Keywords

1 Introduction

Pseudo-Boolean (PB) constraints are Boolean functions over 0/1 Boolean variables \(x_1,\dots ,x_n\) of the form \(\sum _{i = 1}^n w_ix_i\) ‘\(\texttt {op}\)’ \(\theta \) where the \(w_i\) are integer weights, \(\theta \) is an integer threshold and ‘\(\texttt {op}\)’ is a comparison operator \(<, \le ,>\) or \(\ge \). PB-constraints have been studied extensively under different names (e.g. threshold functions [14], Knapsack constraints [13]) due to their omnipresence in many domains of AI and their wide range of practical applications [3, 7, 9, 15, 21].

One way to handle PB-constraints in a constraint satisfaction problem is to translate them into a CNF formula and feed it to a SAT solver. The general idea is to generate a CNF, possibly introducing auxiliary Boolean variables, whose restriction to variables of the constraint is equivalent to the constraint. Two major considerations here are the size of the CNF encoding and its propagation strength. One wants, on the one hand, to avoid the size of the encoding to explode, and on the other hand, to guarantee a good behaviour of the SAT instance under unit propagation – a technique at the very core of SAT solving. Desired propagation strength properties are, for instance, generalized arc consistency (GAC) [4] or propagation completeness (PC) [6]. Several encodings to CNF follow the same two-steps method: first, each constraint is represented in a compact form such as BDD (Binary Decision Diagram) or DNNF (Decomposable Negation Normal Form). Second, the compact forms are turned into CNFs using Tseitin or other transformations. The SAT instance is the conjunction of all obtained CNFs. It is worth mentioning that there are GAC encodings of PB-constraints into polynomial size CNFs that do not follow this two-steps method [5]. However no similar result is known for PC encodings. PC encodings are more restrictive that GAC encodings and may be obtained via techniques requiring compilation to DNNF [17]. Thus the first step is a knowledge compilation task.

Knowledge compilation studies different representations for knowledge [10, 19] under the general idea that some representations are more suitable than others when solving specific reasoning problems. One observation that has been made is that the more reasoning tasks can be solved efficiently with particular representations, the larger these representations get in size. In the context of constraint encodings to SAT, the conversion of compiled forms to CNFs does not reduce the size of the SAT instance, therefore it is essential to control the size of the representations obtained by knowledge compilation.

Several representations have been studied with respect to different encoding techniques with the purpose of determining which properties of representations are sufficient to ensure propagation strength [1, 2, 11, 12, 16, 17]. Popular representations in this context are DNNF and BDD and their many variants: deterministic DNNF, smooth DNNF, OBDD\(\dots \) As mentioned above, a problem occurring when compiling a constraint into such representations is that exponential space may be required. Most notably, it has been shown in [2, 14] that some PB-constraints can only be represented by OBDDs whose size is exponential in \(\sqrt{n}\), where n is the number of variables. Our contribution is the proof of the following theorem where we lift the statement from OBDD to DNNF.

Theorem 1

There is a class of PB-constraints \(\mathcal {F}\) such that for any constraint \(f \in \mathcal {F}\) on \(n^2\) variables, any DNNF representation of f has size \(2^{\varOmega (n)}\).

Since DNNFs are exponentially more succinct than OBDDs [10], our result is a generalisation of the result in [2, 14]. The class \(\mathcal {F}\) is similar to that used in [2, 14], actually the only difference is the choice of the threshold for the PB-constraints. Yet, adapting proofs given in [2, 14] for OBDD to DNNF is not straightforward, thus our proof of Theorem 1 bears very little resemblance.

It has been shown in [18] that there exist sets of PB-constraints such that the whole set (so a conjunction of PB-constraints) requires exponential size DNNF to represent. Our result is a generalisation to single PB-constraints.

2 Preliminaries

Conventions of Notation. Boolean variables are seen as variables over \(\{0,1\}\), where 0 and 1 represent false and true respectively. Via this 0/1 representation, Boolean variables can be used in arithmetic expressions over \(\mathbb {Z}\). For notational convenience, we keep the usual operators \(\lnot \), \(\vee \) and \(\wedge \) to denote, respectively, the negation, disjunction and conjunction of Boolean variables or functions. Given X a set of n Boolean variables, assignments to X are seen as vectors in \(\{0,1\}^n\). Single Boolean variables are written in plain text (x) while assignments to several variables are written in bold (\(\mathbf {x}\)). We write \(\mathbf {x} \le \mathbf {y}\) when the vector \(\mathbf {y}\) dominates \(\mathbf {x}\) element-wise. We write \(\mathbf {x} < \mathbf {y}\) when \(\mathbf {x} \le \mathbf {y}\) and \(\mathbf {x} \ne \mathbf {y}\). In this framework, a Boolean function f over X is a mapping from \(\{0,1\}^n\) to \(\{0,1\}\). f is said to accept an assignment \(\mathbf {x}\) when \(f(\mathbf {x}) = 1\), then \(\mathbf {x}\) is called a model of f. The function is monotone if for any model \(\mathbf {x}\) of f, all \(\mathbf {y} \ge \mathbf {x}\) are models of f as well. The set of models of f is denoted \(f^{\text { -1}}(1)\). Given f and g two Boolean functions over X, we write \(f \le g\) when \(f^{\text { -1}}(1) \subseteq g^{\text { -1}}(1)\). We write \(f < g\) when the inclusion is strict.

Pseudo-Boolean Constraints. Pseudo-Boolean (PB) constraints are inequalities the form \(\sum _{i=1}^n w_i x_i\) ‘\(\texttt {op}\)’ \(\theta \) where the \(x_i\) are 0/1 Boolean variables, the \(w_i\) and \(\theta \) are integers, and ‘\(\texttt {op}\)’ is one of the comparison operator <, \(\le \), > or \(\ge \). A PB-constraint is associated with a Boolean function whose models are exactly the assignments to \(\{x_1, \dots , x_n\}\) that satisfy the inequality. For simplicity we directly consider PB-constraints as Boolean functions – although the same function may represent different constraints – while keeping the term “constraints” when referring to them. In this paper, we restrict our attention to PB-constraints where ‘\(\texttt {op}\)’ is \(\ge \) and all weights are positive integers. Note that such PB-constraints are monotone Boolean functions. Given a sequence of positive integer weights \(W = (w_1, \dots , w_n)\) and an integer threshold \(\theta \), we define the function \(w : \{0,1\}^n \rightarrow \mathbb {N}\) that maps any assignment to its weight by \(w(\mathbf {x}) = \sum _{i=1}^n w_i x_i\). With these notations, a PB-constraint over X for a given pair \((W,\theta )\) is a Boolean function whose models are exactly the \(\mathbf {x}\) such that \(w(\mathbf {x}) \ge \theta \).

Example 1

Let \(n = 5\), \(W = (1,2,3,4,5)\) and \(\theta = 9\). The PB-constraint for \((W,\theta )\) is the Boolean function whose models are the assignments such that \(\sum _{i = 1}^5 ix_i \ge ~9\). E.g. \(\mathbf {x} = (0,1,1,0,1)\) is a model of weight \(w(\mathbf {x}) = 10\).

For notational clarity, given any subset \(Y \subseteq X\) and denoting \(\mathbf {x}\vert _Y\) the restriction of \(\mathbf {x}\) to variables of Y, we overload w so that \(w(\mathbf {x}\vert _Y)\) is the sum of weights activated by variables of Y set to 1 in \(\mathbf {x}\).

Decomposable . A circuit in negation normal form (NNF) is a single output Boolean circuit whose inputs are Boolean variables and their complements, and whose gates are fanin-2 AND and OR gates. The size of the circuit is the number of its gates. We say that an NNF is decomposable (DNNF) if for any AND gate, the two sub-circuits rooted at that gate share no input variable, i.e., if x or \(\lnot x\) is an input of the circuit rooted at the left input of the AND gate, then neither x nor \(\lnot x\) is an input of the circuit rooted at the right input, and vice versa. A Boolean function f is encoded by a DNNF D if the assignments of variables for which the output of D is 1 (true) are exactly the models of f.

Rectangle Covers. Let X be a finite set of Boolean variables and let \(\varPi = (X_1, X_2)\) be a partition of X (i.e., \(X_1 \cup X_2 = X\) and \(X_1 \cap X_2 = \emptyset \)). A rectangle r with respect to \(\varPi \) is a Boolean function over X defined as the conjunction of two functions \(\rho _1\) and \(\rho _2\) over \(X_1\) and \(X_2\) respectively. \(\varPi \) is called the partition of r. We say that the partition and the rectangle are balanced when \(\frac{\vert X \vert }{3} \le \vert X_1 \vert \le \frac{2\vert X \vert }{3}\) (thus the same holds for \(X_2\)). Whenever considering a partition \((X_1,X_2)\), we use for any assignment \(\mathbf {x}\) to X the notations and . And for any two assignments \(\mathbf {x}_1\) and \(\mathbf {x}_2\) to \(X_1\) and \(X_2\), we note \((\mathbf {x}_1, \mathbf {x}_2)\) the assignment to X whose restrictions to \(X_1\) and \(X_2\) are \(\mathbf {x}_1\) and \(\mathbf {x}_2\). Given f a Boolean function over X, a rectangle cover of f is a disjunction of rectangles over X, possibly with different partitions, equivalent to f. The size of a rectangle cover is the number of its rectangles. A cover is called balanced if all its rectangles are balanced.

Example 2

Going back to Example 1, consider the partition , and define and . Then is a rectangle w.r.t. this partition that accepts only models of the PB-constraint from Example 1. Thus it can be part of a rectangle cover for this constraint.

Any function f has at least one balanced rectangle cover as one can create a balanced rectangle accepting exactly one chosen model of f. We denote by C(f) the size of the smallest balanced rectangle cover of f. The following result from [8] links C(f) to the size of any DNNF encoding f.

Theorem 2

Let D be a DNNF encoding a Boolean function f. Then f has a balanced rectangle cover of size at most the size of D.

Theorem 2 reduces the problem of finding lower bounds on the size of DNNFs encoding f to that of finding lower bounds on C(f).

3 Restriction to Threshold Models of PB-Constraints

The strategy to prove Theorem 1 is to find a PB-constraint f over n variables such that C(f) is exponential in \(\sqrt{n}\) and then use Theorem 2. We first show that we can restrict our attention to covering particular models of f with rectangles rather than the whole function. In this section X is a set of n Boolean variables and f is a PB-constraint over X. Recall that we only consider constraints of the form \(\sum _{i = 1}^n w_i x_i \ge \theta \) where the \(w_i\) and \(\theta \) are positive integers.

Definition 1

The threshold models of f are the models \(\mathbf {x}\) such that \(w(\mathbf {x}) = \theta \).

Threshold models should not be confused with minimal models (or minimals).

Definition 2

A minimal of f is a model \(\mathbf {x}\) such that no \(\mathbf {y} < \mathbf {x}\) is a model of f.

For a monotone PB-constraint, a minimal model is such that its sum of weights drops below the threshold if we remove any element from it. Any threshold model is minimal, but not all minimals are threshold models. There even exist constraints with no threshold models (e.g. take even weights and an odd threshold) while there always are minimals for satisfiable constraints.

Example 3

The minimals of the PB-constraint from Example 1 are (0, 0, 0, 1, 1), (0, 1, 1, 1, 0), (1, 0, 1, 0, 1) and (0, 1, 1, 0, 1). The first three are threshold models.

Let \(f^*\) be the Boolean function whose models are exactly the threshold models of f. In the next lemma, we prove that the smallest rectangle cover of \(f^*\) has size at most C(f). Thus, lower bounds on \(C(f^*)\) are also lower bounds on C(f).

Lemma 1

Let \(f^*\) be the Boolean function whose models are exactly the threshold models of f. Then \(C(f) \ge C(f^*)\).

Proof

Let be a balanced rectangle with \(r \le f\) and assume r accepts some threshold models. Let be the partition of r. We claim that there exist two integers \(\theta _1\) and \(\theta _2\) such that \(\theta _1 + \theta _2 = \theta \) and, for any threshold model \(\mathbf {x}\) accepted by r, there is \(w(\mathbf {x_1}) = \theta _1\) and \(w(\mathbf {x_2}) = \theta _2\). To see this, assume by contradiction that there exists another partition \(\theta = \theta '_1 + \theta '_2\) of \(\theta \) such that some other threshold model \(\mathbf {y}\) with \(w(\mathbf {y}_1) = \theta '_1\) and \(w(\mathbf {y}_2) = \theta '_2\) is accepted by r. Then either \(w(\mathbf {x_1}) + w(\mathbf {y_2}) < \theta \) or \(w(\mathbf {y_1}) + w(\mathbf {x_2}) < \theta \), but since \((\mathbf {x}_1,\mathbf {y}_2)\) and \((\mathbf {y}_1,\mathbf {x}_2)\) are also models of r, r would accept a non-model of f, which is forbidden. Now let \(\rho ^*_1\) (resp. \(\rho ^*_2\)) be the function whose models are exactly the models of \(\rho _1\) (resp. \(\rho _2\)) of weight \(\theta _1\) (resp. \(\theta _2\)). Then is a balanced rectangle whose models are exactly the threshold models accepted by r.

Now consider a balanced rectangle cover of f of size C(f). For each rectangle r of the cover, if r accepts no threshold model then discard it, otherwise construct \(r^*\). The disjunction of these new rectangles is a balanced rectangle cover of \(f^*\) of size at most C(f). Therefore \(C(f) \ge C(f^*)\).   \(\square \)

4 Reduction to Covering Maximal Matchings of \(K_{n,n}\)

We define the class of hard PB-constraints for Theorem 1 in this section. Recall that for a hard constraint f, our aim is to find an exponential lower bound on C(f). We will show, using Lemma 1, that the problem can be reduced to that of covering all maximal matchings of the complete \(n \times n\) bipartite graph \(K_{n,n}\) with rectangles. In this section, X is a set of \(n^2\) Boolean variables. For presentability reasons, assignments to X are written as \(n \times n\) matrices. Each variable \(x_{i,j}\) has the weight . Define the matrix of weights and the threshold . The PB-constraint f for the pair \((W,\theta )\) is such that \(f(\mathbf {x}) = 1\) if and only if \(\mathbf {x}\) satisfies

$$\begin{aligned} \sum _{1 \le i,j \le n} \left( \frac{2^{i} + 2^{j+n}}{2}\right) x_{i,j} \ge 2^{2n}-1 \text { .} \end{aligned}$$
(1)

Constraints of this form constitute the class of hard constraints of Theorem 1. One may find it easier to picture f writing the weights and threshold as binary numbers of 2n bits. Bits of indices 1 to n form the lower part of the number and those of indices \(n+1\) to 2n form the upper part. The weight \(w_{i,j}\) is the binary number where the only bits set to 1 are the ith bit of the lower part and the jth bit of the upper part. Thus when a variable \(x_{i,j}\) is set to 1, exactly one bit of value 1 is added to each part of the binary number of the sum.

Assignments to X uniquely encode subgraphs of \(K_{n,n}\). We denote \(U = \{u_1, \dots , u_n\}\) the nodes of the left side and \(V = \{v_1, \dots , v_n\}\) those of the right side of \(K_{n,n}\). The bipartite graph encoded by \(\mathbf {x}\) is such that there is an edge between the \(u_i\) and \(v_j\) if and only if \(x_{i,j}\) is set to 1 in \(\mathbf {x}\).

Example 4

Take \(n = 4\). The assignment \( \mathbf {x} = \small \begin{pmatrix} 1 &{} 1 &{} 0 &{} 1\\ 0 &{} 0 &{} 0 &{} 0\\ 0 &{} 1 &{} 0 &{} 0\\ 0 &{} 1 &{} 0 &{} 0\\ \end{pmatrix} \) encodes

Definition 3

A maximal matching assignment (or maximal matching model) is an assignment \(\mathbf {x}\) to X such that

  • for any \(i \in [n]\), there is exactly one k such that \(x_{i,k}\) is set to 1 in \(\mathbf {x}\),

  • for any \(j \in [n]\), there is exactly one k such that \(x_{k,j}\) is set to 1 in \(\mathbf {x}\).

As the name suggests, the maximal matching assignments are those encoding graphs whose edges form a maximal matching of \(K_{n,n}\) (i.e., a maximum cardinality matching). One can also see them as encodings for permutations of [n].

Example 5

The maximal matching model \( \mathbf {x} = \small \begin{pmatrix} 0 &{} 0 &{} 1 &{} 0\\ 1 &{} 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 &{} 1\\ 0 &{} 1 &{} 0 &{} 0\\ \end{pmatrix} \) encodes

For a given \(\mathbf {x}\), define \(var _{k}\left( \mathbf {x}\right) \) by when \(1 \le k \le n\) and by when \(n+1 \le k \le 2n\). \(var _{k}\left( \mathbf {x}\right) \) stores the index of variables in \(\mathbf {x}\) that directly add 1 to the kth bit of \(w(\mathbf {x})\). Note that a maximal matching model is an assignment \(\mathbf {x}\) such that \(\vert var _{k}\left( \mathbf {x}\right) \vert = 1\) for all k. It is easy to see that maximal matching models are threshold models of f: seeing weights as binary numbers of 2n bits, for every bit of the sum the value 1 is added exactly once, so exactly the first 2n bits of the sum are set to 1, which gives us \(\theta \). Note that not all threshold models of f are maximal matching models, for instance the assignment from Example 4 does not encode a maximal matching but one can verify that it is a threshold model. Recall that \(f^*\) is the function whose models are the threshold models of f. In the next lemmas, we prove that lower bounds on the size of rectangle covers of the maximal matching models are lower bounds on \(C(f^*)\), and a fortiori on C(f).

Lemma 2

Let be a partition of X. Let and be maximal matching assignments. If \((\mathbf {x}_1, \mathbf {y}_2)\) and \((\mathbf {y}_1, \mathbf {x}_2)\) both have weight then both are maximal matching assignments.

Proof

It is sufficient to show that \(\vert var _{k}\left( \mathbf {x}_1,\mathbf {y}_2\right) \vert = 1\) and \(\vert var _{k}\left( \mathbf {y}_1,\mathbf {x}_2\right) \vert = 1\) for all \(1 \le k \le 2n\). We prove it for \((\mathbf {x}_1,\mathbf {y}_2)\) by induction on k. First observe that since \(\vert var _{k}\left( \mathbf {x}\right) \vert = 1\) and \(\vert var _{k}\left( \mathbf {y}\right) \vert = 1\) for all \(1 \le k \le 2n\), the only possibilities for \(\vert var _{k}\left( \mathbf {x}_1,\mathbf {y}_2\right) \vert \) are 0, 1 or 2.

  • For the base case \(k = 1\), if \(\vert var _{1}\left( \mathbf {x}_1,\mathbf {y}_2\right) \vert \) is even then the first bit of \(w(\mathbf {x}_1) + w(\mathbf {y}_2)\) is 0 and the weight of \((\mathbf {x}_1,\mathbf {y}_2)\) is not \(\theta \). So \(\vert var _{1}\left( \mathbf {x}_1,\mathbf {y}_2\right) \vert = 1\).

  • For the general case \(1 < k \le 2n\), assume that \(\vert var _{1}\left( \mathbf {x}_1,\mathbf {y}_2\right) \vert = \dots = \vert var _{k-1}\left( \mathbf {x}_1,\mathbf {y}_2\right) \vert = 1\). So the kth bit of \(w(\mathbf {x}_1) + w(\mathbf {y}_2)\) depends only on the parity of \(\vert var _{k}\left( \mathbf {x}_1,\mathbf {y}_2\right) \vert \): the kth bit is 0 if \(\vert var _{k}\left( \mathbf {x}_1,\mathbf {y}_2\right) \vert \) is even and 1 otherwise. \((\mathbf {x}_1,\mathbf {y}_2)\) has weight \(\theta \) so \(\vert var _{k}\left( \mathbf {x}_1,\mathbf {y}_2\right) \vert = 1\).

The argument applies to \((\mathbf {y}_1, \mathbf {x}_2)\) analogously.    \(\square \)

Lemma 3

Let f be the PB-constraint (1) and let \(\hat{f}\) be the function whose models are exactly the maximal matching assignments. Then \(C(f) \ge C(\hat{f})\).

Proof

By Lemma 1, it is sufficient to prove that \(C(f^*) \ge C(\hat{f})\). We already know that \(\hat{f} \le f^*\). Let be a balanced rectangle of partition with \(r \le f^*\), and assume r accepts some maximal matching assignment. Let \(\hat{\rho }_1\) (resp. \(\hat{\rho }_2\)) be the Boolean function over \(X_1\) (resp. \(X_2\)) whose models are the \(\mathbf {x}_1\) (resp. \(\mathbf {x}_2\)) such that there is a maximal matching assignment \((\mathbf {x}_1, \mathbf {x}_2)\) accepted by r. We claim that the balanced rectangle accepts exactly the maximal matching models of r. On the one hand, it is clear that all maximal matching models of r are models of \(\hat{r}\). On the other hand, all models of \(\hat{r}\) are threshold models of the form \((\mathbf {x}_1, \mathbf {y}_2)\), where \((\mathbf {x}_1, \mathbf {x}_2)\) and \((\mathbf {y}_1, \mathbf {y}_2)\) encode maximal matchings, so by Lemma 2, \(\hat{r}\) accepts only maximal matching models of r.

Now consider a balanced rectangle cover of \(f^*\) of size \(C(f^*)\). For each rectangle r of the cover, if r accepts no maximal matching assignment then discard it, otherwise construct \(\hat{r}\). The disjunction of these new rectangles is a balanced rectangle cover of \(\hat{f}\) of size at most \(C(f^*)\). Therefore \(C(f^*) \ge C(\hat{f})\).    \(\square \)

5 Proof of Theorem 1

Theorem 1

There is a class of PB-constraints \(\mathcal {F}\) such that for any constraint \(f \in \mathcal {F}\) on \(n^2\) variables, any DNNF encoding f has size \(2^{\varOmega (n)}\).

\(\mathcal {F}\) is the class of constraints defined in (1). Thanks to Theorem 2 and Lemma 3, the proof boils down to finding exponential lower bounds on \(C(\hat{f})\), where \(\hat{f}\) is the Boolean function on \(n^2\) variables whose models encode exactly the maximal matchings of \(K_{n,n}\) (or equivalently, the permutations of [n]). \(\hat{f}\) has n! models. The idea is now to prove that rectangles covering \(\hat{f}\) must be relatively small, so that covering the whole function requires many of them.

Lemma 4

Let \(\varPi = (X_1, X_2)\) be a balanced partition of X. Let r be a rectangle with respect to \(\varPi \) with \(r \le \hat{f}\). Then \(\vert r^{\text { -1}}(1) \vert \le n!/\left( {\begin{array}{c}n\\ n\sqrt{2/3}\end{array}}\right) \).

The function \(\hat{f}\) has already been studied extensively in the literature, often under the name \(\text {PERM}_n\) (for permutations on [n]), see for instance Chap. 4 of [22] or Sect. 6.2 of [20] where a statement similar to Lemma 4 is established. With Lemma 4 we can give the proof of Theorem 1.

Proof

(Theorem 1). Let \(\bigvee _{k = 1}^{C(\hat{f})} r_k\) be a balanced rectangle cover of \(\hat{f}\). We have \(\sum _{k = 1}^{C(\hat{f})} \vert r_k^{\text { -1}}(1) \vert \ge \vert \hat{f}^{\text { -1}}(1) \vert = n!\). Lemma 4 gives us \((C(\hat{f})n!)/\left( {\begin{array}{c}n\\ n\sqrt{2/3}\end{array}}\right) \ge n!\), thus

$$\begin{aligned} \begin{aligned} C(\hat{f}) \ge \left( {\begin{array}{c}n\\ n\sqrt{2/3}\end{array}}\right) \ge \left( \frac{n}{n\sqrt{2/3}}\right) ^{n\sqrt{2/3}} = \left( \frac{3}{2}\right) ^{n\frac{\sqrt{2/3}}{2}} \ge 2^{n\frac{\sqrt{2/3}}{4}} = 2^{\varOmega (n)} \end{aligned} \end{aligned}$$

where we have used \(\left( {\begin{array}{c}a\\ b\end{array}}\right) \ge (a/b)^b\) and \(3/2 \ge \sqrt{2}\). Using Lemma 3 we get that \(C(f) \ge C(\hat{f}) \ge 2^{\varOmega (n)}\). Theorem 2 allows us to conclude.    \(\square \)

All that is left is to prove Lemma 4.

Fig. 1.
figure 1

Partition of maximal matching

Full size image

Proof

(Lemma 4).

Let and . Recall that and are the nodes from the left and right part of \(K_{n,n}\) respectively. Define and . Define \(U_2\) and \(V_2\) analogously (this time using \(X_2\) and \(\rho _2\)). Figure 1 illustrates the construction of these sets: Fig. 1a shows a partition \(\varPi \) of the edges of \(K_{4,4}\) (full edges in \(X_1\), dotted edges in \(X_2\)) and Fig. 1b shows the contribution of a model of r to \(U_1\), \(V_1\), \(U_2\), and \(V_2\) after partition according to \(\varPi \).

Models of \(\rho _1\) are clearly matchings of \(K_{n,n}\). Actually they are matchings between \(U_1\) and \(V_1\) by construction of these sets. We claim that they are maximal. To verify this, observe that \(U_1 \cap U_2 = \emptyset \) and \(V_1 \cap V_2 = \emptyset \) since otherwise r has a model that is not a matching. Thus if \(\rho _1\) were to accept a non-maximal matching between \(U_1\) and \(V_1\) then r would accept a non-maximal matching between U and V. So \(\rho _1\) accepts only maximal matchings between \(U_1\) and \(V_1\), consequently \(\vert U_1 \vert = \vert V_1 \vert \). The argument applies symmetrically for \(V_2\) and \(U_2\). We note . It stands that \(U_1 \cup U_2 = U\) and \(V_1 \cup V_2 = V\) as otherwise r accepts matchings that are not maximal. So \(\vert U_2 \vert = \vert V_2 \vert = n-k\). We now have \(\vert \rho _1^{\text { -1}}(1) \vert \le k!\) and \(\vert \rho _2^{\text { -1}}(1) \vert \le (n-k)!\), leading to \(\vert r^{\text { -1}}(1) \vert \le k!(n-k)! = n!/\left( {\begin{array}{c}n\\ k\end{array}}\right) \).

Up to \(k^2\) edges may be used to build matchings between \(U_1\) and \(V_1\). Since r is balanced we obtain \(k^2 \le 2n^2/3\). Applying the same argument to \(U_2\) and \(V_2\) gives us \((n-k)^2 \le 2n^2/3\), so \(n(1-\sqrt{2/3}) \le k \le n\sqrt{2/3}\). Finally, the function \(k \mapsto n!/\left( {\begin{array}{c}n\\ k\end{array}}\right) \), when restricted to some interval \([\![n(1-\alpha ), \alpha n]\!]\), reaches its maximum at \(k = \alpha n\), hence the upper bound \(\vert r^{\text { -1}}(1) \vert \le n!/\left( {\begin{array}{c}n\\ n\sqrt{2/3}\end{array}}\right) \).    \(\square \)