Sample Complexity of the Boolean Multireference Alignment Problem

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by  Emmanuel Abbe

Sample Complexity of the Boolean Multireference
Alignment Problem
Emmanuel Abbe Joa˜o Pereira Amit Singer
Princeton University Princeton University Princeton University
[email protected] [email protected] [email protected]
Abstract—TheBooleanmultireferencealignmentproblemcon- complexity of multireference alignment in the real-valued
7
1 sists in recovering a Boolean signal from multiple shifted and Gaussian noise case, which is a topic of ongoing research
noisy observations. In this paper we obtain an expression for
0 [6], [7].
the error exponent of the maximum A posteriori decoder. This
2 In BMA the search space is finite, and the maximum A
expression is used to characterize the number of measurements

b needed for signal recovery in the low SNR regime, in terms of posteriori decoder (MAP) minimizes the probability of error.
e higherorderautocorrelationsof thesignal.Thecharacterization Our main contributionis an expressionfor the error exponent
F isexplicit forvarious signal dimensions,such asprimeand even of MAP, in the low SNR regime,given in TheoremsIII.2and
dimensions.
2 III.3. Our results imply how many measurements are needed,
as a function of the SNR, in order to accurately estimate the
] I. INTRODUCTION
signal.
T
The Boolean multireference alignment (BMA) problem
I The expression depends on the autocorrelations of the
s. consists of estimating an unknownsignal x∈ZL2, from noisy signal, defined in (6). Our results connect the order of au-
c cyclically shifted copies Y1,...,YN ∈ZL2, i.e., tocorrelations needed to reconstruct the signal to the number
[
Yi =RSix Zi, i 1,...,N , (1) ofmeasurementsneededtoestimatethesignal.Thishassome
2 ⊕ ∈{ } connections with previous theoretical work on uniqueness of
v where the error Zi Ber(p)L, the product measure of L the bispectrum [8].
0 Bernoullivariablesw∼ith parameterp, denotesadditionmod We also consider some generalizations of the original
4 ⊕
2, R is the index cyclic shift operator that shifts a vector one problem in order to model some aspects of multireference
5
7 elementtotheright(x1,...,xN) (xN,x1,...,xN−1),RSi alignment that arise in applications, such as the introduction
7→
0 correspondstoapplyingSi timestheoperatorRandtheshifts of deletions.
. S (Z ), the uniform distribution in Z .
1 i ∼U L L
0 The motivation to study this problem comes from the II. BMA PROBLEM
7 classical multireference alignment problem, where the signal
In the BMA problem, the errors are i.i.d. Bernoulli of
1 andobservationsarerealvaluedvectors,andtheerrorisGaus-
parameter p. If p = 1, then the observations Y Ber(1)L,
v: sian white noise. Several algorithms were recently proposed 2 i ∼ 2
regardlessoftheoriginalsignal,andsignalrecoveryisimpos-
i to solve the problem, including angular synchronization [1],
X sible. This corresponds to the case when SNR = 0. On the
semidefinite program relaxations of the maximum likelihood
r otherhand,p=0or1correspondstothenoiselesscase.Thus
a decoder [2] and reconstruction using the bispectrum [3]. This we define
problem is also an instance of a larger class of problems, 2
1
calledNon-UniqueGames,whichalsoincludestheorientation SNR:= p . (2)
− 2
estimation problem in cryo-electron microscopy [4]. (cid:18) (cid:19)
Despitetheseadvancementsinalgorithmicdevelopment,not In contrast to proposing an algorithm to solve the BMA
muchprogresshasbeenmadeinunderstandingthefundamen- problem, our paper focuses on its sample complexity, in the
tal limits of signal recovery.The recentpaper [5] investigated regime when p 1 and SNR 0.
→ 2 →
fundamental limits of shift recovery in multireference align- Note that the observations Y , i [N], given the signal
i
∈
ment,butnotthoseofsignalrecovery.Wenotethatestimating x, are i.i.d., since both the shifts S and the errors Z are
i i
the shifts is impossible at low signal-to-noise ratio (SNR) i.i.d. For that reason we will dropthe index i when it is more
even if an oracle presents us with the true signal. Also, the convenient.We rewrite (1), denotingbyx(j) the j-th entryof
goal of many applications is signal recovery rather than shift x.
estimation. Our paper aims to fill the gap on signal recovery, Y(j)=x(S+j) Z(j), j Z , (3)
L
by studying the Boolean case. We show here that signal ⊕ ∈
recoveryispossibleatarbitrarilylowSNR,ifsufficientlymany where ’+’ is addition mod L.
measurements are available, and quantify this tradeoff. We Our paper also considers the sample complexity of the
do not consider here the problem of determining the sample following variations of the basic BMA problem:
• BMAProblemwithconsecutivedeletions:Inthiscasethe probability of error (9). We obtain results that do not depend
measurements Y ,...,Y are in ZK, with K L, and on the prior distribution, they depend only on its support.
1 N 2 ≤
Y(j)=x(S+j) Z(j), j Z . (4) Theorem III.2. Consider the BMA problem with known
K
⊕ ∈ deletions Z V and shift distribution ξ. Let ZL be the
When K =L we obtain the original BMA problem. L\ X ⊂ 2
support of the prior distribution of the signals and µ the
x
• BMA Problem with known deletions: Let V ZL be an conditional distribution in ZK of the observations Y given
⊂ 2
ordered set of non-deletions, i.e. the set of deletions is
the signal x, where K = V . The probability of error of the
Z V. Now the measurements Y ,...,Y are in ZK, | |
L\ 1 N 2 MAP estimator, denoted by Pe, has the following asymptotic
with K = V , and:
behavior
| |
Y(j)=x(S+V ) Z(j), j Z , (5) 1
where Vj denotesthe j-thjele⊕mentof∀V.∈WhKen V =[K] Nl→im∞N logPe =xx1m,1x6=2inx∈2XC(µx1,µx2), (10)
werecovertheBMAproblemwithconsecutivedeletions.
with
• BMA Problem (and variations) with non uniform rota-
tions:Similartothepreviousproblems,butnowtheshifts
C(µ ,µ )=
follow some distribution ξ in ZL. x1 x2
24t−3 2
muTlthieresfeervenarcieatiaolingsnmareent.mTohtievacteadseboyf pproosbsilbelmesdesliemtiiolanrs tios t! SNRt Aξ,k(x1)−Aξ,k(x2) +O(SNRt+1),
kX∈Vt(cid:16) (cid:17)
intended to model instances where the observations are only (11)
partial, whereas the extension to non-uniform shifts attempts
to represent a non-symmetric version of the problem. and t=tξ,V(x1,x2).
The theorem implies that the exponenton SNR is t ( ).
III. RESULTS ξ,V X
In the original problem, with uniform shifts and no deletions,
We start by introducing the following notion of autocorre-
the recovery of the original signal is possible only up to a
lation of a signal that is central to our main results. shift, i.e. we can only recover Rkx, where x is the original
DefinitionIII.1. The(ξ,k)-autocorrelationofx, withrespect signal,andk issomeshiftinZL.Forthatreason,weconsider
to a distribution ξ in Z and k = (k ,k ,...,k ) Zd is to have exactly one element of each class of all the shifts
defined as L 1 2 d ∈ L Xof a signal, i.e., there are no two elements in where one is
X
a shift of the other (for example, if L is prime, then there are
L 2L 2 such elements).
Aξ,k(x):= ξ(s)x(k1+s) x(kd+s). (6) −
···
s=1 CorollaryIII.3. Considertheoriginalproblem,withV =[L],
X
ξ (Z ) and as defined above. By inspection one can
We refertod= k astheorderoftheauto-correlation.When L
∼ U X
ξ (ZL),wes|im|plywritek-autocorrelationandAk.Notice obtain the error exponent for L ≤ 5. For L ≥ 6, we either
Ak∼iUs shift invariant, that is Ak(x) = Ak(Rsx), and in this have
case we may assume k1 =0. 1 210SNR3+O(SNR4)
We define the minimum autocorrelation order necessary to lim logP = L (12)
e
distinguish x and x under ξ and V as N→∞N ( O(SNR4)
1 2
tξ,V(x1,x2):=inf d:Aξ,k(x1)=Aξ,k(x2),k Vd , (7) Also, the first case occurs when L is prime, and the second
{ 6 ∈ }
whenL 12andiseven.TheothervaluesofLremainopen.
where Vd denotes the vectors in Zd with entries in V. ≥
2
The minimum autocorrelation order necessary to describe all
IV. PROOF TECHNIQUES
signals in is defined as
X Proof of Theorem III.2: The proof consists of two main
tξ,V( ):= max tξ,V(x1,x2). (8) parts. The nexttheorem givesa formulato the error exponent
X xx1,1x6=2x∈2X and claim IV.2 makes the connection with autocorrelations.
Given a prior distribution on the signals PX, with support Theorem IV.1. Consider the BMA problem with known dele-
GXi,vednenaontealgboyriXthmthfeorrBanMdoAmthveapriraobblaebiwliittyhodfiesrtrroibruitsiodnefiPnXed. tsipoancseZofLp\oVssiabnledsisghnifatlsdaisntrdibµuti:o=nPξ. Le(t xX)t⊂hecZo2Lndbiteionthael
x Y|X
as distribution in ZK of the observations g·i|ven the signal x.
2
P(Xˆ 6=X)= P(Xˆ 6=xi)PX(xi), (9) The probability of error of the MAP estimator (Pe) has the
xXi∈X following asymptotic behavior
where Xˆ is the answer given by the algorithm. In the BMA 1
lim logP = min C(µ ,µ ), (13)
problem the search space is finite, thus MAP minimizes the N→∞N e x16=x2∈X x1 x2
with the correspondingelementof x with 1 values, where Σ :=
2
±
1,1 ,andv :=1 2y.InanalogytotheHammingweight,
C(µx1,µx2)= w{−e defi}ne −
2
12 −p 2s µ(xs1) y;12 −µ(xs2) y;21 W(u):= L u(s)=L 2w(x). (20)
(cid:0) 8(s!)(cid:1)2 yX∈ZK2 (cid:16) (cid:0) µx(cid:1)1 y;21 (cid:0) (cid:1)(cid:17) Xs=1 −
With this we rewrite (18)
(cid:0) 1(cid:1) 2s+2
+O p , (14)
L
(cid:18)2 − (cid:19) µu(v;p)= ξ(s)(1 p)K2+W(v⊕u2(s+V))pK2−W(v⊕u2(s+V)),
whereµx(m)(y;p)denotesthem-thderivativeofµx(y;p)inp, Xs=1 − (21)
i.e. the derivative of the conditional distribution in y given x
where µ (v;p):=µ (y;p). For simplicity of notation denote
in order of the Bernoulli parameter p, and u x
W :=W(v u(s+V)).
1 1 v,u,s
s(x ,x ):=inf m:µ(m) y; =µ(m) y; ,y ZK . ⊕
1 2 x1 2 6 x2 2 ∈ 2 The claim is now proved by induction on n. By properties of
(cid:26) (cid:18) (cid:19) (cid:18) (cid:19) (cid:27)
Jacobi polynomials [11] we have
This theorem follows from Theorems 1 and 2 in [9].
Theorem1is a corollaryofSanovTheorem[10], whichleads (m)
to (13). However the expression obtained by Theorem 1 is pK2−2b(1−p)K2+2b p=1 =(−2)m−KPm(b),
rather complex and not very interpretable. In Theorem 2 [9] (cid:16) (cid:17)| 2
where P is a polynomial with the following property
we Taylor expand (13) and obtain a useful characterizationin m
instances where the SNR is small. We use this expression to P (b)=bm+Q (b), (22)
m m
obtain (14).
whereQ hasdegreeatmostm 1,andQ Q 0.Thus
Claim IV.2. If µ(m) y;1 = µ(m) y;1 for all m< n and m − 0 ≡ 1 ≡
x1 2 x2 2 L
y ZK, then the following expressions are equal: 1
∈ 2 (cid:0) (cid:1) (cid:0) (cid:1) µ(um) v;2 =(−2)m−K ξ(s)Pm(Wv,u,s). (23)
µ(n) y;1 µ(n) y;1 2 (cid:18) (cid:19) Xs=1
x1 2 − x2 2 (15) Then when m=1
yX∈ZK2 (cid:16) (cid:0) µx(cid:1)1 y;12 (cid:0) (cid:1)(cid:17) µ(1) v;1 µ(1) v; 1 2
and (cid:0) (cid:1) u1 2 − u2 2
24nn! Aξ,k(x1)−Aξ,k(x2) 2. (16) vX∈ΣK2 (cid:16) (cid:0) µu(cid:1)1 v;12 (cid:0) (cid:1)(cid:17)
kX∈VL(cid:16) (cid:17) (cid:0) (cid:1)L 2
squInarfeasc,tt,hseinccleaitmheimexpplrieessstihoantst(ξ1,V5)(xa1n,dx(21)6=)asr(exb1,oxth2)s,uwmhoaft =22−KvX∈ΣK2 "Xs=1ξ(s)(Wv,u1,s−Wv,u2,s)# .
concludes the proof of theorem III.2.
Proof of Claim IV.2: Denote by x(V) the vector in ZK2 fNoorwa,llbky thenind1u,ctvionΣhyKpothesis if µ(uk1) v;21 =µ(uk2) v;21
(K = V ) that consists of the values of x with indices in V, ≤ − ∈ 2 (cid:0) (cid:1) (cid:0) (cid:1)
| |
i.e. the j-th element of x(V) is x(Vj). Also, given s ZL L L
∈
denote by s+V the ordered set corresponding to the sum of ξ(s)Qn(Wv,u1,s)= ξ(s)Qn(Wv,u2,s),
each element in V with s mod L. Equation (5) can then be s=1 s=1
X X
rewritten, as for all v ΣK since Q has degree at most n 1. Thus by
Y =x(S+V) Z (17) (22) and∈(23)2 n −
⊕
Then since Z Ber(p)L, we have 2
∼ µ(n) v;1 µ(n) v;1
µ (y;pS =s)=(1 p)K−w(y⊕x(s+V))pw(y⊕x(s+V)), u1 2 − u2 2 =
wherxe w d|enotes the Ha−mming weight, and since S ξ vX∈ΣK2 (cid:16) (cid:0) µu(cid:1)1 v;21 (cid:0) (cid:1)(cid:17)
∼ (cid:0) L (cid:1) 2
L 22n−K ξ(s) Wn Wn (24)
µx(y;p)= ξ(s)(1 p)K−w(y⊕x(s+V))pw(y⊕x(s+V)). " v,u1,s− v,u2,s #
s=1 − vX∈ΣK2 Xs=1 (cid:0) (cid:1)
X (18)
NowsplittingthesquareofthesumontheRHSintoaproduct
In the statement of the theorem we have x ZL, however it
∈ 2 of two sums and expanding, we obtain terms of the form
is convenient for the proof to consider the entries of x to be
L L
1,1, changed by the rule: a 1 2a. We will call
− 7→ − ξ(s )ξ(s )( 1)α+β Wn Wn , (25)
u:=1−2x∈ΣL2 (19) sX1=1sX2=1 1 2 − vX∈ΣK2 v,uα,s1 v,uβ,s2
where α and β are 1 or 2. By Lemma IV.3 we get This together with (24) and (28) concludes the proof.
Wn Wn = Lemma IV.3. ForanypartitionA= a1,...,a|A| ofthe set
v,uα,s1 v,uβ,s2 1,2,...,m ,denotebya thej-the{ntryofa an}dM the
vX∈ΣK2 s{et of all suc}h partitions. Iifju ,...,u ΣK i [m]
|A| K |ai| 1 m ∈ 2
2K CA  uaij(k), (26) W(u1 v) W(um v)=
⊕ ··· ⊕
AA∈XiMse[v2enn] Yi=1Xk=1jY=1  vX∈ΣK2
Where u is u (s +V) if a n, and is u (s +V) |A| K |ai|
aij α 1 ij ≤ β 2 2K C u (k) , (30)
otherwise. So, since ai is even, as A is an even partition, A  aij 
and the entries of uai|j a|re ±1, AA∈XisMe[vmen] iY=1kX=1jY=1 
K |ai| where A is even if all ai are even for i 1,..., A .
uaij(k)= uα(s1+k)uβ(s2+k) Moreover,CA isa consta|nt|thatdependsonly∈on{thepart|iti|o}n
k=1j=1 k∈V A and is always 1 if a =2 for all i 1,..., A .
XY X i
| | ∈{ | |}
if ai [n] is odd, and it is K otherwise. Then K
| ∩ | Proof: Recall (20). We have W(u v)= u(k)v(k)
Wn Wn = ⊕ k=1
v,uα,s1 v,uβ,s2 P
vX∈ΣK2 W(u1 v) W(um v)
⊕ ··· ⊕
R u (s +k)u (s +k) , vX∈ΣK2
n α 1 β 2 K K
!
kX∈V = u1(k1) um(km) v(k1) v(km)
··· ··· ···
where R is a polynomial with degree n (with coefficients
possibly ndependingon K and n), and R (b)=2kb. It cannot kX1=1 kXm=1 vX∈ΣK2
1 K |A| |ai| |A|
ohfav[e2nd]e.gFroeer int +to1besiancpeow|Ae|r≤ofno,rdsienrcne,itwiesnaeneedveAn p=artnit,ioson = uaij(ki) v(ki)|ai|,
ai = 2 for i=1,...,n, thus CA =1, by the Le|mm| a. Also A∈XM[m]k1a,l.l.X.d,kis|tAin|c=t 1iY=1jY=1 vX∈ΣK2 Yi=1
|a | [n] must be odd for all i, thus a [n] = 1. There (31)
i i
| ∩ | | ∩ |
are exactly n! partitions with this property, so the leading The last sum is 2K when A is even,and 0 otherwise. Using a
coefficient of R is 2Kn!. We also have
n combinatorial argument we can rewrite (31) without the ’all-
L L n distinct’ condition, at the cost of a constant C , which is 1
A
ξ(s1)ξ(s2) uα(s1+k)uβ(s2+k) when ai =2 for i 1,..., A . We get
! | | ∈{ | |}
sX1=1sX2=1 kX∈V
L L n K |A| |ai|
= ξ(s1)ξ(s2) uα(s1+ki)uβ(s2+ki) 2K CA uaij(ki)=
sX1=1sX2=1 kX∈VnYi=1 A∈XM[m] k1,..X.,k|A|=1iY=1jY=1
Aiseven
= Aξ,k(uα)Aξ,k(uβ), (27) |A| K |ai|
kX∈Vn =2K C u (k)
Mimicingtheargumentused in(24), theequationwill betrue A  aij 
forn=1,sinceR1(b)=2kb,andbytheinductionhypothesis AA∈XisMe[vmen] Yi=1Xk=1jY=1 
onlytheleadingcoefficientofR isofinterest,sincetheother
n
terms will cancel with each other. We get Proof of Corollary III.3: We first prove equation (12).
Recall (6), and denote by
L 2
ξ(s) Wn Wn = 2
" v,u1,s− v,u2,s # Bm(x1,x2):= Ak(x1)−Ak(x2)
vX∈ΣK2 Xs=1 (cid:0) (cid:1) kX∈ZmL (cid:16) (cid:17)
2kn! (Aξ,k(u1) Aξ,k(u2))2 (28) and
−
k∈Vn B (L):= min B (x ,x )
X m m 1 2
Now through some algebraic manipulation, and using again x16=x2∈X
the argument of the leading coefficient, if k =n, then Note that Bm(x1,x2) = 0 if m < tξ,V(x1,x2) by (7).
| |
For convenience let B(x ,x ) := B (x ,x ) and
(Aξ,k(u1)−Aξ,k(u2))2 = B(L) := Btξ,V(X)(L) . U1sin2g this nottξa,tVio(xn1,wx2e) re1writ2e (10)
k∈Vn and (11)
X
22n (Aξ,k(x1) Aξ,k(x2))2 (29) 1 24tL−3
− lim logP =B(L) SNRtL +O SNRtL+1
kX∈Vn N→∞N e tL!
(cid:0) (cid:1)
Nowequation(12)isequivalenttohavingt ( ) 3and coefficients of x∗ and x∗, respectively, which are given by
ξ,V X ≥ 1 2
B (L) either 12 or 0. Turns out, for L 6, if we take
3 L ≥ 1 L
rα = x (s)ω−js, α 1,2 ,j Z , (34)
x∗ =(1,1,0,1,0,...,0) and x∗ =(1,0,1,1,0,...,0), j √L α L ∈{ } ∈ L
1 2 s=1
X
L−4zeros L−4zeros = 1 ω−js, (35)
√L L
then tξ,V(X) ≥ tξ,|V(x{z∗1,x}∗2) = 3 and B3(L) ≤ B|(x{z∗1,x}∗2) = s:xαX(s)=1
ei1Lsx2i.psotAsslixstoi1vewa,netdhcexarn2enioinstXhka∗vsue∈c1hLZ23Lth>astuBc1L3h2(L>th)aB>t (A0xk.1∗T,(xhx2i1s))i>m6=p0Al.ieSksi∗n(tchxee2r)iet. hwahsezreerωosL, aisndthrejαLis’t0h oronolyt oiffwuL−nijtyi.sra0αro=ot0ofimthpelipesolxyn∗αoomnilayl
But by definition (6), since ξ(s)= L1, LAk∗(x) is an integer bs (36)
for x∈ZL2, and L2(Ak∗(x1)−Ak∗(x2))2 ∈Z. s:xαX(s)=1
whNeroewσbypetrhmeudteefisntihtieonenwtreiesalsoof hka∗v.eAAlsσo(k,∗f)o(rx)s= AZk∗,(xle)t, However, since L is prime, the minimal polynomial of wL−j
s+k∗ :=(s+k1∗,s+k2∗,s+k3∗), then As+k∗(x)=∈AkL∗(x). ipnolQyn[xo]m,ifaolrmLus>t djiv>ide0,(3is6)1. T+hxus+x∗··a·n+dxxL∗−m1u[s1t2b],estohethailsl
There is 6 permutations and L possible values for s Z , 1 2
L
∈ zerosandallonessignals,butthesesignalsalsodonotsatisfy
so B(x ,x ) is an integer multiple of 6. (we can also have
not triv1ial s2 and σ such that s+k∗ =Lσ(k∗) but that case (33).
also has the property mentioned). However we cannot have ACKNOWLEDGMENT
B(x ,x ) = 6. That means there exists only one k∗ Z3
1 2 L ∈ L A. S. and J. P. were partially supported by Award Number
(with permutations and shifts) such that Ak∗(x1)6=Ak∗(x2). R01GM090200 from the NIGMS, FA9550-12-1-0317 from
Then
AFOSR, Simons Foundation Investigator Award and Simons
Collaborations on Algorithms and Geometry, and the Moore
Ak(x1)−Ak(x2)=6L(Ak∗(x1)−Ak∗(x2))6=0 (32) Foundation Data-Driven Discovery Investigator Award.
kX∈Z3L E. A. is partiallysupportedby NSF CAREER Award CCF-
On the other hand 1552131 and ARO grant W911NF-16-1-0051.
L REFERENCES
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in this case both Ak(u1) and Ak(u2) are 0 when k is odd, samplecomplexity ofmulti-reference alignment,” Inpreparation.
| |
since half of the signal is the symmetric of the other half, [8] R.Kakarala, “Thebispectrum as asource ofphase-sensitive invariants
i.e. u ( 1,...,L )= u ( L+1,...,L ). Now because of forFourierdescriptors: agroup-theoretic approach,” JournalofMathe-
(29) w1e{have A2k}(x1)−= 1Ak{(2x2) when k} = 3, so tL 4, [9] mE.aAticbable,ImJ.aMgin.gPearneidraV,iasniodnA,v.oSli.n4g4e,r,n“oV.e3r,ypnpo.is3y41S–a3n5o3v,’s2t0h1e2o.remand
| | ≥
and B (L)=0. applications toalignment problems,”Preprint,2017.
3
[10] T.M.CoverandJ.A.Thomas,Elementsofinformation theory. John
Finally, letL 6 be prime. We proveby contradictionthat
≥ Wiley&Sons,2006.
tL = 3 and B3(L) = 1L2. If this is not true, then it exists x∗1 [11] G. Szego˝, Orthogonal polynomials. American Mathematical Society,
and x∗2 such that tx∗1,x∗2 >3, so [12] 1N9.7J5a,cvoobls.o2n3,.Lectures in Abstract Algebra: III. Theory of Fields and
Ak(x∗1)=Ak(x∗2), k∈ZnL,n≤3 (33) GaloisTheory. Springer-Verlag, 1964,vol.32.
By Theorem 2 of paper [8], if the Fourier coefficients of x∗
1
and x∗ are non-zero,then equation (33) implies one is a shift
2
of the other. Denote by {rj1}j∈ZL and {rj2}j∈ZL the Fourier