On the number of symbols that forces a transversal
OOn the number of symbols that forces a transversal
Peter Keevash ∗ Liana Yepremyan † May 29, 2018
Abstract
Akbari and Alipour [1] conjectured that any Latin array of order n with at least n / n , and moreover, we show that n / symbols suffice. A Latin square of order n is an n by n square with cells filled using n symbols so that every symbolappears once in each row and once in each column. A transversal in a Latin square is a set of cellsusing every row, column and symbol exactly once. The study of transversals in Latin squares goesback to Euler in 1776; his famous ‘36 officers problem’ is equivalent to showing that there is no Latinsquare of order 6 that can be decomposed into 6 transversals (this was finally solved by Tarry in1900). An even more fundamental question is whether a Latin square always has a transversal. Aquick answer is ‘no’, as shown by the addition table of Z k , but it remains open whether there isalways a transversal when n is odd (a conjecture of Ryser [10]) or whether there is always a partialtransversal size n − n − O (log n ).Given the apparent difficulty of finding transversals in Latin squares, it is natural to ask if theproblem becomes easier in Latin arrays with more symbols (now we fill a square with any numberof symbols such that every symbol appears at most once in each row and at most once in eachcolumn). This problem was considered by Akbari and Alipour [1], who conjectured that any Latinarray of order n with at least n / − √ n symbols suffice) and Bar´at and Nagy [2] (who showed that 3 n / n as a properly edge-coloured complete bipartite graph K n,n , with one partcorresponding to rows, the other part to columns, and the colour of an edge is the symbol in thecorresponding cell of the array. In this language, the Akbari-Alipour conjecture is that if there are atleast n / n in a strong form. Theorem 1.1.
Suppose the complete bipartite graph K n,n is properly edge-coloured using dn colours,where n is sufficiently large and d > n − / . Then there is a rainbow perfect matching. ∗ Mathematical Institute, University of Oxford, Oxford, UK. E-mail: [email protected] supported in part by ERC Consolidator Grant 647678. † Mathematical Institute, University of Oxford, Oxford, UK. E-mail: [email protected]. a r X i v : . [ m a t h . C O ] M a y ere the constant ‘200’ could be somewhat improved, but we have sacrificed some optimisationsfor the sake of readability of our proof, as in any case the best bound we can obtain seems far fromoptimal (it might even be true that n o (1) colours suffice!) Here we give the proof of Theorem 1.1, assuming two lemmas that will be proved later in the paper.Consider the complete bipartite graph K n,n with parts A and B both of size n , and a proper edge-colouring using at least dn colours, where n is sufficiently large and d > n − / .Let G be any subgraph of K n,n obtained by including exactly one edge of each colour. Then e ( G ) ≥ dn . We apply the following lemma, which will be proved in the next section, to find a pair( A , B ) that satisfies Hall’s condition for a perfect matching ‘robustly’, so that it will still satisfyHall’s condition after deleting small sets of vertices from each part. Note that as d > n − / weobtain | A | = | B | > n . / Lemma 2.1.
There is G = G [ A , B ] for some A ⊆ A and B ⊆ B with | A | = | B | ≥ d n/ such that G has minimum degree at least − d | A | , and for any S ⊆ A or S ⊆ B we have | N G ( S ) | ≥ min { | S | , | A | / } . We define a random subgraph G r of K n,n of ‘reserved colours’ as follows. Choose each colourindependently with probability p := n − . . Let G r consist of all edges of all chosen colours. ByChernoff bounds, whp | N G r ( b ) ∩ A | = p | A | ± ( p | A | ) / and | N G r ( b ) \ A | = p | A \ A | ± ( p | A \ A | ) / for all b ∈ B , and similarly with A and B interchanged. Let G ∗ := ( K n,n \ G r ) \ ( A ∪ B ). Thenthe minimum degree in G ∗ satisfies δ ( G ∗ ) ≥ (1 − p )( n − | A | ) − ( pn ) / .Let M be a maximum size rainbow matching in G ∗ := ( K n,n \ G r ) \ ( A ∪ B ). Let A = V ( M ) ∩ A and B = V ( M ) ∩ B . By a result of Gy´arf´as and S´ark¨ozy [5, Theorem 2] we have | A | = | B | ≥ δ ( G ∗ ) − δ ( G ∗ ) / ≥ (1 − p )( n − | A | ), as pn = n . (cid:29) n / .Let G (cid:48) be obtained from G by deleting any edges that use a colour used by M and restrictingto some subsets A (cid:48) ⊆ A and B (cid:48) ⊆ B with | A (cid:48) | = | B (cid:48) | = (1 − − d ) | A | so that G (cid:48) has minimumdegree at least (10 − − · − ) d | A | . To see that this is possible, note that we delete at most n edges from G (cid:48) , so each of A and B has at most 10 n/d | A | < − d | A | vertices at which we deletemore than 10 − d | A | edges.Let A = A \ ( A (cid:48) ∪ A ) and B = B \ ( B (cid:48) ∪ B ). Note that | A | = | B | < · − d | A | , as pn = n . (cid:28) d | A | . The form of our intended rainbow matching is illustrated by the black and/orvertical edges in Figure 1 (the coloured diagonal edges illustrate the augmentation algorithm usedin Section 4).Let A (cid:48) be the set of vertices a in A such that at least | B (cid:48) | / a and B (cid:48) havea colour used by M . Define B (cid:48) similarly, interchanging A and B . We prove the following lemma insection 4. Lemma 2.2.
Both A (cid:48) and B (cid:48) have size at most p | A | / . Now we apply a greedy algorithm to construct a rainbow matching M ∪ M where each edge of M joins A ∪ B to A (cid:48) ∪ B (cid:48) . We start by choosing these edges for vertices in A (cid:48) ∪ B (cid:48) using coloursin G r . This is possible as the number of choices at each step is at least p | A | − ( p | A | ) / > p | A | / · p | A | / A ∪ B .2 ' Figure 1: Proof by pictureThis is possible as the number of choices at each step is at least | A (cid:48) | /
2, of which at most 3 | A | areforbidden due to using a colour or a vertex used at a previous step.Finally, consider G = G [ A , B ] obtained from G (cid:48) by deleting all vertices covered by M and alledges that share a colour with M . Then | A | = | B | = | A (cid:48) | − | A | > (1 − − d ) | A | and G hasminimum degree at least 10 − d | A | − | A | − | M | ≥ − d | A | / G has a perfect matching. To see this we check Hall’s condition. Suppose for acontradiction there is S ⊆ A with | N ( S ) | < | S | . By the minimum degree we have | S | ≥ − d | A | / | S | ≤ | A | /
2, as then | N G ( S ) | ≥ min { | S | , | A | / } − | A | > | S | . However,letting T = B \ N ( S ) we have N ( T ) ⊆ A \ S , so | N ( T ) | = | A | − | S | < | B | − | N ( S ) | = | T | . Thesame argument as for S gives | T | > | A | /
2, contradiction. Therefore G has a perfect matching M .Now M ∪ M ∪ M is a rainbow perfect matching in G , which completes the proof of Theorem 1.1. In this section we prove Lemma 2.1. Let G be a bipartite graph with parts A and B . We say G is ( ε, δ )-dense if for any A (cid:48) ⊆ A and B (cid:48) ⊆ B with | A (cid:48) | ≥ ε | A | and | B (cid:48) | ≥ ε | B | we have e G ( A (cid:48) , B (cid:48) ) ≥ δ | A (cid:48) || B (cid:48) | .We start by applying the following result of Peng, R¨odl and Ruci´nski [9, Theorem 1.3] with ε = 1 / c = 0 .
24 and c (cid:48) = 1 / Lemma 3.1.
Suppose c, c (cid:48) ∈ (0 , with c + c (cid:48) ≤ . Then there are A (cid:48) ⊆ A and B (cid:48) ⊆ B with | A (cid:48) | = | B (cid:48) | ≥ d / log (1+ cε ) n/ so that G (cid:48) = G [ A (cid:48) , B (cid:48) ] is ( ε, c (cid:48) d ) -dense. Let G = G [ A , B ] be obtained from G (cid:48) by the following algorithm. Initially, A = A (cid:48) and B = B (cid:48) . At any step of the algorithm, we update G by deleting a vertex or set of vertices of oneof the following types (choosing arbitrarily if there is a choice).i. v ∈ A or v ∈ B with d G ( v ) ≤ − d | A (cid:48) | ,ii. S ⊆ A or S ⊆ B with | S | < ε | A (cid:48) | and | N G ( S ) | ≤ | S | .Whenever we delete some vertices from A or B we delete an arbitrary set of the same size fromthe other, so that we always maintain | A | = | B | . We stop if no deletion is possible or if we havedeleted at least 2 ε | A (cid:48) | vertices from each side.We claim that the latter option is impossible. Indeed, then without loss of generality we deleted ε | A (cid:48) | vertices from A (cid:48) of type (i) or (ii) as above (at least half of the deleted vertices are deleted fora reason other than maintaining equal part sizes). Let D A = D Ai ∪ D Aii be the deleted vertices in A (cid:48) according to deletions of type (i) or (ii). Note that | D A | < ε | A (cid:48) | , and | N G ( D Aii ) | ≤ | D Aii | < ε | A (cid:48) | . This result follows from their proof; they state the case c = 1 / c (cid:48) = 1 / (1 + ε/ ≥ ε/ B = B \ N G ( D Aii ), so | B | > (1 − ε ) | A (cid:48) | = ε | A (cid:48) | . Now e G (cid:48) ( D A , B ) ≤ | D Ai | · − d | A (cid:48) | < d | D A || B | contradicts ( ε, d/ G (cid:48) , which proves the claim.Thus the algorithm stops with | A | = | B | > (1 − ε ) | A (cid:48) | ≥ d n/ / log (1 . < − d | A (cid:48) | and | N G ( S ) | ≥ | S | for any S ⊆ A or S ⊆ B with | S | < ε | A (cid:48) | .Furthermore, for any S ⊆ A or S ⊆ B with | S | ≥ ε | A (cid:48) | , by ( ε, d/ G (cid:48) we have | N G ( S ) | ≥ | B | − ε | A (cid:48) | ≥ | A | /
3. This proves Lemma 2.1.
In this section we prove Lemma 4.5, which will complete the proof of Theorem 1.1. Suppose it is nottrue, say | A (cid:48) | > p | A | /
4. We will iteratively construct R = R A ∪ R B ⊆ M , where we think of R A and R B as ‘reachable’ from A and B . At some point R A and R B will intersect, which will contradict M being a maximum size rainbow matching in G ∗ . Let θ := n − . , and note that θ | A | > n . / Algorithm 4.1.
Let R A = R B = ∅ and let C be the set of colours not used by M . At step i ≥ R A ∩ R B (cid:54) = ∅ stop, otherwise let R Ai be the set of all uv ∈ M where v ∈ B \ V ( R B ) suchthat at least θ | A | edges in G ∗ from v to A use a colour in C , let C Ai be the set of coloursused by R Ai , update R A by adding R Ai and C by adding C Ai ,ii. if R A ∩ R B (cid:54) = ∅ stop, otherwise let R Bi be the set of all uv ∈ M where u ∈ A \ V ( R A ) suchthat at least θ | A | edges in G ∗ from u to B use a colour in C , let C Bi be the set of coloursused by R Bi , update R B by adding R Bi and C by adding C Bi . Claim 4.2. | R A | ≥ | A | / X of edges in G ∗ with colour in C between A (cid:48) and B . Wehave X ≤ | R A || A (cid:48) | + | B | θ | A | by definition of R A . Also, by definition of A (cid:48) , every vertex in A (cid:48) has atleast (1 − p ) | B | − ( | M | − | B (cid:48) | / ≥ | A | / G ∗ to B with colour in C , so X ≥ | A (cid:48) | · | A | / | A (cid:48) | ≥ p | A | / p − θn = n . (cid:28) pn (cid:28) | A | , we deduce | R A | ≥ | A | / − | A (cid:48) | − | B | θ | A | ≥| A | / − p − θn ≥ | A | /
4, as claimed.
Claim 4.3.
For i ≥
1, we have | R Bi | ≥ | R A | − pn and | R Ai +1 | ≥ | R B | − pn .To see this, we first note that as R A ∩ R B = ∅ , any vertex in B has at least (1 − p ) | A | −| R B | − | M \ ( R A ∪ R B ) | = | R A | − p | A | edges in G ∗ to A \ R B with colour in C . Double-counting such edges as in the previous claim gives | B | ( | R A | − p | A | ) ≤ | R Bi || B | + | B | θ | A | , so | R Bi | ≥ | R A | − p | A | − | B | − | B | θ | A | ≥ | R A | − pn . The proof of the second inequality is similar,so the claim holds. Claim 4.4.
The algorithm terminates at some step i = i + < log n .To see this, we show inductively that if R A ∩ R B = ∅ at step i then | R Bi | ≥ f ( i ) | A | / | R Ai | ≥ ( f ( i ) − − ) | A | / f ( i ) = 2 i − + 2 − . First note that f (1) = 5 / i ≥ (cid:80) i − j =1 ( f ( j ) − − ) = 2 − (2 i −
1) + ( i − − − − ) ≥ f ( i ) − /
16. At step 1 we have | R A | ≥ | A | / > ( f (1) − − ) | A | / | R B | ≥ | R A | − pn > . | A | > f (1) | A | /
3. Supposingthe statement at step i − ≥
1, we have | R Ai | ≥ ( (cid:80) i − j =1 | R Bj | ) − pn ≥ ( (cid:80) i − j =1 f ( j )) | A | / − pn ≥ ( f ( i ) − / | A | / − pn ≥ ( f ( i ) − − ) | A | / | R Bi | ≥ ( (cid:80) ij =1 | R Aj | ) − pn ≥ ( (cid:80) ij =1 f ( j ) − − ) | A | / − pn ≥ ( f ( i + 1) − / | A | / − pn ≥ f ( i ) | A | /
3. Thus the required bounds hold byinduction. While R A ∩ R B = ∅ we deduce (2 f ( i ) − − ) | A | / < | M | < n , so i + < log n , as claimed.4he algorithm terminates by finding some edge ab ∈ R A ∩ R B where a ∈ A and b ∈ B . Wewill obtain a contradiction by modifying M to obtain a larger rainbow matching in G ∗ . Given twocolours c and c (cid:48) in C , we say that c is earlier than c (cid:48) if c was added to C before c (cid:48) . We start byapplying the definition of R A and R B to find edges a b and ab of G ∗ with a ∈ A and b ∈ B where the colours of a b and ab are in C and earlier than that of ab . We modify M to obtain M (cid:48) by deleting ab and adding a b and ab . Thus we obtain a larger matching, but M (cid:48) may not berainbow, due to repeating the colours of a b and ab . While the current matching M (cid:48) is not rainbow,we apply the following ‘trace back’ algorithm (similar to that of [5]). Algorithm 4.5.
At step i ≥ M (cid:48) havingsome colour in C shared with some edge that is still present from M . At step 1 these are a b and ab . If there is an active edge at step i , we choose one arbitarily, call it a i b i , and let a (cid:48) i b (cid:48) i be the edgeof M of the same colour c ∈ C . By construction of C , one of a (cid:48) i or b (cid:48) i , say a (cid:48) i , has at least θ | A | edges to B or A using an earlier colour than c in C . We modify M (cid:48) by deleting a (cid:48) i b (cid:48) i and addingsome such edge a (cid:48) i b i where b i ∈ B is distinct from all previous choices. We say that a i b i is no longeractive. We make a (cid:48) i b i active if its colour is shared with some edge that is still present from M .Algorithm 4.5 is illustrated in Figure 1: the thick black edge represents the edge ab ∈ R A ∩ R B ,at step 1 the green and blue diagonals are active, at step 2 the blue diagonal is active, at step 3 thered diagonal is active, at step 4 the pink diagonal is active, at step 5 there are no active edges sothe algorithm terminates. To see that the algorithm succeeds, note that there are at most 4 log n steps of replacing an active edge by another, and each choice has at least θ | A | > n . / > n options. Thus we obtain a rainbow matching M (cid:48) in G ∗ with | M (cid:48) | > | M | . This contradiction provesLemma 4.5. Postscript.
The Akbari-Alipour conjecture was proved independently and simultaneously by Mont-gomery, Pokrovskiy and Sudakov [8]. Our proof is much simpler than theirs, and gives a better boundon the number of symbols required, whereas their proof applies in a much more general setting, andso has several further applications. Results similar to those in [8] (but not including the Akbari-Alipour conjecture) were independently and simultaneously obtained by Kim, K¨uhn, Kupavskii andOsthus [7].
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