Solving non-linear Boolean equation systems by variable elimination

Abstract

In this paper we study Boolean equation systems, and how to eliminate variables from them while bounding the degree of polynomials produced. A procedure for variable elimination is introduced, and we relate the techniques to Gröbner bases and XL methods. We prove that by increasing the degree of the polynomials in the system by one for each variable eliminated, we preserve the solution space, provided that the system satisfies a particular condition. We then estimate how many variables we need to eliminate in order to solve the resulting system by re-linearization, and show that we get complexities lower than the trivial brute-force $\mathcal {O}(2^n)$ when the system is overdetermined.

Appendix: syzygies between polynomials of degrees \(\ge 2\)

When the degree of some $a_i$’s are greater than 1, there may be other syzygies that are not generated by the Koszul and Boolean syzygies. Let $a^1_1, \ldots a^1_{\ell _1}$ be polynomials of degree $\le 1$. By suitable Gaussian elimination we may assume the initial terms are such that:

$$\begin{aligned} {\text {in}}(a^1_1)> {\text {in}}(a^1_2)> \cdots > {\text {in}}(a^1_{\ell _1}). \end{aligned}$$

(13)

Let $a^d_1, \ldots , a^d_{\ell _d}$ be polynomials of degree $\le d$. We perform reduction operations as follows: If a term of $a^d_i$ is of the form $t \cdot {\text {in}}(a^1_j)$ where t is a monomial of degree $\le d-1$, we replace $a^d_i$ by $a^d_i - t \cdot a^1_j$. We then eventually get:

$$\begin{aligned} \text {No term of}\ a^d_i\ \text {is}\ t \cdot {\text {in}}(a^1_j)\ \text {where} \ t\ \text {is a monomial of degree}\ \le d-1. \end{aligned}$$

(14)

Secondly we may perform Gaussian elimination on the $a^d_i$ such that:

$$\begin{aligned} {\text {in}}(a^d_1)> {\text {in}}(a^d_2)> \cdots > {\text {in}}(a^d_{\ell _d}). \end{aligned}$$

(15)

Suppose we have given $a^d_i$ as above for each $1 \le d \le D$ and $i = 1, \ldots , \ell _d$. Let

$$\begin{aligned} \mathbb {B}^L = \mathbb {B}^{\ell _1} \oplus \cdots \oplus \mathbb {B}^{\ell _D} \end{aligned}$$

where $\mathbb {B}^{\ell _d} = \mathbb {B}\epsilon ^d_1 \oplus \cdots \oplus \mathbb {B}\epsilon ^d_{\ell _d}$ and we set $\epsilon ^d_j$ to have degree d. There is a map

$$\begin{aligned} \mathbb {B}^L \rightarrow \mathbb {B}, \quad \epsilon ^d_i \mapsto a^d_i \end{aligned}$$

and the syzygy module $S \subseteq \mathbb {B}^L$ is the kernel of this map.

Suppose now we have a total order on the terms of $\mathbb {B}$. We make a term order on $\mathbb {B}^L$ by letting terms $s \epsilon ^e_j < t \epsilon ^d_i$ if:

• $e < d$, or

• $e = d$ and $j < i$, or

• $e = d, j = i$ and $s < t$

Theorem 26

Given polynomials $a^d_i$ of degree $\le d$, for each $1 \le d \le D$ and suppose for each d they fulfill Condition (15) above. The following syzygies may exist:

• 1. Koszul syzygies $a^d_j \epsilon ^e_k + a^e_k \epsilon ^d_j$ where $e < d$ or $e = d$ and $k < j$. For given sum $d+e$ denote by $K^{d+e}$ the linear space these syzygies generate.

• 2. Boolean syzygies $(a^d_j + 1) \epsilon ^d_j$. For given d denote by $B^{2d}$ the linear space these syzygies generate.

• 3. For each $\delta \ge 2$ syzygies

  $$\begin{aligned} \mathbf {r}= \underset{\begin{matrix} d = 1, \ldots , \delta \\ i = 1, \ldots , \ell _d \end{matrix}}{\sum } r^{\delta -d}_i \epsilon ^d_i \end{aligned}$$

  where $r^{\delta -d}_i$ has degree $\le \delta -d$ and no term of $\mathbf {r}$ is $\tau \cdot t$ where t is the initial term of a syzygy in $K^e$ or $B^e$ and $\deg (\tau ) + e \le \delta$.

For a given $\delta$ in 3., denote by $R^{\le \delta }$ the linear space of such syzygies.

• a. Then for $\delta \ge 2$ we have:

  $$\begin{aligned} S^{\le \delta } = \sum _{d = 2}^{\delta } S^{\le \delta -d} K^d +\sum _{d = 2}^{\delta } S^{\le \delta -d}B^d + R^{\le \delta }. \end{aligned}$$

  (16)

• b. Suppose in addition the $a^d_i$ fulfill the Condition (14) above. Then we may let $R^{\le \delta }$ be the space of all syzygies of type 3. where the coefficient $r^{\delta -1}_i$ of the $a^1_i$ vanish, and we still have the above identity (16).

Proof

Given a syzygy of degree $\le \delta$

$$\begin{aligned} \mathbf {s}= \underset{\begin{matrix} d = 1, \ldots , \delta \\ i = 1,\ldots , \ell _d \end{matrix}}{\sum } s^{\delta -d}_i\epsilon ^d_i. \end{aligned}$$

If a term in $\mathbf {s}$ is a product $n \cdot t$ where t is the initial term of a syzygy $\mathbf {s}^{\prime }$ in $K^p$ or $B^p$ with $\deg (\tau ) + p \le \delta$, we replace $\mathbf {s}$ by $\mathbf {s}- \tau \cdot \mathbf {s}^{\prime }$. In this way we continue and in the end we get syzygy as in 3. This proves the identity (16) above.

Suppose now the Condition (14) is also fulfilled. Let the following relation be of Type 3.:

$$\begin{aligned} \sum _{i = 1}^{\ell _1} r^{\delta -1}_i a^1_i + \underset{\begin{matrix} d = 2,\ldots , \delta \\ i = 1, \ldots , \ell _d \end{matrix}}{\sum } r^{\delta -d}_i \epsilon ^d_i. \end{aligned}$$

Let $x_1 = {\text {in}}(a^1_1)$. Then no term of any other $a^d_i$ contains $x_1$ and also no $r^{\delta -d}_i$ contains $x_1$. But then the relation above is only possible if $r^1_1 = 0$. In this way we may continue and get all $r^1_i = 0$ except possibly if ${\text {in}}(a^1_j)$ is the constant 1 (in which case we must have i the last index $\ell _1$). But then by the reduction process using $a^1_{\ell _1}$, none of the $a^d_i$ for $d \ge 2$ contains a term of degree $<d$ and similarly no term of the $r^{\delta -d}_j$ contains a term of degree $<\delta -d$. But then in the relation

$$\begin{aligned} r^{\delta -1}_{\ell _1} \cdot 1 + \underset{\begin{matrix} d=2, \ldots , \delta \\ i = 1, \ldots , \ell _d \end{matrix}}{\sum } r^{\delta -d}_i a^d_i, \end{aligned}$$

the left side has degree $\le \delta -1$ while the right side has all terms of degree $\delta$. Hence $r^{\delta -1}_{\ell _1} = 0$. $\square$

We now present the algorithm to compute $R^{\le \delta }$ under the assumption of Conditions (15) and (14).

ALGORITHM TO COMPUTE $R^{\le \delta }$

• 1. Set $KB_{in}^{\le 1}, R_{in}^{\le 1}$ equal to 0. Let $\delta := 2$.

• 2. Let $KB_{in}^{\delta }$ consist of all pairs $(t,\delta )$ where t is the initial term of a Koszul syzygy in $K^{\delta }$ or a Boolean syzygy in $B^{\delta }$.

• 3. $KB_{in}^{\le \delta } = KB_{in}^{\le \delta -1} \cup KB_{in}^{\delta }$.

• 4. If $\delta = 2$ let $R^2 = 0$. If $\delta \ge 3$ then $R^{\delta }$ consist of all syzygies

  $$\begin{aligned} \mathbf {r}= \underset{\begin{matrix} d=2, \ldots , \delta \\ i = 1, \ldots , \ell _d \end{matrix}}{\sum } r^{\delta -d}_i\epsilon ^d_i \end{aligned}$$

  where $r^{\delta -d}_i$ has degree $\le \delta -d$ and no term of $\mathbf {r}$ is a product of monomials $\tau \cdot t$ where:

  • $(t,p) \in R_{in}^{\le \delta -1} \cup KB_{in}^{\le \delta -1}$ and $\tau$ is a monomial such that $\deg (\tau ) + p \le \delta$.

  • $\tau = 1$ and $(t,\delta ) \in KB_{in}^{\delta }$

• 5. Perform Gaussian elimination on $R^{\delta }$ and let $R_{in}^{\delta }$ consists of all pairs $(t,\delta )$ where t is the initial term of a syzygy in $R^{\delta }$.

• 6. $R_{in}^{\le \delta } = R_{in}^{\le \delta -1} \cup R_{in}^{\delta }$.

• 7. If $\delta$ is less than the stop bound then $\delta :=\delta +1$ and go to 2.

As for the actual computation of the syzygies in Step 4, this can be done by taking the $r^{\delta -d}_i$ to be linear combinations of the allowed terms (with unknown coefficients), and then solving a system of linear equations.

Proposition 27

With the algorithm above, then

$$\begin{aligned} R^{\le \delta } = \sum _{d \ge 3} S^{\le \delta -d}R^d. \end{aligned}$$

Proof

This is clear by construction. $\square$

Our applications of Theorem 26 are typically for $\delta = 1$ or 2 (This occurs for sets $F^ 2, \ldots , F^d$ where $d = 3$ or 4.). We are then interested in the syzygies $S^{\le 2}$ and $S^{\le 3}$. These are given as follows:

$$\begin{aligned} S^{\le 2}= & {} K_2 + B_2 \\ {S^{\le 3} }= & {} \langle L^1\rangle K_2 + K_3 +\langle L^1\rangle B_2 + R^3. \end{aligned}$$