successfully to the at most t moves èincluding the initial move mè made in these intervals. Thus Spoiler may aswell play the remaining t, 1moves

Transcription

1 The class of æ 1 ææææ r is z and z 2 R x. The class z of æ 1 ææææ r is in L z and R x. As z 2 L z ë R x, z = z. Thus ë1;i r è under M and ë1;i r è under M have the same Ehrenfeucht value. Thus Duplicator can respond successfully to the at most t moves èincluding the initial move mè made in these intervals. Thus Spoiler may aswell play the remaining t, 1moves on M 1 =é ëi r ;në; ç;u é and M =é 1 ëi r ;n ë; ç;u é. These intervals have lengths n 1 ç n and 3 n ç respectively. But now n M and M 1 are both nice 3 with respect to æ 1 =3æ - the sequence A 1 æææa R still appears inside every interval of length æn ç æ 1 n 1 in M and æ 1 n in M 1. Hence we can apply the same argument for the second move - for convenience still looking at Ehrenfeucht values with respect to the t move game. After t moves we still have nice M t ;Mt with respect to æ t ç 1,2 so the arguments are still valid. But at the end of t rounds Duplicator has won. 2

2 suæces to show that a æxed ëy; y + æ në has such a sequence. Generating the 3 k-intervals from 1 almost surely a k-interval ends after y and before y + æ n. 6 Now we generate a random sequence æ 1 æææ on an interval of length æ n. 6 But constants do not aæect the analysis of x2 and almost surely A 1 æææa R appears. Now oné ënë; ç;u é deæne U r by U r èiè if and only if Uèn +1, rè. U r is the sequence U in reverse order. Call U left nice if U r is right nice. Call U nice if it is right nice and left nice. As all four conditions hold almost surely, the random U n;p is almost surely nice. Let U be nice and let æ 1 ææææ N and æ r 1 ææææn r r denote the sequences of k- values for U and U r respectively and let R x and R x r denote their ç R -classes respectively. èboth exist since the sequences are persistent.è Claim. The values R x and R x r determine the Ehrenfeucht value of nice U. We ærst show that Theorem 2 will follow from the Claim. Let R x ;R x r be any two ç R -classes. Let U be random and consider é ëænë; ç;u é. The sequence of k-values lies in R x with probability P ër x ë+oè1è. The same holds for U r on ëænë. But U r examines U on ëè1, æèn; në soasæ é :5 the values of the ç R -classes are independent and so the joint probability of the values being R x and R x r respectively is P ër x ëp ër x rë+oè1è. Given the Claim é ënë; ç;u é would then have avalue v = vèr x ;R x rè. As X P ërx ëp ër x rë= X P ër x ë X P ër x rë=1æ 1=1 this would give a limiting distribution for the Ehrenfeucht value v on é ënë; ç; U é. Now for the claim. Fix two models M =é ënë; ç;u éand M =é ën ë; ç ;U é, both nice and both with the same values R x ;R x r. Consider the t- move Ehrenfeucht game. For the ærst move suppose Spoiler picks m 2 M. By symmetry suppose m ç n. Let ëi 2 r,1;i r è be one of the k-intervals with, say, :51n ç i r ç :52n. We allow Duplicator a ëfree" move and have him select i r. Let æ 1 ææææ N and æ 1 ææææn be the sequences of k-values for M and M respectively. Let z be the class of æ 1 ææææ r. Since U is nice this sequence already contains A 1 æææa R and hence is persistent soz 2 R x. Let z be the class of æ r+1 ææææ N. By the same argument z is persistent. In M inside of, say, ë:5n; :51në we ænd the block A 1 æææa R. By the universality property we can split this block into a segment inl z and another in R z. Adding more to the left or right doesn't change the nature of this split. Thus there is an interval ëi r,1 ;i r è so that æ 1 ææææ r 2 L z and æ r +1 ææææ N 2 R z. Spoiler plays i r in response to i r. 19

3 4 The Linear Model. We have already remarked in x1 that Zero-One Laws generally do not hold for the linear model é ënë; ç; U é and that P. Dolan has characterized those p = pènè for which they do. Our main object in this section is the following convergence result. Theorem 2. Let k be a positive integer, and S a ærst order sentence. Then there is a constant c = c k;s so that for any p = pènè satisfying we have n, 1 k ç pènè ç n, 1 k+1 lim n!1 PrëU n;p j= Së =c Again we shall æx the quantiæer depth t of S and consider Ehrenfeucht classes with respect to that t. For each æ 2 P k let c æ be the constant deæned in x2 as the limiting probability that a k-interval has k-value æ. Let M be the set of equivalence classes of æp k, a Markov Chain as deæned in x3, and for each ç R -class R x let P ër x ë, as deæned in x3, be the probability that a random sequence æ 1 æ 2 æææeventually falls into R x. In é ënë; ç;u é let æ 1 ææææ N denote the sequence of k-values of the successive k-intervals, denoted ë1;i 1 è; ëi 1 ;i 2 è;..., from 1. Set, with foresight, æ =1,2 3,t. We shall call U on ënë right nice if it satisæes two conditions. The ærst is simply that all the æ 1 ;...;æ N described above are persistent. We know from x2 that this holds almost surely. The second will be a particular universality condition. Let A 1 æææa R be a speciæc sequence in æp k with the property that for every R x and L y there exists a q so that A 1 æææa q 2 L y and A q+1 æææa R 2 R x èwe can ænd such a sequence for a particular choice of R x and L y by taking speciæc sequences in æp k in those classes and concatenating them. The full sequence is achieved by concatenting these sequences for all choices of R x and L y. Note that as some A 1 æææa q 2 L y the full sequence is persistent.è The second condition is that inside any interval ëx; x + ænë ç ë1;në there exist R consecutive k-intervals ëi L ;i L+1 è;...; ëi L+R ;i L+R+1 è whose k-values are, in order, precisely A 1 ;...;A R. We claim this condition holds almost surely. We can cover ë1;në with a ænite numberofintervals ëy; y+ æ në and it 3 suæces to show that almost always all of them contain such a sequence, so it 18

4 cycle to a linear structure a i æææa u a 1 æææa i,1 of the form é ëuë; ç;f é with an Ehrenfeucht value x = x i. Two cycles are equivalent if they yield the same set of values x i 2 M. For every persistent x 2 M let èby è3èè p = p x ;q = s x be such that x = p x +y+s x for all y 2 M. Let P x and S x be æxed sequences èi.e., elements of æaè for these equivalence classes and let R x be the sequence consisting of S x in reverse order followed by P x. If the cycle a 1 æææa u contains R x as a subsequence then selecting a i as the ærst element ofp x gives a linear structure beginning with P x and ending with S x, hence of value p x +y+s x = x. Let R 2 æa be a speciæc sequence given by the concatenation of the above R x for all persistent x 2 M. Then we claim R is a universal sequence in the sense that all a 1 æææa u 2 æa èfor any uè that contain R as a subsequence are equivalent. For every persistent x 2 M there is an a i so that a i æææa i,1 has value x. Conversely every a i belongs to at most one of the R x creating R èmaybe none if a i isn't part of Rè and so there will be an R x not containing that a i. Then in a i æææa i,1 the subsequence R x will appear as an interval. Hence the value of a i æææa i,1 can be written w 1 +x+w 2, which is persistent. That is, the values of a i æææa i,1 are precisely the persistent x and hence the class of a 1 æææa u in the circular t + 1-Ehrenfeucht game is determined. 3.6 Recursion on Cycles. Again let A be a ænite alphabet, M the set of equivalence classes in æa and specify some B ç M. Suppose a cycle a 1 æææa u on A may be decomposed into intervals s 1 ;...;s r with Ehrenfeucht values b 1 æææb r. Then the Ehrenfeucht value of the cycle b 1 æææb r determines the Ehrenfeucht value of a 1 æææa u. The argument is the same as for recursion on intervals. Let a 1 æææa u and a 1 æææa u be decomposed into s 1 æææs r and s 1 æææs r with Ehrenfeucht values b 1 æææb r and b 1 æææb r. Spoiler picks x in some s i. In the game on cycles over B Duplicator can respond b i to b i. Then Duplicator picks an x 2 s i so that he can win the subgame on s i and s i. We apply this is x2 with A = f; 1g and B = P k. Here the æ 2 P k may have more information than the Ehrenfeucht value but this only helps Duplicator. 17

5 A random walk on M, beginning at O, will with probability one eventually reach a minimal closed set R x and then it must stay inr x forever. Let P ër x ë denote the probability that R x is the closed state reached. 3.4 Recursion. Again let A be a ænite alphabet, M the set of equivalence classes of æa and now specify some B ç M. AsB is also a ænite set we can deæne equivalence classes èwith respect to the same constant tè onæb, let M + denote the set of such classes. Now let b 1 æææb u and b 1 æææb u be equivalent sequences of æb. We claim that b b u = b b u as elements of M. Let s 1 ;...;s u ;s 1;...;s u be speciæc elements of æa in the repective b i or b i classes. It suæces to give a strategy for Duplicator with models s s u and s s u. Suppose Spoiler picks an element x in, say, some s i. In the game on æb we know Duplicator has a winning reply to b i of some b i. Now Duplicator will pick some x in s i. To decide the appropriate x in s i to pick Duplicator considers a subgame on s i and s i. As these are equivalent Duplicator will be able to ænd such x for the at most t times that he is required to. This general recursion includes the previous statement that for all j; k ç s and any x 2 M we have jx = kx. Here B = fxg and this says that Duplicator can win the t-move Ehrenfeucht game between a sequence of jx's and a sequence of kx's; that is, that é ëjë; çé and é ëkë; çé are equivalent - a basic result on Ehrenfeucht games. In our argument we apply it inductively with A = P k. We know, inductively, that all k-intervals having the same k-value x 2 P k have the same Ehrenfeucht value. Now the k + 1-interval of i is associated with a sequence x 1 æææx u 2 æp k and a ëtail" y u+1 2 T k. We call two such k + 1-intervals equivalent ifx 1 æææx u and x 1 æææx u are equivalent inæp k and y u+1 = yu.nowx x u = x x u and so the k + 1-intervals have equal Ehrenfeucht value. 3.5 Cycles. Again let M be the set of equivalence classes on æa. Now consider cycles a 1 æææa u èthinking of a 1 following a u è with a i 2 A and consider equivalence classes under the èt + 1è-move Ehrenfeucht game. Here we must preserve the ternary clockwise predicate Cèx; y; zè. Any ærst move a i reduces the 16

6 contain 11 and indeed all blocks of length three. We think of property è3è of persistency as indicating that a persistent sequence is characterized by p, its preæx, and s, its suæx. There are properties such as9 x fèxè =1^:9 y yéx that depend on the left side of the sequence, in this case the value fè1è. There are other properties such as9 x fèxè =1^:9 y xéywhich depend on the right side of the sequence. There will be sequences with values p; s for the left and right side respectively so that the Ehrenfeucht value of the sequence is now determined, regardless of what is placed in the middle. Remarks. Certain sentences Q have the property that if any a 1 æææa u satisæes Q then all extensions a 1 æææa u a u+1 æææa v satisfy Q. The sentence that the ærst term of the sequence is 1 has this property; the sentence that the last term of the sequence is 1 does not have this property. Call such properties unrighteous, as they èroughlyè do not depend on the right hand side of the sequence. Sequences with Ehrenfeucht value in a given R x all have the same truth value for all unrighteous properties. Sequences with Ehrenfeucht value in a given L x would all have the same truth value for all ècorrespondingly deænedè unleftuous properties. 3.3 The Markov Chain. Now consider a probability distribution over A, selecting each a with nonzero probability p a. This naturally induces a distribution over A u, the sequences of length u, assuming each element ischosen independently. This then leads to a distribution over the equivalence classes M. For all u ç, x 2 M let P u èxè be the probability that a random string a 1 æææa u is in class x. OnM we deæne a Markov Chain, for each x the transition probability from x to x + a being p a. In Markov Chain theory the states x 2 M are split into persistent and transient, a state x is persistent if and only if it lies in a minimal closed set. We claim Markov Chain persistency is precisely persistency as deæned by è1è; è2è; è3è. If C is closed and x 2 C then R x ç C and R x is itself closed. If x satisæes è1è then R u = R x for all u = x + y 2 R x so x is Markov Chain persistent. Conversely if x is Markov Chain persistent then R x must be minimal closed so R u = R x for all u = x + y 2 R x and so x satisæes è1è. The Markov Chain M restricted to a minimal closed set R x is aperiodic since x + sa 2 R x and èx + saè +a = x + sa. Hence from Markov Chain theory when x is persistent lim u!1 P u èxè exists. 15

7 exists u 6 with u 5 + u 6 = x. Then R x +èu + u 6 è=fxg so that è3è holds with p = x; s = u + u 6. By reversing addition ènoting that è3è is selfdual while the dual of è1è is è2èè these arguments give that è3è and è2è are equivalent, completing the proof. Let x be persistent and consider v = x+y. Let z be such that x+w+z = x for all w. Then for all wv+w+èz+yè =èx+èy+wè+zè+y = x+y = v and hence v is persistent. Dually,if x is persistent y + x is persistent. Together If x is persistent then w 1 + x + w 2 is persistent for any w 1 ;w 2 2 M. From è1è the relation x ç R u deæned by 9 v èx + v = uè is an equivalence relation on the set of persistent x 2 M. We let R x denote the ç R -class containing x so that R x = fx + v : v 2 Mg From è2è the relation x ç L u deæned by 9 v èv + x = uè is also an equivalence relation on the set of persistent x 2 M. We let L x denote the ç L -class containing x so that L x = fv + x : v 2 Mg. Let x be persistent and let p; s èby è3èè be such that p + z + s = x for all x. Setting z = O, x = p + s. Thus for all z x + z + x =èp + sè+z +èp + sè =p +ès + z + pè+s = x Let R x ;L y be equivalence classes under ç R ; ç L respectively. Then x + y 2 R x ë L y. Let z 2 R x ë L y. Then there exist a; b with x = z + a and y = b + z so that x + y = z +èa + bè+z. But as z is persistent the above argument èwith z as x and a + b as zè gives z +èa + bè+z = z. Thus R x ë L y = fx + yg for all persistent x; y Remarks. Let A = f:1g. A sequence a 1 æææa u is transient if and only if there is a sentence Q of quantiæer depth at most t so that a 1 æææa u fails Q but there is an extension to a 1 æææa u a u+1 æææa v which satisæes Q such that all further extensions a 1 æææa v a v+1 æææa w also satisfy Q. For example, with t = 4, let Q be the existence of a block 11. If a sequence does not satisfy Q then the extension given by adding 11 does satisfy Q and all further extensions will satisfy Q. Thus for a 1 æææa u to be persistent for t = 4 it must 14

8 Example. With A = f; 1g we naturally associate sequences such as 11 with é f1; 2; 3g; ç;f é with fè1è=1;fè2è = ;fè3è = 1. The addition of 11 and 111 is their concatenation èin that orderè The ærst order language has as atomic formulas x ç y, x = y and fèxè =a for each a 2 A. The sentence 9 x 9 y 9 z fèxè =1^ fèyè =^ fèzè =1^ xéy^ yéz is satisæed by 1111 but not by 111 so these are in diæerent equivalence classes with t =3.We could also write that 11 appears as consecutive terms with 9 x 9 y 9 z fèxè =1^fèyè =^fèzè =1^x éy^y éz^:9 w ëèxéw^w éyè_èy éw^w ézèë Informally we would just say 9 x fèxè =fèx +1è= fèx + 2è = 1 but the quantiæer depth is four. 3.2 Persistent and Transient. Deænition and Theorem. We call x 2 M persistent if 8 y 9 z x + y + z = x è1è 8 y 9 z z + y + x = x è2è 9 p 9 s 8 y p + y + s = x è3è These three properties are equivalent. We call x transient if it is not persistent. Proof of Equivalence. è3è è è1è : Take z = s, regardless of y. Then x + y + z =èp + y + sè+y + s = p +èy + s + yè+s = x è1è è è3è : Let R x = fx + v : v 2 Mg. We ærst claim there exists u 2 M with jr x + uj = 1, i.e., all x + y + u the same. Otherwise take u 2 M with jr x + uj minimal and say v; w 2 R x + u. As R x + u ç R x we write v = x + u 1 ;w= x + u 2.From è1è, with y = u 1,wehave x = v + u 3 and thus w = v + u 4 with u 4 = u 3 + u 2. Then w + su 4 = v +ès +1èu 4 = v + su 4 Adding su 4 to R+u sends v; w to the same element sojr+u+su 4 j é jr+uj, contradicting the minimality. Nowsay R x + u = fu 5 g. Again by è1è there 13

9 This sequence contains the sequence A 1 æææa R described in x3.5. But this implies èsee x3.6è that the Ehrenfeuct value is determined, completing the proof. 3 Background. 3.1 The Ehrenfeucht Game. Let A be a æxed ænite alphabet èin application A is P k or f; 1gè and t a æxed positive integer. We consider the space æa of ænite sequences a 1 æææa u of elements of A. We can associate with each sequence a model é ëuë; ç; f é where f :ëuë! A is given by fèiè =a i. For completeness we describe the t-round Ehrenfeucht Games on sequences a 1 æææa u and a 1 æææa u. There are two players, Spoiler and Duplicator. On each round the Spoiler ærst selects one term from either sequencs and then the Duplicator chooses a term from the other sequence. Let i 1 ;...i t be the indices of the terms chosen from the ærst sequence, i q in the q-th round and let i ; 1...i t denote the corresponding indices in the second sequence. For Duplicator to win he must ærst assure that a iq = a i for each q, i.e. that he selects each round the same letter as q Spoiler did. Second he must assure that for all a; b i a éi b, i a éi b and i a = i b, i a = i b èit is a foolish strategy for Spoiler to pick an already selected term since Duplicator will simply pick its already selected counterpart but this possiblity comes in in the Recursion discussed later.è This is a perfect information game so some player will win. Two sequences are called equivalent if Duplicator wins. Ehrenfeucht showed that this is an equivalence class and that two sequences are equivalent if their models have the same truth value on all sentences of quantiæer depth at most t. We let M denote the set of equivalence classes which is known to be a ænite set. æa forms a semigroup under concatenation, denoted +, and this operation ælters to an operation, also denoted +, on M. We use x; y;... to denote elements of M: x + y their sum; O is the equivalence class of the null sequence which acts as identity. We associate a 2 A with the sequence a of length one and its equivalence class èwhich contains only itè,, also called a. We let jx denote x x with j summands. From analysis of the Ehrenfeucht game èsee x3.4è it is known that there exists s èfor deæniteness we may take s =3 t è so that: jx = kx for all j; k ç s; x 2 M 12

10 We have a third claim that is somewhat technical. For any 1ç j ç k let æ 1 ;...;æ u denote the j-values of the successive j-intervals starting at one, where æ u is the last such interval that is in P j.we know that almost surely æ 1 ææææ u is persistent inæp j.weclaim further that almost surely æ 2 æ 3 ææææ u is persistent inæp j. It suæces to show this for any particular j as there are only a ænite number of them. For any integer A we have u, 1 ç A almost surely and the probability that æ 2 ææææ A+1 is transient goes to zero with A so almost surely æ 2 ææææ u is persistent. Let us call ëb; cè a super k-interval èfor a given Uè ifitisak-interval and further for every 1 ç j ç k letting æ 1 ;...;æ u be the successive j-values of the j-intervals beginning at b and stopping with the last persistent value - that then æ 2 æ 3 ææææ u is persistent in æp j. So almost surely the k-interval ë1;i 1 è is a super k-interval. We shall show, for an appropriate sequence A 1 ;...;A R, that all U satisfying the above three claims give models é n;c;u é which have the same Ehrenfeucht value. We ærst need some glue. Call ëa; bè an incomplete k-interval èwith respect to some æxed arbitrary U èifthek-interval beginning at a is not completed by b, 1. Suppose ëa; bè is an incomplete k-interval and ëb; cè is a persistent super k-interval. We claim ëa; cè is a persistent k-interval. The argument is by induction on k. For k =1,ëa; bè must consist of just zeroes while ëb; cè consists of at least s zeroes followed by a one. But then so does ëa; cè. Assume the result for k and let ëa; bè be an incomplete k + 1-interval and ëb; cè be a persistent k + 1-interval. We split ëa; bè into a èpossibly emptyè sequence x 1 ;x 2 ;...;x r of persistent k-intervals followed by èpossibly nullè incomplete k-interval ëa + ;bè with value, say, y. We split ëb; cè èrenumbering for convenienceè into a sequence x r+1 ;...;x s ;y s+1 of k-intervals, all persistent except the last which is transient. Then, by induction, y+x r+1 is a persistent k-interval with some value x r+1. Then ëa; cè splits into k-intervals with values x 1 ;...;x r ;x r+1;x r+2 ;...;x s ;y s+1. By the super-persistency x r+2 æææx s is persistent inæp k and hence èsee x3.2è so is x 1 æææx r x x r+1 r+2 æææx s and therefore ëa; cèisapersistent k + 1-interval. Now let é ënë;c;u é be any model that meets the three claims above, all of which hold almost surely for p in this range. We set i = i = 1 and ænd successive k-intervals ëi ;i 1 è; ëi 1 ;i 2 è;...until ëi u,1 ;i u è and then U on ëi u ;në gives an incomplete k-interval. By the third claim ë1;i 1 è is superpersistent and so the ëinterval" ëi u ;nëëë1;i 1 è ègoing around the cornerè is k-persistent. Hence we have split ënë ènow thinking of it as a cycle with 1 following n è into k-persistent intervals with k-values x 1 ;x 2 ;...;x u. The k-value for x 1 may be diæerent from that for ë1;i 1 è but the others have remained the same. 11

11 probability 1, oè1è so that each of these t-intervals will have length at least æ t p,t with probability at least c t, oè1è. As the lengths are independent with conditional probability at least :99 at least c t æ=2 of the intervals have length at least æ t p,t.thus with probability at least, say :4 the total length L t+1 is at least c t ææ t p,t =2 which isæ t+1 p,èt+1è for an appropriate constant æ t+1, completing the induction. Up to now the relation between p and n, the number of integers, has not appeared. Recall that p! and n!1so that np k!1but np k+1!. Now begin at i = i = 1 and generate the k-interval ëi ;i 1 è. Then generate the k-interval ëi 1 ;i 2 è beginning at i 1 and continue. èwe do this with k æxed. Even if one of the intervals is transient we simply continue with k- intervals. Again we imagine continuing forever through the integers.è Let N be that maximal u for which i u, 1 ç n, so that we have split ënë into Nk-intervals plus some excess. As each sequence of k ones deænitely will end a k-interval N is at least the number of disjoint subintervals of k ones. Simple expectation and variance calculations show that Né:99np k almost surely. On the other side set, with foresight, c =4c,1 k æ,1 k. If N é cnp k then the sum of the lengths of the ærst cnp k k-intervals would be less than n. But these lengths are independent identically distributed variables and each length is at least æ k p,k with probability at least c k so that almost surely at least c k cnp k =2 of them would have length at least æ k p,k and thus their total length would be at least ècc k æ k =2èn én. That is, almost surely C 1 np k énéc 2 np k where C 1 ;C 2 are absolute constants. Let æ 1 ;...;æ N be the k-values of the k-intervals generated by this procedure. Now we make two claims about this procedure. We ærst claim that almost surely none of the æ i are transient. Each æ i has probability ç cp of being transient so the probability that some æ i,1ç i ç C 2 np k is transient is at most ç ècpèc 2 np k =æènp k+1 è=oè1è. And almost surely NéC 2 np k, proving the claim. Let A 1 æææa R be any æxed sequence of elements of P k. The second claim is that almost surely A 1 æææa r appears as a subsequence of the æ sequence, more precisely that almost surely there exists i with 1 ç i ç N, R so that æ i+j = A j for 1 ç j ç R. èfor technical reasons we want the subsequence not to start with æ 1.è As each æ i has a positive probability of being any particular x 2 P k and the æ i are independent and C 1 np k!1almost surely this æxed sequence will appear in the ærst C 1 np k æ's. And almost surely NéC 1 np k, proving the claim. 1

12 Then, as P 1 u= Lè1, p æ è u p æ = L, j PrëW = æë, Lj çæ 1X u=u è1, p æ è u p æ +èl +1è X çuéu è1, p æ è u p æ For æxed U the second sum is oè1è èas p æ! è while the ærst sum is less than æ so the entire expression is less than 2æ for p suæciently small. As æ was arbitrary this gives the claim. Recall that the k + 1-value of the full k + 1-interval is a pair consisting of the Ehrenfeucht value W just discussed and the k-value of the ærst transient type y u+1. The transient type's value has a limiting distribution which is independent ofw, for conditional on any L u the distribution on y u+1 is the same. All possible y 2 T k have a limiting probability d y 2 è; 1è. Hence the probability ofak + 1-value being æ = æy is simply the product of the probabilities and hence approaches a constantifæ, and hence æ, is persistent and is ç cp if æ, and hence æ is transient. This completes the inductive argument for the limiting probabilities of the k-values of the k-intervals. Wenow let L = L k be the length of the k-interval of i and ænd bounds on the distribution of L. A simple induction shows that if the sequence 1 æææ1 of k ones appears after i then the k-interval of i ends with this sequence or possibly before. Thus we get the crude bound so that asymptotically PrëL ékaë é è1, p k è a PrëL éæp,k ë ée,cæ where c is a positive constant. In fact, this gives the correct order of magnitude, L is èspeaking roughlyè almost always on the order of p,k.we claim that there are positive constants æ t ;c t so that PrëL t éæ t p,t ë éc t The argument isby induction, for t = 1 the random variable L 1 is simply the number of trials until a success which occurs with probability p and the distribution is easily computable. Assume this true for t and let èas previously shownè e t p be the asymptotic probability that a t-interval will be transient. Pick f t positive with f t e t é:5. With probability at least :5, the ærst æ = f t p,1 t-intervals after i will be persistent. Conditioning on an interval being persistent is conditioning on an event that holds with 9

13 For any æxed ç uéu we have lim p! fèu; æè=f o X èu; æè so that PçuéU lim fèu; æèè1, pæ è u p æ = cf o èu; æè p! p çuéu With U suæciently large this may be made within æ=2 ofc P 1 f o èu; æè. But this holds for æ arbitrarily small, giving the claimed asymptotics of PrëW = æë. Remark. The rough notion here is that the probability of having a transient k + 1-value is dominated by having few persistent k-intervals and then a transient k-interval. The transient 2-intervals all had at most s persistent 1-intervals. The situation changes with 3-intervals. Recall Ba i consisted of at least s ones each preceeded by at least s zeroes and then two ones i apart. Consider an arbitrarily long grouping of 2-intervals of 2-value Ba i but, say, with none of the form Ba 3, i.e., 11 not appearing and then, say, follow the last one, say Ba 1, with a one so that the 3- interval ends 111. For every u there is a ç c u p probability of this being the 3-interval with u such 2-intervals and c u é but all such 3-intervals would be considered transient since a persistent sequence in æp 2 must surely contain every value in P 2. Now suppose æ is persistent. Again we have the precise formula PrëW = æë = 1X u= fèu; æèè1, p æ è u p æ only this time it is the tail of the sum that dominates. As æ is persistent there is a limiting probability L = lim u!1 f o èu; æè with Lé and furthermore the Mèpè approach M o in the sense that We claim L = lim lim fèu; æè p! u!1 PrëW = æë =L + oè1è For any æé there exist æ and U so that for p ç æ and u ç U we have L, æéfèu; æè él+ æ 8

14 æ 2 P k let p + æ = p æ=è1, p æ è, the conditional probability that V k = æ given that V k is persistent. Note that èas p! è p + æ ç p æ ç c æ Conditioning on L u the x 1 ;...;x u are mutually independent with distributions given by the p + æ. Deæne on M a Markov Chain èsee x3.3è with transition probability p + æ from and æ to æ + æ. We let Mèpè denote this Markov Chain. Observe that the set of states M is independent ofp and the nonzeroness of the transition probabilities is independent ofp 2 è; 1è though the actual transition probabilities do depend on p. There is a particular state O representing the null sequence. Let f èu; æè be the probability of being at state æ at time u, beginning at O at time zero. Then fèu; æè is precisely the conditional distribution for æ given L u. But therefore, letting W denote the Ehrenfeucht equivalence class, PrëW = æë = 1X u= fèu; æèè1, p æ è u p æ Let M o be the Markov Chain on the same set with transition probability c æ from æ to æ + æ and let f o èu; æè be the probability of going from O to æ in u steps under M o. Observe that M o is the limit of Mèpè asp! in that taking the limit of any è1-stepè transition probability inmèpè asp!1 gives the transition probability inm o. Now we need some Markov Chain asymptotics. Assume æ is transient. We claim èrecall p æ ç cpè " 1X è PrëW = æë ç c f o èu; æè p and that the interior sum converges. In general the probability of remaining in a transient state drops exponentially in u so there exist constants K; æ so that f o èu; æè ékè1, æè u for all u giving the convergence. Moreover there exists æ 1 ;æ 2 ;K 1 so that for all épéæ 1 we bound uniformly fèu; æè é K 1 è1, æ 2 è u for all u. Pick æ 3 ç æ 1 so that for épéæ 3 wehave p æ ç 2cp. For any positive æ we ænd U so that for épéæ 3 P 1u=U fèu; æèè1, p æ è u p 1X æ é K 1 è1, æ 2 è u p æ p p ç 2cK 1 è1, æ 2 è U é æ æ 2 2 u=u u= 7

15 persistent state èas deæned in x3.2èin æp 2 ; otherwise we call æ transient and place it in T 3.Ifx 1 æææx u and x 1 æææx u are equivalent as strings in P 2 then x x u and x x u have èas shown in x3.4è the same Ehrenfeucht value. So the 3-value of i determines the Ehrenfeucht value of the 3-interval of i though possibly it has more information. What are the probabilities for the 3-value of i? Again we get a string of 2-values z 1 z 2... whose values are mutually independent and we stop when we hit a transient 2-value. We shall see èin the course of the full induction argumentè that the probability of having 3-value æ is ç c æ for persistent æ and ç c æ p for transient æ. Now let us deæne k-interval and k-value, including the split into persistent and transient k-values by induction on k. Suppose E k ;P k ;T k have been deæned. Beginning at i = i let ëi ;i 1 è be the k-interval and then take succesive k-intervals ëi 1 ;i 2 è;...; ëi u,1 ;i u èuntil reaching a k-interval ëi u ;i u+1 è with transient k-value. Then ëi; i u+1 è is the k + 1-interval of i. èincidentally, suppose Uèiè. Then ëi; i + 1è is the 1-interval of i which is transient. But then ëi; i + 1è is the 2-interval of i and is transient. For all k ëi; i +1èis the k-interval of i and is transient.è Let x 1 ;...;x u ;y u+1 be the succesive k-values of the intervals. Let æ be the equivalence class of x 1 æææx u in æp k. Then i has k + 1-value æ = æy u+1. This value is persistent ifæ is persistent and transient ifæ is transient. This deænes E k+1 ;P k+1 ;T k+1, completing the induction. Our construction has assured that the k-value of i determines the Ehrenfeucht value of the k-interval of i, though it may have even more information. Now let us æx i and look at the distribution of its k-value V k. We show, by induction on k, that for every persistent æ PrëV k = æë =c æ + oè1è while for every transient æ PrëV k = æë =èc æ + oè1èèp. Here each c æ is a positive constant. Assume the result for k and set p æ = PrëV k = æë for all æ 2 E k. Let p æ be the probability that V k is transient so that p æ ç cp, c a positive constant. Let x 1 ;...;x u ;y u+1 be the successive k- values of the k-intervals beginning at i, stopping at the ærst transient value. We can assume these values are taken independently from the inductively deæned distribution on E k. The distribution of the ærst transientvalue is the conditional distribution of V k given that V k is transient P so the probability that it is some transient y is d y + oè1è where d y = c y = c y, the sum over all transient y. Note all d y are positive constants. The key to the argument is the distribution for the Ehrenfeucht equivalence class æ for the ænite sequence x 1 æææx u 2 æp k. Let M be the set of all such equivalence classes. Let L u be the event that precisely u persistent x's are found and then a transient y. Then PrëL u ë=è1, p æ è u p æ precisely. For 6

16 this value the 1-value of i. The probability of the 1-value being any particular a i is ç p while the probability of it being b is ç 1. èall asymptotics are as p!.è We let E 1 denote this set of possible 1-values and we split E 1 = P 1 ë T 1 with P 1 = fbg and T 1 = fa 1 ;...;a s g. The 1-values in T 1 we call 1-transient, the 1-value in P 1 we call 1-persistent. Now èwith an eye toward inductionè we deæne the 2-interval of i = i. Take the 1-interval of i, sayëi ;i 1 è. Then take the 1-interval of i 1,sayëi 1 ;i 2 è. Continue until reaching a 1-interval ëi u ;i u+1 è whose 1-value is 1-transient. èof course, this could happen with the very ærst interval.è We call ëi; i u+1 è the 2-interval of i. Nowwe describe the possible 2-values for this 2-interval. In terms of Ehrenfeucht value we can write the interval as b + b +...+b + a i where there are u èpossibly zeroè b's. Any b +...+b with at least s addends b has èsee x3.4è the same value, call it B. Let jb denote the sum of jb's. We deæne the transient 2-values T 2 as those of the form jb + a i with ç jés and the persistent 2-values P 2 as those of the form B + a i. For example, let t = 5 and s =3 5 = 243. Then i has 2-value 6b + a 22 if, starting at i, six times there are at least 243 zeroes before a one and after the sixth one there are 21 zeroes and then a one. The 2-value is B + a 5 if at least 243 times there are at least 243 zeroes before the next one and the ærst time two ones appear less than 243 apart they are exactly 5 apart. What are the probabilities for i = i having any particular 2-value? The ærst 1-interval ëi ;i 1 è has distribution for 1-value as previously discussed: ç p for each a i and ç 1 for b. Having determined the ærst 1-interval the values starting at i 1 have not yet been examined. Hence the 1-value of the second 1-interval will be independent of the 1-value of the ærst and, in general, the sequence of 1-values will be of mutually independent values. Then the transient 2- values jb+ a i each have probability ç p while the persistent 2-values B + a i will each have probability 1 s + oè1è. We let P 2 denote the set of persistent 2-values, T 2 the set of transient 2-values and E 2 = P 2 ët 2 the set of 2-values. The 3-value will contain all the notions of the general case. Beginning at i = i take its 2-interval ëi ;i 1 è. Then take successive 2-intervals ëi 1 ;i 2 è;...; ëi u,1 ;i u èuntil reaching an interval ëi u ;i u+1 è whose 2-value is transient. The 3-interval for i is then ëi; i u+1 è. Let x 1 ;...;x u ;y u+1 be the 2-values for the successive intervals. From the procedure all x i 2 P 2 while y u+1 2 T 2.Nowconsider èsee x3.1èthe Ehrenfeucht equivalence classes èagain with respect to a t-move gameè over æp 2. èæa is the set of strings over alphabet A.è Let æ be the equivalence class for the string x 1 æææx u, then the 3-value of i is deæned as the pair æ = æy u+1. We let E 3 be the set of all such pairs and we call æ persistent èand place it in P 3 èifæ is a 5

17 t either all models é ëuë;c;u é that contain A 1 æææa R as a subsequence satisfy A or no such models satisfy A. èé ëuë;c;u é contains A 1 æææa R as a subsequence if for some 1 ç j ç u for all 1 ç i ç R we have Uèj + iè if and only if x i = 1, with j + i deæned modulo u.è For pènè in this range é ëuë;c;u é almost surely contains any such æxed sequence A 1 æææa R as a subsequence and hence the Zero-One Law is satisæed. This leaves us with only one case in Theorem 1, and that will be the object of the next section. 2 The Main Case. Here we let k be a positive integer and assume n, 1 k ç pènè ç n, 1 k+1 Our object is to show that pènè satisæes the Zero-One Law for circular unary predicates. We shall let t be a æxed, though arbitrary large, positive integer. We shall examine the equivalence class under the t-move Ehrenfeucht game of the circular model. For the most part, however, we shall examine linear models. We deæne èas Ehrenfeucht didè an equivalence class on models M =é n; ç;u é,two models M; M being equivalent if they satisfy the same depth t sentences or, equivalently, if the t-move Ehrenfeucht game on M; M is won by the ëduplicator". The addition of models èwith M on ënë, M on ën ëwe deæne M +M on ën+n ëè yields an addition of equivalence classes. We shall denote the equivalence classes by x; y;... and the sum by x + y. Results from the beautiful theory of these classes are given in Section 3. Let us consider a random unary function U deæned on all positive integers 1; 2;... and with PrëUèièë = p for all i, these events mutually independent. èin the end only the values of Uèiè for 1 ç i ç n will ëcount" but allowing U to be deæned over all positive integers allows for a ëæctitious play" that shall simplify the analysis.è Now for any starting point i examine i; i +1;... until reaching the ærst j èperhaps i itselfè for which Uèjè. Call ëi; jë the 1-interval of i. èwith probability one there will be such aj; æctitious play allows us to postpone the analysis of those negligible cases when no j is found before j é n.è What are the possible Ehrenfeucht values of é ëi; jë; ç;u é? The model must have a series of zeroes èi.e., :Uè followed by one one èi.e., Uè. There is an s ès =3 t will doè so that all such models with at least s zeroes have the same Ehrenfeuct value. We can write these values as a 1 ;...a s and b èa i having i, 1 zeroes, b having s zeroesè. Call 4

18 or 1, pènè ç n,1 Then pènè satisæes the Zero-One Law for circular unary predicates. Inversely if pènè falls into none of the above categories then it does not satisfy the Zero-One Law for circular unary predicates. The inverse part is relatively simple. Let A k be the sentence that there exist k consecutive elements x 1 ;...;x k 2 U. èx; y are consecutive iffor no z is Cèx; z; yè. For k = 2 this is example C. è Then PrëA k ë is èfor a given nè a monotone function of p. When pènè ç cn,1=k and c a positive constant the probability PrëA k ë approaches a limit strictly between zero and one. èroughly speaking, n,1=k is a threshold function for A k.è Thus for pènè to satisfy the Zero-One law wemust have pènè ç n,1=k or pènè ç n,1=k. Further èreplacing U with :Uè, the same holds with pènè replaced by 1, pènè. For pènè to fall between these cracks it must be in one of the above æve categories. Remark. Dolan ë??ë has shown that pènè satisæes the Zero-One Law for linear unary predicates if and only if pènè ç n,1 or n,1 ç pènè ç n,1=2 or 1, pènè ç n,1 or n,1 ç 1, pènè ç n,1=2.for n,1=2 ç pènè =oè1è he considered the following property: D : 9 x Uèxè^ëUèx+1è_Uèx+2èë^:9 y ëuèyè^ëuèy+1è_uèy+2èë^y éxë^uèx+1è èaddition is not in our language but we write x + 1 as shorthand for that z for which xézbut there is no w with xéwéz.è In our zero-one formulation D basically states that the ærst time we have 11 comes before the ærst time we have 11. This actually has limiting probability :5. This example illustrates that limiting probability for linear unary predicates can depend on edge eæects and not just edge eæects looking at U on a æxed size set 1;...;k or n; n, 1;...;n, k. We defer our results for linear unary predicates to section 4. When pènè ç n,1 the Zero-One Law is trivially satisæes since almost surely there is no x for which Uèxè. Also, if pènè satisæes the Zero-One Law so does 1, pènè. Suppose p = pènè satisæes pènè ç n,æ and 1, pènè ç n,æ for all æé. We show in a section 3 that for every t there is a sequence A 1 æææa R with the property that for any sentence A of quantiæer depth 3

19 directly after n, Cèx; y; zè is the relation that x to y to z goes in a clockwise direction. For any sentence S in this new language we can again deæne PrëU n;p j= Së only in this case Ehrenfeucht's results give a Zero-One Law: for any constant p and sentence S lim n!1 PrëU n;p j= Së=or1 We shall call the ærst language the linear language and the second language the circular language. As a general guide, the circular language will tend to Zero-One Laws while the linear language, because of edge eæects, will tend to limit laws. We shall not restrict ourselves to p constant but rather consider p = pènè as a function of n. Wehave in mind the ëevolution of Random Graphs" as ærst developed by Erdíos and Rçenyi. Here as p = pènè evolves from zero to one the unary predicate evolves from holding for no x to holding for all x. Analogously èbut without formal deænitionè we have threshold functions for various properties. For example, pènè =n,1 is a threshold property for A. When pènè ç n,1 almost surely A fails while when pènè ç n,1 almost surely A holds. In Shelah,Spencer ë??ë we showed that when p = n,æ with æ 2 è; 1è, irrational then a Zero-One Law held for the random graph Gèn; pè and in èluczak, Spencer ë??ë we found a near characterization of those p = pènè for which the Zero-One Law held. The situation with random unary predicates turns out to be somewhat simpler. Let us say p = pènè satisæes the Zero-One Law for circular unary predicates if for every sentence S in the circular language lim PrëU n;pènè j= Së =or1 n!1 Here is our main result. Theorem 1. Let p = pènè besuch that pènè 2 ë; 1ë for all n and either or for some positive integer k or for all æé or for some positive integer k pènè ç n,1 n, 1 k ç pènè ç n, 1 k+1 n,æ ç pènè and n,æ ç 1, pènè n, 1 k ç 1, pènè ç n, 1 k+1 2

20 RANDOM SPARSE UNARY PREDICATES 1 Introduction. Let n be a positive integer, ç p ç 1. The random unary predicate U n;p is a probability space over predicates U on ënë = f1;...; ng with the probabilities determined by PrëUèxèë = p; 1 ç x ç n and the events Uèxè being mutually independent over 1 ç x ç n. Informally, we think of æipping a coin for each x to determine if Uèxè holds, the coin coming up ëheads" with probability p. We shall examine the ærst order language é ënë; ç; U é with equality, a unary predicate U and a binary predicate ç. Examples of sentences in this language are: A : 9 x Uèxè B : 9 x Uèxè ^8 y :yéx C : 9 x;y Uèxè ^ Uèyè ^8 z :ëx éz^ zéyë èé; ç; é are natuarally deænable from ç and equality.è For any such sentence S we have the probability PrëU n;p j= Së While the use of unary predicates is natural for logicians there are two other equivalent formulations that will prove useful. We may think of U as a subset of ënë and speak about i 2 U rather than Uèiè. Second we may associate with U a sequence of zeroes and ones where the i-th term is one if Uèiè and zero if :Uèiè. Thus we may talk of starting at i and going to the next one. We shall use all three formulations interchangably. Ehrenfeucht ë??ë showed that for any constant p and any sentence S in this language lim n!1 PrëU n;p j= Së exists. In the case of sentences A and C the limiting probability is one. But sentence B eæectively states 1 2 U, hence its limiting probability isp. We get around these edge eæects with a new language, consisting of equality, a unary predicate U, and a ternary predicate C. We consider C as a built in predicate on ënë with Cèx; y; zè holding if and only if either xéyéz or yézéxor z é x é y. Thinking of ënë as a cycle, with 1 coming 1