1 THE LOGIC IN COMPUTER SCIENCE COLUMN by 2 Yuri GUREVICH Electrical Engineering and Computer Science UniversityofMichigan, Ann Arb or, MI 48109-2122, USA [email protected] Zero-One Laws Quisani: I heard you talking on nite mo del theory the other day.Itisinteresting indeed that all those famous theorems ab out rst-order logic fail in the case when only nite structures are allowed. I can understand that for those, likeyou, educated in the tradition of mathematical logic it seems very imp ortantto ndoutwhichof the classical theorems can b e rescued. But nite structures are to o imp ortant all by themselves. There's got to b e deep nite mo del theory that has nothing to do with in nite structures. Author: \Nothing to do" sounds a little extremist to me. Sometimes in nite ob jects are go o d approximations of nite ob jects. Q: I do not understand this. Usually, nite ob jects approximate in nite ones. A: It may happ en that the in nite case is cleaner and easier to deal with. For example, a long nite sum may b e replaced with a simpler integral. Returning to your question, there is indeed meaningful, inherently nite, mo del theory. One exciting issue is zero- one laws. Consider, say, undirected graphs and let be a propertyofsuch graphs. For example, may b e connectivity. What fraction of n-vertex graphs have the prop erty ? It turns out that, for many natural prop erties , this fraction converges to 0 or 1asn grows to in nity. If the fraction converges to 1, the prop erty is called almost sure. In that sense, almost all graphs are connected, hamiltonian, not 3-colorable, rigid, etc. [BH, Co2, and references there]. 1 \CurrentTrends in Theoretical Computer Science", Eds. G. Rozenb erg and A. Salomaa, World Scienti c, Series in Computer Science, Volume 40, 1993 2 Partially supp orted by NSF grant CCR 89-04728 and ONR grant N00014-91-J-11861. 1 Q: What is exactly the fraction of n-vertex graphs with prop erty ? What do es it mean that a graph is rigid? A: A graph G is called rigid if the identity map is the only automorphism of G (i.e. the only isomorphism from G to G). Now, ab out that fraction. There are two di erent de nitions in the literature: Lab eled Case: l ( )isthenumber L ( )of graphs on f1;:::;ng divided by n n n(n1)=2 2 , the total numb er of graphs on f1;:::;ng. In other words, l ( )= n L ( )=L ( ) where is the trivial prop ertythatevery graph has. n n Unlab eled Case: u ( ) is the number U ( ) of isomorphism typ es of n-vertex n n graphs divided by the total number U ( )ofisomorphismtyp es of n-vertex n graphs. Of course, the prop erty in question is supp osed to b e preserved by isomorphisms. Somewhat lo osely, p eople say that ob eys the lab eled (resp. unlab eled) zero-one law if l ( ) (resp. u ( )) converges to 0 or 1. Let me saythattwonumerical sequences n n and are asymptotical ly equivalent if they converge to the same numb er or else n n they b oth diverge. I will use the sign to denote asymptotic equivalence. Notice that the asymptotic equivalence is weaker than the asymptotic equality n n ; the latter means that the fraction = converges to 1. For example, 1=n n n n n 2 is asymptotically equivalent but not asymptotically equal to 1=n . It turns out that l ( ) u ( ). The reason for the asymptotic equivalence is that almost all graphs n n are rigid. Q: In what sense are most graphs rigid? A: Rigidity,which I will denote by , ob eys either of the two zero-one laws: l () n u () ! 1. Notice that each rigid n-vertex graph has exactly n! distinct lab elings, so n that L ( ^ )=U ( ^ ) n!. The desired asymptotic equivalence l ( ) u ( ) n n n n follows easily.Doyou see how to deriveit? Q: I think I do. Since l () ! 1, n L ( ) L ( ^ )+L ( ^:) L ( ^ ) L ( ^ ) n n n n n = : L ( ) L ( ) L ( ) L () n n n n U ( ) U ( ^ ) U ( ^ ) L ( ^ ) n n n n Similarly, b ecause u () ! 1. But = . n U ( ) U () U () L () n n n n A: Very go o d. Q: Let me verify for myself that rigid graph has n! distinct lab elings. There are n! one-to-one functions f from the universe V of an n-vertex graph G onto f1;:::;ng. Eachsuch f gives a graph G on f1;:::;ng,sothatf is an f 1 isomorphism from G onto G .IfG = G then h f is an automorphism of G. f f h Thus, all n! graphs G are distinct if G is rigid. Hence L ()=U (). f n n 2 I think we are getting somewhere. If G is not rigid and is a nontrivial automor- phism of G, then each graph G equals G .Thus, a nonrigid n-vertex graph has at f h most n!=2 distinct lab elings, so that L (:) U (:) n!=2 <U (:) n!. Wehave n n n U () U () n! L () n n n u ()= = < = l () n n U ()+U (:) U () n!+U (:) n! L ()+L (:) n n n n n n and therefore l () ! 1ifu () ! 1. However, I do not see why the latter is true. n n A: There is a go o d reason for the diculty.Until now, weworked in great generality. We could sp eak as well ab out directed graphs or groups, etc. But nowwe need information sp eci c to graphs. The following simple formula of Polya [HP, Section 9.1] is worth rememb ering: L ( ) n : U ( ) n n! By Polya's formula, U ()+U (:)=2 U ()n!+ U (:)n!=2 L ()+L (:) L ( ) n n n n n n n = = ! 1: U ( ) U ( )n! U ( )n! U ( )n! n n n n U ()+U (:) U (:)=2 U ()+U (:)=2 n n n n n Thus, 1 u (:)=2= = ! 1and n U ( ) U ( ) U ( ) n n n therefore u (:) ! 0. n It is often easier to deal with lab eled graphs. The fraction l ( ) can b e seen as n the probabilityof in the sample space of graphs with universe f1;:::;ng and the uniform probability distribution. Q: Similarly, the fraction u ( )canbeinterpreted as the probabilityof in an n appropriate sample space. A: True, but the sample space of lab eled graphs is easier to deal with. In the case of lab eled graphs, for example, the probabilityofanevent\fi; j g is an edge" is 1/2 whenever i 6= j .Further, anytwosuchevents are pairwise indep endent. The sample space can b e viewed as the set of outcomes of the following exp eriment: for each pair fi; j g of distinct elements of f1;:::;ng, toss a fair coin to decide whether the pair is an edge. Q: How do es logic come in? A: It turns out that every rst-order prop erty satis es the zero-one law [GKLT, and indep endently Fa1]. Q: Wow! A: Wemaysay that rst-order logic ob eys the zero-one law. Q: Do you mean the lab eled or unlab eled law? Mayb e you are talking ab out prop er- ties of graphs only.Weknow that l ( ) u ( )forany graph prop erty preserved n n by isomorphisms. 3 A: I am talking ab out rst-order logic with equality and unary,binary, ternary, etc. predicate symb ols, but without function symb ols or individual constants. Q: Sure. If c is an individual constant then the probability of a sentence P (c) is 1/2 in the case of lab eled structures. A: Right. It turns out that l ( ) u ( )forevery prop erty of structures of n n a given signature . Let me say this more carefully. First of all, I consider only nite signatures. Given a nite collection of predicate symb ols of sp eci ed arities, consider a class K of nite structures of signature . A prop erty of K structures can b e viewed as a function on K that assigns true or false to each structure A in K . We will consider only those prop erties which are preserved under isomorphisms. Q: You got me worried. Do you mean that K is a class and not necessarily a set? A: No, no. I do not intend to involve us in set theory. We are only interested in isomorphism typ es of structures and their lab elings. Generalize the de nitions of l n and u from graphs to structures in K and consider the case when K is the class of n all structures. For every prop erty of structures, l ( ) u ( ). n n Q: It is clear that l ( ) u ( ) if unlab eled -structures are almost surely rigid. n n Are they? A: Not if comprises only unary predicates. But the case of unary predicates is easily analyzable directly and wemay ignore it. If contains at least one predicate of arity > 1 then the unlab eled zero-one law for the rigidityof structures follows from the unlab eled zero-one law for graph rigidity. Q: How do es it follow? A: For simplicity, I supp ose that contains a binary predicate P .Given a structure A, construct a graph G on the universe of A such that any pair fa; bg of distinct elements of A is an edge if and only if either b oth P (a; b)and P (b; a) are true in A or else b oth P (a; b) and P (b; a) are false in A.Every automorphism of A is also an automorphism of G.