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The Spectrum of the Singular Values of Z-Shaped Graph Matrices

Wenjun Cai, Aaron Potechin

TL;DR

This work determines the limiting spectrum of singular values for Z-shaped graph matrices under dense random inputs, and extends the analysis to m-layer Z-shaped shapes. The authors employ the trace power method and a combinatorial constraint-graph framework to reduce moment calculations to counting dominant constraint graphs, which follow Fuss-Catalan type recurrences. They prove that the even moments converge to the Fuss-Catalan sequence C'_k = 1/(2k+1) binom{3k}{k} and identify the limiting spectral density g_{\alpha_Z} with explicit support and expression on [0,a], where a = 3√3/2. The paper also develops a generalization to m-layer Z-shapes, with moments governed by D(m,k) and a scalable recurrence, laying groundwork for further spectral characterizations of broader graph-matrix families.

Abstract

Graph matrices are a type of matrix which has played a crucial role in analyzing the sum of squares hierarchy on average case problems. However, except for rough norm bounds, little is known about graph matrices. In this paper, we take a step towards better understanding graph matrices by determining the limiting distribution of the spectrum of the singular values of Z-shaped graph matrices. We then give a partial generalization of our results for $m$-layer Z-shaped graph matrices.

The Spectrum of the Singular Values of Z-Shaped Graph Matrices

TL;DR

This work determines the limiting spectrum of singular values for Z-shaped graph matrices under dense random inputs, and extends the analysis to m-layer Z-shaped shapes. The authors employ the trace power method and a combinatorial constraint-graph framework to reduce moment calculations to counting dominant constraint graphs, which follow Fuss-Catalan type recurrences. They prove that the even moments converge to the Fuss-Catalan sequence C'_k = 1/(2k+1) binom{3k}{k} and identify the limiting spectral density g_{\alpha_Z} with explicit support and expression on [0,a], where a = 3√3/2. The paper also develops a generalization to m-layer Z-shapes, with moments governed by D(m,k) and a scalable recurrence, laying groundwork for further spectral characterizations of broader graph-matrix families.

Abstract

Graph matrices are a type of matrix which has played a crucial role in analyzing the sum of squares hierarchy on average case problems. However, except for rough norm bounds, little is known about graph matrices. In this paper, we take a step towards better understanding graph matrices by determining the limiting distribution of the spectrum of the singular values of Z-shaped graph matrices. We then give a partial generalization of our results for -layer Z-shaped graph matrices.

Paper Structure

This paper contains 12 sections, 7 theorems, 4 equations, 2 figures.

Table of Contents

  1. Introduction
  2. Definitions
  3. Graph matrices
  4. Spectrum of the singular values of a matrix
  5. Generalized Catalan numbers
  6. Our results
  7. Proof overview and paper outline
  8. Techniques
  9. Convergence of matrix spectra
  10. Proving Theorem \ref{['thm:precisemainresult']} via the trace power method
  11. Analyzing tr((Ma MaT)k) using constraint graphs
  12. Expressing tr((Ma MaT)k) using the graph H(aZ,2k)$ We first show how to express$tr$$M_α_ZM_α_Z^T$^k$$in terms of maps$ϕ$from the vertices of a graph$H(α_Z,2k)$to$[n]$. We will then use constraint graphs to visualize these maps. Given a shape $\alpha$ and a $k \in \mathbb{N}$, we define $H(\alpha,2k)$ to be the multi-graph which is formed as follows: We take $k$ copies $\alpha_1,\ldots,\alpha_{k}$ of $\alpha$ and $k$ copies $\alpha^{T}_1,\ldots,\alpha^{T}_k$ of $\alpha^{T}$, where $\alpha^{T}$ is the shape obtained from $\alpha$ by switching the role of $U_{\alpha}$ and $V_{\alpha}$.We compose the shapes $\alpha_1,\alpha^{T}_1,\ldots,\alpha_k,\alpha^{T}_k$ by setting $V_{\alpha_{i}} = U_{\alpha^{T}_i}$ for all $i \in [k]$ and setting $V_{\alpha^{T}_i} = U_{\alpha_{i+1}}$ for all $i \in [k-1]$. We then set $V_{\alpha^{T}_k} = U_{\alpha_{1}}$. We define $V(\alpha,2k) = V(H(\alpha,2k))$ and we define $E(\alpha,2k) = E(H(\alpha,2k))$. See \ref{['fig:H-a-2k']} for an illustration. $H(\alpha,2k)$ is defined as a multi-graph because edges will be duplicated if $U_{\alpha}$ or $V_{\alpha}$ contains one or more edges or $k = 1$ and $\abs{E$H(,2k)$} > 0$. If $U_{\alpha}$ and $V_{\alpha}$ do not contain any edges and $k > 1$ then $H(\alpha,2k)$ will be a graph. Illustration of \ref{['def:copies']}: $\alpha$ and $H(\alpha,2k)$ where $k=4$. We say that a map $\phi:V$,2k$\to [n]$ is piecewise injective if $\phi$ is injective on each piece $V(\alpha_i)$ and each piece $V$^{T}_i$$ for all $i \in [k]$. In other words, $\phi(u) \neq \phi(v)$ whenever $u,v \in V$_i$$ for some $i \in [k]$ or $u,v \in V$^T_i$$ for some $i \in [k]$. As observed in AMP20, with these definitions$E[tr((M_αM_α^T)^k)]$can be reexpressed as follows For all shapes $\alpha$ and all $k \in \mathbb{N}$, $\mathbb{E}\left[\mathop{\mathrm{tr}}\nolimits\left((M_{\alpha}M_{\alpha}^T)^k\right)\right] = \sum_{\substack{\phi: V(\alpha,2k) \to [n]: \\ \phi \text{ is piecewise injective}}}{\mathbb{E}\chi_{\phi$E(\alpha,2k)$}(G)}\,.$ To analyze this expression, we use constraint graphs$C$where an edge${u,v} ∈ E(C)$represents the fact that$ϕ(u) = ϕ(v)$. We define a constraint graph $C$ on a set of vertices $V$ to be an acyclic graph with vertices $V(C) = V$. Given a set of vertices $V$ and a constraint graph $C$ on $V$, we say that two vertices $u,v \in V$ are constrained together by $C$ if $u = v$ or there is a path from $u$ to $v$ in $C$. We denote this by $u \longleftrightarrow_{C} v$. When $C$ is clear from context, we just write $u \longleftrightarrow v$. Similarly, we say that two edges $e = \{u,v\}$ and $e' = \{u',v'\}$ are constrained together by $C$ if either $u \longleftrightarrow_{C} u'$ and $v \longleftrightarrow_{C} v'$ or $u \longleftrightarrow_{C} v'$ and $v \longleftrightarrow_{C} u'$. We denote this by $e \longleftrightarrow_{C} e'$. When $C$ is clear from context, we just write $e \longleftrightarrow e'$. We say that two constraint graphs $C,C'$ are equivalent if for any pair of vertices $u,v \in V$, $u \longleftrightarrow_C v$ if and only if $u \longleftrightarrow_{C'} v$. We denote this by $C \equiv C'$. Given a multi-graph $H$, a constraint graph $C$ on $V(H)$, and a set of vertices $V \subseteq V(H)$, we define the induced constraint graph $C'$ on $V$ to be the constraint graph such that $V(C') = V$ and for all $u,v \in V$, $u \longleftrightarrow v$ in $C'$ if and only if $u \longleftrightarrow v$ in $C$. Each map$ϕ: V(α,2k) [n]$has an associated constraint graph$C_ϕ$. Given a map $\phi: V(\alpha,2k) \to [n]$, we define $C_{\phi}$ to be the constraint graph such that $u \longleftrightarrow_{C_{\phi}} v$ if and only if $\phi(u) = \phi(v)$. Equivalently, $C_{\phi}$ consists of a disjoint union of trees where $u$ and $v$ are in the same tree if and only if $\phi(u) = \phi(v)$ (note that all such constraint graphs are equivalent). Note that if$ϕ,ϕ'$are two maps with the same constraint graph then$ϕ$and$ϕ'$are the same up to permuting the indices in$[n]$. We define $\mathcal{C}_{(\alpha,2k)} = \left\{C_{\phi}: \text{The map } \phi: V(\alpha,2k) \to [n] \text{ is piecewise injective}\right\}$ to be the set of all possible constraint graphs on $V(\alpha,2k)$ which come from a piecewise injective map $\phi: V(\alpha,2k) \to [n]$. Given a constraint graph $C \in \mathcal{C}_{(\alpha,2k)}$, we make the following definitions: We define $N(C) = \abs{\left\{\phi: V(\alpha,2k) \to [n]: \phi \text{ is piecewise injective}, C_{\phi} \equiv C\right\}}$.We define $\mathop{\mathrm{val}}\nolimits(C) = \mathbb{E}[\chi_{\phi$E(,2k)$}(G)]$ where $\phi: V(\alpha,2k) \to [n]$ is any piecewise injective map such that $C_{\phi} \equiv C$. We say that a constraint graph $C \in \mathcal{C}_{(\alpha,2k)}$ is nonzero-valued if $\mathop{\mathrm{val}}\nolimits(C)\neq 0$. We now define a simple shape which we call the \emph{line shape} and use it to give examples for \ref{['defn:constraint-graph-on-H']}. Let $\alpha_{0}$ be the bipartite shape with vertices $V(\alpha_{0})=\{u,v\}$ and a single edge $\{u,v\}$ with distinguished tuples of vertices $U_{\alpha_{0}}=(u)$ and $V_{\alpha_{0}}=(v)$. We call $\alpha_0$ the line shape. Illustration of \ref{['defn:line-shape']} and \ref{['defn:line-shape-H-label']}: $\alpha_0$ is the line shape. $H(\alpha_0,2k)$ is a cycle of length $2k$. Here $k=4$. Let $H(\alpha_0,2k)$ be the multi-graph given by \ref{['def:copies']}. We label the vertices of $H(\alpha_0,2k)$ in two different but closely related ways. Letting $(\alpha_0)_1,\ldots,(\alpha_0)_k$ be the copies of $\alpha_0$ and letting $$_0^T$_1,\ldots,$_0^T$_k$ be the copies of $\alpha_0^T$, we label the vertices of $V$(_0)_i$$ as $\{u_i,v_i\}$ and the vertices of $V($_0^T$_i)$ as $\{a_i,b_{i+1}\}$ where we take $a_{k+1} = a_1$. This gives the labeling $a_1,b_1,a_2,b_2,\ldots,a_k,b_k$ for the vertices of $H(\alpha_0,2k)$.For convenience, we also label the vertices of $H(\alpha_0,2k)$ as $x_1,\ldots,x_{2k}$. In other words, for all $j \in [k]$, we take $x_{2j-1} = a_j$ and $x_{2j} = b_j$. Illustration of \ref{['eg:val-C']}. Let $\alpha$ be the line shape with $U(\alpha) = \{u\}$, $V(\alpha) = \{v\}$ and $E(\alpha) = \{\{u,v\}\}$ and consider its corresponding graph $H(\alpha,2k)$. Let $k=4$. If $E(C) = \left\{\{x_2,x_8\}, \{x_3,x_5\}, \{x_5, x_7\}\right\}$ then $\mathop{\mathrm{val}}\nolimits(C) = 1$ and $N(C) = \dfrac{n!}{(n-5)!}$ since there are $8-\abs{E(C)} = 8-3 = 5$ distinct vertices whose values $\phi(v)$ can be chosen from $[n]$.If $E(C) = \left\{\{x_2,x_8\}, \{x_3,x_5\}, \{x_4,x_6\}\right\}$ then $\mathop{\mathrm{val}}\nolimits(C) = 0$ and $N(C) = \dfrac{n!}{(n-5)!}$. As observed in AMP20, with these definitions$E[tr($M_{\alpha}M_{\alpha}^T$^k)]$can be re-expressed as follows. For all shapes $\alpha$ and all $k \in \mathbb{N}$, $\mathbb{E}\left[\mathop{\mathrm{tr}}\nolimits\left((M_{\alpha}M_{\alpha}^T)^k\right)\right] = \sum_{C \in \mathcal{C}_{(\alpha,2k)}}{N(C)\mathop{\mathrm{val}}\nolimits(C)}.$ Given a constraint graph$C ∈ C_(α,2k)$, it can be useful to consider the graph obtained by starting with$H(α,2k)$and contracting the edges of$C$. Illustration of \ref{['defn:H/C']}: $H(\alpha,2k)/C$ Given a multi-graph $H$ and a constraint graph $C$ on $V(H)$, we define $H/C$ to be the graph obtained from $H$ by contracting the edges in $C$. More precisely, we can define $H/C$ using the following map. Letting $V(H) = v_1,\ldots,v_m$ be the vertices of $H$, define $\pi: V(H) \to V(H)$ to be the map such that $\pi(v_i) = v_{\min\{j \in [m]:v_i \longleftrightarrow v_j \text{ or } i = j\}}$. We define $H/C$ to be the multi-graph with vertices $V(H/C) = \{v_i: \pi(v_i) = v_i\}$ and edges $E(H/C) = \left\{\{\pi(v_i),\pi(v_j)\}: \{v_i,v_j\} \in E(G)\right\}$. See \ref{['fig:H-constraint-graph-apply']} for an illustration. For technical reasons, the following definition is convenient. Let $H$ be a multi-graph and let $C$ be a constraint graph on $V(H)$. Given two vertices $u,v \in V(H)$, we say that the edge $e = \{u,v\}$ appears with odd multiplicity in $H/C$ if $\{\pi(u),\pi(v)\}$ appears with odd multiplicity in $H/C$. Otherwise, we say that $e = \{u,v\}$ appears with even multiplicity in $H/C$. Note that \ref{['def:appearingmultiplicity']} applies even if${u,v} ∉ E(H)$or$u,v ∉ V(H/C)$. We now analyze$val(C)$and$N(C)$for constraint graphs$C ∈ C_(α,2k)$. We observe that the sum$ ∑_C ∈ C_(α,2k)N(C)val(C)$is dominated by nonzero-valued constraint graphs with the minimum possible number of edges. For every constraint graph $C \in \mathcal{C}_{(\alpha,2k)}$, $\mathop{\mathrm{val}}\nolimits(C) = 1$ if every edge in $\phi$E(,2k)$$ appears an even number of times and $\mathop{\mathrm{val}}\nolimits(C) = 0$ otherwise (where $\phi: V(\alpha,2k) \to [n]$ is any piecewise injective map such that $C_{\phi} \equiv C$). Equivalently, $\mathop{\mathrm{val}}\nolimits(C)=1$ if every edge in $H(\alpha,2k)/C$ appears an even number of times and $\mathop{\mathrm{val}}\nolimits(C) = 0$ otherwise. For every constraint graph $C \in \mathcal{C}_{(\alpha,2k)}$, $N(C) = \frac{n!}{$n - {V(,2k)} + {E(C)}$!}$. Observe that choosing a piecewise injective map $\phi$ such that $C(\phi) \equiv C$ is equivalent to choosing a distinct element of $[n]$ for each of the $\abs{V(\alpha,2k)} - \abs{E(C)}$ connected components of $C$. Since the number of constraint graphs in$C_(α,2k)$depends on$k$but not on$n$, as$n ∞$we only care about the nonzero-valued constraint graphs in$C_(α,2k)$which have the minimum possible number of edges. We call such constraint graphs dominant constraint graphs. we say a constraint graph $C \in \mathcal{C}_{(\alpha,2k)}$ is a dominant constraint graph if $\mathop{\mathrm{val}}\nolimits(C) \neq 0$ and $\abs{E(C)}=\min\left\{\abs{E(C')}: C' \in \mathcal{C}_{(\alpha,2k)}, \mathop{\mathrm{val}}\nolimits(C') \neq 0\right\}$. More generally, given a multi-graph $H$ and a constraint graph $C$ on $V(H)$, we say that $C$ is a dominant constraint graph on $H$ if every edge of $H/C$ appears with even multiplicity and $C$ has the minimum possible number of edges (i.e., for all constraint graphs $C'$ on $V(H)$ such that $\abs{E(C')} < \abs{E(C)}$, In order to determine the order of magnitude of$ E[tr($M_{\alpha}M_{\alpha}^T$^k)] = ∑_C ∈ C_(α,2k)N(C)val(C)$, we need to know how many edges dominant constraint graphs in$C_(α,2k)$have. It turns out that the number of edges in dominant constraint graphs in$C_(α,2k)$only depends on$V(α)$,$|U_α|$,$|V_α|$,$|U_α ∩ V_α|$, and the minimum size of a vertex separator of$α$. We say that $S \subseteq V(\alpha)$ is a vertex separator of $\alpha$ if every path from a vertex $u \in U_{\alpha}$ to a vertex $v \in V_{\alpha}$ contains at least one vertex in $S$. Note that this includes paths of length $0$ so we automatically have that $U_{\alpha} \cap V_{\alpha} \subseteq S$. Given a shape $\alpha$, define $s_{\alpha}$ to be the minimum size of a vertex separator of $\alpha$. For any shape $\alpha$, for any nonzero-valued $C \in \mathcal{C}_{(\alpha,2k)}$, $\abs{E(C)} \geq k\abs{V$$\setminus$U_{} V_{}$} + (k-1)(s_{\alpha} - \abs{U_{\alpha} \cap V_{\alpha}})$. Moreover, there exists a nonzero-valued $C \in \mathcal{C}_{(\alpha,2k)}$ such that $\abs{E(C)} = k\abs{V$$\setminus$U_{} V_{}$} + (k-1)(s_{\alpha} - \abs{U_{\alpha} \cap V_{\alpha}})$ so this bound is tight. In AMP20, this result was only proved for well-behaved constraint graphs (see Definition \ref{['defn:wellbehaved']}). That said, using the ideas in Appendix B of AMP20, it can be shown for all constraint graphs $C \in \mathcal{C}_{(\alpha,2k)}$. For all bipartite shapes $\alpha$, for all dominant constraint graphs $C \in \mathcal{C}_{(\alpha,2k)}$, $\abs{E(C)} = (k-1)s_{\alpha}$. We can now reduce the problem of estimating$E[tr$$M_αM_α^T$^k$]$to the problem of counting the number of dominant constraint graphs in$C_(α,2k)$. For all bipartite shapes $\alpha$ and all $k \in \mathbb{N}$, $\mathbb{E}\left[\mathop{\mathrm{tr}}\nolimits M_{\alpha}M_{\alpha}^T^k\right] = \abs{\left\{C \in \mathcal{C}_{(\alpha,2k)}: C \text{ is dominant}\right\}}n^{k|V(\alpha)| - s_{\alpha} + s_{\alpha}} \pm On^{k|V(\alpha)| - s_{\alpha}+ s_{\alpha} - 1}.$ By \ref{['prop:exp value diffrep']}, $\mathbb{E}[\mathop{\mathrm{tr}}\nolimits\left($M_{}M_{}^T$^k\right)] = \sum_{C \in \mathcal{C}_{(\alpha,2k)}}{N(C)\mathop{\mathrm{val}}\nolimits(C)}$. By Propositions \ref{['prop:nonzero iff even multi-edges']} and \ref{['prop: num of phi maps under a constr graph']}, for each dominant $C \in \mathcal{C}_{(\alpha,2k)}$, $\mathop{\mathrm{val}}\nolimits(C) = 1$ and $N(C) = \frac{n!}{n - k\abs{V(\alpha)} + (k-1)s_{\alpha}!} = n^{k|V(\alpha)| - s_{\alpha} + s_{\alpha}} - On^{k|V(\alpha)| - s_{\alpha}+ s_{\alpha} - 1}$ as $|V(\alpha,2k)| = k|V(\alpha)|$ and $\abs{E(C)} = (k-1)s_{\alpha}$. Thus, the contribution to $\mathbb{E}[\mathop{\mathrm{tr}}\nolimits$$M_{\alpha}M_{\alpha}^T$^k$]$ from $C$ is $n^{k$|V()| - s_{}$+ s_{\alpha}} - O$n^{k$|V(\alpha)| - s_{\alpha}$ + s_{} - 1}$$. For each non-dominant $C \in \mathcal{C}_{(\alpha,2k)}$, either $\mathop{\mathrm{val}}\nolimits(C) = 0$ or $N(C)$ is $O$n^{k$|V(\alpha)| - s_{\alpha}$ + s_{} - 1}$$ so the contribution to $\mathbb{E}[\mathop{\mathrm{tr}}\nolimits$$M_{\alpha}M_{\alpha}^T$^k$]$ from $C$ is $O$n^{k$|V(\alpha)| - s_{\alpha}$ + s_{} - 1}$$. Applying \ref{['cor:expected trace for bipartite shapes']} to$α_Z$gives the following estimate for$E[tr$$M_α_ZM_α_Z^T$^k$]$. $\mathbb{E}[\mathop{\mathrm{tr}}\nolimits$$M_{\alpha_Z}M_{\alpha_Z}^T$^k$] = \abs{\left\{C \in \mathcal{C}_{(\alpha_Z,2k)}: C \text{ is dominant}\right\}}n^{2k + 2} \pm O$n^{2k+1}$$. In order to count the number of dominant constraint graphs for$H(α,2k)$, it is useful to find sets of constraint edges which partition$H(α,2k)$into smaller graphs which can be analyzed recursively. In this subsection, we develop tools for doing this. Given a multi-graph $H$, let $E_{odd}(H)$ be the set (not the multi-set) of edges of $H$ which have odd multiplicity and let $E_{even}(H)$ be the set of edges of $H$ which have even multiplicity. Let $H$ be a multi-graph, let $E$ be a set of constraint edges which does not have any cycles, and let $C_E$ be the constraint graph on $V(H)$ with edges $E(C_E) = E$. We say that the set of constraint edges $E$ partitions $H$ into subgraphs $H_1,\ldots,H_j$ (which may actually be multi-graphs) if the following conditions hold: $H_1,\ldots,H_j$ are subgraphs of $H/{C_E}$ and $E_{odd}(H_1),\ldots,E_{odd}(H_j)$ give a partition of $E_{odd}(H/{C_E})$. More precisely, letting $\pi: V(H) \to V(H)$ be the map describing $H/{C_E}$ (see \ref{['defn:H/C']}) and letting $V(H/{C_E}) = \{v \in V(H): \pi(v) = v\}$ be the vertices of $H/{C_E}$, each subgraph $H_i$ is specified by a map $\pi_i: V(H) \to V(H/{C_E}) \cup \emptyset$ and these maps $\{\pi_i: i \in [j]\}$ satisfy the following properties: For each vertex $u \in V(H/{C_E})$ and each $i \in [j]$, we have that either $\pi_i(u') = u$ for all vertices $u' \in V(H)$ such that $u' \longleftrightarrow_{C_E} u$ or $\pi_i(u') = \emptyset$ for all vertices $u' \in V(H)$ such that $u' \longleftrightarrow_{C_E} u$.For each edge $e = \{u,v\} \in E(H)$ such that $e$ appears with odd multiplicity in $H/C_E$, there is exactly one $i \in [j]$ such that $\pi_i(u) \neq \emptyset$ and $\pi_i(v) \neq \emptyset$ (i.e., $\pi_i(e)$ exists). When this property is satisfied, we say that $e$ appears in $E(H_i)$. We then take $V(H_i) = \{v \in V(H/{C_E}): \pi_i(v) \neq \emptyset\}$ and $E(H_i) = \{\{\pi_i(u),\pi_i(v)\}: e = \{u,v\} \in E(H), \pi_i(u) \neq \emptyset, \pi_i(v) \neq \emptyset\}$ (note that $E(H_i)$ is a multi-set).There is a bijection between dominant constraint graphs $C$ on $H$ such that $E \subseteq E(C)$ and dominant constraint graphs $C_1,\ldots,C_j$ on $H_1,\ldots,H_j$. More precisely, for all dominant constraint graphs $C_1,\ldots,C_j$ on $H_1,\ldots,H_j$, if we take $E(C) =$_{i=1}^{j}{E(C_i)}$\cup E$ then $E(C)$ does not contain any cycles, $\abs{E(C)} = |E| + \sum_{i=1}^{j}{\abs{E(C_i)}}$, and if we take $C$ to be the constraint graph on $H$ with constraint edges $E(C)$ then $C$ is a dominant constraint graph on $H$. Conversely, if $C$ is a dominant constraint graph on $H$ and $E \subseteq E(C)$ then if we take $C_i = \pi_i(C)$ to be the constraint graph on $H_i$ such that for all pairs of vertices $u,v \in V(H_i)$, $u \longleftrightarrow_{C_i} v$ if and only if $u \longleftrightarrow_{C} v$, we have that for all $i \in [j]$, $C_i$ is a dominant constraint graph on $H_i$. As observed in \ref{['defn:line-shape-H-label']}, $H$_0,2k$$ is the graph with vertices $x_1,\ldots,x_{2k}$ and edges $\{\{x_i,x_{i+1}\}: i \in [2k-1]\} \cup \{x_{2k},x_1\}$. Example 1 Example 2 Illustration of \ref{['eg:partition']} For all $i,j \in [2k]$ such that $i < j$ and $j-i$ is even, the constraint edge $\{x_i,x_j\}$ partitions $H$_0,2k$$ into $H_1 \simeq H$_0,2k+i-j$$ and $H_2 \simeq H$_0,j-i$$ where $V(H_1) = \{x_1,\ldots,x_{i}\} \cup \{x_{j+1},\ldots,x_{2k}\}$ and $V(H_2) = \{x_{i},x_{i+1},\ldots,x_{j-1}\}$.For all $j \in [2,k-1]$, the set of constraint edges $E = \{\{x_1,x_{2j+1}\}, \{x_2,x_{2j}\}\}$ partitions $H$_0,2k$$ into $H_1 \simeq H$_0,2(j-1)$$ and $H_2 \simeq H_{\alpha_0,2(k-j)}$ where $V(H_1) = \{x_2,\ldots,x_{2j-1}\}$ and $V(H_2) = \{x_{2j+2},x_{2j+3}\ldots,x_{2k}\} \cup \{x_1\}$. In order to use Definition \ref{['defn:Hpartitions']}, it is helpful to have sufficient conditions for ensuring that the second condition of Definition \ref{['defn:Hpartitions']} holds, i.e., there is a bijection between dominant constraint graphs$C$on$H$such that$E ⊆ E(C)$and dominant constraint graphs$C_1,…,C_j$on$H_1,…,H_j$. Let $H$ be a multi-graph, let $E$ be a set of constraint edges, let $C_E$ be the constraint graph on $H$ with edges $E(C_E) = E$, and let $H_1,\ldots,H_j$ be subgraphs of $H/{C_E}$ satisfying the first condition of Definition \ref{['defn:Hpartitions']}. If the following four conditions are satisfied then the second condition of Definition \ref{['defn:Hpartitions']} is satisfied as well. Letting $H'_{final}$ be the graph with vertices $V(H/C_E)$ and edges $E_{odd}(H/C_E)$, $H'_{final}$ can be constructed by starting with $H' = H_i$ for some $i \in [j]$ and then iteratively applying the following operation. Take the current multi-graph $H'$ and choose $i,i'$ in $[j]$ such that $H_{i}$ has been added to $H'$ but $H_{i'}$ has not yet been added to $H'$.$V(H') \cap V(H_{i'}) = V(H_i) \cap V(H_{i'})$. We then add the vertices and edges of $H_{i'}$ to $H'$.For all $i,i' \in [j]$ and all dominant constraint graphs $C_i$ and $C_{i'}$ on $H_i$ and $H_{i'}$, for any pair of distinct vertices $u,v \in V(H_i) \cap V(H_{i'})$, there is no path from $u$ to $v$ in either $C_i$ or $C_{i'}$.For all dominant constraint graphs $C$ on $H$ such that $E \subseteq E(C)$, for all $i,i' \in [j]$ and all $u,v \in V(H_i) \cap V(H_{i'})$, there is no path from $u$ to $v$ in $C$.For all dominant constraint graphs $C$ on $H$ such that $E \subseteq E(C)$, if $e_1,e_2 \in E(H)$ appear with odd multiplicity in $H/C_E$ and $e_1 \longleftrightarrow_{C} e_2$ then $e_1$ and $e_2$ appear in the same subgraph $H_i$ (i.e, if we let $i,i' \in [j]$ be the unique indices such that $\pi_i(e_1)$ and $\pi_{i'}(e_2)$ exist then $i' = i$). When we use \ref{['lem:Hpartitionconditions']}, we will generally have that the first condition of \ref{['lem:Hpartitionconditions']} is easy to check directly and the second and third conditions of \ref{['lem:Hpartitionconditions']} follow from the fact that all dominant constraint graphs on $H/C_E$, $H_i$, and $H_{i'}$ are well-behaved (see \ref{['defn:wellbehaved']}). Thus, we will just need to verify the fourth condition of \ref{['lem:Hpartitionconditions']}. To prove this lemma, we first show the following statement. Let $C_1,\ldots,C_j$ be constraint graphs on $H_1,\ldots,H_j$ such that for all $i,i' \in [j]$ and all pairs of distinct vertices $u,v \in V(H_i) \cap V(H_{i'})$, there is no path from $u$ to $v$ in either $C_i$ or $C_{i'}$. Letting $E(C') = \bigcup_{i \in [j]}{E(C_i)}$, $E(C')$ does not contain any cycles and $\abs{E(C')} = \sum_{i \in [j]}{\abs{E(C_i)}}$. Moreover, if we take $C'$ to be the constraint graph on $H/C_E$ with constraint edges $E(C')$, for all $i \in [j]$ and distinct vertices $u,v \in V(H_i)$, $C'$ contains a path between $u$ and $v$ if and only if $C_i$ contains a path between $u$ and $v$. To prove this lemma, we use induction to show that at each step in the construction of $H/C_E$, letting $E(C') = \bigcup_{i: H_i \text{ has been added to } H'}{E(C_i)}$, $\abs{E(C')} = \sum_{i: H_i \text{ has been added to } H'}{\abs{E(C_i)}}$ and $E(C')$ does not contain any cycles.Taking $C'$ to be the constraint graph on $H'$ with constraint edges $E(C')$, for all $i$ such that $H_i$ has been added to $H'$, for all vertices $u,v \in V(H_i)$, $C'$ contains a path between $u$ and $v$ if and only if $C_i$ contains a path between $u$ and $v$. The base case $H' = H_i$ and $C' = C_i$ is trivial. For the inductive step, assume that the statement holds for the current $H'$. Let $i,i' \in [j]$ be the indices such that the next step of the construction attaches $H_{i'}$ to $H'$ at the vertices $V(H') \cap V(H_{i'}) = V(H_i) \cap V(H_{i'})$. Let $H"$ be the resulting graph after adding $H_{i'}$ to $H'$ and let $E(C") = E(C') \cup E(H_{i'})$. For all $u,v \in V(H') \cap V(H_{i'}) = V(H_i) \cap V(H_{i'})$, there is no path from $u$ to $v$ in $E(C_{i'})$ so we must have that $E(C') \cap E(C_{i'}) = \emptyset$ and thus $\abs{E(C")} = \abs{E(C_{i'})} + \abs{E(C')}$. By the inductive hypothesis, $\abs{E(C')} = \sum_{i: H_i \text{has been added to } H'}{\abs{E(C_i)}}$ so $\abs{E(C")} = \abs{E(C_{i'})} + \sum_{i: H_i \text{ has been added to } H'}{\abs{E(C_i)}} = \sum_{i: H_i \text{ has been added to } H"}{\abs{E(C_i)}}.$ To show that $E(C")$ does not contain a cycle, observe that since $C'$ and $C_{i'}$ do not contain any cycles, any cycle in $E(C")$ would have to contain a path in $C_{i'}$ between two distinct vertices $u,v \in V(H') \cap V(H_{i'}) = V(H_i) \cap V(H_{i'})$. However, $C_{i'}$ contains no such paths. We now take $C"$ to be the constraint graph on $V(H")$ with constraint edges $E(C")$. For all vertices $u,v \in V(H')$, $C"$ contains a path between $u$ and $v$ if and only if $C'$ contains a path from $u$ to $v$. To see this, assume that there are vertices $u,v \in V(H')$ such that $C"$ contains a path $P$ from $u$ to $v$ but $C'$ does not contain a path from $u$ to $v$. If so, then $P$ must contain a path in $C_{i'}$ between two distinct vertices in $V(H') \cap V(H_{i'}) = V(H_{i}) \cap V(H_{i'})$. This gives a contradiction as $C_{i'}$ contains no such paths. Similarly, for all vertices $u,v \in V(H_{i'})$, $C"$ contains a path between $u$ and $v$ if and only if $C_{i'}$ contains a path from $u$ to $v$. To see this, assume that there are vertices $u,v \in V(H_{i'})$ such that $C"$ contains a path $P$ from $u$ to $v$ but $C_{i'}$ does not contain a path from $u$ to $v$. If so, then $P$ must contain a path in $C'$ between two distinct vertices $u'$ and $v'$ in $V(H') \cap V(H_{i'}) = V(H_{i}) \cap V(H_{i'})$. This gives a contradiction as there is no path from $u'$ to $v'$ in $C_i$ and thus by the inductive hypothesis, there is no path from $u'$ to $v'$ in $C'$. Let $C_1,\ldots,C_j$ be constraint graphs on $H_1,\ldots,H_j$ such that for all $i,i' \in [j]$ and all pairs of distinct vertices $u,v \in V(H_i) \cap V(H_{i'})$, there is no path from $u$ to $v$ in either $C_i$ or $C_{i'}$. Letting $E(C) = \left(\bigcup_{i \in [j]}{E(C_i)}\right) \cup E$, $E(C)$ does not contain any cycles and $\abs{E(C)} = |E| + \sum_{i \in [j]}{\abs{E(C_i)}}$. Moreover, if we take $C$ to be the constraint graph on $H$ with constraint edges $E(C)$, for all $i \in [j]$ and distinct vertices $u,v \in V(H_i)$, $C$ contains a path between $u$ and $v$ if and only if $C_i$ contains a path between $u$ and $v$. Let $C'$ be the constraint graph given by Lemma \ref{['lem:Hpartitionconditionssublemma']}. Observe that we can obtain $C$ from $C'$ by simply adding the vertices $\{v \in V(H): \pi(v) \neq v\}$ (i.e., the vertices of $H$ which are not in $H/C_E$) and the edges $E$. This does not add any cycles and does not affect whether there is a path between any two vertices $u,v \in V(H/C_E)$. We now use Lemma \ref{['cor:Hpartitionconditionssublemmacorollary']} to prove Lemma \ref{['lem:Hpartitionconditions']}. Under the conditions of Lemma \ref{['lem:Hpartitionconditions']}, if $C$ is a dominant constraint graph on $H$ such that $E \subseteq E(C)$ then if we take $C_i$ to be the constraint graph on $H_i$ such that for all vertices $u,v \in V(H_i)$, $u \longleftrightarrow_{C_i} v$ if and only if $u \longleftrightarrow_C v$, Taking $E(C') =$_{i=1}^{j}{E(C_i)}$\cup E$, $E(C')$ has no cycles, $\abs{E(C')} = |E| + \sum_{i=1}^{j}{\abs{E(C_i)}}$, and if we take $C'$ to be the constraint graph on $H$ with constraint edges $E(C')$ then $C \equiv C'$.For all $i \in [j]$, $C_i$ is a dominant constraint graph on $H_i$. We first observe that condition 4 of Lemma \ref{['lem:Hpartitionconditions']} implies that for all $i \in [j]$, $C_i$ is a nonzero-valued constraint graph on $H_i$. To see this, assume that there is an $i \in [j]$ and an edge $e = \{u,v\} \in E(H_i)$ which appears with odd multiplicity in $H_i/C_i$. Since $C$ is a nonzero-valued constraint graph on $H$, $e$ must appear with even multiplicity in $H/C$ which implies that there must be an edge $e' = \{u',v'\} \in E(H)$ such that $e' \longleftrightarrow_{C} e$ and $e'$ appears with odd multiplicity in $H/C_E$ but $\pi(u') \notin V(H_i)$ or $\pi(v') \notin V(H_i)$. However, this contradicts condition 4 of \ref{['lem:Hpartitionconditions']}. By condition 3 of \ref{['lem:Hpartitionconditions']}, for all $i,i' \in [j]$ and all $u,v \in V(H_i) \cap V(H_{i'})$, there is no path from $u$ to $v$ in $C$ so there is no path from $u$ to $v$ in either $C_i$ or $C_{i'}$. By Corollary \ref{['cor:Hpartitionconditionssublemmacorollary']}, $E(C') = \left(\bigcup_{i \in [j]}{E(C_i)}\right) \cup E$ has no cycles and $\abs{E(C')} = |E| + \sum_{i=1}^{j}{\abs{E(C_i)}}$. Since $C_i$ is a nonzero-valued constraint graph on $H_i$ for all $i \in [j]$, $C'$ is a nonzero-valued constraint graph on $H$. Now observe that for any $e = \{u,v\} \in E(C')$, $u \longleftrightarrow_{C} v$. This implies that for all $u,v \in V(H)$, if $u \longleftrightarrow_{C'} v$ then $u \longleftrightarrow_{C} v$. If there were $u,v \in V(H)$ such that $u \longleftrightarrow_{C} v$ but $C'$ does not contain a path from $u$ to $v$ then we would have that $\abs{E(C')} < \abs{E(C)}$ which gives a contradiction as $C$ is a dominant constraint graph on $H$. Thus, we must have that $u \longleftrightarrow_{C'} v$ if and only if $u \longleftrightarrow_{C} v$ and thus $C' \equiv C$. To show that $C_i$ is a dominant constraint graph on $H_i$ for all $i \in [j]$, assume there is an $i' \in [j]$ and a dominant constraint graph $C'_{i'}$ on $H_{i'}$ such that $\abs{E(C'_{i'})} < \abs{E(C_{i'})}$. Letting $E(C") = \left(\bigcup_{i \in [j] \setminus \{i'\}}{E(C_i)}\right) \bigcup E$C'_{i'}$\bigcup E$, by \ref{['cor:Hpartitionconditionssublemmacorollary']}, $E(C")$ does not contain any cycles and $\abs{E(C")} = \abs{E} + \abs{E(C'_{i'})} + \sum_{i \in [j] \setminus \{i\}}{\abs{E(C_i)}} < \abs{E(C)}$. Since $C_i$ is a nonzero-valued constraint graph on $H_i$ for all $i \in [j] \setminus \{i\}$ and $C'_{i'}$ is a nonzero-valued constraint graph on $H_{i'}$, $C"$ is a nonzero-valued constraint graph on $H$. However, this contradicts the assumption that $C \equiv C'$ is a dominant constraint graph on $H$. Under the conditions of Lemma \ref{['lem:Hpartitionconditions']}, if $C_1,\ldots,C_j$ are dominant constraint graphs on $H_1,\ldots,H_j$ and we take $E(C) =$_{i=1}^{j}{E(C_i)}$\cup E$ then $E(C)$ does not contain any cycles and $\abs{E(C)} = |E| + \sum_{i=1}^{j}{\abs{E(C_i)}}$.The constraint graph $C$ with constraint edges $E(C)$ is a dominant constraint graph on $H$. The first statement follows immediately from Corollary \ref{['cor:Hpartitionconditionssublemmacorollary']}. To show that $C$ is a nonzero-valued constraint graph on $H$, observe that since $C_i$ is a nonzero-valued constraint graph on $H_i$ for all $i \in [j]$, all edges in $H/C = (H/C_E)/C'$ appear with even multiplicity which implies that $C$ is a nonzero-valued constraint graph on $H$. To show that $C$ is a dominant constraint graph, assume that there is a dominant constraint graph $C'$ on $H$ such that $\abs{E(C')} < \abs{E(C)}$. By Lemma \ref{['lem:Hpartitionconditionsdirectionone']}, if we take $C'_i$ to be the constraint graph on $H_i$ such that for all vertices $u,v \in V(H_i)$, $u \longleftrightarrow_{C'_i} v$ if and only if $u \longleftrightarrow_C' v$ then $\abs{E(C')} = \abs{E} + \sum_{i=1}^{j}\, {\abs{E(C'_i)}}$ and $C'_i$ is a dominant constraint graph on $H_i$ for all $i \in [j]$. Now observe that $|E| + \sum_{i=1}^{j}{\abs{E(C_i)}} = \abs{E(C)} < \abs{E(C')} = |E| + \sum_{i=1}^{j}{\abs{E(C'_i)}}$ so there must be some $i \in [j]$ such that $\abs{E(C'_i)} < \abs{E(C_i)}$ which contradicts the assumption that $C_i$ is a dominant constraint graph on $H_i$. In this section, we show that the number of dominant constraint graphs on $H(\alpha,2k)$ is $C'_k$. For all $k \in \mathbb{N}$, $\abs{\left\{C \in \mathcal{C}_{(\alpha_Z,2k)}: C \text{ is dominant}\right\}} = C'_k = \frac{1}{2k+1}\binom{3k}{k}$. Combined with \ref{['cor:Zshapeexpectedtrace']}, this shows that $\mathbb{E}[\mathop{\mathrm{tr}}\nolimits$$M_{\alpha_Z}M_{\alpha_Z}^T$^k$] = C'_k\cdot n^{2k + 2} \pm O$n^{2k+1}$$, which is the first part of \ref{['thm:zshape-trace-convergence']}. After proving \ref{['thm:num-zshape-constraint-graph']}, we show that $\mathop{\mathrm{Var}}\nolimits (\mathop{\mathrm{tr}}\nolimits$$M_{\alpha_Z}M_{\alpha_Z}^T$^k$)$ is $O$n^{4k+2}$$, which completes the proof of \ref{['thm:zshape-trace-convergence']}. One of the key ingredients for proving \ref{['thm:num-zshape-constraint-graph']} is the following recurrence relation on $C'_k$. For all $k \in \mathbb{N} \cup \{0\}$, $C'_{k+1} = \sum_{a,b,c\geq 0: a+b+c=k} {C'_a}{C'_b}{C'_c} =\sum_{a=0}^{k}{ {C'_a} \sum_{b=0}^{k-a} {C'_b}{C'_{k-a-b}}}.$ To prove this recurrence relation, we consider walks on grids. This proof is a generalization of the third proof in the Wikipedia article on Catalan numbers (rukavicka2011generalized). Given $x,x',y,y' \in \mathbb{Z}$, a grid walk from $(x,y)$ to $(x',y')$ is a sequence of $(x'-x) + (y'-y) + 1$ coordinates $$z_0,z_1,,z_{(x'-x) + (y'-y)}$$ where $z_i=(x_i,y_i)$ where $x_i\in [x,x']$ and $y_i\in [y,y']$ for each $i \in [x+y]$.$z_0=(x,y)$ and $z_{(x'-x) + (y'-y)}=(x',y')$.$z_{i}-z_{i-1}=(1,0)$ or $(0,1)$ for all $i\in[(x'-x) + (y'-y)]$. Pictorially, a grid is a walk from $(x,y)$ to $(x',y')$ that steps on integer coordinates and only moves straight up or straight right. Given a line $l$ of the form $y = kx + b$ where $k > 0$, we say that a grid walk $w = (z_0,z_1,\ldots,z_{j})$ is weakly below $l$ if $y_i \leq kx_i + b$ for all $i \in [j] \cup \{0\}$. An important special case is that a grid walk $(z_0,z_1,\dots, z_{x+y})$ from $(0,0)$ to $(x,y)$ is weakly below the diagonal if ${y_i/x_i\leq y/x}$ for all $i \in [x+y]$. It is often useful to apply translations to grid walks. Given a grid walk $w = (z_0,z_1,\ldots,z_{j})$ and $x,y \in \mathbb{Z}$, we define $w + (x,y)$ to be the grid walk $w + (x,y) = (z_0 + (x,y),z_1 + (x,y),\ldots,z_{j} + (x,y))$ With these definitions, we can now prove \ref{['thm:catalan2-recurrence-relation']}. Let $W_k$ be the set of all grid walks from $(0,0)$ to $(k,2k)$ weakly below the diagonal and let $d_k=|W_k|$. We will prove that $d_k = C'_k = \frac{1}{2k+1}\binom{3k}{k}$ and $d_k$ satisfies the recurrence relation $d_{k+1}=\sum_{a,b,c\geq 0: a+b+c=k} {d_a}{d_b}{d_c}$. For all $k \in \mathbb{N}$, $d_{k}= \sum_{a=1}^{k}{\sum_{b=1}^{a}{d_{b-1}d_{a-b}d_{k-a}}} = \sum_{\substack{a,b,c\geq 0:\\ a+b+c=k-1}} {d_a}{d_b}{d_c}$ We will establish a bijection between $W_{k}$ and $W'_{k}:=\bigcup_{a,b: 1 \leq b \leq a \leq k}{W_{b-1} \times W_{a-b} \times W_{k-a}}$. Let $w=(z_0,z_1,\dots,z_{3k})$ be a grid walk from $(0,0)$ to $(k,2k)$ weakly below the diagonal. We construct the walks $w'_1,w'_2,w'_3$ in reverse order. Consider the first point $(a,2a)$ after the starting point $(0,0)$ where $w$ touches the diagonal. In other words, let $a$ be the smallest element of $[k]$ such that $z_{3a} = (a,2a)$. Then $w_3=(z_{3a},z_{3a+1},\dots,z_{3k})$ is a grid walk from $(a,2a)$ to $(k,2k)$ weakly below the diagonal. Taking $w'_3 = w_3 - (a,2a)$, $w'_3\in W_{k-a}$. Let $l$ be the line parallel to the diagonal which passes through $(a,2a-1)$. In other words, $l$ is the line $y = 2x - 1$. Since $z_{3a}$ is the first point touching the diagonal, $(z_1,\dots, z_{i-1})$ is weakly below $l$. Let $z_{3b-1}=(b,2b-1)$ be the first point touching $l$. Then $w_2=(z_{3b-1},\dots,z_{3a-1})$ is a grid walk from $(b,2b-1)$ to $(a,2a-1)$ weakly below $l$. Taking $w'_2 = w_2 - (b,2b-1)$, $w'_2\in W_{a-b}$. Let $l'$ be the line parallel to the diagonal which passes through $(b,2b-2)$. In other words, $l$ is the line $y = 2x - 2$. Since $z_{3b-1}$ is the first point touching $l$ and $z_1 = (1,0)$, $(z_1,\dots, z_{3b-2})$ is weakly below $l'$. Observe that $w_3=(z_1,\dots, z_{3b-2})$ is a grid walk from $(1,0)$ to $(b,2b-2)$ weakly below $l'$. Taking $w'_1 = w_1 - (1,0)$, $w'_1 \in W_{b-1}$. Thus from $w\in W_k$ we get a tuple $(w'_1,w'_2,w'_3)\in W_{b-1} \times W_{a-b} \times W_{k-a}$ where $a,b$ are uniquely determined by $w$ and $1 \leq b \leq a \leq k$. Conversely, given $a,b$ such that $1 \leq b \leq a \leq k$ and $(w'_1,w'_2,w'_3)\in W_{b-1} \times W_{a-b} \times W_{k-a}$, let $w=(0,0), w_1+(1,0), w_2+(b,2b-1), w_3+(a,2a)$ It is not hard to check that $w \in W_k$ and this is a bijection. proof part 1 proof part 2 ${d_k = C'_k = \frac{1}{2k+1}\binom{3k}{k}}$. For $r\in\{0,1,\dots,2k\}$, let $V_r$ be the set of grid walks from $(0,0)$ to $(k,2k)$ that have $r$ vertical steps above the diagonal. In other words, for each $w=(z_0,z_1,\dots, z_{3k})\in V_r$, writing $z_j=(x_j,y_j)$, there are $r$ indices $j \in [3k]$ such that $y_j >2x_j$ and $z_j - z_{j-1} = (0,1)$. Let $G_k$ be the set of all grid walk from $(0,0)$ to $(k,2k)$. We have that $\abs{G_n}=\binom{3k}{k}$. Note that $V_0=W_k$ and $\bigcup_{r=0}^{2k} V_r=G_k$. We will prove that $|V_r|=|V_{r-1}|$ for all $r\in [2k]$. This implies that $\abs{V_0}=d_k=\frac{1}{2k+1}\binom{3k}{k}$ as needed. $|V_{r}| = |V_{r-1}|$ for all $r\in [2k]$. For each $r \in [2k]$, we have the following bijection between $V_{r}$ and $V_{r-1}$. Given $w=(z_0,\dots, z_{3k}) \in V_r$, let $z_{j}$ be the last point where the walk is on the diagonal and then takes a step upwards. In other words, $j\in [3k-1] \cup \{0\}$ is the largest index such that $z_j=(a,2a)$ for some $a\in[k]$ and $z_{j+1}-z_{j}=(0,1)$. Let $w_1=(z_0,\dots, z_{j})$ and let $w_2=(z_{j+1},\dots, z_{3n})$. Let $w'=(w'_1,w'_2)$ where $w'_1 = w_2-(a,2a+1)$ and $w'_2 = w_1+(k-a,2k-2a))$ (see \ref{['fig:grid walk']}, $w'$ exchanges the green and blue part of $w$). Observe that $w_2'$ has the same number of steps above the diagonal as $w_1$ does. Since $z_j$ is the last point such that $w$ is on the diagonal and then takes a step upwards, $w'_1$ has the same number of vertical steps above the diagonal as $w_2$. However, the vertical step from $z_j = (a,2a)$ to $z_{j+1} = (a,2a+1)$ in $w$ (i.e., the step between the end of $w_1$ and the start of $w_2$) is above the diagonal while the vertical step from $(k-a,2(k-a)-1)$ to $(k-a,2(k-a))$ in $w'$ (i.e., the step between the end of $w'_1$ and the start of $w'_2$) is below the diagonal. Thus, $w'$ has one less vertical step above the diagonal than $w$ so $w'\in V_{r-1}$.Let $w = (z_0,\dots, z_{3k})\in V_{r-1}$. Let $z_j$ be the first point such that $w$ touches the diagonal from below. In other words, $j \in [3k]$ is the first index such that $z_j=(b,2b)$ for some $b \in [k]$ and $z_j-z_{j-1}=(0,1)$. Let $w_1=(z_0,\dots,z_{j-1})$ and $w_2=(z_{j},\dots,z_{3k})$. Let $w'=(w_1',w_2')$ where $w_1' = w_2-(b,2b)$ and $w_2' = w_1 + (k-b,2k-2b+1)$. Observe that $w_1'$ has the same number of steps above the diagonal as $w_2$ does. Since $z_j$ is the first point where $w$ arrives at the diagonal after a vertical step, $w'_2$ has the same number of vertical steps above the diagonal as $w_1$. However, the vertical step $z_{j-1} = (b,2b-1)$ to $z_{j} = (b,2b)$ in $w$ (i.e., the step between the end of $w_1$ and the start of $w_2$) is below the diagonal while the vertical step from $((k-b,2(k-b))$ to $((k-b,2(k-b) + 1)$ in $w'$ (i.e., the step between the end of $w'_1$ and the start of $w'_2$) is below the diagonal. Thus, $w'$ has one less vertical step above the diagonal than $w$ so $w'\in V_{r-1}$. It is not hard to check that this gives a bijection. \ref{['thm:catalan2-recurrence-relation']} follows directly from \ref{['lem:gridwalkrecurrencerelation']} and \ref{['lem:gridwalkcounting']}. By \ref{['lem:gridwalkcounting']}, ${d_k = C'_k}$. By \ref{['lem:gridwalkrecurrencerelation']}, $d_{k} = \sum_{\substack{a,b,c\geq 0:\\ a+b+c=k-1}} {d_a}{d_b}{d_c}$. Thus, $C'_{k+1} = \sum_{\substack{a,b,c\geq 0:\\ a+b+c=k}} {C'_a}{C'_b}{C'_c} =\sum_{a=0}^{k}{ {C'_a}$_{b=0}^{k-a} {C'_b}{C'_{k-a-b}}$}$, as needed. In order to count the number of dominant constraint graphs in $\mathcal{C}_{$_Z,2k$}$, we need a few properties of these constraint graphs. As a warm-up, we first analyze dominant constraint graphs on a cycle of length $2k$. This is essentially the trace power analysis for Wigner matrices (wigner1993characteristic). Recall the line shape $\alpha_0$ and its corresponding $H$_0, 2k$$ from \ref{['defn:line-shape']} and \ref{['defn:line-shape-H-label']}. When analyzing constraint graphs, it is often useful to consider vertices which are not incident to any constraint edges. We call such vertices isolated. Given a multi-graph $H$ and a constraint graph $C$ on $H$, we say that a vertex $v \in V(C) = V(H)$ is isolated if $v$ is not incident to any edge in $E(C)$. If $C$ is a nonzero-valued constraint graph on $H(\alpha_0,2k)$ and $C$ has an isolated vertex $x_j$, then $x_{j-1}\longleftrightarrow x_{j+1}$. In the cases when $j=1$ or $j=2k$, we take $x_{0}=x_{2k}$ and $x_{2k+1} = x_1$ respectively. Recall that by \ref{['prop:nonzero iff even multi-edges']}, $C$ is nonzero-valued if and only if each edge in $H(\alpha_0,2k)/C$ appears an even number of times. Since $x_j$ is isolated, the only way this can happen is if $x_{j-1}\longleftrightarrow x_{j+1}$. We now describe several properties of dominant constraint graphs in $\mathcal{C}_{(\alpha_0,2k)}$. We say that a constraint graph $C \in \mathcal{C}_{(\alpha_0,2k)}$ is well-behaved if all constraint edges in $C$ are either between two copies of $u$ (i.e., two vertices in $\{u_1,\ldots,u_k\}$) or between two copies of $v$ (i.e., two vertices in $\{v_1,\ldots,v_k\}$). Equivalently, $C$ is well-behaved if whenever $x_i \longleftrightarrow x_j$, $j-i$ is even. We say that a constraint graph $C \in \mathcal{C}_{(\alpha_0,2k)}$ is non-crossing if there are no indices $i_1,i_2,i_3,i_4 \in [k]$ such that $i_1 < i_2 < i_3 < i_4$, $x_{i_1} \longleftrightarrow x_{i_3}$, $x_{i_2} \longleftrightarrow x_{i_4}$, and there is no path from $x_{i_2}$ to $x_{i_3}$ in $C$. Equivalently, $C$ is non-crossing if there is a constraint graph $C'$ such that $C' \equiv C$ and if we draw $C'$ so that the vertices $x_1,\ldots,x_{2k}$ are on a circle and the edges of $C'$ are chords of this circle, no two edges of $C'$ cross. For all $k \in \mathbb{N}$, all dominant constraint graphs in $\mathcal{C}_{(\alpha_0,2k)}$ are well-behaved, non-crossing, and have $k-1$ edges. We first show that all dominant constraint graphs in $\mathcal{C}_{(\alpha_0,2k)}$ have $k-1$ edges. To show this, we make the following observations: The constraint graph $C_0$ with constraint edges $E(C_0) = \left\{\{x_{2j-1},x_{2j+1}\}: j \in [k-1]\right\}$ has value $1$ so dominant constraint graphs in $\mathcal{C}_{(\alpha_0,2k)}$ have at most $k-1$ edges.If $C \in \mathcal{C}_{(\alpha_0,2k)}$ and $\mathop{\mathrm{val}}\nolimits(C) \neq 0$ then every edge in $H(\alpha_0,2k)/C$ has even multiplicity so $H(\alpha_0,2k)/C$ has at most $k$ distinct edges. Since $H(\alpha_0,2k)/C$ is connected, this implies that $H(\alpha_0,2k)/C$ has at most $k+1$ vertices so $|E(C)| = \abs{V$H(_0,2k)$} - \abs{V$H(_0,2k)/C$} \geq k-1$. We prove that all dominant constraint graphs in $\mathcal{C}_{(\alpha_0,2k)}$ are well-behaved and non-crossing by induction. The base case $k = 1$ is trivial. For the inductive step, assume that all dominant constraint graphs in $\mathcal{C}_{(\alpha_0,2(k-1))}$ are well-behaved and non-crossing. Let $C \in \mathcal{C}_{(\alpha_0,2k)}$ be a dominant constraint graph. There are two cases to consider. $C$ does not have any isolated vertices. In this case, $C$ must have at least $k$ edges as $|V(C)| = |V(H(\alpha_0,k))| = 2k$. This is a contradiction as dominant constraint graphs in $\mathcal{C}_{(\alpha_0,2(k-1))}$ have $k-1$ edges.$C$ has an isolated vertex $x_j$. In this case, by \ref{['prop:line shape isolated vertex']}, $x_{j-1} \longleftrightarrow x_{j+1}$ (where $x_0 = x_{2k}$ and $x_{2k+1} = x_1$). Without loss of generality, we can assume that $\{x_{j-1},x_{j+1}\} \in E(C)$. Let $H'$ be the multi-graph obtained from $H$ by deleting the vertex $x_j$, deleting the edges $\{x_{j-1},x_j\}$ and $\{x_j,x_{j+1}\}$, and merging the vertices $x_{j-1}$ and $x_{j+1}$. Similarly, let $C'$ be the constraint graph obtained from $C$ by deleting the vertex $x_j$, deleting the constraint edge $\{x_{j-1},x_{j+1}\}$, and merging the vertices $x_{j-1}$ and $x_{j+1}$. Observe that $H'$ is isomorphic to $H(\alpha_0,k-1)$ so we can view $C'$ as a constraint graph on $H(\alpha_0,k-1)$. Moreover, $\mathop{\mathrm{val}}\nolimits(C') = \mathop{\mathrm{val}}\nolimits(C) \neq 0$ and $|E(C')| = |E(C)| - 1 = k-2$, so $C'$ is a dominant constraint graph on $H(\alpha_0,k-1)$. By the inductive hypothesis, $C'$ is well-behaved and non-crossing and it is not hard to check that $C$ is well-behaved and non-crossing as well. We now show that if $C_k = \abs{C \in \mathcal{C}_{(\alpha_0,2k)}: C \text{ is dominant}}$ then $C_k = \sum_{j=0}^{k-1}\,{C_{j}C_{k-1-j}}$. Since $C_0 = C_1 = 1$, this implies that $C_k = \frac{1}{k+1}\binom{2k}{k}$ is the $k$th Catalan number. The following sets of constraint edges partition $H(\alpha_0,2k)$ into subgraphs $H_1,H_2$. If $E = \{\{x_1,x_3\}\}$ then $E$ partitions $H(\alpha_0,2k)$ into $H_1$ and $H_2$ where $V(H_1) = \emptyset$ and $V(H_2) = \{x_1\} \cup \{x_4,\ldots,x_{2k}\}$. Observe that $H_1 \simeq H(\alpha_0,0)$ and $H_2 \simeq H(\alpha_0,k-1)$If $E = \{\{x_1,x_{2j+3}\},\{x_2,x_{2j+2}\}\}$ for some $j \in [k-2]$ then $E$ partitions $H(\alpha_0,2k)$ into $H_1$ and $H_2$ where $V(H_1) = \{x_2,\ldots,x_{2j+1}\}$ and $V(H_2) = \{x_1\} \cup \{x_{2j+4},\ldots,x_{2k}\}$. Observe that $H_1 \simeq H(\alpha_0,j)$ and $H_2 \simeq H(\alpha_0,k - 1 - j)$.If $E = \{\{x_2,x_{2k}\}\}$ then $E$ partitions $H$ into $H_1$ and $H_2$ where $V(H_1) = \{x_2,\ldots,x_{2k-1}\}$ and $V(H_2) = \emptyset$. Observe that $H_1 \simeq H(\alpha_0,k-1)$ and $H_2 \simeq H(\alpha_0,0)$. See \ref{['fig:line-shape-split']} for an illustration. This follows from \ref{['lem:Hpartitionconditions']}. For each of these sets of edges $E$, the first condition of \ref{['defn:Hpartitions']} and the first condition of \ref{['lem:Hpartitionconditions']} can be seen directly. The second and third conditions of \ref{['lem:Hpartitionconditions']} are trivial as $V(H_1) \cap V(H_2) = \emptyset$. The fourth condition of \ref{['lem:Hpartitionconditions']} is trivial in cases $1$ and $3$. In case $2$, the fourth condition of \ref{['lem:Hpartitionconditions']} follows from \ref{['lem:line-shape-catalan-lemma']} which says that all dominant constraint graphs $C$ on $H(\alpha_0,2k)$ are non-crossing and well-behaved. To see this, let $C$ be a dominant constraint graph on $H(\alpha_0,2k)$ and observe that for each pair of edges $e_1,e_2 \in E(H)$ such that $e_1,e_2$ appear with odd multiplicity in $H/{E_C}$, $e_1$ appears in $H_1$, and $e_2$ appears in $H_2$, one of the endpoints of $e_1$ must be $x_{i}$ for some odd $i \in [3,2j+1]$ and one of the endpoints of $e_2$ must be $x_{i'}$ for some odd $i' \in [2j+3,2k] \cup \{1\}$. Since $C$ contains the edge $\{x_2,x_{2j+1}\}$, is well-behaved, and is non-crossing, we cannot have that $x_{i} \longleftrightarrow_{C} x_{i'}$ so we cannot have that $e_1 \longleftrightarrow_{C} e_2$. Illustration of \ref{['lem:lineshapepartitioning']} case 2: $E = \{\{x_1,x_{2j+3}\},\{x_2,x_{2j+2}\}\}$. This is also an illustration of \ref{['cor:line-shape-recurrence']}, case 2: $j\in[1,k-2]$ is the smallest index such that $x_1 \longleftrightarrow_C x_{2j+3}$. Letting $C_k = \abs{C \in \mathcal{C}_{(\alpha_0,2k)}: C \text{ is dominant}}$, $C_k = \sum_{j=0}^{k-1}\,{C_{j}C_{k-1-j}}$ We partition the dominant constraint graphs $C \in \mathcal{C}_{(\alpha_0,2k)}$ based on the first index $j' \in [2k]$ such that $j' > 1$ and $x_{1} \longleftrightarrow_C x_{j'}$. If no such $j'$ exists then we set $j' = 2k+1$. We must have that $j'$ is odd and $j' \geq 3$ so we can write $j' = 2j + 3$ for some $j \in \{0,1,\ldots,j-1\}$. We have the following cases: If $j = 0$ then this gives case $1$ of \ref{['lem:lineshapepartitioning']}. There are $C_{0}C_{k-1}$ such dominant constraint graphs $C \in \mathcal{C}_{(\alpha_0,2k)}$.If $0 < j < k-1$ then since $C$ is non-crossing, we must have that $x_2 \longleftrightarrow_C x_{2j + 2}$ as otherwise the edges $\{x_1,x_2\}$ and $\{x_{2j+2},x_{2j+3}\}$ would appear with odd multiplicity in $H/C$ and we would have that $\mathop{\mathrm{val}}\nolimits(C) = 0$. This gives case $2$ of \ref{['lem:lineshapepartitioning']} and there are ${C_j}C_{k-1-j}$ such dominant constraint graphs $C \in \mathcal{C}_{(\alpha_0,2k)}$.If $j = k-1$ then we must have that $x_2 \longleftrightarrow_C x_{2k}$ as otherwise the edges $\{x_1,x_2\}$ and $\{x_{2k},x_{1}\}$ would appear with odd multiplicity in $H/C$ and we would have that $\mathop{\mathrm{val}}\nolimits(C) = 0$. This gives case $3$ of \ref{['lem:lineshapepartitioning']} and there are $C_{k-1}C_0$ such dominant constraint graphs $C \in \mathcal{C}_{(\alpha_0,2k)}$. See \ref{['fig:line-shape-split']} for an illustration. Now that we have analyzed dominant constraint graphs in $\mathcal{C}_{(\alpha_0,2k)}$, we can analyze dominant constraint graphs in $\mathcal{C}_{$_Z,2k$}$. Let $\alpha_Z$ be the Z-shape as defined in \ref{['defn:z-shape']} and let $H$_Z,2k$$ be the multi-graph as defined in \ref{['def:copies']}. We label the vertices of $V_{$_Z$_i}$ as $\{a_{i1},a_{i2},b_{i1},b_{i2}\}$ and the vertices of $V_{$_Z^T$_i}$ as $\{b_{i1},b_{i2},a_{(i+1)1},a_{(i+1)2}\}$. We call the induced subgraph of $H$_Z,2k$$ on vertices $\left\{a_{i1},b_{i1}:i\in[k]\right\}$ the outer wheel $W_1$ and the induced subgraph on vertices $\{a_{i2},b_{i2}:i\in[k]\}$ the inner wheel $W_2$. We denote the vertices of $W_i$ as $V_i$ and edges as $E_i$. We label the "middle edges" of $H(\alpha,2k)$ in the following way: let $e_{2i-1}=\{a_{i2},b_{i1}\}$ and $e_{2i}=\{b_{i1},a_{(i+1)2}\}$ for $i \in [k]$. We call the edges $\{e_i: i \in [2k]\}$ the spokes of $H$_Z,2k$$. See \ref{['fig:zshape-H-2k']} for an illustration. $H(\alpha_{Z},2k)$ where $k=4$. Let $C$ be a constraint graph in $\mathcal{C}_{$_Z,2k$}$. For $i=1,2$, we take $C_i$ to be the induced constraint graph of $C$ on the vertices $V_i$. Each wheel $W_i$ can be viewed as $H(\alpha_0,2k)$. The induced constraint graph $C_i$ can be viewed as a constraint graph on $H(\alpha_0,2k)$. The key property that we need about dominant constraint graphs in $\mathcal{C}_{$_Z,2k$}$ is that they are well-behaved. This implies that the induced constraint graphs $C_1,C_2$ are dominant constraint graphs in $\mathcal{C}_{(\alpha_0,2k)}$. Given a shape $\alpha$, we say that a constraint graph $C \in \mathcal{C}_{(\alpha,2k)}$ is well-behaved if whenever $u \longleftrightarrow_C v$, $u$ and $v$ are copies of the same vertex in $\alpha$ or $\alpha^T$. All dominant constraint graphs in $\mathcal{C}_{$_Z,2k$}$ are well-behaved. This theorem is surprisingly tricky to prove, so we defer its proof to the appendix. This theorem is not true for all shapes $\alpha$. In particular, this theorem is false for the bipartite shape $\alpha$ with $U_{\alpha} = (u_1,u_2)$, $V_{\alpha} = (v_1,v_2)$, $V(\alpha) = U_{\alpha} \cup V_{\alpha}$, and $E(\alpha) = \left\{\{u_1,v_1\}, \{u_1,v_2\}, \{u_2,v_1\}, \{u_2,v_2\}\right\}$. If $C \in \mathcal{C}_{$_Z,2k$}$ is a dominant constraint graph then $E(C) = E(C_1) \cup E(C_2)$ and the induced constraint graphs $C_1$ and $C_2$ are dominant constraint graphs on $W_1$ and $W_2$. Since $C$ is well-behaved, $E(C) = E(C_1) \cup E(C_2)$. We must have that $C_1$ and $C_2$ are well-behaved as otherwise we would have that $|E(C)| > 2k-2$, which is too large. We now observe that if $C \in \mathcal{C}_{$_Z,2k$}$ is a dominant constraint graph then the corresponding constraints for which spokes are equal to each other are non-crossing (see \ref{['prop:noncrossingspokes']} below). We will then use this observation to prove a key fact about dominant constraint graphs in $\mathcal{C}_{$_Z,2k$}$. If $C \in \mathcal{C}_{$_Z,2k$}$ is a dominant constraint graph then for all $i_1,i_2,i_3,i_4 \in [2k]$ such that $i_1 < i_2 < i_3 < i_4$, if $e_{i_1} \longleftrightarrow_C e_{i_3}$ and $e_{i_2} \longleftrightarrow_C e_{i_4}$ then $e_{i_1} \longleftrightarrow_C e_{i_2} \longleftrightarrow_C e_{i_3} \longleftrightarrow_C e_{i_4}$ Observe that $e_i$ goes between $a_{(\lfloor{\frac{i}{2}}\rfloor + 1)2}$ and $b_{{\lceil{\frac{i}{2}}\rceil}1}$. Thus, if $e_{i_1} \longleftrightarrow e_{i_3}$ and $e_{i_2} \longleftrightarrow e_{i_4}$ then letting $C_1$ and $C_2$ be the restrictions of $C$ to $W_1$ and $W_2$ respectively, we have that $a_{(\lfloor{\frac{i_1}{2}}\rfloor + 1)2} \longleftrightarrow a_{(\lfloor{\frac{i_3}{2}}\rfloor + 1)2}$ and $a_{(\lfloor{\frac{i_2}{2}}\rfloor + 1)2} \longleftrightarrow a_{(\lfloor{\frac{i_4}{2}}\rfloor + 1)2}$. Since $C_2$ is non-crossing, $a_{(\lfloor{\frac{i_1}{2}}\rfloor + 1)2} \longleftrightarrow a_{(\lfloor{\frac{i_2}{2}}\rfloor + 1)2} \longleftrightarrow a_{(\lfloor{\frac{i_3}{2}}\rfloor + 1)2} \longleftrightarrow a_{(\lfloor{\frac{i_4}{2}}\rfloor + 1)2}.$$b_{{\lceil{\frac{i_1}{2}}\rceil}1} \longleftrightarrow b_{{\lceil{\frac{i_3}{2}}\rceil}1}$ and $b_{{\lceil{\frac{i_2}{2}}\rceil}1} \longleftrightarrow b_{{\lceil{\frac{i_4}{2}}\rceil}1}$. Since $C_1$ is non-crossing, $b_{{\lceil{\frac{i_1}{2}}\rceil}1} \longleftrightarrow b_{{\lceil{\frac{i_2}{2}}\rceil}1} \longleftrightarrow b_{{\lceil{\frac{i_3}{2}}\rceil}1} \longleftrightarrow b_{{\lceil{\frac{i_4}{2}}\rceil}1}.$ This implies that $e_{i_1} \longleftrightarrow_C e_{i_2} \longleftrightarrow_C e_{i_3} \longleftrightarrow_C e_{i_4}$. If $C \in \mathcal{C}_{$_Z,2k$}$ is a dominant constraint graph then For all $i < j \in [k]$, if $a_{i1} \longleftrightarrow_C a_{j1}$ then $a_{i2} \longleftrightarrow_C a_{j2}$ and the spokes $\{e_{i'}: i' \in [2i-1,2j-2]\}$ can only be made equal to each other.Similarly, for all $i < j \in [k]$, if $b_{i2} \longleftrightarrow_C b_{j2}$ then $b_{i1} \longleftrightarrow_C b_{j1}$ and the spokes $\{e_{i'}: i' \in [2i,2j-1]\}$ can only be made equal to each other. $a_{i1} \longleftrightarrow_C a_{j1} \implies a_{i2} \longleftrightarrow_C a_{j2}$ $b_{i2} \longleftrightarrow_C b_{j2} \implies b_{i1} \longleftrightarrow_C b_{j1}$ Illustration of \ref{['lem:zshape-spoke-constraint']}: a spoke can only be made equal to spokes of the same color. We only prove the first statement as the proof of the second statement is similar. Let $C_1$ and $C_2$ be the restrictions of $C$ to $W_1$ and $W_2$ respectively. To see that the spokes $\{e_{i'}: i' \in [2i-1,2j-2]\}$ can only be made equal to each other, observe that since $a_{i1} \longleftrightarrow_C a_{j1}$ and $C_1$ is non-crossing, we cannot have that $b_{{\lceil{\frac{i'}{2}}\rceil}1} \longleftrightarrow b_{{\lceil{\frac{i"}{2}}\rceil}1}$ for any $i',i"$ such that $i' \in [2i-1,2j-2]$ and $i" \in [2k] \setminus [2i-1,2j-2]$. To show that $a_{i2} \longleftrightarrow_C a_{j2}$, we apply the following lemma to $\{e_{i'}: i' \in [2i-1,2j-2]\}$. For all $m \in \mathbb{N}$, if $M$ is a perfect matching on the indices $[2m]$ such that no two edges of $M$ cross (i.e. there is no pair of edges $\{i_1,i_2\},\{i_3,i_4\} \in M$ such that $i_1 < i_3 < i_2 < i_4$) then either $\{1,2m\} \in M$ or there is a sequence of indices $i_1 < \ldots < i_l$ such that $i_1 \geq 1$, $i_l < m$, and $\{1,2{i_1}\} \in M$ and $\{2{i_l}+1,2m\} \in M$.For all $j \in [l-1]$, $\{2i_{j} + 1,2{i_{j+1}}\} \in M$. See Figure \ref{['fig:perfect-matching']} for an illustration. We prove this by induction on $m$. The base case $m = 1$ is trivial. For the inductive step, assume the result is true for $m$ and consider a matching $M$ on $[2m+2]$ such that no edges of $M$ cross. If $\{1,2m+2\} \in M$ then we are done so we can assume that $\{1,2m+2\} \notin M$. Choose $s < t \in [2m+2]$ such that $\{s,t\} \in M$ and $t - s$ is minimized. We claim that $t = s+1$. To see this, assume that $t > s+1$. Since $M$ is a perfect matching, $\{s+1,x\} \in M$ for some $x \in [2m+2]$. Since no two edges of $M$ cross, we must have that $s+1 < x < t$. However, this implies that $x - (s+1) < t-s$, contradicting our choice of $s$ and $t$. Now consider the matching $M'$ obtained from $M$ by deleting the indices $s,s+1$ and decreasing all indices greater than $s+1$ by $2$. By the inductive hypothesis, either $\{1,2m\}\in M$, or there is a sequence $i'_1 < \ldots < i'_l$ such that $i'_1 \geq 1$, $i'_l < m$, $\{1,2i'_1\} \in M'$, $\{2{i'_l}+1,2m\} \in M'$, and for all $j \in [l-1]$, $\{2i'_{j} + 1,2i'_{j+1}\} \in M'$. In the latter case we can modify this sequence for $M'$ as follows to obtain the desired sequence for $M$: Increase all indices in this sequence which are greater than or equal to $s$ by $2$.If $s$ is odd and $i'_j = \frac{s-1}{2}$ for some $j \in [l]$, insert $i'_j + 1 = \frac{s+1}{2}$ after $i'_j$. This accounts for the fact that $2i'_j + 1 = s$ was shifted to $s+2$ so we need to add the edge $\{s,s+1\}$ between $2i'_j = s-1$ and $2i'_j + 1 + 2 = s+2$. If $\{1,2m\} \in M'$, then there are three cases: If $1<s<2m+1$ then $\{1,2m+2\} \in M$.If $s=1$ then $\{1,2\}, \{3,2m+2\} \in M$ so we can take the sequence with the single element $i_1 = 1$.If $s = 2m+1$ then $\{1,2m\}, \{2m+1,2m+2\} \in M$ so we can take the sequence with the single element $i_1 = m$. Applying \ref{['lem:matchinglemma']} to $\{e_{i'}: i' \in [2i-1,2j-2]\}$, we have that either $e_{2i-1} \longleftrightarrow_C e_{2j-2}$ or there is a sequence of indices $i_1 < \ldots < i_l$ such that $i \leq i_1$, $i_l < j-1$, and $e_{2i-1} \longleftrightarrow_C e_{2i_{1}}$ and $e_{2i_{l} + 1} \longleftrightarrow_C e_{2j-2}$For all $j' \in [l-1]$, $e_{2i_{j'} + 1} \longleftrightarrow_C e_{2i_{(j'+1)}}$ As before, we observe that $e_{i'}$ goes between $a_{(\lfloor{\frac{i'}{2}}\rfloor + 1)2}$ and $b_{{\lceil{\frac{i'}{2}}\rceil}1}$. Thus, if $e_{2i-1} \longleftrightarrow_C e_{2j-2}$ then $a_{i2} \longleftrightarrow a_{j2}$. If we have a sequence of $i_1 < \ldots < i_l$ satisfying the above properties then we have that $a_{i2} \longleftrightarrow_C a_{({i_1}+1)2}$ and $a_{(i_l + 1)2} \longleftrightarrow_C a_{j2}$For all $j' \in [l-1]$, $a_{(i_{j'} + 1)2} \longleftrightarrow_C a_{(i_{j'+1} + 1)2}$ Together, these statements imply that $a_{i2} \longleftrightarrow_C a_{j2}$, as needed. Illustration of Lemma \ref{['lem:matchinglemma']}: solid lines are matchings. Illustration of Lemma \ref{['lem:zshape-spoke-constraint']}: $a_{11}\longleftrightarrow a_{j1}$ implies $a_{12}\longleftrightarrow a_{j2}$. Illustration of \ref{['lem:matchinglemma']} and \ref{['lem:zshape-spoke-constraint']} Illustration of \ref{['rmk:zshape-spoke-constraint-rmk']}. If we only know that each edge of $H(\alpha_Z,2k)/C$ has multiplicity at least two rather than the stronger statement that each edge of $H(\alpha_Z,2k)/C$ has even multiplicity then $\alpha_{11}\longleftrightarrow_C a_{i1}$ does not imply $a_{12}\longleftrightarrow_C a_{i2}$. \ref{['lem:zshape-spoke-constraint']} relies on the fact that $\mathop{\mathrm{val}}\nolimits(C) = 0$ whenever there is an edge in $H(\alpha_Z,2k)/C$ which has odd multiplicity. There are constaint graphs $C \in \mathcal{C}_{$_Z,2k$}$ which do not satisfy the conclusion \ref{['lem:zshape-spoke-constraint']} where each edge in $H(\alpha_Z,2k)/C$ appears with multiplicity at least $2$. For an example of this, see \ref{['fig:zshape-spoke-constraint-rmk']}. We are now ready to prove the main result of this section, \ref{['thm:num-zshape-constraint-graph']}. For all $k \in \mathbb{N}$ and $j \in \{0,1,2\}$, we define $H_j$_Z,2k$$ to be the graph obtained by starting with $H$_Z,2k$$, merging the vertices $b_{1j'}$ and $b_{kj'}$ for all $j' \in [j]$, deleting the vertices $\left\{a_{1j'}: j' \in [j]\right\}$ and all edges incident to these vertices, and deleting the spokes incident to $a_{1(j+1)}$. Equivalently, letting $E_j = \left\{\{b_{1j'},b_{kj'}\}: j' \in [j]\right\}$, $H_j$_Z,2k$$ is the graph obtained by taking $H$_Z,2k$/{E_j}$, deleting all multi-edges with even multiplicity, and then deleting all isolated vertices. Illustration of \ref{['defn:H_j_Z']}. Note that for $j = 1$, we have the red constraint edge between $b_{11}$ and $b_{k1}$ but we do not have the blue constraint edge beween $b_{12}$ and $b_{k2}$. For all $k \in \mathbb{N}$ and all $j \in \{0,1,2\}$, define $D(k,j)$ to be the number of dominant constraint graphs on $H_j$_Z,2k$$. Equivalently, $D(k,j)$ is the number of dominant constraint graphs $C \in \mathcal{C}_{$_Z,2k$}$ such that for all $j' \in [j]$, $b_{1j'} \longleftrightarrow_{C} b_{kj'}$. We set $D(0,0) = 1$. Note that $D(k,0) = \abs{C \in \mathcal{C}_{$_Z,2k$}: C \text{ is dominant}}$ and $D(k,2) = D(k-1,0)$. For all $k \in \mathbb{N}$ and $j \in \{0,1\}$, $D(k,j) = \sum_{i=0}^{k}{D(i,0)D(k-i,j+1)}$ Given a dominant constraint graph $C$ on $H_j$_Z,2k$$, let $i \in [k]$ be the last index such that $i=1$ or $a_{1(j+1)} \longleftrightarrow_{C} a_{i(j+1)}$. We have that $i = 1$ if and only if $b_{1(j+1)} \longleftrightarrow_{C} b_{k(j+1)}$. There are $D(k,j+1)$ dominant constraint graphs on $H_j$_Z,2k$$ for which this is true. If $i > 1$ then we make the following observations: By \ref{['lem:zshape-spoke-constraint']}, for all $j' > j+1$, $a_{1j'} \longleftrightarrow_{C} a_{ij'}$.$b_{i(j+1)} \longleftrightarrow_C b_{k(j+1)}$ as otherwise the edges $\left\{a_{i(j+1)},b_{i(j+1)}\right\}$ and $\left\{b_{k(j+1)},a_{1(j+1)}\right\}$ would only appear once in $H_j$_Z,2k$/C$. By \ref{['lem:zshape-spoke-constraint']}, for all $j' \leq j$, $b_{ij'} \longleftrightarrow_C b_{kj'}$. Illustration of \ref{['lem:z-shape-split-1']} and \ref{['lem:z-shape-split-2']}: case $j=0$ $E = \left\{\{a_{1j'},a_{ij'}\}: j' \geq j+1\right\} \bigcup \left\{\{b_{ij'},b_{kj'}\}: j' \leq j+1\right\}$ partitions $H_j$_Z,2k$$ into two parts $H_1$ and $H_2$ where $H_1 \cong H$_Z,2(i-1)$$ and $H_2 \cong H_{j+1}$_Z,2(k+1-i)$$. Let $m = 2$. We take $H_1$ to be the induced subgraph of $H_j$_Z,2k$$ on the vertices $V(H_1) = \left\{a_{i'j'}: i' \in [2,i], j' \in [m]\right\} \cup \left\{b_{i'j'}: i' \in [i-1], j' \in [m]\right\}$ and we take $H_2$ to be the induced subgraph of $H_j$_Z,2k$$ on the vertices V(H_2) =\left\{a_{i'j'}: i' \in [i+1,k], j' \in [m]\right\} \cup \left\{a_{ij'}: j' \in [j+2,m]\right\}\cup \left\{b_{i'j'}: i' \in [i+1,k], j' \in [m]\right\} \cup \left\{b_{ij'}: j' \in [j+2,m]\right\} To see that $H_2 \cong H_{j+1}$_Z,2(k+1-i)$$, note that by the above observations, for all $j' \leq j+1$, $b_{ij'} \longleftrightarrow_C b_{kj'}$ and for all $j' \in [j+1,m]$, $a_{1j'} \longleftrightarrow_C a_{ij'}$. Subtracting $i-1$ from all of the $i'$ indices of $H_2$ gives $H_{j+1}$_Z,2(k+1-i)$$. To see that $H_1 \cong H$_Z,2(i-1)$$, observe that for all $j' \in [m]$, all edges and spokes in the range between $b_{1j'}$ and $a_{ij'}$ are present. For the edges $\left\{\{a_{ij'},b_{1j'}\}: j' \in [m]\right\}$ and the spokes $\left\{\{a_{i(j')}, b_{1(j'-1)}\}: j' \in [2,m]\right\}$, observe that If $j' \leq j$ then the edge $\left\{a_{ij'},b_{1j'}\right\}$ comes from the edge between $a_{ij'}$ and $b_{ij'}$ as $b_{1j'} \longleftrightarrow_C b_{kj'} \longleftrightarrow_C b_{ij'}$. Similarly, if $j' \leq j+1$ then the spoke $\left\{a_{i(j')}, b_{1(j'-1)}\right\}$ comes from the spoke $\left\{a_{i(j')}, b_{i(j'-1)}\right\}$If $j' \geq j+1$ then the edge $\left\{a_{ij'},b_{1j'}\right\}$ comes from the edge between $a_{1j'}$ and $b_{1j'}$ as $a_{1j'} \longleftrightarrow_C a_{ij'}$. Similarly, if $j' \geq j+2$ then the spoke $\left\{a_{ij'},b_{1(j'-1)}\right\}$ comes from the spoke $\left\{a_{1(j')}, b_{1(j'-1)}\right\}$. Note that the spoke $\left\{a_{i(j+1)}, b_{1(j)}\right\}$ cannot come from the spoke $\left\{a_{1(j+1)}, b_{1j}\right\}$ as this spoke was deleted in $H_j$_Z,2k$$. Since all of the needed edges are present and no extra edges are present, $H_1 \cong H$_Z,2(i-1)$$. Finally, we use \ref{['lem:Hpartitionconditions']} to confirm that $E$ partitions $H_j$_Z,2k$$ into the two parts $H_1$ and $H_2$. The first condition of \ref{['defn:Hpartitions']} and the first condition of \ref{['lem:Hpartitionconditions']} can be seen directly. For the second and third conditions of \ref{['lem:Hpartitionconditions']}, we observe that $V(H_1) \cap V(H_2) = \left\{a_{1j'}: j' \in [j+2,m]\right\} \cup \left\{b_{1j'}: j' \in [j]\right\}$. Since all dominant constraint graphs on $H_j$_Z,2k$$, $H_1$, and $H_2$ are well-behaved, none of these constraint graphs can have a path between any two vertices in $V(H_1) \cap V(H_2) = \left\{a_{1j'}: j' \in [j+2,m]\right\} \cup \left\{b_{1j'}: j' \in [j]\right\}$. To see that the fourth condition of \ref{['lem:Hpartitionconditions']} holds, assume that there are edges $e_1,e_2 \in E(H)$ which appear with odd multiplicity in $H/{C_E}$ such that $e_1$ appears in $H_1$, $e_2$ appears in $H_2$, and $e_1 \longleftrightarrow_{C} e_2$. Since $C$ is well-behaved, we must either have that $e_1$ and $e_2$ are edges in $W_i$ for some $i \in [m]$ or $e_1$ and $e_2$ are spokes between $W_i$ and $W_{i+1}$ for some $i \in [m-1]$. Using similar logic as we used to prove \ref{['lem:lineshapepartitioning']}, we can rule out the possibility that $e_1$ and $e_2$ are both in $W_i$ for some $i \in [m]$. The possibility that $e_1$ and $e_2$ are spokes between wheels $W_i$ and $W_{i+1}$ for some $i \in [m-1]$ is ruled out by \ref{['lem:zshape-spoke-constraint']} which implies that spokes in $H_1$ are not made equal to spokes in $H_2$. $D(k,0) = \sum_{\substack{i_1, i_2, i_3\geq 0: \\ i_1+i_2+i_3 = k-1}} Di_1,0 Di_2,0 Di_3, 0.$ By \ref{['lem:z-shape-split-1']} and \ref{['defn:D-k-j']}, D(k,0)= \sum_{i_1=0}^{k}D(i_1,0)D(k-i_1,1)= \sum_{i_1=0}^{k}D(i_1,0)\sum_{i_2=0}^{k-i_1} D(i_2,0)D(k-i_1-i_2, 2)= \sum_{i_1=0}^{k} D(i_1,0) \sum_{i_2=0}^{k-i_1} D(i_2,0)D k-i_1-i_2-1, 0= \sum_{\substack{i_1, i_2, i_3\geq 0: \\ i_1+i_2+i_3 = k-1}} Di_1,0 Di_2,0 Di_3, 0. By \ref{['cor:z-shape-recurrence-relation']} and \ref{['thm:catalan2-recurrence-relation']}, $D(k,0)$ has the same recurrence relation as $C'_k = \dfrac{1}{2k+1}\binom{3k}{k}$, thus \ref{['thm:num-zshape-constraint-graph']} is proved. We now sketch how to show that $\mathop{\mathrm{Var}}\nolimits (\mathop{\mathrm{tr}}\nolimits$$M_{\alpha_Z}M_{\alpha_Z}^T$^k$)$ is $O$n^{4k+2}$$. For each $k$, $\mathop{\mathrm{Var}}\nolimits (\mathop{\mathrm{tr}}\nolimits$$M_{\alpha_Z}M_{\alpha_Z}^T$^k$)$ is $O$n^{4k+2}$$. Observe that \mathop{\mathrm{Var}}\nolimits\mathop{\mathrm{tr}}\nolimitsM_{\alpha_Z}M_{\alpha_Z}^T^k = \mathbb{E}\mathop{\mathrm{tr}}\nolimits$$M_{\alpha_Z}M_{\alpha_Z}^T$^k$^2 - \mathbb{E}\mathop{\mathrm{tr}}\nolimits$$M_{\alpha_Z}M_{\alpha_Z}^T$^k$^2=\sum_{A_1,B_1,A'_1,B'_1,\ldots,A_k,B_k,A'_k,B'_k \subseteq \binom{[n]}{2}}\, (\mathbb{E}\prod_{i=1}^{k}\, M_{\alpha_Z}$A_i,B_i$M_{\alpha_Z}^T$B_i,A_{i+1}$ M_{\alpha_Z}$A_i',B_i'$ M_{\alpha_Z}^T$B_i',A_{i+1}'$\space - \mathbb{E}\prod_{i=1}^{k}\, M_{\alpha_Z}$A_i,B_i$M_{\alpha_Z}^T$B_i,A_{i+1}$ \cdot \mathbb{E}\prod_{i=1}^{k}\, M_{\alpha_Z}$A_i',B_i'$ M_{\alpha_Z}^T$B_i',A_{i+1}'$) To analyze the terms in the summation, we consider two copies $H_1$ and $H_2$ of $$_Z, 2k$$ with vertices $\{A_i: i \in [k]\} \cup \{B_i: i \in [k]\}$ and $\{A'_i: i \in [k]\} \cup \{B'_i: i \in [k]\}$ respectively, where the constraint edges of a constraint graph $C$ can now have edges between $H_1$ and $H_2$ as well as edges within $H_1$ and edges within $H_2$. Denote \mathop{\mathrm{val}}\nolimits(C)= \mathbb{E}[\prod_{i=1}^{k}\, M_{\alpha_Z}A_i,B_i M_{\alpha_Z}^TB_i,A_{i+1} M_{\alpha_Z}A_i',B_i' M_{\alpha_Z}^TB_i',A_{i+1}'] -\mathbb{E}[\prod_{i=1}^{k}\, M_{\alpha_Z}A_i,B_i M_{\alpha_Z}^TB_i,A_{i+1}] \mathbb{E}[\prod_{i=1}^{k}\, M_{\alpha_Z}A_i',B_i' M_{\alpha_Z}^TB_i',A_{i+1}']. Observe that $\mathop{\mathrm{val}}\nolimits(C) = 1$ if under $C$, each edge appears an even number of times in $H_1 \cup H_2$, but there is at least one edge that appears an odd number of times when restricted to only $H_1$ or $H_2$ (call this special edge the "odd" edge). $\mathop{\mathrm{val}}\nolimits(C) = 0$ in all other situations. It suffices to prove that for the constraint graphs $C$ such that $\mathop{\mathrm{val}}\nolimits(C) = 1$, $(H_1 \cup H_2)/C$ contains at most $4k+2$ distinct vertices as this implies that $\sum_{C} N(C)\mathop{\mathrm{val}}\nolimits(C) = O_k$n^{4k+2}$$. In this proof sketch, we will only prove the above claim for well-behaved constraint graphs $C$. For the full analysis, see \ref{['thm:variancewellbehaved']}. For well-behaved constraint graphs $C$, there are two cases: The odd edge $e$ is among one of the wheels of $H_1$ or $H_2$. Let $W_{i,j}$ denote the $j^{th}$ wheel of $H_i$. Observe that for each $j=1,2$, $W_{1,j}\cup W_{2,j}$ has $4k$ edges, thus $(W_{1,j}\cup W_{2,j})/C$ has at most $2k$ multi-edges (since each edge has multiplicity at least $2$). W.L.O.G. assume $e$ is in $W_{1,1}$, the outer wheel of $H_1$. In this case, the following statements are true. $(W_{1,1}\cup W_{2,1})/C$ is a connected graph: since $e$ is an odd edge, it is constrained to some edge in $W_{2,1}$.$(W_{1,1}\cup W_{2,1})/C$ has at least one cycle: each vertex in $W_{1,1}/C$ has an even degree and the multi-edge containing $e$ has an odd multiplicity, thus there must be at least one cycle. This implies that $(W_{1,1}\cup W_{2,1})/C$ has at most $2k$ vertices. On the other hand, $(W_{1,2}\cup W_{2,2})/C$ has at most $2k+2$ vertices because it has at most two connected components. Thus, all together there are at most $2k + (2k+2) = 4k+2$ distinct vertices.The odd edge $e$ is among the spokes. Without loss of generality, assume $e\in H_1$. Then $e$ is constrained to some $e'$ in $H_2$. This implies that for each $j=1,2$, $(W_{1,j}\cup W_{2,j})/C$ is connected, and thus has at most $2k+1$ vertices. Thus, all together there are at most $2(2k+1) = 4k+2$ distinct vertices. In this section, we show how to use \ref{['thm:zshape-trace-convergence']} to derive the limiting distribution of the singular values of $\frac{M_{\alpha_Z}}{n}$ as $n \to \infty$. Before we derive the limiting distribution of the spectrum of the singular values of Z-shaped graph matrices, we will show a simpler but similar analysis for deriving Wigner's Semicircle Law from the trace power moments. A Wigner matrix $M$ is a random Hermitian matrix where entries $M_{ij}$ for $i<j$ are i.i.d. complex random variables with mean $0$ and variance $1$, $M_{ij} = \overline{M_{ji}}$ for $i>j$, and the diagonal entries are i.i.d real random variables with bounded mean and variance. The semi-circle law $F_{sc}(x)$ is a distribution function whose density function is $f_{sc}(x) = F_{sc}'(x) = \;\frac{1}{2\pi} \sqrt{4-x^2}\text{ if } |x|\leq 2\;0\text{ otherwise}$ If $M_n$ is a sequence of $n\times n$ Wigner matrices whose entries have bounded moments then the distribution of eigenvalues of $\frac{M_n}{\sqrt{n}}$ weakly converges almost surely to $F_{sc}(x)$. The analysis in Sections \ref{['sec:dominantcycleconstraintgraphs']} and \ref{['sec:cyclerecurrencerelation']} for the $\pm{1}$ random matrix $M_{\alpha_0}$ can be easily generalized to Wigner matrices so we have the following theorem. If $M_n$ is a sequence of $n\times n$ Wigner matrices whose entries have bounded moments then $\lim_{n\to\infty} \dfrac{\,1\,}{n} \mathbb{E}\, \mathop{\mathrm{tr}}\nolimits$$\frac{M_n}{\sqrt{n}}$^{2k}$ = C_k$ where $C_k = \dfrac{1}{k+1}\binom{2k}{k}$ is the $k^{th}$ Catalan number. One can verify that $\int_{-2}^{2}\, \dfrac{1}{2\pi}\sqrt{4-x^2}\cdot x^{2k} \,dx = C_k$ and conclude \ref{['thm:wigner']} after checking the other conditions in \ref{['lem:moment-convergence']}. Here, we will instead show how to derive $f_{sc}(x)$ by deriving a differential equation for $f_{sc}(x)$ from the recurrence relation for $C_k$. Assume $f(x)$ is a function satisfying that for all $k \in \mathbb{N}$, $\int_{-2}^{2} x^{2k}\cdot f(x)\,dx = C_k$ and moreover, $f(x)$ is continuously differentiable on $(-2,2)$.$\lim_{x\to 2^{-}} f(x) = \lim_{x\to -2^{+}} f(x) = 0$. Then $f(x)$ satisfies the following differential equation on $(-2,2)$: $(4-x^2)f'(x) + xf(x) = 0.$ We use the following recurrence relation for the Catalan numbers. $\dfrac{C_{k+1}}{C_{k}} = \dfrac{2(2k+1)}{k+2}.$ \dfrac{C_{k+1}}{C_{k}} = \dfrac{\dfrac{1}{k+2}\cdot \dfrac{(2k+2)!}{(k+1)!(k+1)!}}{\dfrac{1}{k+1}\cdot \dfrac{(2k)!}{k!k!}} = \dfrac{(k+1)\cdot (2k+2)(2k+1)}{(k+2)\cdot (k+1)^2} = \dfrac{2(2k+1)}{k+2}. For any $k\geq 1$, using integration by parts and the boundary condition that $\lim_{x\to 2^{-}} f(x) = \lim_{x\to -2^{+}} f(x) = 0$, \int_{-2}^2 x^{2k+1}\cdot f'(x) \,dx= x^{2k+1}\cdot f(x)_{-2}^2 - (2k+1)\cdot \int_{-2}^2 f(x)\cdot x^{2k} \,dx = -(2k+1) C_k We have the following relations together with \ref{['eqn:rel-C_k']}. $\int_{-2}^2 x^{2k}\cdot f(x) \,dx = C_k$$\int_{-2}^2 x^{2k+2}\cdot f(x) \,dx = C_{k+1}$$\int_{-2}^2 x^{2k+1}\cdot f'(x) \,dx = -(2k+1)C_k$$\int_{-2}^2 x^{2k+3}\cdot f'(x) \,dx = -(2k+3)C_{k+1}$$(k+2) C_{k+1} = 2(2k+1) C_k$ Multiplying v. by $-2$ and rewriting using i. to iv., we get -2(k+2)C_{k+1} = -(2k+3)C_{k+1} - C_{k+1} = -4(2k+1)C_k\implies\int_{-2}^2 x^{2k+3}\cdot f'(x) \,dx - \int_{-2}^2 x^{2k+2}\cdot f(x) \,dx = 4 \int_{-2}^2 x^{2k+1}\cdot f'(x) \,dx\implies\int_{-2}^2 x^{2k+1}\cdot (4-x^2)f'(x) + xf(x) \, dx = 0. Using a similar argument as in \ref{['lem:zero even moments']}, we can conclude that $(4-x^2)f'(x) + xf(x) = 0.$ Finally we can solve the ODE for the solution of $f(x)$. Assume $f(x)$ satisfies \ref{['eqn:ODE-wigner']} and $\int_{-2}^2 f(x)\,dx = 1$. Then $f(x) = \dfrac{1}{2\pi}\sqrt{4-x^2}$ on $(-2,2)$. (4-x^2)f'(x) + xf(x) = 0 \implies \dfrac{f'(x)}{f(x)} = -\dfrac{x}{4-x^2}\implies\int \dfrac{f'(x)}{f(x)} \, dx = \int -\dfrac{x}{4-x^2} \, dx \implies \ln\abs{f(x)} = \dfrac{\,1\,}{2}\ln\abs{4-x^2} + C\impliesf(x) = A\cdot \sqrt{4-x^2} \text{ for some constant } A\impliesf(x) = \dfrac{1}{2\pi} \sqrt{4-x^2} \text{ since } \int_{-2}^2 f(x)\,dx = 1 We now find the limiting distribution of the spectrum of the singular values of $\frac{M_{\alpha_Z}}{n}$ as $n \to \infty$. Let $a = \dfrac{3\sqrt{3}}{2}$ and let $g_{\alpha_Z}: (0,\infty) \to \mathbb{R}$ be the function such that $g_{\alpha_Z}(x) = \dfrac{i}{\pi}\cdot \sqrt{3}\cdot \sin\dfrac{1}{3}\cdot\arctan\dfrac{3}{\sqrt{4x^2/3-9}}+ \cos\dfrac{1}{3}\cdot\arctan\dfrac{3}{\sqrt{4x^2/3-9}}$ if $x \in (0,a]$ and $g_{\alpha_Z}(x) = 0$ if $x > a$ where we take $\arctan(ix) = \dfrac{i}{2}\ln${1+x}{1-x}$$ for all real $x$, we take $ln(-x) = {\pi}i + ln(x)$ for all $x > 0$, and we take $sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$ and $cos(x) = \frac{e^{ix} + e^{-ix}}{2}$ for all complex $x$. With probability $1$, the spectrum of the singular values of $\frac{1}{n}M_{\alpha_Z}$ approaches $g_{\alpha_Z}$ weakly as $n\to\infty$. By \ref{['thm:zshape-trace-convergence']}, it suffices to prove that for all $k \in \mathbb{N}$, $\mathbb{E}_{x\sim g_{\alpha_Z}(x)} [x^{2k}] = \int_{x = 0}^{a}{x^{2k}g_{\alpha_{Z}}(x) \,dx} = C'_k.$ To prove that $\int_{x = 0}^{a}{x^{2k}g_{\alpha_{Z}}(x) \,dx} = C'_k$, we proceed as follows: We derive a differential equation for $g_{\alpha_{Z}}$ based on a recurrence relation for $C'_k$ (see Theorem \ref{['thm:ODE for f(x) of Z-shape']}).We prove that if $g_{\alpha_{Z}}$ satisfies this differential equation and some conditions at $x = 0$ and $x = a$ then $\int_{x = 0}^{a}{x^{2k}g_{\alpha_{Z}}(x)dx} = C'_k$ (see Theorem \ref{['thm:verify ODE for Z-shape spec']}).We verify that $g_{\alpha_{Z}}$ satisfies the required conditions (see Theorem \ref{['thm:solution to ODE for Z-shape']}). Technically, only steps 2 and 3 are needed. We include the first step because it gives better intuition for where the differential equation comes form. Let $C'_k = \dfrac{1}{2k+1}\binom{3k}{k}$ and $a=\lim_{k\to\infty} C'_{k+1}/C'_k = 3\sqrt{3}/2$. Assume $f(x)$ is an function satisfying that for all $k \in \mathbb{N}$, $\int_{0}^{a} x^{2k}\cdot f(x)\mathop{\mathrm{dx}}\nolimits = C'_k$ and moreover, $f(x)$ is twice continuously differentiable on $(0,a)$.$\lim_{x\to 0^{+}}xf(x)=0$ and $\lim_{x\to 0^{+}}x^2f'(x)=0$.$\lim_{x\to a^{-}}f(x)=0$ and $\lim_{x\to a^{-}}f'(x)(4x^2-27)=\lim_{x \to a^{-}}8af'(x)(x-a)=0$.$\lim_{x \to 0^{+}}{x^3f"(x)} = 0$ and $\lim_{x\to a^{-}}(a-x)^{2}f"(x) = 0$. Then $f(x)$ satisfies the following differential equation on $(0,a)$: $(4x^4-27x^2)f"(x)+(8x^3-27x)f'(x)+3f(x)=0.$ To prove this, we use the following recurrence relation for $C'_k = \dfrac{1}{2k+1}\binom{3k}{k}$. $\dfrac{{C'_k}}{{C'_{k-1}}} = \dfrac{3(3k-1)(3k-2)}{2k(2k+1)}.$ Observe that \dfrac{C'_k}{C'_{k-1}}= \frac{2k-1}{2k+1} \cdot \frac{(3k)!(2k-2)!(k-1)!}{(3k-3)!(2k)!k!}= \frac{2k-1}{2k+1} \cdot \frac{(3k)(3k-1)(3k-2)}{(2k)(2k-1)k} = \dfrac{3(3k-1)(3k-2)}{2k(2k+1)}. We also need the following relationship between the moments of $f$ and the moments of its derivatives. For all $j \in \{0,1,2\}$ and $k \in \mathbb{Z}$ such that $k \geq j$, we define $A(j,k)$ to be $A(j,k):=\int_{0}^{a} f^{(j)}(x)\cdot x^{k}\mathop{\mathrm{dx}}\nolimits$ where $f^{(j)}(x)$ denotes the $j^{th}$ derivative of $f$. Notice that $A(0,2k) = C'_k$. For all $j \in \{1,2\}$ and $k \in \mathbb{Z}$ such that $k \geq j$, $A(j,k) = f^{(j-1)}(x) \cdot x^k_0^{a} - kA(j-1,k-1).$ Using integration by parts, we have that A(j,k) = \int_{0}^{a}{f^{(j)}(x)\cdot x^{k}\mathop{\mathrm{dx}}\nolimits}= f^{(j-1)}(x) \cdot x^k_0^{a} - \int_{0}^{a} kf^{(j-1)}(x)\cdot x^{k-1}\mathop{\mathrm{dx}}\nolimits= f^{(j-1)}(x) \cdot x^k_0^{a} - kA(j-1,k-1). If $\lim_{x\to 0^{+}}xf(x)=0$ and $\lim_{x\to a^{-}}f(x)=0$ then For all $k \in \mathbb{N}$, $A(1,k) = kA(0,k-1)$For all $k \in \mathbb{N}$ such that $k \geq 2$, $A(2,k) = f'(x) \cdot x^k_0^{a} - kA(1,k-1) = f'(x) \cdot x^k_0^{a} + k(k-1)A(0,k-2).$ Using Corollary \ref{['cor:relationsbetweenA']}, Proposition \ref{['prop:Drecurrence']}, and the fact that $A(0,2k) = {C'_k}$, we have that for all $k \in \mathbb{N}$, A(2,2k+2)= f'(x)\cdot x^{2k+2}_{0}^{a} - 2kA(1,2k+1) - 2A(1,2k+1)= f'(x)\cdot x^{2k+2}_{0}^{a} + (2k)(2k+1)A(0,2k) - 2A(1,2k+1)= f'(x)\cdot x^{2k+2}_{0}^{a} + 3(3k-1)(3k-2)A(0,2k-2) - 2A(1,2k+1). Multiplying both sides by $4$ and repeatedly applying Corollary \ref{['cor:relationsbetweenA']}, we get that 4A(2,2k+2)= 4f'(x)\cdot x^{2k+2}_{0}^{a} + 27(2k)(2k-1)A(0,2k-2) + (-54k+24)A(0,2k-2) - 8A(1,2k+1)= 4f'(x)\cdot x^{2k+2}_{0}^{a} + 27\cdot-\begin{equation*} f'(x)\cdot x^{2k} \end{equation*}_{0}^{a} + A(2,2k) - 27(2k-1)A(0,2k-2)- 3A(0,2k-2) - 8A(1,2k+1)= x^{2k}f'(x)\cdot$4x^2-27$_0^{a} + 27A(2,2k) + 27A(1,2k-1) - 3A(0,2k-2) - 8A(1,2k+1)= 27A(2,2k) + 27A(1,2k-1) - 3A(0,2k-2) - 8A(1,2k+1). where the last equality holds because $\lim_{x \to a^{-}}(4x^2-27)f'(x)=0$ and $\lim_{x \to 0^{+}}x^2f'(x)=0$ by assumption. Writing the $A(j,k)$'s above as integrals, we get that for all $k \in \mathbb{N}$ \int_{0}^{a} 4f"(x)\cdot x^4 - 27 f"(x)\cdot x^{2} - 27 f'(x) \cdot x + 8 f'(x) \cdot x^{3} + 3 f(x)\cdot x^{2k-2} \mathop{\mathrm{dx}}\nolimits = 0. One way for this equation to be true is if $(4x^4-27x^2)f"(x)+(8x^3-27x)f'(x)+3f(x)=0$ on $(0,a)$. As shown by the following lemma and corollary, this is the only way for this equation to be true for all $k \in \mathbb{Z}$, which completes the proof of Theorem \ref{['thm:ODE for f(x) of Z-shape']}. Let $a$ be some positive constant. If $f$ is continuous on $[0,a]$ and $\int_{0}^{a} f(x)x^{2k}\mathop{\mathrm{dx}}\nolimits=0$ for all nonnegative integers $k$, then $f=0$ on $(0,a)$. Let $M>0$ be an upper bound of $f$ on $[0,a]$. For an arbitrary $\epsilon>0$, let $p(x)$ be a polynomial such that $\abs{p(x)-f(\sqrt{x})}<\dfrac{\epsilon}{M\cdot a}$ for all $x\in(0,a^2)$. Taking $p_\epsilon(x)=p(x^2)$, $p_\epsilon$ is a linear combination of monomials of even power and $\abs{p_\epsilon(x)-f(x)}<\dfrac{\epsilon}{M\cdot a}$ for all $x\in(0,a)$. Thus $\int_{0}^a$f(x)-p_(x)$\cdot f(x)\mathop{\mathrm{dx}}\nolimits <\epsilon$. On the other hand, since all even moments of $f(x)$ are zero, $\int_{0}^a f(x)-p_\epsilon(x)\cdot f(x)\mathop{\mathrm{dx}}\nolimits = \int_{0}^a f(x)^2 - p_\epsilon(x)f(x) \mathop{\mathrm{dx}}\nolimits = \int_{0}^a f(x)^2\mathop{\mathrm{dx}}\nolimits.$ Thus $\int_0^a f(x)^2\mathop{\mathrm{dx}}\nolimits < \epsilon$ for all $\epsilon>0$ and we conclude that $f(x)=0$ on $(0,a)$. Let $a$ be some positive constant. If $f$ is continuous on $(0,a)$, $\lim_{x\to 0^{+}}x^{2}f(x)=0$, $\lim_{x \to a^{-}}{(a-x)^2f(x)} = 0$, and $\int_{0}^{a} f(x)x^{2k}\mathop{\mathrm{dx}}\nolimits=0$ for all nonnegative integers $k$, then $f=0$ on $(0,a)$. This follows by applying Lemma \ref{['lem:zero even moments']} to the function $f(x)x^2(a - x)^2$. We now confirm that if $f$ satisfies the differential equation $(4x^4-27x^2)f"(x)+(8x^3-27x)f'(x)+3f(x)=0$, the conditions of Theorem \ref{['thm:ODE for f(x) of Z-shape']}, and the condition that $\int_{0}^a f(x)\mathop{\mathrm{dx}}\nolimits =1$, then $\int_{0}^{a} x^{2k}\cdot f(x)\mathop{\mathrm{dx}}\nolimits = {C'_k}$. Let $a=\lim_{k\to\infty} {C'_k}/{C'_{k-1}} = 3\sqrt{3}/2$. Let $f$ be a function satisfying the following ODE on $(0,a)$ $(4x^4-27x^2)f"(x)+(8x^3-27x)f'(x)+3f(x)=0$ and the first three conditions in Theorem \ref{['thm:ODE for f(x) of Z-shape']}, i.e. $f(x)$ is twice continuously differentiable on $(0,a)$.$\lim_{x\to 0^{+}}xf(x)=0$ and $\lim_{x\to 0^{+}}x^2f'(x)=0$.$\lim_{x\to a^{-}}f(x)=0$ and $\lim_{x\to a^{-}}f'(x)(4x^2-27)=\lim_{x \to a^{-}}8af'(x)(x-a)=0$. Moreover, assume that $\int_{0}^a f(x)\mathop{\mathrm{dx}}\nolimits =1$. Then for all $k \in \mathbb{N} \cup \{0\}$, $\int_{0}^{a} x^{2k}\cdot f(x)\mathop{\mathrm{dx}}\nolimits = C'_k \,.$ Notice that $\int_{0}^{a}f(x)\mathop{\mathrm{dx}}\nolimits=1 = {C_0}'$ by assumption. We aim to prove that for all $k \in \mathbb{N} \cup \{0\}$, $(2k+3)(2k+2)\int_{0}^{a} x^{2k+2}\cdot f(x)\mathop{\mathrm{dx}}\nolimits = 3(3k+2)(3k+1)\int_{0}^{a} x^{2k}\cdot f(x)\mathop{\mathrm{dx}}\nolimits.$ If so, then since $(2k+3)(2k+2){C'_{k+1}} = 3(3k+2)(3k+1){C'_{k}}$, we can prove Theorem \ref{['thm:verify ODE for Z-shape spec']} by induction on $k$. We multiply \ref{['eqn: ODE for spectrum of Z-shape']} by $x^{2k}$ and integrate from $0$ to $a$: 0=\int_{0}^{a}(4x^4-27x^2)f"(x)\cdot x^{2k} + (8x^3-27x)f'(x)\cdot x^{2k} + 3x^{2k}f(x) \mathop{\mathrm{dx}}\nolimits= \begin{equation*} f'(x)(4x^4-27x^2)x^{2k} \end{equation*}_{0}^a - \int_{0}^a f'(x)4(2k+4)x^{2k+3} - 27(2k+2)x^{2k+1}\mathop{\mathrm{dx}}\nolimits\space+ \int_{0}^a (8x^{2k+3}-27x^{2k+1})f'(x)\mathop{\mathrm{dx}}\nolimits + \int_{0}^a 3x^{2k}f(x)\mathop{\mathrm{dx}}\nolimits= -\int_{0}^a 8(k+1)x^{2k+3} - 27(2k+1)x^{2k+1} f'(x)\mathop{\mathrm{dx}}\nolimits + \int_{0}^a 3x^{2k}f(x)\mathop{\mathrm{dx}}\nolimits= -f(x)$8(k+1)x^{2k+3} - 27(2k+1)x^{2k+1}$_0^a\space+ \int_{0}^a 8(k+1)(2k+3)x^{2k+2} - 27(2k+1)^2 x^{2k} f(x)\mathop{\mathrm{dx}}\nolimits + \int_{0}^a 3x^{2k}f(x)\mathop{\mathrm{dx}}\nolimits= \int_{0}^a 8(k+1)(2k+3)x^{2k+2} - 3(36k^2+36k+8) x^{2k} f(x)\mathop{\mathrm{dx}}\nolimits= \int_{0}^a 8(k+1)(2k+3)x^{2k+2} - 12(3k+1)(3k+2) x^{2k} f(x)\mathop{\mathrm{dx}}\nolimits as $f'(x)(4x^4-27x^2)x^{2k}_{0}^a$ and $f(x)$8(k+1)x^{2k+3} - 27(2k+1)x^{2k+1}$_0^a$ are zero by the assumed conditions on $f$. Rearranging the last step we get $(2k+2)(2k+3)\int_0^a f(x)x^{2k+2} \mathop{\mathrm{dx}}\nolimits = 3(3k+1)(3k+2)\int_{0}^a f(x)x^{2k}\mathop{\mathrm{dx}}\nolimits$ as needed. Using WolframAlpha to solve the above ODE and analyzing the constant coefficient by the imposed initial conditions, we can get an explicit solution for $f(x)$. We verify the solution below. Let $a=3\sqrt{3}/2$ and $f: (0,a) \to \mathbb{R}$ be the function such that $f(x)= \dfrac{i}{\pi}\cdot \sqrt{3}\cdot \sin\dfrac{1}{3}\cdot\arctan\dfrac{3}{\sqrt{4x^2/3-9}}+ \cos\dfrac{1}{3}\cdot\arctan\dfrac{3}{\sqrt{4x^2/3-9}}$ where we take $\arctan(ix) = \dfrac{i}{2}\ln${1+x}{1-x}$$ for all real $x$, we take $ln(-x) = {\pi}i + ln(x)$ for all $x > 0$, and we take $sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$ and $cos(x) = \frac{e^{ix} + e^{-ix}}{2}$ for all complex $x$. Then $f(x)$ is an solution to the ODE \ref{['eqn:ODE for f(x) of Z-shape']} on $(0,a)$. Moreover, $f$ satisfies the conditions listed in Theorem \ref{['thm:verify ODE for Z-shape spec']}. We first verify that this $f(x)$ satisfies the ODE \ref{['eqn:ODE for f(x) of Z-shape']} $(4x^4-27x^2)f"(x)+(8x^3-27x)f'(x)+3f(x)=0.$ on $(0,a)$. For simplicity, let $g(x)=\dfrac{1}{3}\cdot\arctan${3}{{4x^2/3-9}}$$. Then $f(x) = \dfrac{i}{\pi}${3}(g(x))+(g(x))$$.$f'(x)=\dfrac{i}{\pi}${3}(g(x))-(g(x))$\cdot g'(x) = \dfrac{i}{\pi}${3}(g(x))-(g(x))$\dfrac{-1}{x\sqrt{4x^2/3-9}}$.f"(x)= \dfrac{i}{\pi}-\sqrt{3}\sin g(x)-\cos g(x)\cdot (g'(x))^2 + \dfrac{i}{\pi}\sqrt{3}\cos g(x)-\sin g(x)\cdot g"(x)= \dfrac{-i}{\pi}\sqrt{3}\sin g(x)+\cos g(x)\cdot \dfrac{1}{x^2(4x^2/3-9)}+\dfrac{i}{\pi} \sqrt{3}\cos g(x)-\sin g(x)\cdot\dfrac{8x^2/3-9}{x^2(4x^2/3-9)^{3/2}}. Plugging the above into the LHS of \ref{['eqn:ODE for f(x) of Z-shape']}, one can verify that $(4x^4-27)f"(x)+(8x^3-27x)f'(x)+3f(x)=0$. We now check the conditions listed in Theorem \ref{['thm:verify ODE for Z-shape spec']}. For this purpose, it is more convenient to re-write $f(x)$ as a function of all real terms. Let $y = \dfrac{3}{\sqrt{9-4x^2/3}}$ and let $z=\dfrac{y-1}{y+1}=\dfrac{27-2x^2-9\sqrt{9-4x^2/3}}{2x^2}$. Note that when $0 < x < \frac{3\sqrt{3}}{2}$, $y$ and $z$ are real variables, $y \geq 1$, and $0 \leq z \leq 1$. We make the following observations: $g(x)=\dfrac{1}{3}\cdot\arctan(-iy)=\dfrac{i}{6}\ln${1-y}{1+y}$= \dfrac{i}{6}ln(-z) = \dfrac{i}{6}ln(z) - \frac{\pi}{6}$$sin(g(x)) = \frac{i}{2}\left(e^{-i(iln(z)/6 - {\pi}/6)} - e^{i(iln(z)/6 - {\pi}/6)}\right) = \frac{i}{2}\left(\left(\frac{\sqrt{3} + i}{2}\right)z^{1/6} - \left(\frac{\sqrt{3} - i}{2}\right)z^{-1/6}\right)$$cos(g(x)) = \frac{1}{2}\left(e^{-i(iln(z)/6 - {\pi}/6)} + e^{i(iln(z)/6 - {\pi}/6)}\right) = \frac{1}{2}\left(\left(\frac{\sqrt{3} + i}{2}\right)z^{1/6} + \left(\frac{\sqrt{3} - i}{2}\right)z^{-1/6}\right)$ Plugging in the above equations to $f(x)$ and simplifying, we get that f(x)=\dfrac{i}{\pi}\sqrt{3}\sin(g(x))+\cos(g(x))=\dfrac{-1}{\pi}\cdotz^{1/6}-z^{-1/6},f'(x)=\dfrac{i}{\pi}\sqrt{3}\cos(g(x))-\sin(g(x))\cdot\dfrac{-1}{x\sqrt{4x^2/3-9}} = \dfrac{-1}{\pi}\cdotz^{1/6}+z^{-1/6}\cdot\dfrac{1}{x\sqrt{9-4x^2/3}}. Recall that $y = \dfrac{3}{\sqrt{9-4x^2/3}}$ and $z=\dfrac{y-1}{y+1}$. Observe that As $x \to 0^{+}$, $y \approx 1 + \dfrac{2x^2}{27}$. Thus, $\lim_{x \to 0^{+}}{\dfrac{z}{x^2}} = \dfrac{1}{27}$.As $x \to a^{-}$, $y \to \infty$. Thus, $\lim_{x \to a^{-}}{z} = 1$. Thus, $f$ is twice differentiable on $(0,a)$.$\lim_{x\to 0^{+}}xf(x)= \lim_{x\to 0}x\cdot${z^{-1/6}}{}$= 0$.$\lim_{x\to 0^{+}}x^{2}f'(x)= \lim_{x\to 0}x\cdot${-z^{-1/6}}{3}$= 0$.$\lim_{x\to a^{-}} f(x)=\dfrac{-1}{\pi}(1-1)=0$.$\lim_{x\to a^{-}}(4x^2-27)f'(x)=\lim_{x\to a^{-}}\dfrac{1}{\pi}(z^{1/6}+z^{-1/6})\cdot${{3(27-4x^2)}}{x}$= 0$. We now prove the last piece of this Theorem: $\int_{0}^a f(x)\mathop{\mathrm{dx}}\nolimits=1$. We have that $a=3\sqrt{3}/2$, $z=\dfrac{y-1}{y+1}=\dfrac{27-2x^2-9$9-4x^2/3$^{1/2}}{2x^2}=\dfrac{27-27$1-x^2/a^2$^{1/2}}{2x^2}-1$, and $f(x)=\dfrac{-1}{\pi}\cdot$z^{1/6}-z^{-1/6}$$. Let $x=a\sin\theta$. Then $z=\dfrac{27-27\cos\theta}{2a^2\sin^2\theta}-1=\dfrac{2(1-\cos\theta)}{\sin^2\theta}-1=\dfrac{1-\cos\theta}{1+\cos\theta}$. Expressing $\cos\theta$ in terms of $z$, we get $\cos\theta = \dfrac{1-z}{1+z}$, thus $\sin\theta=\dfrac{2\sqrt{z}}{1+z}$. Moreover, \mathop{\mathrm{dz}}\nolimits= \dfrac{1-\cos\theta}{1+\cos\theta}'d\theta = \dfrac{2\sin\theta}{(1+\cos\theta)^2}d\theta = \dfrac{2\sin\theta(1-\cos\theta)}{\sin^2\theta(1+\cos\theta)} = \dfrac{2z}{\sin\theta}d\theta \implies d\theta = \dfrac{\sqrt{z}}{z(1+z)}\mathop{\mathrm{dz}}\nolimits. Thus \int_{0}^a f(x)\mathop{\mathrm{dx}}\nolimits= \dfrac{-1}{\pi}\int_{0}^{\pi/2} \dfrac{1-\cos\theta}{1+\cos\theta}^{1/6} -\dfrac{1-\cos\theta}{1+\cos\theta}^{-1/6} a\cos\theta d\theta= \dfrac{-a}{\pi}\int_{0}^{1} z^{1/6}-z^{-1/6}\cdot\dfrac{1-z}{1+z}\cdot\dfrac{\sqrt{z}}{z(1+z)} \mathop{\mathrm{dz}}\nolimits= \dfrac{-a}{\pi}\int_{0}^1 \dfrac{(1-z)(z^{2/3}-z^{1/3})}{z(1+z)^2}\mathop{\mathrm{dz}}\nolimits. Let $w=z^3$, then \int_{0}^a f(x)\mathop{\mathrm{dx}}\nolimits=\dfrac{-a}{\pi}\int_{0}^1 \dfrac{(1-w^3)(w-1)}{(1+w^3)^2}\mathop{\mathrm{dw}}\nolimits= \dfrac{-a}{\pi}\int_{0}^1 \dfrac{-4/3}{(1+w)^2} + \dfrac{2w}{(w^2-w+1)^2} +\dfrac{-5/3}{w^2-w+1}\mathop{\mathrm{dw}}\nolimits= \dfrac{-a}{\pi}\dfrac{4}{3}\begin{equation*} \dfrac{1}{1+w} \end{equation*}_0^1 + \int_{0}^1 \dfrac{2w-1}{(w^2-w+1)^2}\mathop{\mathrm{dw}}\nolimits + \int_{0}^1 \dfrac{1}{(w^2-w+1)^2}\mathop{\mathrm{dw}}\nolimits + \int_0^1\dfrac{-5/3}{w^2-w+1}\mathop{\mathrm{dw}}\nolimits= \dfrac{-a}{\pi}-\dfrac{2}{3} + \begin{equation*} \dfrac{-1}{(w^2-w+1)} \end{equation*}_0^1 + \int_{0}^1 \dfrac{1}{(w^2-w+1)^2}\mathop{\mathrm{dw}}\nolimits + \int_0^1\dfrac{-5/3}{w^2-w+1}\mathop{\mathrm{dw}}\nolimits= \dfrac{-a}{\pi}-\dfrac{2}{3} + \int_{0}^1 \dfrac{1}{(w-\frac{1}{2})^2+\frac{3}{4}^2}\mathop{\mathrm{dw}}\nolimits + \int_0^1\dfrac{-5/3}{w^2-w+1}\mathop{\mathrm{dw}}\nolimits. For any $b\neq 0$, $\int\dfrac{1}{(x^2+b^2)^2}\mathop{\mathrm{dx}}\nolimits = \dfrac{1}{2b^2}\int \dfrac{1}{x^2+b^2} \mathop{\mathrm{dx}}\nolimits - \dfrac{x}{x^2+b^2}.$ \int\dfrac{1}{(x^2+b^2)^2}\mathop{\mathrm{dx}}\nolimits= \dfrac{1}{b^2}\int\dfrac{x^2+b^2}{(x^2+b^2)^2} + \dfrac{-x^2}{x^2+b^2}\mathop{\mathrm{dx}}\nolimits= \dfrac{1}{b^2}\int\dfrac{1}{x^2+b^2}\mathop{\mathrm{dx}}\nolimits + \int \dfrac{-x}{2}d\dfrac{1}{x^2+b^2}= \dfrac{1}{b^2}\int\dfrac{1}{x^2+b^2} \mathop{\mathrm{dx}}\nolimits - \dfrac{x}{2(x^2+b^2)} + \int\dfrac{-\frac{1}{2}}{x^2+b^2}\mathop{\mathrm{dx}}\nolimits= \dfrac{1}{2b^2}\int\dfrac{1}{x^2+b^2} \mathop{\mathrm{dx}}\nolimits - \dfrac{x}{x^2+b^2}. Apply the lemma to the $\int_{0}^1 \dfrac{1}{$(w-1/2)^2+3/4$^2}\mathop{\mathrm{dw}}\nolimits$ term, we get that \int_{0}^a f(x)\mathop{\mathrm{dx}}\nolimits= \dfrac{-a}{\pi}-\dfrac{2}{3} +\dfrac{2}{3}\int_{0}^1 \dfrac{1}{(w-1/2)^2+3/4}\mathop{\mathrm{dw}}\nolimits -\dfrac{w-1/2}{w^2-w+1}_{0}^1 + \int_0^1\dfrac{-5/3}{(w-1/2)^2+3/4} \mathop{\mathrm{dw}}\nolimits= \dfrac{-a}{\pi}-\dfrac{2}{3} +\dfrac{2}{3} + \int_0^1\dfrac{-1}{(w-1/2)^2+3/4} \mathop{\mathrm{dw}}\nolimits=\dfrac{a}{\pi}\dfrac{1}{\sqrt{3}/2}\arctan$\dfrac{w-1/2}{\sqrt{3}/2}$_{0}^1 = \dfrac{3\sqrt{3}/2}{\pi}\cdot\dfrac{2\pi}{3\sqrt{3}} = 1. Figure \ref{['fig:z-shape spec with sampling']} shows some graphs of $g_{\alpha_Z}(x)$ and samplings of singular values of $M_{\alpha_Z}$ for $n=20,30,70$. We can see that the sampled distribution of the singular values of $M_{\alpha_Z}$ gets closer to $g_{\alpha_Z}(x)$ as $n$ gets bigger. The spectrum of singular values Sampling of singular values of $M_{\alpha_Z}$ where $n=20$ Sampling of singular values of $M_{\alpha_Z}$ where $n=30$ Sampling of singular values of $M_{\alpha_Z}$ where $n=70$ The Spectrum of singular values of the Z-shape graph matrix and some samplings of the Z-shape graph matrices with random input graphs on $n$ vertices, for $n=20,30,70$. In this section, we consider a generalization of the Z-shape $\alpha_Z$ which which we call the $m$-layer Z-shape. The $m-$layer Z-shape $\alpha_{Z(m)}$ Let $\alpha_{Z(m)}$ be the bipartite shape with vertices $V(\alpha_{Z(m)})=\{u_1,\dots,u_m,v_1,\dots, v_m\}$ and edges $E(\alpha_{Z(m)})=\left\{\{u_i,v_i\}: i\in[m]\right\}\cup \left\{\{u_{i+1},v_{i}\}: i\in[m-1]\right\}$ with distinguished tuples of vertices $U_{\alpha_{Z(m)}}=(u_1,\dots,u_m)$ and $V_{\alpha_{Z(m)}}=(v_1,\dots, v_m)$. See \ref{['fig:m-zshape']} for an illustration. We refer to $\alpha_{Z(m)}$ as the $m$-layer Z-shape. Note that $\alpha_{Z(2)}$ is the Z-shape $\alpha_Z$ as in \ref{['defn:z-shape']}. For the $m$-layer Z-shape $\alpha_{Z(m)}$, $s_{\alpha_{Z(m)}}=m$ where $s_{\alpha_{Z(m)}}$ is the size of the minimum separator. By \ref{['lem:min constr edges for bipartite shape']}, for any nonzero-valued constraint graph $C\in\mathcal{C}_{$_{Z(m)},2k$}$, $\abs{E(C)}\geq m\cdot(k-1)$. By \ref{['cor:dominantnumberofedges']}, dominant constraint graphs $C\in\mathcal{C}_{$_{Z(m)},2k$}$ have $m\cdot(k-1)$ edges. Applying $s_{\alpha_{Z(m)}}$ to \ref{['cor:expected trace for bipartite shapes']}, we get a similar result as \ref{['cor:Zshapeexpectedtrace']}. $\mathbb{E}[\mathop{\mathrm{tr}}\nolimits M_{\alpha_{Z(m)}}M_{\alpha_{Z(m)}}^T^k] = \abs{\left\{C \in \mathcal{C}_{\alpha_{Z(m)},2k}: C \text{ is dominant}\right\}}\cdot n^{mk + m} \pm On^{mk+m-1}.$ Before stating our main result, we need one more definition, which is an extension of $C'_k = \dfrac{1}{2k+1}\binom{3k}{k}$ from \ref{['section:zshape']}. For $m,n$ positive integers, $D(m,n)=\frac{1}{m\cdot n+1}\binom{(m+1)\cdot n}{n}\,.$ $C'_k$ from \ref{['section:zshape']} is equal to $D(2,k)$ here. $D(m,n)$ is also $C_{m+1,1}(n)$ where the generalized Catalan number $C_{m,r}(k)$ is defined in \ref{['defn:generalized-calatan-number']}. An analogous theorem to \ref{['thm:zshape-trace-convergence']} for $\alpha_{Z(m)}$ is stated below. For all $k \in \mathbb{N} \cup \{0\}$, $\mathbb{E}[\mathop{\mathrm{tr}}\nolimits$$M_{\alpha_{Z(m)}}M_{\alpha_{Z(m)}}^T$^k$] = D(m,k)\cdot n^{mk + m} \pm O$n^{mk+m-1}$$.$\mathop{\mathrm{Var}}\nolimits (\mathop{\mathrm{tr}}\nolimits$$M_{\alpha_{Z(m)}}M_{\alpha_{Z(m)}}^T$^k$)$ is $O$n^{2(mk+m-1)}$$. For part 2, the variance trace bound, its proof is very similar to \ref{['subsection:z-shape-variance-trace-bound']} and one can find the complete proof in \ref{['thm:variancewellbehaved']}. We will instead just focus on the first part. Below is the main result of this section. Let $\alpha_{Z(m)}$ be the $m$-layer Z-shape as in \ref{['defn:m-zshape']}. Then the number of dominant constraint graphs $C\in\mathcal{C}_{$_{Z(m)},2k$}$ is $D(m,k)$. When $m=2$, $D(m,k)=D(2,k)=C'_k$, $\alpha_{Z(m)}=\alpha_Z$, and this theorem is exactly \ref{['thm:num-zshape-constraint-graph']}. Combined with \ref{['cor:mzshape-expected-trace']}, this shows that $\mathbb{E}[\mathop{\mathrm{tr}}\nolimits$$M_{\alpha_{Z(m)}}M_{\alpha_{Z(m)}}^T$^k$] = D(m,k)\cdot n^{mk + m} \pm O$n^{mk+m-1}$$, which is the first part of \ref{['thm:mzshape-trace-convergence']}. To prove the main result for this section, We need the following crucial recurrence relation for $D(m,n)$. $D(m,n+1)=\sum_{\substack{ i_0,\dots, i_{m}\geq 0: \\ i_0+\dots+i_{m}=n}} D(m,i_0)\dots D(m,i_{m})\,.$ The proof is similar to the proof of \ref{['thm:catalan2-recurrence-relation']}. Let $W_{m,n}:=$ the set of all grid walks from $(0,0)$ to $(n,mn)$ that are weakly below the diagonal and $d_{m,n}=\abs{W_{m,n}}$. We will prove that $d_{m,n}=D(m,n)$ and that $d_{m,n}$ satisfies the recurrence relation in the theorem. $d_{m,n}=D(m,n)$: For $r\in\{0,1,\dots,mn\}$, let $W_{m,n}(r):=$ the set of grid walks from $(0,0)$ to $(n,mn)$ that are $r$ steps above the diagonal. Then ${\bigcup_{r=0}^{mn} W_{m,n}(r)}$ is the set of all grid walks from $(0,0)$ to $(n,mn)$, which has cardinality $\binom{(m+1)\cdot n}{n}$. Also $|W_{m,n}(0)|=|W_{m,n}|=d_{m,n}$. By the same proof as in the \ref{['thm:catalan2-recurrence-relation']}, there is a bijection between $W_{m,n}(r)$ and $W_{m,n}(r-1)$ for each $r\in[mn]$. Thus ${d_{m,n}=\dfrac{1}{mn+1} \binom{(m+1)\cdot n}{n}}$. Illustration of the proof part 2 for \ref{['thm:generalized-calatan-recurrence-relation']}.${d_{m,n}=\sum_{\substack{ i_0,\dots, i_{m+1}\geq 0: \\ i_0+\dots+i_{m+1}=n-1}} d_{m,i_1}\dots d_{m,i_{m+1}}}$: Similar to the proof of \ref{['thm:catalan2-recurrence-relation']}, we now will find a bijection between $W_{m,n}$ and ${\bigcup_{\substack{ i_0,\dots, i_{m}\geq 0: \\ i_0+\dots+i_{m}=n-1}} W_{m,i_0}\times\dots\times W_{m,i_{m}} }$. Let $d_k$ be the line that is shifted $k$ vertical grids down from the diagonal. i.e. $d_k$ is the line $y=m\cdot x - k$. Let $w=$z_1,,z_{n(m+1)}$\in W_{m,n}$. Let $z_{i_0}=(a_0, m\cdot a_0)$ be the first point where $w$ touches the diagonal. Then $w_0:=$z_{i_0},,z_{n(m+1)}$$ can be viewed as an element in $W_{m,n-a_0}$. Moreover, $w':=$z_2,,z_{i_0-1}$$ is strictly below the diagonal $d_0$ and thus weakly below $d_1$.Let $z_{i_1}=(a_1, m\cdot a_1-1)$ be the first point where $w'$ touches $d_1$. Then $w_1:=(z_{i_1},\dots,z_{i_0-1})$ can be viewed as an element in $W_{m,a_0-a_1}$. Moreover, $w':=(z_2,\dots,z_{i_1-1})$ is strictly below $d_1$ and thus weakly below $d_2$.Continuing in this way, we get a sequence of points $z_{i_0},\dots,z_{i_{m-1}}$ and walks $w_0,\dots,w_{m-1}$ where each $z_{i_j}=(a_j,m\cdot a_j-j)$ is a point on $d_j$ and each $w_i$ can be viewed as an element in $W_{m,a_{i-1}-a_i}$.Since $z_{i_{m-1}}$ is the first point touching $d_{m-1}$, $w_{m}=(z_2,\dots, z_{i_{m-1}-1})$ is strictly below $d_{m-1}$ and thus weakly below $d_m$. Since $d_m$ crosses $(1,0)=z_{2}$, $w_m$ can be viewed as an element in $W_{m,a_{m-1}-1}$. Since $(n-a_0)+(a_0-a_1)+\dots+(a_{m-1}-1)=n-1$, we conclude that ${(w_0,\dots,w_m)\in\bigcup_{\substack{ i_0,\dots, i_{m}\geq 0: \\ i_0+\dots+i_{m}=n-1}} W_{m,i_0}\times\dots\times W_{m,i_{m}}}\,.$ Taking these steps in reverse gives the inverse map. it is not hard to confirm that this construction gives a bijection. Combining statements 1 and 2, we conclude that $D(m,n)$ satisfies the recurrence relation. Let $\alpha_{Z(m)}$ be the $m$-layer Z-shape as in \ref{['defn:m-zshape']} and let $H$_{Z(m)},2k$$ be the multi-graph as in \ref{['def:copies']}. We label the vertices of $V_{(\alpha_Z)_i}$ as $\{a_{ij}, b_{ij}:j\in[m]\}$ and the vertices of $V_{(\alpha_Z)_i}^T$ as $\{b_{ij}, a_{(i+1)j}\}$. Let $V_i=\{a_{ij},b_{ij}:i\in[q]\}$. For $j\in[m]$, we call the induced subgraph of $H$_{Z(m)},2k$$ on vertices $V_i$ the $j^{th}$ wheel $W_j$. We label the "middle edges" of $H(\alpha,2k)$ in the following way: let $e_{2i-1,j}=\{a_{i(j+1)},b_{ij}\}$ and $e_{2i,j}=\{b_{ij},a_{(i+1)(j+1)}\}$ for $i=1,\dots,q$. For a fixed $j\in[m]$, we call the edges $e_{i,j}$'s the spokes between wheels $W_j$ and $W_{j+1}$ of $H$_{Z(m)},2k$$. See \ref{['fig:m-z-shape-H labeling']} for an illustration. Illustration of \ref{['defn:m-zshape-H']}: $H$_{Z(m)}, 2k$$ where $m=3$. Let $\alpha_{Z(m)}$ be the $m$-layer Z-shape. Let $C$ be a constraint graph on $H(\alpha_Z,2k)$. For $i=1,2$, We let $C_i$ denote the induced subgraph of $C$ on vertices $V_i$. We call $C_i$ the induced constraint graph of $C$ on $V_i$. Recall that a constraint graph $C\in\mathcal{C}_{(\alpha,2k)}$ is well-behaved if whenever $u\longleftrightarrow v$ in $C$, $u$ and $v$ are copies of the same vertex in $\alpha$ or $\alpha^T$. All dominant constraint graphs in $\mathcal{C}_{$_{Z(m)},2k$}$ are well-behaved. See Appendix \ref{['append:well-behaved']}. If $C \in \mathcal{C}_{$_{Z(m)},2k$}$ is a dominant constraint graph, then $E(C) = \bigcup_{i=1}^{m} E(C_i)$ and the induced constraint graph $C_i$ is a dominant constraint graph on $W_i$ for all $i\in[m]$. Since $C$ is well-behaved, $E(C) = E(C_1) \cup \dots \cup E(C_m)$ and for all $i \in [m]$, $C_i$ is a nonzero-valued constraint graph on $W_i$. Moreover, each $C_i$ must have the minimum possible number of edges as otherwise $C$ would have too many edges. The proof of \ref{['lem:zshape-spoke-constraint']} easily generalizes to $\alpha_{Z(m)}$, yielding the following statement. Let $\alpha_{Z(m)}$ be as in \ref{['defn:m-zshape']} and let $C$ be a dominant constraint graph in $\mathcal{C}_{$_{Z(m)},2k$}$. If $a_{sj}\longleftrightarrow a_{tj}$ for some $s < t \in [k]$ and $j\in[m-1]$, then $a_{s(j+1)}\longleftrightarrow a_{t(j+1)}$. Moreover, the spokes $\left\{e_{x,j}:x \in [2s-1,2t-2]\right\}$ can only be made equal to each other.Similarly, if $b_{sj}\longleftrightarrow b_{tj}$ for some $s< t \in [k]$ and $j\in \{2,3,\dots,m\}$, then $b_{s(j-1)}\longleftrightarrow b_{t(j-1)}$. Moreover, the spokes $\left\{e_{x,j-1}:x \in [2s,2t-1]\right\}$ can only be made equal to each other. We are now ready to prove the main result of this section, \ref{['thm:num-mzshape-constraint-graph']}. For all $k \in \mathbb{N}$ and $j \in \{0,1,...,m\}$, we define $H_j$_{Z(m)},2k$$ to be the graph obtained by starting with $H$_Z,2k$$, merging the vertices $b_{1j'}$ and $b_{kj'}$ for all $j' \in [j]$, deleting the vertices $\left\{a_{1j'}: j' \in [j]\right\}$ and all edges incident to these vertices, and deleting the spokes incident to $a_{1j}$. Equivalently, letting $E_j = \left\{\{b_{1j'},b_{kj'}\}: j' \in [j]\right\}$, $H_j$_Z,2k$$ is the graph obtained by taking $H$_Z,2k$/{E_j}$, deleting all multi-edges with even multiplicity, and then deleting all isolated vertices. For all $k \in \mathbb{N}$ and all $j \in \{0,1,...,m\}$, define $D_m(k,j)$ to be the number of dominant constraint graphs on $H_j$_{Z(m)},2k$$. Equivalently, $D_m(k,j)$ is the number of dominant constraint graphs $C \in \mathcal{C}_{$_{Z(m)},2k$}$ such that for all $j' \in [j]$, $b_{1j'} \longleftrightarrow_{C} b_{kj'}$. We set $D_m(0,0) = 1$. Note that $D_m(k,0) = \abs{C \in \mathcal{C}_{$_{Z(m)},2k$}: C \text{ is dominant}}$ and $D_m(k,m) = D_m(k-1,0)$. We can now generalize \ref{['lem:z-shape-split-1']} and \ref{['lem:z-shape-split-2']} to the case of a general $m$. Since the proofs are analogous to those in \ref{['section:zshape']}, we will just state them. For all $k \in \mathbb{N}$ and $j \in \{0,1,...,m-1\}$, $D_m(k,j) = \sum_{i=0}^{k}{D_m(i,0)D_m(k-i,j+1)}$ The constraint edges $E = \left\{\{a_{1j'},a_{ij'}\}: j' \geq j+1\right\} \bigcup \left\{\{b_{ij'},b_{kj'}\}: j' \leq j+1\right\}$ partition $H_j$_Z,2k$$ into two parts $H_1$ and $H_2$ where $H_1 \cong H$_Z,2(i-1)$$ and $H_2 \cong H_{j+1}$_Z,2(k+1-i)$$. See \ref{['fig:mzshape-split']} for an illustration. Illustration of \ref{['lem:mzshape-split-2']}: here $j=2$. For all $j\in[m]$ and $m\in\mathbb{N}$, $D_m(k,j) = \sum_{\substack{i_1, ..., i_{m-j+1}\geq 0: \\ i_1+\dots+i_{m-j+1} = k-1}} D_mi_1,0\cdot \dots \cdot D_mi_{m-j},0\cdot D_mi_{m-j+1}, 0.$ In particular, when $j=0$, $D_m(k,0) = \sum_{\substack{i_1, ..., i_{m+1}\geq 0: \\ i_1+\dots+i_{m+1} = k-1}} D_mi_1,0\cdot \dots \cdot D_mi_{m},0\cdot D_mi_{m+1}, 0.$ By \ref{['lem:mzshape-split-1']}, D_m(k,j)= \sum_{i_1=0}^{k}D_m(i_1,0)D_m(k-i_1,j+1)= \sum_{i_1=0}^{k}D_m(i_1,0)\sum_{i_2=0}^{k-i_1} D_m(i_2,0)D_m(k-i_1-i_2, j+2)= \ldots= \sum_{\substack{i_1, ..., i_{m-j+1}\geq 0: \\ i_1+\dots+i_{m-j+1} = k}} D_m(i_1,0)D_m(i_2,0)\dots D_mi_{m-j},0 D_mi_{m-j+1}, m= \sum_{\substack{i_1, ..., i_{m-j+1}\geq 0: \\ i_1+\dots+i_{m-j+1} = k}} D_m(i_1,0)D_m(i_2,0)\dots D_mi_{m-j},0 D_mi_{m-j+1}-1, 0= \sum_{\substack{i_1, ..., i_{m-j+1}\geq 0: \\ i_1+\dots+i_{m-j+1} = k-1}} D_m(i_1,0)D_m(i_2,0)\dots D_mi_{m-j},0 D_mi_{m-j+1}, 0 By \ref{['cor:mzshape-recurrence-relation']} and \ref{['thm:generalized-calatan-recurrence-relation']}, $D_m(k,0)$ has the same recurrence relation as the generalized Calatan number $D(m,k) = \dfrac{1}{mk+1}\binom{(m+1)k}{k}$, thus \ref{['thm:num-mzshape-constraint-graph']} is proved. In this section we aim to find the spectrum of the singular values for m-layer Z-shape graph matrices. By \ref{['thm:mzshape-trace-convergence']}, we get the following corollary using similar analysis as in the proof for \ref{['lem:evenmomentsmatch']}. Let $\{G_n:n \in \mathbb{N}\}$ be a sequence of $G$n,{1}{2}$$ graphs and $M_{m,n} = \dfrac{1}{n^{m/2}}{M_{\alpha_{Z(m)}}(G_n)}$. With probability $1$, for all $k \in \mathbb{N} \cup \{0\}$, $\lim_{n\to\infty}{\mathbb{E}_{X \sim D_{M_{m,n}}} [X^{2k}]} = D(m,k)$. Thus if we can find a function $g_m$ such that ${\int_{0}^{\infty} g_m(x)x^{2k} \mathop{\mathrm{dx}}\nolimits = D(m,k)}$, then $g_m$ describes the limiting spectrum of singular values for the m-layer Z-shape graph matrix as $n$ goes to $\infty$. Recall that ${D(m,n)= \dfrac{1}{mn+1} \binom{(m+1)n}{n}}$ as defined in Definition \ref{['defn:D(m,n)']}. In this section we will generalize the arguments for the $m=2$ case in \ref{['section:zshape-spectrum']} to the case $m=3$. The general steps will be: Assume $\int_{0}^{\infty} f(x)x^{2k}\mathop{\mathrm{dx}}\nolimits = D(3,k)$ and derive a differential equation for $f(x)$.Prove that under this differential equation, the moments of $f(x)$ are indeed $D(3,k)$. Let $a=\lim_{n\to\infty} D(3,n+1)/D(3,n)=\dfrac{16}{3\sqrt{3}}$. If $f(x)$ is a function such that $\int_{0}^{a} f(x)x^{2k} dx=D(3,k)$ for all nonnegative integers $k$ and $f$ is three times continuously differentiable on $(0,a)$ and $\lim_{x\to a^-} f(x)=0$,$\lim_{x\to a^-}(x-a)f'(x)=0$,$\lim_{x\to a^-}2(x-a)f"(x)+f'(x)=0$,$\lim_{x\to 0^+}xf(x)=0$, $\lim_{x\to 0^+}x^2f'(x) = 0$, and $\lim_{x\to 0^+}x^3f"(x)=0$. then $f$ satisfies the following ODE on $(0,a)$: $(27x^4-256x^2)f"'(x) + (162x^3-768x)f"(x)+ (177x^2-192x)f'(x)+15xf(x)=0.$ Let $a$ be some positive constant. If $f$ is continuous on $[0,a]$ and $\int_{0}^{a} f(x)x^{2n+1}\mathop{\mathrm{dx}}\nolimits=0$ for all nonnegative integers $n$, then $f=0$ on $(0,a)$. The proof is similar to the proof for \ref{['lem:zero even moments']} for the case of all zero even moments. Here we approximate $f(\sqrt{x})\cdot\sqrt{x}$ by a polynomial $p(x)$, so that the odd polynomial $p(x^2)/x$ ($p(x)$ has $0$ constant term) approximates $f(x)$. Denote the LHS of the ODE by $G(x)$. Let $A(k,n)=\int_{0}^a f^{(k)}(x)\cdot x^n\mathop{\mathrm{dx}}\nolimits$ and $B(k,n)=f^{k}(x)\cdot x^n_0^a$. Repeatedly doing integration by parts we get that A(m,n)= B(m-1,n) - n\cdot A(m-1,n-1)= B(m-1,n)-nB(m-2,n-1) + n(n-1)A(m-2,n-2)= B(m-1,n)-nB(m-2,n-1) + n(n-1)B(m-3,n-2)\space -n(n-1)(n-2)A(m-3,n-3). Also, we have that $\dfrac{D(3,n)}{D(3,n-1)}=\dfrac{A(0,2n)}{A(0,2n-2)}=\dfrac{4(4n-3)(4n-2)(4n-1)}{(3n+1)(3n)(3n-1)}.$ The steps for deducing the ODE for $f(x)$ are very similar to the steps used in the proof of \ref{['thm:ODE for f(x) of Z-shape']} to deduce the ODE for the Z-shaped graph matrix. We first apply $m=3$ and $n=2n+3$ to the first equation and rewrite the term $n(n-1)(n-2)A(m-3,n-3)$ using the second equation \ref{['eqn:D(3,n) ratio']}. We then gradually eliminate the non-constant coefficients in front of $A(m,n)$'s using the first equation. Plugging in $m=3,n=2n+3$ into the first equation, we get 27A(3,2n+3)= 27B(2,2n+3)-27(2n+3)B(1,2n+2)+27(2n+3)(2n+2)B(0,2n+1) - 27(2n+3)(2n+2)(2n+1)A(0,2n). Rewriting the last term on the RHS above and applying the second equation \ref{['eqn:D(3,n) ratio']}, we get 27(2n+3)(2n+2)(2n+1)A(0,2n)= 8(3n+1)(3n)(3n-1)A(0,2n) -\space 81\cdot 2(2n+2)(2n+1)A(0,2n) + 59\cdot 3(2n-1)A(0,2n) -15A(0,2n)= 32(4n-3)(4n-2)(4n-1)A(0,2n-2) -\space 81\cdot 2(2n+2)(2n+1)A(0,2n) + 59\cdot 3(2n-1)A(0,2n) -15A(0,2n) We can rewrite the first term on the RHS as 32(4n-3)(4n-2)(4n-1)A(0,2n-2)= 256(2n+1)(2n)(2n-1)A(0,2n-2) -\space 256\cdot 3(2n)(2n-1)A(0,2n-2) + 128\cdot 3(2n-1)A(0,2n-2). Now apply the first equation to all the $(n+1)A(m,n)$, $(n+2)(n+1)A(m,n)$ and $(n+3)(n+2)(n+1)A(m,n)$ above, group together the $A(m,n)$ terms and $B(m,n)$ terms separately, and rewrite the $B(m,n)$ term using the definition of $B(m,n)$. We get that for all $n\geq 1$, 27A(3,2n+3)-256A(3,2n+1) + 2\cdot 81A(2,2n+2)-3\cdot 256A(2,2n)\space + 3\cdot 59 A(1,2n+1)-3\cdot 128A(1,2n-1) + 15A(0,2n)= $(27x^2-256)f"(x)+27xf'(x)$x^{2n+1}_0^a\space+ $(2n-2)(-27x^2+256)$f'(x)x^{2n}_0^{a} + p(n,x)\cdot f(x)x^{2n-1}_0^a where $p(n,x)$ is some polynomial in terms of $n$ and $x$. Observe that the last term on the RHS is $0$ since $\lim_{x\to 0^+}xf(x)=0$ and $\lim_{x\to a^-}f(x)=0$ by assumption. The second last term is $0$ since $27x^2-256=27(x+a)(x-a)$, $\lim_{x\to a^-} (x-a)f'(x)=0$ and $\lim_{x\to 0^+}x^2f'(x)=0$. The first term top part is $\lim_{x\to a^-}27$(x+a)(x-a)f"(x)+xf'(x)$x^{2n+1} \\= \lim_{x\to a^-} 27$2a(x-a)f"(x)+af'(x)$a^{2n+1} = 0$ since $\lim_{x\to a^-} 2(x-a)f"(x)+f'(x)=0$ by assumption. The bottom part is $0$ since $\lim_{x\to 0^+}x^3f"(x)=0$ by assumption. Thus the RHS is $0$. Expanding out each $A(m,n)$ term by using the definition of $A(m,n)$, we get that $\int_{0}^a G(x)x^{2n+1}=0$ for all $n\geq 1$. By \ref{['lem:zero odd moments']}, $G(x)=0$ on $(0,a)$, which proves that $f$ satisfies the ODE. Let $a=\lim_{n\to\infty} D(3,n)/D(3,n-1) = \dfrac{16}{3\sqrt{3}}$ and let $f$ be a function satisfying the following ODE $(27x^4-256x^2)f"'(x) + (162x^3-768x)f"(x)+ (177x^2-192)f'(x)+15xf(x)=0$ and the conditions listed in \ref{['thm:ODE for f(x) of 3-Z-shape']}. Moreover, assume $\int_{0}^a f(x)\mathop{\mathrm{dx}}\nolimits = 1$. Then for any nonnegative integer $k$, $A(0,2k)=\int_{0}^{a} x^{2k}\cdot f(x)\mathop{\mathrm{dx}}\nolimits = D(3,k).$ The proof is very similar to the proof for \ref{['thm:verify ODE for Z-shape spec']}. We integrate the ODE from $0$ to $a$, do integration by parts and use the conditions for $f$ to eliminate the redundant terms and finally arrive at the ratio between $A(0,2(k-1))$ and $A(0,2k)$ which matches the ratio between $D(3,k-1)$ and $D(3,k)$. By induction on $k$ we conclude \ref{['eqn:A(0,2k)=D(3,k)']}. Let $M_n = \dfrac{1}{n^{3/2}}\cdot M_{\alpha_{Z(3)}}$ where $M_{\alpha_{Z(3)}}$ is the graph matrix with random input graph $G\sim G(n,1/2)$ and $\alpha_{Z(3)}$ is the multi-Z-shape defined in \ref{['defn:m-zshape']}. Let $g(x)$ be $f(x)$ as in \ref{['thm:verify ODE for 3-Z-shape spec']} on $(0,a)$ and $0$ for $x\geq a$. With probability $1$, the spectrum of the singular values of $M_n$ approaches $g(x)$ weakly as $n\to\infty$. For this ODE \ref{['eqn:ODE for 3-Z-shape f(x)']}, WolframAlpha fails to give us an explicit solution. Instead, we solve the ODE numerically by approximating the tail segment of $f(x)$ by $c\cdot(a-x)^{r}$ for some constants $c$ and $r$. We analyze the behaviour of the ODE when $x$ is very close to $a$. Notice that $a=\dfrac{16}{3\sqrt{3}} \iff 27a^2-256=0$. $27x^4-256x^2 = 27x^2(x-a)(x+a) \sim 52a^3(x-a)$.$162x^3-768x = 81x^3 + 3x(27x^2-256) \sim 81a^3$.$177x^2-192 = \dfrac{3}{4}(209x^2+27x^2-256) \sim \dfrac{3\cdot 209a^2}{4}$. Thus when $x$ is very close to $a$, the ODE is $52a^3(x-a)f"'(x) + 81a^3f"(x) + \dfrac{3\cdot 209a^2}{4} f'(x)=0.$ One can check that $f'(x)=C'$(a-x)^{-1/2}+{209}{64{3}}(a-x)^{1/2}$$ is a solution to the above ODE. Thus $f(x)\sim g(x)= C(a-x)^{1/2} + \dfrac{209}{64\cdot3\sqrt{3}}(a-x)^{3/2}$ for some constant $C$ when $x$ is very close to $a$.We approximate the solution of $f(x)$ by approximating the tail segment of $f(x)$ (where $|x-a|<\epsilon$) by $g(x)$ and use this approximation to obtain initial conditions for the ODE for $f$. In particular, we choose a small $\epsilon>0$ and set the initial conditions for the ODE as follows: $f(a-\epsilon)= g(a-\epsilon), f'(a-\epsilon)=g'(a-\epsilon), f"(a-\epsilon)=g"(a-\epsilon).$ We calculate the constant $C$ in $g$ by noticing that the integration of $f$ over $(-a,a)$ should be $1$. Setting $\epsilon=0.01$ and solving the ODE in python, we get the following plot for the solution to the ODE (concatenated with a tail segment where we use $g$ as an approximation): Plot of the ODE solution with the approximated tail segment. Zoom in at the tail segment. The plot of the spectrum where $x>0$. To test this solution experimentally, we can sample from the distribution of singular values of $M_n$ by sampling a random graph $G$, computing the resulting matrix $M_n(G)$, and computing its singular values. See \ref{['fig:sampling zz']} for a plot of the approximated spectrum together with the empirical distribution of the singular values of $M_n$ with $n=10$ and $n=12$, respectively (where we sampled $100$ random graphs $G$). Sampling of the singular values of $M_n$ for $n=10$. Sampling of the singular values of $M_n$ for $n=12$. The approximated spectrum of the singular values of the 3-layer Z-shaped graph matrix and some samplings of the singular values of $M_n$, for $n=10$ and $n=12$. @article{FussCatalan, title={Fuss Catalan Number}, date={Accessed: 2023-09-02}, journal={https://en.wikipedia.org/wiki/Fuss%E2%80%93Catalan_number} }@article{AMP20, author={Ahn, Kwangjun and Medarametla, Dhruv and Potechin, Aaron}, title={Graph Matrices: Norm Bounds and Applications}, date={2016}, journal={arXiv:1604.03423} }@inproceedings{9996747, author={Bafna, Mitali and Hsieh, Jun-Ting and Kothari, Pravesh K. and Xu, Jeff}, title={Polynomial-Time Power-Sum Decomposition of Polynomials}, booktitle={FOCS} }@book{bai2010spectral, author={Bai, Zhidong and Silverstein, Jack W}, title={Spectral Analysis of Large Dimensional Random Matrices}, publisher={Springer}, date={2010}, volume={20} }@inproceedings{BHKKMP16, author={Barak and Hopkins and Kelner and Kothari and Moitra and Potechin}, title={A Nearly Tight Sum-of-Squares Lower Bound for the Planted Clique Problem}, booktitle={FOCS} }@article{CP22, author={Cai, Wenjun and Potechin, Aaron}, title={On Mixing Distributions Via Random Orthogonal Matrices and the Spectrum of the Singular Values of Multi-Z Shaped Graph Matrices}, date={2022}, journal={arXiv:2206.02224} }@inproceedings{pmlr-v40-Deshpande15, author={Deshpande, Yash and Montanari, Andrea}, title={Improved Sum-of-Squares Lower Bounds for Hidden Clique and Hidden Submatrix Problems}, booktitle={COLT}, url={https://proceedings.mlr.press/v40/Deshpande15.html} }@article{fuss1791solutio, author={Fuss, Nikolaus I}, title={Solutio quaestionis, quot modis polygonum n laterum in polygona m laterum, per diagonales resolvi queat}, date={1791}, journal={Nova Acta Academiae Sci. 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Algorithms}, volume={14}, number={3}, url={https://doi.org/10.1145/3178538} }@inproceedings{hop17, author={Hopkins, Samuel B and Kothari, Pravesh K and Potechin, Aaron and Raghavendra, Prasad and Schramm, Tselil and Steurer, David}, title={The power of sum-of-squares for detecting hidden structures}, booktitle={FOCS} }@inproceedings{hsieh_et_al:LIPIcs.ICALP.2023.78, author={Hsieh, Jun-Ting and Kothari, Pravesh K. and Potechin, Aaron and Xu, Jeff}, title={Ellipsoid Fitting up to a Constant}, booktitle={ICALP} }@inproceedings{9719766, author={Jones, C. and Potechin, A. and Rajendran, G. and Tulsiani, M. and Xu, J.}, title={Sum-of-Squares Lower Bounds for Sparse Independent Set}, booktitle={FOCS} }@article{jones2024iterative, author={Jones, Chris and Pesenti, Lucas}, title={Diagram Analysis of Iterative Algorithms}, date={2024}, journal={arXiv:2404.07881} }@inproceedings{10.1145/3564246.3585221, author={Jones, Chris and Potechin, Aaron and Rajendran, Goutham and Xu, Jeff}, title={Sum-of-Squares Lower Bounds for Densest k-Subgraph}, booktitle={STOC} }@article{Kunisky20, author={Kunisky, Dmitriy}, title={Positivity-preserving extensions of sum-of-squares pseudomoments over the hypercube}, date={2020}, journal={arXiv:2009.07269} }@inproceedings{MP16, author={Medarametla, Dhruv and Potechin, Aaron}, title={Bounds on the Norms of Uniform Low Degree Graph Matrices}, booktitle={RANDOM} }@inproceedings{meka2015sum, author={Meka, Raghu and Potechin, Aaron and Wigderson, Avi}, title={Sum-of-squares Lower Bounds for Planted Clique}, booktitle={STOC} }@inproceedings{MRX20, author={Mohanty, Sidhanth and Raghavendra, Prasad and Xu, Jeff}, title={Lifting Sum-of-Squares Lower Bounds: Degree-2 to Degree-4}, booktitle={STOC} }@article{potechin2022subexponential, author={Potechin, Aaron and Rajendran, Goutham}, title={Sub-exponential time Sum-of-Squares lower bounds for Principal Components Analysis}, date={2022}, journal={Advances in Neural Information Processing Systems} }@inproceedings{PTVW23, author={Potechin, Aaron and Turner, Paxton and Venkat, Prayaag and Wein, Alexander S.}, title={Near-optimal fitting of ellipsoids to random points}, booktitle={COLT} }@inproceedings{rajendran2022concentration, author={Rajendran, Goutham and Tulsiani, Madhur}, title={Concentration of polynomial random matrices via Efron-Stein inequalities}, booktitle={SODA} }@article{rukavicka2011generalized, author={Rukavicka, Josef}, title={On generalized Dyck paths}, date={2011}, journal={The Electronic Journal of Combinatorics}, volume={18}, number={1}, pages={Research\ndash Paper} }@article{wigner1958distribution, author={Wigner, Eugene P}, title={On the Distribution of the Roots of Certain Symmetric Matrices}, date={1958}, journal={Annals of Mathematics}, pages={325\ndash 327} }@article{wigner1993characteristic, author={Wigner, Eugene P}, title={Characteristic Vectors of Bordered Matrices With Infinite Dimensions II}, date={1993}, journal={The Collected Works of Eugene Paul Wigner: Part A: The Scientific Papers}, pages={541\ndash 545} } In this section, we prove that dominant constraint graphs on $H(\alpha_{Z_m},2k)$ and $H(\alpha_{Z_m},2k) \cup H(\alpha_{Z_m},2k)$ are well-behaved. More precisely, we show the following theorems. If $C \in \mathcal{C}_{H(\alpha_{Z_m},2k)}$ and $\mathop{\mathrm{val}}\nolimits(C) \neq 0$ then $C$ has at least $m(k - 1)$ edges. Moreover, if $\mathop{\mathrm{val}}\nolimits(C) \neq 0$ and $C$ is not well-behaved then $C$ has at least $m(k-1) + 1$ edges. If $C \in \mathcal{C}_{H(\alpha_{Z_m},2k) \cup H(\alpha_{Z_m},2k)}$ and $\mathop{\mathrm{val}}\nolimits(C) \neq 0$ then $C$ has at least $2m(k - 1)$ constraint edges. Moreover, if $\mathop{\mathrm{val}}\nolimits(C) \neq 0$ but the restriction of $C$ to either copy of $H(\alpha_{Z_m},2k)$ has value $0$ (i.e., $\mathop{\mathrm{val}}\nolimits(C) \neq 0$ because of constraint edges between the two copies of $H(\alpha_{Z_m},2k)$) then $C$ has at least $2m(k - 1) + 2$ edges. The high level idea for proving Theorems \ref{['thm:dominantwellbehaved']} and \ref{['thm:variancewellbehaved']} is as follows. We show that for any constraint graph which has nonzero value and does not have too many constraint edges, we can find a vertex $v \in H(\alpha_{Z_m},2q)$ which is not incident to any constraint edges or spokes of odd multiplicity. This implies that there must be a constraint edge between the two neighbors of $v$ in its wheel. We can then merge the two neighbors of $v$ together, delete $v$ and all edges incident to $v$, and repeat this process. When we are done, we will be left with a cycle of length $2$ for each wheel. For $H(\alpha_{Z_m},2q)$, this implies that the constraint graph is either well-behaved or has an extra edge. For $H(\alpha_{Z_m},2q) \cup H(\alpha_{Z_m},2q)$, this implies that the restriction of the constraint graph to each copy of $H(\alpha_{Z_m},2q)$ has nonzero value. In order to implement this strategy, we need to analyze the more general class of graphs $H$ which can be obtained by starting with $H(\alpha_{Z_m},2q)$, iteratively taking vertices which are not incident to any spokes with odd multiplicity, merging their neighbors together, and deleting the vertex and all edges incident to it. Illustration of \ref{['defn:R-H']}. Define $R(H(\alpha_{Z_m},2q))$ to be the set of graphs $H$ which can be obtained by starting from $H(\alpha_{Z_m},2q)$ and repeatedly applying the following contraction operation: Choose a vertex $x \in V(H)$ such that $x$ is in a wheel with at least $4$ vertices and all spokes incident to $x$ have even multiplicity. Merge the two neighbors of $x$ in its wheel and delete all edges incident to $x$. See \ref{['fig:R-H']} for an illustration. To analyze what happens when we apply contraction operations to $H(\alpha_{Z_m},2k)$, it is useful to consider the $k$ copies of $\alpha_{Z_m}$ and the $k$ copies of $\alpha_{Z_m}^T$ which $H(\alpha_{Z_m},2k)$ is built from. Illustration of \ref{['defn:H-alpha-i']}: $v_{ij} = u_{(i+1)j}$ and $V_i = U_{j+1}$ for $j\in[2q]$. For each $i \in [2q]$, we take $\alpha_i$ to be the following shape: If $i$ is odd then we take $U_{\alpha_i} = \{u_{ij}: j \in [m]\}$, $V_{\alpha_i} = \{v_{ij}: j \in [m]\}$, and $E(\alpha_i) = \{\{u_{i(j+1)},v_{ij}\}: j \in [m-1]\} \cup \{\{u_{ij}, v_{ij}\}: j\in [m]\}$. In other words, we take $\alpha_i$ to be a copy of $\alpha_{Z_m}$.If $i$ is even then we take $U_{\alpha_i} = \{u_{ij}: j \in [m]\}$, $V_{\alpha_i} = \{v_{ij}: j \in [m]\}$, and $E(\alpha_i) = \{\{u_{ij},v_{i(j+1)}\}: j \in [m-1]\} \cup \{\{u_{ij}, v_{ij}\}: j\in [m]\}$. In other words, we take $\alpha_i$ to be a copy of $\alpha_{Z_m}^T$. See \ref{['fig:H-alpha-i']} for an illustration. As noted in \ref{['def:copies']}, $H(\alpha_{Z_m},2q)$ can be constructed by concatenating $\alpha_1,\ldots,\alpha_{2q}$ and then merging $V_{\alpha_{2q}}$ and $U_{\alpha_1}$. More precisely, for each $i \in [2q-1]$ and $j \in [m]$, we set $v_{ij} = u_{(i+1)j}$. We then set $v_{(2q)j} = u_{1j}$ for all $j \in [m]$. We now consider how $\alpha_1,\ldots,\alpha_{2q}$ are affected when we perform the contraction operations. For all $H \in R(H(\alpha_{Z_m},2k))$, we can construct $H$ by concatenating $\alpha'_1,\ldots,\alpha'_{2k}$ and then merging $V_{\alpha'_{2k}}$ and $U_{\alpha'_1}$ where each $\alpha'_i$ is obtained from $\alpha_i$ by taking the following steps: For each $j \in [m]$ such that $u_{ij}$ was deleted but $v_{ij}$ was not deleted, we delete the edges incident to $u_{ij}$ and replace $u_{ij}$ with $v_{ij}$ in $U_{\alpha_i}$. Note that after this replacement, $v_{ij} \in U_{\alpha_i} \cap V_{\alpha_i}$.For each $j \in [m]$ such that $v_{ij}$ was deleted but $u_{ij}$ was not deleted, we delete the edges incident to $v_{ij}$ and replace $v_{ij}$ with $u_{ij}$ in $V_{\alpha_i}$. Note that after this replacement, $u_{ij} \in U_{\alpha_i} \cap V_{\alpha_i}$.For each $j \in [m]$ such that $u_{ij}$ and $v_{ij}$ were both deleted, we delete all edges incident to $u_{ij}$ and $v_{ij}$ and replace $u_{ij}$ and $v_{ij}$ with $u_{i'j}$ in $U_{\alpha_i}$ and $V_{\alpha_i}$ where $i' \in [2k]$ is the largest index such that $i' < i$ and $u_{i'j}$ was not deleted. If this $i'$ does not exist because $u_{1j},\ldots,u_{(i-1)j}$ were all deleted then we instead take $i' \in [2k]$ to be the largest index such that $u_{i'j}$ was not deleted. Note that after this replacement, $u_{i'j} \in U_{\alpha_i} \cap V_{\alpha_i}$. Also note that $u_{i'j} = v_{i"j}$ where $i" \in [2k]$ is the smallest index such that $i" > i$ and $v_{i"j}$ was not deleted. If this $i"$ does not exist because $v_{(i+1)j},\ldots,u_{(2k)j}$ were all deleted then we instead take $i" \in [2k]$ to be the smallest index such that $v_{i'j}$ was not deleted. Illustration of \ref{['prop:R-H-alpha']} and \ref{['defn:range-v']}. For $v$ in the first figure, $l(v) = 4$ and $r(v) = 3$; for $v'$ in the second figure, $l(v') = 6$ and $r(v') = 3$. Given a vertex $v \in V(H)$, we define $l(v) \in [2k]$ to be the index such that $v \in V_{\alpha'_{l(v)}} \setminus U_{\alpha'_{l(v)}}$. Similarly, we define $r(v) \in [2k]$ to be the index such that $v \in U_{\alpha'_{r(v)}} \setminus V_{\alpha'_{r(v)}}$. If $l(v) < r(v)$ then we take the range of $v$ to be $range(v) = [l(v),r(v)]$. If $l(v) > r(v)$ then we take the range of $v$ to be $range(v) = [l(v),2k] \cup [1,r(v)]$. For all $H \in R(H(\alpha_{Z_m},2k))$, for all $j \in [m]$ and all vertices $v$ in the $j$th wheel of $H$, $range(v)$ is well-defined. For all $i \notin range(v)$, $v \notin V(\alpha'_i)$.$v = v_{l(v)j} = u_{r(v)j}$ and $r(v) - l(v) \equiv 1 \mod 2$.$u_{l(v)j} \neq v$ and there is an edge between $u_{l(v)j}$ and $v$ in $\alpha'_{l(v)}$. Similarly, $v \neq v_{r(v)j}$ and there is an edge between $v$ and $v_{r(v)j}$ in $\alpha'_{r(v)}$.For all $i \in range(v) \setminus \{l(v),r(v)\}$, $v \in U_{\alpha'_i} \cap V_{a'_i}$. We prove this proposition by induction on the number of contraction operations. For the base case, observe that for all $i \in [2k-1]$ and all $j \in [m]$, for each vertex $v = v_{ij} = u_{(i+1)j}$, $l(v) = i$, $r(v) = i+1$, $range(v) = [i,i+1]$, and the third and fourth statements hold. Similarly, for all $j \in [m]$, for each vertex $v = v_{2kj} = u_{1j}$, $l(v) = 2k$, $r(v) = 1$, $range(v) = \{2k\} \cup \{1\}$, and the third and fourth statements hold. For the inductive step, assume that the proposition is true for all $H \in R(H(\alpha_{Z_m},2q))$ such that $H$ was obtained from $H(\alpha_{Z_m},2q)$ using at most $k$ contractions. Now consider an $H' \in R(H(\alpha_{Z_m},2q))$ which is obtained from $H(\alpha_{Z_m},2q)$ using $k+1$ contractions, a $j \in [m]$, and a vertex $v$ in the $j$th wheel of $H'$. Let $H \in R(H(\alpha_{Z_m},2q))$ be the graph we obtain after the first $k$ contractions. By the inductive hypothesis, $range(v)$ is well-defined for $H$ and satisfies all of the given properties. Let $i_1 = l(v)$ and $i_2 = r(v)$ (with respect to $H$). Now consider what happens after the final contraction. There are $3$ cases to consider. Neither $u_{{i_1}j}$ nor $v_{{i_2}j}$ is deleted in the final contraction. In this case, $range(v)$ is unchanged and still satisfies all of the given properties.$v_{{i_2}j}$ is deleted in the final contraction. In this case, let $i_3 = r(v_{{i_2}j})$ (with respect to $H$), let $w = v_{{i_3}j}$, and let $i_4 = r(w)$ (with respect to $H$). By the inductive hypothesis, we have that in $H$, For all $i \in range(v) \setminus \{i_1, i_2\}$, $v \in U_{\alpha'_i} \cap V_{\alpha'_i}$.For all $i \in range(v_{{i_2}j}) \setminus \{i_2, i_3\}$, $v_{{i_2}j} \in U_{\alpha'_i} \cap V_{\alpha'_i}$.For all $i \in range(w) \setminus \{i_3, i_4\}$, $w \in U_{\alpha'_i} \cap V_{\alpha'_i}$.$u_{{i_1}j} \neq v$ and $w \neq v_{{i_4}j}$. The contraction merges $v$ and $w$ and deletes $v_{{i_2}j}$. This replaces $v_{{i_2}j}$ with $v = w$ in $V_{\alpha'_{i_2}}$, $U_{\alpha'_{i_3}}$, and $U_{\alpha'_{i}} \cap V_{\alpha'_{i}}$ for all $i \in range(v_{{i_2}j}) \setminus \{i_2, i_3\}$. Thus, after the contraction, $range(v) = [i_1,i_4]$ (or $[i_4,2q] \cup [1,i_1]$ if $i_4 < i_1$) and $v \in U_{\alpha'_{i}} \cap V_{\alpha'_{i}}$ for all $i \in range(v) \setminus \{i_1, i_4\}$. It is not hard to check the remaining properties. Since $i_2 - i_1 \equiv 1 \mod 2$, $i_3 - i_2 \equiv 1 \mod 2$, and $i_4 - i_3 \equiv 1 \mod 2$, $i_4 - i_1 \equiv 1 \mod 2$. Since $i_1 \neq i_4$, neither $u_{{i_1}j}$ nor $v_{{i_4}j}$ is equal to $v$.$u_{{i_1}j}$ is deleted in the final contraction. This case can be handled in a similar way as the previous case. We now analyze the spokes of $H$. For all $H \in R(H(\alpha_{Z_m},2k))$, any $j \in [m]$, and any vertex $v$ in the $j$th wheel of $H$, If $j = m$ and $l(v)$ is odd or $j = 1$ and $l(v)$ is even then $v$ is not incident to any spokes.If $j > 1$ and $l(v)$ is even then all spokes incident to $v$ are between the $(j-1)$th and $j$th wheels of $H$. Moreover, If neither $u_{l(v)(j-1)}$ nor $v_{r(v)(j-1)}$ was deleted and $u_{l(v)(j-1)} \neq v_{r(v)(j-1)}$ then the spokes $\{u_{l(v)(j-1)},v\}$ and $\{v,v_{r(v)(j-1)}\}$ have odd multiplicity and all other spokes incident to $v$ have even multiplicity.If neither $u_{l(v)(j-1)}$ nor $v_{r(v)(j-1)}$ was deleted and $u_{l(v)(j-1)} = v_{r(v)(j-1)}$ then all spokes incident to $v$ have even multiplicity.If at least one of $u_{l(v)(j-1)}$ and $v_{r(v)(j-1)}$ was deleted then both $u_{l(v)(j-1)}$ and $v_{r(v)(j-1)}$ were deleted (in fact, $u_{l(v)(j-1)}$ and $v_{r(v)(j-1)}$ were merged before being deleted) and all spokes incident to $v$ have even multiplicity.If $j < m$ and $l(v)$ is odd then all spokes incident to $v$ are between the $j$th and $(j+1)$th wheels of $H$. Moreover, If neither $u_{l(v)(j+1)}$ nor $v_{r(v)(j+1)}$ was deleted and $u_{l(v)(j+1)} \neq v_{r(v)(j+1)}$ then the spokes $\{u_{l(v)(j+1)},v\}$ and $\{v,v_{r(v)(j+1)}\}$ have odd multiplicity and all other spokes incident to $v$ have even multiplicity.If neither $u_{l(v)(j+1)}$ nor $v_{r(v)(j+1)}$ was deleted and $u_{l(v)(j+1)} = v_{r(v)(j+1)}$ then all spokes incident to $v$ have even multiplicity.If at least one of $u_{l(v)(j+1)}$ and $v_{r(v)(j+1)}$ was deleted then both $u_{l(v)(j+1)}$ and $v_{r(v)(j+1)}$ were deleted (in fact, $u_{l(v)(j+1)}$ and $v_{r(v)(j+1)}$ were merged before being deleted) and all spokes incident to $v$ have even multiplicity. For all $H \in R(H(\alpha_{Z_m},2k))$, any $j \in [m-1]$, and any $i,i' \in [2k]$ such that $i$ is even and $i'$ is odd, if $v_{ij} = u_{i'j}$ then either $v_{i(j+1)} = u_{i'(j+1)}$ or $v_{i(j+1)}$ and $u_{i'(j+1)}$ were merged and then deleted. Similarly, for all $H \in R(H(\alpha_{Z_m},2k))$, any $j \in [2,m]$, and any $i,i' \in [2k]$ such that $i$ is odd and $i'$ is even, if $v_{ij} = u_{i'j}$ then either $v_{i(j-1)} = u_{i'(j-1)}$ or $v_{i(j-1)}$ and $u_{i'(j-1)}$ were merged and then deleted. We prove these two lemmas by induction on the number of contraction operations. The base case $H(\alpha_{Z_m},2k)$ is trivial. For the inductive step, assume that Lemmas \ref{['lem:spokeproperties']} and \ref{['lem:equalityimplications']} are true for all $H \in R(H(\alpha_{Z_m},2k))$ such that $H$ was obtained from $H(\alpha_{Z_m},2k)$ using at most $x$ contractions. Now consider an $H' \in R(H(\alpha_{Z_m},2k))$ which is obtained from $H(\alpha_{Z_m},2k)$ using $x+1$ contractions. There are three ways that Lemmas \ref{['lem:spokeproperties']} and \ref{['lem:equalityimplications']} could potentially fail: For some vertex $v \in V(H)$, the other endpoint of one of the spokes incident to $v$ is merged with another vertex (which may be the other endpoint of a different spoke incident to $v$) and the spokes incident to $v$ no longer satisfy the statements in Lemma \ref{['lem:spokeproperties']}.Some vertex $v \in V(H)$ is merged with another vertex $w$ and the spokes incident to $v = w$ no longer satisfy the statements in Lemma \ref{['lem:spokeproperties']}.Some vertex $v$ in the $j$th wheel of $H$ is merged with another vertex $w$ but the corresponding vertices in the $(j-1)$th or $(j+1)$th wheel of $H$ are not equal/deleted. Let $H \in R(H(\alpha_{Z_m},2q))$ be the graph we obtain after the first $x$ contractions. We show that these three possibilities do not happen. Given $j \in [m]$ and a vertex $v$ in the $j$th wheel of $H$, let $i_1 = l(v)$, and let $i_2 = r(v)$. There are several cases to consider. $i_1 = l(v)$ is odd, $j < m$, and the final contraction deletes a vertex $v'$ in the $(j+1)$th wheel of $H$ and merges its two neighbors $u'$ and $w'$. If $v' = u_{i'(j+1)}$ where $i'$ is odd then $u'$ and $w'$ cannot be endpoints of spokes incident to $v$ but $v'$ may be an endpoint of a spoke incident to $v$. We have the following cases: If $v'$ is equal to $u_{i_1(j+1)}$ or $v_{i_2(j+1)}$ then since $v'$ was not incident to any spokes of odd multiplicity, we must have had that $u_{i_1(j+1)} = v_{i_2(j+1)}$ and this spoke had even multiplicity. Thus, after the contraction, both $u_{i_1(j+1)}$ and $v_{i_2(j+1)}$ are deleted (and they were merged before being deleted).If $v'$ is not equal to $u_{i_1(j+1)}$ or $v_{i_2(j+1)}$ but is an endpoint of a spoke incident to $v$ then deleting $v'$ deletes this spoke. Since this spoke had even multiplicity, the statements in Lemma \ref{['lem:spokeproperties']} still hold.If $v'$ is not an endpoint of a spoke incident to $v$ then the multiplicities of the spokes incident to $v$ are unaffected. If $v' = u_{i'(j+1)}$ where $i'$ is odd then $v'$ cannot be an endpoint of a spoke incident to $v$ but $u'$ and/or $w'$ may be an endpoint of a spoke incident to $v$. We have the following cases: If $u' = u_{i_1(j+1)}$ and $w' = v_{i_2(j+1)}$ or then after the contraction, the spoke $u_{i_1(j+1)} = v_{i_2(j+1)}$ and the spoke $\{u_{i_1(j+1)},v\}$ has even multiplicity.If $u' \notin \{u_{i_1(j+1)},v_{i_2(j+1)}\}$ or $w' \notin \{u_{i_1(j+1)},v_{i_2(j+1)}\}$ then whether or not $u_{i_1(j+1)} = v_{i_2(j+1)}$ and the parities of the multiplcities of the spokes incident to $v$ are not affected by the contraction.$i_1 = l(v)$ is even, $j > 1$, and the final contraction deletes a vertex $v'$ in the $(j-1)$th wheel of $H$ and merges its two neighbors $u'$ and $w'$. This case can be handled in a similar way as the previous case.$i_1 = l(v)$ is odd and the final contraction deletes the vertex $v_{{i_2}j}$. In this case, let $i_3 = r(v_{{i_2}j})$, let $w = v_{{i_3}j}$, and let $i_4 = r(w)$. If $j = m$ then all of the vertices which were merged to form $v$ and all of the vertices which were merged to form $w$ were of the form $v_{im}$ where $i$ is odd. Since none of these vertices were incident to any spokes, after the contraction, $v = w$ is not incident to any spokes. If $j < m$ then by the inductive hypothesis, all spokes incident to $v$ and $w$ are between the $j$th and $(j+1)$th wheels of $H$. Moreover, before the contraction, All spokes incident to $v$ except for $\{u_{{i_1}(j+1)},v\}$ and $\{v,v_{{i_2}(j+1)}\}$ have even multiplicity. Similarly, all spokes incident to $w$ except for $\{u_{{i_3}(j+1)},w\}$ and $\{w,v_{{i_4}(j+1)}\}$ have even multiplicity.Since $v_{{i_2}j} = u_{{i_3}j}$ and $i_2$ is even, either $v_{{i_2}(j+1)} = u_{{i_3}(j+1)}$ or $v_{{i_2}(j+1)}$ and $u_{{i_3}(j+1)}$ were merged and then deleted. There are several possible cases. If $u_{{i_1}(j+1)}$, $v_{{i_2}(j+1)}$, $u_{{i_3}(j+1)}$, or $v_{{i_4}(j+1)}$ was deleted then $u_{{i_1}(j+1)}$, $v_{{i_2}(j+1)}$, $u_{{i_3}(j+1)}$, and $v_{{i_4}(j+1)}$ must have been all merged together and then deleted.If $u_{{i_1}(j+1)} = v_{{i_2}(j+1)}$ and $u_{{i_3}(j+1)} = v_{{i_4}(j+1)}$ then $u_{{i_1}(j+1)} = v_{{i_2}(j+1)} = u_{{i_3}(j+1)} = v_{{i_4}(j+1)}$ and all spokes incident to $v$ have even multiplicity after the contraction.If $u_{{i_1}(j+1)} = v_{{i_2}(j+1)}$ and $u_{{i_3}(j+1)} = v_{{i_4}(j+1)}$ then $u_{{i_1}(j+1)} = v_{{i_2}(j+1)} = u_{{i_3}(j+1)} \neq v_{{i_4}(j+1)}$. In this case, the spokes $\{u_{{i_1}(j+1)},v\}$ and $\{w,v_{{i_4}(j+1)}\}$ are distinct and have odd multiplicity.If $u_{{i_1}(j+1)} \neq v_{{i_2}(j+1)}$ and $u_{{i_3}(j+1)} = v_{{i_4}(j+1)}$ then $u_{{i_1}(j+1)} \neq v_{{i_2}(j+1)} = u_{{i_3}(j+1)} = v_{{i_4}(j+1)}$. In this case, the spokes $\{u_{{i_1}(j+1)},v\}$ and $\{w,v_{{i_4}(j+1)}\}$ are distinct and have odd multiplicity.If $u_{{i_1}(j+1)} \neq v_{{i_2}(j+1)}$ and $u_{{i_3}(j+1)} \neq v_{{i_4}(j+1)}$ then after the contraction, the spoke $\{v,v_{{i_2}(j+1)}\} = \{u_{{i_2}(j+1)},w\}$ has even multiplcity and is distinct from the spokes $\{u_{{i_1}(j+1)},v\}$ and $\{w,v_{{i_4}(j+1)}\}$. In this case, if $u_{{i_1}(j+1)} \neq v_{{i_4}(j+1)}$ then after the contraction, the spokes $\{u_{{i_1}(j+1)},v\}$ and $\{w,v_{{i_4}(j+1)}\}$ are distinct and have odd multiplicity. If $u_{{i_1}(j+1)} \neq v_{{i_4}(j+1)}$ then after the contraction, the spoke $\{u_{{i_1}(j+1)},v\} = \{w,v_{{i_4}(j+1)}\}$ has even multiplicity. Finally, we observe that for all $j > 1$ (including $j = m$), either $v_{i_1(j-1)} = u_{{i_4}(j-1)}$ or $v_{i_1(j-1)}$ and $u_{{i_4}(j-1)}$ were merged and then deleted. To see this, note that since $i_1$ is odd and $v_{{i_1}j} = u_{{i_2}j}$, by the inductive hypothesis, either $v_{{i_1}(j-1)} = u_{{i_2}(j-1)}$ or $v_{{i_1}(j-1)}$ and $u_{{i_2}(j-1)}$ were merged and then deleted. Similarly, since $i_3$ is odd and $v_{{i_3}j} = u_{{i_4}j}$, either $v_{{i_3}(j-1)} = u_{{i_4}(j-1)}$ or $v_{{i_3}(j-1)}$ and $u_{{i_4}(j-1)}$ were merged and then deleted. Finally, since $i_2$ is even, and $v_{{i_2}j} = u_{{i_3}j}$ was deleted, we must have that either $u_{{i_2}(j-1)} = v_{{i_3}(j-1)}$ or $u_{{i_2}(j-1)}$ and $v_{{i_3}(j-1)}$ were merged and then deleted. Putting these observations together, either $v_{i_1(j-1)} = u_{{i_4}(j-1)}$ or $v_{i_1(j-1)}$ and $u_{{i_4}(j-1)}$ were merged and then deleted, as needed.The remaining cases are as follows: $i_1 = l(v)$ is odd and the final contraction deletes the vertex $u_{{i_1}j}$.$i_2 = l(v)$ is even and the final contraction deletes the vertex $v_{{i_2}j}$.$i_1 = l(v)$ is even and the final contraction deletes the vertex $u_{{i_1}j}$. These cases can be handled in a similar way as the previous case. We now prove Theorems \ref{['thm:dominantwellbehaved']} and \ref{['thm:variancewellbehaved']}. To do this, we use the decomposition of $H \in R(H(\alpha_{Z_m},2k))$ as the composition of $\alpha'_1,\ldots,\alpha'_{2k}$ and we use ideas from Appendix B of AMP20. We first define the set of allowable constraint graphs on a multi-graph $H \in R(H(\alpha_{Z_m},2k))$. These constraint graphs are the possible constraint graphs which could result from starting with a constraint graph $C \in \mathcal{C}_{$_Z,2k$}$ and applying the contraction operations described in Definition \ref{['defn:R-H']}. Given an $H \in R(H(\alpha_{Z_m},2k))$, we define $\mathcal{C}_H$ to be the set of constraint graphs on $H$ which keep all of the vertices in each $\alpha'_i$ distinct. More precisely, we require that for all $i \in [2k]$ and all $j,j' \in [m]$, If $j' \neq j$ and $u_{ij},u_{ij'}$ were not deleted (being merged with other vertices is okay) then there is no path of constraint edges between $u_{ij}$ and $u_{ij'}$.If $j' \neq j$ and $v_{ij},v_{ij'}$ were not deleted (being merged with other vertices is okay) then there is no path of constraint edges between $v_{ij}$ and $v_{ij'}$.If $u_{ij},v_{ij'}$ were not deleted (being merged with other vertices is okay) then there is no path of constraint edges between $u_{ij}$ and $v_{ij'}$. For all $H \in R(H(\alpha_{Z_m},2k))$ and $i \in [2k]$, $E(\alpha'_i) \subseteq E(\alpha_i)$. Moreover, for all $C \in \mathcal{C}_H$, even after all of constraint edges are contracted, all of the edges in $E(\alpha'_i)$ are distinct. Observe that the only way a new vertex $u_{i'j}$ can be introduced to $V(\alpha'_i)$ is if $u_{ij}$ and $v_{ij}$ were both deleted first. If so, all edges incident to $u_{ij}$ and $v_{ij}$ were deleted and the new vertex $u_{i'j}$ is not incident to any edges in $E(\alpha'_i)$. The moreover statement follows from the definition of $\mathcal{C}_H$. We now prove Theorem \ref{['thm:dominantwellbehaved']} by proving the following more general theorem. Given $H \in R(H(\alpha_{Z_m},2k))$, if $C \in \mathcal{C}_H$ and $\mathop{\mathrm{val}}\nolimits(C) \neq 0$ then $|E(C)| \geq \frac{|V(H)|}{2} - m$. Moreoever, if $C$ is not well-behaved then $|E(C)| \geq \frac{|V(H)|}{2} - m + 1$ For each $i \in [2k]$, we define the following vertex separator $S_i(C)$ of $\alpha'_i$. These separators are similar to the separators used in Appendix B of AMP20. Given an $H \in R(H(\alpha_{Z_m},2k))$, we define $B = \{v \in V(H): l(v) > r(v)\}$ to be the set of vertices in $H$ which appear in $V_{\alpha'_{2q}} = U_{\alpha'_1}$. Given an $H \in R(H(\alpha_{Z_m},2k))$ and a $C \in \mathcal{C}_H$, for each $i \in [2k]$, we define the set $S_i(C)$ to be the set of vertices in $V(\alpha'_i)$ which are in $B$ or appear both earlier and later in $H$. More precisely, for each $v \in V(\alpha'_i)$, $v \in S_i(C)$ if and only if at least one of the following holds: $v \in U_{\alpha'_i} \cap V_{\alpha'_i}$$v \in B$ or there is a path of constraint edges between $v$ and a vertex $w \in B$.$l(v) = i$ and there is a path of constraint edges between $v$ and a vertex $u$ such that $l(u) < i$.$r(v) = i$ and there is a path of constraint edges between $v$ and a vertex $w$ such that $r(w) > i$. If $\mathop{\mathrm{val}}\nolimits(C) \neq 0$ then for all $i \in [2q]$, $S_i(C)$ is a vertex separator of $\alpha'_{i}$. Assume that there is an edge $e = \{u,v\} \in E(\alpha'_i)$ such that $u,v \notin S_i(C)$. We must have that $u = u_{ij}$ and $v = v_{ij'}$ for some $i \in [2q]$ and $j,j' \in [m]$ such that $|j' - j| \leq 1$. Since $u = u_{ij} \notin S_i(C)$, $r(u) = i$, $l(u) < r(u)$, and there is no path of constraint edges between $u$ and any vertex $u'$ such that $r(u') > r(u)$ or $l(u') > r(u')$. Thus, for all $i' > i$, $u \notin V(\alpha'_{i'})$. Similarly, since $v = v_{ij'} \notin S_i(C)$, $l(v) = i$, $l(v) < r(v)$, and there is no path of constraint edges between $v_{ij'}$ and any vertex $u'$ such that $l(u') < l(v)$ or $l(u') > r(u')$. Thus, for all $i' < i$, $v \notin V(\alpha'_{i'})$. This implies that $e \notin E(\alpha'_{i'})$ for any $i' \neq i$ so $e$ only appears once in $H/C$ and thus $\mathop{\mathrm{val}}\nolimits(C) = 0$, which is a contradiction. These vertex separators are useful because they allow us to upper bound $|V(H/C)|$ which in turn allows us to lower bound $|E(C)|$. $|V(H/C)| = |V(H)| - |E(C)| = \frac{\sum_{i = 1}^{2k}{|V(\alpha'_{i}) \setminus S_i(C)|}}{2} + |B/C|$ where $|B/C|$ is the number of vertices in $B$ after we contract the constrint edges in $C$. Observe that in the sum $\sum_{i = 1}^{2k}{|V(\alpha'_{i}) \setminus S_i(C)|}$, each vertex $v \in V(H/C) \setminus B$ is counted twice, once for the first $i$ such that $v \in V(\alpha'_i)$ and once for the last $i$ such that $v \in V(\alpha'_i)$. Given an $H \in R(H(\alpha_{Z_m},2k))$ and a constraint graph $C$ on $H$, we say that an edge $\{u_{ij},v_{ij}\} \in E(\alpha'_i)$ is a critical edge if $u_{ij},v_{ij} \in S_i(C)$. If $H \in R(H(\alpha_{Z_m},2k))$ and $C \in \mathcal{C}_H$ is a constraint graph such that $\mathop{\mathrm{val}}\nolimits(C) \neq 0$ then $|E(C)| \geq \frac{|V(H)|}{2} - |B/C| + \frac{\# \text{ of critical edges}}{2}$. Observe that $|E(C)| = |V(H)| - \frac{\sum_{i = 1}^{2q}{|V(\alpha'_{i}) \setminus S_i(C)|}}{2} - |B/C|$ and $\sum_{i = 1}^{2q}{|S_i(C)|} = 2qm + \# \text{ of critical edges}$ To prove Theorem \ref{['thm:generaldominantwellbehaved']}, it is sufficient to show that if $H$ does not contain a vertex which is not incident to any constraint edges or spokes of odd multiplicity then either $|B/C| < m$ or there are at least two critical edges. Given $H \in R(H(\alpha_{Z_m},2k))$ and $C \in \mathcal{C}_H$, we say that a vertex $v \in V(H)$ is removable if it is contained in a wheel of $H$ which has at least $4$ vertices and it is not incident to any constraint edges in $C$ or spokes of odd multiplicity in $H$. For all $H \in R(H(\alpha_{Z_m},2k))$ and $C \in \mathcal{C}_H$, for all $i \in [2k]$ and $j \in [m]$, if $l(v_{ij}) = i$, $v_{ij} \in S_i(C)$, and the $j$th wheel of $H$ has at least $4$ vertices then at least one of the following is true: There exists an $i' \in [2,i]$ and a $j' \in [m]$ such that $r(u_{i'j'}) = i'$, $l(u_{i'j'}) < i'$, and $u_{i'j'}$ is a removable vertex. Note that $u_{i'j'} \notin B$.There exists an $i' \in [i]$ and a $j' \in [m]$ such that $\{u_{i'j'},v_{i'j'}\}$ is a critical edge. Similarly, for all $H \in R(H(\alpha_{Z_m},2k))$ and $C \in \mathcal{C}_H$, for all $i \in [2k]$ and $j \in [m]$, if $r(u_{ij}) = i$, $u_{ij} \in S_i(C)$, and the $j$th wheel of $H$ has at least $4$ vertices then at least one of the following is true: There exists an $i' \in [i,2k-1]$ and a $j' \in [m]$ such that $l(v_{i'j'}) = i'$, $r(v_{i'j'}) > i'$, and $v_{i'j'}$ is a removable vertex. Note that $v_{i'j'} \notin B$.There exists an $i' \in [i,2k]$ and a $j' \in [m]$ such that $\{u_{i'j'},v_{i'j'}\}$ is a critical edge. We only prove the first statement as the proof of the second statement is similar. We prove this statement by induction. For the base case $i = 1$, observe that we automatically have that $u_{1j} \in S_1(C)$. Since $l(v_{1j}) = 1$ and $v_{1j} \in S_1(C)$, $\{u_{1j},v_{1j}\}$ is a critical edge. For the inductive step, assume that the statement is true for $i \leq x$ and assume that $i = x+1$. Let $u = u_{ij}$ and let $i' = l(u)$. There are a few cases to consider: If $u = u_{ij} \in S_i(C)$ then $\{u_{ij},v_{ij}\}$ is a critical edge.If $u = v_{i'j} \in S_{i'}(C)$ then the result follows from the inductive hypothesis (we know that $i' < i$ as otherwise we would have that $u \in S_i(C)$).If $u$ is not incident to any constraint edges and is not incident to any spokes with odd multiplicity then we are done as $u$ is the vertex we are looking for.If $u$ is not incident to any constraint edges, $u$ is incident to spokes with odd multiplicity, and $i'$ is odd (which happens if and only if $i$ is even) then by Lemma \ref{['lem:spokeproperties']}, the only spokes incident to $u$ with odd multiplicity are $\{u_{i'(j+1)},v_{i'j}\}$ and $\{u_{ij},v_{i(j+1)}\}$. Since $u = v_{i'j} = u_{ij}$ is not incident to any constraint edges, this means that the vertices $u = u_{i'j}$, $u_{i'(j+1)}$, $v = v_{ij}$, and $v_{i(j+1)}$ are paired up by paths of constraint edges. Let $w = v_{i(j+1)}$ and let $i" = l(w)$. Note that there cannot be a path of constraint edges between $v = v_{ij}$ and $w$ as $v$ and $w$ were both part of $\alpha_i$ so they must be distinct. We must have that $i' \leq l(w) \leq i$ as the spokes $\{u_{i'(j+1)},u\}$ and $\{u,w\}$ are distinct. Thus, regardless of whether there is a path of constraint edges between $w$ and $u_{i'j}$ or a path of constraint edges between $w$ and $u_{i'(j+1)}$, $w \in S_{i"}(C)$. If $i" < i$ then the result follows from the inductive hypothesis (note that wheel $j+1$ must have at least $4$ vertices as otherwise all spokes incident to wheel $j+1$ would have even multiplicity). If $i" = i$ then we can repeat the entire argument using $w$ as the starting vertex (this must eventually terminate as $j$ increases each time the argument is repeated).If $u$ is not incident to any constraint edges, $u$ is incident to spokes with odd multiplcity, and $i'$ is even then we can use a similar argument as the previous case.If $u$ is incident to a constraint edge but $u = v_{i'j} \notin S_{i'}(C)$ and $u = u_{ij} \notin S_i(C)$ then the other endpoint of the constraint edge incident to $u$ must be a vertex $w$ such that $l(w) > l(u)$ and $r(w) < r(u) = i$. Moreover, $w \in S_{l(w)}(C)$ and $w \in S_{r(w)}(C)$. Letting $i" = l(w)$ and $j'$ be the wheel containing $w$, the result follows by applying the inductive hypothesis on $w$ (unless wheel $j'$ only has two vertices in which case $\{u_{i"j'},v_{i"j'}\}$ and $\{u_{r(w)j'},v_{r(w)j'}\}$ are both critical edges). For all $H \in R(H(\alpha_{Z_m},2q))$ such that at least one wheel of $H$ has at least $4$ vertices and all $C \in \mathcal{C}_{H}$, there are a total of at least two of the following: Removable vertices.Critical edges.Paths of constraint edges between vertices of $B$. We first observe that there is always a critical edge or a removable vertex which is not in $B$. To see this, choose a $j$ such that $W_j$ has at least $4$ vertices and let $u$ be the vertex in $W_j \cap B$. Letting $i = l(u)$, observe that $u = v_{ij} \in S_i(C)$. By Lemma \ref{['lem:findingkey']}, we can find a critical edge or a remoavable vertex which is not in $B$, as needed. This implies that if there is a path of constraint edges between two vertices of $B$ then the result is true. Thus, we can assume that $C$ does not have a path of constraint edges between any two vertices of $B$. We now observe that if $C$ contains a path of constraint edges between a vertex $u \in B$ and a vertex $v \notin B$ then there must be at a total of at least two critical edges and/or removable vertices. To see this, let $W_j$ be the wheel containing $v$, let $i_1 = l(v)$ and let $i_2 = r(v)$. Observe that $i_1 < i_2$. If $W_j$ only contains two vertices then both $\{u_{{i_1}j},v_{{i_1}j}\}$ and $\{u_{{i_2}j},v_{{i_2}j}\}$ are critical edges. If $W_j$ contains at least 4 vertices then by Lemma \ref{['lem:findingkey']}, either there exists an $i' \in [2,i_1]$ and a $j' \in [m]$ such that $u_{i'j'}$ is a removable vertex or there exists an $i' \in [i_1]$ and a $j' \in [m]$ such that $\{u_{i'j'},v_{i'j'}\}$ is a critical edge (or both). Similarly, by Lemma \ref{['lem:findingkey']}, either there exists an $i" \in [i_2,2k-1]$ and a $j" \in [m]$ such that $v_{i"j"}$ is a removable vertex or there exists an $i" \in [i_2,2k]$ and a $j" \in [m]$ such that $\{u_{i"j"},v_{i"j"}\}$ is a critical edge (or both). The remaining case is when none of the vertices in $B$ are incident to a constraint edge. In this case, let $u \in B$ be a vertex such that $u$ is contained in a wheel with at least $4$ vertices and $r(u) + 2k - l(u)$ is minimized. If $u$ is not incident to any spokes with odd multiplicity then $u$ is a removable vertex. If $u$ is incident to spokes with odd multiplicity, let $W_j$ be the wheel containing $u$. We have the following cases: If $l(u)$ is even and $r(u)$ is odd then by Lemma \ref{['lem:spokeproperties']}, the spokes of odd multiplicity which $u$ is incident to are $\{u_{l(u)(j-1)},u\}$ and $\{u,v_{r(u)(j-1)}\}$. Since $u$ is not incident to any constraint edges, $u_{l(u)(j-1)}$ and $v_{r(u)(j-1)}$ must both be incident to constraint edges. Letting $u'$ be the vertex in $W_ {j-1}$ which is in $B$, we must have that $l(u) \leq l(u')$ and $r(u') \leq r(u)$. Since we chose $u$ to minimize $r(u) + 2k - l(u)$, we must have that $l(u') = l(u)$ and $r(u') = r(u)$. Thus, we can repeat this argument with $u'$. Since $j$ decreases every time we repeat this argument, we cannot repeat this argument indefinitely so we must eventually reach a removable vertex in $B$. Since we always have either a critical edge or a removable vertex which is not in $B$, the result follows.If $l(u)$ is odd and $r(u)$ is even then we can use a similar argument. By Lemma \ref{['lem:spokeproperties']}, the spokes of odd multiplicity which $u$ is incident to are $\{u_{l(u)(j+1)},u\}$ and $\{u,v_{r(u)(j+1)}\}$. Since $u$ is not incident to any constraint edges, $u_{l(u)(j+1)}$ and $v_{r(u)(j+1)}$ must both be incident to constraint edges. Letting $u'$ be the vertex in $W_ {j+1}$ which is in $B$, we must have that $l(u) \leq l(u')$ and $r(u') \leq r(u)$. Since we chose $u$ to minimize $r(u) + 2k - l(u)$, we must have that $l(u') = l(u)$ and $r(u') = r(u)$. Thus, we can repeat this argument with $u'$. Since $j$ increases every time we repeat this argument, we cannot repeat this argument indefinitely so we must eventually reach a removable vertex in $B$. Since we always have either a critical edge or a removable vertex which is not in $B$, the result follows. we can now prove Theorem \ref{['thm:generaldominantwellbehaved']} by induction on $\frac{|V(H)|}{2} - m$. The base case $\frac{|V(H)|}{2} - m = 0$ is trivial as $|E(C)| \geq 0$ and if $C$ is not well-behaved then $|E(C)| \geq 1$. For the inductive step, assume that Theorem \ref{['thm:generaldominantwellbehaved']} is true for all $H \in R(H(\alpha_{Z_m},2k))$ such that $\frac{|V(H)|}{2} - m = x$. Consider an $H \in R(H(\alpha_{Z_m},2k))$ such that $\frac{|V(H)|}{2} - m = x+1$. If $H$ contains a removable vertex $v$ then there must be a path of constraint edges between the two neighbors of $v$ so we can merge these two vertices together and delete $v$. This gives us a constraint graph $H'$ and a constraint graph $C' \in \mathcal{C}_{H}$ with $|V(H')| = |V(H)| - 2$ and $|E(C')| = |E(C) - 1$. The result now follows from the inductive hypothesis. If $H$ does not contain a removable vertex $v$ then by Corollary \ref{['cor:findingkeycorollary']} and Corollary \ref{['cor:constraintedgeslowerbound']}, $|E(C)| \geq \frac{|V(H)|}{2} - m + 1$, as needed. We now prove Theorem \ref{['thm:variancewellbehaved']} by proving the following more general theorem. Given $H_1,H_2 \in R(H(\alpha_{Z_m},2k))$, if $C$ is a constraint graph on $H_1 \cup H_2$ such that $\mathop{\mathrm{val}}\nolimits(C) \neq 0$ and the restrictions $C_1,C_2$ of $C$ to $H_1$ and $H_2$ are in $\mathcal{C}_{H_1}$ and $\mathcal{C}_{H_2}$ respectively then $|E(C)| \geq \frac{|V(H_1)|}{2} + \frac{|V(H_2)|}{2} - 2m$. Moreover, if $\mathop{\mathrm{val}}\nolimits(C_1) = 0$ or $\mathop{\mathrm{val}}\nolimits(C_2) = 0$ then $|E(C)| \geq \frac{|V(H_1)|}{2} + \frac{|V(H_2)|}{2} - 2m + 2$. We prove this theorem by using the same strategy we used to prove \ref{['thm:generaldominantwellbehaved']}. We adapt this strategy to analyze $H_1 \cup H_2$ as follows We shift the indices of $H_2$ to be in $[2k+1,4k]$ rather than $[1,2k]$. This places all of $H_2$ to the right of $H_1$.We take $B = \{v \in V(H_1) \cup V(H_2): l(v) > r(v)\}$. Note that we can decompose $B$ as $B = B_1 \cup B_2$ where $B_1 = \{v \in V(H_1): l(v) > r(v)\}$ and $B_2 = \{v \in V(H_2): l(v) > r(v)\}$.We define the vertex separator $S_i(C)$ for $\alpha'_i$ in the same way as before.The analogue of Corollary {cor:constraintedgeslowerbound} is that if $H_1,H_2 \in R(H(\alpha_{Z_m},2k))$ and $C \in \mathcal{C}_H$ is a constraint graph such that $\mathop{\mathrm{val}}\nolimits(C) \neq 0$ and the restrictions $C_1,C_2$ of $C$ to $H_1$ and $H_2$ are in $\mathcal{C}_{H_1}$ and $\mathcal{C}_{H_2}$ respectively then $|E(C)| \geq \frac{|V(H_1)| + |V(H_2)|}{2} - |B/C| + \frac{\# \text{ of critical edges}}{2}.$Lemma \ref{['lem:findingkey']} holds for both $H_1$ and $H_2$ (where the $i$ indices are increased by $2k$ for $H_2$). We now show the following analogue of Corollary \ref{['cor:findingkeycorollary']}. Given $H_1,H_2 \in R(H(\alpha_{Z_m},2k))$ such that both $H_1$ and $H_2$ have at least one wheel which has at least $4$ vertices, if $C$ is a constraint graph on $H_1 \cup H_2$ such that $\mathop{\mathrm{val}}\nolimits(C) \neq 0$ and the restrictions $C_1,C_2$ of $C$ to $H_1$ and $H_2$ are in $\mathcal{C}_{H_1}$ and $\mathcal{C}_{H_2}$ respectively then at least one of the following must hold: $H_1$ or $H_2$ contains a removable vertex.$2m - |B/C| + \frac{\# \text{ of critical edges}}{2} \geq 2$. We use similar reasoning as we used to prove Corollary \ref{['cor:findingkeycorollary']}. In particular, we make the following observations. If neither $H_1$ nor $H_2$ contains a removable vertex then By the same reasoning as the first paragraph of the proof of Corollary \ref{['cor:findingkeycorollary']}, both $H_1$ and $H_2$ contain at least one critical edge.If there is a path of constraint edges from a vertex $u \in B$ to a vertex $v \in V(H_1) \setminus B$ then by the same reasoning as the third paragraph of the proof of Corollary \ref{['cor:findingkeycorollary']}, $H_1$ must contain at least two critical edges. Since $H_2$ must contain at least one critical edge and the number of critical edges must be even, there must be at least $4$ critical edges.Similarly, if there is a path of constraint edges from a vertex $u \in B$ to a vertex $v \in V(H_2) \setminus B$ then $H_1$ must contain at least two critical edges. Since $H_1$ must contain at least one critical edge and the number of critical edges must be even, there must be at least $4$ critical edges.If there is a path of constraint edges between two vertices $u$ and $v$ in $B$ then since both $H_1$ and $H_2$ must contain at least one critical edge, we have that $2m - |B/C| + \frac{\# \text{ of critical edges}}{2} \geq 2$.If none of the vertices in $B$ are incident to a constraint edge then following the same reasoning we used in the proof of Corollary \ref{['cor:findingkeycorollary']}, we can find a removable vertex in both $H_1$ and $H_2$. This contradicts the assumption that neither $H_1$ nor $H_2$ has a removable vertex. we can now prove Theorem \ref{['thm:generaldominantwellbehaved']} by induction on $\frac{|V(H_1)|}{2} + \frac{|V(H_2)|}{2} - 2m$. For the base case, observe that if $\frac{|V(H_1)|}{2} + \frac{|V(H_2)|}{2} - 2m = 0$ then the result holds as $|E(C)| \geq \frac{|V(H_1)|}{2} + \frac{|V(H_2)|}{2} - 2m = 0$ and $\mathop{\mathrm{val}}\nolimits(C_1) = \mathop{\mathrm{val}}\nolimits(C_2) = 1$ as all edges in $E(H_1)$ and $E(H_2)$ have even multiplicity. For the inductive step, assume the result is true for $H_1,H_2$ such that $\frac{|V(H_1)|}{2} + \frac{|V(H_2)|}{2} - 2m = x$ and consider an $H_1,H_2$ such that $\frac{|V(H_1)|}{2} + \frac{|V(H_2)|}{2} - 2m = x+1$. If all wheels of $H_1$ only have two vertices then $\frac{|V(H_1)|}{2} = m$ and all edges of $H_1$ appear with even multiplicity. In this case, $\mathop{\mathrm{val}}\nolimits(C_1) = \mathop{\mathrm{val}}\nolimits(C_2) = 1$ as the edges in $H_1$ are already paired up with each other so $C_2$ must pair up the edges of $H_2$. By Theorem \ref{['thm:generaldominantwellbehaved']}, $|E(C)| \geq E(C_2) \geq \frac{|V(H_2)|}{2} - m = \frac{|V(H_1)|}{2} + \frac{|V(H_2)|}{2} - 2m$. Similar logic applies if all wheels of $H_2$ only have two vertices. If both $H_1$ and $H_2$ have a wheel which has at least $4$ vertices and either $H_1$ or $H_2$ has a removable vertex, we can delete this vertex, merge the two neighbors of this vertex, and use the inductive hypothesis. If neither $H_1$ nor $H_2$ has a removable vertex then by Corollary \ref{['cor:findingkeycorollaryanalogue']}, $|E(C)| \geq \frac{|V(H_1)| + |V(H_2)|}{2} - 2m$ and if $\mathop{\mathrm{val}}\nolimits(C_1) \neq 0$ or $\mathop{\mathrm{val}}\nolimits(C_2) \neq 0$, $|E(C)| \geq \frac{|V(H_1)| + |V(H_2)|}{2} - 2m + 2$, as needed. $

Key Result

Proposition 1.10

For all $k \in \mathbb{N}$, $\mathop{\mathrm{tr}}\nolimits$$M{M^T}$^k$= \sum_{i=1}^{r}{\sigma_i^{2k}}$

Figures (2)

  • Figure 1.1: Z-shape $\alpha_Z$
  • Figure 1.2: The limiting distribution of the singular values of $\frac{1}{n}M_{\alpha_Z}$ as $n \to \infty$

Theorems & Definitions (25)

  • Definition 1.1: Fourier characters
  • Definition 1.2: Shapes
  • Definition 1.3: Bipartite Shapes
  • Definition 1.4: Graph Matrices
  • Remark 1.5
  • Remark 1.6
  • Definition 1.7: Singular value decomposition
  • Remark 1.8
  • Definition 1.9
  • Proposition 1.10
  • ...and 15 more