Bipartite Perfect Matching as a Real Polynomial

Proof. Let $`G \in (MC_n \setminus \{K_{n,n}\})`$ and let $`\mathcal{P} = (MC_n \cup \{\hat{0}\}, \subseteq)`$ be the Eulerian matching-covered lattice, where $`\hat{0}`$ is the empty graph. Since $`\mathcal{P}`$ is Eulerian, its Möbius function is multiplicative, thus: $`\forall H \in MC_n, H \supseteq G:\ \mu_{\mathcal{P}}(\hat{0}, H) = \mu_{\mathcal{P}}(\hat{0}, G) \cdot \mu_{\mathcal{P}}(G, H)`$. Therefore by Definition [mobius_func_poset] and Lemma [upper_sum_mobius], we have:

\begin{equation*}
	\begin{split}
	a^\star_G &= (-1)^{|E(G)|}  \sum_{\substack{G \subseteq H \subseteq K_{n,n}\\H \in MC_n}} \mu_P(\hat{0}, H) \\
	&= (-1)^{|E(G)|} \mu_{\mathcal{P}}(\hat{0}, G) \sum_{\substack{G \subseteq H \subseteq K_{n,n}\\H \in MC_n}} \mu_{\mathcal{P}}(G, H) = 0 \qedhere
	\end{split}
\end{equation*}

 ◻

Proof. Let $`n > 1`$ and let $`\mathcal{P} = (MC_n \cup \{\hat{0}\}, \subseteq)`$ be the Eulerian matching-covered lattice, where $`\hat{0}`$ is the empty graph. Let $`G \subseteq K_{n,n}`$, where $`G \ne \hat{0}`$. For any $`\emptyset \ne S \subseteq MC_n`$, denote by $`\bigvee S`$ the join of all graphs in $`S`$. Recall (Proposition [graph_cover_lattice]) that in $`\mathcal{P}`$, the join $`\bigvee S`$ is the union of all graphs in $`S`$. By Lemma [upper_sum_mobius], we have:

a_G^\star = (-1)^{|E(G)|} \cdot \sum_{\substack{G \subseteq H \subseteq K_{n,n}\\H \in MC_n}} \mu_{\mathcal{P}}(\hat{0}, H)

Using the inclusion-exclusion principle on the umbrella of $`G`$, $`\mathcal{U}(G)`$, we obtain:

\begin{equation*}
	\begin{split}
	\sum_{\substack{G \subseteq H \subseteq K_{n,n}\\H \in MC_n}} \mu_{\mathcal{P}}(\hat{0}, H) &= \sum_{\emptyset \ne S \subseteq \mathcal{U}(G)} (-1)^{|S|+1} \sum_{\substack{(\bigvee S) \subseteq H \subseteq K_{n,n}\\H \in MC_n}} \mu_{\mathcal{P}}(\hat{0}, H) \\
	&= \left(\sum_{\substack{\emptyset \ne S \subseteq \mathcal{U}(G)\\(\bigvee S) \subset K_{n,n}}} (-1)^{|S|+1} \sum_{\substack{(\bigvee S) \subseteq H \subseteq K_{n,n}\\H \in MC_n}} \mu_{\mathcal{P}}(\hat{0}, H) + \sum_{\substack{\emptyset \ne S \subseteq \mathcal{U}(G)\\(\bigvee S) = K_{n,n}}} (-1)^{|S|+1} \mu_{\mathcal{P}}(\hat{0}, K_{n,n})\right)
	\end{split}
\end{equation*}

Since $`\mathcal{P}`$ is Eulerian, the sum of Möbius numbers in any nontrivial closed interval is zero (see Lemma [mu_graphs_zero_coeff]). In particular, for any $`S \subseteq \mathcal{U}(G)`$ where $`(\bigvee S) \ne K_{n,n}`$, we have:

\sum_{\substack{(\bigvee S) \subseteq H \subseteq K_{n,n}\\H \in MC_n}} \mu_{\mathcal{P}}(\hat{0}, H) = 0

Therefore:

\begin{equation*}
	\begin{split}
	a^\star_G &= (-1)^{|E(G)|} \cdot \sum_{\substack{\emptyset \ne S \subseteq \mathcal{U}(G)\\\bar{E}(S) = K_{n,n}}} (-1)^{|S|+1} (-1)^{\chi(K_{n,n}) + 1} \\
	&= (-1)^{n+|E(G)|} \cdot \sum_{\substack{\emptyset \ne S \subseteq \mathcal{U}(G)\\\bar{E}(S) = K_{n,n}}} (-1)^{|S|+1} \qedhere
	\end{split}
\end{equation*}

 ◻

Proof. Let $`(a,b) \notin E(G)`$ be a wildcard edge for $`G`$. We show that $`(a,b) \notin \bar{E}(\mathcal{U}(G))`$. Assume towards a contradiction that $`(a,b) \in \bar{E}(\mathcal{U}(G))`$, and let $`H \in \mathcal{U}(G)`$ be a graph such that $`(a,b) \in E(H)`$. Then by the definition of $`(a,b)`$, we have $`H' = (H \setminus \{(a,b)\}) \in MC_n`$, and furthermore $`G \subseteq H' \subset H`$, in contradiction to the fact that $`H \in \mathcal{U}(G)`$. ◻

Proof. Let $`(a,b) \notin E(G)`$ be a surplus edge for $`G`$ and let $`H \in MC_n`$ such that $`H \supseteq G \cupdot \{(a,b)\}`$. Denote $`H' = H \setminus \{(a,b)\}`$. It remains to show that $`H' \in MC_n`$. Assume towards a contradiction that $`H' \notin MC_n`$ and denote by $`C=(A_C \cupdot B_C, E) \in C(H)`$ the connected component of $`H`$ containing the edge $`(a,b)`$. First, note that $`K_2 \ne (C \setminus \{(a,b)\}) \in C(H')`$, since elementary graphs are $`2-connected`$. Since $`C \setminus \{(a,b)\}`$ is not elementary, then by Theorem [hetyei_conditions] there exists $`\emptyset \ne X \subset A_C \subseteq A`$ such that $`|N_{H'}(X)| \le |X|`$.

Observe that $`a \in X`$ and $`b \notin N_{H'}(X)`$. Otherwise, we have $`N_{H'}(X) = N_{H}(X)`$ and since $`H \in MC_n`$ then $`C`$ is elementary and thus $`|N_{H'}(X)| = |N_H(X)| > |X|`$, a contradiction. However, since $`(a,b)`$ is a surplus edge for $`G`$, then for all $`X \subset A`$ such that $`a \in X`$, $`b \notin N_G(X)`$, we have $`|N_G(X)| > |X|`$. In particular, since $`H' \supseteq G`$, then for our $`X`$ we have $`a \in X`$ and $`b \notin N_G(X)`$ and thus $`|N_{H'}(X)| \ge |N_G(X)| > |X|`$, in contradiction to the definition of $`X`$. ◻

Proof. Let $`A = \{a_1, \dots, a_n\}`$, $`B = \{b_1, \dots, b_n\}`$ be two sets, and let $`G =(A \cupdot B, E) \in HVC_n`$, such that $`G`$ is not totally ordered. Thus, there exist two vertices $`a_i, a_j \in A`$ such that:

N(a_i) \not\supseteq N(a_j) \ \ \land\ \  N(a_j) \not\supseteq N(a_i) \ \ \land\ \  |N(a_i)| \ge |N(a_j)|

We will show that $`\forall b_k \in (N(a_j) \setminus N(a_i)):`$ $`(a_i,b_k)`$ is a surplus edge for $`G`$. Let $`b_k \in N(a_j) \setminus N(a_i)`$, $`b_m \in N(a_i) \setminus N(a_j)`$. Since $`G \in HVC_n`$, every edge of $`G`$ is covered by some graph $`K \in HV_n`$, and in particular so are $`(a_i, b_m)`$, $`(a_j, b_k)`$. Thus, there exist $`X_{i,m}, X_{j,k} \subseteq A`$, $`Y_{i,m}, Y_{j,k} \subseteq B`$ such that:

\begin{equation*}
	\begin{split}
	|X_{i,m}| + |Y_{i,m}| &= n+1, \\
	(X_{i,m} \times Y_{i,m}) &\subseteq E(G)
	\end{split}
	\quad\quad
	\begin{split}
	|X_{j,k}| + |Y_{j,k}| &= n+1 \\
	(X_{j,k} \times Y_{j,k}) &\subseteq E(G)
	\end{split}
\end{equation*}

and furthermore, $`a_i \in X_{i,m}`$, $`b_m \in Y_{i,m}`$, $`a_j \in X_{j,k}`$ and $`b_k \in Y_{j,k}`$. Assume towards a contradiction that $`(a_i, b_k)`$ is not a surplus edge for $`G`$. Then, there exists $`X \subset A`$ such that $`a_i \in X`$, $`b_k \notin N(X)`$ and $`|N(X)| \le |X|`$.

Since $`a_i \in X`$ then $`N(X) \supseteq N(a_i)`$ and in particular $`|N(X)| \ge |N(a_i)|`$. Furthermore, observe that $`X \cap X_{j,k} = \emptyset`$, since otherwise $`b_k \in N(X)`$, in contradiction to the definition of $`X`$. Thus, we have $`n - |X_{j,k}| \ge |X|`$. Moreover, recall that by the definition of $`a_i`$ and $`a_j`$, we have $`|N(a_i)| \ge |N(a_j)|`$. Since the edge $`(a_j, b_k)`$ is covered by $`K_{X_{j,k}, Y_{j,k}}`$, then $`N(a_j) \supseteq Y_{j,k}`$. Lastly, by the definition of $`X`$, $`|N(X)| \le |X|`$. Putting all the above inequalities together, we have:

\begin{equation*}
	\begin{split}
	n - |X_{j,k}| \ge |X| \ge |N(X)| \ge |N(a_i)| \ge |N(a_j)| \ge |Y_{j,k}|
	\end{split}
\end{equation*}

Therefore $`|X_{j,k}| + |Y_{j,k}| \le n`$, in contradiction to the fact that $`|X_{j,k}| + |Y_{j,k}| = n+1`$. ◻

Proof. Let $`G=(A \cupdot B, E) \subseteq K_{n,n}`$ be a graph, where $`A=\{a_1, \dots, a_n\}`$ and $`B = \{b_1, \dots, b_n\}`$. The edges of $`G`$ are given by: $`\forall i \in [n]: N_G(a_i) = \{b_1, \dots, b_i\}`$. Observe that $`G`$ is strictly totally ordered, since $`N(a_n) \supsetneq N(a_{n-1}) \supsetneq \dots N(a_1) \supsetneq \emptyset`$.

For every $`k \in [n]`$, denote $`A_k = \{a_k, \dots, a_n\} \subseteq A`$, $`B_k = \{b_1, \dots, b_k\} \subseteq B`$. By the definition of $`G`$, $`\forall k \in [n]`$, $`K_{A_k, B_k} \subseteq G`$. We now show that for any $`K_{X,Y} \in HV_n`$ such that $`K_{X,Y} \subseteq G`$ and $`|Y| = k`$, we necessarily have $`X = A_k`$ and $`Y = B_k`$.

Assume towards a contradiction this is not the case. Let $`K_{X,Y} \in HV_n`$ such that $`K_{X,Y} \subseteq G`$, $`|Y|=k`$ and $`Y \ne B_k`$ or $`X \ne A_k`$. If $`Y = B_k`$, then for any $`a_i \in X`$ where $`a_i \notin A_k`$ (i.e., $`i < k`$), the edge $`(a_i, b_1) \notin E(G)`$ – a contradiction. Otherwise, let $`Y \ne B_k`$ and let $`j > k`$ the maximal index such that $`y_j \in Y`$. By the definition of $`G`$, $`\bigcap_{b \in Y} N_G(b) = N_G(y_j) = A_j`$. Since $`K_{X,Y} \subseteq G`$, then in particular $`X \subseteq A_j`$, and therefore $`|Y| + |X| \le k + |A_j| = n`$, in contradiction to the fact that $`K_{X,Y} \in HV_n`$.

Thus, the only Hall violators appearing in $`G`$ are the set:

\mathcal{H} \eqdef \set{H \in HV_n}{H \subseteq G} = \set{K_{A_k, B_k}}{k \in [n]}

The union of all graphs in $`\mathcal{H}`$ is exactly $`G`$, therefore $`G \in HVC_n`$. However, recall that $`BPM_n^\star`$ is a graph cover function for the set $`HV_n`$, and thus by arithmetizing the formula representing the function (recall Proposition [monomials_graph_cover_functions]), we get that:

BPM_n^\star(x_{1,1}, \dots, x_{n,n}) =  \sum_{G \in HVC_n} \left(\sum_{\substack{\emptyset \ne S \subseteq HVC_n\\ \bar{E}(S)=E(G)}}(-1)^{|S|+1}\right) \prod_{(i,j) \in E(G)} x_{i, j}

Observe that the only set of graphs $`S \subseteq HVC_n`$ whose union is equal to $`G`$ is the set $`\mathcal{H}`$ itself, since by omitting any $`K_{A_k, B_k}`$ we will fail to cover the edge $`(a_k, b_k) \in E(G)`$. Thus $`a^\star_G = (-1)^{|\mathcal{H}| + 1} = (-1)^{n+1}`$, as required.

Lastly, we observe that any strictly totally ordered graph $`G' \subseteq K_{n,n}`$ is equivalent, up to permutations over each bipartition, to $`G`$ (and therefore has the same coefficient). This equivalence can be achieved by sorting the vertices of each bipartition by the cardinality of their neighbour sets, where the left vertices are sorted in ascending order, and the right vertices in descending order. ◻

Proof. Let $`n > 1`$. For the lower bound, let $`G`$ be a strictly totally ordered graph. By Lemma [lower_bound_graph], all strictly totally ordered graphs, and in particlar $`G`$, have $`a_G^\star = (-1)^{n+1}`$. However, since no two right or left vertices of $`G`$ have the same set of neighbours, any pair of permutations over the left and right bipartitions yields a new strictly totally ordered graph $`\tilde{G} \cong G`$, thus completing the lower bound.

For the upper bound, let $`U = \{u_1, \dots, u_{n+1}\}`$, $`V = \{v_1, \dots, v_{n+1}\}`$ be two sets. Denote by $`C_n`$ the set of all graphs $`G \subseteq K_{n,n}`$ that are totally ordered. We begin by showing that:

|C_n| = \sum_{k=1}^{n+1} \left((k-1)! \cdot \stirlingII{n+1}{k}\right)^2

Where the notation $`\stirlingII{n}{k}`$ refers to the Stirling number of the second kind. To prove the equality, let us explicitly construct the set $`C_n`$ as follows; for every $`1 \le k \le n+1`$, let:

\begin{equation*}
	\begin{split}
	U = U_1 \cupdot U_2 \dots \cupdot U_k \quad\quad\quad V = V_1 \cupdot V_2 \dots \cupdot V_k
	\end{split}
\end{equation*}

be partitions of $`U,V`$, respectively, into $`k`$ non-empty subsets, where without loss of generality $`u_{n+1} \in U_k`$ and $`v_{n+1} \in V_k`$. Then, for every $`\pi, \tau \in S_{k-1}`$, consider the graph $`G \in C_n`$, whose edges are given by:

\forall i \in [k-1]:\ \forall u \in U_{\pi(i)}:\ N_G(u) = V_{\tau(1)} \cupdot \dots \cupdot V_{\tau(i)}

Recall that the number of partitions of $`n`$ elements into $`k`$ non-empty subsets is given by $`\stirlingII{n}{k}`$, the Stirling number of the second kind. Thus by the above construction, the cardinality of the set $`C_n`$ satisfies:

|C_n| = \sum_{k=1}^{n+1} \underbrace{\vphantom{\stirlingII{n+1}{k}^2}\left((k-1)!\right)^2}_{\text{Choosing }\pi, \tau} \ \cdot \underbrace{\stirlingII{n+1}{k}^2}_{\text{Partitioning }U,V}

Therefore:

\begin{equation*}
	\begin{split}
	|mon(BPM_n^\star)| \le |C_n| &= \sum_{k=1}^{n+1} \left((k-1)! \cdot \stirlingII{n+1}{k}\right)^2  \\
	&\le \left(\sum_{k=1}^{n+1} k! \cdot \stirlingII{n+1}{k}\right)^2 = (F_{n+1})^2  
	\end{split}
\end{equation*}

Where $`F_n`$ denotes the n’th Fubini number. We now use the upper bound

$`\forall n \ge 1:\ F_n < (n+1)^n`$, thereby concluding the proof. ◻

Proof. Let $`n > 1`$ and without loss of generality assume that $`n = 2k`$ where $`k \in \mathbb{N}^+`$. Let $`A`$,$`B`$ be two sets such that $`|A|=|B|=n`$. The lower bound follows by constructing a graph $`G=(A \cupdot B, E_G) \in HVC_n`$ where $`|E(G)| = n^2 - n/2(n/2 + 1)`$, such that $`\{H \supseteq G\} \subseteq HVC_n`$. First, partition each bipartition $`A`$,$`B`$ into two sets, as follows:

\begin{equation*}
	\begin{split}
	A = (X \cupdot Y): \\
	B = (U \cupdot V): \\
	\end{split}
	\quad
	\begin{split}
	X &= \{a_1, \dots, a_{k}\}\ \ \ , \\
	U &= \{b_1, \dots, b_{k-1}\}\ ,
	\end{split}
	\quad
	\begin{split}
	Y &= \{a_{k + 1}, \dots, a_{2k}\} \\
	V &= \{b_{k}, \dots, b_{2k}\}
	\end{split}
\end{equation*}

The edges of $`G`$ are formed by connecting all edges between $`X`$ and $`B`$, and all edges between $`Y`$ and $`U`$, thus: $`E(G) = (X \times B) \cup (Y \times U)`$. Observe that $`G \in HVC_n`$, since it can be covered by taking $`k`$ copies of $`K_{x,B}`$, one for each $`x \in X`$, and taking another $`k-1`$ copies of $`K_{A,u}`$, one for each $`u \in U`$.

Any missing edge $`(y,v) \notin E(G)`$ (where $`y \in Y`$ and $`v \in V`$) can be covered by $`K_{X \cupdot \{y\}, U \cupdot \{v\}}`$ (the complete bipartite graph connecting $`X \cupdot \{y\}`$ and $`U \cupdot \{v\}`$). Observe that $`K_{X \cupdot \{y\}, U \cupdot \{v\}} \in HV_n`$, since $`|X \cupdot U \cupdot \{y, v\} | = n+1`$. Thus $`\{H \supseteq G\} \subseteq HVC_n`$, as required. ◻

Proof. Let $`M`$ be the aforementioned communication matrix, and let $`\bar{M} = J - M`$, where $`J`$ is the all-ones matrix. The polynomial $`BPM^\star_n`$ induces an (at most) $`|mon(BPM^\star_n)|`$-rank decomposition of $`\bar{M}`$, since each monomial is a rank-1 matrix (see, e.g., ). However, $`rank(M) \le rank(\bar{M}) + 1`$, and by Corollary [counting_monomials_dual], $`|mon(BPM^\star_n)| < (n+2)^{2n + 2}`$, thus concluding the proof. ◻

Proof. Let $`T`$ be an OR-decision tree computing $`f`$ and denote $`d = depth(T)`$. Let $`\mathcal{P}`$ be the set of all root to 0-leaf paths in $`T`$. For any $`P \in \mathcal{P}`$, the indicator function for the path is given by the following multilinear polynomial:

\mathbbm{1}_{P}(x_1, \dots, x_n) = \left(\prod_{\lnot OR(S) \in P} \left(1-\sum_{\emptyset \ne S \subseteq [n]} (-1)^{|S|+1} \prod_{i \in S} x_i\right)\right) \left(\prod_{ OR(S) \in P} \left(\sum_{\emptyset \ne S \subseteq [n]} (-1)^{|S|+1} \prod_{i \in S} x_i\right) \right)

By the definition of the dual function, we can construct $`f^\star`$ by summing the indicators for all paths $`P \in \mathcal{P}`$, where the inputs to each indicator are the negated input bits:

\begin{equation*}
		\begin{split}
			f^\star(x_1, \dots, x_n) &= \sum_{P \in \mathcal{P}} \mathbbm{1}_{P}(1-x_1, \dots, 1-x_n) \\
			&= \sum_{P \in \mathcal{P}} \left(\prod_{OR(S) \in P} \left(1-\prod_{i \in S} x_i\right)\right) \left(\prod_{\lnot OR(S) \in P} \left(\prod_{i \in S} x_i\right) \right)
		\end{split}
\end{equation*}

Therefore, each path $`P`$ making $`k`$ right turns contributes at most $`2^k`$ monomials to $`f^\star`$. In a binary tree of depth $`d`$ there are at most $`{d \choose k}`$ paths making exactly $`k`$ right turns (i.e., by selecting the position in the path at which the right turns are made). Thus, we have:

\begin{equation*}
	\begin{split}
	|mon(f^\star)| \le \sum_{k=0}^{d} {d \choose k} 2^k = 3^d \qedhere
	\end{split}
\end{equation*}

 ◻

Proof. Let $`n > 2`$ and let $`A = \{a_1, \dots, a_n\}`$, $`B = \{b_1, \dots, b_n\}`$ be two sets. Denote:

A_{n-1} = \{a_1, \dots, a_{n-1}\},\quad B_{n-1} = \{b_1, \dots, b_{n-1}\}

For the case $`a^\star_G = 0`$, consider any totally ordered graph such that $`G \in (MC_n \setminus \{K_{n,n}\})`$. For example, let $`G = (A \cupdot B, E)`$ such that $`\forall i \in [n-1]:\ N_G(a_i) = B`$ and $`N_G(a_n) = \{b_1, b_2\}`$. $`G`$ is totally ordered, since $`N_G(a_1) \supseteq N_G(a_2) \supseteq \dots \supseteq N_G(a_n)`$. However, we also have that $`G \in MC_n`$, therefore by Lemma [mu_graphs_zero_coeff], $`a^\star_G = 0`$.

For the case $`a^\star_G = 1`$, consider any graph $`G \in HV_n`$. Observe that $`G`$ is both totally ordered and $`G \in HVC_n`$. Furthermore $`G`$ contains a single Hall violator graph (itself), and is therefore a minterm of $`BPM^\star_n`$, and so $`a^\star_G = 1`$.

Lastly, for the case $`a^\star_G = (n-2)^2`$, consider the graph $`G = K_{A_{n-1}, B_{n-1}}`$. Using Theorem [hetyei_conditions], the set of matching-covered graphs containing $`G`$, which we will denote by $`\mathcal{H}`$, can be partitioned into three sets $`\mathcal{H} = \mathcal{H}_1 \cupdot \mathcal{H}_2 \cupdot \mathcal{H}_3`$, as follows:

\begin{equation*}
		\begin{split}
			\mathcal{H}_1 &= \{G \cupdot \{(a_n, b_n)\}\} \\
			\mathcal{H}_2 &= \set{G \cupdot \{(a_n,b_n)\} \cupdot (U \times \{b_n\}) \cupdot (\{a_n\} \times V)}{\emptyset \ne U \subseteq A_{n-1},\ \emptyset \ne V \subseteq B_{n-1}} \\
			\mathcal{H}_3 &= \set{G \cupdot (U \times \{b_n\}) \cupdot (\{a_n\} \times V)}{U \subseteq A_{n-1},\ V \subseteq B_{n-1},\ |U| \ge 2,\ |V| \ge 2}
		\end{split}
\end{equation*}

By Lemma [upper_sum_mobius], the dual coefficient of $`G`$ is given by:

\begin{equation*}
		\begin{split} 
			a^\star_G &= (-1)^{|E(G)| + 1} \sum_{\substack{H \supseteq G\\H \in MC_n}} (-1)^{\chi(H)} = -\sum_{H \in \mathcal{H}} (-1)^{|E(H) \setminus E(G)| + |C(H)|} 
		\end{split}
\end{equation*}

For the single graph $`H \in \mathcal{H}_1`$, $`|E(H) \setminus E(G)| = 1`$, $`|C(H)| = 2`$, thus contributing $`1`$ to the sum. For each $`H \in \mathcal{H}_2`$, $`|C(H)|=1`$, thus $`\mathcal{H}_2`$’s contribution to the sum is:

\sum_{i=1}^{n-1} \sum_{j=1}^{n-1} {n-1 \choose i} {n-1 \choose j} (-1)^{i+j+1} = -1

Lastly, for each $`H \in \mathcal{H}_3`$, $`|C(H)|=1`$. Thus $`\mathcal{H}_3`$’s contribution to the sum is:

\sum_{i=2}^{n-1} \sum_{j=2}^{n-1} {n-1 \choose i} {n-1 \choose j} (-1)^{i+j} = (n-2)^2

Summing up all the contributions, we get $`a^\star_G = (n-2)^2`$, thus concluding the proof. ◻

Proof. The DNF formula representing the graph cover function is:

\varphi = \bigvee_{H \in \mathcal{H}}\ \bigwedge_{(i,j) \in E(H)}\ x_{i, j}

Since each $`x_{i,j} \in \{0,1\}`$, we have $`\forall k \in \mathbb{N}^+`$, $`x_{i,j}^k = x_{i,j}`$. Therefore, arithmetizing the formula yields the following polynomial representation:

\begin{equation*}
		\begin{split}
			f_{\mathcal{H}}(x_{1,1}, \dots, x_{n,m}) &= 1 - \underset{H \in \mathcal{H}}{\prod}(1 - \underset{(i,j) \in E(H)}{\prod} x_{i, j}) \\
			&= \sum_{\emptyset \ne S \subseteq \mathcal{H}} (-1)^{|S|+1} \prod_{G \in S} \ \prod_{(i,j) \in E(G)} x_{i, j} \\
			&= \sum_{G \in \mathcal{C}(\mathcal{H})} \left(\sum_{\substack{\emptyset \ne S \subseteq \mathcal{H}\\ \bar{E}(S)=E(G)}}(-1)^{|S|+1}\right) \prod_{(i,j) \in E(G)} x_{i, j} \qedhere
		\end{split}
\end{equation*}

 ◻

Proof. The subset relation over the edges is reflexive, transitive and anti-symmetric, thus $`\mathcal{P}`$ is a poset. Furthermore, $`\mathcal{P}`$ is bounded, since $`\hat{0} = (V(\mathcal{H}), \emptyset)`$ and $`\hat{1} = (V(\mathcal{H}), \bar{E}(\mathcal{H}))`$. It remains to show that $`\forall G_1,G_2 \in \mathcal{C}(\mathcal{H})`$ there exists a join (unique supremum) and a meet (unique infimum).

Let $`G_1,G_2 \in \mathcal{C}(\mathcal{H})`$. The meet and join of $`G_1`$ and $`G_2`$ are given by:

\begin{equation*}
	\begin{split}
	E(G_1 \vee G_2) &= \bigcup_{\substack{H \in \mathcal{H}\\(H \subseteq G_1) \lor (H \subseteq G_2)}} E(H) = E(G_1) \cup E(G_2) \\
	E(G_1 \wedge G_2) &= \bigcup_{\substack{H \in \mathcal{H}\\(H \subseteq G_1) \land (H \subseteq G_2)}} E(H)
	\end{split}
\end{equation*}

For the join operator, let $`G \eqdef G_1 \vee G_2`$. By construction, $`G_1 \subseteq G`$ and $`G_2 \subseteq G`$, therefore $`G`$ is a supremum. Assume towards a contradiction that there exists another supremum $`\hat{G} \ne G`$ such that $`G \not\subseteq \hat{G}`$. Let $`x \in E(G) \setminus E(\hat{G})`$. Without loss of generality, assume $`x \in E(G_1)`$. Then $`x \in E(G_1)`$ and $`x \notin E(\hat{G})`$ therefore $`G_1 \not\subseteq \hat{G}`$, in contradiction to the fact that $`\hat{G}`$ is a supremum.

For the meet operator, let $`G \eqdef G_1 \wedge G_2`$. By construction, $`G \subseteq G_1`$ and $`G \subseteq G_2`$, therefore $`G`$ is an infimum. Assume towards a contradiction that there exists another infimum $`\hat{G} \ne G`$ such that $`G \not\supseteq \hat{G}`$. Let $`x \in E(\hat{G}) \setminus E(G)`$. Since $`\hat{G} \in \mathcal{C}(\mathcal{H})`$, there exists $`H_x \in \mathcal{H}`$ such that $`H_x \subseteq \hat{G}`$, $`x \in E(H_x)`$. However, $`\hat{G}`$ is an infimum, thus $`H_x \subseteq \hat{G} \subseteq G_1`$ and $`H_x \subseteq \hat{G} \subseteq G_2`$, thus by construction $`H_x \subseteq G`$ and $`x \in E(G)`$, a contradiction. ◻

Proof. Let $`f`$ be the polynomial $`f(x_{1,1}, \dots, x_{n,m}) = \sum_{G \in \mathcal{C}(\mathcal{H})} {-\mu_P(\hat{0}, G) \cdot \prod_{(i,j) \in E(G)} x_{i, j}}`$, and let $`H \subseteq K_{n,m}`$ be a graph. Denote by $`H'`$ the union of all graphs $`G \in \mathcal{C}(\mathcal{H})`$ such that $`G \subseteq H`$. We now show that $`f`$ agrees with $`f_{\mathcal{H}}`$ on all inputs, and deduce the identity by the uniqueness of the representing polynomial. If $`H' = \hat{0}`$, then indeed $`f(H) = 0`$ as required. Otherwise, we have:

\begin{equation*}
		\begin{split}
			f(H) &= \sum_{G \in \mathcal{C}(\mathcal{H})} {-\mu_P(\hat{0}, G) \cdot \mathbbm{1}\{G \subseteq H\}} \\
				&= \sum_{\substack{
						\hat{0} \subset G \subseteq H' \\ 
						G \in \mathcal{C}(\mathcal{H})}} -\mu_P(\hat{0}, G) \\
				&= \sum_{\substack{
						\hat{0} \subseteq G \subseteq H' \\ 
						G \in \mathcal{C}(\mathcal{H})}} -\mu_P(\hat{0}, G) + \mu_P(\hat{0}, \hat{0})
		\end{split}
\end{equation*}

And by the definition of the Möbius function, $`\mu_P(\hat{0}, \hat{0}) = 1`$ and $`\displaystyle\sum_{\substack{ \hat{0} \subseteq G \subseteq H' \\ G \in \mathcal{C}(\mathcal{H})}} -\mu_P(\hat{0}, G) = 0`$ ◻

Proof. Let $`n > 1`$ and let $`A,B`$ be two sets, where $`|A|=|B|=n`$. Denote by $`G(n,n,p)`$ the distribution over balanced bipartite graphs of order $`2n`$, in which each edge appears i.i.d with probability $`p`$. Recall that by Theorem [hetyei_conditions], $`G=(A \cupdot B, E) \subseteq K_{n,n}`$ is elementary if and only if $`\forall a \in A`$, $`\forall b \in B`$: $`G - a - b`$ has a perfect matching. By the union bound:

\begin{equation*}
	\begin{split}
	\Pr_{G \sim G(n,n,0.5)}\left[\text{G is not elementary}\right] &= \Pr_{G \sim G(n,n,0.5)}\left[\exists a \in A,\ b \in B:\ G - a - b\text{ has no perfect matching}\right] \\
	&\le n^2 \cdot \Pr_{G \sim G(n-1,n-1,0.5)}\left[\text{G has no perfect matching}\right]
	\end{split}
\end{equation*}

By Hall’s Theorem, $`G`$ has a perfect matching if and only if $`\forall X \subseteq A`$: $`|N(X)| \ge |X|`$. Thus $`G`$ has no perfect matching if and only if there exist two sets $`S \subseteq A`$, $`T \subseteq B`$ such that $`|S| + |T| = n+1`$, and none of the edges in $`S \times T`$ appear in $`G`$. Using the union bound again:

\begin{equation*}
	\begin{split}
	\Pr_{G \sim G(n,n,0.5)}\left[\text{G has no perfect matching}\right] \le \sum_{k=1}^{n} {n \choose k} {n \choose k-1} 2^{-k(n-k+1)} \le \frac{n^2}{2^n}
	\end{split}
\end{equation*}

Thus:

\begin{equation*}
	\begin{split}
	\Pr_{G \sim G(n,n,0.5)}\left[G \in MC_n\right] \ge \Pr_{G \sim G(n,n,0.5)}\left[\text{G is elementary}\right] \ge 1 - \frac{2n^4}{2^n} \qedhere
	\end{split}
\end{equation*}

 ◻

Proof. Recall that every face of a polytope is also a polytope. Thus, for any face $`x \in F(Q)`$, we denote its face lattice by $`\mathcal{P}_x`$. The lattice $`\mathcal{P}_x`$ consists of all faces $`y \in F(Q)`$ where $`y \le_P x`$, thus $`\mathcal{P}_x`$ is a sub-lattice of $`\mathcal{P}`$ and $`\mu_{\mathcal{P}}(\hat{0}, x) = \mu_{\mathcal{P}_x}(\hat{0}, x)`$. By the definition of the face lattice, the rank of any face $`y \in F(x)`$ in $`\mathcal{P}_x`$ is given by $`rk(x) = dim(x) + 1`$, and thus agrees with its rank in $`\mathcal{P}`$. Consequently, we denote the rank of any face by $`rk(\cdot)`$.

The proof proceeds by induction. If $`x = \hat{0}`$, the equality follows from the definition of the Möbius function. Otherwise, let $`x \in F(Q)`$, where $`k \eqdef rk(x) \ge 1`$. By the definition of the Möbius function and using the induction hypothesis:

\mu_{\mathcal{P}_x}(\hat{0}, x) =  -\sum_{\substack{y \in F(x) \cupdot \{\hat{0}\} \\ y \neq x}} \mu_{\mathcal{P}_x}(\hat{0}, y) = -\sum_{\substack{y \in F(x) \cupdot \{\hat{0}\} \\ y \neq x}} (-1)^{rk(y)}

Since $`x`$ is a Polytope of dimension $`k-1`$, then by the Euler-Poincaré Formula for Polytopes (see, e.g., ) we have:

\begin{equation*}
	\begin{split}
	1 &= \sum_{j = 0}^{k-1} (-1)^j |\set{y \in F(x)}{dim(y) = j}| \\
	&= \sum_{x \ne y \in F(x)} (-1)^{rk(y)-1} + (-1)^{k-1} \\
	&= -\sum_{\substack{y \in F(x) \cupdot \{\hat{0}\} \\ y \neq x}} (-1)^{rk(y)} + (-1)^{k-1} + 1 = \mu_{\mathcal{P}_x}(\hat{0}, x) + (-1)^{k-1} + 1 \qedhere
	\end{split}
\end{equation*}

 ◻

Proof. Let $`G`$ be an elementary graph and let $`G \ne H \in MC(G)`$. If $`H`$ is elementary, then $`|E(H)| < |E(G)|`$ and $`|C(H)| = |C(G)| = 1`$, thus $`\chi(H) < \chi(G)`$, as required.

Otherwise, observe that the connected components of $`H`$ are joined by edges in $`G`$, since $`G`$ is elementary and thus connected. Furthermore, we claim that every component of $`H`$ must be adjacent to at least 2 edges in $`G`$, one connected to a left vertex of the component, and another to a right vertex. Assume toward a contradiction that this is not the case, then there exists a component $`C \in C(H)`$ which is only adjacent to a single edge $`e \in E(G)`$. Since $`G`$ is elementary, there exists some perfect matching involving $`e`$ (i.e., the edge $`e`$ is allowed). However, upon selecting the edge $`e`$, the component $`C`$ becomes unbalanced, and therefore the perfect matching cannot be extended over $`C`$, a contradiction.

Thus, since each component of $`G`$ has at least two adjacent edges in $`G`$, we have that $`|E(H)| + |C(H)| \le |E(G)|`$ (i.e., if the adjacent edges form a cycle over the components $`C(H)`$). Thus:

\begin{equation*}
		\begin{split}
			\chi(H) &= |E(H)| - |V(H)| + |C(H)| \le |E(G)| - |V(G|| < \chi(G) \qedhere 
		\end{split}
\end{equation*}

 ◻

Proof. If $`H`$ is elementary, the proof follows from Lemma [order_of_subgraphs_of_elementary]. Otherwise, the proof follows from the additivity of $`\chi`$, by applying Lemma [order_of_subgraphs_of_elementary] to each connected component in which $`G`$ and $`H`$ differ. ◻

Proof. Let $`G \in MC_n`$, $`G \notin PM(K_{n,n})`$. Since $`G`$ is not a perfect matching, there exists a component $`C \in C(G)`$ such that $`C \ne K_2`$. $`C`$ is elementary, and therefore there exists a bipartite ear decomposition: $`C = e + P_1 + \dots + P_k`$. Let $`C' = e + P_1 + \dots P_{k-1}`$, and observe that since $`C'`$ has a bipartite ear decomposition, it too is elementary.

If $`P_k`$ is a single edge, then we construct $`H`$ by taking $`G`$, and replacing the component $`C`$ with $`C'`$. Observe that $`H \in MC_n`$, and furthermore $`|C(H)| = |C(G)|`$ and $`|E(H)| = |E(G)| - 1`$. Thus $`\chi(H) = \chi(G) - 1`$, as required.

Otherwise, $`P_k`$ is an ear $`(v_1, u_1, \dots, v_t, u_t)`$. In this case, we construct $`H`$ by taking $`G`$, replacing $`C`$ with $`C'`$, and replacing the ear $`P_k`$ with the edges $`(v_2, u_1), \dots, (v_t, u_{t-1})`$. Once again, $`H \in MC_n`$ (since all its components are elementary). Furthermore $`|C(H)| = |C(G)| + t - 1`$ and $`|E(H)| = |E(G)| - t`$, and thus $`\chi(H) = \chi(G) - 1`$. ◻

Proof. Let $`\mathcal{P}=(MC_n \cup \{\hat{0}\}, \subseteq)`$ be the lattice of matching-covered graphs, and let $`B_n`$ be the Birkhoff Polytope. Since $`BPM_n`$ is a graph cover function for the set $`PM(K_{n,n})`$, then by Proposition [graph_cover_polynomial_mobius] we have:

BPM_n(x_{1,1}, \dots, x_{n,n}) = \sum_{G \in MC_n} -\mu_P(\hat{0}, G) \cdot \prod_{(i,j) \in E(G)} x_{i, j}

By Theorem [elementary_faces_of_birkhoff], $`\mathcal{P}`$ is isomorphic to the face lattice of $`B_n`$, and thus by Corollary [rank_in_mu_lattice] and Lemma [mobius_of_face_lattice], we get:

\forall G \in MC_n:\ \mu_P(\hat{0}, G) = (-1)^{rk(G)} = (-1)^{\chi(G)+1} \qedhere

 ◻

Proof. The bound follows immediately from Proposition [number_of_mu_graphs] and Theorem [bpm_poly]. ◻

Proof. We show that for any Boolean function $`f`$, $`D^{XOR}(f) \ge deg_2(f)`$, where $`deg_2(f)`$ is the degree of the polynomial representing $`f`$ over $`\mathbb{F}_2`$. Thus in particular $`D^{XOR}(BPM_n) = n^2`$. Let $`f: \{0,1\}^n \rightarrow \{0,1\}`$ be a Boolean function and let $`T`$ be a $`XOR`$ decision tree computing $`f`$. Let $`\mathcal{P}`$ be the set of all root to 1-leaf paths in $`T`$. For any path $`P \in \mathcal{P}`$ we construct the indicator over the path, denoted $`\mathbbm{1}_P(x_1, \dots, x_n)`$, by taking the product over any parity along the path (taking the parity itself for any right turn, and adding 1 to the term for any left turn). Observe that $`f(x_1, \dots, x_n) = \sum_{P \in \mathcal{P}} \mathbbm{1}_P(x_1, \dots, x_n)`$, therefore $`deg_2(f) \le \max_{P \in \mathcal{P}} deg_2(\mathbbm{1}_P(x_1, \dots, x_n)) \le depth(T)`$, where the last inequality follows since parities over $`\mathbb{F}_2`$ are linear functionals. ◻

Proof. For any Boolean function $`f`$, $`|\set{x \in \{0,1\}^n}{f(x)=1}| \equiv 1 \pmod 2`$ if and only if the polynomial representing $`f`$ over $`\mathbb{F}_2`$ has full degree. Thus the number of graphs containing a perfect matching is odd. Let $`H \in MC_n`$. Clearly $`H`$ has a perfect matching, therefore:

\begin{equation*}
		\begin{split}
			1 = BPM_n(H) &= \sum_{G \in MC_n} {(-1)^{\chi(G)} \cdot \mathbbm{1}\{G \subseteq H\}} \\
			&= \sum_{G \in MC(H)} {(-1)^{\chi(G)}} \end{split}
\end{equation*}

In particular $`K_{n,n} \in MC_n`$, thus:

\begin{equation*}
		\begin{split}
			1 \equiv BPM_n(K_{n,n}) \pmod 2 \equiv |MC_n| \pmod 2 \qedhere
		\end{split}
\end{equation*}

 ◻

Proof. Let $`T`$ be an AND decision tree computing $`f`$ and denote $`d = depth(T)`$. Let $`\mathcal{P}`$ be the set of all root to 1-leaf paths in $`T`$. For any $`P \in \mathcal{P}`$, construct the indicator function for the path as follows:

\mathbbm{1}_{P}(x_1, \dots, x_n) = \left(\prod_{\lnot AND(S) \in P} \left(1-\prod_{i \in S} x_i\right)\right) \left(\prod_{ AND(S) \in P} \left(\prod_{i \in S} x_i\right) \right)

Notice that the multilinear polynomial of each indicator function of a path $`P`$ making $`k`$ left turns has at most $`2^k`$ monomials. Furthermore, in a binary tree of depth $`d`$ there are at most $`{d \choose k}`$ paths making exactly $`k`$ left turns (i.e., by selecting the position in the path at which the left turns are made). Finally, observe that $`f(x_1, \dots, x_n) = \sum_{P \in \mathcal{P}} \mathbbm{1}_{P}(x_1, \dots, x_n)`$. Thus by the uniqueness of the multilinear polynomial representing $`f`$, we have:

\begin{equation*}
	\begin{split}
	|mon(f)| \le \sum_{P \in \mathcal{P}} |mon(\mathbbm{1}_{P})| \le \sum_{k=0}^{d} {d \choose k} 2^k = 3^d \qedhere
	\end{split}
\end{equation*}

 ◻

Proof. If $`G`$ is elementary, then any graph $`H \supseteq G`$ is also elementary, as its ear decomposition is that of $`G`$, followed by adding single-edge ears for each edge in $`E(H) \setminus E(G)`$. Thus by Theorem [bpm_poly] and Lemma [convert_to_fourier]:

\begin{equation*}
	\begin{split}
	\widehat{BPM^G_n} &= (-1)^{|E(G)| - 1} \sum_{\substack{H \supseteq G\\ H \in MC_n}} \frac{(-1)^{\chi(H)}}{2^{|E(H)| - 1}} \\
	&= \sum_{t=0}^{n^2 - |E(G)|} {n^2 - |E(G)| \choose t} \frac{(-1)^t}{2^{t + |E(G)| - 1}} = 2^{-n^2 + 1} \qedhere
	\end{split}
\end{equation*}

 ◻

Proof. By Proposition [number_of_mu_graphs], the number of elementary graphs is at least:

|\set{G \subseteq K_{n,n}}{\text{G is elementary}}| \ge \left(1 - \frac{2n^4}{2^n}\right) \cdot 2^{n^2} = \left(1 - o_n(1)\right)\cdot 2^{n^2} \qedhere

 ◻

Proof. By Theorem [bpm_poly] and Lemma [convert_to_fourier], the Fourier coefficient of the empty set in $`BPM_n`$ is:

\widehat{BPM^\emptyset_n} = 1 - \sum_{G \in MC_n} \frac{(-1)^{\chi(G)}}{2^{|E(G)|-1}}

Furthermore, for any Boolean function $`f: \{1,-1\}^n \rightarrow \{1,-1\}`$:

\hat{f}_\emptyset = \mathop{\mathbb{E}}_{x \sim \{1,-1 \}^n}\left[f(x)\right] = \mathop{\mathbb{E}}_{x \sim \{1,-1\}^n}\left[-2 \cdot \mathbbm{1}\{f(x)=-1\} + 1\right] = -2 \cdot \Pr_{x \sim \{1,-1\}^n}\left[f(x) = -1\right] + 1

And the equality now follows by rearranging. ◻