A Crevice on the Crane Beach: Finite-Degree Predicates

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by  Michaël Cadilhac

A Crevice on the Crane Beach:
Finite-Degree Predicates
Michaël Cadilhac Charles Paperman
WSI, Universität Tübingen WSI, Universität Tübingen
Email: [email protected] Email: [email protected]
Abstract—First-order logic (FO) over words is shown to expressible in FO[ARB], where ARB denotes all possible
be equiexpressive with FO equipped with a restricted set of numerical predicates (expressing numerical properties of the
7 numericalpredicates,namelytheorder,abinarypredicateMSB0, positions in a word). Further, as all regular neutral letter
01 TanhdeCthreanfienBitee-adcehgrPereopperretdyic(aCtBesP:)F,iOnt[rAodRuB]ce=dFmOor[≤et,hManSBa0d,eFcaINde]. languages of FO[ARB] are star-free [10], i.e., in FO[≤], the
2 ago, is true of a logic if all the expressible languages admitting Crane Beach Conjecture asked:
n a neutral letter are regular. Although it is known that FO[ARB] Are all neutral letter languages of FO[ARB] in FO[≤]?
does not have the CBP, it is shown here that the (strong form Notethatthisechoestheaboveintuitiononuniformity,since
a
J ofthe)CBPholdsforbothFO[≤,FIN]andFO[≤,MSB0].Thus the numerical predicates correspond precisely to the allowed
2 sFtOill[≤ex,pFrIeNss]aexwhiidbeitsvaariefotyrmofolafnlgoucaalgiteys,awnhdilethbeeiCngBPo,naensdimcpalne power to compute the circuit for a given input length [11]. The
1 predicate away from the expressive power of FO[ARB]. The intuition on the logic side is even more compelling: if a letter
counting ability of FO[≤,FIN] is studied as an application. can be introduced anywhere without impacting membership,
]
O thentheonlymeaningfulrelationthatcanrelatepositionsisthe
I. INTRODUCTION linear order. However, first-order logic can “count” up to logn
L
Ajtai [1] and Furst, Saxe, and Sipser [2] showed some 30 (see, e.g., [12]), meaning that even within a word with neutral
.
s yearsagothatParity,thelanguageofwordsover{0,1}having letters, FO[ARB] can assert some property on the number of
c
an even number of 1, is not computable by families of shallow nonneutral letters. This is, in essence, why nonregular neutral
[
circuits, namely AC0 circuits. Since then, a wealth of precise letter languages can be expressed in FO[ARB].
3 expressiveness properties of AC0 has been derived from this In the recent years, a great deal of efforts was put into
v
sole result [3], [4]. Naturally aiming at a better understanding studying the Crane Beach Property in different logics, i.e.,
3
7 ofthecorereasonsbehindthislowerbound,acontinuouseffort whether the definable neutral letter languages are regular.
6 has been made to provide alternative proofs of Parity∈/ AC0. Krebs and Sreejith [13], building on the work of Roy and
2 However, this has been a rather fruitless endeavor, with the Straubing [14], show that all first-order logics with monoidal
0
notable exception of the early works of Razborov [5] and quantifiers and + as the sole numerical predicate have the
.
1 Smolenski [6] that develop a less combinatorial approach with Crane Beach Property. Lautemann et al. [15] show Crane
0 an algebraic flavor. For instance, Koucký et al. [7] foray into Beach Properties for classes of bounded-width branching pro-
7
descriptive complexity and use model-theoretic tools to obtain grams, with an algebraic approach relying on communication
1
Parity ∈/ AC0, but assert that “contrary to [their] original complexity. Some expressiveness results were also derived
:
v hope, [their] Ehrenfeucht-Fraïssé game arguments are not from Crane Beach Properties, for instance Lee [16] shows that
i
X simpler than classical lower bounds.” More recent promising FO[+] is strictly included in FO[≤,×] by proving that only
approaches, especially the topological ones of [8], [9], have the former has the Crane Beach Property. Notably, all these
r
a yet to yield strong lower bounds. logics are quite far from full FO[ARB], and in that sense, fail
A different take originated from a conjecture of Lautemann to identify the part of the arbitrary numerical predicates that
and Thérien, investigated by Barrington et al. [10]: the Crane fit the intuition that they are rendered useless by the presence
Beach Conjecture. They noticed that the letter 0 acts as a of a neutral letter.
neutral letter in Parity, i.e., 0 can be added or removed from In the present paper, we identify a large class of predicates,
any word without affecting its membership to the language. If the finite-degree predicates, and a predicate MSB such that
0
a circuit family recognizes a language with a neutral letter, it any numerical predicate can be first-order defined using them
seems convincing that the circuits for two given input sizes and the order; in symbols, FO[≤,MSB0,FIN] = FO[ARB].
should look very similar, that is: the circuit family must be We show that, strikingly, both FO[≤,MSB0] and FO[≤,FIN]
highly uniform. It was thus conjectured that all neutral letter have the Crane Beach Property, this latter statement being our
languages in AC0 were regular, and this was disproved in [10]. main result. Hence showing that some nonregular neutral letter
This however sparked an interest in the study of neutral languageisnotexpressibleinFO[ARB]couldbedonebyshow-
letter languages, in particular from the descriptive complexity ing that MSB0 may be removed from any FO[≤,MSB0,FIN]
view. Indeed, AC0 circuits recognize precisely the languages formula expressing it.
TheprooffortheCraneBeachPropertyofFO[≤,FIN]relies the length n of u is denoted |u|. We write ε for the empty
on a communication complexity argument different from that word and A≤k for words of length ≤k.
of [15]. It is also unrelated to the database collapse techniques
B. Logic on words
of[10](succinctlyput,nologicwiththeCraneBeachProperty
For an alphabet A, let σ be the vocabulary {a|a∈A} of
has the so-called independence property, i.e., can encode A
unary letter predicates. A (finite) word u=u u ···u ∈
arbitrary large sets). We will show that in fact FO[≤,FIN] 0 1 n−1
A∗ is naturally associated with the structure over σ with
doeshavetheindependenceproperty.Thisprovides,tothebest A
universe [n] and with a interpreted as the set of positions i
of our knowledge, the first example of a logic that exhibits
such that u =a, for any a∈A. A numerical predicate is a
both the independence and the Crane Beach properties. i
k-aryrelationsymboltogetherwithaninterpretationin[n]k for
TheaforementionedcountingpropertyofFO[ARB]ledtothe
each possible universe size n. Given a formula ϕ that relies on
conjecture [10], [16] that a logic has the Crane Beach Property
some numerical predicates and a word u, we write u|=ϕ to
if and only if it cannot count beyond a constant. To the best
mean that ϕ is true of the σ -structure for u augmented with
of our knowledge, neither of the directions is known; we show A
the interpretations of the numerical predicates for the universe
however that FO[≤,FIN] can only count up to a constant, by
of size |u|. A formula ϕ thus defines or expresses the language
showing that it cannot even express very restricted forms of
{u∈A∗ |u|=ϕ}.
the addition. This adds evidence to the “if” direction of the
conjecture. C. Classes of formulas
Structure of the paper. In Section II, we introduce the
WeletARBbethesetofallnumericalpredicates.Givenaset
required notions, although some familiarity with language
N ⊆ARB, we write FO[N] for the set of first-order formulas
theory and logic on words is assumed (see, e.g., [4]). In
built using the symbols from N ∪σ , for any alphabet A.
A
Section III, we show that FO[≤,MSB0,FIN] = FO[ARB]. In Similarly, MSO[N] denotes monadic second-order formulas
Section IV, we present a simple proof, relying on a much
builtwiththosesymbols.WefurtherdefinethequantifiersMaj
harder result from [10], that FO[≤,MSB0] has the Crane and ∃≡, for i∈N, that will only be used in discussions:
Beach Property. The failing of the aforementioned collapse i
• u|=(Majx)[ϕ(x)] iff there is strict majority of positions
technique for FO[≤,FIN] is shown in Section V. We tackle i∈[|u|] such that (cid:104)u,x:=i(cid:105)|=ϕ;
tSheectCiornanVeIB,eaafctherPtrhoepenretyceosfsaFrOy[≤to,oFlsINh]a,voeurbemenainderveseuloltp,eidn. • u |= (∃≡i )[ϕ(x)] iff the number of positions i ∈ [|u|]
verifying (cid:104)u,x:=i(cid:105)|=ϕ is a multiple of i.
Finally, in Section VII, we focus on the counting inabilities of
We will write MAJ[N] and FO+MAJ[N] with the obvious
FO[≤,FIN]. meanings. Further, FO+MOD[N] allows all the quantifiers ∃≡
Previous works. Finite-degree predicates were introduced i
in FO[N] formulas.
by the second author in [17], in the context of two-variable
logics. Therein, it is shown that the two-variable fragment of D. On numerical predicates
FO[≤,FIN] has the Crane Beach Property, and, even stronger, The most ubiquitous numerical predicate here will be the
that the neutral letter languages expressible with k quantifier
binary order predicate ≤. The predicate that zeroes the most
alternations can be expressed without the finite-degree pred-
significant bit (MSB) of a number will also be important:
icates with the same amount of quantifier alternations. The (m,n)∈MSB iffn=m−2(cid:98)logm(cid:99).Notethatbothpredicates
0
techniques used in [17] are specific to two-variable logics,
do not depend on the universe size, and we single out this
relying heavily on the fact that each quantification depends on
concept:
a single previously quantified variable. We thus stress that the
communication complexity argument developed in Section VI Definition 1. A k-ary numerical predicate P is unvaried if
is unrelated to [17]. there is a set E ⊆ Nk such that the interpretation of P on
ThefactthattwosetsofpredicatescanbothverifytheCrane universes of size n is E ∩[n]k. In this case, we identify P
Beach Property while their union does not has already been with the set E. It is varied otherwise.1 We write ARBu for the
witnessed in [10]. Indeed, letting MON be the set of monoidal set of unvaried numerical predicates.
numerical predicates, the Property holds for both FO[≤,+]
Naturally, any varied predicate can be converted to an
and FO[≤,MON] but fails for FO[≤,+,MON], although this unvaried one by turning the universe length into an argument
latter class is less expressive than FO[ARB] (this can be shown andquantifyingthemaximumposition;thisimpliesinparticular
using the same proof as [7, Proposition 5]). that FO[ARB] = FO[ARBu]. This is however not entirely
innocuous, as will be discussed in Section VII.
II. PRELIMINARIES
We will rely on the following class of unvaried predicates,
A. Generalities
generalizing a definition of [17] (see also the older notion of
We write N = {0,1,2,...} for the set of nonnegative “finite formula” [18]):
numbers.Forn∈N,welet[n]={0,1,...,n−1}.Afunction
f: N→N is nondecreasing if m>n implies f(m)≥f(n). 1Therelevanceofthisconcepthasbeennotedinpreviousworks(e.g., [10]),
butwasleftunnamed.Thesecondauthorusedin[17]theterms(non)uniform,
An alphabet A is a finite set of letters (symbols), and we
an unfortunate coinage in this context. We prefer here the less conflicting
write A∗ for the set of finite words. For u = u0u1···un−1, terms(un)varied.
2
Definition2. AnunvariedpredicateP ⊆Nk isoffinitedegree2 The work zone has two salient properties: 1. Checking that
if for all n∈N, n appears in a finite number of tuples in P. a number k ∈[(cid:96)] belongs to it amounts to checking that k has
We write FIN for the class of such predicates. exactly one greater power of two; in particular, two work-zone
positions share the same MSB; 2. Any number in [(cid:96)] outside
Note that this does not imply that there is a N that bounds
the work zone can be obtained by replacing the MSB of a
the number of appearance for all n’s. Some examples:
number in the work zone with some other bits (0, 10, and 11,
• MSB0 is not a finite-degree predicate, as, e.g., (2n,0)∈ for the first, third, and fourth zone, respectively); we call this
MSB for any n, hence 0 appears infinitely often;
0 a translation to a zone, e.g., in our example above, 10101 is
• Any unvaried monadic numerical predicate is of finite the translation of 01101 to the third zone.
degree, this implies in particular that any language over a
More formally, we can define a formula work(x) which is
unary alphabet is expressed by a FO[≤,FIN] formula; true iff x belongs to the work zone, by expressing that there
• The graph of any nondecreasing unbounded function is exactly one power of two strictly greater than x, using the
f: N→N defines a finite-degree predicate, since f−1(n)
monadic predicate true on powers of two. Moreover, we can
is a finite set for all n;
define formulas trans(i)(x,y), 1≤i≤4, which are true if x
• Theorder,sum,andmultiplicationarenotoffinitedegree; is in the work zone and y is its translation to the i-th zone; let
• One can usually “translate” unvaried predicates to make us treat the case i=3, the others being similar. The formula
themfinitedegree;forinstance,thepredicatetrueof(x,y)
trans(3)(x,y) is true if y is obtained by replacing the MSB
if y−x<x<y is of finite degree, see also the proof of
of x with 10, this is expressed using MSB by finding z such
0
Proposition 4.
that MSB (x,z) holds and then checking that y is the first
0
E. Crane Beach Property value z(cid:48) strictly greater than x such that MSB (z(cid:48),z) holds.
0
A language L⊆A∗ is said to have a neutral letter if there The strategy will then be to: 1. Quantify over the work
is a e ∈ A such that adding or deleting e from a word does zone only; 2. Modify the predicates to internally change the
not change its membership to L. Following [15], we say that MSBs according to which zone the variables were supposed
a logic has the Crane Beach Property if all the neutral letter to belong; 3. Compute the translations of the variables for the
languages it defines are regular. We further say that it has the letter predicates. Step 1 relies on work and trans(i), Step 2
strong Crane Beach Property if all the neutral letter languages transforms all numerical predicates to finite-degree ones, and
it defines can be defined using order as the sole numerical Step 3 simply uses trans(i).
predicate. Let ϕ ∈ FO[ARBu]. Step 1. We rewrite ϕ with annotated
variables; with x a variable, we write x(i), 1≤i≤4, to mean
III. FO[ARB]ANDFO[≤,MSB0,FIN]DEFINETHESAME “x translated to zone i”—as all the variables will be quantified
LANGUAGES in the work zone, this is well defined. The following rewriting
Inthissection,weexpressallthenumericalpredicatesusing is then performed:
only finite-degree ones, MSB , and the order. The result is a
0 ∃xψ(x)(cid:32)
variant of [17, Theorem 3], where it is proven for the two-
variable fragment, and on neutral letter languages. ∃x(cid:104)work(x)∧ (cid:95) (cid:104)(∃y)[trans(i)(x,y)]∧ψ(x(i))(cid:3)(cid:105) ,
Theorem 1. FO[ARB] and FO[≤,MSB0,FIN] define the same 1≤i≤4
languages. and mutatis mutandis for ∀.
Step 2. We sketch this step for binary numerical predicates.
Proof. We show that any FO[ARBu] language is definable in
Suppose such a predicate P is used in ϕ. For 1≤i,j ≤4, we
FO[≤,MSB0,FIN]. definethepredicateP(i,j)thatexpectstwowork-zonepositions,
The main idea is to divide the set of word positions in four
translates them to the i-th and j-th zone, respectively, then
contiguous zones and have the variables range over only the
checks whether they belong to P. Crucially, as the inputs
second zone, called the work zone. Given an input of length
are work-zone positions, P(i,j) immediately rejects if they do
(cid:96) = 2n, the set of positions [(cid:96)] is divided in four zones of
not share the same MSB: it is thus a finite-degree predicate.
equal size 2n−2; if the input length is not a power of 2, then
Now every occurrence of P(x(i),y(j)) in ϕ can be replaced
we apply the same split as the closest greater power of two,
by P(i,j)(x,y).
leaving the third and fourth zone possibly smaller than the first
Step3.Theonlyremainingannotatedvariablesappearunder
two.
letter predicates. To evaluate them, we simply have to retrieve
As an example, suppose that the word size is (cid:96) = 11110
the translated position. Hence each a(x(i)) will be replaced by
(here and in the following, we write numbers in binary). The
(∃y)[trans(i)(x,y)∧a(y)], concluding the proof.
four zones of [(cid:96)] will be:
Remark. Theorem 1 can be shown to hold also for
1) 00000→00111; 2) 01000→01111;
3) 10000→10111; 4) 11000→11101=(cid:96)−1 . FO+MAJ[≤,MSB0,FIN],i.e.,thislogicisequiexpressivewith
FO+MAJ[ARB]. The main modification to the proof is to allow
arbitrary quantifications (as opposed to work zone ones only)
2The name stems from the fact that the hypergraph defined by P, with
edgesofsizek,isoffinitedegree. and compute the work zone equivalent of each position before
3
checkingthenumericalpredicates.Thisensuresthatthenumber Proof. Letn>0,anddefinea =2n+2i for0≤i<n.Now
i
of positions verifying a formula is not changed. Likewise, for M ⊆[n], let b =2n+(cid:80) 2i. It holds that i∈M iff
M i∈M
FO+MOD[≤,MSB0,FIN] is equivalent with FO+MOD[ARB]. the binary AND of ai and bM is ai. Consider this latter binary
predicate; its behavior on two arguments that do not share
IV. FO[≤,MSB0]HASTHECRANEBEACHPROPERTY the same MSB is irrelevant, and we can thus decide that such
Following a short chain of rewriting, we will express MSB0 inputs are rejected. Thanks to this, we obtain a finite-degree
using predicates that appear in [10] and conclude that: predicate.Consequently,theformulathatconsistsofthissingle
predicate has the independence property.
Theorem 2. FO[≤,MSB ] has the strong Crane Beach Prop-
0
erty.
VI. FO[≤,FIN]HASTHECRANEBEACHPROPERTY
Proof. Let f: N → N be defined by f(n) = 2((cid:98)logn(cid:99)2), and A. Communication complexity
let F ⊆N2 be its graph. Barrington et al. [10, Corollary 4.14]
We will show the Crane Beach Property of FO[≤,FIN] by a
show that FO[≤,+,F] has the strong Crane Beach Property;
communication complexity argument. This approach is mostly
we show that MSB can be expressed in that logic. First, the
0 unrelated to the use of communication complexity of [15],
monadicpredicateQ={2n |n∈N}isdefinableinFO[≤,F],
[22]; in particular, we are concerned with two-party protocols
since n is a power of two iff f(n−1)(cid:54)=f(n). Second, given
with a split of the input in two contiguous parts, as opposed
n∈N,thegreatestpoweroftwosmallerthannisp=2(cid:98)logn(cid:99),
to worst-case partitioning of the input among multiple players.
whichiseasytofindinFO[≤,Q].Finally,MSB (n,m)istrue
0 We rely on a characterization of [23] of the class of languages
iff m+p=n, and is thus definable in FO[≤,+,F].
expressible in monadic second-order with varied monadic
Remark. From Lange [19], MAJ[≤] and FO+MAJ[≤,+] are numerical predicates. Writing this class MSO[≤,MON], they
equiexpressive, and as MSB is expressible using the unary state in particular the following:
0
predicate {2n | n ∈ N} and the sum, this shows that
Proposition 2 ([23, Theorem 2.2]). Let L ⊆ A∗ and define,
MAJ[≤,FIN] is equiexpressive with FO+MAJ[ARB]. Hence for all p∈N, the equivalence relation ∼ over A∗ as: u∼ v
MAJ[≤,FIN] does not have the strong Crane Beach Property. iff for all w ∈Ap, u·w ∈L⇔v·w ∈Lp. If there is a N ∈pN
V. FO[≤,FIN]HASTHEINDEPENDENCEPROPERTY such that for all p∈N, ∼p has at most N equivalence classes,
In[10],animportanttoolisintroducedtoshowCraneBeach then L∈MSO[≤,MON].
Properties, relying on the notion of collapse in databases, Lemma 1. Let L ⊆ A∗. Suppose there are functions
see [20, Chapter 13] for a modern account. Specifically, let us f : A∗×N×{0,1}∗ →{0,1} and f : A∗×N×{0,1}∗
Alice Bob
define an ad-hoc version of the: and a constant K ∈ N such that for any u,v ∈ A∗, the
sequence, for 1≤i≤K:
Definition 3 (Independence property (e.g., [21])). Let N be
#– #–
a set of unvaried numerical predicates. Let x, y be two • ai =fAlice(u, |u·v|, b1b2···bi−1)
vectors of fi#–rst-#–order variables of size k and (cid:96), respectively. A • bi =fBob (v, |u·v|, a1a2···ai);
formula ϕ(x, y) of FO[N], over a single-letter alphabet, has
the independence property if for all n > 0 there are vectors is such that bK =1 iff u·v ∈L. Then L∈MSO[≤,MON].
#– #– # –
a ,a ,...,a , each of Nk, for which for any M ⊆ [n], Proof. We adapt the (folklore) proof that L is regular iff such
0 1 n−1 # –
there is a vector b ∈N(cid:96) such that:3 functions exist where f and f do not use their second
M Alice Bob
#– #– #– # – parameter.
(cid:104)N,x :=a , y :=b (cid:105)|=ϕ iff i∈M .
i M Let p ∈ N. For any u ∈ A∗, let c(u) be the set of pairs
The logic FO[N] has the independence property if it contains (a1a2···aK,b1b2···bK−1) such that for all 1 ≤ i ≤ K,
such a ϕ. it holds that ai = fAlice(u,|u|+p,b1b2···bi−1). Define the
equivalence relation ≡ by letting u ≡ v iff c(u) = c(v); it
Intuitively, a logic has the independence property iff it can
clearly has a finite number N =N(K) of equivalence classes.
encode arbitrary sets. Barrington et al. [10], relying on a deep
Moreover, if u≡v and w ∈Ap, then (u,w) and (v,w) define
result of Baldwin and Benedikt [21], show that:
the same sequences of a ’s and b ’s, by a simple induction.
i i
Theorem 3 ([10, Corollary 4.13]). If a logic does not have Hence u·w ∈L⇔v·w ∈L. This shows that ≡ refines ∼p,
the independence property, then it has the strong Crane Beach implying, by Proposition 2, that L∈MSO[≤,MON].
Property.
Weshalladopttheclassicalcommunicationcomplexityview
We note that this powerful tool cannot show that the logic here, and consider f and f as two players, Alice and
Alice Bob
we consider exhibits the Crane Beach Property: Bob, that alternate exchanging a bounded number of bits in
order to decide if the concatenation of their respective inputs
Proposition 1. FO[≤,FIN] has the independence property. is in L. To show that L is in MSO[≤,MON], the protocol
3Notethatweevaluateaformulaoveraninfinitedomain;thisiswelldefined between Alice and Bob should end in a constant number of
inourcasesinceweonlyuseunvariedpredicatesandtheletterpredicatesare rounds. We will then rely on the fact that:
irrelevant.
4
∨
a) b)
∃Ax ∃Bx
∃x
∧ ∧
∀y
∀Ay ∀By ∀Ay ∀By
ψ ψ ψ ψ ψ
Quantified by Alice
∨
c)
(cid:87)
∃Bx
x=0 x=1
∧ ∧
∧
(cid:86) (cid:86)
∀By ∀By (cid:86)
∀By
y=1 y=0
y=1
y=0 (cid:62)∧ (cid:62)∧ y=1 (cid:62)∧
y=0
a(x)∧
(cid:62)→⊥ ⊥→⊥ (cid:62)→b(y)
⊥→⊥
(cid:62)∧ (cid:62)∧ (cid:62)∧
a(x)∧
⊥→⊥ (cid:62)→b(y) ⊥→⊥ ψ
⊥→⊥
Fig.1. TheformulaϕasitgetsevaluatedbyAliceandBob.
Theorem4([23,Theorem4.6]). MSO[≤,MON]hastheCrane Alicewillnowexpandherquantifierstorangeoverherword;
Beach Property. she will thus replace, e.g., (∀Ay)[ψ] by (cid:86)1 ψ. Crucially, at
y=0
the leaves of the formula, it is known which variables were
B. A toy example: FO[<]⊆MSO[≤,MON] quantified by each player, and if they are Alice’s, their values.
We will demonstrate how the communication complexity Consider for instance a leaf where Alice substituted y with a
approach will be used with a toy example. Doing so, the numericalvalue.Theletterpredicateb(y)canthusbereplaced
requirements for this protocol to work will be emphasized, and by its truth value. More importantly, the predicate x < y
they will be enforced when showing the Crane Beach Property can also be evaluated: Either Alice quantified x, and it has a
of FO[≤,FIN] in Section VI-C. numerical value, or she did not, and we know for sure that
Let us consider the following formula over A={a,b,c}: x<ydoesnothold,sincexwillbequantifiedbyBob.Applied
to our example, we obtain the tree of Figure 1.c.
ϕ≡(∃x)(∀y)[ψ], with ψ ≡a(x)∧(x<y →b(y)) ,
The resulting formulas at the leaves are thus free from the
depicted as a tree in Figure 1.a. The formula ϕ asserts that the variables quantified by Alice. Moreover, for each internal node
all the letters after the last a are b’s. In this example, Alice of the tree, its children represent subformulas of bounded
will receive u=aa, and Bob v =bb. Naturally, ϕ over words quantifier depth, and there are thus a finite number of possible
of length 4 is equivalent to the formula where ∃x is replaced nonequivalent subformulas. Once only one subformula per
by (cid:87)3 , and ∀y is replaced by (cid:86)3 ; our approach will be equivalenceclassiskept,theresultingtreeisofboundeddepth
x=0 y=0
to split this rewriting between Alice and Bob. and each node has a bounded number of children. Hence the
Considerthevariablex.Tocheckthevalidityoftheformula size of this tree is bounded by a value that only depends on ϕ.
over a u·v, the variable should range over the positions of Alice can thus communicate this tree to Bob. In our example,
both players. In other words, the formula is true if there is simplifying the tree, we obtain the formula:
a position x of Alice verifying (∀y)[ψ] or a position x of
(cid:104) (cid:105)
Bob verifying it—likewise for the universal quantifier. We thus (∀By)[b(y)]∨(∃Bx) a(x)∧(∀By)[ψ] .
“split” the quantifiers by enforcing the domain to be either
Alice’s (∀A,∃A) or Bob’s (∀B,∃B), obtaining Figure 1.b. Finally, Bob can actually quantify his variables, resulting
5
in a formula with no quantified variable, that he can evaluate, Proof. This is easily shown by induction; we prove the first
concluding the protocol. item, the second being similar. For n = 1, this is clear. Let
Takeaway.Thisprotocolreliesonthefactthatpredicatesthat n>1.Ifm=0,thisisimmediatefromFact1,letthusm>0.
involve variables from both Alice and Bob can be evaluated We have that:
by Alice alone. This enables Alice to remove “her” variables
Lm(Rn(p))=L(Lm−1(Rn−1(p(cid:48)))) ,
before sending the partially evaluated tree to Bob, who can
quantify the remainder of the variables. with p(cid:48) =R(p). By induction hypothesis and the fact that L
is nondecreasing, it holds that:
C. The case of FO[≤,FIN]
Theorem 5. FO[≤,FIN]hasthestrongCraneBeachProperty. Lm(Rn(p))≥L(R(p(cid:48)))=q .
Proof. Let ϕ be a formula over an alphabet A in FO[≤,N], Let p(cid:48)(cid:48) = R(p(cid:48)). By definition of L, (q+1,p(cid:48)(cid:48)) is an edge
for some finite subset N of FIN, and suppose ϕ expresses in G. Now by definition of G, if q <p(cid:48), then (p(cid:48),p(cid:48)(cid:48)) should
a language L that admits a neutral letter e. We show that also be an edge in G, which contradicts the definition of p(cid:48)(cid:48).
L∈MSO[≤,MON] using Lemma 1. This concludes the proof Hence q ≥p(cid:48), showing the property. of Fact 2.
since by Theorem 4, L is a neutral letter regular language
Letusnowsupposewehavetwolargepositions|u|(cid:28)(cid:96) (cid:28)
in FO[ARB], and it thus belongs to FO[≤] (see [10]; this is r (cid:28)|v|,therequirementsonwhichwillbemadeclearsho0rtly.
essentially a consequence of Parity∈/ AC0). 0
LetusdeemapositionptobeAlicicifp≤(cid:96) ,Bobicifp≥r ,
Let us write u ∈ A∗ for Alice’s word, and v for Bob’s. 0 0
and Neutral otherwise; we call this the type of the position.
Both players will compute a value N >0 that depends solely
We wish to ensure that two positions of two different types
on ϕ and |u·v|, and the protocol will then decide whether
cannot be linked in G, so that they cannot appear in a tuple of
u·eN·v ∈L,whichisequivalenttou·v ∈Lbyhypothesis.We
a predicate in N. This surely is not the case if the typing of
supposethatalargeenoughN hasbeenpickedfortheprotocol
positionsdoesnotreflectpreviouslytypedpositions,e.g.,(cid:96) −1
to work, and delay to the end of the proof its computation. 0
is Alicic, but (cid:96) is Neutral, and their distance may not be large
Wewillhenceforthsupposethatϕisgiveninprenexnormal 0
enough to ensure that they do not form an edge in G. Thus
form and that all variables are quantified only once:
the boundaries of the zones, (cid:96) and r , will be moving with
0 0
ϕ≡(Q x )(Q x )···(Q x )[ψ] , eachnewtyping.Formally,letT ={Alice,Neutral,Bob}bean
1 1 2 2 k k
alphabet,anddefinethefunctionbounds: T≤k →[|u·eN·v|]2
with ψ quantifier-free and Q ∈{∀,∃}. We again see formulas
i by:
as trees with leaves containing quantifier-free formulas.
Rather than splitting the domain [|u·eN ·v|] at a precise bounds(ε)=(l ,r )
0 0
position, and tasking Alice to quantify over the first half and bounds(t t ···t )=
1 2 i
Bob over the second half, we will rely on a third group, that is 
(Rn((cid:96)),r) if t =Alice
“far enough” from both Alice’s and Bob’s words. The core of  i
this proof is to formalize this notion. Let us first introduce the (Ln((cid:96)),Rn(r)) if ti =Neutral
tools that will enable this formalization: one set of definitions, ((cid:96),Ln(r)) if t =Bob
i
and two facts that will be used later on. with ((cid:96),r)=bounds(t t ···t ) and n=2k−i.
1 2 i−1
Definition 4. LetC bethesetofpairsofintegers(p ,p )that
1 2 Assumption. Wehenceforthassumethatif((cid:96),r)=bounds(h)
appearinasametupleofarelationinN.Definethelinkgraph
for some word h ∈ T≤k, then |u| < (cid:96) < r < |u|+N. This
G=(N,E) as the undirected graph defined by (p ,p )∈E
1 2 will have to be guaranteed by carefully picking N, (cid:96) and r .
iff p =p or there are integers p(cid:48) ≤{p ,p }≤p(cid:48) such that 0 0
1 2 1 1 2 2
(p(cid:48),p(cid:48)) ∈ C. For p ∈ N, L(p) (resp. R(p)) is the greatest The type of a position p under type history t t ···t ∈T∗
1 2 1 2 i
q <p (resp. smallest q >p) which is not a neighbor of p in is computed by first taking ((cid:96),r) = bounds(t t ···t ), and
1 2 i
G. Equivalently, L(p) is the smallest neighbor of p minus 1, reasoning as before: it is Alicic if p≤(cid:96), Bobic if p≥r, and
and R(p) is the greatest neighbor of p plus 1. Neutral otherwise. This is well defined since (cid:96) < r by our
Assumption. The crucial property here is as follows:
Note that L and R are well defined since each vertex of G
has a finite number of neighbors. This directly implies that: Fact 3. Let p ,p ,...,p be positions, and inductively define
1 2 k
the type t of p as its type under type history t t ···t .
Fact 1. The functions L and R are nondecreasing and i i 1 2 i−1
unbounded. Moreover, for any p∈N, L(p)<p<R(p). 1) Two positions with different types do not form an edge
in G;
WritingRn forthefunctionRcomposedntimeswithitself, 2) All Alicic positions are strictly smaller than the Neutral
and similarly for L, we have: ones, which are strictly smaller than the Bobic ones;
3) All Neutral positions are labeled with the neutral letter.
Fact 2. For any position p and n>m≥0:
• Lm(Rn(p))≥R(p); Proof. (Points 1 and 2.) Suppose pi is Alicic and pj is
• Rm(Ln(p))≤L(p). Neutral, with i < j. Let ((cid:96),r) = bounds(t1t2···ti−1),
6
we thus have that p is maximally (cid:96). Let ((cid:96)(cid:48),r(cid:48)) = This is precisely the algorithm that Alice and Bob will
i
bounds(t t ···t ),thenp isminimally(cid:96)(cid:48)+1.Bydefinition, execute. First, Alice will quantify her variables according to
1 2 j−1 j
once the types of p ,p ,...,p are fixed, the smallest (cid:96)(cid:48) that the boundsof thetype historyof eachnode, asin Algorithm1.
1 2 i
can be obtained with the types t is by having all positions At the leaves, she will thus obtain the formula ψ, and have
>i
p , with i<t<j, Neutral. In that case, an easy computation a set of quantified Alicic variables. She can then evaluate ψ
t
shows that (cid:96)(cid:48) would be: partially: if an atomic formula only relies on Alicic variables,
she can compute its value. If an atomic formula uses a mix
L2k−(j−1)(L2k−(j−2)(···(L2k−(i+1)(R2k−i((cid:96))))···)) .
of Alicic and non-Alicic variables, then she can also evaluate
it: if the formula is a numerical predicate, then by Fact 3.1,
That is, L is composed with itself m times with:
it will be valued false; if the formula is of the form x < y,
k then it is true iff x is Alicic, by Fact 3.2. Alice now simplifies
(cid:88)
m=2k−(i+1)+···+2k−(j−1) < 2k−s
her tree: logically equivalent leaves with the same parent are
s=i+1 merged,andinductively,eachinternalnodekeepsonlyasingle
<2k−i =n . occurrence per formula appearing as a child. We remark that
the semantic of the tree is preserved. This results in a tree
Hence (cid:96)(cid:48) is at most Lm(Rn((cid:96))) with m<n, and by Fact 2,
whose size depends solely on ϕ, and the values of N, (cid:96) , and
(cid:96)(cid:48) ≥R((cid:96)). Hence (p ,p ) is not an edge in G, and p <p . 0
i j i j r , and Alice can thus send it to Bob.
The other cases are similar. For instance, if p is Neutral 0
i Bob will now expand the remaining quantifiers (Neutral
andp Bobic,withi<j,then,withthesamenotationasabove,
j and Bobic), respecting the bounds of the type history, as in
(cid:96)(cid:48) can be at most Lm(Rn((cid:96))), and by Fact 2, (cid:96)(cid:48) ≥L((cid:96)).
Algorithm 1. He can then evaluate all the leaves, since, by
(Point 3.) This is a direct consequence of the Assumption.
Fact 3.3, the only letter predicate true of a Neutral position is
Consider((cid:96),r)=bounds(Neutralk);thisprovidestheminimal
that of the neutral letter. This concludes the protocol, which
(cid:96) and maximal r between which a position can be labeled
clearly produces the same result as Algorithm 1.
Neutral. By the Assumption, |u|<(cid:96)<r <|u|+N, hence a
WhatareN,(cid:96) ,r ?WecheckthatAliceandBobcanagree
Neutral position has a neutral letter. of Fact 3. 0 0
onthesevalueswithoutcommunication.Therequirementswere
We are now ready to present the protocol. First, we rewrite made explicit in our Assumption. The values computed by the
quantifiers using Alicic/Neutral/Bobic annotated quantifiers: function bounds are obtained by applying L and R on (cid:96)0 and
• (∀x)[ρ](cid:32)(∀Ax)[ρ] ∧ (∀Nx)[ρ] ∧ (∀Bx)[ρ], r0 at most n = (cid:80)ki=−012i times. From Fact 1, it is clear that
• (∃x)[ρ](cid:32)(∃Ax)[ρ] ∨ (∀Nx)[ρ] ∨ (∃Bx)[ρ]. any ((cid:96),r)=bounds(h), for h∈T≤k, verifies:
Let us further equip each node with the type history of the • (cid:96)min =Ln((cid:96)0)≤(cid:96)≤Rn((cid:96)0)=(cid:96)max;
variables quantified before it; that is, each node holds a string • rmin =Ln(r0)≤r ≤Rn(r0)=rmax.
t t ···t ∈ T≤k where t is the annotation of the i-th
1 2 n i Hence we pick (cid:96) = Rn+1(|u|), ensuring, by Fact 2, that
quantifier from the root to the node, excluding the node itself. 0
(cid:96) > |u|. Next, we pick r to be Rn+1((cid:96) ), ensuring
Now if we were given the entire word u·eN ·v, a way min 0 max
that r > (cid:96) by the same Fact 2. Finally, we pick N =
min max
to evaluate the formula that respects the semantic of “Alicic”,
Rn+1(r ), ensuring, by Fact 1, that N > r , so that in
0 max
“Neutral”, and “Bobic” is as follows:
particular,r <|u|+N.Wethenindeedobtainthat|u|<(cid:96)<
max
r <|u|+N, as required. Note that these computations depend
Algorithm 1 Formula Evaluation
solely on ϕ and the lengths of u and v. of Theorem 5.
1: foreach quantifier node ∀Ax or ∃Ax do
2: ((cid:96),r):=bounds(type history at node)
3: if node is ∀Ax then Remark. It should be noted that the crux of this proof is that a
4: Replace node with (cid:86)(cid:96) relation R(x,y) with x Alicic and y Neutral or Bobic can be
x=0 readily evaluated by Alice. If R were monadic, then it could
(cid:87)
5: Similarly with ∃ becoming
notmixtwopositionsofdifferenttypes,henceAlicecouldstill
6: end
remove all of her variables at the end of her evaluation. The
7: Evaluate the part of the leaves than can be evaluated
rest of the protocol will be similar, with Bob quantifying the
8: foreach quantifier node do
remaining positions. This shows that FO[≤,MON,FIN] also
9: ((cid:96),r):=bounds(type history at node)
has the Crane Beach Property.
10: if node is ∀Nx then
11: Replace node with (cid:86)r−1
x=(cid:96)+1
12: else if node is ∀Bx then VII. ONCOUNTING
13: Replace node with (cid:86)|ueNv|
x=r
(cid:87) A compelling notion of computational power, for a logic, is
14: Similarly with ∃ becoming
the extent to which it is able to precisely evaluate the number
15: end
of positions that verify a formula. This is formalized with the
16: Finish evaluating the tree
following standard definition:
7
Definition 5. For a nondecreasing function f(n)≤n, a logic logn, hence n appears a finite number of time as (n,y) in
is said to count up to f(n) if there is a formula ϕ(c) in this Bit(cid:48). Suppose (x,n)∈Bit(cid:48), then n−f(x)>0, but for x large
logic such that for all n and w ∈{0,1}n: enough, f(x)>n, hence there can only be a finite number of
pairs (x,n) in Bit(cid:48).
w |=ϕ(c) ⇔ c≤f(n)∧c=number of 1’s in w .
Now Bit can be defined in FO[≤,FIN] using ϕ, since
It is known from [10] that if a logic can count up to Bit(x,y) holds iff (∃z)[ϕ(z,y,x)∧Bit(cid:48)(x,z)], a contradiction
log(log(···(logn))), for some number of iterations of log, concluding the proof.
then the logic does not have the Crane Beach Property. It has
Corollary 1. FO[≤,FIN] cannot count beyond a constant.
also been conjectured [10], [16] that a logic has the Crane
Beach Property iff it cannot count beyond a constant. It is not VIII. CONCLUSION
known whether there exists a set of predicates N such that
We showed that FO[≤,FIN] is one simple predicate away
FO[N] can count beyond a constant but not up to logn.
from expressing all of FO[ARB], and that it exhibits the Crane
We define a much weaker ability:
Beach Property. This logic is thus really on the brink of a
Definition 6. For a nondecreasing function f(n)≤n, a logic crevice on the Crane Beach, and exemplifies a diverse set of
is said to sum through f(n) if there is a formula ϕ(a,b,c) in behaviors that fit the intuition that neutral letters should render
this logic such that for all n and w ∈{0,1}n: numerical predicates essentially useless. We emphasize some
future research directions:
w |=ϕ(a,b,c) ⇔ a=b+f(c) .
• As a consequence of our results, one can show that a
This is in general even weaker than being able to sum “up nonregular neutral letter language L is not in AC0 as
to” f(n), that is, having a formula expressing that a=b+c follows.AssumeL∈AC0 foracontradiction,andletϕ∈
and c≤f(n). Naturally, counting and summing are related: FO[≤,MSB0,FIN] be a formula expressing it. Suppose
that one can show that ϕ can be rewritten without the
Proposition3. LetN beasetofunvariednumericalpredicates.
If FO[≤,N] can count up to f(n), it can sum through f(n). predicate MSB0, then L ∈ FO[≤,FIN], and thus L is
regular, a contradiction. We hope to be able to apply this
Proof. Letting ϕ(c) be the formula that counts up to f(n), we strategy in the future.
modify it into ϕ(cid:48)(a,b,c) by changing the letter predicates to • As noted in [14] and [10] and studied in particular
consider that there is a 1 in position p iff b ≤ p < a. This in [13], the interest in circuit complexity calls for the
expresses that a=b+c provided that c≤f(n). study of logics with more sophisticated quantifiers, no-
Next, the graph F of f is obtained as follows. First, modify tably modular quantifiers and, more generally, monoidal
ϕ(c) into ϕ(cid:48)(c,c(cid:48)), by restricting all quantifications to c and quantifiers. Hence the natural question here is whether
replacing the letter predicates to have 1’s in all positions below FO+MOD[≤,FIN] has the Crane Beach Property.
c(cid:48). Second, (c,c(cid:48)) ∈ F iff c(cid:48) is maximal among those that • As asked in [10], can we dispense from our implicit
verify ϕ(cid:48)(c,c(cid:48)). This relies on the fact that N consists solely reliance on the lower bound Parity∈/ AC0? In the cases
of unvaried predicates. of [10], and as noted by the authors, this would be
The logic can then sum through f(n) by: very difficult, as their results imply the lower bound.
ψ(a,b,c)≡(∃c(cid:48))[F(c,c(cid:48))∧a=b+c(cid:48)] . Here, the strong Crane Beach Property for FO[≤,FIN]
does not directly imply the lower bound. To show that
Remark. Proposition 3 depends crucially on the fact that the Parity ∈/ AC0, one could additionally prove that all the
predicates are unvaried to show that the graph of the summing regular, neutral letter languages of FO[≤,MSB0,FIN] are
functionisexpressible.WritingS forthesetofvariedmonadic in FO[≤,FIN]—we know that this statement holds, but
predicates S = (S ) with |S | = 1 for all n, it is easily only thanks to Parity∈/ AC0.
n n≥0 n
shown that FO[≤,+,×,S] can count up to any function ≤ • ArewereallyonthebrinkoffallingofftheCraneBeach?
logn. However, we conjecture that there are functions whose That is, are there unvaried predicates that cannot be
graphs are not expressible in this logic. expressed in FO[≤,FIN] but can still be added to the
logic while preserving the Crane Beach Property? We
Proposition 4. FO[≤,FIN] cannot sum through beyond a noted that all varied monadic predicates can be added
constant.
safely,butalreadyverysimplepredicatesfalsifytheCrane
Proof. Suppose for a contradiction that FO[≤,FIN] can sum Beach Property. For instance, with F the graph of the 2-
throughanondecreasingunboundedfunctionf usingaformula adic valuation, FO[≤,F] is as expressive as FO[≤,+,×]
ϕ(a,b,c). Let Bit be the binary predicate true of (x,y) if the (see [24, Theorem 3]), which does not have the Crane
y-th bit of x is 1. We define a translated version as: Beach Property [10].
• Numerical predicates correspond in a precise sense [11]
Bit(cid:48) ={(x,y)|(x,y−f(x))∈Bit} .
to the computing power allowed to construct circuit
We show that Bit(cid:48) is of finite degree. Let n∈N, and suppose families for a language. Is there a natural way to present
(n,y) ∈ Bit(cid:48). This implies in particular that 0 < y−f(n) < FO[≤,FIN]-uniform circuits?
8
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