ODIn/paper-BagRelationalPDBsAreHard
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Changed the way inequalities were referenced in SampMon Runtime Analysis.

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Aaron Huber 2021-03-15 11:39:39 -04:00
parent 66f792559d
commit 4a6b4db236
1 changed files with 16 additions and 14 deletions
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hardness-app.tex
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@ -702,19 +702,19 @@ If \circuit.\type $= \circplus$, then $\degree(\circuit) = \max\left(\degree(\ci
If \circuit.\type $= \circmult$, then,
substituing values, the following should hold,
\begin{align}
&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}\\
&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}\\
&\qquad\geq 2\degree(\circuit_\linput) \cdot \depth(\circuit_\linput) + 2 \degree(\circuit_\rinput) \cdot \depth(\circuit_\rinput) + 3\label{eq:times-middle} \\
&\qquad\geq 1 + \cost(\circuit_\linput) + \cost(\circuit_\rinput) = \cost(\circuit) (\ref{eq:cost-sampmon})\label{eq:times-rhs}.
\end{align}
To prove the inequality $\Cref{eq:times-lhs} \geq \Cref{eq:times-middle}$, first, let us expand \Cref{eq:times-lhs},
\begin{align}
(\ref{eq:times-lhs}) &= 2\degree(\circuit_\linput)\depth_{\max} + 2\degree(\circuit_\rinput)\depth_{\max} + 2\degree(\circuit_\linput) +\nonumber \\
& 2\degree(\circuit_\rinput) + 1\label{eq:times-lhs-expanded}
\end{align}
To prove (\ref{eq:times-middle}), first, the LHS expands to, %\Cref{eq:times-lhs},
\begin{equation}
%(\ref{eq:times-lhs})
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}
\end{equation}
where $\depth_{\max}$ is used to denote the maximum depth of the two input subcircuits.
Let us simplify the inequality $(\ref{eq:times-lhs}) \geq (\ref{eq:times-middle})$.
Let us now simplify the inequality (\ref{eq:times-middle}).
\begin{align}
&2\degree(\circuit_\linput)\depth_{\max} + 2\degree(\circuit_\rinput)\depth_{\max} + 2\degree(\circuit_\linput) + 2\degree(\circuit_\rinput) + 1 \nonumber\\
&\qquad \geq 2\degree(\circuit_\linput) \cdot \depth(\circuit_\linput) + 2 \degree(\circuit_\rinput) \cdot \depth(\circuit_\rinput) + 3\nonumber\\
@ -722,31 +722,33 @@ Let us simplify the inequality $(\ref{eq:times-lhs}) \geq (\ref{eq:times-middle}
\end{align}
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.
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.
Now to justify (\ref{eq:times-rhs}) which holds for the following reasons. First, the RHS%\Cref{eq:times-rhs}
is the result of \Cref{eq:cost-sampmon} when $\circuit.\type = \circmult$. The LHS %\Cref{eq:times-middle}
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.
For the case when \circuit.\type $= \circplus$, substituting values yields
\begin{align}
&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}\\
&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}\\
&\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}\\
&\qquad \geq 1 + \max(\cost(\circuit_\linput), \cost(\circuit_\rinput)) = \cost(\circuit)\label{eq:plus-rhs}
\end{align}
To prove that $(\ref{eq:plus-lhs-inequality}) \geq (\ref{eq:plus-middle})$, we can rewrite (\ref{eq:plus-lhs-inequality}) as
To prove (\ref{eq:plus-middle}), we can rewrite the LHS as %(\ref{eq:plus-lhs-inequality}) as
\begin{equation}
2\degree_{\max}\depth_{\max} + 2\degree_{\max} + 1.\label{eq:plus-lhs-expanded}
\end{equation}
Since $\degree_{\max} \cdot \depth_{\max} \geq \degree(\circuit_i)\cdot \depth(\circuit_i),$ the following upper bound holds for \Cref{eq:plus-middle}:
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}):
\begin{equation}
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}
\end{equation}
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}):
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}):
\begin{align}
&2\degree_{\max}\depth_{\max} + 2\degree_{\max} + 1 \geq 2\degree_{\max}\depth_{\max} + 2\nonumber\\
&\implies 2\degree_{\max} + 1 \geq 2\label{eq:plus-upper-bound-final}.
\end{align}
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}).
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}).
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 $(\ref{eq:ih-bound-cost})$.
Similar to the case of $\circuit.\type = \circmult$, (\ref{eq:plus-rhs}) follows by equations $(\ref{eq:cost-sampmon})$ and $(\ref{eq:ih-bound-cost})$.
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.