VC-Dimension of Sets of Permutations - Semantic Scholar
VC-Dimension of Sets of Permutations Ran Raz Weizmann Institute Abstract We de ne the VC-dimension of a set of permutations A Sn to be the maximal k such that there exist distinct i1 ; :::; ik 2 f1; :::; ng that appear in A in all possible linear orders, that is, every linear order of fi1 ; :::; ik g is equivalent to the standard order of f(i1 ); :::; (ik )g for at least one permutation 2 A. In other words, the VC-dimension of A is the maximal k such that for some i1 ; :::; ik the restriction of A to fi1 ; :::; ik g contains all possible linear orders. This is analogous to the VC-dimension of a set of strings. Our main result is that there exists a universal constant C such that any set of permutations A Sn with VC-dimension 2 is of size < C n. This is analogous to Sauer's lemma for the case of VC-dimension 2. One corollary of our main result is that any acyclic set of linear orders of f1; :::; ng is of size < C n, (a set A of linear orders on f1; :::; ng is called acyclic if no 3 elements i; j; k 2 f1; :::; ng appear in A in all 3 orders (i; j; k), (k; i; j ) and (j; k; i)). The size of the largest acyclic set of linear orders has interested researchers for many years because it is the largest number of linear orders of n alternatives such that the following is always satis ed: if each one of a set of voters chooses one of these orders as his preference then the majority relation between each two alternatives is transitive.
1 Introduction The VC-dimension of a set A f0; 1gn is the maximal k such that for some distinct i1; :::; ik 2 f1; :::; ng the restriction of A to fi1; :::; ik g contains all 2k possible assignments. This de nition, rst given by Vapnik and Chervonenkis in 1971 [VC], has become a central de nition in combinatorics and in certain areas of theoretical computer science. The following lemma, known as Sauer's lemma, was proven independently by Sauer [Sau], Perles and Shelah, and in a slightly weaker form by Vapnik and Chervonenkis. [email protected], Department of Applied Mathematics, Weizmann Institute, Rehovot 76100, ISRAEL
1
Lemma 1.1 For every n and every A f0; 1gn with VC-dimension k k n! X : jAj i=0 i
Lemma 1.1 gives a tight bound for the size of any set with VC-dimension k. The lemma has numerous of applications in combinatorics and in theoretical computer science. In this paper we extend the de nition of VC-dimension to sets of permutations and we prove that every set of permutations with VC-dimension 2 is of size < C n (for some universal constant C ). This bound is tight (up to determining the exact constant), and is analogous to Lemma 1.1 for k = 2. Denote by Sn the set of all permutations on f1; :::; ng. A permutation 2 Sn corresponds to a linear order of the n elements 1; :::; n in the following way: for every i; j we say that i 20). That's OK because we can assume that C0 20. The proof of the lemma is deferred to the next section. Lemma 3.1 For every n and every A Sn, if v(A) > 20 then there exist two indices s t, a number m (1=16) v(A) and a set I [1; n] with jI j = m such that for every i 2 I : 1. 2. 3. 4.
jA(i)j (1=2) v(A) jA(i) \ [1; s]j (1=8) jA(i)j jA(i) \ [t; n]j (1=8) jA(i)j jA(i) \ [s; t]j (1=68) v(A)3=2 m?1
Fix s; t; m; I from Lemma 3.1. For simplicity we assume that m is larger than some xed constant (say m > 20). That's OK because we can assume that C0 is large enough. Note that so far we have not used the assumption that the VC-dimension of A is 2. This assumption will be used now. For every i 6= j , denote by Ri;j the set of all k 2 [n] n fi; j g such that the projection of A on the 3 elements i; j; k does not contain the linear order (i; k; j ), i.e., the set of all k 2 [n] n fi; j g such that no permutation 2 A satis es (i) < (k) < (j ). For i = j we de ne Ri;j to be the empty set. Since the VC-dimension of A is 2, no triple (i; j; k) appears in all 6 possible linear orders, and hence every triple (i; j; k) contributes 1 to at least one of the 6 sets: Ri;j ,Rj;i,Ri;k ,Rk;i,Rj;k ,Rk;j . Therefore, ! X n jRi;j j 3 : i;j In the same way, we can restrict ourselves to I . We de ne Ri;j (I ) = Ri;j \ I and we get ! X m jRi;j (I )j 3 > (1=8) m3 : (i;j )2I I Now de ne
J = f(i; j ) 2 I I j jRi;j (I )j (1=16) mg; that is, J is the set of all pairs (i; j ) 2 I I such that jRi;j (I )j is not too small. Since for every i; j we have jRi;j (I )j < m, we can conclude X X X (1=8) m3 < jRi;j (I )j = jRi;j (I )j + jRi;j (I )j m2 (1=16) m + jJ j m; (i;j )2I I
that is,
J
(I I )nJ
jJ j > (1=16) m2 : 8
We are now ready to do the second case of the induction.
The second case: Claim 3.4 If there exists (i; j ) 2 J such that PROB2RA [((i) 2 [1; s]) ^ ((j ) 2 [t; n])] 2?10 v(A)?2 then
0 v(A)n 1?1 Y jAj (C0)n @ (r=n)A : r=C0 n+1
Proof:
In this case, we de ne A0 by A0 = f 2 A j ((i) 2 [1; s]) ^ ((j ) 2 [t; n])g; that is, we restrict (i) to be in [1; s] and (j ) to be in [t; n]. We will show that A0 satis es the requirements of Claim 3.2. For every k 2 Ri;j (I ), the projection of A on the 3 elements i; j; k does not contain the linear order (i; k; j ). Therefore after we restrict (i) to be in [1; s] and (j ) to be in [t; n], we cannot have (k) 2 [s; t]. Hence, for every k 2 Ri;j (I ), jA0(k)j jA(k)j ? jA(k) \ [s; t]j jA(k)j ? (1=68) v(A)3=2 m?1 : Since (i; j ) 2 J , jRi;j (I )j (1=16) m: Hence, n n X X 0 jA (k)j jA(k)j ? (1=16) m (1=68) v(A)3=2 m?1: That is,
k=1
k=1
d def = v(A) n ? v(A0) n (1=1088) v(A)3=2 : Since we assumed that PROB2RA[((i) 2 [1; s]) ^ ((j ) 2 [t; n])] 2?10v(A)?2, we have jA0j jAj 2?10v(A)?2; that is, p def = jAj=jA0j 210v(A)2 : and since by v(A) C0 we can assume that v(A) is large enough (say v(A) 210 ) we have ln(p) ln(v(A)3) = 3 ln(v(A)): Thus A0 satis es the requirements of Claim 3.2. 2 We will now assume that the requirements of Claim 3.3 and Claim 3.4 are not satis ed. This will be the third case of the induction.
The third case:
9
Claim 3.5 If for every i 2 f1; :::; ng and k 2 A(i), PROB2R A [(i) = k] v(A)?2 ; and for every (i; j ) 2 J , PROB2RA[((i) 2 [1; s]) ^ ((j ) 2 [t; n])] < 2?10 v(A)?2 then
0 v(A)n 1?1 Y jAj (C0)n @ (r=n)A : r=C0 n+1
Proof:
For every i 2 I de ne
Ji = fj 2 I j (i; j ) 2 J g: Since jJ j > (1=16) m2 , there exists i0 such that
jJi j > (1=16) m (1=256) v(A): 0
W.l.o.g assume that
jJi j = d(1=256) v(A)e; 0
(otherwise discard arbitrarily some elements from Ji0 ). Here we de ne A0 by
A0 = f 2 A j ((i0 ) 2 [1; s]) ^ (for every j 2 Ji0 ; (j ) 62 [t; n])g; We will show that A0 satis es the requirements of Claim 3.2. Since jJi0 j (1=256) v(A), and since for every j 2 Ji0 ,
jA(j )j ? jA0(j )j jA(j ) \ [t; n]j (1=8) jA(j )j (1=16) v(A); we have That is,
n X j =1
jA0(j )j
n X j =1
jA(j )j ? (1=256) v(A) (1=16) v(A):
d def = v(A) n ? v(A0) n (2?12 ) v(A)2: Let us now turn to bounding jAj=jA0j. Since
jA(i0 ) \ [1; s]j (1=8) jA(i0)j (1=16) v(A); and since for every k 2 A(i0 ), PROB2RA[(i0 ) = k] v(A)?2; 10
we have
PROB2RA[(i0 ) 2 [1; s]] (1=16) v(A) v(A)?2 = (1=16) v(A)?1: Also, since for every (i; j ) 2 J PROB2RA[((i) 2 [1; s]) ^ ((j ) 2 [t; n])] < 2?10 v(A)?2;
we have for every j 2 Ji0 , PROB2RA[((j ) 2 [t; n]) j ((i0 ) 2 [1; s])] =
Therefore
PROB2R A[((j ) 2 [t; n]) ^ ((i0 ) 2 [1; s])] = PROB2RA[(i0 ) 2 [1; s]] < 2?10 v(A)?2 = [(1=16) v(A)?1] = 2?6v(A)?1:
PROB2RA[(9j 2 Ji0 s.t., (j ) 2 [t; n]) j ((i0) 2 [1; s])] < jJi0 j 2?6v(A)?1 = d(1=256) v(A)e 2?6v(A)?1 < 2?13; (where we assume again that v(A) is large enough). We can conclude that
jA0j = jAj PROB2RA[(i0 ) 2 [1; s]] PROB2RA[(8j 2 Ji ; (j ) 62 [t; n]) j ((i0) 2 [1; s])] jAj (1=16) v(A)?1 (1 ? 2?13 ) > jAj (1=32) v(A)?1; 0
That is
p def = jAj=jA0j < 32 v(A) < v(A)2;
(for large enough v(A)). Thus A0 satis es the requirements of Claim 3.2.
2
4 Proof of the Technical Lemma In this section we give the proof of Lemma 3.1. De ne
I0 = fi 2 [n] j jA(i)j (1=2) v(A)g: Obviously,
X I0
jA(i)j =
X [n]
jA(i)j ?
X [n]nI0
jA(i)j = n v(A) ?
X [n]nI0
n v(A) ? n (1=2) v(A) = (1=2) n v(A):
The set I will be de ned to be a subset of I0. For every i 2 I0, let si ti be two indices such that 11
jA(i)j
1. jA(i) \ [1; si]j (1=4) jA(i)j; 2. jA(i) \ [ti; n]j (1=4) jA(i)j; 3. jA(i) \ [si; ti]j (1=2) jA(i)j; and de ne
Ag (i) = A(i) \ [si; ti]: X g X jA(i)j (1=2) jA(i)j (1=4) n v(A):
Then
I0
I0
We will de ne the indices s; t to be such that
j[s; t]j = d(1=16) v(A)e: Since [n] can be covered by d16n=v(A)e such intervals, we will be able to nd such s; t that
also satisfy
X g jA(i) \ [s; t]j (1=4) n v(A)=d16n=v(A)e I0
(1=4) n v(A)=(17n=v(A)) = (1=68) v(A)2: We can now de ne I1 I0 by I1 = fi 2 I0 j jAg (i) \ [s; t]j > 0g; and get
X I1
jA(i) \ [s; t]j
X g X jA(i) \ [s; t]j = jAg (i) \ [s; t]j (1=68) v(A)2: I1
I0
The set I will be de ned to be a subset of I1. Every i 2 I1 already satis es the rst 3 requirements of the lemma: 1. By the de nition of I0,
jA(i)j (1=2) v(A): 2. By the de nition of I1 we have jAg (i) \ [s; t]j > 0, and therefore by the de nition of Ag (i) we can conclude that si t. Hence, jA(i) \ [1; s]j jA(i) \ [1; si]j ? j[s + 1; t]j (1=4) jA(i)j ? d(1=16) v(A)e + 1 (1=4) jA(i)j ? (1=16) v(A) (1=4) jA(i)j ? (1=8) jA(i)j = (1=8) jA(i)j: 3. In the same way,
jA(i) \ [t; n]j (1=8) jA(i)j: 12
Since I will be a subset of I1 , every i 2 I will satisfy requirements 1,2,3. Thus we only have to make sure that I is large enough and that every i 2 I satis es requirement 4. We will now split the proof into two cases:
Case A:
jI1j (1=68) v(A)3=2 :
In this case we de ne I = I1 , and since we assumed that v(A) > 20 we have
m = jI1j (1=68) v(A)3=2 (1=16) v(A): For every i 2 I1 requirement 4 is satis ed by jA(i) \ [s; t]j jAg (i) \ [s; t]j 1 (1=68) v(A)3=2 m?1 :
Case B:
jI1j < (1=68) v(A)3=2:
In this case we de ne I by
I = fi 2 I1 j jA(i) \ [s; t]j (1=2) v(A)1=2 g: Then
X I
But for every i
jA(i) \ [s; t]j = X I1
X I1
X I1
jA(i) \ [s; t]j ?
X I1 nI
jA(i) \ [s; t]j
jA(i) \ [s; t]j ? jI1j (1=2) v(A)1=2
jA(i) \ [s; t]j ? (1=68) v(A)3=2 (1=2) v(A)1=2
(1=68) v(A)2 ? (1=136) v(A)2 = (1=136) v(A)2: jA(i) \ [s; t]j j[s; t]j = d(1=16) v(A)e (2=17) v(A);
(where we used again the assumption v(A) > 20). Hence,
jI j (2=17) v(A)
X I
jA(i) \ [s; t]j (1=136) v(A)2;
that is
m = jI j (1=16) v(A): Requirement 4 is satis ed for every i 2 I , since by the de nition of I , jA(i) \ [s; t]j (1=2) v(A)1=2 (1=68) v(A)3=2 m?1 :
2 13
Acknowledgments I would like to thank Mario Szegedy for introducing me to the acyclic linear orders problem and for many important discussions that led to this result. I would like to thank Peter Fishburn for the encouragement in writing up this manuscript.
References [Abe] [Bla] [Con] [Fis] [Rik] [Sau] [VC]
J.M. Abello, \The weak Bruhat order of S , consistent sets, and Catalan numbers", SIAM Journal of Discrete Math., 4, (1991), 1{16. D.J. Black, \The theory of committees and elections", Cambridge Press, London, 1958. Marquis de Condorcet, \Essai sur l'application de l'analyse a la probabilite des decisions rendues a la pluralite de voix", Paris, 1785. P. Fishburn, \Acyclic sets of linear orders", Social Choice and Welfare, 14, (1997), 113{124. W.H. Riker, \Arrow's theorem and some examples of the paradox of voting", Foundation monograph, Southern Methodist University Press, Dallas, 1961. N. Sauer, \On the density of families of sets", Journal of Combinatorial Theory, Series A 13, (1972), 145{147. V.N. Vapnik and A.Ya Chervonenkis, \On the uniform convergence of relative frequencies of events to their probabilities", Theory of Probability Applications, 16, (1971), 264{280.
14
1 Introduction The VC-dimension of a set A f0; 1gn is the maximal k such that for some distinct i1; :::; ik 2 f1; :::; ng the restriction of A to fi1; :::; ik g contains all 2k possible assignments. This de nition, rst given by Vapnik and Chervonenkis in 1971 [VC], has become a central de nition in combinatorics and in certain areas of theoretical computer science. The following lemma, known as Sauer's lemma, was proven independently by Sauer [Sau], Perles and Shelah, and in a slightly weaker form by Vapnik and Chervonenkis. [email protected], Department of Applied Mathematics, Weizmann Institute, Rehovot 76100, ISRAEL
1
Lemma 1.1 For every n and every A f0; 1gn with VC-dimension k k n! X : jAj i=0 i
Lemma 1.1 gives a tight bound for the size of any set with VC-dimension k. The lemma has numerous of applications in combinatorics and in theoretical computer science. In this paper we extend the de nition of VC-dimension to sets of permutations and we prove that every set of permutations with VC-dimension 2 is of size < C n (for some universal constant C ). This bound is tight (up to determining the exact constant), and is analogous to Lemma 1.1 for k = 2. Denote by Sn the set of all permutations on f1; :::; ng. A permutation 2 Sn corresponds to a linear order of the n elements 1; :::; n in the following way: for every i; j we say that i 20). That's OK because we can assume that C0 20. The proof of the lemma is deferred to the next section. Lemma 3.1 For every n and every A Sn, if v(A) > 20 then there exist two indices s t, a number m (1=16) v(A) and a set I [1; n] with jI j = m such that for every i 2 I : 1. 2. 3. 4.
jA(i)j (1=2) v(A) jA(i) \ [1; s]j (1=8) jA(i)j jA(i) \ [t; n]j (1=8) jA(i)j jA(i) \ [s; t]j (1=68) v(A)3=2 m?1
Fix s; t; m; I from Lemma 3.1. For simplicity we assume that m is larger than some xed constant (say m > 20). That's OK because we can assume that C0 is large enough. Note that so far we have not used the assumption that the VC-dimension of A is 2. This assumption will be used now. For every i 6= j , denote by Ri;j the set of all k 2 [n] n fi; j g such that the projection of A on the 3 elements i; j; k does not contain the linear order (i; k; j ), i.e., the set of all k 2 [n] n fi; j g such that no permutation 2 A satis es (i) < (k) < (j ). For i = j we de ne Ri;j to be the empty set. Since the VC-dimension of A is 2, no triple (i; j; k) appears in all 6 possible linear orders, and hence every triple (i; j; k) contributes 1 to at least one of the 6 sets: Ri;j ,Rj;i,Ri;k ,Rk;i,Rj;k ,Rk;j . Therefore, ! X n jRi;j j 3 : i;j In the same way, we can restrict ourselves to I . We de ne Ri;j (I ) = Ri;j \ I and we get ! X m jRi;j (I )j 3 > (1=8) m3 : (i;j )2I I Now de ne
J = f(i; j ) 2 I I j jRi;j (I )j (1=16) mg; that is, J is the set of all pairs (i; j ) 2 I I such that jRi;j (I )j is not too small. Since for every i; j we have jRi;j (I )j < m, we can conclude X X X (1=8) m3 < jRi;j (I )j = jRi;j (I )j + jRi;j (I )j m2 (1=16) m + jJ j m; (i;j )2I I
that is,
J
(I I )nJ
jJ j > (1=16) m2 : 8
We are now ready to do the second case of the induction.
The second case: Claim 3.4 If there exists (i; j ) 2 J such that PROB2RA [((i) 2 [1; s]) ^ ((j ) 2 [t; n])] 2?10 v(A)?2 then
0 v(A)n 1?1 Y jAj (C0)n @ (r=n)A : r=C0 n+1
Proof:
In this case, we de ne A0 by A0 = f 2 A j ((i) 2 [1; s]) ^ ((j ) 2 [t; n])g; that is, we restrict (i) to be in [1; s] and (j ) to be in [t; n]. We will show that A0 satis es the requirements of Claim 3.2. For every k 2 Ri;j (I ), the projection of A on the 3 elements i; j; k does not contain the linear order (i; k; j ). Therefore after we restrict (i) to be in [1; s] and (j ) to be in [t; n], we cannot have (k) 2 [s; t]. Hence, for every k 2 Ri;j (I ), jA0(k)j jA(k)j ? jA(k) \ [s; t]j jA(k)j ? (1=68) v(A)3=2 m?1 : Since (i; j ) 2 J , jRi;j (I )j (1=16) m: Hence, n n X X 0 jA (k)j jA(k)j ? (1=16) m (1=68) v(A)3=2 m?1: That is,
k=1
k=1
d def = v(A) n ? v(A0) n (1=1088) v(A)3=2 : Since we assumed that PROB2RA[((i) 2 [1; s]) ^ ((j ) 2 [t; n])] 2?10v(A)?2, we have jA0j jAj 2?10v(A)?2; that is, p def = jAj=jA0j 210v(A)2 : and since by v(A) C0 we can assume that v(A) is large enough (say v(A) 210 ) we have ln(p) ln(v(A)3) = 3 ln(v(A)): Thus A0 satis es the requirements of Claim 3.2. 2 We will now assume that the requirements of Claim 3.3 and Claim 3.4 are not satis ed. This will be the third case of the induction.
The third case:
9
Claim 3.5 If for every i 2 f1; :::; ng and k 2 A(i), PROB2R A [(i) = k] v(A)?2 ; and for every (i; j ) 2 J , PROB2RA[((i) 2 [1; s]) ^ ((j ) 2 [t; n])] < 2?10 v(A)?2 then
0 v(A)n 1?1 Y jAj (C0)n @ (r=n)A : r=C0 n+1
Proof:
For every i 2 I de ne
Ji = fj 2 I j (i; j ) 2 J g: Since jJ j > (1=16) m2 , there exists i0 such that
jJi j > (1=16) m (1=256) v(A): 0
W.l.o.g assume that
jJi j = d(1=256) v(A)e; 0
(otherwise discard arbitrarily some elements from Ji0 ). Here we de ne A0 by
A0 = f 2 A j ((i0 ) 2 [1; s]) ^ (for every j 2 Ji0 ; (j ) 62 [t; n])g; We will show that A0 satis es the requirements of Claim 3.2. Since jJi0 j (1=256) v(A), and since for every j 2 Ji0 ,
jA(j )j ? jA0(j )j jA(j ) \ [t; n]j (1=8) jA(j )j (1=16) v(A); we have That is,
n X j =1
jA0(j )j
n X j =1
jA(j )j ? (1=256) v(A) (1=16) v(A):
d def = v(A) n ? v(A0) n (2?12 ) v(A)2: Let us now turn to bounding jAj=jA0j. Since
jA(i0 ) \ [1; s]j (1=8) jA(i0)j (1=16) v(A); and since for every k 2 A(i0 ), PROB2RA[(i0 ) = k] v(A)?2; 10
we have
PROB2RA[(i0 ) 2 [1; s]] (1=16) v(A) v(A)?2 = (1=16) v(A)?1: Also, since for every (i; j ) 2 J PROB2RA[((i) 2 [1; s]) ^ ((j ) 2 [t; n])] < 2?10 v(A)?2;
we have for every j 2 Ji0 , PROB2RA[((j ) 2 [t; n]) j ((i0 ) 2 [1; s])] =
Therefore
PROB2R A[((j ) 2 [t; n]) ^ ((i0 ) 2 [1; s])] = PROB2RA[(i0 ) 2 [1; s]] < 2?10 v(A)?2 = [(1=16) v(A)?1] = 2?6v(A)?1:
PROB2RA[(9j 2 Ji0 s.t., (j ) 2 [t; n]) j ((i0) 2 [1; s])] < jJi0 j 2?6v(A)?1 = d(1=256) v(A)e 2?6v(A)?1 < 2?13; (where we assume again that v(A) is large enough). We can conclude that
jA0j = jAj PROB2RA[(i0 ) 2 [1; s]] PROB2RA[(8j 2 Ji ; (j ) 62 [t; n]) j ((i0) 2 [1; s])] jAj (1=16) v(A)?1 (1 ? 2?13 ) > jAj (1=32) v(A)?1; 0
That is
p def = jAj=jA0j < 32 v(A) < v(A)2;
(for large enough v(A)). Thus A0 satis es the requirements of Claim 3.2.
2
4 Proof of the Technical Lemma In this section we give the proof of Lemma 3.1. De ne
I0 = fi 2 [n] j jA(i)j (1=2) v(A)g: Obviously,
X I0
jA(i)j =
X [n]
jA(i)j ?
X [n]nI0
jA(i)j = n v(A) ?
X [n]nI0
n v(A) ? n (1=2) v(A) = (1=2) n v(A):
The set I will be de ned to be a subset of I0. For every i 2 I0, let si ti be two indices such that 11
jA(i)j
1. jA(i) \ [1; si]j (1=4) jA(i)j; 2. jA(i) \ [ti; n]j (1=4) jA(i)j; 3. jA(i) \ [si; ti]j (1=2) jA(i)j; and de ne
Ag (i) = A(i) \ [si; ti]: X g X jA(i)j (1=2) jA(i)j (1=4) n v(A):
Then
I0
I0
We will de ne the indices s; t to be such that
j[s; t]j = d(1=16) v(A)e: Since [n] can be covered by d16n=v(A)e such intervals, we will be able to nd such s; t that
also satisfy
X g jA(i) \ [s; t]j (1=4) n v(A)=d16n=v(A)e I0
(1=4) n v(A)=(17n=v(A)) = (1=68) v(A)2: We can now de ne I1 I0 by I1 = fi 2 I0 j jAg (i) \ [s; t]j > 0g; and get
X I1
jA(i) \ [s; t]j
X g X jA(i) \ [s; t]j = jAg (i) \ [s; t]j (1=68) v(A)2: I1
I0
The set I will be de ned to be a subset of I1. Every i 2 I1 already satis es the rst 3 requirements of the lemma: 1. By the de nition of I0,
jA(i)j (1=2) v(A): 2. By the de nition of I1 we have jAg (i) \ [s; t]j > 0, and therefore by the de nition of Ag (i) we can conclude that si t. Hence, jA(i) \ [1; s]j jA(i) \ [1; si]j ? j[s + 1; t]j (1=4) jA(i)j ? d(1=16) v(A)e + 1 (1=4) jA(i)j ? (1=16) v(A) (1=4) jA(i)j ? (1=8) jA(i)j = (1=8) jA(i)j: 3. In the same way,
jA(i) \ [t; n]j (1=8) jA(i)j: 12
Since I will be a subset of I1 , every i 2 I will satisfy requirements 1,2,3. Thus we only have to make sure that I is large enough and that every i 2 I satis es requirement 4. We will now split the proof into two cases:
Case A:
jI1j (1=68) v(A)3=2 :
In this case we de ne I = I1 , and since we assumed that v(A) > 20 we have
m = jI1j (1=68) v(A)3=2 (1=16) v(A): For every i 2 I1 requirement 4 is satis ed by jA(i) \ [s; t]j jAg (i) \ [s; t]j 1 (1=68) v(A)3=2 m?1 :
Case B:
jI1j < (1=68) v(A)3=2:
In this case we de ne I by
I = fi 2 I1 j jA(i) \ [s; t]j (1=2) v(A)1=2 g: Then
X I
But for every i
jA(i) \ [s; t]j = X I1
X I1
X I1
jA(i) \ [s; t]j ?
X I1 nI
jA(i) \ [s; t]j
jA(i) \ [s; t]j ? jI1j (1=2) v(A)1=2
jA(i) \ [s; t]j ? (1=68) v(A)3=2 (1=2) v(A)1=2
(1=68) v(A)2 ? (1=136) v(A)2 = (1=136) v(A)2: jA(i) \ [s; t]j j[s; t]j = d(1=16) v(A)e (2=17) v(A);
(where we used again the assumption v(A) > 20). Hence,
jI j (2=17) v(A)
X I
jA(i) \ [s; t]j (1=136) v(A)2;
that is
m = jI j (1=16) v(A): Requirement 4 is satis ed for every i 2 I , since by the de nition of I , jA(i) \ [s; t]j (1=2) v(A)1=2 (1=68) v(A)3=2 m?1 :
2 13
Acknowledgments I would like to thank Mario Szegedy for introducing me to the acyclic linear orders problem and for many important discussions that led to this result. I would like to thank Peter Fishburn for the encouragement in writing up this manuscript.
References [Abe] [Bla] [Con] [Fis] [Rik] [Sau] [VC]
J.M. Abello, \The weak Bruhat order of S , consistent sets, and Catalan numbers", SIAM Journal of Discrete Math., 4, (1991), 1{16. D.J. Black, \The theory of committees and elections", Cambridge Press, London, 1958. Marquis de Condorcet, \Essai sur l'application de l'analyse a la probabilite des decisions rendues a la pluralite de voix", Paris, 1785. P. Fishburn, \Acyclic sets of linear orders", Social Choice and Welfare, 14, (1997), 113{124. W.H. Riker, \Arrow's theorem and some examples of the paradox of voting", Foundation monograph, Southern Methodist University Press, Dallas, 1961. N. Sauer, \On the density of families of sets", Journal of Combinatorial Theory, Series A 13, (1972), 145{147. V.N. Vapnik and A.Ya Chervonenkis, \On the uniform convergence of relative frequencies of events to their probabilities", Theory of Probability Applications, 16, (1971), 264{280.
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