An Asymptotically Sharp Bound on the Maximum Number of Independent Transversals

Let $G$ be a multipartite graph with partition $V_1, V_2,\ldots, V_k$ of $V(G)$. Let $d_{i,j}$ denote the edge density of the pair $(V_i, V_j)$. An independent transversal is an independent set of $G$ with exactly one vertex in each $V_i$. In this paper, we prove an asymptotically sharp upper bound on the maximum number of independent transversals given the $d_{i,j}$'s.


Introduction
Let G be a multipartite graph with vertex partition V 1 , V 2 , . . ., V k .Jacob Fox asked the following question: given the edge density between every two vertex parts, what is the asymptotically maximum number of independent transversals in G as |V i | goes to infinity for each i?An independent transversal is an independent set of G with exactly one vertex in each V i .More precisely, for each i = j, let constant d i,j be the edge density between V i , V j , defined as e(V i , V j )/|V i ||V j |.Independent transversals arise naturally in extremal combinatorics and bounding its number appears, for example, in inducibility type problems [2,3].The previous known bound to this question is the following result of Fox, Huang, and Lee, which is an ingredient in [3] to prove a bound on the number of induced copies of a given subgraph in another graph.
Theorem 1.1 (Lemma 4.1 [3]).Let k ≥ 2 be an integer.For each integer pair 1 ≤ i < j ≤ k, let d i,j = d j,i be constants in [0, 1].Let G be a multipartite graph with vertex partition V 1 , V 2 , . . ., V k such that for each pair 1 ≤ i < j ≤ k, the edge density between V i , V j is d i,j .Let |V i | = n i .Then the number of independent transversals in G is at most However, this bound is not sharp in general, or even not asymptotically sharp.This means that there are choices of constants d i,j 's such that the number of independent transversals divided by k i=1 |V i | is strictly less than 1≤i<j≤k (1−d i,j ) ⌊k/2⌋/( k 2 ) as k i=1 |V i | goes to infinity.
In this paper, we prove an asymptotically sharp bound, previously asked by Jacob Fox [1].Before stating our main theorem, we need the following definition.Definition 1.1.An odd cycle decomposition H of the complete graph on k vertices K k is a collection of disjoint multigraphs F 1 , F 2 , . . ., F ℓ satisfying i∈ℓ V (F i ) = V (K k ) such that for all i ∈ [ℓ], F i is an odd cycle, a double edge, or an isolated vertex.A double edge is obtained by adding an additional edge between the ends of an isolated edge.
Note that our definition of odd cycle decomposition may be different from other uses in the literature.See Figure 1 for an example.As a matter of notation, we denote an edge between vertices v i , v j by ij.
Our main result is the following: Theorem 1.2.Let k ≥ 2 be an integer.For each integer pair 1 ≤ i < j ≤ k, let d i,j = d j,i be constants in [0, 1].Let G be a multipartite graph with vertex partition V 1 , V 2 , . . ., V k such that for each pair 1 ≤ i < j ≤ k, the edge density between V i , V j is d i,j .Let Then the number of independent transversals in G is at most where H is the set of all odd cycle decompositions of K k .Furthermore, this bound is asymptotically sharp.
Figure 1: An odd cycle decomposition of K 8 .
By asymptotically sharp we mean that the bound is sharp up to a o(n 1 • • • n k )-term, which refers to a function f (n 1 , . . ., n k ) with the property that lim n 1 ,n 2 ,...,n k →∞ f (n 1 , . . ., n k )/(n 1 • • • n k ) = 0.In particular, given densities d i,j for i = j ∈ [k], we can construct a k-partite graph G that attains the bound in Theorem 1.2 up to a o(n Observe that Theorem 1.2 implies Theorem 1.1 as follows: Let G be a multipartite graph as in the hypotheses of the two theorems.Let {a t b t } ⌊k/2⌋ t=1 with a t , b t ∈ [k] be a set of ⌊k/2⌋ edges corresponding to a matching on K k .This corresponds to an odd cycle decomposition of K k where each subgraph F i is a double edge between a t and b t .Since each edge ij ∈ [k]  2 is in ⌊k/2⌋/ k 2 fraction of all possible matchings as above, the bound in Theorem 1.1 is equal to the geometric mean over all products i=1 n i given by all matchings.This geometric mean is at least the minimum over all possible products, which is at least the bound in Theorem 1.2 on the maximum number of independent transversals in G.

Proof of Theorem 1.2
The proof of Theorem 1.2 comprises of two parts.In Section 2.1, we show that the result is an upper bound.In Section 2.2, we show that the result is asymptotically sharp.

Proof of Upper Bound
The following lemma is a special case of Theorem 1.2 (except that we consider transversal cliques instead of independent transversals).We use this case in the proof of Lemma 2.8, which is key to proving Theorem 1.2.
Lemma 2.1.For each integer pair (1) Then we must prove that the number of transversal cliques in G is at most d 1,2 n 1 n 2 n 3 .However, there are at most d 1,2 n 1 n 2 choices for an edge between V 1 and V 2 , so clearly the number of transversal cliques is upper-bounded by d 1,2 n 1 n 2 n 3 .Thus we may assume subject to Here we still have the assumption that d 1,2 ≤ d 1,3 .
Let O be the set of all optimal solutions to our problem.The set O is nonempty because the objective function ( 2) is continuous on the compact set defined by the constraints (3).We prove a series of claims, using local adjustments, to show that there must be optimal solutions in O that satisfy certain nice conditions.We then use these conditions to get the desired upper bound on the number of transversal cliques in G.

Claim 2.2. There is a nonempty subset
Proof.Suppose that there is O ∈ O where for some i ∈ . This new set of variables still satisfies the constraints (3) while the objective function (2) has the same value.
To obtain a contradiction, we have the following two cases (note that Claim 2.2 implies a i b i ≤ d 2,3 for all i).
First assume that there exists an i By increasing b i by ε, we satisfy the constraints (3) and strictly increase the objective function (2), contradicting the optimality of O.This implies that for all i ∈ [n 1 ] with a i b i < d 2,3 we have a i ≤ b i .Now assume that there exists i ∈ [n 1 ] such that a i b i = d 2,3 and a i > b i .We prove that there exists j ∈ [n 1 ] with a j b j < d 2,3 .To this end, assume for contradiction that for all j ∈ [n 1 ], we have a j b j , which implies d 1,2 > d 2,3 , a contradiction.Therefore we can fix j ∈ [n 1 ] with a j b j < d 2,3 .By the preceding paragraph, we must have a Hence, we can increase the objective function while satisfying the constraints, once again contradicting the optimality of O. Thus for all i ∈ [n 1 ] such that a i b i = d 2,3 , we have Proof.Assume for contradiction that there exists i Then by Claim 2.3, there must exist j ∈ , we can choose such an ε to obtain a solution to the objective function whose value is greater than that of O as long as the constraints are still satisfied.Thus, by choosing ε ∈ (0, min( )), we increase the objective function while satisfying the constraints.This contradicts the optimality of O.
, we have a i = a j and b i = b j .
Proof.Assume for contradiction that there exists i, j ∈ [n 1 ] with a i b i = a j b j = d 2,3 but a i = a j .By symmetry, say a i > a j .Then b i < b j .Choose sufficiently small ε, ε ′ > 0 and let , and in this way the objective function value does not change although a i + b i strictly decreases.This contradicts the definition of O 2 .
For each Proof.Suppose that there exists i ∈ C O with b i = 1.Then we must prove that the objective function is at most Assume for contradiction that there exists j with a j b j > 0 and b j < 1.Then we can decrease a j and increase a i by some sufficiently small ε > 0 to contradict the optimality of O. Thus for all j with a j b j > 0, we must have b , and so To finish the proof, fix By the constraints, the quantity Using this, we can write 3 .This implies that the objective function is at most n 1 d 1,2 d 1,3 d 2,3 , and so the number of transversal cliques in G is at most By Claim 2.7, the only other case that we need to consider is where the second inequality follows from the AM-GM inequality (x + y ≥ 2 √ xy for any x, y ≥ 0).In the last step above, note that d 1,2 d 1,3 n 2 1 − a j b j ≥ 0 because a j ≤ d 1,2 n 1 and b j ≤ d 1,3 n 1 by the constraints.Using the fact that 0 < a j b j < d 2,3 , we can bound the objective function by

This implies that the number of transversal cliques in
We use Lemma 2.1 in the below proof of Lemma 2.8, which is key to the proof of the upper bound in Theorem 1.2.
Lemma 2.8.Let k ≥ 3 be an integer.For each integer pair where the index i + 1 is modulo k.
Proof.We show by induction on k that the statement holds for all k ≥ 3.By Lemma 2.1, the statement holds for k = 3. Assume that the statement holds for k We bound the number of transversal cliques in G containing v i (i.e., the number of transversal cliques in ), and then sum this quantity over all v i ∈ V 1 to get the bound in the lemma.
If a i 1,2 = 0 or a i 1,k = 0, then there are zero transversal cliques in G containing v i , so we need only consider i with a i 1,2 , a i 1,k > 0. Observe that the edge density between , and the edge density between To bound the number of transversal cliques in G, we sum the above quantity over all v i ∈ V 1 .This amounts to bounding In turn, this amounts to bounding Applying the Cauchy-Schwarz inequality, we have Hence, the number of transversal cliques in G is at most as desired.
We now use Lemma 2.8 to prove the below proposition, which is the upper bound in Theorem 1.2.Although we may minimize over all (not necessarily odd) cycle decompositions in the below bound, it turns out that it suffices to minimize over just the odd ones to get that the bound is asymptotically sharp (see Section 2.2).Proposition 2.9.Let k ≥ 2 be an integer.For each integer pair 1 ≤ i < j ≤ k, let d i,j = d j,i be constants in [0, 1].Let G be a multipartite graph with vertex partition V 1 , V 2 , . . ., V k such that for each pair 1 ≤ i < j ≤ k, the edge density between V i , V j is d i,j .Let |V i | = n i .Then the number of independent transversals in G is at most where H is the set of all odd cycle decompositions of K k .
Proof.Since the number of independent transversals in G is equal to the number of transversal cliques in G, we obtain an upper bound for the latter.Note that each edge density d i,j in G corresponds to the edge density 1 By definition of odd cycle decomposition (Definition 1.1), the sets Observe that S 1 , . . ., S l correspond bijectively to V (F 1 ), . . ., V (F l ) and partitions V 1 , V 2 , . . ., V k .We now obtain a bound for each possible G [S i ], that is, each possible subgraph of G induced by S i .If S r = {V 1 , . . ., V m } corresponds to an odd cycle, then the number of transversal cliques in corresponds to an isolated vertex, then the number of transversal cliques in G[S r ] is trivially n 1 .Putting the three cases together, the number of transversal cliques in G is at most 1 − d i,j .Minimizing this over H, we obtain the desired upper bound.
The following section is dedicated to showing that the above upper bound is asymptotically sharp.

Proof of Asymptotic Sharpness
Assume the hypotheses of Theorem 1.2.To show that the upper bound in Theorem 1.2 is asymptotically sharp, we use linear programming and duality to prove the existence of a k-partite graph G that attains the bound up to a o(n 1 • • • n k )-term.Since the number of independent transversals in G equals the number of transversal cliques in G, it suffices to consider the latter.Note that each edge density d i,j in G corresponds to the edge density 1 − d i,j in G. Consider the following construction of G which satisfies our constraints: where a i a j ≤ 1 − d i,j (for all 1 ≤ i < j ≤ k), and a i ∈ (0, 1] (for all i).Suppose that each pair S i , S j (i = j) forms a complete bipartite graph.
In this construction, the number of transversal cliques is at least k i=1 ⌊a i n i ⌋.We will show that some construction described above is close to the maximum number of transversal cliques.This motivates the following optimization problem: If we have an optimal solution (a 1 , . . ., a k ), then k i=1 ⌊a i n i ⌋ is a lower bound on the number of transversal cliques in the above construction in G.
Thus the construction will attain the bound in Theorem 1.2 up to a o(n 1 • • • n k )-term if we can prove that k i=1 a i n i is given by a product over some odd cycle decomposition of K k (see Equation ( 4) for more specificity).
In problem P, note that a i > 0 for all i implies d i,j < 1 for all i, j (if d i,j = 1 is allowed, then the number of transversal cliques in G is zero).Applying the natural log to P, we get an equivalent linear programming problem: Ignoring the constant k i=1 ln(n i ) and setting b i = − ln(a i ), p i,j = − ln(1 − d i,j ), we can rewrite LP1 as an equivalent problem more convenient to work with: Introducing the slack variable y i , LP2-dual becomes We need the following results from linear programming.The first is known as the duality of linear programming.the following holds: If both (P) and (D) have a feasible solution, then both have an optimal solution, and if x * is an optimal solution of (P) and y * is an optimal solution of (D), then That is, the maximum of (P) equals the minimum of (D).
We also need a corollary to the above duality theorem called complementary slackness.Then the following two statements are equivalent: 1. x * is optimal for (P) and y * is optimal for (D).

For all i ∈ [m],
x * satisfies the ith constraint of (P) with equality or y * i = 0; similarly, for all j ∈ [n], y * satisfies the jth constraint of (D) with equality or x * j = 0.
Since LP2-dual ′ and LP2 are both feasible, both have an optimal solution by Theorem 2.10.We will show that there is an optimal solution to LP2-dual ′ which is given by an odd cycle decomposition of K k .Then Theorem 2.11 will show that optimal solutions to LP2 are given by that odd cycle decomposition of K k , which will imply that k i=1 a i n i is given by a product over that odd cycle decomposition of K k , as desired.
More specifically, we define the graph Q on k vertices x 1 , . . ., x k as follows (recall that LP2-dual ′ has variables x i,j ): 1. there exists an edge x i x j ∈ Q if and only if x i,j > 0, 2. there exists a self-loop on vertex x i if and only if y i > 0.
Thus Q can be used to represent elements of the feasible set of LP2-dual ′ (non-edges and non-self-loops correspond to variables being 0).Consider the set of Q's that represent optimal solutions to LP2-dual ′ (this set is nonempty by Theorem 2.10).From this set, consider the Q's with the minimum number of non-loop edges, and then among which choose the ones with the minimum number of self-loops.Call the resulting set Q. Asymptotic sharpness of the bound in Theorem 1.2 is proved in the following proposition.
Proposition 2.12.Let k ≥ 2 be an integer.For each integer pair 1 ≤ i < j ≤ k, let d i,j = d j,i be constants in [0, 1].For any sufficiently large integers n 1 , . . ., n k , there exists a multipartite graph G with vertex partition V 1 , . . ., V k where such that the number of independent transversals in G is at least where H is the set of all odd cycle decompositions of K k , and the edge density between V i , V j is d i,j for each pair 1 ≤ i < j ≤ k.Hence, the bound in Theorem 1.2 is asymptotically sharp.
Proof.Recall that we must prove that k i=1 a i n i is given by a product over some odd cycle decomposition of K k .We show that Claim 2.13 below gives the required odd cycle decomposition.We then prove Claim 2.13 to complete the proof of the proposition.Claim 2.13.For each Q ∈ Q, define a graph Q ′ constructed by adding an additional edge between the ends of isolated edges in Q and removing self-loops from isolated vertices in Q.Then there is a Q ′ constructed from some Q ∈ Q which is an odd cycle decomposition of K k .
Suppose that Claim 2.13 holds.Fix such a Q and Q ′ .Recall that Q represents an optimal solution to LP2-dual ′ .We are done if we can prove Let (b 1 , . . ., b k ) be an optimal solution to LP2.Then Theorem 2.11 (Complementary Slackness) and the construction of Q imply that each edge x i x j in Q corresponds to an equality in LP2's constraints, that is, b i + b j = p i,j .Similarly, each isolated vertex x i of Q corresponds to b i = 0. Converting (b 1 , . . ., b k ) back to an optimal solution (a 1 , . . ., a k ) of P, we get that for each edge x i x j in Q, we have an equality in P's constraints, that is, a i a j = 1 − d i,j ; similarly, for each isolated vertex x i of Q, we have the equation a i = 1.Thus, since Q ′ is an odd cycle decomposition of K k , the objective function of P, a i n i , has the above form, as desired.
To prove Claim 2.13, we first need a series of claims.
Claim 2.14.For any Q ∈ Q, there is at most one vertex with a self-loop in Q.
Proof.Assume for contradiction that there exists Q ∈ Q with at least two vertices with a self-loop.Then by the construction of Q, there exist i < j with y i > 0 and y j > 0.
To contradict the optimality of Q, set ε = min(y i , y j ).Let y ′ i = y i − ε, y ′ j = y j − ε, x ′ i,j = x i,j + ε, and all other primed variables have the same value as the corresponding non-primed variable.Then the objective function of LP2-dual ′ , p i,j x i,j , increases while the constraints are still satisfied.However, the number of self-loops has decreased, contradicting the minimality of self-loops in Q.
Claim 2.15.For any Q ∈ Q, there is no even cycle in Q.
Proof.Assume for contradiction that there exists Q ∈ Q containing an even cycle.Say be the part of the objective function of LP2-dual ′ , p i,j x i,j , involving this even cycle.Without loss of generality, assume Notice that the constraints are still satisfied.If this inequality is strict, then the optimality of Q is contradicted.If there is equality, then, by our choice of ε, at least one of x 2,3 − ε, x 4,5 − ε, . . ., x 1,m − ε is 0. Thus we have obtained an optimal solution by removing an edge from Q, contradicting the minimality of non-loop edges in Q.Now our goal is to show that every connected component of Q ∈ Q with at least three vertices is an induced odd cycle.In the following claim, the degree of a vertex includes the possibility of self-loops.Claim 2.16.For any Q ∈ Q, if C is a connected component of Q with at least three vertices, then C contains an odd cycle.
Proof.Suppose that C is a connected component of Q ∈ Q with at least three vertices.We first prove that C has no vertex of degree 1 in Q. Assume, to the contrary, that C has a vertex x i of degree 1 in Q.Let x k be the neighbor of x i .Then since C is a connected component on at least three vertices, it follows that x k ∈ C and x k = x i .Thus x i does not have a self-loop, and so, by the construction of Q, we have y i = 0. Thus the ith constraint y i + i =j x i,j = 1 in LP2-dual ′ becomes x i,k = 1, and so the kth constraint in LP2-dual ′ implies that x k has degree 1 in Q.Hence, C must be a connected component on only the two vertices x i , x k , a contradiction since C has at least three vertices.
We now prove the claim.By Claim 2.15, it suffices to prove that C contains a cycle.Assume, to the contrary, that C is acyclic.Then since C is connected, acyclic, and |C| ≥ 2, there are at least two vertices x i 1 = x j 1 in C of non-loop degree 1 in C. By the preceding paragraph, x i 1 must have another neighbor x i 2 , and x j 1 must have another neighbor x j 2 .Since C is a connected component, these two neighbors must lie in C, and so we must have x i 2 = x i 1 and x j 2 = x j 1 .In other words, x i 1 , x j 1 have self-loops, contradicting Claim 2.14.Claim 2.17.For any Q ∈ Q, every odd cycle in Q is an induced cycle.
Proof.Let C be an odd cycle in Q ∈ Q. Assume for contradiction that C is not induced.Then there exists a chord in E(C).This chord splits C into two cycles C ′ and C ′′ .One of them is an even cycle, contradicting Claim 2.15.
In the remaining claims, we use the following notation.Define an edge weight function w : V (Q) 2 → R such that for each distinct x i , x j ∈ V (Q), we have w(x i , x j ) = w(x j , x i ) = x i,j and w(x i , x i ) = y i .For convenience, we write w(e) for e ∈ E(Q).
Claim 2.18.For any Q ∈ Q, the odd cycles in Q are vertex disjoint.
Proof.It is easy to see that if there are two distinct odd cycles in Q sharing an edge or at least two vertices, then Q contains an even cycle, contradicting Claim 2.15. Let e ∈ E(C 1 ) ∪ E(C 2 )}.We show that by adjusting the values of the w(e)'s, we can either construct an optimal solution with fewer edges or increase the value of the objective function.In both cases, we reach a contradiction.We define a new edge weight function w ′ as follows (where the indices of u i+1 , v i+1 below are considered modulo 2s + 1, 2s ′ + 1 respectively): w(e) otherwise.
By choice of ε, we know that w ′ (e) ≥ 0 for all e ∈ E(Q).Furthermore, for each fixed x r ∈ V (Q), it is easy to see that r =i w ′ (x r , x i ) = r =i w(x r , x i ).This shows that the constraints of LP2-dual ′ are satisfied.Let ∆ = i<j p i,j (w ′ (x i , x j ) − w(x i , x j )).If ∆ > 0, then the optimality of Q is contradicted.If ∆ < 0, then add 1 to the powers of −1 in the definition of w ′ to contradict the optimality of Q. Assume ∆ = 0.By choice of ε, either there exists a w ′ (e) = 0, or there exists such a zero edge weight after adding 1 to the powers of −1 in the definition of w ′ .Since ∆ = 0, we get an optimal solution with fewer non-loop edges, contradicting the minimality of non-loop edges in Q.
Claim 2.19.For any Q ∈ Q, each connected component of Q contains at most one odd cycle.
Proof.Assume, for the sake of contradiction, that C is a connected component of Q ∈ Q that contains distinct odd cycles C 1 and C 2 .By Claim 2.18, these two cycles are vertex disjoint.Let P be a shortest ( w(e) + (−1) i+1 2ε if e = w i w i+1 ∈ E(P ), w(e) otherwise.
We chose a shortest path P to ensure that P does not share edges with C 1 or C 2 .This makes w ′ well-defined.By choice of ε, we know that w ′ (e) ≥ 0 for all e ∈ E(Q).Furthermore, for each fixed x r ∈ V (Q), it is easy to see that r =i w ′ (x r , x i ) = r =i w(x r , x i ).This shows that the constraints of LP2-dual ′ are satisfied.Let ∆ = i<j p i,j (w ′ (x i , x j ) − w(x i , x j )).If ∆ > 0, then the optimality of Q is contradicted.If ∆ < 0, then add 1 to the powers of −1 in the definition of w ′ to contradict the optimality of Q. Assume ∆ = 0.By choice of ε, either there exists a w ′ (e) = 0, or there exists such a zero edge weight after adding 1 to the powers of −1 in the definition of w ′ .Since ∆ = 0, we get an optimal solution with fewer non-loop edges, contradicting the minimality of non-loop edges in Q.
In the following claim, a pendant path P of an odd cycle C ′ is a path v 1 v 2 . . .v ℓ such that v 1 is the only vertex on P that lies on C ′ and v ℓ has degree 1 in non-loop edges.
Claim 2.20.For any Q ∈ Q, no connected component in Q contains an odd cycle with a pendant path as a subgraph.
Proof.Assume, for the sake of contradiction, that C is a connected component of Q ∈ Q containing an odd cycle C ′ with a pendant path P .Say and ε = δ ′ /2 otherwise.We define a new edge weight function w ′ as follows (where the index of u i+1 below is considered modulo 2s + 1): By choice of ε, we know that w ′ (e) ≥ 0 for all e ∈ E(Q).Furthermore, for each fixed Thus the constraints of LP2-dual ′ are satisfied.Let ∆ = i<j p i,j (w ′ (x i , x j ) − w(x i , x j )).If ∆ > 0, then the optimality of Q is contradicted.If ∆ < 0, then add 1 to the powers of −1 in the definition of w ′ to contradict the optimality of Q. Assume ∆ = 0.By choice of ε, either there exists a w ′ (e) = 0, or there exists such a zero edge weight after adding 1 to the powers of −1 in the definition of w ′ .Since ∆ = 0, we get an optimal solution with fewer edges, contradicting the minimality of edges in Q.
We now have almost all the pieces needed to prove Claim 2.13.Fix Q ∈ Q. Suppose that C is a connected component of Q.If C contains exactly one vertex, then C is an isolated vertex with a self-loop.If C contains exactly two vertices, then C is an isolated edge which, a priori, may have self-loops.Suppose now that C contains at least three vertices.By Claim 2.16, C contains an odd cycle.By Claims 2.17 and 2.19, this cycle in C must be induced and unique.By Claim 2.20, every vertex in C lies on this unique cycle.This implies that C is an odd cycle which may have self-loops.In fact, the following claim proves that isolated edges and connected components with at least three vertices must have no self-loops.

Claim 2.21. For any
Proof.Let Q ∈ Q, and let v ∈ V (Q) be a vertex with a self-loop.Assume, to the contrary, that v is not an isolated vertex.Then v must lie in a connected component C of Q on exactly two vertices or at least three vertices.In the first case, we can use the constraints to deduce that both vertices in C must have a self-loop.This contradicts Claim 2. Now the proof is similar to that of Claim 2.20.We can either increase the objective function or reduce the number of edges, a contradiction.
By Claim 2.21, only isolated vertices in Q have self-loops.Constructing Q ′ from Q, we get that Q ′ is an odd cycle decomposition of K k , thus proving Claim 2.13, as desired.This completes the proof of Proposition 2.12, finishing the proof of Theorem 1.2.
Concluding Remarks: Observe that the o(n 1 • • • n k )-term in the bound of Proposition 2.12 occurs due to divisibility issues.In particular, it is nonzero when an optimal solution (a 1 , . . ., a k ) of problem P is not rational.Here are two examples where the bound is sharp for infinitely many values of n 1 , . . ., n k .
1. Suppose that, in the hypotheses of Theorem 1.2, k ≥ 3 is an odd integer, and for all i = j ∈ [k], 1 − d i,j = a 2 /b 2 ∈ Q.That is, all the 1 − d i,j 's are the same rational perfect square.By Theorem 1.2, the number of independent transversals is at most , where the index i + 1 is modulo k.We now construct a graph G that attains this bound.Note that there are infinitely many integers n such that (a/b)n is an integer.Let n be such an integer, and for all i ∈ [k], let n i = n and S i ⊆ V i be a set of size (a/b)n.For each i = j ∈ [k], put a complete bipartite graph between V i \ S i , V j and between S i , V j \ S j with no other edges occurring between V i , V j .Then G satisfies 1 − d i,j = a 2 /b 2 for all i = j ∈ [k], and the number of independent transversals in G is k i=1 |S i | = (a/b) k n k , showing that G attains the desired bound.
2. Suppose the hypotheses of Theorem 1.2.Let I 1 , . . ., I ℓ be a a partition of [k] such that for all r ∈ [ℓ], |I r | is odd.For all r ∈ [ℓ] and i = j ∈ I r , let 1 − d i,j = a 2 r /b 2 r ∈ Q (if |I r | = 1, let a r /b r = 1); otherwise, if i ∈ I r , j ∈ I r , let 1 − d i,j = 1.We are essentially taking disjoint copies of example 1 above.By Theorem 1.2, the number of independent transversals is at most ℓ r=1 (a r /b r ) |Ir| k i=1 n i .We construct a graph G that attains this bound for infinitely many n 1 , . . ., n k by applying example 1 to each I r .
One could also study stability type of results.One could show that if d i,j = d for all i, j, then, by an argument similar to our proof, if the number of independent transversals in G is close to the optimal value, then G should be close to the construction in Section 2.2 where a i = √ 1 − d for each i.In the case where the d i,j are not all equal to each other, there could be very different optimal constructions.However, one should be able to show that each of those optimal constructions is close to a construction corresponding to some solution of P.

Proof.
Assume for contradiction that |C O | ≥ 2. Without loss of generality, assume indices 1, 2 ∈ C O .By definition of C O and Claim 2.4, we have 0 < a 1 , a 2 < 1.We have three cases.If b 1 < b 2 , then we can decrease a 1 and increase a 2 by some ε ∈ (0, min(a 1 , 1 − a 2 )] to increase the objective function, thereby contradicting the optimality of O. Similarly, if b 1 > b 2 , then we can decrease a 2 and increase a 1 by some ε ∈ (0, min(a 2 , 1 − a 1 )] to contradict the optimality of O.If b 1 = b 2 , then we can assume 0 < b 1 = b 2 < 1 by Claim 2.6 and the definition of C O .Now, without loss of generality, suppose a 1 ≤ a 2 .In this case, we can decrease a 1 , b 1 and increase a 2 , b 2 by the same amount ε ∈ (0, min(a 1 , 1−a 2 , b 1 , 1−b 2 )] to contradict the optimality of O.

Theorem 2 .
10 (Duality of Linear Programming [4]).For the linear programs maximize c T x subject to Ax ≤ b and x ≥ 0 (P) and minimize b T y subject to A T y ≥ c and y ≥ 0 (D),

Theorem 2 .
11 (Complementary Slackness [4]).Let x * = (x * 1 , x * 2 , . . ., x * n ) be a feasible solution of the linear program maximize c T x subject to Ax ≤ b and x ≥ 0 (P), and let y * = (y * 1 , y * 2 , . . ., y * m ) be a feasible solution of the dual linear program minimize b T y subject to A T y ≥ c and y ≥ 0 (D).