Fourier Sparsity of GF(2) Polynomials

Abstract

We study a conjecture called “linear rank conjecture” recently raised in (Tsang et al. [16]), which asserts that if many linear constraints are required to lower the degree of a GF(2) polynomial, then the Fourier sparsity (i.e. number of non-zero Fourier coefficients) of the polynomial must be large. We notice that the conjecture implies a surprising phenomenon that, if the highest degree monomials of a GF(2) polynomial satisfy a certain condition (Specifically, the highest degree monomials do not vanish under a small number of linear restrictions.), then the Fourier sparsity of the polynomial is large regardless of the monomials of lower degrees—whose number is generally much larger than that of the highest degree monomials. We develop a new technique for proving lower bound on the Fourier sparsity of GF(2) polynomials, and apply it to certain special classes of polynomials to showcase the above phenomenon (A full version of this paper is available at http://arxiv.org/abs/1508.02158).

N. Xie—Research supported in part by NSF grant 1423034.

S. Zhang—Research supported in part by RGC of Hong Kong (Project no. CUHK419413).

A Linear Rank of Complete d-uniform Maxonomials

In this section we compute the exact value of the linear rank of a degree d polynomial whose set of maxonomials consists of all $\left( {\begin{array}{c}n\\ d\end{array}}\right)$ degree-d monomials.

Define $\mathcal {C}_{d,n}(x) = \sum _{I \subseteq [n], |I| = d} \prod _{i\in I}x_i$, the summation of all degree-d monomials over variables $x_1, \ldots , x_n\in {\mathbb {F}_2}$. The subscript n is dropped when it is clear from the context. We use the equivalence relation $\equiv _d$ for polynomials with the same maxonomials, i.e. $p \equiv _dq$ if both p and q have ${\mathbb {F}_2}$-degree d and $p+q$ has ${\mathbb {F}_2}$-degree strictly less than d. It is clear that if $p \equiv _dq$, then $\mathsf{{lin\text {-}rank}}(p) = \mathsf{{lin\text {-}rank}}(q)$.

Theorem 6

Let $n\ge d\ge 0$ be integers. Then the following hold:

• 1. If d is odd, then $\mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,n}) = 1$.

• 2. If d is even, then $\mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,n}) = \lfloor \frac{n}{2} \rfloor -\frac{d}{2}+1$, i.e.

  $$\begin{aligned} \mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,n}) = {\left\{ \begin{array}{ll} \frac{n-d}{2}+1 &{} \text {if }n\text { is even},\\ \frac{n-d-1}{2}+1 &{} \text {if }n\text { is odd}. \end{array}\right. } \end{aligned}$$

Proof

The first item follows simply by the factorization $\mathcal {C}_{d,n} \equiv _d\mathcal {C}_{1,n}\mathcal {C}_{d-1,n}$. Indeed, when we multiply $\mathcal {C}_{1,n} = \sum _{i\in [n]} x_i$ and $\mathcal {C}_{d-1,n} = \sum _{|I| = d-1} x_I$, for $i\notin I$, $x_i x_I = x_{I\cup \{i\}}$, and each J with $|J| = d$ comes from d many (i, I). For each $i\in I$, $x_i x_I = x_I$, and each resulting $x_I$ with $|I| = d-1$ comes from $d-1$ many $i \in I$. Thus

$$\begin{aligned} \mathcal {C}_{1,n} \mathcal {C}_{d-1,n}&= d \Big (\sum _{|J| = d} x_J \Big ) + (d-1)\Big (\sum _{|I| = d-1} x_I \Big ) = d\mathcal {C}_{d,n}+(d-1)\mathcal {C}_{d-1,n} \\&= \mathcal {C}_{d,n}, \end{aligned}$$

for all odd d.

Now we consider the second item in the statement and assume from now on that d is even and $d \le n$. The second item follows from the following two claims.

Claim 1

If $\mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,n+1}) = \mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,n})$, then $\mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,n+2}) > \mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,n+1})$.

Claim 2

$\mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,n+2}) \le \mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,n})+1$.

Let us first show Theorem 6 assuming these two claims. We prove by induction on the number of variables that for all $k \ge d/2$,

$$\begin{aligned} \mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,2k}) = \mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,2k+1}) = k-\frac{d}{2}+1. \end{aligned}$$

(8)

which is just a restatement of the second item of Theorem 6.

Base Case $k = d/2$. We have

$$\begin{aligned} \mathcal {C}_{d}(x_1, \ldots , x_{2k}) = \mathcal {C}_{d}(x_1, \ldots , x_{d}) = \mathcal {C}_{d-1}(x_1, \ldots , x_{d-1})\cdot x_{d}, \end{aligned}$$

(9)

so $\mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,2k}) = 1$. For $n=2k+1$, note that

$$\begin{aligned} \mathcal {C}_{d}(x_1, \ldots , x_{2k+1})&= \mathcal {C}_{d}(x_1, \ldots , x_{d+1}) \nonumber \\&= \mathcal {C}_{d-1}(x_1, \ldots , x_{d-1})(x_{d}+x_{d+1})+\mathcal {C}_{d-2}(x_1, \ldots , x_{d-1})x_{d}x_{d+1}, \end{aligned}$$

(10)

Putting restriction $x_d=x_{d+1}$ makes the first summand vanish and decreases the degree of the second summand, hence $\mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,2k+1}) = 1$.

General k. Now we assume that (8) holds for k and will prove the case for $k+1$. The following sequence of inequalities hold.

$$\begin{aligned} k-\frac{d}{2}+1 < \mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,2(k+1)}) \le \mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,2(k+1)+1}) \le k-\frac{d}{2}+2, \end{aligned}$$

where the first inequality follows by Claim 1; the second follows by the facts that $\mathcal {C}_{d,n-1}$ can be obtained from $\mathcal {C}_{d,n}$ by restricting $x_n = 0$ and restriction does not increase $\mathsf{{lin\text {-}rank}}$; and the last inequality follows by Claim 2. Therefore (8) also holds for $k+1$.    $\square$

Now it remains to prove the two claims. We start with Claim 2, which is simpler.

Proof

(Proof of Claim 2 ). We first observe the following identity:

$$\begin{aligned} \mathcal {C}_{d}(x_1, \dots , x_{n+2})&= \mathcal {C}_{d}(x_1, \dots , x_n) + \mathcal {C}_{d-1}(x_1, \dots , x_n)(x_{n+1} + x_{n+2}) \nonumber \\&\quad + \mathcal {C}_{d-2}(x_1, \dots , x_n) x_{n+1} x_{n+2} \nonumber \\&\equiv _d\mathcal {C}_{d}(x_1, \dots , x_n) + \mathcal {C}_{d-1}(x_1, \dots , x_n, x_{n+1})(x_{n+1} + x_{n+2}). \end{aligned}$$

(11)

Therefore the restriction $x_{n+2} = x_{n+1}$ reduces $\mathcal {C}_{d}(x_1, \ldots , x_{n+2})$ to

$$\begin{aligned} \mathcal {C}_{d}(x_1, \ldots , x_{n+2})|_{x_{n+1}=x_{n+2}} \equiv _d\mathcal {C}_{d}(x_1, \dots , x_n). \end{aligned}$$

Since each restriction can reduce $\mathsf{{lin\text {-}rank}}$ by at most 1, we have

$$\begin{aligned} \mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,n+2})-1 \le \mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,n+2}|_{x_{n+2} = x_{n+1}}) = \mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,n}), \end{aligned}$$

as desired.    $\square$

Proof

(Proof of Claim 1 ). For the sake of contradiction, assume that

$$\begin{aligned} \mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,n+2}) = \mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,n+1}) = \mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,n}) = r. \end{aligned}$$

Fix an optimal set of linear restrictions for $\mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,n+2})$. Without loss of generality, we can assume it contains a restriction of the form $x_{n+2} = \ell (x_1, \dots , x_{n+1}) = \ell (x)$ for some linear form $\ell$. It is clear that such restriction will reduce the $\mathsf{{lin\text {-}rank}}$ by exactly 1. So we have

$$\begin{aligned} \mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,n+2}|_{x_{n+2} = \ell (x)}) \le \mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,n+2})-1 = r-1. \end{aligned}$$

(12)

But by the expansion

$$\begin{aligned} \mathcal {C}_{d}(x_1, \dots , x_{m+1}) = \mathcal {C}_d(x_1, \dots , x_m) + \mathcal {C}_{d-1}(x_1, \dots , x_m)x_{m+1}, \end{aligned}$$

we have

$$\begin{aligned} \mathcal {C}_{d}(x_1, \dots , x_{n+2})|_{x_{n+2} = \ell (x)}&= \mathcal {C}_d(x_1, \dots , x_{n+1}) + \mathcal {C}_{d-1}(x_1, \dots , x_{n+1})\ell (x) \nonumber \\&= \mathcal {C}_d(x_1, \dots , x_n) + \mathcal {C}_{d-1}(x_1, \dots , x_n)x_{n+1} \nonumber \\&\quad + \mathcal {C}_{d-1}(x_1, \dots , x_{n+1})\ell (x). \end{aligned}$$

(13)

Now, consider to further restrict $x_{n+1} = x_1 + x_2 + \dots + x_n = \mathcal {C}_{1}(x_1, \dots , x_n)$. By the fact that $\mathcal {C}_{d-1}(x_1, \dots , x_m) \equiv _d\mathcal {C}_{d-2}(x_1, \dots , x_m)\mathcal {C}_{1}(x_1, \dots , x_m)$ for every even $d \ge 4$, the second term on the right of (13) is $\equiv _d$-equivalent to

$$\begin{aligned}&\mathcal {C}_{d-2}(x_1, \dots , x_n)\mathcal {C}_1(x_1, \dots , x_n)x_{n+1}|_{x_{n+1} = \mathcal {C}_1(x_1, \dots , x_n)} \\ =&\mathcal {C}_{d-2}(x_1, \dots , x_n)\mathcal {C}_1^2(x_1, \dots , x_n) \\ =&\mathcal {C}_{d-2}(x_1, \dots , x_n)\mathcal {C}_1(x_1, \dots , x_n) \equiv _d0, \end{aligned}$$

and the last term becomes

$$\begin{aligned} \mathcal {C}_{d-2}(x_1, \dots , x_{n+1})\mathcal {C}_1(x_1, \dots , x_{n+1}) \ell (x)|_{x_{n+1} = \mathcal {C}_1(x_1, \dots , x_n)} = 0. \end{aligned}$$

Plugging these two back to (13),

$$\begin{aligned} \mathcal {C}_{d,n+2}|_{x_{n+2} = \ell (x), x_{n+1} = x_1 + \dots + x_{n}} \equiv _d\mathcal {C}_{d,n}. \end{aligned}$$

As restriction does not increase linear rank, we have from (12) that

$$\begin{aligned} r = \mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,n})&= \mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,n+2}|_{x_{n+2} = \ell (x), x_{n+1} = x_1 + \dots + x_{n}})\\&\le \mathsf{{lin\text {-}rank}}(\mathcal {C}_{d,n+2}|_{x_{n+2} = \ell (x)})\\&\le r-1, \end{aligned}$$

which is a contradiction.    $\square$

As a simple application of Theorem 6, for any symmetric function f, let $r_1$, $r_0$ be the largest and smallest integers such that f(x) is constant or parity on $\{x\in {\{0, 1\}^{n}} : r_0 \le |x| \le n-r_1\}$. The quantity $r := r_0 + r_1$ turns out to be an important complexity measure for symmetric functions. For example, the randomized and quantum communication complexity of symmetric XOR functions is characterized by this r [9, 10, 18], and $\log \Vert \hat{f} \Vert _1 = \varTheta (r\log (n/r))$ for all symmetric functions f [1].

Here we relate this measure to the ${\mathbb {F}_2}$-degree of f. It is clear that we can fix $x_1 = x_2 = \dots = x_{r_0} = 1$ and $x_n = x_{n-1} = \dots = x_{n - r_1 + 1} = 0$ to reduce the degree of f to at most 1. We therefore have the following corollary.

Corollary 1

Let f be a symmetric function with even ${\mathbb {F}_2}$-degree d, then

• 1. $\lfloor \frac{n}{2} \rfloor -\frac{d}{2}+1 \le r_0 + r_1$.

• 2. $\log \Vert \hat{f} \Vert _1 = \varOmega (n/\log n)$, if $d = (1-\varOmega (1))n$.