On factors of independent transversals in $k$-partite graphs

A $[k,n,1]$-graph is a $k$-partite graph with parts of order $n$ such that the bipartite graph induced by any pair of parts is a matching. An independent transversal in such a graph is an independent set that intersects each part in a single vertex. A factor of independent transversals is a set of $n$ pairwise-disjoint independent transversals. Let $f(k)$ be the smallest integer $n_0$ such that every $[k,n,1]$-graph has a factor of independent transversals assuming $n \ge n_0$. Several known conjectures imply that for $k \ge 2$, $f(k)=k$ if $k$ is even and $f(k)=k+1$ if $k$ is odd. While a simple greedy algorithm based on iterating Hall's Theorem shows that $f(k) \le 2k-2$, no better bound is known and in fact, there are instances showing that the bound $2k-2$ is tight for the greedy algorithm. Here we significantly improve upon the greedy algorithm bound and prove that $f(k) \le 1.78k$ for all $k$ sufficiently large, answering a question of MacKeigan.


Introduction
Given a k-partite graph, a transversal is a set of vertices containing a single vertex from each part.
An independent transversal is an independent set which is also a transversal. A factor of transversals is a set of pairwise-disjoint transversals covering all vertices. The problem of finding sufficient conditions for the existence of independent transversals and factors of independent transversals in k-partite graphs was studied by several researchers [1,2,3,4,5,6,7,8,10,12,13,14,15,16,17] not least because it is strongly related to the Hajnal-Szemerédi Theorem and to concepts such as the strong chromatic number and list coloring.
In this paper we consider sufficient conditions for a factor of independent transversals in very sparse k-partite graphs. To this end, define a [k, n, 1]-graph to be a k-partite graph with parts of order n such that the bipartite graph induced by any pair of parts is a matching (more generally, in a [k, n, ∆]-graph the bipartite graph induced by any pair of parts has maximum degree at most ∆).
for every fixed k and sufficiently large n = n(k) by Keevash and Mycroft [10]. However, as all of these proofs require n to be very large compared to k, they do not imply Conjecture 1.1 nor an upper bound close to the conjecture. Lastly, we mention that the author in [16] considered the refinement of f (k) to [k, n, 1]-graphs in which every matching consists of at most s independent edges (so the case s = n is the general case of [k, n, 1]-graphs). Denoting the refined parameter by f (k, s) we observe by the construction above that already f (k, 1) ≥ k and it is proved in [16] that f (k, 1) = f (k, 2) = k. Furthermore, it is conjectured there that f (k, s) = k for all 1 ≤ s ≤ k, which again states that f (k) = k, ignoring Catlin's obstacle (this was also observed in [12]), and so if the modified Fischer Conjecture holds (namely: Catlin's construction is the only obstacle), so does the modified conjecture on f (k, s) = k unless k ≥ 3 is odd and s = k in which case f (k, s) = k + 1.
There is, however, a simple upper bound of f (k) ≤ 2k − 2 observed by MacKeigan [12] which follows by greedily applying Hall's algorithm. Assume that G is a [k, n, 1]-graph with n ≥ 2k − 2. Suppose we have already constructed a factor F of independent transversals on the induced subgraph G ′ of G consisting of all but the last set V k . Then G ′ is a [k − 1, n, 1]-graph. Now consider the bipartite graph B with one part being F and the other part being V k , and there is an edge between I ∈ F and v ∈ V k if v is not a neighbor in G of any vertex of I. Then B has n vertices in each part and minimum degree at least n − (k − 1) ≥ n/2, so B has a perfect matching by Hall's Theorem, implying that F can be extended to a factor of independent transversals of G. Notice that this already shows that f (3) ≤ 4 so together with Catlin's example, f (3) = 4. Unfortunately, there are examples where this greedy algorithm of repeatedly applying Hall's theorem fails to produce a factor of independent transversals if n = 2k − 3. Indeed we can construct examples of F as above such that it has a subset of k − 1 elements F * = {I 1 , . . . , I k−1 }, and such that each of the vertices in V * = {v k,1 , . . . , v k,k−1 } ⊂ V k has exactly one neighbor in G belonging to I j for each 1 ≤ j ≤ k − 1 (for example, any Latin square of order k − 1 corresponds to such a construction). Then in the bipartite graph B, the neighborhood of V * is only of size n − (k − 1) = k − 2, violating Hall's condition, so B has no perfect matching. A problem therefore raised by MacKeigan asks whether it is possible to improve upon the greedy algorithm. Our main result shows that indeed this is the case.
Our main idea in the proof of Theorem 1.2 is to apply semi-randomness to the perfect matchings at each stage of the greedy algorithm (hence it is no longer greedy) so as to guarantee that as we come close to the end of the algorithm (close to Stage k) we can still guarantee Hall's condition with high probability.
The rest of this paper continues as follows. In Section 2 we state and prove some preliminary lemmas that are useful in the proof of the main result. Theorem 1.2 is then proved in Section 3.
In Section 4 we prove that f (4) = 4 thereby verifying Conjecture 1.1 in the first non-elementary case. Throughout the paper, we assume, whenever necessary, that k and n are sufficiently large.

Preliminary setup
Let us first observe that if we can show that every [k, n, 1]-graph has a factor of independent transversals, then f (k) ≤ n. Indeed, if G is a [k, n ′ , 1]-graph with n ′ > n we can just take an induced subgraph consisting of n vertices of each part observing that this subgraph is an [n, k, 1]graph. Being such, it has a (factor of) independent transversal(s), so we can just apply this repeatedly to find n ′ − n pairwise disjoint independent transversals in G. Removing them from G we remain with an induced subgraph G ′ of G which is an [n, k, 1]-graph and find a factor of independent transversals in G ′ , hence, altogether, a factor of independent transversals in G.
In our proof of Theorem 1.2 it will be more convenient to fix the number of vertices in each part and to parameterize on the number of parts. Let, therefore g(n) denote the largest integer k such that every [k, n, 1]-graph has a factor of independent transversals. We therefore have: 562n for all n sufficiently large, then f (k) ≤ 1.78k for all k sufficiently large, so Theorem 1.2 holds.
As we prove in Lemma 3.4 that g(n) ≥ 0.562n for all n sufficiently large, we have by the last corollary that Theorem 1.2 holds.
For a graph G and a set of its vertices X, let N G (X) denote the set of vertices of G that have a neighbor in X. The following is an immediate corollary of Hall's Theorem.
Lemma 2.2. Let B be a bipartite graph with vertex parts X, Y of order m each. Assume that the degree of every vertex is at least m − t and, furthermore, every W ⊆ Y with m − t ≤ |W | ≤ t has |N B (W )| ≥ |W |. Then B has a perfect matching.
Lemma 2.3. Let δ > 0. Then there exists a constant C = C(δ) > 0 such that the following holds. Let B be a bipartite graph with vertex parts X = {x 1 , . . . , x n }, Y = {y 1 , . . . , y n } and suppose the degree of each vertex of X is precisely n − t. Furthermore, suppose that X * ⊆ X and Y * ⊆ Y are given. Let π be a random permutation of the vertices of Y and let M = {(x i , y π(i) )} n i=1 . Then with probability at least 1 − C/n the following hold: (i) The number of pairs in M that are edges of B is at least n − t − δn and at most n − t + δn. (ii) There is a subset M * ⊆ M with |M * | ≥ |X * |(|Y * |−t)/n−δn such that every pair (x i , y π(i) ) ∈ M * is an edge of B and furthermore, x i ∈ X * and y π(i) ∈ Y * . Proof. We start with (ii). Suppose without loss of generality, that the vertices of X * are {x 1 , . . . , x q } where q = |X * |. Let d i be the number of neighbors of x i in Y * and observe that d i ≥ (n − t) − (n − |Y * |) = |Y * | − t. Let R i be the indicator random variable for the event A i = "y π(i) ∈ Y * and (x i , y π(i) ) is an edge of B". Let R = q i=1 R i count the number of pairs that satisfy the condition of being in M * . Now, P r[A i ] = d i /n ≥ (|Y * | − t)/n. It follows that E[R] ≥ q(|Y * | − t)/n. It remains to upper bound the probability that R deviates from its expectation by more than δn. Now, if q ≤ δn then (ii) is trivial since R ≤ q. Otherwise, R is the sum of Θ(n) indicator random variables.
given that A i occurred, there are two possibilities. If y π(i) is a neighbor of x j , then A j occurred with probability (d j −1)/(n−1). If y π(i) is not a neighbor of x j , then A j occurred with probability d j /(n − 1). In any case, . It now follows from Chebyshev's inequality that the probability that R deviates from its expected value by more than δn is O(1/n).
For (i), apply the proof above for the special case that X * = X and Y * = Y (so now q = n and |Y * | = n). Then the corresponding indicator random variable R i has Pr[R i ] = (n − t)/n since the degree of x i is precisely n − t. Hence, E[R] = n − t and, as we have shown, the probability that R deviates from its expectation by more than δn is O(1/n).
We need the next definition that formalizes the following procedure. Suppose we are given a  We emphasize that when constructing B * in Definition 2.4, it will be very important to select the s edges of M at random, as stated, and not just arbitrarily.
In our proof we will need to ascertain that the constant we choose satisfies the following constraint.
Proof. The definite integral equals µ 2 (c 2 + 3c 2 + 1 2 ) + µc 2 ln(1 − µ). Hence, dividing by µ, it suffices to prove that the function So, f is strictly monotone increasing if µ < 1 − c 2 /(c 2 + 3c/2 + 1/2) and indeed this is easily verified to hold for all 0 ≤ µ ≤ 1 1+c . Since f (0) = −c, to prove the non-negativity of f in the specified range it suffices to prove that f ( 1 1+c ) ≤ 0. Now, rearranging the terms it is easily verified that For the second part of the lemma, define γ = h(ǫ) = −f 1−ǫ 1+c and observe that h(0) ≥ 0 and that h is a continuous strictly monotone increasing function, since f is.

Proof of the main result
In this section we assume that G is a [k, n, 1]-graph with vertex parts V 1 , . . . , V k of order n each.
Definition 3.1 (t-partial factor of (independent) transversals). A t-partial factor of transversals of G is a factor of transversals of G t . If each element in the factor is an independent set, it is called a t-partial factor of independent transversals of G.
In our proof we construct a sequence F 1 , . . . , F k such that F t is a t-partial factor of (hopefully independent) transversals of G. Let F t = {I(t, 1), . . . , I(t, n)} where I(t, j) is a (not necessarily independent) transversal of G t . Our sequence will have the property that for every 1 ≤ t ≤ k − 1, To facilitate the construction of F t+1 as an extension of F t we need the following definition.
Definition 3.2 (The auxiliary bipartite graph B t ). Suppose that F t is t-partial factor of transversals. The auxiliary bipartite graph B t is defined by one part being F t (so the vertices of this part are Observe that a perfect matching in B t corresponds to F t+1 by defining I(t+1, j) to be the union of I(t, j) and its match. Notice that indeed F t+1 extends F t and notice that if F t is a t-partial factor of independent transversals, then F t+1 is also a (t + 1)-partial factor of independent transversals.
Proof. Observe first that I(t, i) ∈ F t has t vertices, and each such vertex has at most one neighbor The following is our main lemma.
Lemma 3.4. For all sufficiently large n it holds that g(n) ≥ 0.562n.
Proof. Let 0.778 ≤ c ≤ 1 and recall that the conditions of Lemma 2.5 hold. (Note: our proof works for every c in this range, but we will eventually optimize by using c = 0.778.) Suppose ǫ > 0 is a given small constant. Let γ = γ(ǫ) be the constant from Lemma 2.5 and choose constants 0 < δ ≪ η ≪ γ (by x ≪ y we mean that x is a small function of y, small enough to satisfy the claimed inequalities that will follow). Let C = C(δ) be the constant from Lemma 2.3. Throughout the proof we assume that n is sufficiently large as a function of ǫ. Let k = ⌊n(1 − ǫ)/(c + 1)⌋ and let G be a [k, n, 1]-graph with vertex parts V 1 , . . . , V k of order n each. We will prove that G has a factor of independent transversals.
We construct partial factors of transversals F 1 , . . . , F k starting with F 1 = {I(1, 1), . . . , I(1, n)} where I(1, j) is just a single vertex of V 1 . Trivially, F 1 is a 1-partial factor of independent transversals of G. We next define stages t = 1, . . . , k − 1 where at Stage t we construct (partly using a probabilistic argument) F t+1 as an extension of F t . Note that at this point we only assume that F t is a t-partial factor of transversals. However, we will prove that with positive probability, all elements of F t , for all t = 1, . . . , k are in fact independent sets. Hence we will obtain with positive probability that the final F k is a factor of independent transversals of G, as required. We now describe Stage t in detail, assuming all previous stages have been completed (for completeness, Stage 0 constructs the trivial F 1 as above).
Consider B t in which one part is the already constructed F t = {I(t, 1), . . . , I(t, n)} and the other part is V t+1 = {y 1 , . . . , y n }. Recall by Lemma 3.3 that each vertex of B t has degree at least n − t. So, remove edges from B t until each vertex of the part F t of B t has degree exactly n − t and denote the resulting spanning subgraph by B ′ t (observe that in B ′ t some vertices in the part V t+1 may now have degree smaller than n − t). Randomly select a permutation of V t+1 , denoting it by π t . Now consider the set of all pairs M (π t ) = {(I(t, i), y πt(i) )} n i=1 . Unfortunately, not each pair (I(t, i), y πt(i) ) is necessarily an edge of B t and thus, moreover, not necessarily an edge of B ′ t (recall, to be an edge of B t we should have that all t vertices of I(t, i) are not neighbors in G of y πt(i) ). Let M ′ (π t ) denote the set of elements of M (π t ) that are edges of B ′ t and set m t = |M ′ (π t )|. We say that M (π t ) is good if |m t − (n − t)| ≤ δn. Proof. We apply Lemma 2.3 with B = B ′ t , X = F t , Y = V t+1 and π = π t . According to case (i) of that lemma, with probability at least 1 − C/n, M (π t ) contains at least (n − t) − δn edges of B ′ t and at most (n − t) + δn edges of B ′ t .
It is important to note that the stated probability 1−C/n of M (π t ) being good does not depend on the goodness of M (π ℓ ) for any ℓ < t, as the proof of Lemma 2.3 does not assume anything except that the degree of each vertex of B ′ t is n − t and that the permutation π t is chosen at random. Now, if M (π t ) is good (which happens with very high probability by Claim 3.5) we proceed by performing a (B t , M ′ (π t ), s t )-reshuffling where s t = ⌊ct + ηn⌋ (recall Definition 2.4). To see that the parameters in Definition 2.4 fit, we must show that s t ≤ m t = |M ′ (π t )|. Indeed this holds since Now, if this (B t , M ′ (π t ), s t )-reshuffling is successful (recall again Definition 2.4), then B t has a perfect matching, and we define F t+1 using this perfect matching as shown in the paragraph after Definition 3.2. Recall that F t+1 extends F t and that if F t is a t-partial factor of independent transversals, then F t+1 is also a (t + 1)-partial factor of independent transversals. We can now define the notion of a successful stage. Definition 3.6 (Successful stage). We say that Stage t is successful if M (π t ) is good and furthermore, the corresponding (B t , M ′ (π t ), s t )-reshuffling is successful.
As an immediate corollary we obtain: Corollary 3.7. If all stages are successful then F k is a factor of independent transversals of G. Now, Corollary 3.16 below asserts that indeed, the probability that all stages are successful is positive, so given Corollary 3.16, we have that F k is a factor of independent transversals of G. Now, if we choose c = 0.778 then k ≥ 0.5624n(1 − ǫ), so Lemma 3.4 holds.
It remains to prove Corollary 3.16 below, namely that with positive probability, all stages are successful. To help with the analysis, it would be convenient (i.e. more uniform) to also say how we construct F t+1 as an extension of F t in case that Stage t is not successful, either because M (π t ) is not good or because the (B t , M ′ (π t ), s t )-reshuffling failed. While this may seem artificial (since if some stage failed, why proceed to the next stage?) the analysis becomes more uniform as it will help us lower-bound success probability of stages regardless of successes or failures of previous stages. So, if Stage t is unsuccessful, we simply use M (π t ) (which might certainly not be a perfect matching of B t ) to define F t+1 . In other words, we define I(t + 1, i) = I(t, i) ∪ {y πt(i) } for i = 1, . . . , n and Observe that even if F t is a factor of independent transversals, F t+1 might not be, since if (I(t + 1, i), y πt(i) ) is not an edge of B t , then I(t + 1, i) is not an independent set of G t+1 . Nevertheless, F t+1 is still a factor of transversals that extends F t . We have now defined how to perform all stages 1 ≤ t ≤ k − 1 (some of which may be successful, while others might not be).
Corollary 3.8. With probability at least 1 2 e −C we have that M (π t ) is good for all stages 1 ≤ t ≤ k − 1.
Proof. By Claim 3.5, for each t it holds that M (π t ) is good with probability 1 − C/n and recall that this probability bound holds for stage t regardless of the goodness of stages other than t. Hence, the probability that M (π t ) is good for all 1 ≤ t ≤ k − 1 is at least ( Given that M (π t ) is good, we would like to ascertain some property whose existence guarantees that the corresponding (B t , M ′ (π t ), s t )-reshuffling is successful. One way to do that is to prove that the leftover graph B t * of the (B t , M ′ (π t ), s t )-reshuffling satisfies the conditions of Lemma 2.2 (so B * t has a perfect matching implying that the (B t , M ′ (π t ), s t )-reshuffling is successful, hence the entire Stage t is successful). We start by showing that if t is very small (namely, in the first few stages) then, given that M (π t ) is good, surely (i.e. with probability 1) the corresponding (B t , M ′ (π t ), s t )-reshuffling is successful.
is good, then the corresponding (B t , M ′ (π t ), s t )reshuffling is successful, namely Stage t is successful.
Proof. We need to prove that the leftover graph B * t has a perfect matching. Recall (Definition 2.4) that B * t has n − m t + s t vertices in each part. On the other hand, B * t is an induced subgraph of B t and by Lemma 3.3 the latter has minimum degree at least n − t. Hence the minimum degree of B * t is at least n − m t + s t − t. Since by Hall's Theorem, every bipartite graph with x vertices in each side and minimum degree at least x/2 has a perfect matching, it suffices to prove that n − m t + s t − t ≥ (n − m t + s t )/2. Equivalently, it suffices to prove that t ≤ (n − m t + s t )/2. Indeed, In contrast with Lemma 3.9, proving that (with high probability) the reshuffling corresponding to larger t is successful is more involved. We start with the following lemma that gives a sufficient condition for a reshuffling to be successful when t ≥ ηn/3. Lemma 3.10. Let t ≥ ηn/3 and suppose that M (π t ) is good. If each W ⊂ V t+1 with |W | = ⌊ct⌋ has |N Bt (W )| ≥ n − ⌊ηn/7⌋ then the (B t , M ′ (π t ), s t )-reshuffling is successful.
Proof. As in the proof of Lemma 3.9 we observe that B * t has n − m t + s t vertices in each part and its minimum degree is at least To prove the latter, it suffices to prove the even stronger statement that |N Bt (W )| ≥ t + m t − s t . Now, as M (π t ) is good we have that m t ≤ n − t + δn and recall that s t = ⌊ct + ηn⌋ ≥ ct. Thus, we have: where we have used that c ≥ 1 2 and δ ≪ η. We also have that Hence, if W ⊂ V t+1 with |W | = ⌊ct⌋ has |N Bt (W )| ≥ n − ⌊ηn/7⌋ then B * t has a perfect matching, implying that the (B t , M ′ (π t ), s t )-reshuffling is successful.
Proving that the conditions of Lemma 3.10 hold with high probability, namely that with high probability each W ⊂ V t+1 with |W | = ⌊ct⌋ has |N Bt (W )| ≥ n − ⌊ηn/7⌋, is rather technical and requires a few additional definitions, notations, and lemmas.

For Q ⊆ [n] and for a t-partial factor of transversals
Observe that |Y ℓ (W )| ≥ n − |W | holds for all 1 ≤ ℓ ≤ t + 1 since each vertex of W has at most one neighbor in G that belongs to V ℓ .
We need the notions of F t (Q) and Y ℓ (W ) from the previous paragraph in the following definition, which is an important ingredient in the remainder of the proof. (ii) there are at least r edges of M ′ (π ℓ ) where in each of these r edges, one endpoint is from F ℓ (Q) and the other endpoint is from Y ℓ+1 (W ), (iii) the 2r endpoints of these r edges from (ii) are not selected to the leftover graph B * ℓ .
Until the end of this section, let with |Q| = q. Then with probability at least 1 − 5 −n the following holds. There exists T * ⊆ T with |T * | ≥ t − δn such that for all ℓ ∈ T * , a (Q, ℓ, W, r ℓ,t )-intersection exists.
Proof. Suppose that t, W, Q are given as in the statement of the lemma. Fixing some ℓ ∈ T we estimate the probability that a (Q, ℓ, W, r ℓ,t )-intersection exists. We recap what we are doing in Stage ℓ. At the beginning of that stage, we have F ℓ , the already constructed ℓ-partial factor of transversals (note: we do not assume anything about success or failure of prior stages) as one part of B ℓ and we have V ℓ+1 as the other part. We remove some edges of B ℓ to obtain B ′ ℓ . We then take a random permutation π ℓ of V ℓ+1 yielding M (π ℓ ) and its subset M ′ (π ℓ ) which is a matching of B ′ ℓ . Observe that F ℓ (Q) is a subset of F ℓ of order q, and that Y ℓ+1 (W ) is a subset of V ℓ+1 of order at . We obtain from that lemma that with probability 1 − C/n, both items (i) and (ii) of Lemma 2.3 hold. Recall that item (i) of Lemma 2.3 says that M (π ℓ ) is good, since |m ℓ −(n−ℓ)| ≤ δn where m ℓ = |M ′ (π ℓ )|, so already item (i) of Definition 3.11 holds. Item (ii) of Lemma 2.3 says that there is a subset M * ⊂ M (π ℓ ) of size such that every pair (x i , y π(i) ) ∈ M * is and edge of B ′ t (hence an edge of M ′ (π ℓ )) and furthermore, x i ∈ F ℓ (Q) and y π ℓ (i) ∈ Y ℓ+1 (W ).
Notice that if r ℓ,t ≥ |M * | then Item (ii) of Definition 3.11 holds, but this is not enough for us, as we also want Item (iii) to hold. In the reshuffling, we create B * ℓ by randomly selecting s ℓ out of the m ℓ edges of M ′ (π ℓ ) and add the 2s ℓ endpoints of the selected edges to the vertices not matched by M ′ (π ℓ ). So, the probability that a single edge (and its two endpoints) from M * is selected to B * ℓ is precisely s ℓ /m ℓ . Hence, if Z ⊆ M * denotes the set of edges of M * that are not selected to the leftover graph B * ℓ then E[|Z|] = |M * |(1 − s ℓ m ℓ ). Observe that |Z| has hypergeometric distribution (as its distribution is identical to having |M * | red balls, m t − |M * | blue balls, and we select without replacement m ℓ − s ℓ balls and ask for the number of red balls that are selected). As trivially, |Z| ≤ n, we have by Chebyshev's inequality, that the probability that |Z| deviates from its expected value by more than δn is O(1/n) (since the variance in our hypergeometric distribution is O(n)). So overall, we obtain that with probability 1 − O(1/n), To prove that with probability 1 − O(1/n), a (Q, ℓ, W, r ℓ,t )-intersection exists we just need to make sure that r ℓ,t is not larger than the right-hand side of the last inequality. Indeed, using (1), using Lemma 3.13 below which upper bounds s ℓ /m ℓ and using δ ≪ η we obtain We have proved that with probability 1 − O(1/n), a (Q, ℓ, W, r ℓ,t )-intersection exists but we still need to prove that the claimed T * exists with high probability. First observe that if t ≤ δn then T * exists vacuously so assume that t ≥ δn. For T * not to exist, there should be a set of ⌈δn⌉ indices ℓ with 1 ≤ ℓ ≤ t, such that a (Q, ℓ, W, r ℓ,t )-intersection does not exist. Fix a set L ⊆ T of ⌈δn⌉ indices. We have proved that the probability that for ℓ ∈ L, a (Q, ℓ, W, r ℓ,t )-intersection does not exist is O(1/n) regardless of the outcome of any other ℓ ′ ∈ L. Hence, the probability that all ℓ ∈ L are such that the corresponding (Q, ℓ, W, r ℓ,t )-intersection does not exist is O(1/n) δn . As there are less than 2 n possible sets L to consider, we have by the union bound that the probability that T * does not exist is at most We next prove the upper bound for s ℓ m ℓ that we have used in Lemma 3.12.
Proof. Recall that s ℓ = ⌊cℓ + ηn⌋ and that M (π ℓ ) being good implies that m ℓ ≥ n − ℓ − δn. Also notice that ℓ < k < n/(1 + c) ≤ 2n/3 and that δ ≪ η. Hence, Lemma 3.15 below proves that if all stages satisfy a certain condition, then all stages are successful. The next lemma describes this condition and proves that it holds with positive probability.
Lemma 3.14. Let A denote the intersection of the two following events: (i) M (π t ) is good for all stages 1 ≤ t ≤ k − 1.
Then A holds with positive probability.
Proof. By Corollary 3.8, (i) holds with probability at least 1 2 e −C . As for (ii), observe that the number of choices for Q is less than 2 n and that for a given t, the number of possible choices for W is less than 2 n . Hence, the number of possible triples (t, Q, W ) is less than n4 n . For each such triple, the probability that there exists T * as specified is at least 1 − 5 −n by Lemma 3.12. So, the probability that (ii) is satisfied is at least 1 − n4 n 5 −n > 1 − 1/n. Hence the event A holds with probability at least 1 2 e −C − 1/n > 0. Proof. Assume by contradiction that A holds, yet there is an unsuccessful stage t. First observe that we must have t ≥ ηn/3, since by Lemma 3.9, for smaller t, Stage t is successful as M (π t ) is good (since we assume that A holds).
As M (π t ) is good, the only reason that Stage t is unsuccessful is because the (B t , M ′ (π t ), s t )reshuffling is not successful. Hence, by Lemma 3.10, there exists W ⊂ V t+1 with |W | = ⌊ct⌋ such that |N Bt (W )| < n − ⌊ηn/7⌋. This means that there are at least q = ⌊ηn/7⌋ elements of F t that are non-neighbors in B t of all elements of W . Let, therefore, Q ⊂ [n] with |Q| = q be such that F t (Q) = {I(t, i) | i ∈ Q} ⊂ F t and for each pair (I(t, i), y) where I(t, i) ∈ F t (Q) and y ∈ W , I(t, i) and y are non-adjacent in B t . Being non-adjacent in B t means that for each y ∈ W and for each I(t, i) ∈ F t (Q), there is at least one vertex x ∈ I(t, i) such that xy is an edge of G. Let P = ∪ i∈Q I(t, i) and let P * ⊆ P be such that x ∈ P * if and only if there exists y ∈ W such that xy is an edge of G. So we have that |P | = qt and since every y ∈ W has at least q neighbors in G that belong to P * , and these neighbors are unique for y since G is an [n, k, 1]-graph (every vertex in P * has exactly one neighbor in W ). We will now count the number of elements of P * in another way, and obtain that it is less than q(ct − 1), hence reaching a contradiction with (2). For every 1 ≤ ℓ ≤ t, let P ℓ = P ∩ V ℓ and P * ℓ = P * ∩ V ℓ . Clearly |P ℓ | = q so |P * ℓ | ≤ q, but we can obtain a better upper bound for |P * ℓ | for most values of ℓ. In order to show this, recall that as we assume that A holds, there exist T * ⊆ {1, . . . , t} with |T * | ≥ t − δn such that for all ℓ ∈ T * , a (Q, ℓ, W, r ℓ,t )-intersection exists.
Consider first ℓ ∈ T \ T * . For such ℓ we will use the trivial bound |P * ℓ+1 | ≤ q but since there are at most δn such ℓ, we have that Consider next ℓ ∈ T * in which case a (Q, ℓ, W, r ℓ,t )-intersection exists. First observe that by the definition of P ℓ+1 , when we perform Stage ℓ extending F ℓ to F ℓ+1 , the elements of F ℓ (Q) are matched with the vertices of P ℓ+1 to obtain F ℓ+1 (Q Altogether, we have Observe that in the last inequality we are summing over all 1 ≤ ℓ ≤ t − 1 although it would have been enough to sum over the elements of T * . To obtain the desired contradiction to (2) it therefore suffices to prove that Dividing both sides by q, our remaining task is to prove Rearranging the terms, it suffices to show that Recall that t ≥ ηn/3 and that δ ≪ η so ct − 2 − δn − 5ηt > t(c − 6η). Now, define µ = t/n and define x = ℓ/n. So it suffices to prove that But now µ = t/n ≤ k/n ≤ 1−ǫ 1+c and η ≤ γ/6 so the last inequality indeed holds by Lemma 2.5.
From the last two lemmas we immediately obtain: Here we provide a computer-assisted proof that f (4) = 4. Recall from the introduction that f (4) ≥ 4 and recall from the first paragraph in Section 2 that to prove f (4) ≤ 4 it suffices to prove that every [4, 4, 1]-graph has a factor of independent transversals. Consider the set of [4, 4, 1]-graphs with vertex parts V 1 , V 2 , V 3 , V 4 and V i = {x i,1 , x i,2 , x i,3 , x i,4 }.
As adding edges only makes the problem more difficult, we may assume that each of the 6 pairs (V i , V j ) with 1 ≤ i < j ≤ 4 induces a perfect matching M i,j . By relabeling vertices, we may assume that the perfect matching (V 1 , V j ) for j = 2, 3, 4 is Checking each constructed graph for a factor of independent transversals proceeds as follows. Take three permutations π 2 , π 3 , π 4 of V 2 , V 3 , V 4 respectively, and check if the following factor {{x 1,j , x 2,π 2 (j) , x 3,π 3 (j) , x 4,π 4 (j) } | j = 1, 2, 3, 4} is a factor of independent transversals. Note that there are only (4!) 3 = 13824 choices for π 2 , π 3 , π 4 , so there are only 13824 checks to perform for each of the 13824 constructed graphs. It turns out that each of these graphs has a factor of independent transversals. A simple program implementing this search and verifying its conclusion appears in Appendix A.