New bounds on Simonyi s conjecture

Transcription

1 New bouns on Simonyi s conjecture Daniel Soltész soltesz@math.bme.hu Department of Computer Science an Information Theory, Buapest University of Technology an Economics arxiv: v1 [math.co] 6 Oct 015 May 17, 018 Abstract We say that a pair (A,B) is a recovering pair if A an B are set systems on an n element groun set, such that for every A,A A an B,B B we have that (A\B = A \B impliesa = A ) ansymmetrically (B\A = B \A implies B = B ). G. Simonyi conjecture that if (A,B) is a recovering pair, then A B n. For the quantity A B thebestknownupperbounis.364 n uetokörneranholzman. In this paper we improve this upper boun to.84 n. Our proof is combinatorial. Keywors sanglass, recovering pair, cancellative 1 Introuction The main subject of this paper is Simonyi s conjecture. Conjecture 1 (Simonyi). [7] Let A an B be set systems on an n element groun set. If for every A,A A an B,B B we have that then A B n. A\B = A \B A = A B \A = B \A B = B A pair of set systems (A,B) that satisfy the conitions of the conjecture is calle a recovering pair. Weefine the size of a pair (A,B)to be the quantity A B.If Conjecture 1 is true, it is best possible as one can take an arbitrary set C [n], an let A be every set containe in C an B be every set containe in the complement of C. The best known upper boun on the size of a recovering pair is.364 n ue to Körner an Holzman [6], for the proof they use the subaitivity of the entropy function. Their proof assumes only that A\B = A \B A = A Research partially supporte by the Hungarian Founation for Scientific Research Grant (OTKA) No

2 B \A = B \A B = B. A pair of set systems satisfying these conitions is calle cancellative. There are cancellative pairs of size larger than n, see [6]. Another weakening of the conition is if we only assume that for the pair (A,B) we have that A \ B = A \ B A = A, but we o not assume the ual conition. The size of such a half recovering pair can also be larger than n see [7]. Conjecture 1 was verifie up to n = 8 in [1]. There is a lattice version of Conjecture 1, which roughly asserts that if instea of the boolean lattice we consier a lattice that is the prouct of chains, a similar construction is optimal. For etails see [7], one can also fin results about the lattice version in [] [8] [4]. There was an unpublishe 1, can question of Aharoni, that instea of the size of the recovering pair, A B = A i A A i B j be also boune by n. This was not conjecture, it was more the quantity A i A of an invitation to prouce a counterexample [5]. In this paper we present such a counterexample, see Corollary. In the case when for a fixe k we have that A i = B j = k for every A i A an B j B an n is sufficiently large, Simonyi an Sali prove that Conjecture 1 is true, see []. Their result is very close to the general case, we will show that the case where A i = B j = c n is equivalent to the general case. This paper is organize as follows. In the first section we present a new combinatorial approach. This is not enough to improve the Körner Holzman boun, but it is very short, it significantly improves the trivial 3 n boun. Here we present the example that answers Aharoni s question. In the secon section we fine tune our approach an introuce a secon upper boun. In the thir section we show that the two bouns combine yiel an improvement of.84 n. Preliminaries an a new proof for a weaker boun Let us first present the easy upper boun of 3 n, an some motivation for Aharoni s question. Claim 1. If (A,B) is a recovering pair, then A B 3 n. Proof. The pairs (A i \B j,b j \A i ) are ifferent since we can recover A i from A i \B j, an B j from B j \A i. But there can be at most 3 n pairs of isjoint sets from [n]. With a slight moification of the above proof, we can prove more. Claim (Aharoni [5]). If (A,B) is a recovering pair, then A B = A i A 1 A i A A i B j 3 n. Proof. The equality an the first inequality is trivial. The last inequality can be proven as follows. For each A i an B j an each subset S of A i B j, the pairs (A i \B j S,B j \A i ) are ifferent, since we can recover B j from B j \A i, then from A i \B j S we can subtract B j an from the result we can recover A i, an if we know both A i an B j, we can easily recover S too. Thus there are at most 3 n such pairs as before an the proof is complete.

3 Thus with a slight refinement of the argument we coul boun A i A A i B j instea A i B j n woul hol, it woul of the size of the recovering pair. Note that if A i A immeiately followthat theonlyrecovering pairsthat have size n aretheonesmentione in the introuction. We present our counterexample in the en of this section. For our new approach, we nee the following efinition. Definition 1. Let us enote by f(n) the maximal number of solutions of the equation A i B j = [n] where A i A an B j B such that the maximum is taken over every recovering pair (A,B) on a groun set of size n. Observation: Note that given a recovering pair, if A i B i = A j B j then A i A j an B i B j otherwise we woul have B i \ A i = B j \ A i or the other way aroun. So for each A i there can be at most one B j such that A i B j = [n]. Thus one can think of the solutions as isjoint pairs (A 1,B 1 ),(A,B ),...,(A f(n),b f(n) ). Lemma 1. f(n) n Proof. The existence of a single pair such that A 1 B 1 = [n] is enough to prove this. All the sets in B must be ifferent on the complement of A 1, an similarly all the sets in A must be ifferent on the complement of B 1. So A B Ac 1 + Bc 1 n hols in every system realizing f(n). Since f(n) is realize by isjoint pairs, f(n) min{ A, B } A B n. Theorem 1. If (A,B) is a recovering pair then A B (1+ ) n. Proof. Let us count the (A i,b j ) pairs accoring to their unions. A B = A i A 1 = C [n] n k=0 ( ) n f(k) k {(A i,b j ) A i A,B j B,A i B j = C} n k=0 ( ) n k = (1+ ) n. k This proof is the starting point of our results. It relies heavily on the estimate of f(n). Before proving our mainresult, let us present a still simple proof for a better upper boun on f(n). For the upper boun we nee the following lemma. Lemma. Let (A,B) be a recovering pair an A 1,A A an B 1,B B. If A 1 B 1 = A B then A 1 B 1 A B. Proof. Suppose to the contrary that A 1 B 1 = A B an A 1 B 1 = A B. We will get a contraiction by observing that A 1 \B = B 1 \A, this is emonstrate in Figure 1 where the first column contains the elements in A 1 \B 1, the secon column contains the elements in A 1 B 1 an the thir one the elements in B 1 \A 1. The meaning of the rows is similar. The fact that we o not nee a complete Venn iagram with four sets follows from A 1 B 1 = A B. The emptiness of four of the areas in the iagram follows from A 1 B 1 = A B. 3

4 A 1 B 1 B x A Figure 1: A 1 \B = x = A \B 1 We will use the following well known estimate of the orer of magnitue of binomial coefficients. Lemma 3. [3] Let k [0,1/] an h(x) = xlog (x) (1 x)log (1 x) the binary entropy function, then we have that 1 8nk(1 k) h(k)n kn i=0 ( ) n h(k)n. i Since h(x) is unimoular, it has two inverses, we will enote the increasing an the ecreasing inverse of h(x) by h 1 i (x) an h 1 (x) respectively. Note that 0 h 1 i (x) 1/ an h 1 (x) = 1 h 1 i (x). In all the cases where we will use Lemma 3, we will only use that ( n kn) h(k)n. Now we are reay to improve the upper boun on f(n). Lemma 4. Let s = s(n) be such that f(n) = sn. Then 0 1 s h 1 i (s), in particular f(n) n. Proof. By Lemma we know that there are at least sn ifferent intersections of type A i B i where A i B i = [n]. Let the pair with the largest such intersection be (A j,b j ), an let m be such that A j B j = mn. Since the sets in A must be ifferent on the complement of B j an similarly the sets in B must be ifferent on the complement of A j we have that sn = f(n) A B Bc j + Ac j = A j B j = (1 m)n From this it follows that m 1 s. If 1/ m we have that s 1/4 an it is easy to check that the statement of the lemma hols as 1 s h 1 i (s) is monotone ecreasing in s an 0 < 1 1/ h 1 i (1/4). If m 1/ we have that mn ( ) n sn h(m)n. i From this it follows that Which implies the inequality i=0 h 1 i (s) m. h 1 i (s) m 1 s Where we have equality if s is approximately 0.455, thus s is smaller than

5 Corollary 1. The boun of Theorem 1 can be improve from (1 + ) n.414 n to.3685 n. Note that this boun is still slightly weaker than that of Körner an Holzman. We present its proof because of its simplicity, an because we feel that the easiest way to improve our results is to provie a better upper boun on f(n). However to emonstrate the limits of this metho, we will show that f(n) is exponential in n. Before proviing a lower boun for f(n), we nee the following lemma, which will be heavily use uring subsequent proofs. Lemma 5 (MultiplyingLemma). Let(A 1,B 1 ) an (A,B ) be recoveringpairs on isjoint groun sets of size n 1 an n respectively. Let (A 3,B 3 ) be a set system on the union of the two groun sets, efine as follows: Then (A 3,B 3 ) is also a recovering pair A 3 := {A i A j A i A 1,A j A } B 3 := {B i B j B i B 1,B j B }. Proof. Each set from the family (A 3,B 3 ) consists of two parts, one from (A 1,B 1 ) an the other from (A,B ). By symmetry it is enough to show that we can recover A k from A k \B l where A k A 3 an B l B 3. From [n 1 ] (A k \B l ) we can recover the part of A k which comes from A 1 since (A 1,B 1 ) is a recovering pair. Similarly from [n ] (A k \B l ) we can recover the part that comes from A since (A,B ) is a recovering pair. Remark. We have that A 3 B 3 = A 1 B 1 A B. It is also true that if there are exactly f(n 1 ) solutions of the equation A i B j = [n 1 ] in (A 1,B 1 ), an exactly f(n ) solutions of A i B j = [n ]in(a,b ),thenthereareexactlyf(n 1 )f(n )solutionsofa i B j = [n 1 +n ] in (A 3,B 3 ). Claim 3. f(6n) 3 n n Proof. It is enough to show that 3 f(6) since multiplying such a pair with itself we get the esire boun. Let us efine the recovering pair (A 6,B 6 ) as follows A 6 := {{1,} c,{3,4} c,{5,6} c } B 6 := {{,3} c,{4,5} c,{6,1} c } It is left to the reaer to verify that this is inee a recovering pair, an that there are three solutions of the equation A i B j = [6] where A i A 6 an B j B 6. Although we mention that it is faster to verify that the complements of the sets in (A 6,B 6 ) satisfy the complementary properties of recovering systems. Corollary. The pair (A 6,B 6 ) := (A 6 { },B 6 { }) is also a recovering pair, an it answers the question of Aharoni negatively since A i B j = 67 > 64 = 6. A i A 6 6 Note that by blowing up the pair (A 6,B 6 ) we get a lower boun on f(n). Thus our knowlege about f(n) can be summarize as /6 lim n (f(n)) 1/n

6 3 The new upper boun 3.1 The combinatorial ieas Now we are aiming to improve the Körner-Holzman boun. We will often multiply a recovering pair with itself, thus let us enote the r-fol prouct of (A,B) with itself by (A r,b r ). First we prove that subexponential factors can be ignore in the upper bouns of A B. Claim 4. If we have that for every recovering pair A B g(n)c n for some c > 1 an a fixe g(n) such that g(n) is subexponential (log(g(n)) = o(n)) then for every recovering pair A B c n. Proof. Suppose that we have a recovering pair (A,B) on a groun set of size n 1 such that A B > c n 1. Let be such that A B = n 1. For a large enough r we have a contraiction by A r B r = n 1r > g(n 1 r)c n 1r. By the following lemma, we can assume some convenient properties of recovering pairs. Definition. Let us call a recovering pair (A,B) uniform if there exists a k such that for any A i A an B j B we have that A i = B j = k, an completely uniform if it is uniform an A = B also hols. Lemma 6. If there exists a c > 1 such that for all n, we have that for any completely uniform recovering pair (A u,b u ) on a groun set of size n we have that A B c n, then for any recovering pair on a groun set of size n we have that A B c n. Proof. For the sake of contraiction assume that we have a recovering pair (A,B) on a groun set of size n 1 such that A B = n 1 > c n 1. Let us remove all but the sets with the most frequent size among the elements of A r an B r to get (A r f,br f ). By the pigeon hole principle, there must be at least A r /(rn 1 ) sets with the most frequent size in A r, an similarly at least B r /(rn 1 ) in B r. Now let us multiply (A r f,br f ) with (Br f,ar f ) to get a completely uniform recovering pair (A r g,bg) r on a groun set of size n 1 r such that which is a contraiction for large r. n 1r (n 1 r) 4 Ar g B r g c n 1r From now on we will assume that the recovering pair (A,B) is completely uniform. To improve the Körner-Holzman boun, we will fine tune the approach in the previous chapter. We will introuce two parameters u,t [0,1] of a recovering pair, that will control its size A r B r. Thus knowing that the size is large, we will gain information about the parameters. Both t an u are functions of the recovering pair, but since it will not cause any confusion we always omit this epenence in the notation. Definition 3. Let u(r) be efine as follows. Take the size of every union A i B j such that A i A r an B j B r. Let u(r) be such that the number that is attaine the most often (if there are more such numbers pick one arbitrarily) among these union sizes be equal to u(r)nr. Let u := lim r u(r). 6

7 It is easy to see that u(r) converges using Hoeffing s inequality [9]. Let X r enote the probability istribution that takes two sets A i,b j from A r an B r uniformly at ranom, an attains the value A i B j /(nr). We can think of u(r) as the moe of X r, an u as the expecte value of X r (or just the expecte value of X 1, it oes not epen on r). Definition 4. Let t(r) be efine as follows. Average the number of solutions of the equations A i B j = C for every set C of size u(r)nr, where A i A r an B j B r. Let t(r) be such that this average be equal to t(r)u(r)nr. Formally t(r) := 1 u(r)nr log C [n] C =u(r)nr {(A i,b j ) A i B j = C,A i A r,b j B r } ) ( nr u(r)nr t := lim r t(r) The limit exists, it is easy to see this using the bouns in the subsequent proof of Theorem. The efinitions are motivate by the following theorem. Theorem. If (A,B) is a recovering pair on a groun set of size n, then A B = (h(u)+ut)n. Proof. A r B r = A r B r = nr C [nr] C =u(r)nr 1 = A i A r r C [nr] {(A i,b j ) A i A r,b j B r,a i B j = C} {(A i,b j ) A i A r,b j B r,a i B j = C} = nr ( ) nr = nr t(r)u(r)nr nr (h(u(r))+t(r)u(r))nr u(r)nr C [nr] C =u(r)nr t(r)u(r)nr = Taking r-th roots an letting r ten to infinity yiels A B (h(u)+tu)n. For the lower boun we work similarly. A r B r = A r B r = C [nr] C =u(r)nr = 1 = A i A r r C [nr] {(A i,b j ) A i A r,b j B r,a i B j = C} {(A i,b j ) A i A r,b j B r,a i B j = C} = ( ) nr t(r)u(r)nr 1 u(r)nr u(r)nr u(r)nr i=0 C [nr] C =u(r)nr ( ) nr t(r)u(r)nr i 1 u(r)nr 8nru(r)(1 u(r)) (h(u(r))+t(r)u(r))nr. t(r)u(r)nr Again taking r-th roots an letting r ten to infinity we establishe the lower boun an the proof is complete. 7

8 Note that the main ieas behin this last proof are essentially the same as there in the proof of Theorem 1, but here we have more information about the recovering pairs with large A B, in terms of u an t. It is trivial that u [0,1] an from the inequality establishe in Lemma 4 it follows that t [0,0.455]. In this region, the function h(u)+tu attains a single maximum, which is by no surprise.3685, but now we know that there is a single choice of parameters u an t at which the function can attain this value. So if we manage to push these parameters away from this location, our upper boun on A B will improve. We will o this by introucing another upper boun for A B in terms of u an t. The basic iea behin the following upper boun is that if for a fixe C there are many solutions of the equation A i B j = C, then among all A 0 A use in these solutions there must be many small ifferences of type A 0 \B. But these must be ifferent for ifferent A A, an there is not enough space for too many small A\B in [n]. Since our recovering pair is uniform, we can fin small ifferences by fining large intersections. Definition 5. Let c be the relative size of the sets in (A,B), i.e. c is such that cn is the size of the sets in A (an also in B as the pair is completely uniform). The size of the sets in the pair (A r,b r ) is exactly cnr. Let us efine m Sr := c u(r) an let the symmetric intersection size be efines as m S = lim r m Sr. Note that m Sr is the size of the intersection of every pair of sets A i,b j, for which A i B j = u(r)nr. We call m Sr the symmetric intersection size, since for a fixe set C 0 of size u(r)nr, the solutions of the equation A i supb j = C 0 come in pairs (A 1,B 1 ), (A,B ),..., (A w,b w ) an among the w possible intersections, m Sr enotes the size of the ones where the inices are the same. These are the smallest intersections. We are aiming to fin pairs with large intersections, but the size of the smallest ones will also play an important role. The next step will be to efine an asymmetric intersection size which will be strictly larger than. To o this first we nee a lemma that roughly states that a large enough proportion of the sets in A an in B is use as a solution of the equation A i B j = C 0 where C 0 is such that there are a lot of solutions of this equation. Definition 6. We say that a set C [nr] of size u(r)nr is crowe if there are at least t(r)u(r)nr 1 solutions to the equation A i B j = C in (A r,b r ). If for a set A 1 A r there is a set B j B r such that A i B j = C we say that A i is use in C. Lemma 7. Among the sets A 0 A r there are at least A r /(nr) ones, such that they are use in a crowe C. Proof. By the efinition of t(r), there are on average t(r)u(r)n solutions for each set of size u(r)nr. Let us forget those (A i,b j ) pairs that have a union C of size u(r)nr such that there are at most t(r)u(r)nr 1 solutions for A i B j = C. Then the average number of solutions can not ecrease to less than t(r)u(r)nr 1. Thus at least half of the pairs which has their union of size u(r)nr are use as a solution for a C that has at least t(r)u(r)nr 1 solutions. Since by the efinition of u(r) an t(r) we have that 1 nr Ar B r ( nr u(r)nr ) t(r)u(r)nr 1 at least 1 nr Ar of the sets in A r have to be use to prouce this many pairs. 8

9 Now we are reay to efine the asymmetric intersection size. Definition 7. For each A j that is use in a crowe C, fix such a C with solutions (A 1,B 1 ),...,(A t(r)u(r)nr 1,B t(r)u(r)nr 1)... Let m Aj r be such that the number that appears the most often among the numbers A j B 1,..., A j B t(r)u(r)n be equal to m Aj rnr (if there are more such numbers, pick one arbitrarily). Let m Ar be the number that appears the most often among the numbers m Aj r where A j is use in a crowe C (if there are more such numbers, pick one arbitrarily). Finally we efine the asymmetric intersection size m A as m A := liminf r m Ar. Note that we o not know anything about the convergence of m Ar. Our subsequent arguments work if we choose any accumulation point of the sequence m Ar instea of the smallest one. Later we will prove lower bouns of m S an m A, but we will not nee them before we are trying to quantify our results. We woul like to fin large intersections, to have small ifferences. Our last lemma before the proof of the secon upper boun roughly says that for a set A 0 A we not only have intersections of size m A, but there are exponentially many ifferent such intersections, forcing exponentially many ifferent ifferences. Lemma 8. Let A 1 A r be such that it is use as a solution in a crowe C. Then there are at least (t(r)u(r) m A 1 r+m Sr )nr 1 ifferent intersections of size m A1 r of type A 1 B. Proof. There are t(r)u(r)nr 1 pairs (A i,b i ) such that their union is C, we will use only these sets. Among these B i, let B i1,...,b ik enote those that have the same intersection I of size m A1 r with the set A 1. Consier the recovering pair that consists of these B ik, an the corresponing A ik for which B ik A ik = C. We claim that the system A := {A ik \(A 1 \I) k [K]} B := {B ik k [K]} is a recovering pair, on (u(r) c+m A1 r)nr elements, since the set A 1 \I is isjoint from all B ik an is containe by all A ik. Every A ik must be ifferent on the complement of B i1, thus K (u(r) c+m A 1 r c)nr = (m A 1 r m Sr )nr an the proof is complete. Theorem 3. If (A,B) is a completely uniform recovering pair, then ( ) n A B (m A tu m S )n (h(c m A)+m A tu m S ). (c m A )n Proof. Since the pair is completely uniform it is enough to boun A. By Lemma 7 we have that A r = A r nr {A i A i A,A i is use in a crowe set} n r {A i A i A,A i is use in a crowe set,m Ai r = m Ar }. By Lemma 8, every such A i has at least (t(r)u(r) m Ar+m Sr )nr 1 ifferent intersections of size m Ar nr. Thus every such A i has at least (t(r)u(r) m Ar+m Sr )nr 1 ifferent ifferences of 9

10 size (c m Ar )nr, which must be ifferent by the efinition of a recovering pair. Thus we have that n r {A i A i A,A i is use in a crowe set,m Ai r = m Ar } ( ) n r nr (t(r)u(r)+m Ar m Sr )nr+1. (c m Ar )nr Taking r-th roots an letting r ten to infinity finishes the proof. Note that the proof of Theorem 3 is rather straightforwar once we have the appropriate efinitions. Intuitively speaking we feel that the whole argument is about that we woul like to have a large subset of sets in A such that they have small ifferences with some set from B, an since all these must be ifferent, there can not be too many of them, as there is not enough space. It is not the proof of Theorem 3 that is important, but the fact that we can prove effective bouns on the parameters use there. To show that Theorem an Theorem 3 together improve on the Körner-Holzman boun, we will present lower bouns on m S an m A in the next section. 4 Quantifying the upper boun To improve the Körner-Holzman boun, we nee the following lower bouns on m S an m A. Lemma 9. h 1 i (t)u m S. Proof. Since the average number of solutions of A i B j = C where A i,b j are from A,B an C = u(r)n is t(u)u(r)n, there must be at least one set C such that there are at least t(u)u(r)n solutions. Let us fix such a C. By Lemma, for each solution (A i,b i ) of the equation A i B j = C, we have that A i B i is unique. Thus ( ) ( ) u(r)n u(r)n t(u)u(r)n = m m Sr n Sr u(r)n h(m Sr u(r))u(r)n u(r) an by taking r to infinity we obtain an the proof is complete. t(u)u(r) h ( ) msr u(r) u(r) h 1 i (t(u))u(r) m Sr h 1 (t(u))u(r) h 1 i (t)u m S Lemma 10. ( ) tu c h 1 () m A 10

11 Proof. Fix a set A j that is use in a crowe C. Since all the sets B 1,...,B t(r)u(r)nr 1 that areuse in C areifferent outsie of A j, they must iffer on a set of size (u(r) c)nr. The most frequent size among B 1 \A j,..., B t(r)u(r)nr 1\A j is by the efinition of m Aj r equal to (c m Aj r)nr. Thus ( ) (u(r) c)nr t(r)u(r)nr 1 (u(r) c)nr (u(r) c)nr h (c m Aj r)nr for all r an all A j thus tu h ( ) c ma () ( c maj r u(r) c hols. After a simple rearrangement, the statement of the lemma follows: ( ) ( ) tu tu h 1 i () c m A h 1 () ( ) tu c h 1 () m A ) (u(r) c)nr Now we procee with the proof of the following claim, that finishes our proof, that for any recovering pair A B.84 n. Claim 5. For any u [0,1] an t [0,0.455] we have that or equivalently min{ (h(u)+tu)n, (h(c m A)+m A tu m S )n }.84 n h(u)+tu 1.19 or h(c m A )+m A tu m S Note that the boun.84 can be improve to.815 by using more avance computer calculations. Here we only present a proof of.84 which uses a computer only to evaluate a function. Claim 6. For any a fixe t [0,0.455] we have that for u (1 + t ) 1 the function h(u)+tu is monotone increasing in u an for u (1+ t ) 1 is is monotone ecreasing in u. Proof. The erivative of h(x) is log ( x ). For fixe t the function h(u)+tu is unimoular, since the erivative of h(u) is monotone ecreasing in the interval u [0,1]. For 1 x fixe t, the maximum of h(u)+tu is at u = (1+ t ) 1. We will call h(u)+tu the first an h(c m A )+m A tu m S the secon boun. Since the first boun is a function of two variables, an the secon is a function of four (since c = (u+m S )/) we are aiming to eliminate m A an m S from the secon boun. Then we will establish certain monotonicity properties of these bouns, such that evaluating them on 16 places will yiel our claim. Note that evaluating them on more places woul yiel better bouns, but the improvement is in the thir ecimal igit. We start our work with narrowing the range of parameters using the first boun. 11

12 Claim 7. Outsie of the rectangle u [0.4400,0.7100]ant [0.3600,0.455] we have that h(u)+tu Proof. The function h(u)+tu is trivially increasing in t an unimoular in u. For t = 0.36 by Claim 6 its maximum (as a function of u) is attaine at 1/( ) an it is less than 1.19 so when t 0.36 it is smaller than 1.19 for every value of u. For t = it is less than 1.19 outsie of the interval u [0.4400,0.7100] an we are one by unimoularity in u an monotonicity in t. From now on we will assume that u [0.4400,0.7100] an t [0.3600,0.455]. The following lemma will be useful in the proofs that m A an m S shoul be minimize. Lemma 11. ( ) tu A(u,t,m S ) := h 1 u ms u m S < 1/3 Proof. Since h 1 (x) is ecreasing, A(u,t,m S) is trivially monotone ecreasing in m S so after using h 1 i (t)u m S we get ( ) A(u,t,m S ) h 1 t u(1 h 1 i (t)) 1 h 1 i (t) which is trivially monotone ecreasing in t an monotone increasing in u so substituting u = an t = an checking that ( ) h (1 h 1 i (0.36)) 1 h 1 < 1 i (0.36) 3 finishes the proof. We procee with showing that in the secon boun m A shoul be minimize. Note that for this it is Claim 8. For fixe u [0.44,0.71] an t [0.36,0.455] an m S, the boun h(c m A )+ m A tu m S is maximal if m A is minimal. Proof. To maximize h(c m A )+m A tu m S we have to minimize m A if an only if its erivative with respect to m A is negative. Thus we nee that h (c m A )+1 0 which, since h (x) is ecreasing an it attains one at x = 1/3, is equivalent to c m A 1/3. By rearranging the statement of Lemma 10 we have that ( ) tu c m A h 1 () = which we can rewrite using = (u m S )/ to get ( ) tu = h 1 u ms u m S which is smaller than 1/3 by Lemma 11. 1

13 Thus the secon boun became ( ( ) ) tu h h 1 () ( ( ) ) tu h h 1 u ms u m S ( tu +c h 1 + u m S h 1 ) () tu m S = ( tu u m s ) u ms tu = h(a(u,t,m S )) A(u,t,m S )+ u m S Now we show that m S shoul be minimize. tu. Claim 9. For fixe u,t the boun h(a(u,t,m S )) A(u,t,m S )+ u m S tu is maximal if m S is minimal. Proof. The u m S tu part is monotone ecreasing in m S, for the h(a(u,t,m S )) A(u,t,m S ) part observe that A(u,t,m S ) is monotone ecreasing in m S an the function h(x) x is increasing if x 1/3 which inequality is guarantee by Lemma 11. h So the secon boun takes the form ( ) ) t u 1 h 1 i (t) (1 h 1 i (t)) ( h 1 h 1 ( ) t u 1 h 1 i (t) (1 h 1 i (t))+ u (1 h 1 i (t) t). From the proof of Claim 9 we see that the secon boun is monotone ecreasing in t, an is monotone increasing in u since the positivity of 1 h 1 i (t) t is guarantee by Lemma 4. Note that h(u)+tu is trivially increasing in t. By the monotonicity properties of our bouns, evaluating the first an the ouble of the secon boun alternately (starting with the secon) in the points below, we can euce that for any u [0.4400,0.7100] an t [0.3600,0.455] one of the bouns is smaller than.84. Note that if we connect the points, the resulting shape resembles a staircase. The points (u i,t i ) are: (0.5893,0.36), (0.5893, 0.364),(0.599, 0.364),(0.599, 0.367),(0.607, 0.367),(0.607, 0.37),(0.615, 0.37), (0.615, 0.374),(0.67, 0.374),(0.67, 0.38),(0.645, 0.38),(0.645, 0.39),(0.688, 0.39), (0.688, 0.43),(0.71, 0.43),(0.71, 0.455). 5 Concluing remarks an open problems Note that in the last section we prove that for a certain range of parameters, A B < (h(u)+tu)n (where the secon boun is stronger than the first one) which contraicts Theorem showing that this range of parameters is impossible to achieve by a recovering pair. Note that there are completely uniform recovering pairs with parameters u [0, 1] an t = 0, but we o not know whether t > 0 is possible at all. From a proof that t > 0 is impossible, or a proof that presents an upper boun that contraicts Theorem for a large enough range of parameters u an t, woul follow Simonyi s conjecture. A less ambitious way of improving our results woul be to improve the upper boun on f(n). Question 1. lim n (f(n))1/n =? 13

14 We know that this quantity is in the interval [1.009,1.3685], we call this less ambitious, as the lower boun shows that this metho can not prove Simonyi s conjecture without aitional ieas. Intuitively speaking, f(n) prevents the concentration of the unions on a single set. It woul be interesting to have a lemma that prevents the concentration of the unions on some sets close to each other. Another interesting way to prevent the concentration of sets is to punish large intersections, the sum propose by Aharoni oes exactly this. Its asymptotic exponent woul also be of inepenent interest. Question. What is the maximal asymptotic exponent of the sum where (A,B) is a recovering pair? A i A A i B j We know that this quantity is somewhere in the interval [.0153, 3]. Let us finish with a question that is more of an invitation to prouce a counterexample. Can t be larger than zero? From a negative answer to this question woul immeiately follow Conjecture 1, an a positive answer woul be interesting too, since in any counterexample to Conjecture 1, t is necessarily positive. 6 Acknowlegements The author woul like to thank Gábor Simonyi, Márton Zubor an Ron Aharoni for stimulating iscussions. References [1] Dustin Styner, A Collection of Results on Simonyi s Conjecture, MSc thesis, 01. [] Gábor Simonyi an Attila Sali, Recovering Set Systems an Graph Entropy, Combinatorics, Probability an Computing 6 (1997), [3] Imre Csiszár an János Körner, Information Theory: Coing Theorems for Discrete Memoryless Systems, Cambrige University Press, Cambrige, 011. [4] Rita Csákány, Some results on the sanglass conjecture, Electronic Notes in Discrete Mathematics 5 (000), [5] Ron Aharoni, Private communication (015). [6] Ron Holzman an János Körner, Cancellative Pairs of Families of Sets, European Journal of combinatorics 16 (1995), [7] Ruolf Ahlswee an Gábor Simonyi, Note on the optimal structure of recovering set pairs in lattices: the sanglass conjecture, Discrete Mathematics 18 (1994), [8] Shinnyih Huang an Hoa Bikhori, Strongly cancellative an recovering sets on lattices, arxiv: [math.co] (1997). [9] W. Hoeffing, Probability inequalities for sums of boune ranom variables, Journal of the American Statistical Association 58 (1963),