Sample Complexity of the Boolean Multireference Alignment Problem

Proof of Theorem III.2

The proof consists of two main parts. The next theorem gives a formula to the error exponent and claim IV.2 makes the connection with autocorrelations.

Proof of Claim IV.2

The claim is proved by induction on n. Note that condition (15) is equivalent to (16) vanishing for m < n, which implies (17) also vanishes by applying the claim with m. In general (17) is a function of (16) and lower order terms, which vanish when we enforce condition (15). The rigorous proof follows.

Denote by x(V) the vector in ${\mathbb{Z}}_2^K(K=\mid V\mid )$ that consists of the values of x with indices in V, i.e. the j-th element of x(V) is x(V_j). Also, given s ∈ ℤ_L denote by s + V the ordered set corresponding to the sum of each element in V with s mod L. Equation (5) can then be rewritten, as

Then since Z ~ Ber(p)^L, we have

where w denotes the Hamming weight, and since S ~ ξ

In the statement of the theorem we have $x\in {\mathbb{Z}}_2^L$, however it is convenient for the proof to consider the entries of x to be −1, 1, changed by the rule: a ↦ 1 − 2a. We will call

the corresponding element of x with ±1 values, where Σ₂ := {−1, 1}, and v := 1 − 2y. In analogy to the Hamming weight, we define

With this we rewrite (19)

where μ_u(v; p) := μ_x(y; p). For simplicity of notation denote

By properties of Jacobi polynomials [11] we have

where P_m is a polynomial with the following property

where Q_m has degree at most m − 1, and Q₀ ≡ Q₁ ≡ 0. Thus

Then when m = 1

Now, by the induction hypothesis if ${\mu }_{u_1}^{(k)}\left(v;\frac{1}{2}\right)={\mu }_{u_2}^{(k)}\left(v;\frac{1}{2}\right)$ for all k ≤ n − 1, $v\in {\mathrm{\sum }}_2^K$

for all $v\in {\mathrm{\sum }}_2^K$ since Q_n has degree at most n − 1. Thus by (23) and (24)

Now splitting the square of the sum on the RHS into a product of two sums and expanding, we obtain terms of the form

where α and β are 1 or 2. By Lemma IV.3 we get

Where u_aij is u_α(s₁ + V) if a_ij ≤ n, and is u_β(s₂ + V) otherwise. So, since |a_i| is even, as A is an even partition, and the entries of u_aij are ±1,

if |a_i ∩ [n]| is odd, and it is K otherwise. Then

where R_n is a polynomial with degree n (with coefficients possibly depending on K and n), and R₁(b) = 2^kb. It cannot have degree n + 1 since |A| ≤ n, since it is an even partition of [2n]. For it to be a power of order n, we need |A| = n, so |a_i| = 2 for i = 1, …, n, thus C_A = 1, by the Lemma. Also |a_i ∩ [n]| must be odd for all i, thus |a_i ∩ [n]| = 1. There are exactly n! partitions with this property, so the leading coefficient of R_n is 2^K n!. We also have

The equation will be true for n = 1, since R₁(b) = 2^kb. As in (25), the induction hypothesis implies the lower order terms in R_n cancel and only the leading coefficient is of interest. We get

Now through some algebraic manipulation, and using again the argument of the leading coefficient, if |k| = n, then

This together with (25) and (29) concludes the proof.

Proof

Recall (21). We have $W(u\oplus v)={\displaystyle \sum _{k=1}^K}u(k)v(k)$

The last sum is 2^K when A is even, and 0 otherwise. Using a combinatorial argument we can rewrite (32) without the ‘all-distinct’ condition, at the cost of a constant C_A, which is 1 when |a_i| = 2 for i ∈ {1, …, |A|}. We get

Proof of Corollary III.3

We first prove equation (12). Recall (6), and denote by

and

Note that B_m(x₁, x₂) = 0 if m < t_ξ,V (x₁, x₂) by (7). For convenience let B(x₁, x₂) := B_tξ,V(x1,x2)(x₁, x₂) and B(L) := Bt_ξ,V(𝒳)(L). Using this notation we rewrite (10) and (11)

Now equation (12) is equivalent to having t_ξ,V(𝒳) ≥ 3 and B₃(L) either $\frac{12}{L}$ or 0. Turns out, for L ≥ 6, if we take

then $t_{\xi ,V}(\mathcal{X})\ge t_{\xi ,V}(x_1^{\ast },x_2^{\ast })=3$ and $B_3(L)\le B(x_1^{\ast },x_2^{\ast })=\frac{12}{L}$. Also we cannot have $\frac{12}{L}>B_3(L)>0$. This implies there exists x₁ and x₂ in 𝒳 such that $\frac{12}{L}>B(x_1,x_2)>0$. Since it is positive, there is ${\mathbf{k}}^{\ast }\in {\mathbb{Z}}_L^3$ such that A_k^*(x₁) ≠ A_k^*(x₂). But by definition (6), since $\xi (s)=\frac{1}{L}$, LA_k^*(x) is an integer for $x\in {\mathbb{Z}}_2^L$, and L²(A_k^*(x₁) − A_k^*(x₂))² ∈ ℤ.

Now by the definition we also have A_σ(k^*)(x) = A_k^*(x), where σ permutes the entries of k^*. Also, for s ∈ ℤ_L, let $s+{\mathbf{k}}^{\ast }:=(s+k_1^{\ast },s+k_2^{\ast },s+k_3^{\ast })$, then A_s+k^*(x) = A_k^*(x). There is 6 permutations and L possible values for s ∈ ℤ_L, so B(x₁, x₂) is an integer multiple of $\frac{6}{L}$. (we can also have not trivial s and σ such that s + k^* = σ(k^*) but that case also has the property mentioned). However we cannot have $B(x_1,x_2)=\frac{6}{L}$. That means there exists only one ${\mathbf{k}}^{\ast }\in {\mathbb{Z}}_L^3$ (with permutations and shifts) such that A_k^*(x₁) ≠ A_k^* (x₂). Then

On the other hand

where A₀ denotes k-autocorrelation with k = 0. Since t_L > 1, A₀(x₁) = A₀(x₂), so equation (33) must be 0, and equation (12) follows by contradiction. Now if L ≥ 12 is even, choose

and $x_2^{\ast }$ the vector obtained by reversing the entries of $x_1^{\ast }$. Since one is the reverse of the other, they have same 1 and 2 order autocorrelations. Recall (20) and (6) and notice that in this case both A_k(u₁) and A_k(u₂) are 0 when |k| is odd, since half of the signal is the symmetric of the other half, i.e. $u_1(\{1,\dots ,\frac{L}{2}\})=-u_1(\{\frac{L}{2}+1,\dots ,L\})$. Now because of (30) we have A_k(x₁) = A_k(x₂) when |k| = 3, so t_L ≥ 4, and B₃(L) = 0.

Finally, let L ≥ 6 be prime. We prove by contradiction that t_L = 3 and $B_3(L)=\frac{12}{L}$. If this is not true, then it exists $x_1^{\ast }$ and $x_2^{\ast }$ such that $t_{x_1^{\ast },x_2^{\ast }}>3$, so

By Theorem 2 of paper [8], if the Fourier coefficients of $x_1^{\ast }$ and $x_2^{\ast }$ are non-zero, then equation (34) implies one is a shift of the other. Denote by ${\{r_j^1\}}_{j\in {\mathbb{Z}}_L}$ and ${\{r_j^2\}}_{j\in {\mathbb{Z}}_L}$ the Fourier coefficients of $x_1^{\ast }$ and $x_2^{\ast }$, respectively, which are given by

where ω_L is the L’th root of unity. $r_0^{\alpha }=0$ implies $x_{\alpha }^{\ast }$ only has zeros, and $r_j^{\alpha }$ is 0 only if $w_L^{-j}$ is a root of the polynomial

However, since L is prime, the minimal polynomial of $w_L^{-j}$ in ℚ[x], for L > j > 0, is 1 + x + ··· + x^L−1 [12], so this polynomial must divide (37). Thus $x_1^{\ast }$ and $x_2^{\ast }$ must be the all zeros and all ones signals, but these signals also do not satisfy (34).