aa r X i v : . [ m a t h . C O ] S e p Random Perfect Graphs
Colin McDiarmid [email protected]
Department of StatisticsUniversity of Oxford Nikola Yolov [email protected]
Department of Computer ScienceUniversity of OxfordAugust 29, 2018
Abstract
We investigate the asymptotic structure of a random perfect graph P n sampled uniformly from the set of perfect graphs on vertex set { , . . . , n } .Our approach is based on the result of Pr¨omel and Steger that almostall perfect graphs are generalised split graphs, together with a method togenerate such graphs almost uniformly.We show that the distribution of the maximum of the stability num-ber α ( P n ) and clique number ω ( P n ) is close to a concentrated distribution L ( n ) which plays an important role in our generation method. We alsoprove that the probability that P n contains any given graph H as aninduced subgraph is asymptotically 0 or or 1. Further we show thatalmost all perfect graphs are 2-clique-colourable, improving a result ofBacs´o et al from 2004; they are almost all Hamiltonian; they almost allhave connectivity κ ( P n ) equal to their minimum degree; they are almostall in class one (edge-colourable using ∆ colours, where ∆ is the max-imum degree); and a sequence of independently and uniformly sampledperfect graphs of increasing size converges almost surely to the graphon W P ( x, y ) = ( [ x ≤ /
2] + [ y ≤ / Keywords:
Perfect graphs, Edge-colouring, Clique-colouring, Hamilto-nian, Graph limits
A graph is perfect if the chromatic number equals the clique number in eachof its induced subgraphs. Perfect graphs have formed a central field in graphtheory for some decades, partly because of the challenging open problems andpartly because of the connections to polyhedral combinatorics, linear optimi-sation and computational complexity, see for example [RAR01]. Fundamentalclasses of graphs that are all perfect include bipartite graphs, chordal graphs,comparability graphs and interval graphs.Generalised split graphs form another important class of perfect graphs. Agraph G is unipolar if for some k ≥ V ( G ) can be partitioned1nto k + 1 cliques C , C , . . . C k , so that there are no edges between C i and C j for 1 ≤ i < j ≤ k . We call C the central clique , and the C i for i ≥ sidecliques ; and we call the pair ( G, C ) a unipolar arrangement of order v ( G ). Agraph G is co-unipolar if its complement G is unipolar; and it is a generalisedsplit graph if it is unipolar or co-unipolar. We denote the classes of perfect,unipolar, co-unipolar and generalised split graphs by P , GS + , GS − and GS respectively. Given a class G of graphs, we let G n denote the set of graphs in G on vertex set [ n ] := { , . . . , n } . We say that a sequence of events ( A n ) n ≥ holds with high probabilty (or whp) if P ( A n ) = 1 − o (1), and that it holds with veryhigh probabilty (or wvhp) if P ( A n ) = 1 − e − Ω( n ) .Our goal is to describe the asymptotic properties of perfect graphs. Werely on the key theorem of Pr¨omel and Steger [PS92] that almost all perfectgraphs are generalised split graphs. We present and analyse an almost uniformgeneration process for the graphs in GS n . This generation process, togetherwith the Pr¨omel-Steger theorem, yields a powerful method for working withuniformly sampled perfect graphs. The usefulness of the method will be seenwhen we prove a range of results about the asymptotic behaviour of perfectgraphs, described in the next subsection. In § GS n and the uniformly sampledperfect graph P n is e − Θ( n ) . We introduce a family of concentrated distributions L ( n ) which are used in the generation process, and give a number of resultsessential for our analysis, including surveying relevant results about randompartitions, but we defer the detailed proofs concerning the generation processto § § α ( P n ) and clique number ω ( P n ). Weshow that the minimum of the two numbers is asymptotically normally dis-tributed (with specified mean and variance), and the distribution of the maxi-mum is very close to the distribution L ( n ) mentioned above; see Theorem 3.7.In § H ∈ P \ GS , (ii) H ∈ GS \ GS − or H ∈ GS \ GS + and (iii) H ∈ GS + ∩ GS − , the probability that P n contains an induced copy of H isrespectively (i) e − Θ( n ) , (ii) 1 / ± e − Ω( n ) and (iii) 1 − e − Ω( n ) . For example, thecomplete bipartite graph K , ∈ GS − \ GS + , so the probability that P n containsan induced K , is well-estimated by .In § +
04] from 2004 that almostall perfect graphs are 3-clique colourable.2n § L ( n ) appearsagain. We show that, if X ∼ L ( n ), the probability that P n is Hamiltonian is1 − P ( X > n/ ± e − Ω( n ) = 1 − − (1+ o (1)) log n = 1 − o (1) . (1)The precise statement is given in Theorem 6.1.In § P n the connectivity κ ( P n ) equalsthe minimum degree wvhp, Theorem 7.1; and the chromatic index equals themaximum degree whp, Theorem 7.2.In § P n . A graphon is a symmetric measurable function W : [0 , → [0 , ,
1] asvertices. The cut distance , δ (cid:3) , is a metric over the space of graphons, W , gen-eralising the maximum absolute value of a rectangular difference of two matricesto graphons. The space ( W , δ (cid:3) ) has many important properties. In particular,it is compact, under the assumption that graphons with cut distance 0 are iden-tified. This result has interesting corollaries, including some of the strongestgeneralisations of the Szemer´edi regularity lemma. Theorem 8.1 states that P ( δ (cid:3) ( P n , W P ) ≤ n − / ) = 1 − e − Ω( √ n log n ) and P ( δ (cid:3) ( P n , W P ) ≤ (log n ) − ) = 1 − e − Θ( n ) , where W P ( x, y ) is the graphon ( [ x ≤ /
2] + [ y ≤ / W P and provides an analytic tool for estimatingsubgraph densities of P n .In § §
2; and finally in §
10 we make a few concluding remarks andmention some open problems.
Before we go any further, let us check that generalised split graphs are perfect.The weak perfect graph theorem states that the complement of a perfect graphis perfect [Lov72a, Lov72b], so it suffices to consider a unipolar graph G , andshow it is perfect. Also, each induced subgraph of a unipolar graph is unipolar,so it suffices to show that χ ( G ) = ω ( G ).Consider a unipolar arrangement ( G, C ), where C is the central clique,and let C , . . . , C k be the side cliques. Let G j denote the induced subgraph G [ C ∪ C j ]. Bipartite graphs are perfect, so co-bipartite graphs are perfect,and thus χ ( G j ) = ω ( G j ) for each j . Therefore we can colour each G j withmax j ω ( G j ) colours, and then make sure that the different colourings agree onthe central clique C . It is easy to see that ω ( G ) = max j ω ( G j ) and it followsthat χ ( G ) = ω ( G ), as required. 3e next discuss some probability distributions on the set G n of all graphson [ n ]. Probably the most well-known is that of the binomial random graph G ( n, p ), which assigns probability p e ( G ) (1 − p ) e ( G ) to each graph G in G n . Weare predominantly interested in the following probability distribution on G n : P P n ( G ) = ( / |P n | if G ∈ P n G n : P GS n ( G ) = ( / |GS n | if G ∈ GS n |GS n | = (1 − e − Ω( n ) ) |P n | . (2)In other words, the class of GS -graphs is a very good asymptotic approximationto the class of perfect graphs, as all GS -graphs are perfect and almost all perfectgraphs are GS -graphs. The bound in (2) is sharp. Fix any graph U ∈ P \ GS ,for example take the disjoint union of K , and K , and consider the family F U of the graphs that are a disjoint union of U and a perfect graph. We have F U ⊆ P \ GS and |F Un | ≥ |P n − v ( U ) | , which combined with estimates on the sizeof P n , e.g. |P n | / |P n − | = 2 (1+ o (1)) n/ , implies |GS n | = (1 − e − Θ( n ) ) |P n | . (3)We can also express (3) using total variation distance: d T V ( P GS n , P P n ) = sup A ⊆G n | P GS n ( A ) − P P n ( A ) | = P P n ( P n \ GS n )= |P n \ GS n ||P n | = e − Θ( n ) . (4)How many n -vertex perfect graphs are there? The number is very close tothe number of n -vertex GS -graphs, which is approximately twice the numberof n -vertex unipolar graphs. The last class is easy to estimate: for each k between 0 and n there are (cid:0) nk (cid:1) ways to choose which k vertices to include in thecentral clique, B n − k ways to specify the partition of the remaining vertices and2 k ( n − k ) ways to specify the edges between the central clique and the remainingvertices. Here B n stands for the n -th Bell number, which is the number of waysto partition an n -element set. Note that the last calculation is not precise, sinceunipolar graphs may be part of more than one unipolar arrangement. If wedenote (cid:0) nk (cid:1) k ( n − k ) B n − k by ℓ n,k , we would expect that the number of unipolargraphs is close to L n := P nk =0 ℓ n,k . This turns out to be a very precise estimateand we show in § |GS n | = 2 L n (1 − e − Ω( n ) ) . (5)(See Lemma 2.2 of [PS92] for an asymptotic estimate of |GS n | .)4 efinition 2.1. For each n let L ( n ) be the discrete integer-valued distributionwith probability mass function p L ( n ) ( x ) = ℓ n,x / L n for x = 0 , . . . , n. Now we describe our method to sample almost uniformly from the set of gener-alised split graphs on [ n ]. Definition 2.2.
For each n let Gen ( n ) be a random quadruple ( B, E, ( k, σ ) , π ) ,such that1. B is a {− , } -valued random variable, taking each value with probabilityhalf,2. E is the set of edges of an Erd˝os-R´enyi random graph G ( n, ) ,3. ( k, σ ) is the result of the following random experiment: choose a number k ∈ [0 , n ] with distribution L ( n ) , set W = { k + 1 , . . . , n } , and then sample σ uniformly at random from Π( W ) , the set of all partitions of W ,4. π is a uniformly sampled permutation from Sym ([ n ]) ,5. the components of the quadruple are independent. Let ρ be the (deterministic) map from such quadruples to G n defined asfollows. Let U = ( B, E, ( k, σ ) , π ) be an outcome of Gen ( n ). Start with an n -vertex empty graph G = ([ n ] , ∅ ). It is convenient to write C = [ k ] and S = V ( G ) \ C . Create G by adding edges to G , so that C induces a clique,and S induces a disjoint union of cliques corresponding to σ , i.e. ij ∈ E ( G )for 1 ≤ i < j ≤ k ; and ij ∈ E ( G ) whenever i and j are contained in thesame part of σ for k < i < j ≤ n . Let G be obtained from G by adding theedges of E between C and S , and let G be G if B = 1 and G otherwise.Finally let G be obtained by permuting the vertices of G according to π , i.e. π ( i ) π ( j ) ∈ E ( G ) iff ij ∈ E ( G ). We define ρ ( U ) = G for each such vector U .Consider U = ( B, E, ( k, σ ) , π ) ∼ Gen ( n ) and the corresponding randomgraph G = ρ ( U ). It is easy to see that G is unipolar if B = 1 and co-unipolar if B = −
1. Indeed, (
G, π ( C )) is a unipolar arrangement if B = 1 and ( G, π ( C )) isa unipolar arrangement otherwise. We often refer to the appropriate pair as the induced arrangement of ρ ( U ) by the generation. Whenever we want to test if G has some isomorphism-closed graph property, π does not affect the membership,so we can assume that π is the identity permutation. All graph properties inwhich we are interested are isomorphism-closed, so π will largely be ignored.We define our last probability distribution, P Gen ( n ) on G n , as follows: P Gen ( n ) ( G ) = P ( ρ ( Gen ( n )) = G ) .
5y the above, P Gen ( n ) ( G ) > G ∈ GS n . We shall show that d T V ( P GS n , P Gen ( n ) ) = e − Ω( n ) . (6)It is not hard to prove (6) from (5). However, both equations have technicalproofs, so they are postponed to § Theorem 2.3.
The uniform measure P P n and the generated measure P Gen ( n ) satisfy d T V ( P P n , P Gen ( n ) ) = e − Θ( n ) . Proof.
From the triangle inequality for total variation distance, (4) and (6), wehave d T V ( P P n , P Gen ( n ) ) ≤ d T V ( P P n , P GS n ) + d T V ( P GS n , P Gen ( n ) ) = e − Θ( n ) . A lower bound follows from arguments similar to these for (3).Theorem 2.3 describes the precision of our almost-uniform sampling. Notethat any approach relying on (3) cannot hope to achieve better precision. Theproof is rather technical, and is postponed to Section 9. The properties of thegenerated graph ρ (( B, E, ( k, σ ) , π )) depend a lot on the value of B . It is oftenthe case that we break the analysis into different cases depending on the valueof B . We introduce two new random 4-tuples - Gen + ( n ) and Gen − ( n ), whichare the same as Gen ( n ), but with overwritten values for B : Definition 2.4.
For each n define Gen + ( n ) to be (1 , E, ( k, σ ) , π ) and Gen − ( n ) to be ( − , E, ( k, σ ) , π ) , where E , ( k, σ ) and π are the same as in the definitionof Gen ( n ) . The following observation is easily verified: Let Q be an arbitrary graphproperty. If G ∼ ρ ( Gen ( n )), G + ∼ ρ ( Gen + ( n )) and G − ∼ ρ ( Gen − ( n )), then P ( G ∈ Q ) = 12 P ( G + ∈ Q ) + 12 P ( G − ∈ Q ) . (7)Still with G ∼ ρ ( Gen ( n )), we shall see in Section 9 that wvhp G is not bothunipolar and co-unipolar, that is P ( G ∈ GS + ∩ GS − ) = e − Ω( n ) . (8) Theorem 2.3 will prove to be very useful. Let P n ∈ u P n . Whenever we wantto estimate q n = P P n ( Q ) = P ( P n ∈ Q ) for some graph property Q , we caninstead estimate q ′ n = P Gen ( n ) ( Q ) = P ( ρ ( Gen ( n )) ∈ Q ), and use the fact that | q n − q ′ n | = e − Ω( n ) . The approximate sampling, Gen ( n ), is a convenient methodfor estimating probability, because it decomposes the resulting graph into threeindependent parts – a central clique, a partition of the remaining vertices anda bipartite part in between. 6 .2.1 Central clique An essential part of the generation process involves determining the size of thecentral clique. For this purpose we introduced the distribution L ( n ). Recall thatif X ∼ L ( n ) and x is an integer between 0 and n , then P ( X = x ) = ℓ n,x / L n ,where ℓ n,x = (cid:0) nx (cid:1) x ( n − x ) B n − x and L n = P nk =0 ℓ n,k . The term 2 x ( n − x ) dwarfsthe other two terms and heavily concentrates the distribution. The term B n − x breaks the symmetry around n/ log n , where throughout the paper log means log and ln denotes the naturallogarithm, log e . Theorem 2.5.
Suppose X ∼ L ( n ) . For sufficiently large n we have − ( x +1) − ≤ P ( | X − µ | ≥ x ) ≤ − ( x − +2 + n − n , for each x > , where µ = µ ( n ) = n − log n + log ln n . The upper bound in the theorem above is essential for handling the gener-ation process. Indeed, Theorem 2.3 will always be used in combination withTheorem 2.5. Most often we take x = Ω( √ n ) in order to have an exponentialdecay. However, sometimes there is a phase transition when the central cliquebecomes bigger than the remaining graph, most notably when we look at Hamil-tonian cycles in co-unipolar graphs, and then we take x to be n/ − µ ∼ log n .The proof of Theorem 2.5 is rather technical, and is postponed to § X ∼ L ( n ), then | E X − µ ( n ) | < Let U ∼ Gen ( n ) and let ( G, C ) be the arrangement induced by the generationprocess. The edges between C and C are best seen as a random bipartite graphwith colour classes C and C and edges in between, each present with probability1 / C and C are not necessarilyidentical, but Theorem 2.5 shows that their difference is O ( √ n ) wvhp. Randombipartite graphs share many properties with binomial random graphs G ( n, p ),which have been exhaustively studied. Let Π n be the set of all partitions of [ n ]. The elements of each partition willbe referred to as parts . We turn Π n into a probability space by taking eachpartition with equal probability, namely 1 / | Π n | . Let σ = σ ( n ) be an outcomefrom this probability space, i.e. σ ∈ u Π n . The structure of σ can vary fromhaving a single part with n elements to having n parts with one element each.However, σ has a well-defined shape with high probability.7et r = r ( n ) be the unique root of the equation re r = n and let B n = | Π n | be the n th Bell number. It can be found in [DB70] that r ( n ) = ln n − (1 + o (1)) ln ln n, and (9) B n = 1 + o (1) √ r e n ( r − /r ) − . (10)There is a sizeable literature for random partitions, see in particular [Pit97].Let | π | be the number of parts in a partition π . If X is a random variable wewill use b X to denote ( X − E X ) / √ Var X . One of the earliest results on randompartitions is by Harper: Theorem 2.6. (Harper [Har66]) For σ ∈ u Π n c | σ | d → N (0 , ,2. E | σ | = B n +1 B n − o (1)) nr = (1 + o (1)) n ln n ,3. Var | σ | = B n +2 B n − (cid:16) B n +1 B n (cid:17) − nr ( r +1) − o (1) = (1 + o (1)) n (ln n ) . Let Y t ( π ) be the number of parts of size t of an arbitrary partition π ∈ Π n ,so that | π | = P nt =1 Y t ( π ) and n = P nt =1 tY t ( π ). Suppose that σ ∈ u Π n fromnow on. Theorem 2.7. (Pittel [Pit97]) Given a partition π and an interval I = [ k , k ] where ≤ k ≤ k ≤ n , let Y I ( π ) = X j ∈ I Y j ( π ) , λ I = X j ∈ I r j j ! . Then, uniformly for I = [ k , k ] and ξ ≥ , P ( | Y I ( σ ) − λ I | ≥ ξ ) = O ( e − min( ξ λI , ξ ) ) . The last theorem is useful for small deviations, but rather crude for devia-tions proportional to the mean. A counting argument yields a better bound:
Lemma 2.8.
Let ǫ > be fixed and let λ = n/r = e r . Then for large n P (cid:0)(cid:12)(cid:12) | σ | − λ (cid:12)(cid:12) ≥ ǫλ (cid:1) < ne − n ( ǫ − ln( ǫ +1)) . In particular, (cid:12)(cid:12) | σ | − λ (cid:12)(cid:12) < ǫλ wvhp.Proof. For each 0 ≤ k ≤ n we have P ( | σ | = k ) B n ≤ k n k ! ≤ k n (cid:18) ke (cid:19) − k = exp( n ln k − k ln k + k ) . k = (1+ c ) λ for some c = c ( n ) > −
1. Substituting in the inequalityabove we get: P ( | σ | = (1 + c ) λ ) ≤ B − n exp( n ln(1 + c ) λ − (1 + c ) λ ln(1 + c ) λ + (1 + c ) λ )= B − n exp { n ln λ − λ ln λ + λ + n ln(1 + c ) − cλ ln λ + λ ( c − ln(1 + c ) − c ln(1 + c )) }≤ exp( n { ln(1 + c ) − c } + λ { c − ln(1 + c ) − c ln(1 + c ) } )for sufficiently large n . Here we have used λe λ = n ; and that (10) implies thatwe have B − n exp( n ( r − /r )) < n. Next, note that c − ln(1 + c ) − c ln(1 + c ) ≤ c > −
1, so we can write P ( | σ | = (1 + c ) λ ) ≤ exp( n { ln(1 + c ) − c } )for large n . Let f ( x ) : ( − , + ∞ ) → R be defined as f ( x ) = ln(1 + x ) − x , sothat P ( | σ | = (1 + c ) λ ) ≤ exp( nf ( c )) . The following properties are easily verified by taking derivatives: f ( x ) ≤ f ( | x | ); f (0) = 0; f ( x ) < x = 0; and f is strictly increasing in ( − ,
0] andstrictly decreasing in [0 , ∞ ). We deduce that for every I ⊆ ( − , ∞ ] we havesup x ∈ I f ( x ) ≤ f (inf x ∈ I | x | ). Let I δ = [0 , n ] \ [ ⌊ (1 − δ ) λ ⌋ , ⌈ (1 + δ ) λ ⌉ ], for δ > P (cid:0)(cid:12)(cid:12) | σ | − λ (cid:12)(cid:12) ≥ ǫλ (cid:1) = P ( | σ | ∈ I ǫ ) ≤ max k ∈ I ǫ P ( | σ | = k ) < n exp { f ( ǫ ) n } . For π ∈ Π n let L ( π ) be the maximum size of a part. Lemma 2.9.
Suppose ≤ x ≤ n/ and let σ ∈ u Π n . We have P ( L ( σ ) ≥ x ) < (1 + o (1)) e − x (ln x − ln r − n . Proof.
We may assume that x ≥ e r (since otherwise the RHS is ≥ E x be the event that σ contains a part of size ⌊ x ⌋ . Then by (10) and noting that r n − x − r n − x ≤ r n − r n (this holds because f ( t ) = t − t − and r t areincreasing in t ∈ [1 , ∞ )) we get P ( E x ) ≤ (cid:18) n ⌊ x ⌋ (cid:19) B n −⌊ x ⌋ B n ≤ (cid:16) enx (cid:17) x (1 + o (1)) r r n r n − x e − x ( r n − /r n ) ≤ (1 + o (1)) (cid:16) enx (cid:17) x (cid:18) ner n (cid:19) − x = (1 + o (1)) e − x (ln x − ln r − . P ( L ( σ ) ≥ x ) = P [ z ≥ x E z ≤ X z ≥ x P ( E z ) < n max z ≥ x P ( E z ) ≤ (1 + o (1)) e − x (ln x − ln r − n , since ln x − ln r − ≥ Lemma 2.10.
Suppose a = a ( n ) = o ( n ) is integral for all n . Then B n − a B n = exp( − ar n − a + o ( n )) . Proof.
We have B n − a B n = (1 + o (1)) exp (cid:18) ( n − a ) (cid:18) r n − a − r n − a (cid:19) − n (cid:18) r n − r n (cid:19)(cid:19) = exp((1 + o (1)) n ln(1 − a/n ) − a ( r n − a −
1) + o ( n ))= exp((1 + o (1))( − a ) − a ( r n − a −
1) + o ( n ))= exp( − ar n − a + o ( n )) . Corollary 2.11.
We have P ( L ( σ ) ≥ n/ log n ) = e − Θ( n ) . (11) and further, if ǫ ( n ) = o (1) , then P ( L ( σ ) ≥ ǫ ( n ) n/ log n ) = e − o ( n ) . (12) Proof.
Let a = ⌈ n/ log n ⌉ . Then P ( L ( σ ) ≥ n/ log n ) is at least B n − a B n ; and isat most (cid:0) na (cid:1) n B n − a B n (where the factor n is to account for choosing a part (orthe empty set) to add to the chosen part of size a ). Now we can use the lastlemma.To close this section on generating random perfect graphs, let us brieflyconsider the graph G output by the generation process, and how to recover aunipolar arrangement for G or its complement (or how to seek such arrange-ments for the random perfect graph P n ). On input an arbitrary n -vertex graph H , in O ( n )-time we can test if H is unipolar or co-unipolar, and if it is eitherthen output a corresponding vertex partition. This result is due the present au-thors [MY15], improving on earlier O ( n )-time approaches. The method is notvery complicated, but in the present case there is a trivial O ( n )-time heuristic.Consider G . Let A be the set of vertices with degree at most n/
2, and let B consist of the remaining vertices. Then wvhp exactly one of the following twoevents happens: (a) B is a clique and A is a disjoint union of cliques (forming aunipolar arrangement for G ), or (b) A is stable and B is a disjoint union of stable10ets (forming a unipolar arrangement for G ); and so, since by symmetry the twoevents have the same probability, each holds with probability ± e − Ω( n ) . Givena unipolar arrangement of G or G , we can efficiently calculate graph invariantslike α , ω and χ [MY15].To check this, let 0 < ǫ < /
4. We have seen that, in the generation process,wvhp | k − n/ | < ǫn/ σ has size < ǫn/ B = 1), by standard Chernoff bounds,wvhp each vertex in the central clique has degree in (3 n/ − ǫn, n/ ǫn ) andeach other vertex has degree in ( n/ − ǫn, n/ ǫn ), so (a) holds. Similarly, inthe co-unipolar case, wvhp each vertex in the central stable set has degree in( n/ − ǫn, n/ ǫn ) and each other vertex has degree in (3 n/ − ǫn, n/ ǫn ),so (b) holds. Also, by (8) wvhp (a) and (b) cannot both hold. In this section we discuss the stability number α ( P n ) and clique number ω ( P n )of the random perfect graph P n . Our main result is Theorem 3.7. Of courseeverything we say also applies to the clique covering number and chromaticnumber, as they equal the former two parameters for perfect graphs. Lemma 3.1.
Let U = (1 , E, ( k, σ ) , π ) be an arbitrary outcome of Gen + ( n ) andlet G + = ρ ( U ) . Then ≤ α ( G + ) − | σ | ≤ .Proof. Is is easy to find a stable set of size | σ | from the side cliques. For theother direction, note that G + can be covered by | σ | + 1 cliques.The last lemma pins down α ( G + ) to only two values, namely | σ | and | σ | + 1.We next show that the second outcome is very unlikely. Lemma 3.2.
Let U = (1 , E, ( k, σ ) , π ) ∼ Gen + ( n ) , let G + = ρ ( U ) , and let < δ < − ln 2 ≈ . . Then P ( α ( G + ) = | σ | + 1) = O ( e − n δ ) . We give an informal proof first. Condition on the induced unipolar ar-rangement ( G + , C ) and on the partition σ . We have α ( G + ) = | σ | + 1 if andonly if there is a vertex from C whose neighbourhood does not contain anyside clique in C . Let A v be the event that v is such a vertex, and hence { α ( G + ) = | σ | + 1 } ≡ S v A v . For a fixed vertex u ∈ C and a fixed side clique R we have P ( R N ( u )) = 1 − −| R | . Therefore P ( A v ) = Y R (cid:16) − −| R | (cid:17) , where that product is taken over the side cliques in C . Up until now everythingis rigorous. Observe from Theorem 2.7 that whp the number of side cliques ofsize i is close to r i /i !, where r is the unique root of the equation n − k = re r .11or a moment assume that the number of side cliques of size i equals this value,for each i . Then P ( A v ) = n Y i =1 Y R : | R | = i (1 − − i ) = n Y i =1 (1 − − i ) r i /i ! ≤ n Y i =1 (cid:16) e − − i (cid:17) r i /i ! since 1 + x ≤ e x = exp − n X i =1 ( r/ i /i ! ! ≤ exp (cid:16) − e r/ + 2 (cid:17) = exp (cid:16) − p ( n − k ) /r + 2 (cid:17) = exp (cid:16) − (1 + o (1)) p n/ (2 ln n ) (cid:17) . For the last step we assumed that k ∼ n/
2, which is easily justifiable from theconcentration theorem, Theorem 2.5. We use the union bound to finalise ourarguments: P ( α ( G + ) = | σ | + 1) ≤ X v P ( A v ) ≤ exp (cid:16) − (1 + o (1)) p n/ (2 ln n ) (cid:17) = O (exp( − n / − ǫ )) . We observe that δ from the statement is 1 / − ǫ in this case, which may be thebest possible. Proof of Lemma 3.2.
Let J = [ n/ − √ n, n/ √ n ]. We have P ( k ∈J ) ≥ − − n from Theorem 2.5. Condition σ on k = k for some k ∈ J . Weuse Theorem 2.7 for σ with I = [0 , r + r / ], λ I = e r P [ P o ( r ) ≤ r + r / ] ≤ e r and ξ = 2( e r ) / , where r is the unique root of n − k = re r . The theorem saysthat P ( Y I ( σ ) ≥ λ I − ξ (cid:12)(cid:12) k = k ) ≥ − c exp ( −√ e r ) , for some constant c . In order to give a lower bound on λ I , we use a stan-dard Chernoff bound for a Poisson random variable: P ( P o ( r ) ≥ (1 + η ) r ) < exp ( − rη / < η ≤
1. Taking η = r − / we obtain: λ I − ξ > e r (1 − exp ( −√ r/ − e r ) − / ) = e r (1 − o (1)) . Define the events A v as in the preceding informal discussion. The event { Y I ( σ ) ≥ λ I − ξ } implies that in σ there are at least e r (1 + o (1)) parts of size at most12 (1 + o (1)), hence P ( A v (cid:12)(cid:12) ( Y I ( σ ) ≥ λ I − ξ ) ∩ ( k = k )) ≤ (1 − − r (1+ o (1)) ) e r (1+ o (1)) ≤ exp {− − r (1+ o (1)) e r (1 + o (1)) } = exp {− ( e/ r (1+ o (1)) } = exp {− exp [ r · (1 − ln 2)(1 + o (1))] } = exp { (( n − k ) /r ) (1 − ln 2)(1+ o (1)) } since e r = ( n − k ) /r = O (exp ( − n δ )) . From the union bound we get P ( α ( G + ) = | σ | + 1 (cid:12)(cid:12) k = k ) ≤ P ( Y I ( σ ) < λ I − ξ (cid:12)(cid:12) k = k )+ X v P ( A v (cid:12)(cid:12) ( Y I ( σ ) ≥ λ I − ξ ) ∩ ( k = k ))= O (exp ( − n δ )) . Hence P ( α ( G + ) = | σ | + 1) ≤ P ( k
6∈ J ) + max k ∈J P ( α ( G + ) = | σ | (cid:12)(cid:12) k = k )= O (exp ( − n δ )) . Note that the bound for δ in the last proof is 1 − ln 2 ≈ .
3, which is slightlyworse than the constant in the informal discussion.We shorten G + ∼ ρ ( Gen + ( n )) to G + ∼ Gen + ( n ). Proposition 3.3.
Let G + ∼ Gen + ( n ) , and denote α ( G + ) by X . Then, as n → ∞ , b X d → N (0 , , E [ X ] ∼ n n , and Var[ X ] ∼ n n . We write Y for | σ | and Y m for | σ m | when σ m ∈ u Π m . To prove the lastresult we will use the following claim. Claim 3.4. E Y n +1 = E Y n + o (1)ln n .Proof. From Theorem 2.6, E Y n = B n +1 B n − Y n = B n +2 B n − (cid:16) B n +1 B n (cid:17) − Y n = B n +2 B n +1 B n +1 B n − (cid:18) B n +1 B n (cid:19) − E Y n +1 + 1)( E Y n + 1) − ( E Y n + 1) − . Therefore ( E Y n +1 + 1) − ( E Y n + 1) = Var Y n + 1 E Y n + 1 = 1 + o (1)ln n . roof of Proposition 3.3. Recall that σ is a random partition created by firstchoosing a value for k ∼ L ( n ), and then taking σ ∈ u Π n − k . It suffices to showthat b Y d → N (0 , Y ≤ X ≤ Y +1 , Var( Y ) →∞ , b Y d → N (0 , b X d → N (0 , x ) be the CDF of N (0 , x and ǫ > n | P ( b Y < x ) − Φ( x ) | < ǫ. Find ℓ = ℓ ( ǫ ) such that P ( k
6∈ I ) < ǫ/ n − n for I = I ( n ) = [ ⌊ E k − ℓ ⌋ , ⌈ E k + ℓ ⌉ ]. Such ℓ exists by Theorem 2.5. We emphasise that the length of I is at most 2 ℓ + 1, which is independent of n . (cid:12)(cid:12)(cid:12) P ( b Y < x ) − Φ( x ) (cid:12)(cid:12)(cid:12) = n X c =0 (cid:12)(cid:12)(cid:12) ( P ( b Y < x (cid:12)(cid:12) k = c ) − Φ( x )) P ( k = c ) (cid:12)(cid:12)(cid:12) < X c ∈I (cid:12)(cid:12)(cid:12) ( P ( b Y < x (cid:12)(cid:12) k = c ) − Φ( x )) (cid:12)(cid:12)(cid:12) P ( k = c ) + ǫ/ n − n ≤ max c ∈I (cid:12)(cid:12)(cid:12) P ( b Y < x (cid:12)(cid:12) k = c ) − Φ( x ) (cid:12)(cid:12)(cid:12) + ǫ/ n − n . Fix any integer valued function q ( n ) with q ( n ) ∈ I ( n ). We shorten q ( n ) to q for readability. (cid:12)(cid:12) E Y − E Y n − q (cid:12)(cid:12) = (cid:12)(cid:12) E (cid:8) E (cid:2) Y (cid:12)(cid:12) k (cid:3)(cid:9) − E Y n − q (cid:12)(cid:12) ≤ n X i =0 (cid:12)(cid:12) E Y n − i − E Y n − q (cid:12)(cid:12) P ( k = i ) ≤ X i ∈I (cid:12)(cid:12) E Y n − i − E Y n − q (cid:12)(cid:12) P ( k = i ) + ǫ/ n − n ≤ max i ∈I (cid:12)(cid:12) E Y n − i − E Y n − q (cid:12)(cid:12) + ǫ/ n − n ≤ (1 + o (1)) 2 ℓ + 12 ln n + ǫ/ n − n = ǫ/ o (1) . (13)We proceed with the variance in a similar fashion. (cid:12)(cid:12) Var Y − Var Y n − q (cid:12)(cid:12) ≤ max i ∈I (cid:12)(cid:12) Var Y n − i − Var Y n − q (cid:12)(cid:12) + ǫ/ n − n = o (Var Y n − q ) . For the last equality we have used part 3 of Theorem 2.6. We concludeVar Y = (1 + o (1)) Var Y n − q . (14)Write y = ( x (Var Y ) / + E Y − E Y n − q )(Var Y n − q ) − / , so that P ( b Y < x (cid:12)(cid:12) k = q ) = P ( [ Y n − q < y ) .
14y Theorem 2.6 part 3, we have Var Y → ∞ as n → ∞ ; and so by (13) and (14)it follows that lim n →∞ y = x . Find δ >
0, so that Φ( x ) ≤ Φ( x − δ ) + ǫ/
6. FromTheorem 2.6 we have P ( [ Y n − q < x − δ ) → Φ( x − δ ) if n − q → ∞ . For large n P ( [ Y n − q < y ) ≥ P ( [ Y n − q < x − δ ) ≥ Φ( x − δ ) − ǫ/ ≥ Φ( x ) − ǫ/ . We can do the same in the other direction and deduce that for large n | P ( b Y < x (cid:12)(cid:12) k = q ) − Φ( x ) | < ǫ/ . But q was an arbitrary function, hence for large n max c ∈I (cid:12)(cid:12)(cid:12) P ( b Y < x (cid:12)(cid:12) k = c ) − Φ( x ) (cid:12)(cid:12)(cid:12) < ǫ/ , and the proposition is proven.We continue with ω ( G + ). The overview is that usually there is a uniqueclique of maximum size in G + – the central clique. The vertices in the centralclique have degree (1+ o (1))3 n/ o (1)) n/ C from the generation is easily recognisablein the resulting graph (and in the 4-tuple U , as well as k and B , we know σ upto relabelling of C , we know a little about E and we know π ([ k ]) = C ). Lemma 3.5.
Suppose G + = ρ ( U ) , where U = ( B, E, ( k, σ ) , π ) ∼ Gen + ( n ) .Then both the probabilty that C is not the unique clique of maximum size and P ( ω ( G + ) = k ) are − ( + o (1)) n .Proof. As usual, let I = [ n/ − √ n, n/ √ n ], and recall that k ∈ I wvhp byTheorem 2.5. Let k ∈ I , and condition on k = k .We first give a lower bound for P ( ω ( G + ) = k ). With probability at least2 − k there is a vertex outside the central clique which is adjacent to all verticesfrom the central clique, and hence P ( ω ( G + ) > k ) ≥ − ( + o (1)) n .We now prove an upper bound. Of course, P ( ω ( G + ) = k ) is at most theprobabilty that C is not the unique clique of maximum size: we upper boundthe latter probability. Let ( G + , C ) be the unipolar arrangement induced by U .There are two cases in which G + contains a clique M = C of size at least k .The first case is when | C ∩ M | = 1, when also | C \ M | ≤
1. The probability ofthat happening is at most ( n − k ) k − ( k − = 2 − ( + o (1)) n .Let us focus on the second case; | C ∩ M | >
1. This implies that there areat least two vertices from C with degree at least k . If we substitute n/ √ log n for x in Lemma 2.9, we get that the probability that there is a part in σ of sizemore than n/ √ log n is less than e − n for large n . Assume there is no such partin σ .For every v ∈ C write d ( v ) = d C ( v ) + d C ( v ). We have d C ( v ) ∼ Bin ( k , / d C ( v ) ≤ n/ √ log n −
1. Let E v be the event { d C ( v ) ≥ k − n/ √ log n } .Next we use the Chernoff bound in the form: P ( X ≥ mp + t ) ≤ e − t /m , for X ∼ Bin ( m, p ) and t ≥
0. We obtain P ( E v ) ≤ e − (1+ o (1)) k / . E v and E u are independent for u = v , hence P ( E v ∩ E u ) ≤ e − (1+ o (1)) k for every pair u = v . We deduce that the probability that there are two verticesfrom C with degree at least k is at most n e − (1+ o (1)) k + e − n = e − ( + o (1)) n .Combining the two cases completes the proof of the upper bound. Lemma 3.6.
For G + ∼ Gen + ( n ) and a fixed ǫ > we have wvhp (1 − ǫ ) n n < α ( G + ) < (1 + ǫ ) n n , and n − ǫ √ n < ω ( G + ) < n ǫ √ n. In particular, α ( G + ) < ω ( G + ) wvhp.Proof. The first line follows from Lemma 3.1, Lemma 2.8 and Theorem 2.5; thesecond line follows from Theorem 2.5 and Lemma 3.5.Observe that the proof of Theorem 2.5 will complete the proof of Lemma 3.6.For every graph G we have ω ( G ) = α ( G ) and therefore everything we showedfor α ( G + ) and ω ( G + ) directly translates to ω ( G − ) and α ( G − ). Let h ( G ) =min { α ( G ) , ω ( G ) } and H ( G ) = max { α ( G ) , ω ( G ) } for every graph G . (Here h and H come from homogeneous.) Theorem 3.7.
For P n ∈ u P n ,1. h ( P n ) is asymptotically normal with mean ∼ n n and variance ∼ n n ,2. d T V ( H ( P n ) , L ( n )) = e − Ω( n ) .Proof. Let U = ( B, E, ( k, σ ) , π ) ∼ Gen ( n ). We couple G n , G + n and G − n bywriting U + = (1 , E, ( k, σ ) , π ), U − = ( − , E, ( k, σ ) , π ), G n = ρ ( U ), G + n = ρ ( U + )and G − n = ρ ( U − ). We have d T V ( h ( P n ) , α ( G + n )) ≤ d T V ( h ( P n ) , h ( G n )) + d T V ( h ( G n ) , α ( G + n )) ≤ d T V ( P P n , P Gen ( n ) ) + P ( h ( G n ) = α ( G + n )) . But d T V ( P P n , P Gen ( n ) ) = e − Ω( n ) from Theorem 2.3; and, since h ( G n ) = h ( G + n ), P ( h ( G n ) = α ( G + n )) = P ( ω ( G + n ) < α ( G + n )) = e − Ω( n ) from Lemma 3.6. Thus d T V ( h ( P n ) , α ( G + )) = e − Ω( n ) . The first part in thestatement now follows from Proposition 3.3.Similar arguments establish the second part of the theorem. Much as abovewe may write d T V ( H ( P n ) , L ( n )) ≤ d T V ( H ( P n ) , H ( G n ))+ d T V ( H ( G n ) , ω ( G + n ))+ d T V ( ω ( G + n ) , L ( n )) . The first and third terms in the upper bound here are e − Ω( n ) by Theorem 2.3and Lemma 3.5. Since H ( G n ) = H ( G + n ), the second term is at most P ( α ( G + n ) >ω ( G + n )), which is e − Ω( n ) by Lemma 3.6. Thus d T V ( H ( P n ) , L ( n )) = e − Ω( n ) , asrequired. 16he last result shows that the distribution L ( n ) is not only needed for theapproximate generation process, but also corresponds to an important propertyof perfect graphs.Finally here in this section on stability and clique numbers, let us brieflyconsider algorithmic aspects, following the comments at the end of the lastsection, and using Lemma 3.6.Given a random perfect graph P n , let A be the set of vertices with degree atmost n/
2, and let B consist of the remaining vertices. Then wvhp exactly oneof the following two cases holds.(a) B is a clique and indeed is the unique maximum clique (by Lemma 3.5),so ω = | B | ; and A is a disjoint union of cliques, so α is either the numberof parts (cliques) in A or this number plus 1 (by Lemma 3.1), and we caneasily tell which in O ( n )-time.(b) A is the unique maximum stable set, so α = | A | ; and B induces a completemultipartite graph, so ω is either the number of parts or this number plus1, and we can easily tell which in O ( n )-time.In the first case, to tell if α is the number of parts plus 1 we just test if somevertex in B is non-adjacent to at least one vertex in each side clique; and simi-larly for the second case. In particular, wvhp in O ( n ) time we can determine ω and α (and know they are correct). Suppose P n ∈ u P n . From Lemma 3.6 and Theorem 2.3 we have ω ( P n ) ≥ (1 − ǫ ) n n wvhp, so we can of course find any fixed graph H as a subgraph of P n wvhp.We use H ⊆ i G to denote that H is an induced subgraph of G . We observedalready that the generated graphs, ρ ( Gen ( n )), are symmetric with respect totaking complements. The same is true for perfect graphs, that is P n ∼ P n , bythe weak perfect graph theorem [Lov72a, Lov72b]. Thus we see that, if H isfixed and P n ∈ u P n , then P ( H ⊆ i P n ) = P ( H ⊆ i P n ) . (15) Lemma 4.1.
Suppose that G + n ∼ Gen + ( n ) and that H is fixed. Then P ( H ⊆ i G + n ) = ( if H / ∈ GS + , − e − Ω( n ln n ) if H ∈ GS + . To prove this lemma, we first establish two preliminary claims.
Claim 4.2.
Suppose σ ∈ u Π n , and let a = a ( n ) ≤ n δ for some < δ < . Theprobability that σ contains at most a parts is at most exp {− (1 − δ + o (1)) n ln n } .Proof. The probability of the event in the statement is at most a n /B n , and thebound follows, since ln B n = (1 + o (1)) n ln n .17 laim 4.3. Suppose σ ∈ u Π n , and let l ≥ be a fixed integer. The probabilitythat all parts of σ are of size at most l is at most exp {− ( l + o (1)) n ln n } .Proof. We call a finite sequence ( a i ) ki =1 type if P ki =1 a i = n . We say that σ hastype ( a i ) if ( a i ) corresponds to the sizes of the parts of σ taken in any order.The number of partitions with type ( a i ) with a i ≤ l for all i is n ! k ! Q ki =1 a i ! ≤ n ! k ! ≤ e √ n √ πk · n n e − n k k e − k = n n ( nl ) nl O ( n ) = exp (cid:26) (1 − l ) n ln n + O ( n ) (cid:27) . The number of types ( a i ) with a i ≤ l for all i is at most l n = exp( O ( n )), andhence the number of partitions with parts at most l is exp (cid:8) (1 − l ) n ln n + O ( n ) (cid:9) .The statement in the lemma follows after dividing by B n . Corollary 4.4.
For fixed δ > and l ≥ , the probability that a random parti-tion σ ∈ u Π n contains at most n δ parts of size more than l is exp( − Θ( n ln n )) .Proof. Let Π ′ Q and Π ′′ Q be the classes of partitions of Q with at most n δ partsand with parts of size at most l respectively. The number of partitions describedin the statement can be bounded by P S ⊆ [ n ] | Π ′ S || Π ′′ S | = exp( − O ( n ln n )) by theprevious two claims. For a lower bound consider the partition π with n parts; P ( σ = π ) = B − n = exp( − Θ( n ln n )). Proof of Lemma 4.1.
Since GS + is a hereditary class of graphs and G + n takesas values only GS + -graphs with non-zero probability, we get that if H / ∈ GS + ,then P ( H ⊆ i G + n ) = 0 for any n .Now suppose that H ∈ GS + . Fix a unipolar arrangement ( H, A ). Supposethat π ∈ Π A corresponds to H [ A ], and let l = L ( π ) be the maximum size of apart in π . Let I n be the interval [ n/ − n / , n/ n / ]. Recall that we use k todenote the size of the central clique in the generation process. By Theorem 2.5, P ( k I n ) ≤ n − n for n sufficiently large; so we may condition on k ∈ I n . Let( G + n , C ) be the unipolar arrangement induced by the generation process.By Corollary 4.4 (with 1 / < δ <
1) there are at least | π | n / side cliques in C of size at least l with probability 1 − exp( − Θ( n ln n )), so we may conditionon this event. Hence we can find a collection of at least t = ⌈ n / ⌉ disjoint sets T j ⊆ C such that G + n [ T j ] ∼ = H [ A ]. Pick t = ⌈ n/ (3 | A | ) ⌉ disjoint subsets S i of C ,each of size | A | . Then G + n [ S i ] ∼ = H [ A ] for every i , and hence for 1 ≤ i ≤ t and1 ≤ j ≤ t we have p = P ( G + n [ S i ∪ T j ] ∼ = H ) ≥ −| A || A | . The probability thatno such pair induces H is (1 − p ) t t = exp( − Θ( n / )). The failure probabilityis dominated by the term exp( − Θ( n ln n )).The bound in Lemma 4.1 cannot be improved. For suppose H is the 4-edgepath: then P ( H i G + n ) ≥ P ( σ has only parts of size 1) ≥ /B n = e − O ( n ln n ) .
18t is now immediate that for a co-unipolar graph H we have P ( H ⊆ i G − n ) = P ( H ⊆ i G + n ) = 1 − e Ω( n ln n ) , and hence we arrive at the following theorem: Theorem 4.5.
Let H be any (fixed) graph, and let P n ∈ u P n . Then P ( H ⊆ i P n ) = e − Ω( n ) if H / ∈ GS , / ± e − Ω( n ) if H ∈ GS \ ( GS + ∩ GS − ) , − e − Ω( n ) if H ∈ ( GS + ∩ GS − ) . Proof.
Consider Q n = { G ∈ G n : H ⊆ i G } and P P n ( Q n ). Now the result followsfrom Lemma 4.1, Equation (7) and Theorem 2.3.Note that Theorem 4.5 accords with (15). It also follows from Theorem 4.5that for every graph H : P (cid:16) ( H ⊆ i P n ) [ ( H ⊆ i P n ) (cid:17) = ( e − Ω( n ) if H / ∈ GS , − e − Ω( n ) if H ∈ GS . and P (cid:16) ( H ⊆ i P n ) \ ( H ⊆ i P n ) (cid:17) = ( e − Ω( n ) if H / ∈ GS + ∩ GS − , − e − Ω( n ) if H ∈ GS + ∩ GS − . A j - clique colouring of a graph is a colouring of the vertices with j colours sothat no maximal clique is monochromatic (ignoring any isolated vertices). Weimprove Corollary 6 of [BGG + Theorem 5.1.
Almost all perfect graphs are -clique-colourable. Corollary 6 of [BGG +
04] is proved by showing that all generalised splitgraphs are 3-clique-colourable, and then using the theorem of Pr¨omel and Stegerdiscussed earlier. In the same article it is shown that there are generalised splitgraphs which are not 2-clique-colourable. There are several other subclasses ofperfect graphs for which the clique-chromatic number is known to be at most 3,and it was conjectured in [DSSW91] that perfect graphs have bounded clique-chromatic number. This was recently disproved in [CPTT16]. See [MMP16] forrecent work on clique-colourings of binomial random graphs and of geometricgraphs. We prove Theorem 5.1 using Theorem 2.3. We consider unipolar andco-unipolar graphs separately. 19 .1 Clique colouring of unipolar graphs
We start with a deterministic lemma. We say that a vertex sees a set of verticesif it has a neighbour from the set.
Lemma 5.2.
Let G be a unipolar graph, with given central clique C and thusgiven side cliques. If there is a vertex x ∈ C which sees each side clique whichis a maximal clique in G , then G is -clique-colourable.Proof. We may suppose that G contains at least 2 vertices. Let x be the vertexof C that sees each maximal side clique. Choose a vertex y Q ∈ N ( x ) ∩ Q foreach side clique Q with N ( x ) ∩ Q = ∅ . Define a colouring c : V ( G ) → { , } inthe following way: c ( v ) = v = x, v ∈ C − x, v = y Q for some side clique Q, . We claim that c is a proper clique colouring. Note first that N ( x ) = ∅ : thisis trivial if | C | ≥
2; and if C = { x } then there is a side clique Q , and Q mustcontain a neighbour of x (whether it is a maximal clique or not).Suppose for a contradiction that M is a monochromatic maximal clique.Since M is a clique, it must lie in the co-bipartite graph induced by C and a sideclique Q . Further M C : for if so then M = C , and C is not monochromaticif | C | ≥ C is not a maximal clique if C = { x } .If N ( x ) ∩ Q is empty, then M contains a vertex of colour 1 in Q , does notcontain x , and since M is monochromatic, M is contained in Q . By the definitionof x , Q is not maximal; hence M is not maximal, but this is a contradiction.Thus we may assume that N ( x ) ∩ Q is nonempty, and so contains y Q . Theneither (1) M ⊆ { x } ∪ ( Q \ { y Q } ) or (2) M ⊆ ( C \ { x } ) ∪ { y Q } . In case (1),we could add y Q to M , and in case (2) we could add x . Thus again M is notmaximal, and we have our final contradiction.Let t = t ( n ) = log n − n ( t is for threshold). Given a unipolararrangement ( G, C ) of order n , we use t to split the side cliques of ( G, C ) intotwo categories - the big side cliques with at least t vertices, and the small sidecliques with less than t vertices. We shall see in the next two lemmas that, for G + ∼ Gen + ( n ), whp the big side cliques can all be seen by some x ∈ C andthe small side cliques are not maximal, so we can apply the above deterministiclemma to deduce that whp G + is 2-clique-colourable. Lemma 5.3.
Let G + ∼ Gen + ( n ) , with induced unipolar arrangement ( G, C ) .Then with probability − − Ω(log n ) there is a vertex x ∈ C which sees each bigside clique.Proof. Condition on k = k ∈ I = [ n/ −√ n, n/ √ n ], as usual. Condition alsoon the partition σ of W = [ n ] \ [ k ]; and assume wlog that the permutation π is theidentity, so C = [ k ]. Let B be the set of all big side cliques, so |B| ≤ ( n − k ) /t .20or v ∈ C and Q ∈ B , the probability that v does not see Q is2 −| Q | ≤ − t = log nn . Thus the probability that v sees each Q ∈ B is at least (cid:18) − log nn (cid:19) n − k t = exp (cid:18) − log nn n − k t + O ( log nn ) (cid:19) = exp (cid:18) − ( 12 + o (1)) log n (cid:19) = n − α + o (1) where α = 1 / (2 log e ) ≈ .
35. Then, the probability that for each vertex v ∈ C there is some Q ∈ B which v fails to see is at most(1 − n − α + o (1) ) k ≤ e − n − α + o (1) , which is easily at most 2 − Ω(log n ) . Lemma 5.4.
Let G + ∼ Gen + ( n ) , with induced unipolar arrangement ( G, C ) .Then with probability − − Ω(log n ) no small side clique is a maximal clique.Proof. Condition on k = k ∈ I = [ n/ − √ n, n/ √ n ], as usual. Let us saythat a vertex v ∈ C extends a side clique Q if Q ⊆ N ( v ). For v ∈ C and Q ∈ B ,the probability that v does not extend Q is1 − −| Q | ≤ − − t = 1 − log nn ≤ e − log2 nn . Thus the probability that no v ∈ C extends Q is at most (cid:18) e − log2 nn (cid:19) k = e − ( + o (1)) log n . But there are at most n small side cliques, so the same upper bound holds forthe probability that some small side clique is a maximal clique. Lemma 5.5. G + ∼ Gen + ( n ) is -clique-colourable with probability − − Ω(log n ) .Proof. This follows from the last three lemmas.
Co-unipolar graphs are easier to clique-colour than unipolar graphs.
Lemma 5.6.
The random graph G − ∼ Gen − ( n ) is -clique-colourable wvhp.Proof. We first prove an auxiliary claim.21 laim 5.7.
Let G be a co-unipolar graph, with a unipolar arrangement ( G, C ) for G . Suppose that C is non-empty, and every vertex of C has neighbours inat least two side independent sets of C . Then G is -clique-colourable.Proof. Since C is non-empty, there are at least two side independent sets in C .Let Q be a side independent set. Colour the vertices in C ∪ Q with 1 and theremaining vertices with 2. A maximal clique contained completely in C containsa vertex from each side independent set and therefore is not monochromatic.There are no maximal cliques contained completely in C . A monochromaticclique which intersects both C and C must contain precisely two vertices – c ∈ C and q ∈ Q . However, such a clique can always be extended with aneighbour of c from a different side independent set.Let us check that G − satisfies the condition of the claim wvhp. By Theo-rem 2.5 we may assume k ∈ I = [ n/ − √ n, n/ √ n ]. Let ( G − , C ) be theco-unipolar arrangement induced by the generation. By Lemma 2.9 wvhp themaximum size of an independent set of C is at most n/ √ ln n = o ( n ). On theother hand every vertex of C has degree at least n/ C isbigger than the maximum size of a side independent set C , then every vertex of C must see at least two side independent sets.Theorem 5.1 now follows from Equation (7), Lemma 5.5, Lemma 5.6 andTheorem 2.3. Indeed we see that the probability that P n fails to be 2-clique-colourable is 2 − Ω(log n ) . Further, we may see from the proofs that there is an O ( n )-time algorithm that on input P n first generates a unipolar representationfor the graph or its complement, and then finds a 2-clique-colouring, with failureprobability 2 − Ω(log n ) . P n A Hamilton cycle in a graph is a cycle visiting every vertex exactly once. Recallthe distribution L ( n ) in Definition 2.1. In this section we prove the followingtheorem. Theorem 6.1.
Almost all perfect graphs are Hamiltonian. Indeed, for P n ∈ u P n and X ∼ L ( n ) , P ( P n is Hamiltonian ) = 1 − P ( X > n/ ± e − Ω( n ) . It follows from this result and Theorem 2.5 that P ( P n is Hamiltonian) = 1 − − ( + o (1)) log n = 1 − o (1) , which will complete the proof of (1). Indeed, we may see from the proofs thatthere is an O ( n )-time algorithm that succeeds wvhp on input P n , and either22utputs a Hamilton cycle, or a stable set of size > n/ Definition 6.2.
Given a graph G and two disjoint sets of vertices, S and T ,we say that an S – T matching M is a complete matching from S to T if theedges of M cover S . Lemma 6.3.
Suppose that ( G, C ) is a unipolar arrangement. Let T , T ⊆ C each contain exactly one vertex from each side clique and be such that T ∩ T consists of the vertices of C forming side cliques of size one. Let A and A bedisjoint subsets of C . Suppose that in G there are complete matchings M from T into A and M from T into A . Then G is Hamiltonian.Proof. Fix some ordering of the side cliques of C . There is a Hamilton cyclewhich enters each clique from an edge of M , visits all vertices inside and thenleaves the clique from an edge of M . After all side cliques are visited, theHamilton cycle visits any remaining vertices in C and finishes at the initialvertex. Definition 6.4.
Let G n,m, denote a random n × m bipartite graph with theedges present independently with probability / . Lemma 6.5.
Suppose that ≤ n ≤ m . Then G = ( V , V , E ) ∼ G n,m, contains a complete matching from V to V with probability − − m (1+ o m (1)) .Proof. We can find an isolated vertex in V with probability at least 2 − m , hence2 − m is a lower bound on the probability of failing to have a complete matching(from V to V ). Hall’s marriage theorem states that G contains a completematching if and only if | N ( S ) | ≥ | S | for every S ⊆ V . Let E S be the event that | N ( S ) | < | S | . We have P ( E S ) ≤ (cid:0) mm − s +1 (cid:1) − s ( m − s +1) , where s = | S | . Thus P ( ∪ S E S ) ≤ X S P ( E S ) = X | S |≤ P ( E S ) + X ≤| S |≤ n − P ( E S ) + X | S |≥ n − P ( E S ) . Clearly X | S |≤ P ( E S ) ≤ n − m + (cid:18) n (cid:19) m − m − ≤ n − m + O (1) . For 3 ≤ s ≤ min { n, m − } we have 2 − s ( m − s +1) ≤ − m − , so X ≤| S |≤ n − P ( E S ) ≤ X ≤| S |≤ min { n,m − } P ( E S ) ≤ n + m − m − ≤ − m + O (1) . m ≥ n + 2 we are done, so assume that m = n + δ where δ is 0 or 1, andconsider s = n − s = n . We have ( n − m − n +2) = ( m − δ − δ +2) ≥ m for m ≥
3, and n ( m − n + 1) = ( m − δ )( δ + 1) ≥ m for m ≥
2. Hence X | S |≥ n − P ( E S ) ≤ n − m for m ≥
3, which completes the proof.
Lemma 6.6.
The random graph G + ∼ Gen + ( n ) is Hamiltonian wvhp.Proof. Condition on | k − n/ | < √ n , which happens with wvhp by Theorem 2.5.By Lemma 2.8, the number of side cliques is at most n/
16 wvhp. Condition onthis event and on the arrangement ( G + , C ) induced by the generation process.Partition C into two near equal parts A, B ⊆ C with || A | − | B || ≤
1, so that | A | ∼ | B | ∼ n/
4. Find T A , T B ⊂ C as in Lemma 6.3. By Lemma 6.5, we canfind complete matchings between T A and A and between T B and B wvhp. Wenow see that G + satisfies the conditions of Lemma 6.3 wvhp and therefore G + is Hamiltonian wvhp. Hamilton cycles in random bipartite graphs have been studied in Frieze [Fri85]and in Bollob´as and Kohayakawa [BK91]. However, we are interested in thedense case and seek an exponentially small failure probability, which does notappear to have been done before. We shall show:
Theorem 6.7. If G ∼ G n,n, then G is Hamiltonian with probability − − (1+ o (1)) n . This theorem will follow from the deterministic Theorem 6.10 below, basedon ‘Posa flips’, but first we need some preparation.
Definition 6.8.
Suppose G is a bipartite graph. We define e α ( G ) to be themaximum integer k such that we can find k -sets S ⊆ V and S ⊆ V with E ( S , S ) = ∅ for a bipartition ( V , V ) of G . We call the pair ( S , S ) a bipartitehole of G . We denote the minimum degree of a graph G by δ ( G ). Lemma 6.9.
Suppose G is a bipartite graph such that δ ( G ) > e α ( G ) . Then G is ( δ ( G ) − e α ( G )) -connected.Proof. Let S ⊆ V ( G ) be a set of less than δ ( G ) − e α ( G ) vertices. We must showthat G ′ = G [ V ( G ) \ S ] is connected. Let ( V , V ) be a bipartition of G .Let v ∈ V \ S and v ∈ V \ S . Then | N ( v ) \ S | ≥ e α ( G )+1 and | N ( v ) \ S | ≥ e α ( G )+1; and hence either v and v are adjacent, or a neighbour of v is adjacentto a neighbour of v in G ′ . Thus v and v are connected in G ′ by a path oflength at most 3. If v , v ∈ V \ S , then v has a neighbour u ∈ V \ S , andwe have already seen that v and u must be connected in G ′ , so v and v areconnected in G ′ . The same conclusion holds if v , v ∈ V \ S .24 heorem 6.10. An n × n bipartite graph G with n ≥ is Hamiltonian if δ ( G ) ≥ e α ( G ) + 1 .Proof. All paths in this proof have their terminal vertices labelled as start and end . Suppose P = { v , . . . , v l } is a path with start v and end v l , and suppose v l is adjacent to v i . A flip of P around v i is the path { v , . . . , v i , v l , . . . , v i +1 } over V ( P ), starting at v and ending at v i +1 . (If i = l − P = { v , . . . , v l } be a path starting at v , ending at v l , and havingmaximum length of any path in G . For k = 0 , , W k be the set of endsof paths obtained from P by at most k flips. Clearly W = { v l } . Since P ismaximal, v l is adjacent to vertices of P only, hence W = { v i +1 : v i v l ∈ E ( G ) } satisfies | W | ≥ δ ( G ).Assume that δ ( G ) ≥ e α ( G ) + 1, and let ( V , V ) be a bipartition of G .Without loss of generality suppose that v l ∈ V . Thus each set W k ⊆ V . Tocomplete the proof it is sufficient to show that every vertex of V is adjacent to avertex of W . Indeed, suppose this is the case. If v ∈ V , then P can be closedto a cycle, which must be Hamiltonian since G is connected by Lemma 6.9 and P has maximum length. If v ∈ V , then there is a vertex u ∈ V \ V ( P ). But thiscontradicts the maximality of P , because there is a path Q with V ( Q ) = V ( P )ending at a neighbour of u obtained from P by two flips, and Q can be extendedto contain u , yielding a longer path.We claim that | W | ≥ n − e α ( G ) . (16)Assuming this, and recalling that δ ( G ) ≥ e α ( G ) + 1, we see that every vertexof V must be adjacent to a vertex of W , and we are done. So it remains toestablish the claim (16).Suppose v i +1 ∈ W is adjacent to v j = v i . Then v j +1 ∈ W if j < i ; (17) v j − ∈ W if j > i. (18)More precisely, the path obtained from P by first flipping around v i and then v j ends with v j +1 or v j − depending on whether v j comes before v i in P .Since | W | ≥ e α ( G ) + 1, we can find an integer t such that v t +1 ∈ W andif we write F = { v i ∈ V ( G ) : i > t + 1 } and F = { v i ∈ V ( G ) : i < t + 1 } , then | F ∩ W | ≥ e α ( G ) and | F ∩ W | ≥ e α ( G ). Let M = ( F ∩ W ) ∪ { v t +1 } and M = ( F ∩ W ) ∪ { v t +1 } . It follows from (18) that if v j ∈ F has a neighbour in M , then its predecessor, v j − , is contained in W ; and from (17) it follows thatif v j ∈ F , v j = v t has a neighbour in M , then its successor, v j +1 , is containedin W . But v t +1 ∈ W , and hence v t +1 ∈ W , so no special care is required forthe case v j = v t . For k = 1 , N k = N [ M − k ] ∩ F k and N ck = ( F k ∩ V ) \ N k (and hence F k ∩ V = N k ∪ N ck ); and let N − be the set of predecessors in P ofvertices in N , and let N +2 be the set of successors of N . We see that25. N − ∪ N +2 ⊆ W \ { v l } ;2. | N − | = | N i | , | N +2 | = | N | ;3. N − ∩ N +2 = { v t +1 } ;4. v l ∈ W .Since F k ∩ V = N k ∪ N ck and v t +1 ∈ V , we observe that n = | V | = | N | + | N c | + | N | + | N c | , and therefore | W | ≥ | N − ∪ N +2 | + 1 = ( | N − | + | N +2 | −
1) + 1= | N | + | N | = n − ( | N c | + | N c | ) . However, E ( M − k , N ck ) = ∅ (by the definition of N k ) and | M k | > e α ( G ), so | N ck | ≤ e α ( G ), and hence | W | ≥ n − e α ( G ). Thus we have proved (16), and theproof is complete.Theorem 6.7 now follows because if G ∼ G n,n, and t = √ n , then the events { δ ( G ) < t + 1 } and { e α ( G ) > t } both have probability 2 − (1+ o (1)) n . As a lowerbound consider the event that G contains an isolated vertex. For similar results,concerning non-bipartite graphs, we refer to [MY17]. The paper [MY17] explainshow Theorem 6.7 could be extended to state that we can find n δ , 0 < δ < ,edge-disjoint Hamilton cycles with probability 1 − − (1+ o (1)) n . Lemma 6.11.
Suppose that G − ∼ Gen − ( n ) and X ∼ L ( n ) . Then G − isHamiltonian with probability P ( L ( n ) ≤ n/ − e − Ω( n ) .Proof. Recall that
Gen − ( n ) is a random quadruple ( − , E, ( k, σ ) , π ). If k > n/ α ( G − ) > n/
2, which makes it impossible to contain a Hamilton cycle. Weshow that conditional on k ≤ n/ G − is Hamiltonian wvhp, which will completethe proof.From Theorem 2.5 and Lemma 2.8 we may condition on the event that k ≥ n/ −√ n and the event that the number of parts in σ is at least n/ (3 ln n ) (sinceboth hold wvhp). Condition also on ( G − , C ) being the co-unipolar arrangementinduced by the generation. Partition C as Q ∪ T , so that | C | = | Q | = k , | T | = n − k ≤ √ n , and T contains vertices only from different parts of σ .By Corollary 6.7, there is a Hamilton cycle in the induced subgraph G − [ C ∪ Q ]wvhp. Condition on this event and on H being a Hamiltonian cycle. If T isempty we are done, so assume not.Now we have a Hamilton cycle H in G − [ C ∪ Q ], and a non-empty set T ofvertices inducing a clique. Let u, v be vertices in T , and note that there is a u – v Hamilton path in G − [ T ] (where u = v if | T | = 1). We may assume that themaximum size of a side stable set is at most n/ log n by Lemma 2.9. Let S u , S v be the side stable sets containing u , v respectively, and let R be the set of vertices26n Q \ S u . Then u is adjacent to each vertex in R , and | R | ≥ k −| S u | ≥ n/ | S v | for n sufficiently large.The vertices in R have at least | R | − | S v | ≥ n/ H whichare not in S v , and if v is adjacent to any of these neighbours then we have aHamilton cycle in the whole graph. But the probability that this fails is at most2 − n/ , since any neighbour in C must be adjacent to v and so we may assumethere are none. A set of vertices in a graph G is a cutset if removing these vertices leaves adisconnected graph. The (vertex) connectivity of G is the minimum size of acutset, except that, if G is complete, then by convention κ ( G ) = | V ( G ) | − δ ( G ) to denote the minimum vertex degree. Observe that δ ( G ) isnatural upper bound for κ ( G ). We prove the following theorem: Theorem 7.1.
The connectivity of almost all perfect graphs equals their mini-mum degree. Indeed, wvhp κ ( P n ) = δ ( P n ) , where P n ∈ u P n .Proof. We first consider unipolar graphs. Let ( G + , C ) be the unipolar ar-rangement induced by the generation of Gen + ( n ). Then, for each fixed ǫ with0 < ǫ < , G + satisfies each of the following properties wvhp:(a1) ∀ u ∈ C , d C ( u ) ≤ (1 + ǫ ) n .(a2) ∀ u ∈ C , d C ( u ) ≤ ǫn .(a3) ∀ u, v ∈ C , | ( N ( v ) ∪ N ( u )) ∩ C | ≥ (1 − ǫ ) n .(a4) | C | > (1 − ǫ ) n .Chernoff’s inequality is sufficient to prove (a1) and (a3); for (a2) we can useLemma 2.9, and (a4) follows from the concentration theorem, Theorem 2.5.These conditions are sufficient to ensure κ ( G + ) = δ ( G + ). Note first that if C = ∅ then G + is a clique and the result holds; so we may assume that C = ∅ .Suppose for contradiction that Q ⊂ V = V ( G + ) is a separator of G + with | Q | < δ ( G + ). Note that δ ( G + ) ≤ (1 + 5 ǫ ) n ≤ | C | , and so C Q . We claimthat every vertex of V \ Q is connected to a vertex in C \ Q , which will yieldour contradiction. This is obvious for vertices u ∈ C \ Q . If u ∈ C , then since | Q | < δ ( G + ), u must have a neighbour v outside Q . If v ∈ C \ Q we are done, sosuppose that v ∈ N ( u ) ∩ C . From (a3), u and v see together at least (1 − ǫ ) n vertices of C , while | Q | is at most (1 + 5 ǫ ) n , so there must be a vertex from C \ Q connected to either u or v .The situation is similar for co-unipolar graphs. Let ( G − , C ) be the co-unipolar arrangement induced by the generation of Gen − ( n ). Then, for eachfixed ǫ with 0 < ǫ < , G − satisfies the following conditions wvhp.27b1) ∀ u ∈ C , d ( u ) ≤ (1 + ǫ ) n .(b2) Each side independent set in C has size at most ǫn .(b3) | C | > (1 − ǫ ) n .To prove these, we can use a Chernoff bound for (b1), Lemma 2.9 for (b2), andTheorem 2.5 for (b3).These conditions are sufficient to ensure κ ( G − ) = δ ( G − ). Note first that if C = ∅ then G − is a complete multipartite graph and the result holds; so we mayassume that C = ∅ . Thus δ ( G − ) ≤ (1 + ǫ ) n . Suppose that Q ⊂ V = V ( G − )is a cutset of G − and | Q | < δ ( G − ). Fix some vertex w ∈ V \ Q . We claim thatevery vertex of V \ Q is connected to w .First consider u ∈ C . Then w and u are not neighbours (if and) only if theyare contained in the same side independent set I . But then there is a vertex x ∈ C \ Q in a different side independent set, since | C \ Q | > (1 − ǫ ) n ≥ | I | ; andthe path u, x, w connects u to w . Now consider u ∈ C . Since | Q | < δ ( G ), u musthave a neighbour in C , and we just seen that all vertices in C are connectedto w .We have now shown that P Gen ( n ) ( κ ( G ) = δ ( G )) = 1 − e − Ω( n ) , and Theo-rem 2.3 completes the proof. The chromatic index of a graph G , χ ′ ( G ), is the minimum number of coloursrequired to colour the edges, so that no edges with the same colour share avertex. Vizing proved in [Viz64] that χ ′ ( G ) = ∆( G ) or χ ′ ( G ) = ∆( G ) + 1. If χ ′ ( G ) = ∆( G ), then G is said to be class one, otherwise it is said to be classtwo.In this subsection we prove Theorem 7.2.
Almost all perfect graphs are class one graphs.
Class one graphs have a natural connection with perfect graphs. Denotethe line graph of a graph G by L ( G ). Then, for graphs G with ∆( G ) ≥ G ) = ω ( L ( G )) and χ ′ ( G ) = χ ( L ( G )); and hence G is a class one graph iff ω ( L ( G )) = χ ( L ( G )). Vizing showed in [Viz65] that each graph of class two hasat least 3 vertices of maximum degree (and indeed the vertices of maximumdegree induce a subgraph with a cycle). Erd˝os and Wilson proved in [EW77]that almost all graphs have a unique vertex of maximum degree, and thus areclass one, see also [FJMR88]. We shall show that almost all perfect graphshave a unique vertex of maximum degree, and thus are class one and can be∆-edge-coloured in O ( n ) time. (If G has a unique vertex v of maximum degreethen we can ∆-edge-colour G in O ( nm )-time as follows. Pick an edge e incidentwith v , use for example the algorithm of [MG92] to ∆( G )-edge-colour G \ e , andthen use a standard Vizing fan iteration to colour e .)Let d j = d j ( G ) be the j th largest degree in the list of all v ( G ) degrees.Theorem 15 in Chapter III (The degree sequence) of the book [Bol01] concerns28he first few gaps d i − d i +1 for a random graph G n,p . The next lemma is a mucheasier result that may be proved along similar lines; we do not spell a proofhere. Lemma 7.3.
Whp B := G n,m, / with m/n → is such that d ( B ) − d ( B ) ≥ n / log n . where d ( B ) and d ( B ) are the largest and second largest degrees of vertices inthe first colour class. Let ( G + , C ) be the unipolar arrangement induced by the generation of Gen + ( n ), and let B denote the corresponding bipartite graph with parts C and C . With high probability, d ( u ) ≥ (1 − ǫ ) n for all u in C , and d ( v ) ≤ (1 + ǫ ) n for all v in C . Hence, by Lemma 7.3, whp d ( G + ) − d ( G + ) = d ( B ) − d ( B ) ≥ n / n , and so whp G + has a unique vertex of maximum degree.Let ( G − , C ) be the co-unipolar arrangement induced by the generation of Gen − ( n ), and let B denote the corresponding bipartite graph with parts C and C (in this order). With high probability, d ( u ) ≥ (1 − ǫ ) n for all u in C , and d ( v ) ≤ (1 + ǫ ) n for all v in C . Also, by Lemma 2.9, with high probability | d C ( u ) − d C ( v ) | <
10 ln n for each u, v ∈ C , and therefore d ( G − ) − d ( G − ) ≥ d ( B ) − d ( B ) −
10 ln n ≥ (1 − ǫ ) n / n −
10 ln n, and so whp G − has a unique vertex of maximum degree.We have now seen that whp both G + and G − have a unique vertex ofmaximum degree, which implies they are class one; Theorem 7.2 now followsfrom Theorem 2.3. We show that a sequence of uniformly and independently sampled n -vertexperfect graphs converges with probability one to the graphon W P ( x, y ) = [ x ≤ /
2] + [ y ≤ / .1 Notation We use the notation for left limits from [Lov12]. Suppose throughout (as usual)that P n is a uniformly sampled perfect graph on vertex set [ n ]. Let λ be theLebesgue measure on R . A kernel W : [0 , → R is an a.e. bounded symmet-ric measurable function. A graphon W is a kernel with 0 ≤ W ≤ cutnorm , || . || (cid:3) , is a norm on the vector space of kernels defined by || W || (cid:3) = sup S,T ⊆ [0 , (cid:12)(cid:12)(cid:12)(cid:12)Z S × T W dλ (cid:12)(cid:12)(cid:12)(cid:12) . The supremum is taken over all measurable
S, T ⊆ [0 , S [0 , denotethe set of invertible measure preserving maps on [0 , W ϕ ( x, y ) := W ( ϕ ( x ) , ϕ ( y )) for ϕ ∈ S [0 , . The cut distance , δ (cid:3) , is defined by δ (cid:3) ( U, W ) = inf ϕ ∈ S [0 , || U − W ϕ || (cid:3) for kernels U, W . For a graph G we define the graphon W G by partitioning[0 ,
1] = S ⊔ . . . ⊔ S v ( G ) , λ ( S i ) = 1 /v ( G ), and writing W ( x, y ) = ij ∈ E ( G ) if ( x, y ) ∈ S i × S j . Let hom( F, G ) and hom inj ( F, G ) be the number of homo-morphisms (edge preserving maps), injective homomorphisms respectively, fromgraph F to graph G . Let t ( F, G ) = hom(
F,G ) n k and let t inj ( F, G ) = hom inj ( F,G ) n ( n − ... ( n − k +1) ,where k = v ( F ) and n = v ( G ). For a graph F and kernel W define t ( F, W ) = Z [0 , v ( F ) Y ij ∈ E ( G ) W ( x i , x j ) Y i ∈ V ( G ) dx i . We have t ( F, G ) = t ( F, W ϕG ) for each pair of graphs F and G and each ϕ ∈ S [0 , .A sequence of graphs, G , G , . . . , is said to converge to a graphon W , denotedby G n → W , if W G n converges to W with respect to δ (cid:3) . Theorem 8.1.
We have δ (cid:3) ( W P n , W P ) ≤ n − / with probability − e − Ω( √ n log n ) , (19) δ (cid:3) ( W P n , W P ) ≤ (log n ) − with probability − e − Ω( n ) . (20)Before we prove Theorem 8.1 we discuss a few corollaries. Suppose each P n issampled independently. Corollary 8.2.
With probability one we have P n → W P .Proof. This follows from Theorem 8.1 (either part), a Borel-Cantelli lemma,and Theorem 11.22 of [Lov12]. 30 orollary 8.3.
For a fixed graph F | t inj ( F, P n ) − t ( F, W P ) | ≤ (cid:18) v ( F )2 (cid:19) n − + e ( F ) n − / with probability at least − e − Ω( √ n log n ) .Proof. It is easy to see that for every graph G we have | t inj ( F, G ) − t ( F, G ) | ≤ (cid:18) v ( F )2 (cid:19) v ( G ) . Theorem 10.23 in [Lov12] states that | t ( F, W ) − t ( F, W ) | ≤ e ( F ) · δ (cid:3) ( W , W )for every two graphons W and W . The rest follows from part 1 of Theorem 8.1.The notation in the second corollary may be misleading, in the sense that thestatement has nothing to do with graph limits. Indeed, the value t ( F, W P )can be calculated explicitly and then used as an estimate on the number ofsubgraphs F of P n . For example, if F is the single edge K then t ( F, W P ) = so the random perfect graph P n contains (1+ O ( n − / )) n edges with probabilityat least 1 − e − Ω( √ n log n ) ; and similarly P n contains (1 + O ( n − / )) n triangleswith probability at least 1 − e − Ω( √ n log n ) . Using an inclusion-exclusion argumentwe can estimate the densities of induced graphs and in particular Corollary 8.3implies that the split graphs are the only graphs with induced subgraph densitybounded away from zero.We can be more precise for example concerning the number e ( P n ) of edgesof P n . Note that the expected value is exactly (cid:0) n (cid:1) . Also, whp e ( G + n ) = n + ( + o (1)) n ln n and so whp e ( G − n ) = n − ( + o (1)) n ln n ; and hencewhp e ( P n ) = n + O ( n log n ), with a bimodal distribution. We consider the unipolar case G + ∼ Gen + ( n ) and the co-unipolar case G − ∼ Gen − ( n ) separately, and start with the former. Suppose Gen + ( n ) = (1 , E, ( k, σ ) , π ).Let ϕ ∈ S [0 , be a measure preserving map, mapping each vertex of G + to aninterval of the form V i = [ i − n , in ) where i ∈ [ n ], and in addition let ϕ map thecentral clique of W G + to [0 , k/n ) and the side cliques to [ k/n, S, T ⊆ [0 , δ (cid:3) ( G + , W P ) ≤ ǫ it is sufficient to showthat (cid:12)(cid:12)(cid:12)(cid:12)Z S × T ( W ϕG + − W P ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ ǫ. laim 8.4. We can find S ′ , T ′ ⊆ [0 , of the form S ′ = ∪ i ∈ I s V i and T ′ = ∪ i ∈ I t V i , where I s , I t ⊆ [ n ] , such that (cid:12)(cid:12)(cid:12)(cid:12)Z S × T ( W ϕG + − W P ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12)Z S ′ × T ′ ( W ϕG + − W P ) (cid:12)(cid:12)(cid:12)(cid:12) . Proof.
Initially set S ′ = S and T ′ = T . For i = 1 , . . . , n in turn, check if R T ′ ( W ϕG + − W P )( i − n , y ) dy has the same sign as R S ′ × T ′ ( W ϕG + − W P ) dλ , and ifso, add the entire interval V i to S ′ , or remove V i from S ′ otherwise. Theseoperations can only increase the absolute value of the integral. Now repeat thesame procedure for T ′ . It is clear that in the end we obtain S ′ and T ′ with thedesired properties.Every such pair ( S ′ , T ′ ) is measurable. We may assume that initially we aregiven S and T of this form.Let A = [0 , / B = [1 / , A = [0 , k/n ), B = [ k/n, A = A ∩ A , B = B ∩ B and I = A ∪ B . By Theorem 2.5, wvhp both λ ( A ) and λ ( B ) arecontained in h − n − / , i . Condition on this event, and now we have (cid:12)(cid:12)(cid:12)(cid:12)Z S × T ( W ϕG + − W P ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z ( S × T ) ∩ ( I × I ) ( W ϕG + − W P ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + n − / . Let O = ( S × T ) ∩ ( A × A ), O = ( S × T ) ∩ ( A × B ), O = ( S × T ) ∩ ( B × A ), O = ( S × T ) ∩ ( B × B ), and w i = (cid:12)(cid:12)(cid:12)R O i ( W ϕG + − W P ) (cid:12)(cid:12)(cid:12) .We have W ϕG + = W P = 1 over O , so clearly w = 0. The restriction of W ϕG + over O is a step function corresponding to a random bipartite graph withdensity 1 /
2, while W P is uniformly 1 /
2. This classical example motivated thestudy of the cut distance, but we give full details here.
Claim 8.5.
Suppose M is a random {− , } -valued m × m matrix, where eachentry has expected value and is independent from the others. Then the maxi-mum absolute value of a rectangular sum in M is at most m / wvhp.Proof. Let R ⊆ [ m ] be a subset of the rows of M , and let Q ⊆ [ m ] be a subsetof the columns of M . Clearly M Q,R := P r ∈ R,q ∈ Q M r,q ∼ X − | Q || R | , where X ∼ Bin ( | Q || R | , / P ( | M Q,R | ≥ x ) = P (cid:18)(cid:12)(cid:12)(cid:12)(cid:12) X − | Q || R | (cid:12)(cid:12)(cid:12)(cid:12) ≥ x (cid:19) ≤ e − x m . Taking x = m / , and the union bound over all 2 m possible choices for R, Q completes the proof.Setting m = ⌈ n/ ⌉ and rescaling in Claim 8.5 shows that w and w are boundedby n − / wvhp.Finally, w is bounded by twice the number of edges in the partition partof G + divided by n . Lemma 2.9 states that the maximum size of a part in a ran-dom partition is at most x with probability at least 1 − (1+ o (1)) e − x (ln x − ln ln n − n .32he maximum size of a part bounds the maximum degree. Here we have tomake a compromise between low cut distance and high probability. To favourthe former, use Lemma 2.9 with x = √ n and deduce that (cid:12)(cid:12)(cid:12)(cid:12)Z S × T ( W ϕG + − W P ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ √ n with probability 1 − e − Ω( √ n log n ) .Now let us aim for a bound that holds wvhp. Let log n ≤ x ≤ n/
2. Observethat if x vertices are contained in parts of σ each of size at least s , then x vertices are contained in at most ⌈ x/s ⌉ parts. Thus, assuming s ≤ x and so ⌈ x/s ⌉ ≤ x/s , the probability that some set of x points is contained in parts of σ each of size at least s is at most (cid:18) nx (cid:19) (cid:16)l xs m(cid:17) x n B n − x B n ≤ (cid:18) enx · xs (cid:19) x (1 + o (1)) (cid:18) r n r n − x (cid:19) n exp {− x ( r n − /r n ) } = (cid:18) ens (cid:19) x exp {− x (ln n − (1 + o (1)) log log n ) } = exp {− x (ln s − (1 + o (1)) log log n ) } . The factor n in the first bound above arises since the ‘last part’ for the x -set mayneed to be amalgamated with some part for the ( n − x )-set. Set x = ⌈ n/ (3 log n ) ⌉ and s = n/ log n , to see that wvhp at most n/ (3 log n ) points are containedin parts of σ each of size at least n/ log n . Since also wvhp no part has sizegreater than n/ log n by (11), we see that wvhp the number of edges in G + [ S ]is at most ( 12 + o (1)) n · n log n + n n n log n so wvhp w ≤ (log n ) − . Hence wvhp δ (cid:3) ( G + , W P ) ≤ (log n ) − .Let J be the graphon with J ( x, y ) = 1 for each x, y ∈ [0 , J − W P may be written as W ψP for some ψ ∈ S [0 , . Then Z S × T ( W ϕG − − W P ψ ) = − Z S × T ( W ϕG + − W P ) . Hence, the bounds for G + transfer to G − . Theorem 2.3 completes the proof.Using (12) we may see that the bound in (20) is best possible, in the sensethat, if ǫ ( n ) = o (1), then δ (cid:3) ( W P n , W P ) ≥ ǫ ( n )(log n ) − holds with probability e − o ( n ) . We conclude this paper with the proofs of Theorems 2.3 and 2.5, which weredeferred to here. Of course, the proofs of these theorems do not depend on anyof the earlier work which used them. 33 .1 Proof of Theorem 2.3
We shall prove (5) and (6) and complete the proof of Theorem 2.3. For realnumbers p ≥ L p norm of X in the probability space ( G n , G n , µ )by k X k µp . Recall that k X k µp is non-decreasing in p . (To see this, let 1 ≤ p < q and note that f ( x ) = x q/p is convex for x >
0: hence by Jensen’s inequality, k X k µp = ( E | X | p ) /p = ( f ( E | X | p )) /q ≤ ( E f ( | X | p )) /q = ( E | X | q ) /q = k X k µq . )Let ν and µ be discrete probability measures on (all subsets of) G n such that ν ( G ) > µ ( G ) >
0. In this case ν is said to be absolutely continuous with respect to µ , written ν ≪ µ . The Radon–Nikodym derivative dνdµ is arandom variable given by dνdµ ( G ) = ν ( G ) µ ( G ) µ ( G ) > . We may express the totalvariation distance between ν and µ in terms of dνdµ and the L norm k·k µ : wehave2 d T V ( ν, µ ) = X G | ν ( G ) − µ ( G ) | = X G (cid:12)(cid:12)(cid:12)(cid:12) dνdµ ( G ) − (cid:12)(cid:12)(cid:12)(cid:12) µ ( G ) = (cid:13)(cid:13)(cid:13)(cid:13) dνdµ − (cid:13)(cid:13)(cid:13)(cid:13) µ . (The same holds for general, not necessarily discrete, probability measures, pro-vided that ν ≪ µ .) Also note that for each Gd P Gen ( n ) d P P n ( G ) = P ( ρ ( Gen ( n )) = G ) |P n | . The main result in this section is stronger than Theorem 2.3.
Theorem 9.1.
For each real p ≥ we have d T V ( P Gen ( n ) , P P n ) = (cid:13)(cid:13)(cid:13)(cid:13) d P Gen ( n ) d P P n − (cid:13)(cid:13)(cid:13)(cid:13) P P n ≤ (cid:13)(cid:13)(cid:13)(cid:13) d P Gen ( n ) d P P n − (cid:13)(cid:13)(cid:13)(cid:13) P P n p = e − Θ( n ) . We have just noted the first two (in)equalities in the statement of the theorem:the non-trivial part is the final equality. To prove it we start from the triangleinequality (cid:13)(cid:13)(cid:13)(cid:13) d P Gen ( n ) d P P n − (cid:13)(cid:13)(cid:13)(cid:13) P P n p ≤ (cid:13)(cid:13)(cid:13)(cid:13) d P Gen ( n ) d P P n − d P GS n d P P n (cid:13)(cid:13)(cid:13)(cid:13) P P n p + (cid:13)(cid:13)(cid:13)(cid:13) d P GS n d P P n − (cid:13)(cid:13)(cid:13)(cid:13) P P n p . It is straightforward to show from (3) that the second term on the right is e − Ω( n ) . For the first term on the right, routine manipulations yield (cid:13)(cid:13)(cid:13)(cid:13) d P Gen ( n ) d P P n − d P GS n d P P n (cid:13)(cid:13)(cid:13)(cid:13) P P n p = (cid:13)(cid:13)(cid:13)(cid:13) d P Gen ( n ) d P GS n − (cid:13)(cid:13)(cid:13)(cid:13) P GS n p (cid:18) |P n ||GS n | (cid:19) − /p . Since |P n ||GS n | = 1 + e − Ω( n ) by (3), it follows that it is sufficient to show that (cid:13)(cid:13)(cid:13) d P Gen ( n ) d P GS n − (cid:13)(cid:13)(cid:13) P GS n p = e − Ω( n ) to complete the proof of Theorem 9.1.We need a few definitions to continue.34 efinition 9.2. Let
CGS + n (from coloured generalised split graphs) be the setof all unipolar arrangements of order n and let CGS n = CGS + n × {− , } . Recall that L n was introduced just before Definition 2.1. We see that |CGS + n | = L n and |CGS n | = 2 L n . A closer look at the definitions revealsthat if (( G, C ) , B ) ∈ u CGS n is uniformly selected, if we set H = G if B = 1and H = G otherwise, and set H ∼ ρ ( Gen ( n )), then H and H are equal indistribution. Therefore P Gen ( n ) ( G ) = 1 |CGS n | X C ⊆ [ n ] (cid:8) [( G, C ) ∈ CGS + n ] + [( G, C ) ∈ CGS + n ] (cid:9) . To simplify the notation we define R + ( G ) = P C ⊆ [ n ] [( G, C ) ∈ CGS + n ], R − ( G ) = R + ( G ) and R ( G ) = R + ( G ) + R − ( G ), so that P G R ( G ) = |CGS n | and P Gen ( n ) ( G ) = R ( G ) |CGS n | . Note that R ( G ) > G ∈ GS n . Now we see that (for each G ) d P Gen ( n ) d P GS n = R |CGS n | |GS n | = R |GS n ||CGS n | . It is clear from the definitions that k R − k P GS n = |CGS n ||GS n | −
1. We have (cid:13)(cid:13)(cid:13)(cid:13) d P Gen ( n ) d P GS n − (cid:13)(cid:13)(cid:13)(cid:13) P GS n p = (cid:13)(cid:13)(cid:13)(cid:13) R |GS n ||CGS n | − (cid:13)(cid:13)(cid:13)(cid:13) P GS n p = (cid:13)(cid:13)(cid:13)(cid:13) R − |CGS n ||GS n | (cid:13)(cid:13)(cid:13)(cid:13) P GS n p |GS n ||CGS n |≤ (cid:13)(cid:13)(cid:13)(cid:13) R − − |CGS n ||GS n | (cid:13)(cid:13)(cid:13)(cid:13) P GS n p ≤ k R − k P GS n p + (cid:12)(cid:12)(cid:12)(cid:12) − |CGS n ||GS n | (cid:12)(cid:12)(cid:12)(cid:12) = k R − k P GS n p + k R − k P GS n ≤ k R − k P GS n p . Thus (cid:13)(cid:13)(cid:13)(cid:13) d P Gen ( n ) d P GS n − (cid:13)(cid:13)(cid:13)(cid:13) P GS n p ≤ k R − k P GS n p . (21)Further simplifications lead us to k R − k P GS n p = k ( R + − GS + n ) + ( R − − GS − n ) + ( GS + n + GS − n − k P GS n p ≤ k R + − GS + n k P GS n p + k GS + n GS − n k P GS n p = 2 k R + − k P GS + n p (cid:18) |GS + n ||GS n | (cid:19) /p + k GS + n GS − n k P GS n p ≤ k R + − k P GS + n p + k GS + n GS − n k P GS n . (22)The only combinatorial argument we need to complete this proof is phrasedas a lemma with a proof in the next section:35 emma 9.3. For every integral j ≥ we have k R j + k P GS + n ≤ k R + k P GS + n (1 + 2 − n/ o ( n ) ) . From now on until equation (23), all norms are for µ = P GS + n , and we drop thesuperscript P GS + n for legibility. By combining the lemma for j = 2 with Jensen’sinequality we get k R + k ≤ k R k ≤ k R + k (1 + e − Ω( n ) ) , and so k R + k ≤ e − Ω( n ) . Hence, by the last lemma, we have k R j + k ≤ e − Ω( n ) . Now 1 ≤ R j + ≤ R j +1+ a.s. (that is, for all G ∈ GS + n ), and therefore k R j +1+ − R j + k = k R j +1+ k − k R j + k = e − Ω( n ) . Finally, we see that for each j ≥ k ( R + − j +1 k = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ( R + − j X i =0 (cid:18) ji (cid:19) R i + ( − j − i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≤ j X i =0 (cid:18) ji (cid:19) (cid:13)(cid:13) ( R + − R i + (cid:13)(cid:13) ≤ j (cid:13)(cid:13)(cid:13) ( R + − R j + (cid:13)(cid:13)(cid:13) ≤ e − Ω( n ) . Therefore k R + − k P GS + n p = e − Ω( n ) (23)for each real p ≥ P GS + n ( α ( G ) ≥ ω ( G )) = e − Ω( n ) and P GS − n ( α ( G ) ≤ ω ( G )) = e − Ω( n ) . Partition the set of graphs which are both unipolar and co-unipolar into U = { G ∈ GS + n ∩ GS − n : α ( G ) > ω ( G ) } and U = { G ∈ GS + n ∩GS − n : α ( G ) ≤ ω ( G ) } . We see from Lemma 3.6 that U consists of rare unipolargraphs and U consists of rare co-unipolar graphs; or more formally, | U | = e − Ω( n ) |GS + n | and | U | = e − Ω( n ) |GS − n | . Hence k GS + n GS − n k P GS n = |GS + n ∩ GS − n | / |GS n | = e − Ω( n ) . (24)Now (22), (23) and (24) imply that k R − k P GS n p = e − Ω( n ) . (25)We now have all the pieces to complete the story, assuming Lemma 9.3.By (23), L n / |GS + n | − k R + − k P GS + n = e − Ω( n ) , so using also (24) |GS n | = 2 |GS + n | − |GS + n ∩ GS − n | = 2 L n (1 − e − Ω( n ) ) , and we have proved (5). Also, by (21) and (25), (cid:13)(cid:13)(cid:13)(cid:13) d P Gen ( n ) d P GS n − (cid:13)(cid:13)(cid:13)(cid:13) P GS n p = e − Ω( n ) , G ∼ Gen ( n ) P ( G ∈ GS + n ∩ GS − n ) ≤ (1 + e − Ω( n ) ) |GS + n ∩ GS − n | / |GS n | = e − Ω( n ) , proving (8). Fix j ≥
1. Suppose throughout this section that G ∈ GS + n , C , . . . , C j ⊆ V ( G )and ( G, C i ) is a unipolar arrangement of order n for each i ∈ [ j ]. (For this given G , the number of choices for C , . . . , C j is R + ( G ) j , which we want to upperbound.) Further, let C = ∩ ji =1 C i , e C = ∩ ji =1 C i , and let l = n − | C | − | e C | . Thefollowing claim is easy to verify. Claim 9.4.
The pair ( G [ C ∪ e C ] , C ) is a unipolar arrangement of order | C ∪ e C | . Now let C ′ i = C i \ C and Q = V ( G ) \ ( C ∪ e C ), so that Q = ∪ ji =1 C ′ i . Lemma 9.5.
Let i ∈ [ j ] . Then the pair ( G [ Q ] , C ′ i ) is a unipolar arrangementof order l with at most j − side cliques.Proof. Without loss of generality suppose that i = 1. It is easy to see that G [ Q ]is a generalised split graph with a central clique C ′ , since C ′ ⊆ C , Q \ C ′ ⊆ C and ( G, C ) is a unipolar arrangement. Further, Q = ∪ ji =1 C ′ i , so every vertexin Q \ C ′ is covered by a clique C ′ i , where 2 ≤ i ≤ j . Hence G [ Q \ C ′ ] is adisjoint union of cliques, covered by j − G [ Q ] , C ′ ) contains at most j − C ′ i is complete to C and Q = ∪ ji =1 C ′ i , it follows that Q is completeto C .We now focus on C ′ . Observe that, for each vertex v ∈ C ′ , since v C wecan find C ′ i such that v C ′ i . Thus C ′ can be expressed as a disjoint union C ′′ ∪ · · · ∪ C ′′ j where C ′′ i ∩ C ′ i = ∅ for each i = 2 , . . . , j . Let C denote the familyof sets { C ′ i \ C ′ } ji =2 ∪{ C ′′ i } ji =2 . Note that the union of these sets is Q . Lemma 9.6.
For each S ∈ C , either S is complete to a side clique of ( G [ C ∪ e C ] , C ) and non-adjacent to all other side cliques, or S is non-adjacent to e C .Proof. In the unipolar arrangement (
G, C ), each C ′ i \ C ′ is contained in a sideclique, so it is either complete to a side clique of ( G [ C ∪ e C ] , C ), and non-adjacentto the others, or non-adjacent to any vertex of e C . Similarly, in the unipolararrangement ( G, C i ), C ′′ i is contained in a side clique, hence it has the sameproperty.Let f be the function mapping each set S in C to the side clique S ′ of( G [ C ∪ e C ] , C ) such that S is complete to S ′ , or to ∅ if no vertex in S is adjacentto e C . 37 emma 9.7. The edge set E ( Q, e C ) can be reconstructed from f .Proof. For each v ∈ Q either v ∈ C ′ and then v lies in some set S ∈ { C ′′ i } ji =2 ,or v lies in some set S ∈ { C ′ i \ C ′ } ji =2 . In either case, S is in C and the set ofneighbours of v in e C is precisely the set f ( S ).The edges of G can be reconstructed from G [ C ∪ e C ] – a unipolar graph, G [ Q ]– a unipolar graph with at most k − E ( Q, C ∪ e C ), which inturn can be reconstructed from f , since Q is complete to C . It will turn outthat the number of choices for G [ C ∪ e C ] dominates the rest, and this numberis maximised whenever Q = ∅ . However, the vertex labels are a nuisance, andadditional information is required to correctly recover G . The precise statementis given below. Lemma 9.8.
The j +1 tuple ( G, C , . . . , C j ) can be uniquely reconstructed from:1. the pair ( G [ C ∪ e C ] , C ) seen as a unipolar ( n − l ) -arrangement over thevertex set [ n − l ] so that the order of the labels of the vertices is preserved;2. the pair ( G [ Q ] , C ′ ) seen as a unipolar l -arrangement over the vertex set [ l ] with at most j − side cliques so that the order of the labels of the verticesis preserved;3. the sets C ′ , . . . , C ′ j ⊆ [ l ] ;4. the l -subset of V ( G ) specifying the original labels of Q ;5. the function f . Observe that there are at most j l l choices for the pair ( G [ Q ] , C ′ ) above.Recall that L n = |CGS + n | . Let T n = 2 n + n log n − n log ln n − n log e for n ≥ T n = 1 for n = 0 , ,
2. Pr¨omel and Steger show in [PS92] that L n ≤ T n × O ( n ln ln( n +3) ), hence for some c we have |CGS + n | ≤ T n × cn ln ln( n +3) . It followsthat U n,l := (cid:16) T n − l × cn ln ln( n +3) (cid:17) × (cid:16) j l l (cid:17) × (cid:16) ( j − l (cid:17) × (cid:18) nl (cid:19) × (cid:16) ( n + 1) j − (cid:17) is an upper bound on the number of choices for ( G, C , . . . , C j ) when | Q | = l , and U n = P nl =1 U n,l is an upper bound on the number of choices for ( G, C , . . . , C j )when Q = ∅ . Lemma 9.9.
The number U n,l is maximised subject to ≤ l ≤ n when l = 1 .Proof. We may prove this by thinking of l as a continuous variable and takingthe derivative of U n,l with respect to l . The analysis is straightforward butunpleasant: we leave the details to the reader. Corollary 9.10.
We have U n = |CGS + n | − n/ o ( n ) . roof. Since T n − = T n − n/ o ( n ) , we see that U n, |CGS + n | = 2 − n/ o ( n ) , and hence |CGS + n | − n/ o ( n ) = U n, ≤ U n ≤ nU n, = |CGS + n | − n/ o ( n ) . Finally, the number of choices for the j + 1 tuple ( G, C , . . . , C j ) when Q = ∅ is clearly |CGS + n | , so we have k R j + k P GS + n = X G ∈GS + n R j + ( G ) 1 |GS + n | ≤ (cid:0) U n + |CGS + n | (cid:1) |GS + n | = |CGS + n | (1 + 2 − n/ o ( n ) ) 1 |GS + n | = k R + k P GS + n (1 + 2 − n/ o ( n ) ) . This completes the proof of Lemma 9.3, Theorem 9.1 and Theorem 2.3.
Let µ = ( n − log n + log ln n ) / n ≥ µ = ⌈ µ ⌉ and let ˇ µ = ˆ µ −
1. As before, let ℓ n,k = (cid:0) nk (cid:1) k ( n − k ) B n − k . We fix n and drop it in the subscript for legibility, i.e. write ℓ k instead of ℓ n,k . Lemma 9.11.
Using the notation above, there is an n such that ℓ k /ℓ ˆ µ ≤ n − n − as long as | k − ˆ µ | ≥ n / for n ≥ n .Proof. Let x = k − ˆ µ , we have ℓ k ℓ ˆ µ = (cid:0) nk (cid:1) k ( n − k ) B n − k (cid:0) n ˆ µ (cid:1) ˆ µ ( n − ˆ µ ) B n − ˆ µ = ( n − ˆ µ )!( n − ˆ µ − x )! (cid:18) (ˆ µ + x )!ˆ µ ! (cid:19) − x ( n − µ ) − x B n − ˆ µ − x B n − ˆ µ . We can bound each of the terms ( n − ˆ µ )!( n − ˆ µ − x )! , (cid:16) (ˆ µ + x )!ˆ µ ! (cid:17) − , 2 x ( n − µ ) and B n − ˆ µ − x B n − ˆ µ with n | x | , so we get ℓ k ℓ ˆ µ < n | x | n | x | −| x | n | x | = 2 | x | log n −| x | ≤ n / log n − n / ≤ n − n − for large n .The bound is rather crude, and it can be improved, but we do not need abetter bound for our purposes. From now we focus on k = ˆ µ + x with | x | ≤ n / .This implies k ∼ n/
2, which allows us to use asymptotics for k as n grows, forinstance ( n − k ) / ( k + 1) → B n − k − B n − k = (1 + o n (1)) ln( n − k ) n − k = (1 + o n (1)) 2 ln nn , where the error term is uniform over k , provided that | x | ≤ n / . Lemma 9.12.
There is an n such that the following holds for each n ≥ n .Suppose ≤ x ≤ n / is integral and r ∈ { ℓ ˇ µ − x ℓ ˇ µ , ℓ ˆ µ + x ℓ ˆ µ } . Then − x − x ≤ r ≤ − x +2 x . roof. Define the continuous, strictly decreasing function d ( y ) = 2 n − y ln nn , sothat for k ∼ n/ ℓ k +1 ℓ k = n − kk + 1 2 n − k − (1 + o n (1)) 2 ln nn = (1 + o n (1)) d ( k ) . Observe that d ( k ) = 1 ⇐⇒ k = n + log ln n − log n ⇐⇒ k = µ . We have d ( µ + x ) = d ( µ ) · − x = 2 − x , and ℓ ˆ µ + x = ℓ ˆ µ x − Y i =0 ℓ ˆ µ + i +1 ℓ ˆ µ + i = ℓ ˆ µ x − Y i =0 (1 + o n (1)) d (ˆ µ + i ) . For each i we use the bounds2 − i − = d ( µ + 1 + i ) ≤ d (ˆ µ + i ) ≤ d ( µ + i ) = 2 − i , while the error terms, 1 + o n (1), we bound uniformly by and 2, provided n ≥ n . By multiplying the terms we get − (cid:0) x (cid:1) − x and − (cid:0) x (cid:1) + x in the exponentfor the lower and upper bounds respectively, which matches the statement ofthe lemma for the case r = ℓ ˆ µ + x /ℓ ˆ µ . The second case is analogous. Claim 9.13.
For a > we have P ∞ x = a − x < − ( a − a −
1) ln 2 .Proof.
Since − x is decreasing, we have ∞ X x = a +1 − x < Z ∞ a − x dx ≤ a Z ∞ a x − x dx = 1 − a ln 2 (cid:18) − x (cid:12)(cid:12)(cid:12)(cid:12) x = ∞ x = a (cid:19) = 2 − a a ln 2 . In the claim below and its proof, for simplicity we have written n / ratherthan ⌊ n / ⌋ . Claim 9.14.
For each integer k ≥ we have n / X i = k ℓ ˇ µ − i ≤ ℓ ˇ µ − ( k − (2+ 1( k −
1) ln 2 ) and n / X i = k ℓ ˆ µ + i ≤ ℓ ˆ µ − ( k − (2+ 1( k −
1) ln 2 ) . Proof.
Using Lemma 9.12 we see that n / X i = k ℓ ˇ µ − i ≤ ℓ ˇ µ n / X i = k − i +2 i = 2 ℓ ˇ µ n / X i = k − ( i − ≤ ℓ ˇ µ − ( k − + ∞ X i = k − i ! = 2 ℓ ˇ µ − ( k − + 2 − ( k − k −
1) ln 2 ! . The last inequality follows from Claim 9.13. The second part of the statementis similar. 40ow we put everything together to complete the proof of Theorem 2.5. Sup-pose n ≥ X ∼ L ( n ) and x >
2, then (noting the trivial bound L n ≥ ℓ ˆ µ + ℓ ˇ µ ) P ( | X − µ | ≥ x ) ≤ P (ˇ µ − n / ≤ X ≤ ˇ µ − x + 1)+ P (ˆ µ + x − ≤ X ≤ ˆ µ + n / )+ P ( | X − µ | ≥ n / ) ≤ ℓ ˆ µ ℓ ˆ µ + ℓ ˇ µ · − ( x − · ℓ ˇ µ ℓ ˆ µ + ℓ ˇ µ · − ( x − · n − n = 2 − ( x − +2 + n − n , and P ( | X − µ | ≥ x ) ≥ P ( X = ˆ µ + x ) + P ( X = ˇ µ − x ) ≥ − x − x ( ℓ ˆ µ + ℓ ˇ µ )(3 +
12 ln 2 )( ℓ ˆ µ + ℓ ˇ µ ) + n − n ˆ µ ≥ − x − x − .
10 Concluding remarks
In this paper we presented a generation model for perfect graphs which yieldssuch graphs almost uniformly, with error e − Θ( n ) (in terms of total variationdistance); and we used this approach to investigate several questions aboutsuch graphs.We aimed to investigate the most natural questions about random perfectgraphs, but of course there are further natural open problems. For example,concerning induced subgraphs H , we found the limiting probability that P n hasan induced subgraph H , but we did not discuss the distribution of the numberof such induced subgraphs.Our methods do not let us approach the following question about automor-phisms. It is well known that almost all graphs G n on vertex set [ n ] have nonon-trivial automorphisms, and easy to see that in fact the probability that G n has a non-trivial automorphism is 2 − (1+ o (1)) n . It is also not hard to see that, if R n ∼ Gen ( n ), then the probability that R n has a non-trivial automorphism is2 − ( + o (1)) n . It is natural to expect that the probability for P n is similar to thatfor R n , but we do not know: we would need a finer result than (3). Similarly,if H is a fixed graph in GS + ∩ GS − and R n ∼ Gen ( n ), then by Lemma 4.1 theprobability that R n fails to have an induced copy of H is e − Ω( n log n ) . Does sucha result hold for P n ?A long standing open problem is to describe the structure of a random per-fect graph G with a prescribed number of edges, say e ( G ) ∼ v ( G ) log v ( G ). Thisproblem has been discussed before [PS92], and even directly approached [BTW12],but to the best of our knowledge a complete answer is a distant possibility.Let C -free denote the class of graphs with no induced subgraph C . Then G S ⊂ P ⊂ C − free, and the key result (2) from Theorem 2.4 of Pr¨omel and41teger [PS92] in fact extends as follows: |GS n | = (1 − e − Ω( n ) ) | ( C − free) n | . (26)Thus our results refer also to random C -free graphs. For related general resultssee [ABBM11, BB11] and references there. References [ABBM11] N Alon, J Balogh, B Bollob´as, and R Morris. The structure ofalmost all graphs in a hereditary property.
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