Changed the way inequalities were referenced in SampMon Runtime Analysis.
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@ -702,19 +702,19 @@ If \circuit.\type $= \circplus$, then $\degree(\circuit) = \max\left(\degree(\ci
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If \circuit.\type $= \circmult$, then,
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substituing values, the following should hold,
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\begin{align}
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&2\left(\degree(\circuit_\linput) + \degree(\circuit_\rinput)\right) \cdot \left(\max(\depth(\circuit_\linput), \depth(\circuit_\rinput)) + 1\right) + 1 \label{eq:times-lhs}\\
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&2\left(\degree(\circuit_\linput) + \degree(\circuit_\rinput)\right) \cdot \left(\max(\depth(\circuit_\linput), \depth(\circuit_\rinput)) + 1\right) + 1 \nonumber\\%\label{eq:times-lhs}\\
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&\qquad\geq 2\degree(\circuit_\linput) \cdot \depth(\circuit_\linput) + 2 \degree(\circuit_\rinput) \cdot \depth(\circuit_\rinput) + 3\label{eq:times-middle} \\
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&\qquad\geq 1 + \cost(\circuit_\linput) + \cost(\circuit_\rinput) = \cost(\circuit) (\ref{eq:cost-sampmon})\label{eq:times-rhs}.
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\end{align}
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To prove the inequality $\Cref{eq:times-lhs} \geq \Cref{eq:times-middle}$, first, let us expand \Cref{eq:times-lhs},
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\begin{align}
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(\ref{eq:times-lhs}) &= 2\degree(\circuit_\linput)\depth_{\max} + 2\degree(\circuit_\rinput)\depth_{\max} + 2\degree(\circuit_\linput) +\nonumber \\
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& 2\degree(\circuit_\rinput) + 1\label{eq:times-lhs-expanded}
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\end{align}
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To prove (\ref{eq:times-middle}), first, the LHS expands to, %\Cref{eq:times-lhs},
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\begin{equation}
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%(\ref{eq:times-lhs})
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2\degree(\circuit_\linput)\depth_{\max} + 2\degree(\circuit_\rinput)\depth_{\max} + 2\degree(\circuit_\linput) + 2\degree(\circuit_\rinput) + 1\label{eq:times-lhs-expanded}
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\end{equation}
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where $\depth_{\max}$ is used to denote the maximum depth of the two input subcircuits.
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Let us simplify the inequality $(\ref{eq:times-lhs}) \geq (\ref{eq:times-middle})$.
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Let us now simplify the inequality (\ref{eq:times-middle}).
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\begin{align}
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&2\degree(\circuit_\linput)\depth_{\max} + 2\degree(\circuit_\rinput)\depth_{\max} + 2\degree(\circuit_\linput) + 2\degree(\circuit_\rinput) + 1 \nonumber\\
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&\qquad \geq 2\degree(\circuit_\linput) \cdot \depth(\circuit_\linput) + 2 \degree(\circuit_\rinput) \cdot \depth(\circuit_\rinput) + 3\nonumber\\
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@ -722,31 +722,33 @@ Let us simplify the inequality $(\ref{eq:times-lhs}) \geq (\ref{eq:times-middle}
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\end{align}
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Note that by the \emph{reduced} invariant of \reduce, a circuit \circuit with $\depth(\circuit) \geq 1$ will always have at least one input with $\degree(\circuit_i) \geq 1$. Thus, \Cref{eq:times-lhs-middle-step1} follows, and the inequality is upheld.
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Now to justify the inequality (\ref{eq:times-middle}) $\geq$ (\ref{eq:times-rhs}) which holds for the following reasons. First, \Cref{eq:times-rhs} is the result of \Cref{eq:cost-sampmon} when $\circuit.\type = \circmult$. \Cref{eq:times-middle} is then produced by substituting the upperbound of (\ref{eq:ih-bound-cost}) for each $\cost(\circuit_i)$, trivially establishing an upper bound to (\ref{eq:times-rhs}). This proves \Cref{eq:strict-upper-bound} for the $\circmult$ case.
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Now to justify (\ref{eq:times-rhs}) which holds for the following reasons. First, the RHS%\Cref{eq:times-rhs}
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is the result of \Cref{eq:cost-sampmon} when $\circuit.\type = \circmult$. The LHS %\Cref{eq:times-middle}
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is then produced by substituting the upperbound of (\ref{eq:ih-bound-cost}) for each $\cost(\circuit_i)$, trivially establishing the upper bound of (\ref{eq:times-rhs}). This proves \Cref{eq:strict-upper-bound} for the $\circmult$ case.
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For the case when \circuit.\type $= \circplus$, substituting values yields
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\begin{align}
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&2\max(\degree(\circuit_\linput), \degree(\circuit_\rinput)) \cdot \left(\max(\depth(\circuit_\linput), \depth(\circuit_\rinput)) + 1\right) +1\label{eq:plus-lhs-inequality}\\
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&2\max(\degree(\circuit_\linput), \degree(\circuit_\rinput)) \cdot \left(\max(\depth(\circuit_\linput), \depth(\circuit_\rinput)) + 1\right) +1\nonumber\\%\label{eq:plus-lhs-inequality}\\
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&\qquad \geq \max\left(2\degree(\circuit_\linput) \cdot \depth(\circuit_\linput) + 1, 2\degree(\circuit_\rinput) \cdot \depth(\circuit_\rinput) +1\right) + 1\label{eq:plus-middle}\\
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&\qquad \geq 1 + \max(\cost(\circuit_\linput), \cost(\circuit_\rinput)) = \cost(\circuit)\label{eq:plus-rhs}
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\end{align}
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To prove that $(\ref{eq:plus-lhs-inequality}) \geq (\ref{eq:plus-middle})$, we can rewrite (\ref{eq:plus-lhs-inequality}) as
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To prove (\ref{eq:plus-middle}), we can rewrite the LHS as %(\ref{eq:plus-lhs-inequality}) as
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\begin{equation}
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2\degree_{\max}\depth_{\max} + 2\degree_{\max} + 1.\label{eq:plus-lhs-expanded}
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\end{equation}
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Since $\degree_{\max} \cdot \depth_{\max} \geq \degree(\circuit_i)\cdot \depth(\circuit_i),$ the following upper bound holds for \Cref{eq:plus-middle}:
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Since $\degree_{\max} \cdot \depth_{\max} \geq \degree(\circuit_i)\cdot \depth(\circuit_i),$ the following upper bound holds for the RHS of (\ref{eq:plus-middle}):
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\begin{equation}
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2\degree_{\max}\depth_{\max} + 2 \geq \max\left(2\degree(\circuit_\linput) \cdot \depth(\circuit_\linput) + 1, 2\degree(\circuit_\rinput) \cdot \depth(\circuit_\rinput) +1\right) + 1.\label{eq:plus-middle-expanded}
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\end{equation}
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Substituting the upperbound of (\ref{eq:plus-middle-expanded}) in for (\ref{eq:plus-middle}) we obtain the following for the inequality (\ref{eq:plus-lhs-inequality}) $\geq$ (\ref{eq:plus-middle}):
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Substituting the upperbound (LHS) of (\ref{eq:plus-middle-expanded}) in for the RHS of (\ref{eq:plus-middle}) we obtain the following for (\ref{eq:plus-middle}):
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\begin{align}
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&2\degree_{\max}\depth_{\max} + 2\degree_{\max} + 1 \geq 2\degree_{\max}\depth_{\max} + 2\nonumber\\
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&\implies 2\degree_{\max} + 1 \geq 2\label{eq:plus-upper-bound-final}.
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\end{align}
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As in the $\circmult$ case the \emph{reduced} invariant of \reduce implies that $\degree_{\max} \geq 1$, and (\ref{eq:plus-upper-bound-final}) follows. This proves (\ref{eq:plus-lhs-inequality}) $\geq$ (\ref{eq:plus-middle}).
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As in the $\circmult$ case the \emph{reduced} invariant of \reduce implies that $\degree_{\max} \geq 1$, and (\ref{eq:plus-upper-bound-final}) follows. This proves (\ref{eq:plus-middle}).
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Similar to the case of $\circuit.\type = \circmult$, the inequality $(\ref{eq:plus-middle}) \geq (\ref{eq:plus-rhs})$ follows by equations $(\ref{eq:cost-sampmon})$ and $(\ref{eq:ih-bound-cost})$.
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Similar to the case of $\circuit.\type = \circmult$, (\ref{eq:plus-rhs}) follows by equations $(\ref{eq:cost-sampmon})$ and $(\ref{eq:ih-bound-cost})$.
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This proves (\ref{eq:strict-upper-bound}) for the $\circplus$ case and thus the claimed $O(k\log{k}\cdot\depth(\circuit))$ runtime for $k = \degree(\circuit)$ follows.
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