2r 1. Definition (Degree Measure). Let G be a r-graph of order n and average degree d. Let S V (G). The degree measure µ(s) of S is defined by,

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1 Theorem Simple Containers Theorem) Let G be a simple, r-graph of average egree an of orer n Let 0 < δ < If is large enough, then there exists a collection of sets C PV G)) satisfying: i) for every inepenent set I [n], there exists C C with I C, ii) eg[c]) < δeg) for every C C, iii) C 2 βn where β = Degree Measure Definition Let G be a hypergraph For v V G) we efine v) = {e EG) : v e} Definition Degree Measure) Let G be a r-graph of orer n an average egree Let S V G) The egree measure µs) of S is efine by, µs) = n v) Note that µ is probability measure, so particularly aitive an µv G)) = The egree measure has several useful properties, which are state below v S First property tell us that µs) small implies eg[s]) small eg[s]) r v S v) = n µs) = µs)eg) r eg[s]) µs)eg) Let ea, B) enote number of eges between sets A, B V G) We have r )nµv G)\S) = r ) v / S v) r )ev G)\S, S) v S v) reg[s]) That is, So finally, r ) µs)) = r )µv \S) µs) r n eg[s]) eg[s]) n r rµs) r + ) = µs) + ) n 2) r

Definition Degree Measure) Let G be a r-graph of orer n an average egree Let S V G) The egree measure µs) of S is efine by, µs) = n v) Note that µ is probability measure, so particularly aitive an µv

2 If I is an inepenent set in G From 2) we get, 0 = eg[i]) µi) + ) n r r µi) If G is -regular, inequality eg[s]) < δeg) with 2) implies: δ n = δeg) > eg[s]) µs) + ) n = r r S n + r ) n What gives us S < r + δ ) n r 2 Proof of Theorem For a proof of Theorem, we will nee following lemma Lemma Let G be a simple, r-graph of average egree an of orer n If is large enough, then there exists a collection of sets C PV G)) satisfying: i) for every inepenent set I [n], there exists C C with I C, ii) µc) < 4r 2 ) for every C C, iii) C 2 αn where α = cr 2 log Statement of the lemma is quite similar to the statement of the Theorem, conition i) is exactly the same However ii) is expresse in terms of egree measure We can suspect that it will be use to boun number of eges in G[C] using property ) of µ In iii) carinality of the container family is smaller than in the Theorem, provie sufficiently large Definition Let G be a r-graph For a given 0 j r an sets R, S V G) we efine Γ j R, S), by Γ j R, S) = { v V G) : e = {v} f g, f e EG) ) R, g j S r j )} Proof Let V = V G), E = EG) For a given 0 j r an sets R, S, T V G) we efine C j R, S, T ), by { V \Γj R, S)\T ) if µγ C j R, S, T ) = j R, S)\T ) otherwise 4r 2, Since = µv \Γ j R, S)\T )) + µγ j R, S)\T ), if C j R, S, T ) woul be a container conition ii) will be satisfie 2

of sets C PV G)) satisfying: i) for every inepenent set I [n], there exists C C with I C, ii) µc) < 4r 2 ) for every C C, iii) C 2 αn where α = cr 2 log Statement of the lemma is quite similar to the

3 Let u = 6r 3r 2 an q = 5ru = 5 3r 6r 2 By the efinition q is small if is large Now we will efine a family C which will be caniate for a containers family, C = {C j R, S, T ) : 0 j r, R, S, T } Then ) 3 n C r) 3 ) 7 n 2 ) 7 = 2 log 2 7/ 2)) We will now focus on the exponent of 2, log 2 7nq/ 2 log 6r ) = 7 2 log 2 = 7 5 3r = n 7 5 3r6r/ 2) 2) log 2 } {{ } cr) ) 3 )) 7 en e = 5 3r 6r 2 2 log log 2 log 2 = So C 2 αn, α = cr 2 log We will use probabilistic metho, to show that the probability that C is not satisfying conitions i),ii),iii) is smaller that For a given inepenent set I we will efine sets R, S, T, each not bigger than such that for some j, C j R, S, T ) is a container for I Fix an inepenent set I, For a subset A V with I A an for 0 j r we efine E j A) = {e E : e A, e A j} P 0): Now, let P j) be a statement: A V I A µa) + I A µa) ) A V egree measure we have E 0 A) = eg[a]) Hence the statement P 0) is true j ) 2 E j A) nuj E 0 A) n From property 2) of µa) + ) n = n r 3

log 2 = So C 2 αn, α = cr 2 log We will use probabilistic metho, to show that the probability that C is not satisfying conitions i),ii),iii) is smaller that For a given inepenent set I we will efine

4 P r): I A µa) ) E r A) nur So A = V Since I cannot A V contain any eges, 0 = E r A) nur > 0 Hence the statement P r) is false Therefore there exists 0 j r such that the statement P j) is true an P j + ) is false Fix a set A witnessing falsity of P j + ) Such set exists since µa) + j+ 2 + j Thus 2 I A µa) + j + 2 E j+a) < nuj+ Let p = 6r/u j /r ) Since u < an 0 j r, we have p = 6r r 6r r u j u = 3r = 3r u2 u = 3ru = q 5 6r 3r 2 ) 2 u = Let R I an S A\I be ranom sets, where each vertex if I an A\I respectively) is inclue inepenently with probability p By Markov s inequality an Pr R > ) E R ) Pr S > ) E S ) = I p np 5, = A\I p np 5 Now, let T = Γ j R, S) I Provie µγ j R, S)\T ) /4r 2, C j R, S, T ) is a container for I Inee, if v I, then v / Γ j R, S)\T, by efinition of T So clearly v V \Γ j R, S)\T ) = C j R, S, T ) To complete the proof we have to show that with the aequate probabilities T > an µγ j R, S)\T ) < /4r 2 hols Let v I If v Γ j R, S) then, by efinition v T This woul happen if there exists e EG) such that e = {v} f g, f ) R j, g S r j ) Since e I = j+ we have that e E j+ A) There are j+ partitions e = {v} f g, v I, f ) I j, g A\I r j ), for each such partition, the probability that both f ) R j an g S r j ) is p r Observation Let us consier fixe R an S Let Y enote pairs v, e) such that v I an e E j+ A) with e = {v} f g, v I, f ) R j, g S r j ) Then T Y j + ) E j+ A) < j + ) nuj+ 2 nuj+ 4

$3r = 3r u2 u = 3ru = q 5 6r 3r 2 ) 2 u = Let R I an S A\I be ranom sets, where each vertex if I an A\I respectively) is inclue inepenently with probability p By Markov s inequality an Pr R > ) E R )$

5 Taking expectations, by observation we obtain E T ) E Y ) < 2 nuj+ p r = 6r nuj+ 2 u j = 3run = 5 Applying Markov s inequality Pr T > ) E T ) = 5 = 5 Now we will show PrµΓ j R, S)\T ) < /4r 2 ) 5 For v A, let F j v) = {e E : v e, e E j A), e / E j+ A)} = {e : v e A, e I = j}, an D = { v A\I : F j v) } uj u) Note that I A\D Consier an ege e E j A\D) an e / E j+ A) Then e A\D an e I = j Since j < r there is v A\D)\I an v e Note that E j A\D) E j A), so e F j v) By the efinition of D an facts that v / D an v / I we have F j v) < uj u) Therefore E j A\D)\E j+ A) < A\D uj u) nuj u) By the choice of A, E j+ A) < nuj+, hence E j A\D) E j A\D)\E j+ A) + E j+ A) < < nuj u) + nuj+ = nuj However, we know that the statement P j) is true, so it has to be µa\d) < + j On the other han A was chosen such that µa) 2 + j+ 2 Hence + j 2 > µa\d) = µa) µd) + j + 2 µd), an so µd) > 2 By the efinition of D there is D I =, in particular D T = Let D = D\Γ j R, S) Then D\D = D Γ j R, S) Γ j R, S)\T So µγ j R, S)\T ) µd\d ) = µd) µd ) Now, let v D, then F j v) uj u) 2 5r uj since u is small) Each e F j v) has a partition e = f g with f I j ), g A\I r j) where v g since v / I For every ege the probability that f R an g\{v} S is p r Moreover these events are inepenent over all e F j v), because G is simple Hence 5

$there is v A\D)\I an v e Note that E j A\D) E j A), so e F j v) By the efinition of D an facts that v / D an v / I we have F j v) < uj u) Therefore E j A\D)\E j+ A) < A\D uj u) nuj u) By the choice$

6 Prv D ) Pr e {v} f g, v D, f e F jv) = p r ) Fjv) exp 2 5r uj p r } {{ } =6r R j ), g = exp ) ) R = r j { 2 } < 5 0 Now, µd ) = n v D v) = n v D v)i v, where I v is an inicator of event v D Since EI v ) = Prv D ) < 0, EµD )) = v)ei v ) < µd n 0 v D Applying Markov s inequality, we get Pr µd ) > µd) ) 2EµD )) 2 µd) Therefore to Pr < 2µD) 0µD) = 5 µγ j R, S)\T ) µd) µd ) µd) > 2 4r 2 Finally, the probability that C j R, S, T ) is not the container for I is equal R > S > T > µγ j R, S)\T ) ) 4r 2 Pr R > ) + Pr S > ) + Pr R > ) + PrµΓ j R, S)\T ) 4r 2 ) 4 5 < Proof of Theorem The proof strategy is as follows We will iteratively apply lemma, starting with G If we will get a container, say C, with at least δeg) eges, we will apply the lemma to G[C], an so on to each of the new containers, until we obtain a collection of containers each with less than δeg) eges If C is a container obtaine after applying lemma to G, an eg[c]) δeg), if C stans for athe average egree of G[C], we have C = C r v C v) = C eg[c]) δeg) = δ C So we can apply lemma to G[C] with C δ, we will obtain a collection C of containers for G[C] with 6

$probability that C j R, S, T ) is not the container for I is equal R > S > T > µγ j R, S)\T ) ) 4r 2 Pr R > ) + Pr S > ) + Pr R > ) + PrµΓ j R, S)\T ) 4r 2 ) 4 5 < Proof of Theorem The proof strategy$

7 i) for every inepenent set I V C), there exists C C with I C, ii) µc ) < 4r 2 for every C C, iii) C 2 α n where α = cr C 2 log C Let I IG) an I C, it is equivalent to I IG[C]) Hence every inepenent set containe in C will be also containe in some member C of C Now, by ) we have eg[c ]) µc )eg[c]) µc )µc)eg) 4r 2 )2 eg) One can see that each application of the lemma ecreases the fraction of eges by /4r 2 Note that new containers are smaller concerning the number of eges, they o not necessarily have much less vertices Let m be the number of levels of iteration From above observation we ant to have /4r 2 ) m < δ Note that /4r 2 ) m < δ m log So we nee at most m = log δ log 4r 2 4r 2 ) ) < log δ m > log δ log 4r 2 ) + < c log δ levels of iteration, where c > 0 is some positive constant By iterative applications of the lemma the fraction of eges in each container ecreases, however the carinality of our family of containers can grow rapily We must assure that it will be not bigger than 2 βn, as state in Theorem Let α max enote maximum over all applications Then α max = cr) max 2 2 log max < cr) log δ m Since max = δ a, for some 0 a m We want to show m α max < β Let /δ < log, we have m α max β < m cr) δ m = cr)m δ m < cr)c log δ 2 log = 2 ) c δ ) 2) log < 2 log δ < cr)c log log log ) c 2 log log < C log ) C2 log log Hence, m α max < β for sufficiently large ) C3 0 ) 2) log < ) 2) log < 7

by /4r 2 Note that new containers are smaller concerning the number of eges, they o not necessarily have much less vertices Let m be the number of levels of iteration From above observation we ant to

8 3 List chromatic number Theorem Let G be a simple -regular r-graph Then, as, χ L G) + o)) log 2 log r Note: Originally 202), χ L G) + o) 4r log log Proof First, since G is simple, -regular, r-graph, we can apply Theorem to G Then we have a collection C of containers of size not greater than 2 βn Moreover, since G is -regular, for every container C we have C < /r + δ/r)n where 0 < δ < Let c = δ)/r, k = / 2), with this values we can apply C-compability theorem theorem 2 from 200) to a family C We obtain collection of lists L, each of size l, which are not C-compatible, where as k, l +o)) log k log/c) Hence, log k χ L G) + o)) log/c) Since δ was arbitrary log / 2) = + o)) logr/ δ)) = + o) log 2 log r 8

C we have C < /r + δ/r)n where 0 < δ < Let c = δ)/r, k = / 2), with this values we can apply C-compability theorem theorem 2 from 200) to a family C We obtain collection of

A Generalization of Sauer s Lemma to Classes of Large-Margin Functions

A Generalization of Sauer s Lemma to Classes of Large-Margin Functions Joel Ratsaby University College Lonon Gower Street, Lonon WC1E 6BT, Unite Kingom J.Ratsaby@cs.ucl.ac.uk, WWW home page: http://www.cs.ucl.ac.uk/staff/j.ratsaby/