So $f = f_j = f_i = f_j'$ and $g = g_j = g_i = g_j'$ on the set $I_j' = I_i \cup I_j$.

Finally, we claim that $W_{\ii'} = W$.
Indeed, this follows directly from Lemma <ref> and the fact that $W = W_\ii$.
Since $\ii'$ is a partition of $[0,1]$ and since $f,g$ are unit eigenfunctions for $W$ Equation (<ref>) holds, and Equations (<ref>) and (<ref>) hold for all $k\in S'$.
This completes the proof of Claim A.

Next, we prove the following claim.

Claim B:
If $S$ satisfies the criteria of Proposition <ref>, then without loss of generality the following holds.

If there exists some $i\in S$ such that $N_i = S$, then $i = 1$.

$S\cap \{1,2,3,4\}\neq\emptyset$.

$S\cap \{2,3,4\}$ is one of $\emptyset, \{4\}, \{2,4\}$, and $\{2,3,4\}$.

$S\cap \{5,6,7\}$ is one of $\{7\}, \{5,7\}$, and $\{5,6,7\}$.

Since $N_1\cap S = S = N_i$, item (<ref>) follows from Claim A applied to the pair $(i,1)$.
Since $f,g$ are orthogonal and $f$ is positive on $[0,1]$, $g$ is positive on a set of positive measure, so item (<ref>) holds.

To prove item (<ref>), we have $4$ cases.
If $S\cap \{2,3,4\} = \{2\}$, then $N_2\cap S = N_1\cap S$ and we may apply Claim A to the pair $(2,1)$.
If $S\cap \{2,3,4\} = \{3\}$ or $\{3,4\}$,  then $N_3\cap S = N_4\cap S$ and we may apply Claim A to the pair $(3,4)$.
If $S\cap \{2,3,4\} = \{2,3\}$, then $N_2\cap S = N_1\cap S$ and we may apply Claim A to the pair $(2,1)$.
So item (<ref>) holds.
For item (<ref>), we reduce $S\cap \{5,6,7\}$ to one of $\emptyset, \{7\}, \{5,7\}$, and $\{5,6,7\}$ in the same fashion.
To eliminate the case where $S\cap \{5,6,7\} = \emptyset$, we simply note that since $f$ and $g$ are orthogonal and $f$ is positive on $[0,1]$, $g$ is negative on a set of positive measure.
This completes the proof of Claim B.

After repeatedly applying Claim B, we may replace $S$ with one of the cases found in Table <ref>.
Let $\mathcal{S}_{21}$ denote the sets in Table <ref>.
By definition,

\begin{align*}
\mathcal{S}_{21}
\sset
\bigcup \left\{
\{4,7\}, \{2,4,7\}, \{2,3,4,7\}, \{2,3,4,5,7\}
\right\}
\end{align*}

Finally, we eliminate the $4$ cases in $\mathcal{S}_{21}\setminus \sset$.
If $S = \{4,7\}$, then $W$ is a bipartite graphon, hence $\spr(W) \leq 1$, a contradiction since $\max_{U\in\ww}\spr(W) > 1$.

For the three remaining cases, let $\tau$ be the permutation on $\{2,\dots,7\}$ defined as follows.
For all $i\in \{2,3,4\}$, $\tau(i) := i+3$ and $\tau(i+3) := i$.
If $S$ is among $\{2,4,7\}, \{2,3,4,7\}, \{2,3,4,5,7\}$, we apply $\tau$ to $S$ in the following sense.
Replace $g$ with $-g$ and replace $(\alpha_i, I_i, f_i, g_i)_{i\in S}$ with $(\alpha_{\tau(i)}, I_{\tau(i)}, f_{\tau(i)}, -g_{\tau(i)})_{i\in \tau(S)}$.
By careful inspection, it follows that $\tau(S)$ satisfies the criteria from Proposition <ref>.
Since $\tau(\{2,4,7\}) = \{4,5,7\}$, $\tau(\{2,3,4,7\}) = \{4,5,6,7\}$, and $\tau(\{2,3,4,5,7\}) = \{2,4,5,6,7\}$, this completes the proof.

§.§ Proof of Lemma <ref>

Let $(\alpha_4, \alpha_5, \alpha_7)$ be a solution to $\SPR_{457}$.

First, let $T :=
\{(\varepsilon_1,\varepsilon_2)\in(-1/3, 2/3)\times (-2/3, 1/3) : \varepsilon_1+\varepsilon_2 \in (0,1)\}$, and for all $\varepsilon = (\varepsilon_1,\varepsilon_2)\in T$, let

\begin{align*}
\left[\begin{array}{ccc}
2/3-\varepsilon_1 & 0 & 1/3-\varepsilon_2\\
0 & 0 & 1/3-\varepsilon_2\\
2/3-\varepsilon_1 & \varepsilon_1 + \varepsilon_2 & 0
\end{array}\right]
\end{align*}

As a motivation, suppose $\mu,\nu$, and $(\alpha_4,\alpha_5,\alpha_7)$ are part of a solution to $\SPR_{\{4,5,7\}}$.
Then with $\varepsilon := (\varepsilon_1,\varepsilon_2) = (2/3-\alpha_5, 1/3-\alpha_4)$, $\varepsilon\in T$ and $\mu,\nu$ are the maximum and minimum eigenvalues of $M(\varepsilon)$, respectively.
By the end of the proof, we show that any solution of $\SPR_{\{4,5,7\}}$ has $\alpha_7 = 0$.

To proceed, we prove the following claims.

Claim A:
For all $\varepsilon\in T$, $M(\varepsilon)$ has two distinct positive eigenvalues and one negative eigenvalue.

Since $M(\varepsilon)$ is diagonalizable, it has $3$ real eigenvalues which we may order as $\mu\geq \delta\geq \nu$.
Since $\mu\delta\nu = \det(M(\varepsilon)) = -\alpha_4\alpha_5\alpha_7\neq 0 < 0$, $M(\varepsilon)$ has an odd number of negative eigenvalues.
Since $0 < \alpha_5 = \mu + \delta + \nu$, it follows that $\mu \geq \delta > 0 > \nu$.
Finally, note by the Perron-Frobenius Theorem that $\mu > \delta$.
This completes the proof of Claim A.

Next, we define the following quantities, treated as functions of $\varepsilon$ for all $\varepsilon\in T$.
For convenience, we suppress the argument “$\varepsilon$” in most places.
Let $k(x) = ax^3+bx^2+cx+d$ be the characteristic polynomial of $M(\varepsilon)$.
By inspection,

\begin{align*}
\varepsilon_1-\dfrac{2}{3}
\\
\dfrac{
\dfrac{
\end{align*}

Continuing, let

\begin{align*}
\dfrac{
\dfrac{
\\
\dfrac{-p}
\dfrac{
\\
\arccos\left(
\dfrac{
\right).
\end{align*}

Let $S(\varepsilon)$ be the difference between the maximum and minimum eigenvalues of $M(\varepsilon)$.
We show the following claim.

Claim B:
For all $\varepsilon\in T$,

\begin{align*}
\sqrt{3}\cdot A(\varepsilon)\cdot\cos\left(
\dfrac{2\phi(\varepsilon) - \pi}{6}
\right).
\end{align*}

Moreover, $S$ is analytic on $T$.

Indeed, by Viéte's Formula, using the fact that $k(x,y)$ has exactly $3$ distinct real roots, the quantities $a(\varepsilon),\dots,\phi(x,y)$ are analytic on $T$.
Moreover, the eigenvalues of $M(\varepsilon)$ are $x_0, x_1, x_2$ where, for all $k\in\{0,1,2\}$,

\begin{align*}
A(\varepsilon)\cdot \cos\left(
\dfrac{\phi + 2\pi\cdot k}{3}
\right)
+ B(\varepsilon)
\end{align*}

Moreover, $x_0(\varepsilon),x_1(\varepsilon),x_2(\varepsilon)$ are analytic on $T$.
For all $k,\ell\in \{1,2,3\}$, let

\begin{align*}
\cos\left(
x+\dfrac{2\pi k}{3}
\right)
- \cos\left(
x+\dfrac{2\pi \ell}{3}
\right)
\end{align*}

For all $(k,\ell)\in \{(0,1), (0,2), (2,1)\}$, note the trigonometric identities

\begin{align*}
\sqrt{3}\cdot\left\{\begin{array}{rl}
\cos\left(
x - \dfrac{\pi}{6}
\right),
& (k,\ell) = (0, 1)
\\
\cos\left(
x + \dfrac{\pi}{6}
\right),
& (k,\ell) = (0, 2)
\\
\sin(x),
& (k,\ell) = (2, 1)
\end{array}\right.
\end{align*}

By inspection, for all $x\in (0,\pi/3)$,

\begin{align*}
\max\left\{
D(0,2), D(2,1)
\right\}
\geq
\min\left\{
D(0,2), D(2,1)
\right\}
\geq
\end{align*}

Since $A > 0$ and $\phi \in (0,\pi/3)$, the claimed equality holds.
Since $x_0(\varepsilon),x_1(\varepsilon)$ are analytic, $S(\varepsilon)$ is analytic on $T$.
This completes the proof of Claim B.

Next, we compute the derivatives of  $S(\varepsilon)$ on $T$.
For convenience, denote by $A_i, \phi_i, $ and $S_i$ for the partial derivatives of $A$ and $\phi$ by $\varepsilon_i$, respectively, for $i\in \{1,2\}$.
Furthermore, let

\begin{align*}
\psi(\varepsilon)
\dfrac{2\phi(\varepsilon)-\pi}{6}.
\end{align*}

The next claim follows directly from Claim B.

Claim C:
For all $i\in T$, then on the set $T$, we have

\begin{align*}
3A_i\cdot \cos\left(
\psi
\right)
\cdot A\phi_i \sin\left(
\psi
\right)
\end{align*}

Moreover, each expression is analytic on $T$.

Finally, we solve $\SPR_{\{4,5,7\}}$.

Claim D:
If $(\alpha_4,\alpha_5,\alpha_7)$ is a solution to $\SPR_{\{4,5,7\}}$, then $0\in\{\alpha_4,\alpha_5,\alpha_7\}$.

With $(\alpha_4,\alpha_5,\alpha_7) := (1/3-\varepsilon_2, 2/3-\varepsilon_1,  \varepsilon_1+\varepsilon_2)$ and using the fact that $S$ is analytic on $T$, it is sufficient to eliminate all common zeroes of $S_1$ and $S_2$ on $T$.
With the help of a computer algebra system and the formulas for $S_1$ and $S_2$ from Claim C, we replace the system $S_1 = 0$ and $S_2 = 0$ with a polynomial system of equations $P = 0$ and $Q = 0$ whose real solution set contains all previous solutions.

\begin{align*}
9\varepsilon_1^3 + 18\varepsilon_1^2\varepsilon_2 + 54\varepsilon_1\varepsilon_2^2 + 18\varepsilon_2^3 - 15\varepsilon_1^2 - 33\varepsilon_1\varepsilon_2 - 27\varepsilon_2^2 + 5\varepsilon_1 + \varepsilon_2
\end{align*}

and $Q = 43046721\varepsilon_1^{18}\varepsilon_2+\cdots + (-532480\varepsilon_2)$ is a polynomial of degree $19$, with coefficients between $-184862311457373$ and $192054273812559$.
For brevity, we do not express $Q$ explicitly.

To complete the proof of Claim D, it suffices to show that no common real solution to $P = Q = 0$ which lies in $T$ also satisfies $S_1 = S_2 = 0$.
Again using a computer algebra system, we first find all common zeroes of $P$ and $Q$ on $\RR^2$.
Included are the rational solutions $(2/3, -2/3), (-1/3, 1/3), (0,0), (2/3, 1/3), $ and $(2/3, -1/6)$ which do not lie in $T$.
Furthermore, the solution $(1.2047\dots, 0.0707\dots)$ may also be eliminated.
For the remaining $4$ zeroes, $S_1, S_2\neq 0$.  A notebook showing these calculations can be found at <cit.>.

Claim E:
If $\mu, \nu$, and $\alpha = (\alpha_4,\alpha_5,\alpha_7)$ is part of a solution to $\SPR_{\{4,5,7\}}$ such that $\mu-\nu\geq 1$, then $\alpha_7 = 0$.

By definition of $\SPR_{\{4,5,7\}}$, $\mu$ and $\nu$ are eigenvalues of the matrix

\begin{align*}
\left[\begin{array}{ccc}
\alpha_5 & 0 & \alpha_4\\
0 & 0 & \alpha_4\\
\alpha_5 & \alpha_7 & 0
\end{array}\right].
\end{align*}

Furthermore, $N(\alpha)$ has characteristic polynomial

\begin{align*}
- \alpha_5 x^2
\end{align*}

Recall that $\alpha_4+\alpha_5+\alpha_7 = 1$.
By Claim D, $0\in\{4,5,7\}$, and it follows that $p\in\{p_4, p_5, p_7\}$ where

\begin{align*}
x^2\cdot (x-\alpha_5),
\\
x\cdot (x^2-\alpha_4(1-\alpha_4)),
\text{ and }
\\
x\cdot (x^2-(1-\alpha_4)x-\alpha_4(1-\alpha_4)).
\end{align*}

If $p = p_4$, then $\mu-\nu = \alpha_5\leq 1$, and if $p = p_5$, then $\mu-\nu = 2\sqrt{\alpha_4(1-\alpha_4)} \leq 1$.
So $p = p_7$, which completes the proof of Claim E.

This completes the proof of Lemma <ref>.

§.§ Proof of Lemma <ref>

First, we find $S_z(\varepsilon_1,\varepsilon_3)$ using Viète's Formula.
In doing so, we define functions  $k_z(\varepsilon_1,\varepsilon_2;x),\dots,\delta_z(\varepsilon_1,\varepsilon_2)$.
To ease the burden on the reader, we suppress the subscript $z$ and the arguments $\varepsilon_1,\varepsilon_2$ when convenient and unambiguous.
Let $k(x) = ax^3+bx^2+cx+d$ be the characteristic polynomial of $M_z(\varepsilon_1,\varepsilon_2)$.
By inspection,

\begin{align*}
\varepsilon_1+z-\dfrac{2}{3}
\\
\dfrac{
\dfrac{
\end{align*}

Continuing, let

\begin{align*}
\dfrac{
\dfrac{
\\
\dfrac{-p}
\dfrac{
\\
\arccos\left(
\dfrac{
\right).
\end{align*}

By Viète's Formula, the roots of $k_z(\varepsilon_1,\varepsilon_2;x)$ are the suggestively defined quantities:

\begin{align*}
\cos\left(
\dfrac{
\phi
\right)
\cos\left(
\dfrac{
\phi
\right)
\\
\delta
\cos\left(
\dfrac{
\phi
\right)
\end{align*}

First, We prove the following claim.

Claim A:
If $(\varepsilon_1,\varepsilon_2,z)$ is sufficiently close to $(0,0,0)$, then

\begin{align}\label{eq: spread trig formula}
\cdot\cos\left(
\dfrac{2\phi_z(\varepsilon_1,\varepsilon_2)-\pi}{6}
\right)
\end{align}

Indeed, suppose $z>0$ and $z\to 0$.
Then for all $(\varepsilon_1,\varepsilon_2)\in (-3z,3z)$, $\varepsilon_1,\varepsilon_2\to 0$.
With the help of a computer algebra system, we substitute in $z=0$ and $\varepsilon_1,\varepsilon_2=0$ to find the limits:

\begin{align*}
\left(
\dfrac{-2}{3},
\dfrac{-2}{9},
\right)
\\
\left(
\dfrac{-10}{27},
\dfrac{-52}{729}
\right)
\\
\left(
\dfrac{2\sqrt{10}}{9},
\dfrac{2}{9},
\arccos\left(
\dfrac{13\sqrt{10}}{50}
\right)
\right).
\end{align*}

Using a computer algebra system, these substitutions imply that

\begin{align*}
\to
\left(
\right)
\end{align*}

So for all $z$ sufficiently small, $S = \mu-\nu$.
After some trigonometric simplification,

\begin{align*}
\mu - \nu
\cos\left(
\dfrac{\phi}{3}
\right)
\dfrac{\phi+2\phi}{3}
\right)
\right)
\cdot\cos\left(
\dfrac{2\phi-\pi}{6}
\right)
\end{align*}

and Equation (<ref>).
This completes the proof of Claim A.

Now we prove the following claim.

Claim B:
There exists a constants $C_0'>0$ such that the following holds.
If $|z|$ is sufficiently small, then $S_z$ is concave-down on $[-C_0,C_0]^2$ and strictly decreasing on $[-C_0,C_0]^2\setminus [-C_0z, C_0z]^2$.

First, we define

\begin{align*}
:= \restr{\left(
\dfrac{\partial^2 S_z}{\partial\varepsilon_1^2}
\cdot \dfrac{\partial^2 S_z}{\partial\varepsilon_2^2}
- \left(\dfrac{\partial^2 S_z}{\partial\varepsilon_1\partial\varepsilon_2}\right)^2
\right)}
\end{align*}

As a function of $(\varepsilon_1,\varepsilon_2)$, $D_z$ is the determinant of the Hessian matrix of $S_z$.
Using a computer algebra system, we note that

\begin{align*}
\quad \text{ and }
\\
\restr{\left(
\dfrac{
\partial^2S
\partial \varepsilon_1^2
\dfrac{
\partial^2S
\partial \varepsilon_1
\partial \varepsilon_2
\dfrac{
\partial^2S
\partial \varepsilon_2^2
\right)
\left(
\right).
\end{align*}

Since $S$ is analytic to $(0,0,0)$, there exist constants $C_1, C_2>0$ such that the following holds.
For all $z\in [-C_1,C_1]$, $S_z$ is concave-down on $[-C_1,C_1]^2$.
This completes the proof of the first claim.
Moreover for all $z\in [-C_1,C_1]$ and for all $(\varepsilon_1,\varepsilon_2)\in [-C_1,C_1]^2$,

\begin{align*}
\restr{\max\left\{
\dfrac{\partial^2S_z}
\dfrac{\partial^2S_z}
\dfrac{\partial^2S_z}
\right\}}
&\leq -C_2.
\end{align*}

to complete the proof of the second claim, note also that since $S$ is analytic at $(0,0,0)$, there exist constants $C_3,C_4>0$ such that for all $z\in [-C_3,C_3]$ and all $(\varepsilon_1,\varepsilon_2)\in [-C_3, C_3]^2$,

\begin{align*}
\dfrac{\partial^2 S}{\partial z\partial\varepsilon_i} \leq C_4.
\end{align*}

Since $(0,0)$ is a local maximum of $S_0$,

\begin{align*}
\restr{
\dfrac{\partial S}
{\partial \varepsilon_i}
\restr{
\dfrac{\partial S}
{\partial \varepsilon_i}
\restr{
\dfrac{\partial^2 S}
{\partial z\partial \varepsilon_i}
\\
&\quad + \int_{ {\bf u} = (0,0)}^{ (\varepsilon_1,\varepsilon_2) }
\restr{\dfrac{
\partial^2 S
\partial {\bf u}\partial \varepsilon_i
{({\bf u}, z)}
d{\bf u}
\\
C_4\cdot z
- C_2 \cdot \|(\varepsilon_1,\varepsilon_2)\|_2.
\end{align*}

Since $C_2, C_4>0$, this completes the proof of Claim B.

Next, we prove the following claim.

Claim C:
If $z$ is sufficiently small, then $\pp_{z,C_0}$ is solved by a unique point $(\varepsilon_1^*,\varepsilon_2^*) = (\varepsilon_1^*(z),\varepsilon_2^*(z))$.
Moreover as $z\to0$,

\begin{align}\label{eq: optimal epsilon approximation}
\left(
\varepsilon_1^*, \varepsilon_2^*
\right)
\left(
\dfrac{7z}{30},
\dfrac{-z}{3}
\right).
\end{align}

Indeed, the existence of a unique maximum $(\varepsilon_1^*, \varepsilon_2^*)$ on $[-C_0,C_0]^2$ follows from the fact that $S_z$ is strictly concave-down and bounded on $[-C_0, C_0]^2$ for all $z$ sufficiently small.
Since $S_z$ is strictly decreasing on $[-C_0,C_0]^2\setminus (-C_0z, C_0z)^2$, it follows that $(\varepsilon_1^*, \varepsilon_2^*)\in (-C_0z, C_0z)$.
For the second claim, note that since $S$ is analytic at $(0,0,0)$,

\begin{align*}
0 &=
\restr{
\dfrac{\partial S}
{\partial \varepsilon_i}
{(\varepsilon_1^*, \varepsilon_2^*, z)}
\sqrt{3}\cdot \left(
\dfrac{\partial A}{\partial\varepsilon_i}\cdot
\cos\left(
\dfrac{2\phi-\pi}{6}
\right)
- \dfrac{A}{3}\cdot
\dfrac{\partial\phi}{\partial\varepsilon_i}\cdot
\sin\left(
\dfrac{2\phi-\pi}{6}
\right)
\right)
\end{align*}

for both $i = 1$ and $i = 2$.

\begin{align*}
\tau_i :=
\dfrac{
3\cdot \dfrac{\partial A}{\partial\varepsilon_i}
A\cdot \dfrac{\partial \phi}{\partial \varepsilon_i}
\end{align*}

for both $i = 1$ and $i = 2$
Then by Equation (<ref>),

\begin{align*}
\restr{
\arctan(\tau_i)
\restr{
\dfrac{2\phi-\pi}{6}
\end{align*}

for both $i=1$ and $i=2$.
We first consider linear approximation of the above quantities under the limit  $(\varepsilon_1,\varepsilon_2,z)\to (0,0,0)$.
Here, we write $f(\varepsilon_1,\varepsilon_2,z) \sim g(\varepsilon_1,\varepsilon_2,z)$ to mean that

\begin{align*}
\left(
\right\}\right)
\right)
\cdot g(\varepsilon_1,\varepsilon_2,z).
\end{align*}

With the help of a computer algebra system, we note that

\begin{align*}
\arctan\left(
\tau_1
\right)
\dfrac{
\dfrac{1}{3}
\right)
\\
\arctan\left(
\tau_2
\right)
\dfrac{
\dfrac{1}{3}
\right)
\\
\dfrac{2\phi-\pi}
\dfrac{
\dfrac{13\sqrt{10}}{50}
\right)
\end{align*}

By inspection, the constant terms match due to the identity

\begin{align*}
\dfrac{1}{3}
\right)
\dfrac{1}{3}
\arccos\left(
\dfrac{13\sqrt{10}}{50}
\right)
\end{align*}

Since $\max\left\{|\varepsilon_1^*|, |\varepsilon_2^*|\right\}\leq C_0z$, replacing $(\varepsilon_1,\varepsilon_2)$ with $(\varepsilon_1^*,\varepsilon_2^*)$ implies that

\begin{align*}
\dfrac{-78\varepsilon_1^*-96\varepsilon_2^*-3z}{2}
&= (1+o(z))\cdot
(36\varepsilon_1^*+27\varepsilon_2^*+6z), \quad \text{ and }
\\
\end{align*}

as $z\to 0$.
After applying Gaussian Elimination to this $3$-variable system of $2$ equations, it follows that

\begin{align*}
(\varepsilon_1^*, \varepsilon_2^*)
\left(
(1+o(z))\cdot \dfrac{7z}{30},
(1+o(z))\cdot \dfrac{-z}{3}
\right).
\end{align*}

This completes the proof of Claim C.

For the next step, we prove the following claim.
First, let $\qq_{n}$ denote the program formed from $\pp_{n^{-1}, C_0}$ subject to the added constraint that $\textstyle n\cdot( \frac{2}{3}-\varepsilon_1 ),n\cdot( \frac{1}{3}-\varepsilon_2 )\in\mathbb{Z}$.

Claim D:
For all $n$ sufficiently large, $\qq_{n}$ is solved by a unique point $(n_1^*, n_3^*)$ which satisfies $n_1^* + n_3^* = n$.

Note by Lemma <ref> that for all $n$ sufficiently large,

\begin{align*}
\max\left\{
\left|\dfrac{n_1}{n}-\dfrac{2}{3}\right|,
\left|\dfrac{n_3}{n}-\dfrac{1}{3}\right|
\right\}
&\leq C_0.
\end{align*}

Moreover, by Claim C, $\pp_{n^{-1}}$ is solved uniquely by

\begin{align*}
(\varepsilon_1^*, \varepsilon_2^*)
\left(
(1+o(z))\cdot \dfrac{7}{30n},
(1+o(z))\cdot \dfrac{-1}{3n}
\right).
\end{align*}

\begin{align*}
\dfrac{2n}{3}-n\cdot \varepsilon_1^*
\dfrac{2n}{3}-(1+o(1))\cdot \dfrac{7}{30}
\end{align*}

and $7/30 < 1/3$, it follows for $n$ sufficiently large that $2n/3-n\cdot\varepsilon_1^*\in I_1$ where

\begin{align*}
I_1 &:=
\left\{\begin{array}{rl}
\left(
\dfrac{2n}{3}
\dfrac{2n}{3}
\right),
3 \mid n
\\
\left(
\left\lfloor
\dfrac{2n}{3}
\right\rfloor,
\left\lceil
\dfrac{2n}{3}
\right\rceil
\right),
3 \nmid n
\end{array}\right.
\end{align*}

Similarly since

\begin{align*}
n\cdot(\varepsilon_1^* + \varepsilon_2^*)
\cdot \left(
\dfrac{7}{30}-\dfrac{1}{3}
\right)
\cdot
\dfrac{-1}{10}
\end{align*}

and $1/10 < 1/3$, it follows that $n\cdot (\varepsilon_1^*+\varepsilon_2^*)\in (-1,0)$.

\begin{align*}
\left(
\dfrac{2n}{3}-n\cdot \varepsilon_1,
\right)
&\in I_1\times (-1, 0).
\end{align*}

Note that to solve $\qq_{n}$, it is sufficient to maximize $S_{n^{-1}}$ on the set $[-C_0, C_0]^2 \cap \{ (n_1/n, n_3/n) \}_{u,v\in\mathbb{N}}$.
Since $S_{n^{-1}}$ is concave-down on $I_1\times (-1, 0)$, $(n_1^*, n-n_1^*-n_3^*)$ is a corner of the square $I_1\times (-1,0)$.
So $n_1^*+n_2^* = n$, which implies Claim D.
This completes the proof of the main result.

For completeness, we prove this result directly.
Let $G = K_{n_1}\vee K_{n_2}^c$.
Then $A_G$ has the same spread as the “reduced matrix” defined as

\begin{align*}
\left[\begin{array}{cc}
n_1-1 & n_2\\
n_1 & 0
\end{array}\right].
\end{align*}

By inspection, $A(n_1,n_2)$ has characteristic polynomial $x^2 - (n_1 - 1)x + -n_1n_2$ and thus its eigenvalues are

\begin{align*}
\dfrac{
\pm \sqrt{
n_1^2 + 4n_1n_2 - 2n_1+1
\end{align*}

\begin{align*}
\spr(G)
\sqrt{
\end{align*}

Making the substitution $(n_1, n_2) = (rn, (1-r)n)$ and simplifying, we see that

\begin{align*}
\spr(G)^2
\\
\cdot\left(
\left(
\right)^2
\right),
\end{align*}

which is maximized when $r = n_1/n$ is nearest $2/3-1/(3n)$, or equivalently, when $n_1$ is nearest $(2n-1)/3$.
After noting that

\begin{align*}
\left\lfloor
\dfrac{2n}{3}
\right\rfloor
\dfrac{2n-1}{3}
\left\{\begin{array}{rl}
1/3, & n\equiv0\pmod3\\
-1/3, & n\equiv1\pmod3\\
0, & n\equiv2\pmod3
\end{array}\right.
\end{align*}

the desired claim follows.

§ A COMPUTER-ASSISTED PROOF OF LEMMA <REF>

In this appendix, we derive a number of formulas that a stepgraphon corresponding to some set $S \subseteq \{1,2,3,4,5,6,7\}$ in Lemma <ref> satisfies, and detail how these formulas are used to provide a computer-assisted proof of Lemma <ref>.

§.§ Formulas

In this subsection, we derive the formulas used in our computer-assisted proof, from the equations described in Section <ref>.
First, we define a number of functions which will ease the notational burden in the results that follow. Let

\begin{align*}
F_1(x) &:= (\mu+\nu)x + 2 \mu \nu,
\\
F_2(x) &:= 2 ( \mu \nu + (\mu+\nu) x)^2 + (\mu+\nu) x^3,
\\
F_3(x) &:=
4\mu^2\nu^2\cdot (\mu\nu + (\mu+\nu)x)^2
\\
&\quad - 2(\mu+\nu)x^3\cdot ((\mu+\nu)x + \mu\nu)((\mu+\nu)x+3\mu\nu)
\\
&\quad - (\mu+\nu)x^5\cdot (2\mu\nu + (\mu+\nu)x),
\\
F_4(x) &:=
4\mu^2\nu^2x\cdot ((3(\mu+\nu)x + \mu\nu)\cdot (2(\mu+\nu)x+\mu\nu)-\mu\nu(\mu+\nu)x)
\\
&\quad +4(\mu+\nu)x^4\cdot (((\mu+\nu)x+\mu\nu)^2 + (\mu+\nu)^2\cdot ((\mu+\nu)x+4\mu\nu))
\\
&\quad + (\mu+\nu)^2x^7.
\end{align*}

Letting  $S := \{i\in\{1,\dots,7\} : \alpha_i>0\}$, we prove the following six formulas.

Let $i\in \{1,2,5\} \cap S$ and $j\in \{3,4,6,7\} \cap S$ be such that $N_i \cap S = (N_j\cap S)\dot{\cup}\{j\}$. Then

\begin{align*}
&= \dfrac{(\alpha_j+2\nu)\mu}{F_1(\alpha_j)},
\quad \quad g_j^2
= \dfrac{(\alpha_j+2\mu)\nu}{F_1(\alpha_j)}
\end{align*}

\begin{align*}
f_i &= \left( 1 + \frac{\alpha_j}{\mu} \right) f_j,
\quad \quad g_i = \left( 1 + \frac{\alpha_j}{\nu} \right) g_j.
\end{align*}

Moreover, $F_1(\alpha_j)$ and $\alpha_j+2\nu$ are negative.

By Lemma <ref>,

\begin{align*}
\mu f_i^2 - \nu g_i^2 &= \mu-\nu\\
\mu f_j^2 - \nu g_j^2 &= \mu-\nu.
\end{align*}

By taking the difference of the eigenvector equations for $f_i$ and $f_j$ (and also $g_i$ and $g_j$), we obtain

\begin{align*}
\alpha_j f_j &= \mu(f_i - f_j)\\
\alpha_j g_j &= \nu(g_i - g_j),
\end{align*}

or, equivalently,

\begin{align*}
f_i &= \left( 1 + \frac{\alpha_j}{\mu} \right) f_j\\
g_i &= \left( 1 + \frac{\alpha_j}{\nu} \right) g_j.
\end{align*}

This leads to the system of equations

\begin{align*}
\left[\begin{array}{cc}
\mu
& -\nu\\
\mu\cdot \left(1+\dfrac{\alpha_j}{\mu}\right)^2
& -\nu\cdot \left(1+\dfrac{\alpha_j}{\nu}\right)^2
\end{array}\right]
\cdot
\left[
\begin{array}{c}
\end{array}
\right]
\left[\begin{array}{c}
\mu-\nu\\
\mu-\nu
\end{array}\right].
\end{align*}

If the corresponding matrix is invertible, then after  substituting the claimed formulas for $f_j^2,g_j^2$ and simplifying, it follows that they are the unique solutions.
To verify that $F_1(\alpha_j)$ and $\alpha_j+2\nu$ are negative, it is sufficient to inspect the formulas for $f_j$ and $g_j$, noting that $\nu$ is negative and both $\mu$ and $\alpha_j+2\mu$ are positive.

Suppose the matrix is not invertible. By assumption $\mu, \nu \ne 0$, and so

$$\bigg(	1+\frac{\alpha_j}{\mu}\bigg)^2 = \bigg(	1+\frac{\alpha_j}{\nu}\bigg)^2.$$

But, since $i\in \{1,2,5\}$ and $j\in \{3,4,6,7\}$,

\begin{align*}
1 > f_j^2g_i^2
f_i^2g_i^2\cdot \left(
\right)^2
f_i^2g_i^2\cdot \left(
\right)^2
f_i^2g_j^2 > 1,
\end{align*}

a contradiction.

Let $i\in \{1,2,5\} \cap S$ and $j\in \{3,4,6,7\} \cap S$ be such that $N_i \cap S = (N_j\cap S)\dot{\cup}\{i\}$. Then

\begin{align*}
&= \dfrac{(\alpha_i-2\nu)\mu}
\quad \quad
= \dfrac{(\alpha_i-2\mu)\nu}
\end{align*}

\begin{align*}
&= \left( 1- \frac{\alpha_i}{\mu} \right) f_i,
\quad \quad
= \left( 1- \frac{\alpha_i}{\nu} \right) g_i.
\end{align*}

Moreover, $-F_1(-\alpha_i)$ is positive and $\alpha_i-2\mu$ is negative.

The proof of Proposition <ref>, slightly modified, gives the desired result.

Suppose $i,j,k\in S$ where $(i,j,k)$ is either $(2,3,4)$ or $(5,6,7)$. Then

\begin{align*}
f_k &= \dfrac{ \mu f_j-\alpha_if_i }{\mu}, \quad \quad
g_k = \dfrac{ \nu g_j-\alpha_ig_i }{\nu},
\end{align*}

$$\alpha_i = \frac{2 \mu^2 \nu^2 \alpha_j}{ F_2(\alpha_j)}.$$

Using the eigenfunction equations for $f_j,f_k$ and for $g_j,g_k$, it follows that

\begin{align*}
f_k &= \dfrac{ \mu f_j-\alpha_if_i }{\mu}, \quad \quad
g_k = \dfrac{ \nu g_j-\alpha_ig_i }{\nu}.
\end{align*}

Combined with Lemma <ref>, it follows that

\begin{align*}
0 &= \mu f_k^2 - \nu g_k^2 - (\mu-\nu) \\
\mu\left(
\dfrac{ \mu f_j-\alpha_if_i }{\mu}
\right)^2
\dfrac{ \nu g_j-\alpha_ig_i }{\nu}
\right)^2
- (\mu-\nu).
\end{align*}

After expanding, we note that the right-hand side can be expressed purely in terms of $\mu, \nu, \alpha_i, f_i^2, f_if_j, f_j^2, g_i^2, g_ig_j,$ and $g_j^2$.
Note that Proposition <ref> gives explicit formulas for $f_i^2, f_if_j$, and $f_j^2$, as well as $g_i^2, g_ig_j$, and $g_j^2$, purely in terms of $\mu, \nu$, and $\alpha_j$.
With the help of a computer algebra system, we make these substitutions and factor the right-hand side as:

\begin{align*}
0 &=
(\mu-\nu)\cdot \alpha_i
\cdot \dfrac{
2\mu^2\nu^2\cdot \alpha_j
- F_2(\alpha_j)\cdot \alpha_i
\mu^2\nu^2 \cdot F_1(\alpha_i)
\end{align*}

Since $\alpha_i, (\mu-\nu) \neq 0$, the desired claim holds.

Suppose $1,i,j,k \in S$ where $(i,j,k)$ is either $(2,3,4)$ or $(5,6,7)$.

\begin{align*}
f_1 &= \dfrac{\mu f_i + \alpha_k f_k}{\mu}, \quad\quad g_1 = \dfrac{\nu g_i + \alpha_k g_k}{\nu},
\end{align*}

\begin{align*}
\alpha_k
\dfrac{
\alpha_j\cdot F_2(\alpha_j)^2
\end{align*}

Using the eigenfunction equations for $f_1, f_i, f_j, f_k$ and for $g_1, g_i, g_j, g_k$, it follows that

\begin{align*}
f_1 &= \dfrac{\mu f_i + \alpha_k f_k}{\mu}, \quad\quad g_1 = \dfrac{\nu g_i + \alpha_k g_k}{\nu},
\end{align*}

\begin{align*}
f_k &= \dfrac{ \mu f_j-\alpha_if_i }{\mu}, \quad \quad
g_k = \dfrac{ \nu g_j-\alpha_ig_i }{\nu}.
\end{align*}

\begin{align*}
f_1 &=
\dfrac{\mu^2 f_i + \alpha_k( \mu f_j - \alpha_if_i )}{\mu^2},
\quad\quad
g_1 &=
\dfrac{\nu^2 g_i + \alpha_k( \nu g_j - \alpha_ig_i )}{\nu^2}
\end{align*}

Combined with Lemma 3.6, it follows that

\begin{align*}
0 &= \mu f_1^2-\nu g_1^2 - (\mu-\nu)
\\
\mu\left( \dfrac{\mu^2 f_i + \alpha_k( \mu f_j - \alpha_if_i )}{\mu^2} \right)^2
-\nu\left( \dfrac{\nu^2 g_i + \alpha_k( \nu g_j - \alpha_ig_i )}{\nu^2} \right)^2
\end{align*}

After expanding, we note that the right-hand side can be expressed purely in terms of $\mu,\nu, f_i^2, f_if_j, f_j^2, g_i^2, g_ig_j,$ and $\alpha_i$.
Note that Proposition <ref> gives explicit formulas for $f_i^2, f_if_j, f_j^2, g_i^2, g_ig_j,$ and $g_j^2$ purely in terms of $\mu, \nu$, and $\alpha_j$.
With the help of a computer algebra system, we make these substitutions and factor the right-hand side as:

\begin{align*}
2\alpha_k\cdot (\mu-\nu)\cdot
\dfrac{
\alpha_j\cdot F_2(\alpha_j)^2 - \alpha_k\cdot F_3(\alpha_j)
F_1(\alpha_j)\cdot F_2(\alpha_j)^2
\end{align*}

So the desired claim holds.

Suppose $1,i,k\in S$ and $j\notin S$ where $(i,j,k)$ is either $(2,3,4)$ or $(5,6,7)$.

\begin{align*}
f_1 &= \dfrac{\mu f_i + \alpha_k f_k}{\mu}, \quad\quad g_1 = \dfrac{\nu g_i + \alpha_k g_k}{\nu},
\end{align*}

\begin{align*}
\alpha_k
\dfrac{2\alpha_i \mu^2\nu^2}
\end{align*}

Using the eigenfunction equations for $f_1, f_i, f_j, f_k$ and for $g_1, g_i, g_j, g_k$, it follows that

\begin{align*}
f_1 &= \dfrac{\mu f_i + \alpha_k f_k}{\mu}, \quad\quad g_1 = \dfrac{\nu g_i + \alpha_k g_k}{\nu},
\end{align*}

\begin{align*}
f_k &= \dfrac{ \mu f_i-\alpha_if_i }{\mu}, \quad \quad
g_k = \dfrac{ \nu g_i-\alpha_ig_i }{\nu}.
\end{align*}

\begin{align*}
f_1 &=
\dfrac{\mu^2 f_i + \alpha_k( \mu f_i - \alpha_if_i )}{\mu^2},
\quad\quad
g_1 &=
\dfrac{\nu^2 g_i + \alpha_k( \nu g_i - \alpha_ig_i )}{\nu^2}
\end{align*}

Combined with Lemma 3.6, it follows that

\begin{align*}
0 &= \mu f_1^2-\nu g_1^2 - (\mu-\nu)
\\
\mu\left( \dfrac{\mu^2 f_i + \alpha_k( \mu f_i - \alpha_if_i )}{\mu^2} \right)^2
-\nu\left( \dfrac{\nu^2 g_i + \alpha_k( \nu g_i - \alpha_if_i )}{\nu^2} \right)^2
\end{align*}

After expanding, we note that the right-hand side can be expressed purely in terms of $\mu,\nu, f_i^2, f_if_j, f_j^2, g_i^2, g_ig_j,$ and $\alpha_i$.
Note that Proposition <ref> gives explicit formulas for $f_i^2, f_if_j, f_j^2, g_i^2, g_ig_j,$ and $g_j^2$ purely in terms of $\mu, \nu$, and $\alpha_j$.
With the help of a computer algebra system, we make these substitutions and factor the right-hand side as:

\begin{align*}
2\alpha_k\cdot (\mu-\nu)\cdot
\dfrac{
\alpha_j\cdot F_2(\alpha_j)^2 - \alpha_k\cdot F_3(\alpha_j)
F_1(\alpha_j)\cdot F_2(\alpha_j)^2
\end{align*}

So the desired claim holds.

Suppose $1\notin S$ and $i,j,k,\ell\in S$ where $(i,j,k,\ell)$ is either $(2,3,4,7)$ or $(5,6,7,4)$.

\begin{align*}
\alpha_k &= \dfrac{F_4(x)}{F_3(x)}.
\end{align*}

Using the eigenfunction equations for $f_\ell, f_i, f_j, f_k$ and for $g_\ell, g_i, g_j, g_k$, it follows that

\begin{align*}
&= \dfrac{\alpha_if_i + \alpha_jf_j + \alpha_k f_k}{\mu},
\quad\quad
g_1 = \dfrac{\alpha_ig_i + \alpha_jg_j + \alpha_k g_k}{\nu},
\end{align*}

\begin{align*}
f_k &= \dfrac{ \mu f_j-\alpha_if_i }{\mu}, \quad \quad
g_k = \dfrac{ \nu g_j-\alpha_ig_i }{\nu}.
\end{align*}

\begin{align*}
\dfrac{ \mu \alpha_if_i + \alpha_jf_j + \alpha_k (\mu f_j-\alpha_if_i) }{\mu^2},
\quad\quad
\dfrac{ \nu \alpha_ig_i + \alpha_jg_j + \alpha_k (\nu g_j-\alpha_ig_i) }{\nu^2}
\end{align*}

Combined with Lemma <ref>, it follows that

\begin{align*}
0 &= \mu f_\ell^2 - \nu g_\ell^2 - (\mu-\nu)\\
\mu\left(
\dfrac{
\mu \alpha_if_i + \alpha_jf_j + \alpha_k (\mu f_j-\alpha_if_i)
\right)^2-\nu\left(
\dfrac{
\nu \alpha_ig_i + \alpha_jg_j + \alpha_k (\nu g_j-\alpha_ig_i)
\right)^2\\
&\quad -(\mu-\nu)
\end{align*}

After expanding, we note that the right-hand side can be expressed purely in terms of $\mu,\nu, f_i^2, f_if_j, f_j^2, g_i^2, g_ig_j,$ and $\alpha_i$.
Note that Proposition <ref> gives explicit formulas for $f_i^2, f_if_j, f_j^2, g_i^2, g_ig_j,g_j^2,\alpha_i,\alpha_j,\alpha_k$ purely in terms of $\mu, \nu$, and $\alpha_j$.
With the help of a computer algebra system, we make these substitutions and factor the right-hand side as:

\begin{align*}
\dfrac{
F_4(\alpha_j)-\alpha_k\cdot F_3(\alpha_j)
{F_1(\alpha_j)\cdot F_2(\alpha_j)^2}
\end{align*}

Suppose $2,4,5,7\in S$ and let $\alpha_{\ne 4,7}:=\sum_{\substack{i\in S,\\ i \ne 4,7}} \alpha_i$. Then

\begin{align*}
\alpha_4 &= \frac{ (1-\alpha_{\ne 4,7}) f_7 - \mu (f_2 - f_7)}{f_4 +f_7}, \\
\alpha_7 &= \frac{ (1-\alpha_{\ne 4,7}) f_4 - \mu (f_5 - f_2)}{f_4 +f_7},
\end{align*}

\begin{align*}
\alpha_4 &= \frac{ ((1-\alpha_{\ne 4,7}) g_7 - \nu (g_2 - g_7)}{g_4 +g_7}, \\
\alpha_7 &= \frac{ (1-\alpha_{\ne 4,7}) g_4 - \nu (g_5 - g_2)}{g_4 +g_7}.
\end{align*}

Taking the difference of the eigenvector equations for $f_2$ and $f_5$, and for $g_2$ and $g_5$, we have

$$\alpha_7 f_7 -\alpha_4 f_4 = \mu(f_2 - f_5), \qquad \alpha_7 g_7 -\alpha_4 g_4 = \nu(g_2 - g_5).$$

Combining these equalities with the equation $\alpha_4 + \alpha_7 = 1-\alpha_{\ne 4,7}$ completes the proof.

§.§ Algorithm

In this subsection, we briefly detail how the computer-assisted proof of Lemma <ref> works. This proof is via interval arithmetic, and, at a high level, consists largely of iteratively decomposing the domain of feasible choices of $(\alpha_3,\alpha_6,\mu,\nu)$ for a given $S$ into smaller subregions (boxes) until all subregions violate some required equality or inequality. We provide two similar, but slightly different computer assisted proofs of this fact, and both of which can be found at the spread_numeric GitHub repository <cit.>. The first, found in folder interval$1$, is a shorter and simpler version, containing slightly fewer formulas, albeit at the cost of overall computation and run time. The second, found in the folder interval$2$, contains slightly more formulas and makes a greater attempt to optimize computation and run time. Below, we further detail the exact output and run time of both versions (exact output can be found in <cit.>), but for now, we focus on the main aspects of both proofs, and consider both together, saving a more detailed discussion of the differences for later.

These algorithms are implemented in Python using the PyInterval package. The algorithms consists of two parts: a main file containing useful formulas and subroutines and $17$ different files used to rule out each of the $17$ cases for $S$. The main file, casework_helper, contains functions with the formulas of Appendix Subsection <ref> (suitably modified to limit error growth), and functions used to check that certain equalities and inequalities are satisfied. In particular,  casework_helper contains formulas for

* $\alpha_2$, assuming $\{2,3,4\} \subset S$ (using Proposition <ref>)

* $\alpha_4$, assuming $\{1,2,3,4\} \subset S$ (using Proposition <ref>)

* $\alpha_4$, assuming $\{2,3,4,7\} \subset S$, $1 \not \in S$ (using Proposition <ref>)

* $\alpha_4$, assuming $\{1,2,4\}\subset S$, $3 \not \in S$ (using Proposition <ref>)

* $f_3$ and $g_3$, assuming $\{2,3\}\subset S$ (using Proposition <ref>)

* $f_2$ and $g_2$, assuming $\{2,3\} \subset S$ (using Proposition <ref>)

* $f_4$ and $g_4$, assuming $\{2,3,4\} \subset S$ (using Proposition <ref>)

* $f_1$ and $g_1$, assuming $\{1,2,4\} \subset S$ (using Propositions <ref> and <ref>)

* $f_2$ and $g_2$, assuming $\{2,4\} \subset S$, $3 \not\in S$ (using Proposition <ref>)

* $f_4$ and $g_4$, assuming $\{2,4\} \subset S$, $3 \not\in S$ (using Proposition <ref>)

as a function of $\alpha_3$, $\mu$, and $\nu$ (and $\alpha_2$ and $\alpha_4$, which can be computed as functions of $\alpha_3$, $\mu$, and $\nu$). Some of the formulas are slightly modified compared to their counterparts in this Appendix, for the purpose of minimizing accumulated error. Each formula is performed using interval arithmetic, while restricting the resulting interval solution to the correct range. In addition, we recall that we have the inequalities

* $\alpha_i \in [0,1]$, for $i \in S$

* $|g_2|,|f_3|\le 1$, $|f_2|,|g_3| \ge 1$, for $\{2,3\}\subset S$

* $|f_4|\le 1$, $|g_4|\ge 1$, for $4 \in S$

* $|f_1|\ge 1$, $|g_1|\le 1$, for $\{1,2,4\}\in S$

* $|f_4|,|g_2| \le 1$, $|f_2|,|g_4|\ge 1$, for $\{2,4\} \in S$, $3 \not \in S$

* $\alpha_3 + 2 \nu \le 0$, for $\{2,3\} \in S$ (using Proposition <ref>)

* $\alpha_2 - 2 \mu \le 0$, for $\{2,4\} \in S$, $3 \not \in S$ (using Proposition <ref>).

These inequalities are also used at various points in the algorithms. This completes a brief overview of the casework_helper file. Next, we consider the different files used to test feasibility for a specific choice of $S \subset \{1,...,7\}$, each denoted by case$\{\text{elements of S}\}$, i.e., for $S = \{1,4,5,7\}$, the associated file is case$1457$. For each specific case, there are a number of different properties which can be checked, including eigenvector equations, bounds on edge density, norm equations for the eigenvectors, and the ellipse equations. Each of these properties has an associated function which returns FALSE, if the property cannot be satisfied, given the intervals for each variable, and returns TRUE otherwise. The implementation of each of these properties is rather intuitive, and we refer the reader to the programs themselves (which contain comments) for exact details <cit.>. Each feasibility file consists of two parts. The first part is a function is_feasible(mu,nu,a3,a6) that, given bounding intervals for $\mu$, $\nu$, $\alpha_3$, $\alpha_6$, computes intervals for all other variables (using interval arithmetic) and checks feasibility using the functions in the casework_helper file. If any checked equation or inequality in the file is proven to be unsatisfiable (i.e., see Example <ref>), then this function outputs `FALSE', otherwise the function outputs `TRUE' by default. The second part is a divide and conquer algorithm that breaks the hypercube

$$ (\mu, \nu, \alpha_3,\alpha_6) \in [.65,1] \times [-.5,-.15] \times [0,1] \times [0,1]$$

into sub-boxes of size $1/20$ by $1/20$ by $1/10$ by $1/10$, checks feasibility in each box using is_feasible, and subdivides any box that does not rule out feasibility (i.e., subdivides any box that returns `TRUE'). This subdivision breaks a single box into two boxes of equal size, by subdividing along one of the four variables. The variable used for this subdivision is chosen iteratively, in the order $\alpha_3,\alpha_6,\mu, \nu, \alpha_3,...$. The entire divide and conquer algorithm terminates after all sub-boxes, and therefore, the entire domain

$$ (\mu, \nu, \alpha_3,\alpha_6) \in [.65,1] \times [-.5,-.15] \times [0,1] \times [0,1],$$

has been shown to be infeasible, at which point the algorithm prints `infeasible'. Alternatively, if the number of subdivisions reaches some threshold, then the algorithm terminates and outputs `feasible'.

Next, we briefly detail the output of the algorithms casework_helper/intervals$1$ and casework_helper/intervals$2$. Both algorithms ruled out 15 of the 17 choices for $S$ using a maximum depth of 26, and failed to rule out cases $S = \{4,5,7\}$ and $S = \{1,7\}$ up to depth 51. For the remaining 15 cases, intervals$1$ considered a total of 5.5 million boxes, was run serially on a personal computer, and terminated in slightly over twelve hours. For these same 15 cases, intervals$2$ considered a total of 1.3 million boxes, was run in parallel using the Penn State math department's `mathcalc' computer, and terminated in under 140 minutes. The exact output for both versions of the spread_numeric algorithm can be found at <cit.>.