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Linear combination for G_3 added.

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Aaron Huber 2020-07-13 19:16:19 -04:00
parent 481915528c
commit 0f88083d62
2 changed files with 34 additions and 21 deletions
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3
macros.tex
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@ -30,12 +30,13 @@
\newcommand{\wbit}{w}
\newcommand{\expct}{\mathop{\mathbb{E}}}
\newcommand{\pbox}[1]{\left[#1\right]}
\newcommand{\pbrace}[1]{\left(#1\right)}
\newcommand{\vct}[1]{\textbf{#1}}
%using \wVec for world bit vector notation
\newcommand{\poly}{Q}
\newcommand{\rpoly}{\widetilde{Q}}%r for reduced as in reduced 'Q'
\newcommand{\out}{output}%output aggregation over the output vector
\newcommand{\numocc}[1]{\#\left(G, #1\right)}
\newcommand{\numocc}[2]{\#\left(#1, #2\right)}
%Graph Symbols
\newcommand{\ed}{|}
\newcommand{\twodis}{\|}

52
poly-form.tex
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@ -105,13 +105,13 @@ If we can compute $\poly(\wElem_1,\ldots, \wElem_\numTup) = q_E(\wElem_1,\ldots,
\begin{Lemma}\label{lem:qE3-exp}
When we expand $\poly(\wElem_1,\ldots, \wElem_N) = q_E(\wElem_1,\ldots, \wElem_\numTup)^3$ out and assign all exponents $e \geq 1$ a value of $1$, we have the following,
\begin{align}
&\rpoly(\prob,\ldots, \prob) = \numocc{\ed}\prob^2 + 6\numocc{\twopath}\prob^3 + 6\numocc{\twodis} + 6\numocc{\tri}\prob^3 +\nonumber\\
&\qquad\qquad6\numocc{\oneint}\prob^4 + 6\numocc{\threepath}\prob^4 + 6\numocc{\twopathdis}\prob^5 + 6\numocc{\threedis}\prob^6.\label{claim:four-one}
&\rpoly(\prob,\ldots, \prob) = \numocc{G}{\ed}\prob^2 + 6\numocc{G}{\twopath}\prob^3 + 6\numocc{G}{\twodis} + 6\numocc{G}{\tri}\prob^3 +\nonumber\\
&\qquad\qquad6\numocc{G}{\oneint}\prob^4 + 6\numocc{G}{\threepath}\prob^4 + 6\numocc{G}{\twopathdis}\prob^5 + 6\numocc{G}{\threedis}\prob^6.\label{claim:four-one}
\end{align}
\end{Lemma}
\AH{The warm-up below is fine for now, but will need to be removed for the final draft}
First, let us do a warm-up by computing $\rpoly(\wElem_1,\dots, \wElem_\numTup)$ when $\poly = q_E(\wElem_1,\ldots, \wElem_\numTup)$. Before doing so, we introduce a notation. Let $\numocc{H}$ denote the number of occurrences that $H$ occurs in $G$. So, e.g., $\numocc{\ed}$ is the number of edges ($m$) in $G$.
First, let us do a warm-up by computing $\rpoly(\wElem_1,\dots, \wElem_\numTup)$ when $\poly = q_E(\wElem_1,\ldots, \wElem_\numTup)$. Before doing so, we introduce a notation. Let $\numocc{G}{H}$ denote the number of occurrences that $H$ occurs in $G$. So, e.g., $\numocc{G}{\ed}$ is the number of edges ($m$) in $G$.
\AH{We need to make a decision on subgraph notation, and number of occurrences notation. Waiting to hear back from Oliver before making a decision.}
@ -125,7 +125,7 @@ First, let us do a warm-up by computing $\rpoly(\wElem_1,\dots, \wElem_\numTup)$
We can compute $\rpoly(\prob,\ldots, \prob)^2$ in O(m) time.
\end{Claim}
\begin{proof}
The proof basically follows by definition. When we expand $\poly^2$, and make all exponents $e = 1$, substituting $\prob$ for all $\wElem_i$ we get $\rpoly_2(\prob,\ldots, \prob) = \numocc{\ed} \cdot \prob^2 + 2\cdot \numocc{\twopath}\cdot \prob^3 + 2\cdot \numocc{\twodis}\cdot \prob^4$.
The proof basically follows by definition. When we expand $\poly^2$, and make all exponents $e = 1$, substituting $\prob$ for all $\wElem_i$ we get $\rpoly_2(\prob,\ldots, \prob) = \numocc{G}{\ed} \cdot \prob^2 + 2\cdot \numocc{G}{\twopath}\cdot \prob^3 + 2\cdot \numocc{G}{\twodis}\cdot \prob^4$.
\begin{enumerate}
\item First note that
\begin{align*}
@ -137,12 +137,12 @@ We can compute $\rpoly(\prob,\ldots, \prob)^2$ in O(m) time.
\begin{equation*}
\rpoly^2(\wVec) = \sum_{(i, j) \in E} \wElem_i\wElem_j + \sum_{\substack{(i, j), (j, \ell) \in E\\s.t. i \neq \ell}}\wElem_i\wElem_j\wElem_\ell + \sum_{\substack{(i, j), (k, \ell) \in E\\s.t. i \neq j \neq k \neq \ell}} \wElem_i\wElem_j\wElem_k\wElem_\ell\label{eq:part-1}
\end{equation*}
Notice that the first term is $\numocc{\ed}\cdot \prob^2$, the second $\numocc{\twopath}\cdot \prob^3$, and the third $\numocc{\twodis}\cdot \prob^4.$
Notice that the first term is $\numocc{G}{\ed}\cdot \prob^2$, the second $\numocc{G}{\twopath}\cdot \prob^3$, and the third $\numocc{G}{\twodis}\cdot \prob^4.$
\item Note that
\AH{We need the correct formula for two-matchings below.}
\begin{align*}
&\numocc{\ed} = m,\\
&\numocc{\twopath} = \sum_{u \in V} \binom{d_u}{2} \text{where $d_u$ is the degree of vertex $u$}\\ &\numocc{\twodis} = \textbf{\textit{a correct formula}}
&\numocc{G}{\ed} = m,\\
&\numocc{G}{\twopath} = \sum_{u \in V} \binom{d_u}{2} \text{where $d_u$ is the degree of vertex $u$}\\ &\numocc{G}{\twodis} = \textbf{\textit{a correct formula}}
\end{align*}
\end{enumerate}
Thus, since each of the summations can be computed in O(m) time, this implies that by \cref{eq:part-1} $\rpoly(\prob,\ldots, \prob)$ can be computed in O(m) time.\qed
@ -153,13 +153,13 @@ We are now ready to state the claim we need to prove \cref{lem:const-p} and \cre
Let $\poly(\wVec) = q_E(\wVec)^3$.
\begin{Claim}\label{claim:four-two}
If one can compute $\rpoly(\prob,\ldots, \prob)$ in time T(m), then we can compute the following in O(T(m) + m):
\[\numocc{\tri} + \numocc{\threepath} \cdot \prob - \numocc{\threedis}\cdot(\prob^2 - \prob^3).\]
\[\numocc{G}{\tri} + \numocc{G}{\threepath} \cdot \prob - \numocc{G}{\threedis}\cdot(\prob^2 - \prob^3).\]
\end{Claim}
\begin{proof}
We have either shown or will show that the following subgraph cardinalities can be computed in $O(m)$ time:
\[\numocc{\ed}, \numocc{\twopath}, \numocc{\twodis}, \numocc{\oneint}, \numocc{\twopathdis} + \numocc{\threedis}.\]
\[\numocc{G}{\ed}, \numocc{G}{\twopath}, \numocc{G}{\twodis}, \numocc{G}{\oneint}, \numocc{G}{\twopathdis} + \numocc{G}{\threedis}.\]
By definition we have that
\[\poly(\wElem_1,\ldots, \wElem_\numTup) = \sum_{\substack{(i_1, j_1),\\ (i_2, j_2),\\ (i_3, j_3) \in E}} \prod_{\ell = 1}^{3}\wElem_{i_\ell}\wElem_{j_\ell}.\]
@ -173,10 +173,10 @@ We have either shown or will show that the following subgraph cardinalities can
\AH{This proof I think could some reorganization. We really don't need the warm-up anymore, but we can use the formulas for case 1 and case 2.}
It has already been shown previously that $\numocc{\ed}, \numocc{\twopath}, \numocc{\twodis}$ can be computed in O(m) time. Here are the arguments for the rest.
\[\numocc{\oneint} = \sum_{u \in V} \binom{d_u}{3}\]
$\numocc{\twopathdis} + \numocc{\threedis} = $ the number of occurrences of three distinct edges with five or six vertices. This can be counted in the following manner. For every edge $(u, v) \in E$, throw away all neighbors of $u$ and $v$ and pick two more distinct edges.
\[\numocc{\twopathdis} + \numocc{\threedis} = \sum_{(u, v) \in E} \binom{m - d_u - d_v + 1}{2}\] The implication in \cref{claim:four-two} follows by the above and \cref{lem:qE3-exp}.
It has already been shown previously that $\numocc{G}{\ed}, \numocc{G}{\twopath}, \numocc{G}{\twodis}$ can be computed in O(m) time. Here are the arguments for the rest.
\[\numocc{G}{\oneint} = \sum_{u \in V} \binom{d_u}{3}\]
$\numocc{G}{\twopathdis} + \numocc{G}{\threedis} = $ the number of occurrences of three distinct edges with five or six vertices. This can be counted in the following manner. For every edge $(u, v) \in E$, throw away all neighbors of $u$ and $v$ and pick two more distinct edges.
\[\numocc{G}{\twopathdis} + \numocc{G}{\threedis} = \sum_{(u, v) \in E} \binom{m - d_u - d_v + 1}{2}\] The implication in \cref{claim:four-two} follows by the above and \cref{lem:qE3-exp}.
\AH{Justify the last sentence.}
\end{proof}
@ -188,7 +188,7 @@ $\numocc{\twopathdis} + \numocc{\threedis} = $ the number of occurrences of thre
%\AR{Also you can modify the text of \textsc{Proof} by using the following latex command \texttt{\\begin\{proof\}[Proof of Lemma 2]} and Latex will typeset this as \textsc{Proof of Lemma 2}, which is what you really want.}
\cref{claim:four-two} says that if we know $\rpoly_3(\prob,\ldots, \prob)$, then we can know in O(m) additional time
\[\numocc{\tri} + \numocc{\threepath} \cdot \prob - \numocc{\threedis}\cdot(\prob^2 - \prob^3).\] We can think of each term in the above equation as a variable, where one can solve a linear system given 3 distinct $\prob$ values, assuming independence of the three linear equations. In the worst case, without independence, 4 distince values of $\prob$ would suffice...because Atri said so, and I need to ask him for understanding why this is the case, of which I suspect that it has to do with basic result(s) in linear algebra.\AR{Follows from the fact that the corresponding coefficient matrix is the so called Vandermonde matrix, which has full rank.}
\[\numocc{G}{\tri} + \numocc{G}{\threepath} \cdot \prob - \numocc{G}{\threedis}\cdot(\prob^2 - \prob^3).\] We can think of each term in the above equation as a variable, where one can solve a linear system given 3 distinct $\prob$ values, assuming independence of the three linear equations. In the worst case, without independence, 4 distince values of $\prob$ would suffice...because Atri said so, and I need to ask him for understanding why this is the case, of which I suspect that it has to do with basic result(s) in linear algebra.\AR{Follows from the fact that the corresponding coefficient matrix is the so called Vandermonde matrix, which has full rank.}
\AH{This Vandermonde matrix I need to research.}
\end{proof}
@ -198,7 +198,7 @@ $\numocc{\twopathdis} + \numocc{\threedis} = $ the number of occurrences of thre
\begin{proof}[Proof of \cref{lem:const-p}]
The argument for \cref{lem:gen-p} cannot be applied to \cref{lem:const-p} since we have that $\prob$ is fixed. We have hope in the following: we assume that we can solve this problem for all graphs, and the hope would be be to solve the problem for say $G_1, G_2, G_3$, where $G_1$ is arbitrary, and relate the values of $\numocc{H}$, where $H$ is a placeholder for the relevant edge combination. The hope is that these relations would result in three independent linear equations, and then we would be done.
The argument for \cref{lem:gen-p} cannot be applied to \cref{lem:const-p} since we have that $\prob$ is fixed. We have hope in the following: we assume that we can solve this problem for all graphs, and the hope would be be to solve the problem for say $G_1, G_2, G_3$, where $G_1$ is arbitrary, and relate the values of $\numocc{G}{H}$, where $H$ is a placeholder for the relevant edge combination. The hope is that these relations would result in three independent linear equations, and then we would be done.
The following is an option.
\begin{enumerate}
@ -206,12 +206,12 @@ The following is an option.
\item Build $G_2$ from $G_1$, where each edge in $G_1$ gets replaced by a 2 path.
\end{enumerate}
Then $\numocc{\tri}_2 = 0$, and if we can prove that
Then $\numocc{G}{\tri}_2 = 0$, and if we can prove that
\begin{itemize}
\item $\numocc{\threepath}_2 = 2 \cdot \numocc{\twopath}_1$
\item $\numocc{\threedis}_2 = 8 \cdot \numocc{\threedis}_1 + 6 \cdot \numocc{\twopathdis}_1 + 4 \cdot \numocc{\oneint}_1 + 4 \cdot \numocc{\threepath}_1 + 2 \cdot \numocc{\tri}_1$
\item $\numocc{G}{\threepath}_2 = 2 \cdot \numocc{G}{\twopath}_1$
\item $\numocc{G}{\threedis}_2 = 8 \cdot \numocc{G}{\threedis}_1 + 6 \cdot \numocc{G}{\twopathdis}_1 + 4 \cdot \numocc{G}{\oneint}_1 + 4 \cdot \numocc{G}{\threepath}_1 + 2 \cdot \numocc{G}{\tri}_1$
\end{itemize}
we solve our problem for $q_E^3$ based on $G_2$ and we can compute $\numocc{\threedis}$, a hard problem.
we solve our problem for $q_E^3$ based on $G_2$ and we can compute $\numocc{G}{\threedis}$, a hard problem.
\end{proof}
\AH{Proving the above linear combination for 3-matchings in $G_2$ always holds for an arbitrary $G_1$.}
@ -248,7 +248,7 @@ The fact that there is a \textit{fixed} number of possible subgraphs that can be
Earlier, we claimed the following.
\[\numocc{\threedis}_2 = 8 \cdot \numocc{\threedis}_1 + 6 \cdot \numocc{\twopathdis}_1 + 4 \cdot \numocc{\oneint}_1 + 4 \cdot \numocc{\threepath}_1 + 2 \cdot \numocc{\tri}_1\]
\[\numocc{G}{\threedis}_2 = 8 \cdot \numocc{G}{\threedis}_1 + 6 \cdot \numocc{G}{\twopathdis}_1 + 4 \cdot \numocc{G}{\oneint}_1 + 4 \cdot \numocc{G}{\threepath}_1 + 2 \cdot \numocc{G}{\tri}_1\]
Beginning with the leftmost of RHS terms and proceeding to the consecutive rightmost terms, let us show this to be the case.
@ -268,3 +268,15 @@ For Tri, note that it is the case that the graph $G_2$ is a 'triangle of two pat
\end{proof}
\qed
In a similar way we can count the number of 3-matchings in graph $G_3$, where each edge in a given $G_1$ gets replaced with a disjoint 3-path, disjoint meaining that no other 3-path intersects another 3-path, except at its endpoints as in the original graph. The linear combination for 3-matchings in $G_3$ follows.
Because of $G_3$ construction, we now need to also account for two paths in $G_1$.
\begin{align*}
\numocc{G}{\threedis}_3 = &4\pbrace{\numocc{G_1}{\twopath}} + 6\pbrace{\numocc{G_1}{\twodis}} + 30\pbrace{\numocc{G_1}{\tri}} + 35\pbrace{\numocc{G_1}{\threepath}}\\
&+ 40\pbrace{\numocc{G_1}{\twopathdis}} + 32\pbrace{\numocc{G_1}{\oneint}} + 45\pbrace{\numocc{G_1}{\threedis}}
\end{align*}
\AH{Justification next.}