ and $n$, a set of representatives for

[TABLE]

can be seen as matrices $(U^{\prime},V^{\prime})\in\mathcal{M}_{d,m}\times\mathcal{M}_{d,n}$ in reduced column echelon form and for which there exists matrices $(U,V)\in\mathcal{M}_{d,d}^{2}$, representing an element of $\Omega^{\prime}_{d}$, obtained by adding $d-m$ an $d-n$ zero columns to $U^{\prime}$ and $V^{\prime}$, respectively, or by removing zero columns if $d<m$ or $d<n$.

Our strategy consists in deducing $\Omega_{d}$ from the computation of $\Omega_{d-1}$. Algorithm 7 describes this strategy: for each vector space $W$ of $\Omega_{d-1}$, we extend it to a vector space of $\mathcal{L}(K^{d},K^{d};K)$ by padding with zeros, and we consider the vector spaces $W\oplus\operatorname{Span}(\{\phi\})$ that can be obtained by adding an element $\phi$ of rank one. We remove from the set of $W\oplus\operatorname{Span}(\{\phi\})$ the vector spaces that are isomorphic via an isomorphism test. We determine whether two vector spaces $W^{\prime}$ and $W$ are isomorphic if there exists a basis of $W^{\prime}$ of rank-one bilinear forms such that the corresponding couple of matrices $(U^{\prime},V^{\prime})$ in reduced column echelon form is equal to $(U,V)$. The complexity of this approach depends on the number of bases of rank-one bilinear forms of $W$, which, compared to $d$, is not large generically. However, there are degenerate cases for which the nomber of bases is very large (exponential in $d^{2}$). These cases require specific code to recognize them and to treat them separately.

The naive algorithm which checks for each pair of elements of the set $\mathcal{L}$ whether or not they are isomorphic, computed in Line 11 of Algorithm 7, can be improved. Indeed, we propose to compute invariants for the group action induced by $\operatorname{GL}({K^{d}})\times\operatorname{GL}({K^{d}})$ and to compare subspaces having the same invariants. For example, for $W\in\mathscr{S}_{{d,d,d}}$, we consider the polynomial $P_{W}=\sum_{0\leq t\leq d}p_{t}(W)X^{t}$ such that

[TABLE]

Therefore, for any $\sigma\in\operatorname{GL}({K^{d}})\times\operatorname{GL}({K^{d}})$, $P_{W\circ\sigma}=P_{W}$.

We have been able to compute $\Omega_{d}$ for $d\in\left\{1,\ldots,8\right\}$ and $K=\mathbb{F}_{2}$ with an implementation in Magma V2.21-3 [5]111The code of this implementation can be found at the address http://karancode.gforge.inria.fr. The timings are described in Table 2.

It would be interesting to obtain an upper bound on $\#\Omega_{d}$ with the good order of magnitude. Indeed, we are able to say for instance that $\#\Omega_{d}$ is bounded by the quantity

[TABLE]

corresponding to the number of possible rank-one bilinear forms that we add to elements of $\Omega_{d-1}$ to obtain an element of $\Omega_{d}$. This formula leads recursively to the following bound:

[TABLE]

However, this upper bound differs by a huge factor from the true cardinality of $\Omega_{d}$ and cannot consequently be used in a complexity analysis.

To conclude, we show in Figure 3 how the subspaces of $\Omega_{3}$ over $\mathbb{F}_{2}$ are related to $\Omega_{2}$ and $\Omega_{1}$ by using its partially ordered set structure. Each element of $\Omega_{d}$ is represented by the corresponding couple of matrices $(U,V)$ of $\mathcal{M}_{d,d}^{2}$.

7 Experimental results

An implementation in Magma V2.21-3 [5] of the algorithms presented in the previous sections has been done1. We compare in this section the timings obtained from various instances of the bilinear rank problem for these different algorithms. Our Magma implementation of the algorithm described in [1] is clearly slower than the original C version. However, since we are interested in the speed-up obtained from our work, we need a fair approach. We show in particular that Algorithm BDEZStab, although it is neither multithreaded nor written in C, improves considerably on the timings estimated in [1]. The new algorithm proposed in the current article is denoted by CoveringSetsMethod: compared to Algorithm BDEZStab, it constitutes a huge speed-up on particular instances of the bilinear rank problem among which the matrix product, discussed in Sections 7.2 and 7.3, and the short product, discussed in Section 7.4. All the timings presented in this section have been done on a single core of a 3.3 GHz Intel Core i5-4590 processor.

7.1 Recursive approach

We need a few notations to denote the various bilinear maps we are interested in:

$\textsf{MatProduct}_{(p,q,r)}$ denotes the product of matrices $p\times q$ by $q\times r$,

• 2.

$\textsf{ShortProduct}_{\ell}$ denotes the product of polynomials modulo $X^{\ell}$,

• 3.

$\textsf{CirculantProduct}_{\ell}$ denotes the product of polynomials modulo $X^{\ell}-1$.

We give in Table 3 timings for various bilinear maps and for the implementations of BDEZ and BDEZStab. The number of tests represents the number of calls to HasRankOneBasis.

It is possible to estimate the time it would take to obtain a result for a bilinear rank problem out of reach for BDEZ or BDEZStab. We denote by $\mathcal{N}_{t}$ the number of calls to HasRankOneBasis in these algorithms when the input $r$ is equal to $\ell+t$. ($\ell$ is the dimension of the vector space $T$ corresponding to the bilinear map). Since when $r$ is too large, BDEZ is too expensive, there is a practical limit on the known values of $\mathcal{N}_{t}$, $t$ being a positive integer. We consider the ratio $\lceil\frac{\mathcal{N}_{t}}{\mathcal{N}_{t-1}}\rceil$ to estimate $\mathcal{N}_{t+1}$. Assuming that this ratio decreases with $t$, which seems to hold empirically, we have

[TABLE]

$t$ being a positive integer of $\left\{1,\ldots,r-\ell\right\}$.

Thus, we are able to predict timings for bilinear maps indicated in Table 3 via to this assumption, which allows us to compare Algorithm BDEZ to other approaches for problems of larger sizes. We estimate the number of tests by computing

[TABLE]

where $r-\ell$ is the difference $\operatorname{rk}(T)-\dim(T)$ for $T$ representing a bilinear map and $t$ is the largest integer for which we are able to compute $\mathcal{N}_{t}$. The time can be estimated with a similar technique. We observe that the speed-up seems to match with $\#\operatorname{Stab}(T)$, as expected. The estimated values in Table 3 relying on BDEZStab have not been effectively done because the implementation of CoveringSetsMethod allowed us to obtain more results, more efficiently. The estimations rely on the heuristic given by the Inequality 1. In the global strategy, we increase progressively the lower bound $r$ on the rank, before running BDEZ, BDEZStab or CoveringSetsMethod. For $r<\operatorname{rk}(T)$, the time spent in those algorithms is negligible, because of the exponential growth of their complexity.

It is not clear how to estimate timings for our approach CoveringSetsMethod beyond what has been done and reported in Table 3. However, for the set of bilinear maps for which CoveringSetsMethod allows one to compute all the optimal formulae, we observe a clear speed-up compared to BDEZStab.

In order to compute bilinear maps of larger degrees using this method, we need to be able to compute and store all the elements of

[TABLE]

for $\textsf{ShortProduct}_{6}$ (and even more for other bilinear maps), which has not been done yet and requires a specific effort for an optimized implementation of the algorithm described in Section 6.3. Moreover, being able to decompose a matrix product of larger dimensions, such as $3\times 3$ by $3\times 3$, requires to improve on the theoretical aspect of our strategy, since the size of the required set

[TABLE]

is expected to be too large, based on the apparent exponential growth of the progression of the sets described in Table 2.

In the following, we describe how we computed optimal formulae for bilinear maps given in Table 3 via our approach using the stems. We provide some technical details, specific to each bilinear map, necessary for an implementation.

7.2 Matrix product $3\times 2$ by $2\times 3$

We give in this section the timings obtained with our approach for computing the bilinear rank of the matrix product $(3,2,3)$ over $\mathbb{F}_{2}$. We use the same notations as in Section 5.3. We recall that we denote by $\boldsymbol{\Phi}_{3,2,3}$ the bilinear map

[TABLE]

We denote by $\Phi_{i,j}$ the bilinear forms such that $\Phi_{i,j}(A,B)$ is the coefficient $(i,j)$ of $\boldsymbol{\Phi}_{3,2,3}(A,B)$. The subspace $T_{3,2,3}$ is defined by

[TABLE]

As described in Section 6.1, we need to precompute the quotients

[TABLE]

for $k\in\left\{1,2,3\right\}$, and, given the stem that is used, we can restrict the enumeration to subspaces containing at least one element of rank $6$. The techniques for computing theses subsets are described in Section 6.3.

The intermediate sets, corresponding to the quotient $\mathcal{Q}$ computed using IntermediateSetViaQuotientComputation in Section 6, were computed in $1.6\cdot 10^{5}$ seconds. They are defined as the following sets: $\tilde{\mathcal{E}}_{0}=\tilde{\mathscr{S}}_{7}(\{\Phi_{0,0}+\Phi_{1,1}+\Phi_{2,2}\})$, $\tilde{\mathcal{E}}_{1}=\tilde{\mathscr{S}}_{8}(\{\Phi_{0,0}+\Phi_{1,1},\Phi_{0,1}+\Phi_{2,2}\})$, $\tilde{\mathcal{E}}_{2}=\tilde{\mathscr{S}}_{8}(\{\Phi_{0,0}+\Phi_{1,1},\Phi_{1,1}+\Phi_{2,2}\})$, $\tilde{\mathcal{E}}_{3}=\tilde{\mathscr{S}}_{8}(\{\Phi_{0,0}+\Phi_{1,1},\Phi_{2,2}\})$, $\tilde{\mathcal{E}}_{4}=\tilde{\mathscr{S}}_{9}(\{\Phi_{0,0},\Phi_{1,1},\Phi_{2,2}\})$. For the set $\tilde{\mathcal{E}}_{4}$, we actually used an additional trick, described in B, which allowed us to consider only a much smaller subset $\tilde{\mathcal{E}}^{\prime}_{4}$.

We give in Table 4 the time required to compute the second step of Section 6.1, which corresponds to $\mathcal{Q}\circ\mathcal{L}$ calls to HasRankOneBasis.

In conclusion, we are able to decompose $\boldsymbol{\Phi}_{3,2,3}$ over $\mathbb{F}_{2}$ and to give all the possible optimal decompositions. We have a speed-up of $10^{4}$ compared to our implementation of Algorithm BDEZStab. Although the rank of this bilinear map was already known thanks to Hopcroft and Kerr [13], determining all the possible optimal decompositions was not a well studied problem to our knowledge.

We prove with our algorithm that there is only one class of equivalence of vector spaces $W\in\mathscr{S}_{{6,6,15}}$ containing $T_{3,2,3}$, for the group action induced by $\operatorname{Stab}(T_{3,2,3})$. It is interesting to note that this is also the case for $T_{2,2,2}$. We do not have this kind of result for the short product for example.

7.3 Matrix product $2\times 3$ by $3\times 2$

We denote by $\boldsymbol{\Phi}_{2,3,2}$ the bilinear map

[TABLE]

and $\Phi_{i,j}$ its coefficients.

We compute the following sets, corresponding to the quotient $\mathcal{Q}$ computed with IntermediateSetViaQuotientComputation in Section 6, within $1.5\cdot 10^{6}$ seconds:

$\tilde{\mathcal{E}}_{0}=\tilde{\mathscr{S}}_{9}(\{\Phi_{0,0}+\Phi_{1,1},\Phi_{0,0}+\Phi_{0,1}+\Phi_{1,0}\})$,

• 2.

$\tilde{\mathcal{E}}_{1}=\tilde{\mathscr{S}}_{9}(\{\Phi_{0,0}+\Phi_{1,1},\Phi_{0,0}\})$.

We used, in particular, the fact that for any basis $\mathcal{B}$ of $T_{2,3,2}$, there exist two elements $\Phi$ and $\Psi$ of $\mathcal{B}$ such that there exists an element in $\operatorname{Span}(\{\Phi,\Psi\})$ whose decomposition over $(\Phi_{0,0},\Phi_{1,1},\Phi_{0,1},\Phi_{1,0})$ has the following shape:

[TABLE]

The timings for the second step of the method proposed in Section 6 are described in Table 5.

We obtained a speed-up of $10^{9}$ compared to our implementation of BDEZStab, and we found $1{,}096{,}452$ elements of $\mathscr{S}_{{11}}(T_{2,3,2})$, divided in $196$ equivalence classes of solutions with respect to the action of $\operatorname{Stab}(T_{2,3,2})$. The computations described in Table 5 used an improved basic test HasRankOneBasis specialized for $T_{2,3,2}$. This test uses the fact that, given a subspace $W$ of $\tilde{\mathcal{E}}_{0}$ or $\tilde{\mathcal{E}}_{1}$, we have two elements $t_{0}$ and $t_{1}$ in $T_{2,3,2}$ such that there exist $w_{0},w_{1}\in W$ such that $t_{0}-w_{0}$ and $t_{1}-w_{1}$ have rank one. We enumerate the elements $w\in W$ such that the rank of $t_{0}-w$ or $t_{1}-w$ is one, instead of enumerating the whole set of rank-one bilinear forms.

7.4 Short product

We present in this section the timings obtained with our method for the decomposition of the short product. We managed to obtain all the elements of $\mathscr{S}_{{r}}(T)$, where $T$ is the vector space generated by the bilinear forms associated to $\textsf{ShortProduct}_{\ell}$ for $\ell=4$ and $\ell=5$ and $r=\operatorname{rk}(T)$.

The last column of Table 6 describes the number of equivalence classes of vector spaces in $\mathscr{S}_{{r}}(T)$, with respect to the group $\operatorname{Stab}(T)$.

7.5 Circulant product

We present in this section how to find, with our approach, optimal decompositions of the polynomial product modulo $(X^{5}-1)$. We denote by $T$ the target space spanned by the coefficients $\Phi_{i}$ of the bilinear map

[TABLE]

The structure of $T$ allows us to gain an interesting speed-up. Indeed, $T$ has the following structure: there exists, up to a constant mulitplicative factor, a unique element of rank one $\phi=\Phi_{0}+\Phi_{1}+\Phi_{2}+\Phi_{3}+\Phi_{4}$ and a hyperplane $H$ such that $H$ contains all the elements of rank $4$ and such that all the elements of rank $5$ are included in $\operatorname{Span}(\{\phi\})\oplus H$. Moreover, the action of $\operatorname{Stab}(T)$ on $H-\{0\}$ is transitive (proved by an exhaustive enumeration in $\mathbb{F}_{2}$), which means that all the elements of rank $4$ are in the same orbit. Consequently, it is also transitive on $\operatorname{Span}(\{\phi\})\oplus H$ and all the elements of rank $5$ are in the same orbit.

Let $\mathcal{B}=\{\Phi_{0},\ldots,\Phi_{4}\}$ be a basis of $T$. We distinguish then $2$ cases: either there exists $i$ such that $\Phi_{i}$ has rank $5$, or there is no such $i$, which implies that $\phi\in\mathcal{B}$. We deduce from these observations the following sets to compute:

$\tilde{\mathcal{E}}_{0}=\tilde{\mathscr{S}}_{6}(\{\Phi_{4}\})$ and

• 2.

\mathcal{E}_{1}=\hbox{\mathscr{S}_{{9}}(\operatorname{Span}(H))}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{Stab}(H)} (any element $V\in\mathcal{E}_{1}$ satisfies $T\subset V+\operatorname{Span}(\{\phi\})\in\mathscr{S}_{{10}}$).

We obtain the set $\mathcal{E}_{1}$ via the computation of a covering of $\mathcal{E}_{1}$ obtained with

[TABLE]

We have in Table 7 the timings for the second step of the procedure described in Section 6.1. The set $\mathscr{S}_{{10}}(T)$ contains $2025$ elements divided in $9$ equivalence classes of solutions. Interestingly, the set $\tilde{\mathcal{E}}_{0}$ does not correspond to any element of $\mathscr{S}_{{10}}(T)$. It means that, for a basis $\mathcal{B}$ of bilinear forms of rank one containing $\phi$ and generating a subspace of $\mathscr{S}_{{10}}(T)$, the coordinate of the elements of rank $4$ on $\phi$ is zero.

8 Conclusions

One of the most challenging problems in the field of bilinear complexity is the decomposition of the bilinear map given by the product of $3\times 3$ matrices. Currently, our approach cannot be used to tackle this problem. However, we believe that it could be approached by further research in the direction of the Hamming weight idea developed in B. An important obstacle is the fact that, assuming that the rank is $21$, it would require to compute \hbox{\mathscr{S}_{{15}}}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{GL}({K^{9}})\times\operatorname{GL}({K^{9}})}, which is very large.

Another aspect which is not well understood currently for our approach is to establish a realistic complexity analysis. It requires a theoretical understanding of how the cardinality of the quotients \hbox{\mathscr{S}_{{d}}}\kern 0.0pt/\kern-1.00006pt\lower 1.50696pt\hbox{\operatorname{GL}({K^{d}})\times\operatorname{GL}({K^{d}})} behave asymptotically and a classification of their representatives.

Further research could focus on symmetric decompositions of bilinear maps, which have applications for the multiplication of polynomials over “small” finite fields (such as $\mathbb{F}_{2}$). Especially, we can improve on the upper bounds on the rank of the product of two polynomials of fixed degrees by improving on the bilinear complexity of the multiplication algorithms used in the Chudnovsky-Chudnovsky approach [8, 23, 22].

Finally, the approach proposed in this work allows one to compute exhaustively the optimal formulae for new bilinear maps, which was not feasible with [1]. Moreover, it uses combinatorial objects which are not well documented in the litterature, which may rekindle curiosity for them.

Acknowledgements

The author is grateful to Jérémie Detrey and Emmanuel Thomé for their helpful comments and suggestions.