Full rainbow matchings in equivalence relations
aa r X i v : . [ m a t h . C O ] F e b Full rainbow matchings in equivalence relations
David Munh´a Correia ∗ Liana Yepremyan † February 24, 2020
Abstract
We show that if a multigraph G with maximum edge-multiplicity of at most √ n log n , is edge-coloured by n colours such that each colour class is a disjoint union of cliques with at least2 n + o ( n ) vertices, then it has a full rainbow matching, that is, a matching where each colourappears exactly once. This asymptotically solves a question raised by Clemens, Ehrenm¨ullerand Pokrovskiy, and is related to problems on algebras of sets studied by Grinblat in [Grinblat2002]. This paper is motivated by a question of Grinblat which demonstrates a beautiful interplay betweenmeasure theory and combinatorics. Recall that an algebra A on a set X is a family of subsets ofthis set closed under the operations of union and difference of two subsets. In his book [6] andalso later [7, 8] Grinblat investigated necessary and sufficient conditions under which the unionof at most countably many algebras on X equals to P ( X ), the power set of X . In particular,one of the questions he studied, as observed by Nivasch and Omri [12], can be phrased aboutequivalence relations as follows. Let X be a finite set and let A be an equivalence relation on X ,define the kernel of A , ker ( A ) to be the set of all elements in X , which have non-trivial equivalenceclasses. Define ν ( n ) to be the minimal number such that if A , . . . , A n are equivalence relationswith ker ( A i ) ≥ ν ( n ) for all i ∈ [ n ], then A . . . , A n contains a rainbow matching , that is, a set of2 n distinct elements x , y , . . . , x n , y n ∈ X with x i ∼ y i ∈ A i for each i ∈ [ n ].In [6] Grinblat showed that 3 n − ≤ ν ( n ) ≤ n/ p n/
3, and he asked whether the lowerbound is the correct answer for all n ≥
4. Nivasch and Omri [12] improved Grinblat’s upper boundon ν ( n ) to 16 n/ O (1). Finally, Clemens, Ehrenm¨uller, Pokrovskiy [3] improved the bound on ν ( n ) to asymptotically best possible, that is, ν ( n ) = (3+ o (1)) n , using the graph theoretic language.If A , . . . , A n are equivalence relations on a set X , let the vertices of an edge-coloured multigraphbe the elements of X and, for each i ∈ [ n ], let x, y be an edge of colour i if x ∼ A i y . Each equivalencerelation A i then corresponds to the colour class i in the multigraph G and, each colour class is adisjoint union of non-trivial cliques. So, Grinblat’s original question can be reformulated as follows:we are given a multigraph G whose edges are coloured with n colours and each subgraph inducedby a colour class has at least 3 n − G contains a full rainbow matching , i.e. a set of n disjoint edges, which allhave distinct colours? The authors in [3] showed that for sufficiently large n , if each colour class ∗ St. Hugh’s College, University of Oxford, UK, e-mail:[email protected] † Department of Mathematics, Statistics, and Computer Science, University of Illinois at Chicago, Chicago,USA, London School of Economics, Department of Mathematics, London, UK, e-mail:[email protected],[email protected] o (1)) n many vertices then such a rainbow matching exists. Note that this isasymptotically the best bound, as if we take a disjoint union of n − n , one edge per colour, then there is no rainbow matching of size n . If n = 3, then ν (3) = 9 > n −
2, demonstrated by a 3-factorization of two disjoint K ’s, as shown by Grinblat [6]and also observed by Nivasch and Omri [12].In [3] the authors proposed to study Grinblat’s original problem when every pair of distinctelements belongs to at most one equivalence relation (Problem 2, in [3]), that is, to determine ν ′ ( n ) such that if A , A , . . . , A n are equivalence relations with ker ( A i ) ≥ ν ′ ( n ) and A i ∩ A j ⊆{ ( x, x ) | x ∈ X } then A , A , . . . , A n contain a rainbow matching. In the graph theoretic languagethis is equivalent to finding the minimum ν ′ ( n ) such that every simple graph G whose edges arecoloured with n colours, each subgraph induced by a colour class has at least ν ′ ( n ) vertices andis the disjoint union of non-trivial complete graphs, contains a full rainbow matching. The trivialupper bound is ν ′ ( n ) ≤ ν ( n ) ≤ (3 + o (1) n . As for the lower bound, the graph composed of n − n edges, one edge per colour, exhibits that ν ′ ( n ) > n − n is the correct answer, in fact, the same result holds for multigraphswith bounded edge multiplicity. Our main result follows. Theorem 1.
For every δ > there exists n such that for all n ≥ n the following holds. If G is amultigraph whose edges are coloured with n colours, such that each colour class is a disjoint unionof non-trivial cliques with at least (2 + δ ) n vertices, and the edge-multiplicity of G is at most √ n log n then G contains a full rainbow matching. We suspect that improving this upper bound on ν ′ ( n ) to something not of an asymptotic formwill be hard. That is because our problem is closely related to the Brualdi-Ryser-Stein conjectureon Latin squares and its generalizations. A Latin square of order n is an n × n square with cells filled using n symbols so that every symbolappears once in each row and once in each column. A partial transversal of size k of a Latin squareis a set of k entries in the square which all come from distinct rows and columns, and containdistinct symbols. If k = n , where n is the order of the Latin square, the partial transversal issimply called a transversal . Conjecture 1.1 (Brualdi-Ryser-Stein) . Every Latin square of order n has a partial transversal ofsize n − and moreover, if n is odd, it has a transversal. The history of the conjecture is as follows. In 1967, Ryser [15] conjectured that the number oftransversals in a Latin square of order n has the same parity as n , so any Latin square of odd orderhas a transversal (see also [1]). Note that for even n this is not true; for example, the additiontable of Z n is a Latin square with no transversal. Brualdi [2] conjectured that every Latin squareof order n has a partial transversal of size n − n is odd, it has a transversal.Stein [16] conjectured that a stronger statement holds and the same outcome should hold even inan n × n array filled with the numbers 1 , , . . . , n such that every number occuring exactly n times.Very recently this was disproved by Pokrovskiy and Sudakov [14]. The current best bound on thesize of the partial transversal in Brualdi-Ryser-Stein conjecture is n − O (log n ) proved by Hatamiand Shor [9].To every Latin square one can assign an edge-colouring of the complete bipartite graph K n,n bycolouring the edge ij by the symbol in the cell ( i, j ). This is a proper colouring, i.e., one in which2ny edges which share a vertex have distinct colours. Identifying the cell ( i, j ) with the edge ij ,a partial transversal corresponds to a rainbow matching of the same size. So Conjecture 1.1 saysthat any proper edge-colouring of K n,n contains a rainbow matching of size n −
1, and a perfectrainbow matching, when n is odd. Aharoni and Berger conjectured the following generalization ofConjecture 1.1. Conjecture 1.2.
Let G be a bipartite multigraph that is properly edge-coloured with n colours andhas at least n + 1 edges of each colour. Then G has a rainbow matching using every colour. Note that Conjecture 1.2 would imply that ν ′′ ( n ) ≤ n + 2 where ν ′′ ( n ) can be be defined justas ν ′ ( n ) but restricted to only bipartite graphs (however, Conjecture 1.2 is much stronger, becauseit allows any edge multiplicity). Pokrovskiy [13] showed that Conjecture 1.2 is asymptotically true,in that the conclusion holds if there are at least n + o ( n ) edges of each colour. Keevash and Yepre-myan [10] considered the same question without the bipartiteness assumption and obtained a resultsomewhat analogous to Pokrovskiy’s. They showed that any multigraph with edge multiplicities o ( n ) that is properly edge-coloured by n colours with at least n + o ( n ) edges of each colour containsa rainbow matching of size n − C , for some large absolute constant C >
0. A similar result wasalso obtained independently by Gao, Ramadurai, Wanless and Wormald [5]. They showed thatevery properly edge-coloured multigraph with edge-multiplicity at most O ( √ n/ log n ) such thateach colour class has at least 2 n + o ( n ) vertices has a full rainbow matching (note that here thecolouring is proper while in our case it is not). Our Theorem 1 is a generalization of this result. We further generalize the approach developed in [5]. If a multigraph G has a certain structurewe construct a full rainbow matching via some randomized algorithm. This consists of finding analmost full rainbow matching by a sequence of random iterations and then completing it to a fullrainbow matching by the greedy algorithm. To show that the last part is possible, the so-calleddifferential equation method is used. Informally saying, one analyzes the random method to showthat the graph left at the very end behaves “nicely” enough to contain such a matching. To describehow degrees of vertices, edges and other variables are changing in the left-over graph after eachrandom choice, differential equations are used. The following theorem gives the types of multigraphto which the algorithm can be applied.For a multigraph whose edges are coloured, let, for a colour c , n c denote the number of verticesin the colour class of c and e c denote the number of edges. Let also, d v denote the number of edgesincident to the vertex v . Theorem 2.
For any < σ < σ , there exists n such that for all n ≥ n the following is true.Suppose G is a multigraph whose edges are coloured with n colours, such that each colour class is adisjoint union of cliques of order at most three. If the edge-multiplicity of G is at most √ n log n , andmoreover, • n c ≤ n and e c ≥ σ n , for every colour c , • d v ≤ σ n , for every vertex v ,then G contains a full rainbow matching. In order to prove our main result, Theorem 1, we show that every multigraph we are consideringcontains a subgraph with the structure given in Theorem 2. This is done via a careful randomsampling. 3
Some Probability Tools
In this section we gather some classic probability results which we use throughout our proofs.
Proposition 1 ( Chernoff bound , [4]) . Let X be a sum of n independent [0 , -valued randomvariables. Then, for all t ≥ , P [ | X − E [ X ] | > t ] ≤ e − t n Proposition 2 ( McDiarmid’s Inequality , [11]) . Let k ∈ N and Π : [ k ] → [ k ] be a permuta-tion chosen uniformly at random. Let also h be a non-negative real-valued function on the set ofpermuations of { , . . . , k } and define the random variable Z := h (Π) and its median M . Supposethat there exist constants c, r > such that the following two items occur for any (deterministic)permutation π : • Swapping two coordinates in the permutation π changes h ( π ) by at most c . • If h ( π ) = s , then there is a set of at most rs coordinates such that h ( π ′ ) ≥ s for any otherpermutation π ′ which agrees with π on these coordinates.Then, for all t ≥ , P [ | Z − M | ≥ t ] ≤ (cid:18) − t rc ( M + t ) (cid:19) Proposition 3 ( Azuma’s Inequality, [4] ) . Let X , . . . , X n be independent random variableswhich take values in some set S and let f be real-valued on S n such that there exists a constant c > such that for any x = ( x , . . . , x n ) ∈ S n , changing a coordinate of x deviates f ( x ) by at most c . Then, defining Y = f ( X , . . . , X n ) , we have for every t > , P [ | Y − E [ Y ] | ≥ t ] ≤ (cid:18) − t c n (cid:19) . Proof.
Note that we only need to consider the case when the cliques mentioned in the statementare of order at most three. This is because any other case can be reduced to this one by deletingedges in such a way that all monochromatic K t ’s with t ≥ K ’s and K ’s without reducing the number of vertices in a colour class. We can also assumethat for every colour c , n c ≤ ⌈ (2 + δ ) n ⌉ + 2. Finally, without loss of generality, we may assume δ <
2. For any colour c , let a c , b c be the number of K , K components, respectively, in the colourclass of c . Note that n c = 3 a c + 2 b c and e c = 3 a c + b c .We construct a random subgraph H ⊆ G in the following manner: • Independently, for each monochromatic K , either delete two of its edges (transforming it into K ), each pair of edges having probability of being deleted or keep the K with probability .Define for every vertex v , d tr v , d line v to be respectively, the number of edges incident to v whichbelong to a monochromatic K , K . Note that d v = d tr v + d line v and Cd v = d tr v + d line v , where Cd v is4olour degree of the vertex v . We will now show that with positive probability, H will satisfy theconditions of Theorem 2, thus H will have having a full rainbow matching, and so will G .For every colour c , let X c denote the number of K components in the colour class c whichremain unchanged, so that n c ( H ) = 2( a c + b c ) + X c and e c ( H ) = a c + b c + 2 X c . Note that X c ∼ Bin( a c , ). Using the Chernoff bound, we have that for any ε ≥ • P ( e c ( H ) ≤ n c − εa c ) ≤ e − ε ac Similarly, for every vertex v , let Y v denote the number of edges incident to v , belonging to amonochromatic K , which aren’t deleted. We note that d v ( H ) = d line v + Y v and Y v ∼ Bin( d tr v , ).Again by the Chernoff bound, for any ε ≥ • P ( d v ( H ) ≥ Cd v + εd tr v ) ≤ e − ε d tr v These two facts allow us to get the following bounds on the structure of the subgraph H . Let ε = δ δ ) and ε = δ .For colours c such that a c ≤ ε ⌈ (2+ δ ) n ⌉ , note that e c ( H ) ≥ a c + b c = n c − a c ≥ ⌈ (2 + δ ) n ⌉ − a c ≥ (3 − ε ) ⌈ (2 + δ ) n ⌉ n is large, a c ≥ − ε ) ⌈ (2 + δ ) n ⌉ ≤ n c − ε ⌈ (2 + δ ) n ⌉ ≤ n c − ε (cid:0) a c − (cid:1) ≤ n c − ε a c , where in the third inequality we used 3 a c ≤ n c ≤ ⌈ (2 + δ ) n ⌉ + 2. Therefore, P (cid:18) e c ( H ) ≤ (3 − ε ) ⌈ (2 + δ ) n ⌉ (cid:19) ≤ P (cid:18) e c ( H ) ≤ n c − a c ε (cid:19) = o (cid:18) n (cid:19) . Hence, with probability 1 − o (1), we have that that e c ( H ) ≥ (3 − ε ) ⌈ (2+ δ ) n ⌉ for every colour.Similarly, for vertices v with ε n < d tr v , since Cd v ≤ n and d tr v ≤ n , we have P ( d v ( H ) ≥ (1 + ε ) n ) ≤ P (cid:0) d v ( H ) ≥ Cd v + ε d tr v (cid:1) = o (cid:18) n (cid:19) For the rest of the vertices, note that d v ( H ) ≤ (1 + ε ) n . Hence, since v ( G ) ≤ ( ⌈ (2 + δ ) n ⌉ + 2) n = O ( n ), we have that with probability 1 − o (1), d v ( H ) ≤ (1 + ε ) n for every vertex.To conclude, we have that with probability at least 1 − o (1), for every colour c , by the choiceof ε , ε , • e c ( H ) ≥ (3 − ε ) ⌈ (2+ δ ) n ⌉ ≥ (2 + δ )( − ε ) n = (1 + δ ) n , for every colour c , • d v ( H ) ≤ (1 + ε ) n = (1 + δ ) n , for every vertex v ,Then, if n is sufficiently large, with positive probability H will satisfy the conditions of Theorem 2and contain a full rainbow matching. 5 Informal Treatment of Theorem 2
The proof of Theorem 2 is technical, here we give a heuristic argument why Theorem 2 is true. Wewill formalize this in the next section. Make a note of the following greedy deterministic algorithmwhich we will use throughout the paper. Take a multigraph which is edge-coloured such that eachcolour class is a disjoint union of cliques and whose number of vertices is at least four times thenumber of colours. The greedy way of finding a full rainbow matching consists of picking an edgeof each colour and deleting its vertices from the graph at each iteration. Indeed, we can check thatwhen we do so, the number of vertices in each of the colour classes of those colours that aren’t yetin the matching decreases by at most four. Given the initial assumption on the multigraph, thiswill then always produce a full rainbow matching. We now describe the randomized algorithm weuse. Take a multigraph G which satisfies the conditions stated in Theorem 2.Informally, the algorithm goes in the following manner: we first randomly order the colours inorder to put them into chunks C , ..., C τ of size εn (except maybe the last one), so that τ = ⌈ ε ⌉ -we will take ε = ε ( n ) to be of a specific order so that the formalities work out; at each iteration1 ≤ i ≤ τ − C i , that is, we construct a rainbow matching with the colours in C i and add it to the rainbow matching we have by the previous iterations in order to get a rainbowmatching with the colours in C ∪ . . . ∪ C i ; finally, iteration τ will consist of greedily finding arainbow matching with the colours in C τ and adding it to the previous rainbow matching with thecolours in C ∪ . . . ∪ C τ − , thus constructing a full rainbow matching.At each iteration 1 ≤ i ≤ τ −
1, we will process chunk C i in such a way that after finishing iteration i , we expect to have:1) For every vertex v that survived and j > i , d C j v ( i ) ≤ εd ( iε ) σ n , where d C j v ( i ) denotes thenumber of edges of colours in C j that are incident to v after finishing iteration i .2) For every unprocessed colour c , the number of edges in its colour class graph is ≥ e ( iε ) n .These functions d, e will be defined in Section 5, but here, we will informally guess what they shouldbe. We first note that we should be able to have d (0) = 1 and e (0) = σ , given our assumptions on G and since we randomly order the colours at the beginning of the algorithm.Let’s describe how we will process chunk C i +1 in iteration 1 ≤ i + 1 ≤ τ − C i +1 . Let’s say these arethe edges that are chosen and that their vertices are marked. From these, we delete (as wellas their vertices) from our graph the non-colliding ones and process them into our rainbowmatching. Let’s call these edges killed.2) We then zap (that is, delete from the graph) each vertex that survives the previous step,independently with some probability (specific to each vertex), so that overall, every vertexhas the same probability p i (which will be defined later in Section 5) of being marked orzapped (let’s call this condemned).3) Finally, we greedily process the rest of the colours that were involved in collisions in 1). In theend, we will have constructed a rainbow matching with the colours in C i +1 and we add it tothe previous rainbow matching with colours in C ∪ . . . ∪ C i given by the previous iterations.We will prove in the next section that the effect of the collisions occurring in 1) will be negligibleand so, we will be able to perform 3).Because of this negligible effect of the collisions, for this informal analysis, we can think of avertex being deleted as being equivalent to a vertex being condemned.6ote that for a vertex v, we have P (v is chosen in step 1) ≤ d C i +1 v ( i )min c ∈ C i +1 e c ( i ) ≤ εσ d ( iε ) e ( iε )where e c ( i ) denotes the number of edges in the colour class graph of c after iteration i . Hence, weshould be able to take p i := εσ d ( iε ) e ( iε ) .Let’s now take a look at what we expect to happen to our parameters during iteration i + 1.Let a c ( i ) , b c ( i ) denote, respectively, the number of K ’s, K ’s in the colour class graph of c afteriteration i . We ignore cases where 2 vertices in the same component are deleted, since again, wewill show this to be negligible in the next section. We can then see that we should have:1) For every unprocessed colour c , a c ( i +1) ≈ a c ( i )(1 − p i ) and b c ( i +1) ≈ b c ( i )+ p i (3 a c ( i ) − b c ( i ))2) For every vertex v surviving the iteration and j > i + 1, d C j v ( i + 1) ≈ d C j v ( i )(1 − p i )Note that step 2) tells us that we should be able to have d (( i + 1) ε ) = d ( iε )(1 − p i ). Further,note that for every unprocessed colour, the number of edges in its colour class after iteration i + 1is e c ( i + 1) = 3 a c ( i + 1) + b c ( i + 1) ≈ (3 a c ( i ) + b c ( i ))(1 − p i ) = e c ( i )(1 − p i ). Hence, we shouldalso be able to have e (( i + 1) ε ) = e ( iε )(1 − p i ).Since we will choose ε = ε ( n ) so that it tends to 0 as n tends to ∞ we should be able to takethe derivatives of d, e at x = iε where 0 ≤ i ≤ τ −
1. We then can get, given our choice of p i , that: • d ′ = − σ d e • e ′ = − σ d We can solve this to get d ( x ) = 1 − ( σ σ ) x and e ( x ) = σ d ( x ) .We are now in position to see if we expect to indeed be able to greedily process the last chunk C τ . For this, we need to check the values of these functions at x = ( τ − ε ∈ [1 − ε, d (( τ − ε ) ≥ − σ σ >
0, by assumption.The number of vertices in the colour class of a colour c ∈ C τ after iteration τ − e (( τ − ε ) n ≥ σ (1 − σ σ ) n . Since ε →
0, if n is large then thiswill be larger than 4 εn ≥ | C τ | and therefore, by the discussion in the beginning of this section, wewill be able to greedily find a rainbow matching with the colours in C τ . Concluding, we get a fullrainbow matching in G . Proof.
We first give some notation and define in detail the algorithm. Our setup is a graph Gwith the conditions stated. We will also take ε = ε ( n ) ∈ [
12 log log n , n ] and moreover, such that εn ∈ N . Note this exists provided that n is large enough.Initially, take a random permutation of the colours, which results in a partition of them into sets(which we shall call ”chunks”) C , ..., C τ , where τ = ⌈ ε ⌉ and every chunk has size εm except formaybe the last chunk (which has size at most εm ). We then start with our iterations.Let at the start M = ∅ , G = G and denote the random permutation we performed above asiteration 0. At iteration 1 ≤ i ≤ τ −
1, we ”process” the chunk C i in the following way. We look atthe graph G i − that we have, that is, the one that we are left with after iteration i −
1; For each7ertex v in this graph, there is a probability P i − ( v ) ∈ [0 ,
1] such that if we pick independentlyand u.a.r an edge of each colour in C i , the probability that v is incident to one of these edges is P i − ( v ); Given this, define for every vertex v the probability Q i − ( v ) to be such that P i − ( v ) + Q i − ( v )(1 − P i − ( v )) = p i − , which we will later define; If there exists such a Q i − ( v ) ∈ [0 ,
1] forevery vertex v , we continue - if not, the algorithm breaks; We now randomly assign to each vertexa bit Z v ∈ { , } ; We do this independently across all vertices and such that Z v ∼ Ber( Q i − ( v ));We then proceed with the following steps: Step 1
First, pick independently and u.a.r an edge of each colour in C i . Denote these edges bythe chosen ones and say that their vertices are the marked vertices. Note by before that P i − ( v ) = P ( v is marked) for every vertex v . Step 2
From the chosen edges, say that two of them collide if they share a vertex. Delete from G i − , along with their vertices, those chosen edges which don’t collide with any other. Saythat these are the killed edges and their vertices are also killed . Add the killed edges to M and say that the colours of these edges were processed into M . Step 3
For each vertex v ∈ V ( G i − ) that survived Step 2 (i.e, wasn’t killed), we look at the valueof Z v . If this is 1, then we zap (i.e, delete from G i − ) the vertex v . Otherwise, we do nothing. Step 4
Let Φ i ⊆ C i denote the set of colours which haven’t yet been processed into M because theirchosen edges collide with others. Greedily find a rainbow matching of these colours and deleteit from the graph along with its vertices. Add the rainbow matching to M , thus processingthe colours in Φ i . Denote the graph resulting from these 4 steps by G i .Say that a vertex v was condemned if it was zapped or marked . Note then that P ( v condemned) = P ( v marked) + P ( v zapped | v not marked) P ( v not marked) = P i − ( v ) + P ( Z v = 1)(1 − P i − ( v )) = P i − ( v ) + Q i − ( v )(1 − P i − ( v )) = p i − .To finish the algorithm, iteration τ consists of greedily finding a rainbow matching in G τ − ofthe colours in C τ and adding it to M , thus processing the colours in C τ .Note that if the algorithm is successful, M will be a full rainbow matching. However, there aremany things that can break the algorithm. Specifically, each of the iterations 1 ≤ i ≤ τ − Q i − ( v ) and perform Step 4. Moreover, for thealgorithm not to break, we also need to be able to perform iteration τ .Let’s now give some notation which we will need in the analysis of the algorithm: Define γ = σ σ < d ( x ) = 1 − γx ∈ [1 − γ,
1] for 0 ≤ x ≤
1. For each colour c , let e c ( x ) = e c (0) d ( x ) where e c (0) := e c n (note that 4 ≥ e c (0) ≥ σ ). Let e c ( i ) denote the number of edgesof colour c in the graph G i , that is in the graph we have after finishing iteration i . Define also e i := max c ∈ C j ,j>i | e c ( iε ) n − e c ( i ) | . For a vertex v ∈ V ( G i ), that is, one which survived itera-tions 1 through i , let d C j v ( i ) denote the degree of v in G i with respect to colours in C j . Let then d i := max j>i,v ∈ G i ( d C j v ( i ) − εσ d ( iε ) n ) + . Note that 2 εn ≥ d v and so, d i ≤ (2 + 4 σ ) εn . Take for0 ≤ j ≤ τ − p j := εγd ( jε ) + c j , where c j = d j d ( jε )+2 εγe j σ d ( jε ) n ≥
0. Let also µ := √ n log n , which by8ssumption is an upper bound on the edge multiplicity between each pair of vertices.We now begin with the analysis of the algorithm. Lemma 1.
There exist constants t j = t j ( σ , σ ) > , ( ≤ j ≤ and some n ( σ , σ ) ∈ N , suchthat if n ≥ n ( σ , σ ) then for every ≤ i ≤ τ − , if we have just finished iteration i with e i ≤ t n ,then with positive probability we can perform iteration i + 1 (that is we can successfully process thechunk C i +1 ) and get, at the end, • d i +1 ≤ d i + t ε n • e i +1 ≤ e i (1 + t εd ( iε ) ) + t ε n + d i σ Proof.
See Section 6.
Lemma 2.
There exists some n ′ ( σ , σ ) ∈ N such that if n ≥ n ′ ( σ , σ ) , with positive probabilitywe have d ≤ ε n Proof.
Note that for a vertex v and 1 ≤ j ≤ τ , d C j v (0) is determined by the random permutationof the colours performed in the beginning. Note by swapping two colours in that permutation wechange d C j v (0) by at most 2. Further, if d C j v (0) = s , there is a set of at most s colours, which forany other permutation in which these are left unchanged, guarantee that d C j v (0) ≥ s . Then, byMcDiarmid’s inequality, we have that if M is the median of d C j v (0), then for t ≥ P ( | d C j v (0) − M | ≥ t ) ≤ e − t M + t ) (1)Note then, by integrating, we have that E [ | d C j v (0) − M | ] ≤ R ∞ P ( | d C j v (0) − M | ≥ t ) dt ≤ R ∞ e − t M + t ) dt ≤ R M e − t M dt + 4 R ∞ M e − t dt ≤ √ πM + 512 e − M ≤ √ πM + 512 π . Now, note this in par-ticular, gives us that M − E [ d C j v (0)] ≤ √ πM + 512 π and thus,( √ M − √ π ) − π ≤ E [ d C j v (0)]Since E [ d C j v (0)] ≤ εd v ≤ εσ n , by assumption, we can see then that if n is large enough, M ≤ εσ n + O ( √ εσ n ). Hence, by (1) with t = √ εn log n , we get P ( d C j v (0) ≥ √ εn log n + εσ n + O ( √ εσ n )) ≤ e − εn log2 nO ( εn ) = o ( n − )Concluding, we get with probability 1 − o ( m − ), d C j v (0) ≤ εσ n + O ( √ εn log n )Since we have O ( n ) pairs v, j , then, with positive probability, we have d = O ( √ εn log n ) ≤ ε n if n is large enough. Lemma 3.
Take the positive constants t = t , s = t and r = t + t +1) σ . There exists some n ′′ ( σ , σ ) ∈ N such that if n ≥ n ′′ ( σ , σ ) , then with positive probability, we can perform thealgorithm up to iteration τ − (that is, we can successfully process all but the last chunk) and get, ∀ ≤ i ≤ τ − , • d ≤ ε n d i +1 ≤ d i + tε n • e i +1 ≤ e i (1 + sεd ( iε ) ) + rε n Proof.
Take n ′′ ( σ , σ ) ≥ n ′ ( σ , σ ) , n ( σ , σ ) and moreover, such that we have n ′′ ≤ t (1 − γ ) sγ r .We prove the lemma inductively on k ≥ −
1. The statement of the lemma is the case k = τ −
2. Thebase case k = − τ − > k ≥ −
1. Thatis, if n ≥ n ′′ ( σ , σ ) then with positive probability, we can perform the algorithm up to iteration k + 1 (that is, we can successfully process all chunks C j with j ≤ k + 1) and get, ∀ ≤ i ≤ k , • d ≤ ε n • d i +1 ≤ d i + tε n • e i +1 ≤ e i (1 + sεd ( iε ) ) + rε n Note that by iterating the above bounds we get that e k +1 ≤ ( k + 1) rε n Q kj =1 (1 + sεd ( jε ) ) ≤ ( k + 1) rε n exp ( sε P kj =1 1 d ( jε ) ). By comparing the sum to the integral, this is at most ( k +1) rε n exp ( s R ( k +1) εε dx − γx ) which since ε, ( k + 1) ε ∈ (0 , rεn exp ( sγ log( − γ )) = rεn (1 − γ ) sγ ≤ t n as ε ≤ t (1 − γ ) sγ r . Also by iterating, we get d k +1 ≤ ( t + 1) ε n .Then, by Lemma 1, with positive probability (conditional on the previous iterations), we cansuccessfully perform iteration k + 2 and get • d k +2 ≤ d k +1 + tε n • e k +2 ≤ e k +1 (1 + sεd (( k +1) ε ) ) + t ε n + d k +1 σ ≤ e k +1 (1 + sεd (( k +1) ε ) ) + rε n Thus, case k + 1 is also true and therefore, the Lemma follows by induction.Note that Theorem 2 will follow immediately from this last Lemma 3. Indeed, we get that if n is large enough, with positive probability, the algorithm is successful in processing all chunks C j with j ≤ τ −
1. Further, by the same argument as in the proof of Lemma 3, we can also have that e τ − ≤ rεn (1 − γ ) sγ . That is, for every colour c ∈ C τ e c ( τ − ≥ e c (( τ − ε ) n − rεn (1 − γ ) sγ ≥ σ (1 − γ ) n − rεn (1 − γ ) sγ Hence, we can take n large enough so that e c ( τ − ≥ εn for every colour, implying that the lastchunk can be greedily processed. Theorem 2 then follows. Proof.
Take t = σ (1 − γ ) . Let us recall that the setup is that we just finished iteration i with e i ≤ t n . We then work with the graph G i we have after iteration i . Therefore, when we talk aboutvertices, we always mean a vertex in this graph. We are performing iteration i + 1. Lemma 4.
There exists a constant s = s ( σ , σ ) > such that for every vertex v , s ε ≥ p i ≥ εσ d ( iε ) n + d i σ d ( iε ) n − e i ≥ P i ( v )10 roof. Note that by a union bound on the events that a specific edge incident to v that has coloursin C i +1 is chosen in Step 1 and the definitions of d i , e i , we get P i ( v ) ≤ max v d C i +1 v ( i )min c ∈ C i +1 e c ( i ) ≤ εσ d ( iε ) n + d i σ d ( iε ) n − e i Note since e i ≤ t n ≤ σ d ( iε ) n then σ d ( iε ) n − e i > εσ d ( iε ) n + d i σ d ( iε ) n − e i − εγd ( iε ) = d i d ( iε ) + εγe i d ( iε )( σ d ( iε ) n − e i ) ≤ c i where we are using that γ = σ σ and that e i ≤ σ d ( iε ) n . Hence, p i = c i + εγd ( iε ) ≥ εσ d ( iε ) n + d i σ d ( iε ) n − e i Finally, note that by recalling that d i ≤ (2 + 4 σ ) εn and using that e i ≤ σ d ( iε ) n and that d ( iε ) ∈ [1 − γ, c i = 2 d i d ( iε ) + 2 εγe i σ d ( iε ) n ≤ ( 2(2 + 4 σ ) + γσ σ (1 − γ ) ) ε Thus, take s = σ )+ γσ σ (1 − γ ) + γ − γ to get p i = c i + εγd ( iε ) ≤ s ε .Note this Lemma 4 shows that if n is large enough, we can find probabilities Q i ( v ) for everyvertex v . Indeed, by the Lemma 4, we get that ≥ p i ≥ max v P i ( v ) and so, we can always find Q i ( v ) uniquely for every vertex v . Lemma 5.
There exists a constant s = s ( σ , σ ) > such that with probability − o (1) , | Φ i +1 | ≤ s ε n Proof.
For every vertex u , define X u to be the number of chosen edges (in Step 1 of iteration i + 1)incident to u . Let also Y u := X u X u ≥ . This counts the number of edge collisions that the vertex u creates. Since | Φ i +1 | counts the number of colours whose chosen edge collides with others, we havethat | Φ i +1 | ≤ P u Y u where the sum is over all vertices u . Moreover, note that for every such u , bydefinition, Y u ≤ X u ( X u −
1) giving that | Φ i +1 | ≤ P u X u ( X u − C i +1 which are incident to u (after iteration i ) byΓ C i +1 u ( i ), so that d C i +1 u ( i ) = | Γ C i +1 u ( i ) | . Then, note that for distinct edges e , e in Γ C i +1 u ( i ), P ( e , e are both chosen) ≤ c ∈ C i +1 e c ( i )) ≤ t n where we are using that e c ( i ) ≥ e c ( iε ) n − e i ≥ σ (1 − γ ) n − e i ≥ σ (1 − γ ) n − t n = t n bydefinition of t . Further, X u ( X u −
1) counts the number of ordered pairs of distinct chosen edgesin Γ C i +1 u ( i ). Thus, by a union bound on the events that a specific pair is chosen, we can see that E [ X u ( X u − ≤ d Ci +1 u ( i ) t n . Using that d C i +1 u ( i ) ≤ εn (which we have since | C i +1 | ≤ εn and eachcolour class is a disjoint union of K ’s and K ’s), we can also see that E [ X u ( X u − ≤ ( εt n ) d C i +1 u ( i ).Summing up over the vertices we get E [ | Φ i +1 | ] ≤ εt n X u d C i +1 u ( i ) ≤ εt n × εn × max c ∈ C i +1 e c ( i ) ≤ t ε n e c ( i ) ≤ e c ≤ n c ≤ n .We now prove concentration. Note that | Φ i +1 | is a function of the εn chosen edges in Step 1.Further, changing a chosen edge of a certain colour in C i +1 will deviate | Φ i +1 | by at most 3. Hence,by Azuma’s Inequality, we have P ( || Φ i +1 | − E [ | Φ i +1 | ] | ≥ √ εn log n ) ≤ n log n Since √ εn log n = o ( ε n ) (as ε ≍ n ), we have with probability 1 − o (1), | Φ i +1 | ≤ s ε n where s := t + 1.Let’s introduce some notation for the next Lemma. Define for a colour c , e c ( i + ) to be thenumber of edges of that colour that still remain after Step 3 of iteration i + 1. Similarly, define d C j v ( i + ). Lemma 6.
There exists a constant s = s ( σ , σ ) > such that with probability − o (1) , we have: • | e c ( i + ) − e c ( i )(1 − p i ) | ≤ s ( ε n + | Φ i +1 | ) for every unprocessed colour c after Step 3, thatis every c ∈ Φ i +1 ∪ ∪ τj = i +2 C j . • d C j v ( i + ) ≤ d C j v ( i )(1 − p i ) + s ε n for every vertex v surviving Steps 1,2 and 3 (of iteration i + 1 ) and j > i + 1 .Proof. This is the most technical part of the proof and we include it in the Appendix.
Lemma 7.
There exists a constant s = s ( σ , σ ) > such that with probability − o (1) , we cansuccessfully perform iteration i + 1 and get • | e c ( i + 1) − e c ( i )(1 − p i ) | ≤ s ε n for every c ∈ ∪ τj = i +2 C j . • d C j v ( i + 1) ≤ d C j v ( i )(1 − p i ) + s ε n for every vertex v that survives iteration i + 1 (that is, v ∈ V ( G i +1 ) ) and j > i + 1 .Proof. We first need to show that Step 4 is successfully performed (note by Lemma 4, we havealready shown that the algorithm doesn’t break initially when we are assigning the probabilities Q i ( u )). Indeed, by Lemmas 5 and 6, we have that with probability 1 − o (1), • | e c ( i + ) − e c ( i )(1 − p i ) | ≤ ( s + s s ) ε n for every c ∈ Φ i +1 ∪ ∪ τj = i +2 C j . • d C j v ( i + ) ≤ d C j v ( i )(1 − p i ) + s ε n for every vertex v that survives Steps 1,2 and 3 and j > i + 1.Note that from the first bound, using Lemma 4 (which in particular also shows that for n large p i ≤ s ε ≤ ) and that e c ( i ) ≥ t n (which follows by e i ≤ t n as we’ve seen in the proof of Lemma5), we can get that e c ( i + ) ≥ e c ( i )(1 − p i ) − ( s + s s ) ε n ≥ t n (1 − p i ) − ( s + s s ) ε n ≥ t n − (2 s t + s ε + s s ε ) εn .Therefore, since ε → n → ∞ we have that if n is large enough, then e c ( i + ) ≥ εn ≥ | Φ i +1 | and so, we can greedily process the colours in Φ i +1 , that is, Step 4 is successfully performed andso, the iteration is successfully performed. 12urther, when we do perform Step 4, we greedily delete 2 | Φ i +1 | vertices from the graph. There-fore, we have that d C j v ( i + 1) ≤ d C j v ( i + ) for every vertex v surviving the iteration and j > i + 1as well as | e c ( i + 1) − e c ( i + ) | ≤ | Φ i +1 | ≤ s ε n (by Lemma 5) for every colour c ∈ ∪ τj = i +2 C j .Therefore, Lemma 7 follows by taking s := 4 s + s + s s .To conclude the proof of Lemma 1, define t = s , t = σ − > t = s + 4 γ . Notethen that if the bounds in Lemma 7 are satisfied, then we have the following:i) Take a vertex v surviving iteration i + 1 and j > i + 1. Then, d C j v ( i + 1) − εσ d (( i + 1) ε ) n ≤ d C j v ( i )(1 − p i ) + s ε n − εσ d ( iε ) n (1 − p i ) + εσ n ( γε − p i d ( iε )) ≤ d i (1 − p i ) + s ε n , where wehave used that p i ≥ εγd ( iε ) and d (( i + 1) ε ) = d ( iε ) − γε . Therefore, d i +1 ≤ d i + s ε n .ii) Take a colour c ∈ ∪ τj = i +2 C j . Note that by the bound in Lemma 7 and the triangle inequality, | e c ( i + 1) − e c (( i + 1) ε ) n | ≤ s ε n + | e c ( i )(1 − p i ) − e c (( i + 1) ε ) n | . Further, this secondterm is equal to | ( e c ( i ) − e c ( iε ) n )(1 − p i ) − p i e c ( iε ) n + ( e c ( iε ) − e c (( i + 1) ε )) n | . Since e c ( iε ) − e c (( i + 1) ε ) = e c (0) γε (2 d ( iε ) − γε ), p i = εγd ( iε ) + c i and e c (0) n = e c ≤ n c ≤ n this isat most | ( e c ( i ) − e c ( iε ) n )(1 − p i ) − c i e c ( iε ) n | + 4 γ ε n ≤ e i (1 − p i ) + 2 c i e c ( iε ) n + 4 γ ε n .Recalling the definition of c i , note that2 c i e c ( iε ) n = 4 e c (0) d i σ + 4 εγe c (0) e i σ d ( iε ) ≤ d i σ + 16 σ εγe i d ( iε )Therefore, since p i ≥ εγd ( iε ) , we have e i +1 ≤ e i (1 + ( 16 σ − εγd ( iε ) ) + ( s + 4 γ ) ε n + 16 d i σ By how t , t , t where defined, this gives us what we need, thus finishing the proof of Lemma 1. In this paper we showed that every multigraph with maximum edge-multiplicity at most √ n log n ,edge-coloured by n colours such that each colour class is a disjoint union of cliques with at least2 n + o ( n ) vertices has a full rainbow matching. It would be interesting to know what is the rightmultiplicity bound? For general multigraphs, the graph composed of n disjoint triangles with eachedge of multiplicity n −
1, and all edges of a i th triangle being of colour i shows that there is nofull rainbow matching. We suspect that our bound on the multipliclity bound on the edges is closeto the right answer up to the logarithmic factor. In particular, we would like to pose the followingproblem. Problem 1:
Is there a multigraph G with maximum edge-multiplicity of at most √ n , edge-coloured by n colours such that each colour class has at least 2 n + o ( n ) vertices and is a disjointunion of non-trivial cliques, contains no full rainbow matching?Another problem is to improve the asymptotic error term on the number of vertices in a colourclass. We note that our proof can be modified so that the result holds when the size of eachcolour class is 2 n + n − α , for some absolute α >
0. So we would like to ask for a sub-polynomialimprovement. The question below is natural to ask because of the known lower bound [9] on theBrualdi-Ryser-Stein conjecture. 13 roblem 2:
Is it true that for some constant
C > G edge-coloured by n colours such that each colour class has at least 2 n + C log n vertices and is a disjoint union ofnon-trivial cliques, contains a full rainbow matching? The first author would like to thank the support provided by the London Mathematical Societyand the Mathematical Institute at the University of Oxford through an Undergraduate ResearchBursary. The second author would like to thank Alexey Pokrovskiy for introducing her to Grinblat’soriginal problem on multigraphs and for valuable discussions on the topic.
References [1] D. Best and I. M Wanless,
What did Ryser conjecture? , arXiv preprint arXiv:1801.02893(2018).[2] R. A Brualdi, H. J. Ryser, et al.,
Combinatorial matrix theory , vol. 39, Springer, 1991.[3] D. Clemens, J. Ehrenm¨uller, and A. Pokrovskiy,
On sets not belonging to algebras and rainbowmatchings in graphs , Journal of Combinatorial Theory, Series B (2017), 109 – 120.[4] D. P. Dubhashi and A. Panconesi,
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Theorems on sets not belonging to algebras , Electronic Research Announcements ofthe American Mathematical Society (2004), no. 6, 51–57.[8] , Families of sets not belonging to algebras and combinatorics of finite sets of ultrafilters ,Journal of Inequalities and Applications (2015), 1–19.[9] P. Hatami and P. W. Shor, A lower bound for the length of a partial transversal in a Latinsquare , Journal of Combinatorial Theory, Series A (2008), no. 7, 1103–1113.[10] P. Keevash and L. Yepremyan,
Rainbow matchings in properly-coloured multigraphs , SIAMjournal on Discrete Mathematics (2018), no. 3, 1577–1584.[11] C. McDiarmid, Concentration for independent permutations , Combinatorics, Probability andComputing (2002), no. 2, 163–178.[12] E. Nivasch and E. Omri, Rainbow matchings and algebras of sets , Graphs and Combinatorics , no. 2, 473–484.[13] A. Pokrovskiy, (2020), Private communication.[14] A. Pokrovskiy and B. Sudakov, A counterexample to Stein’s equi- n -square conjecture , Proceed-ings of AMS. 1415] H. J. Ryser, Neuere probleme der kombinatorik , Vortr¨age ¨uber Kombinatorik, Oberwolfach (1967), 91.[16] S. K. Stein, Transversals of Latin squares and their generalizations , Pacific J. Math. (1975),567–575. Proof.
Take a colour c ∈ Φ i +1 ∪ ∪ τj = i +2 C j . Let us recall that we are in iteration i + 1. With thatin mind, let T (1) c , T (2) c , L c be respectively, the number of K , K , K components in the colour classof c with at least 1,2,1 condemned vertices. From now on, denote a c ( i ) , b c ( i ) by respectively, thenumber of K ’s, K ’s in the colour class of colour c after iteration i . Similarly as before, define a c ( i + ) , b c ( i + ) to be respectively, the number of K ’s, K ’s in the colour class of c after Steps1,2 and 3.Note that a vertex doesn’t survive the first 3 steps (i.e, is deleted in the first 3 steps) if andonly if it is killed (in Step 2) or zapped (in Step 3). Moreover, this is equivalent to the vertex beingcondemned with the exception that it can’t be simultaneously marked, not killed and not zapped.However, by the definition of the algorithm, this in particular implies that such a vertex which ismarked, not killed and not zapped is incident to a chosen edge from a colour in Φ i +1 , that is, achosen edge which is involved in collisions. Thus, by noting that a monochromatic clique survivesif none of its vertices are deleted and that a monochromatic K becomes a K if exactly one ofits vertices is deleted, we can get the following bounds: | a c ( i + ) − ( a c ( i ) − T (1) c ) | ≤ | Φ i +1 | and | b c ( i + ) − ( b c ( i ) − L c + T (1) c − T (2) c ) | ≤ | Φ i +1 | . Hence, since e c ( i ) = 3 a c ( i ) + b c ( i ) and similarlyfor e c ( i + ), then (cid:12)(cid:12)(cid:12)(cid:12) e c ( i + 12 ) − ( e c ( i ) − L c − T (1) c − T (2) c ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ | Φ i +1 | (2) Lemma 8.
Let u, v be two distinct vertices. Then if n is large enough, the probability that they areboth condemned is at most p i .Proof. Note that by a union bound over the pairs of edges ( e , e ) of colours in C i +1 where u ∈ e and v ∈ e , we have that P ( u, v are both marked) is bounded above by P e ,e P ( e , e are both chosen) ≤ (max w d Ci +1 w ( i )) (min c ∈ Ci +1 e c ( i )) + µ min c ∈ Ci +1 e c ( i ) ≤ p i + √ nt n log n ≤ p i + ( εγd ( iε ) ) ≤ p i if n is large enough (here weare using Lemma 4, that e c ( i ) ≥ t n (which follows by e i ≤ t n as seen in the proof of Lemma 5), d ( iε ) ∈ [1 − γ,
1] and ε ≍ n ).Moreover, by independence of the zapping, P (one of u, v is condemned and the other is zapped) ≤ p i ( Q i ( u ) + Q i ( v )) ≤ p i . Now the claim follows by a union bound.We are now able to calculate the expectations of L c , T (1) c , T (2) c . Indeed, by Lemma 8 and somesimple use of the Inclusion-Exclusion principle, we can see thati) | E [ L c ] − p i b c ( i ) | ≤ p i b c ( i ) ≤ s ε n ii) | E [ T (1) c ] − p i a c ( i ) | ≤ p i a c ( i ) ≤ s ε n iii) E [ T (2) c ] ≤ p i a c ( i ) ≤ s ε n where we are using that b c ( i ) ≤ n c ≤ n and a c ( i ) ≤ n c ≤ n . Now we need prove concentrationfor these random variables. 15 emma 9. There exists a constant r = r ( σ , σ ) > such that for every c ∈ Φ i +1 ∪ ∪ τj = i +2 C j ,with probability at least − o ( n − ) we have | L c − p i b c ( i ) | ≤ r ε n. Proof.
First, recall that a vertex v is condemned if it is marked or zapped. However, note that thisis equivalent to saying that the vertex v is marked or Z v = 1.Fix a colour c ∈ Φ i +1 ∪ ∪ τj = i +2 C j . Let then L ′ c denote the number of K components in thecolour class of c with a vertex v such that Z v = 1. Let L ′′ c denote the number of K componentswith a marked vertex and let L ′′′ c denote the number of K components with both a marked vertexand vertex v with Z v = 1. We can then write L c = L ′ c + L ′′ c − L ′′′ c .Note first that L ′ c is sum of b c ( i ) ≤ n independent Bernoulli random variables. Hence, by aChernoff bound, P ( | L ′ c − E [ L ′ c ] | > √ n log n ) ≤ (cid:18) − n log nb c ( i ) (cid:19) ≤ n log n Moreover, note that E [ L ′ c ] ≤ p i b c ( i ) (3)since for every vertex u , P ( Z u = 1) = Q i ( v ) ≤ p i .Secondly, we have that L ′′ c is a function of the εn chosen edges. Further, changing a chosen edgeof a certain colour in C i +1 will deviate L ′′ c by at most 2. Thus, by Azuma’s inequality, for y > P ( | L ′′ c − E [ L ′′ c ] | > √ εn log n ) ≤ n log n Finally, we deal with L ′′′ c . First note that since the assignment of the random variables Z x isindependent of the choice of edges in Step 1, we have that for a set { u, v } of 2 vertices, P ( { u, v } has both a marked vertex and a vertex w with Z w = 1) ≤ p i Therefore, E [ L ′′′ c ] ≤ p i b c ( i ) (4)Now, take the random set of vertices Z := { x : Z x = 1 } and let S be some (deterministic) setof vertices. Since L ′ c is σ ( Z )-measurable, let L ′ c ( S ) denote the value L ′ c takes when Z = S . Then, E [ L ′′′ c | Z = S ] ≤ p i L ′ c ( S ) (since each vertex is marked with probability at most p i ). Moreover, notethat given { Z = S } , L ′′′ c is a function of the εn chosen edges and it deviates by at most 2 when wechange one of these edges. Thus, by Azuma’s inequality, P ( | L ′′′ c − E [ L ′′′ c | Z = S ] | > √ εn log n | Z = S ) ≤ n log n (5)We will now combine all of these bounds we have gotten so far. Define the events A := {| L ′ c − E [ L ′ c ] | ≤ √ n log n } , B := {| L ′′ c − E [ L ′′ c ] | ≤ √ εn log n } and C := { L ′′′ c ≤ √ εn log n +2 p i ( E [ L ′ c ]+ √ n log n ) } . By the discussion above, we have that P ( A ) ≥ − n log n and P ( B ) ≥ − n log n . Also,note that when { Z = S } ⊆ A , we have that E [ L ′′′ c | Z = S ] ≤ p i L ′ c ( S ) ≤ p i ( E [ L ′ c ] + √ n log n ).Thus, by our application of Azuma’s inequality in (5) above, we have P ( C | A ) ≥ − n log n . Finally,this gives us that P ( A ∩ B ∩ C ) ≥ P ( B ) − P ( A ∩ C ) = P ( B ) − (1 − P ( A ) P ( C | A )) = 1 − o ( n − ) bythe bounds we got. 16inally, note that using (3) and (4), we can see that the event A ∩ B ∩ C implies | L c − E [ L c ] | ≤ p i b c ( i ) + | L c − E [ L ′ c + L ′′ c ] | ≤ p i b c ( i ) + √ n log n + 2 √ εn log n + 2 p i (2 p i b c ( i ) + √ n log n ) ≤ s ε n if n is large enough.Concluding, since we have the bound in i), we get that | L c − p i b c ( i ) | ≤ | L c − E [ L c ] | + 8 s ε n ≤ s ε n The Lemma follows by then taking r := 25 s . Lemma 10.
There exists a constant r = r ( σ , σ ) > such that for every c ∈ Φ i +1 ∪ ∪ τj = i +2 C j ,with probability − o ( n − ) we have | T (1) c − p i a c ( i ) | ≤ r ε n Proof.
The proof is almost exactly the same as the proof of Lemma 9 and so, we omit it.
Lemma 11.
There exists a constant r = r ( σ , σ ) > such that for every c ∈ Φ i +1 ∪ ∪ τj = i +2 C j ,with probability − o ( n − ) we have T (2) c ≤ r ε n Proof.
Fix a colour c ∈ Φ i +1 ∪ ∪ τj = i +2 C j . Note that by a similar reasoning as in the proof of Lemma9, we can see that T (2) c = T (2) ′ c + T (2) ′′ c where T (2) ′ c denotes the number of K components in the colour class of c with at least 2 distinctvertices v, u such that Z v = Z u = 1 and T (2) ′′ c denotes the number of K components with at most1 vertex v with Z v = 1 and at least 2 condemned vertices. Let also, T (2) ′′′ c denote the number of K components with exactly 1 vertex v such that Z v = 1.Note first that T (2) ′ c is sum of a c ( i ) ≤ n independent Bernoulli random variables and that E [ T (2) ′ c ] ≤ p i a c ( i ) ≤ s ε n . Hence, by a Chernoff bound, P ( T (2) ′ c > √ n log n + 4 s ε n ) ≤ − n log na c ( i ) ) ≤ n log n Secondly note that E [ T (2) ′′′ c ] ≤ p i a c ( i ) ≤ s εn and that T (2) ′′′ c is also a sum of a c ( i ) ≤ n independent Bernoulli random variables so that we also get P ( T (2) ′′′ c > √ n log n + 4 s εn ) ≤ n log n Finally, take the random set of vertices Z as defined in the proof of Lemma 9. Since T (2) ′′′ c is σ ( Z )-measurable, let T (2) ′′′ c ( S ) denote its value when Z = S for a (deterministic) set of vertices S .We note then that E [ T (2) ′′ c | Z = S ] ≤ p i T (2) ′′′ c ( S ) + 6 p i a c ( i ) ≤ s ε ( T (2) ′′′ c ( S ) + 4 s εn )where we are using that the probability that a vertex is marked is at most p i and that, by the proofof Lemma 8, the probability that two distinct vertices are marked is at most 2 p i . Moreover, given Z = S , T (2) ′′ c is a function of the εn chosen edges and deviates by at most 2 when we change oneof these edges. Thus, by Azuma’s inequality, 17 ( T (2) ′′ c > √ εn log n + E [ T (2) ′′ c | Z = S ] | Z = S ) ≤ n log n Joining these observations in a similar fashion as in the proof of Lemma 9, we get that withprobability 1 − o ( n − ), T (2) c = T (2) ′ c + T (2) ′′ c ≤ ( √ n log n + 4 s ε n ) + ( √ εn log n + 2 s ε ( √ n log n +8 s εn )) ≤ s ε n if n is large. Therefore, the Lemma follows by taking r := 21 s .Note then that by (2) before, we get for each colour c ∈ Φ i +1 ∪ ∪ τj = i +2 C j , with probability1 − o ( n − ), | e c ( i + 12 ) − e c ( i )(1 − p i ) | ≤ ( r + 2 r + r ) ε n + 10 | Φ ′ i +1 | We now deal with the degrees.Let’s first define for vertices u, v , µ C j ( k )( uv ) to be the edge multiplicity of uv with respect tocolours in C j after iteration k . Take then a vertex v surviving steps 1,2 and 3 (of iteration i + 1)and j > i + 1. We note that d C j v ( i + ) = d C j v ( i ) − D v,j + D v,j , where • D v,j = P w ∈ Γ Cjv ( i ) µ C j ( i )( vw ) 1 w is condemned • D v,j = P w ∈ Γ Cjv ( i ) µ C j ( i )( vw ) 1 w is condemned but not deleted We first look at D v,j . Note that D v,j ≤ D v,j where • D v,j = P w ∈ Γ Cjv ( i ) µ C j ( i )( vw ) 1 w is marked but not killed To calculate the expectation of this random variable we require the following simple Lemma:
Lemma 12.
For a vertex w , the probability that it is marked but not killed is at most p i .Proof. Note that w is marked but not killed if and only if one of the following situations occurs:1) Two edges incident to w are chosen;2) Two edges wu, uy are chosen, where y = w .The first situation has probability at most d Ci +1 w ( i ) (min c ∈ Ci +1 e c ( i )) ≤ p i (by Lemma 4) of occurring, whilethe second situation has probability at most d C i +1 w ( i ) × max u d C i +1 u ( i )(min c ∈ C i +1 e c ( i )) ≤ p i which follows also by Lemma 4). Thus, the lemma follows.Equipped with Lemma 12, we can now say that E [ D v,j ( i )] ≤ p i d C j v ( i ) ≤ s ε n , where we areusing the trivial bound d C j v ( i ) ≤ εn and Lemma 4. Moreover, note that D v,j is a function of the εn chosen edges and deviates by at most 4 µ when we change a chosen edge of a certain colour.Thus, by Azuma’s inequality we have P ( | D v,j − E [ D v,j ] | > µ √ εn log n ) ≤ n log n and so, since µ √ εn log n = o ( ε n ), with probability 1 − o ( n − ) we have D v,j ≤ D v,j ≤ (4 s + 1) ε n Finally, we deal with D v,j . We need the following Lemma.18 emma 13. There exists a constant r = r ( σ , σ ) > such that for every vertex v survivingsteps 1,2 and 3 and j > i + 1 , with probability − o ( n − ) we have | D v,j − p i d C j v ( i ) | ≤ r ε n Proof.
The proof follows exactly the same reasoning as the proofs of Lemmas 9 and 10 and further,exploits the fact that µ = √ n log n .Recall that d C j v ( i + ) = d C j v ( i ) − D v,j + D v,j . But then, by this and Lemma 13, we can nowsee that for every vertex v surviving steps 1,2 and 3 and j > i + 1, with probability 1 − o ( n − ) wehave d C j v ( i + 12 ) ≤ d C j v ( i )(1 − p i ) + (4 s + r + 1) ε n To finish off the proof of Lemma 6, note that we have at most n colours in Φ i +1 ∪ ∪ τj = i +2 C j andat most n ε = O ( n ) pairs ( v, j ) where τ ≥ j > i + 1 and v is a vertex. We then get, by setting s := max { , r + 2 r + r , s + r + 1 } , that with probability 1 − o (1), • | e c ( i + ) − e c ( i )(1 − p i ) | ≤ s ( ε n + | Φ i +1 | ) for every colour c ∈ Φ i +1 ∪ ∪ τj = i +2 C j , • d C j v ( i + ) ≤ d C j v ( i )(1 − p i ) + s ε n for every vertex v surviving step 3 and j > ij > i