Minor fixes to proof of lemma 3 (fixed p).
This commit is contained in:
parent
b3c492c97b
commit
341adb0d68
|
|
@ -194,33 +194,16 @@ The implication in \cref{claim:four-two} follows by the above and \cref{lem:qE3-
|
|||
|
||||
\qed
|
||||
|
||||
\AH{Below is only a transcription of the notes. The claims need to be verified and further worked out.}
|
||||
|
||||
\AH{{\Huge\bf Changes start here.}}
|
||||
\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 $\graph{1}, \graph{2}, \graph{3}$, where $\graph{1}$ 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.
|
||||
Here is the outline of the proof.
|
||||
|
||||
The following is an option.
|
||||
\begin{enumerate}
|
||||
\item Let $\graph{1}$ be an arbitrary graph
|
||||
\item Build $\graph{2}$ from $\graph{1}$, where each edge in $\graph{1}$ gets replaced by a 2 path.
|
||||
\end{enumerate}
|
||||
Let us show that for arbitrary graph $G$, which we refer to as $\graph{1}$, that we can construct graphs $\graph{2}$ and $\graph{3}$, which we will define later.
|
||||
|
||||
Then $\numocc{\graph{2}}{\tri} = 0$, and if we can prove that
|
||||
\begin{itemize}
|
||||
\item $\numocc{\graph{2}}{\threepath} = 2 \cdot \numocc{\graph{1}}{\twopath}$
|
||||
\item $\numocc{\graph{2}}{\threedis} = 8 \cdot \numocc{\graph{1}}{\threedis} + 6 \cdot \numocc{\graph{1}}{\twopathdis} + 4 \cdot \numocc{\graph{1}}{\oneint} + 4 \cdot \numocc{\graph{1}}{\threepath} + 2 \cdot \numocc{\graph{1}}{\tri}$
|
||||
\end{itemize}
|
||||
we solve our problem for $q_E^3$ based on $\graph{2}$ and we can compute $\numocc{G}{\threedis}$, a hard problem.
|
||||
\end{proof}
|
||||
We then show that for any subgraph $S$ for which $\numocc{G}{S}$ is known to be hard to compute, we can use linear combinations in terms of $\graph{1}$ to compute $\numocc{\graph{i}}{S}$, where $i$ in $\{2, 3\}$. Then, we can combine all three linear combinations into a linear system, solving for $\numocc{\graph{1}}{S}$.
|
||||
|
||||
\AH{Proving the above linear combination for 3-matchings in $\graph{2}$ always holds for an arbitrary $\graph{1}$.}
|
||||
|
||||
Consider graph $\graph{2}$, constructed from an arbitrary graph $\graph{1}$. We wish to show that the number of 3-matchings in $\graph{2}$ will always be the linear combination above, regardless of the construction of $\graph{1}$.
|
||||
|
||||
\AR{I did not make a pass on the above since it looks incomplete and does not seem to have changed for a while. Also it would be good to define $\graph{2}$ and $\graph{3}$ outside of the proofs below.}
|
||||
|
||||
\AH{{\Huge\bf Changes start here. 07-20-2020.}}
|
||||
First, let us introduce some useful definitions.
|
||||
|
||||
\subsubsection{$f_k$ and $\graph{k}$}
|
||||
|
||||
|
|
@ -229,9 +212,9 @@ Consider graph $\graph{2}$, constructed from an arbitrary graph $\graph{1}$. We
|
|||
For $k > 1$, let graph $\graph{k}$ be a graph generated from an arbitrary graph $\graph{1}$, by replacing every edge $e$ of $\graph{1}$ with a $k$-path, such that all $k$-path replacement edges are disjoint in the sense that they only intersect at the original intersection endpoints as seen in $\graph{1}$.
|
||||
\end{Definition}
|
||||
|
||||
For any graph $\graph{k}$, we denote its edges to be a pair $(e, b)$, such that $b \in \{0,\ldots, k-1\}$ and $e\in E_1$.
|
||||
The set of edges in $\graph{k}$ is written as $E_k$. For any graph $\graph{k}$, we denote its edges to be a pair $(e, b)$, such that $b \in \{0,\ldots, k-1\}$ and $e\in E_1$.
|
||||
|
||||
Before defining $f_k$, the following notation will be useful. Let $\binom{S}{t}$ denote the set of subsets in $S$ with exactly $t$ edges. In a similar manner, $\binom{S}{\leq t}$ is used to mean the subsets of $S$ with size $\leq t$. The set of edges in $\graph{k}$ is written as $E_k$.
|
||||
Before defining $f_k$, the following notation will be useful. Let $\binom{S}{t}$ denote the set of subsets in $S$ with exactly $t$ edges. In a similar manner, $\binom{S}{\leq t}$ is used to mean the subsets of $S$ with size $\leq t$.
|
||||
\begin{Definition}\label{def:fk}
|
||||
Define $f_k: \binom{E_k}{3} \mapsto \binom{E_1}{\leq3}$. For $S \in \binom{E_3}{3}$, such that $S = \pbrace{(e_1, b_1), (e_2, b_2), (e_3, b_3)}$, $f_k\left(\pbrace{(e_1, b_1), (e_2, b_2), (e_3, b_3)}\right) = \pbrace{e_1, e_2, e_3}$.
|
||||
\end{Definition}
|
||||
|
|
@ -266,7 +249,7 @@ The number of $3$-matchings in graph $\graph{2}$ satisfies the following identit
|
|||
\begin{proof}
|
||||
\AR{TODO for {\em later}: I think the proof will be much easier to follow with figures: just drawing out $S\times \{0,1\}$ along with the $(e_i,b_i)$ explicity notated on the edges will make the proof much easier to follow.}
|
||||
|
||||
Given any $S \in \binom{E_1}{\leq3}$, we consider $f_2^{-1}(S)$, which is the set of all possible sets of edges in $S \times \{0, 1\}$ which $f_2$ maps to $S$. Then we count the number of $3$-matchings in the $3$-edge subgraphs of $\graph{2}$ in $f_2^{-1}(S)$. We start with $S \in \binom{E_1}{3}$, where $S$ is composed of the edges $e_1, e_2, e_3$ and $f_2^{-1}(S)$ is set of all $3$-edge subsets of the set $\{(e_1, 0), (e_1, 1), (e_2, 0), (e_2, 1), (e_3, 0), (e_3, 1)\}$.
|
||||
Given any $S \in \binom{E_1}{\leq3}$, we consider $f_2^{-1}(S)$, which is the set of all possible sets of $3$-edge subgraphs in $S \times \{0, 1\}$ which $f_2$ maps to $S$. Then we count the number of $3$-matchings in the $3$-edge subgraphs of $\graph{2}$ in $f_2^{-1}(S)$. We start with $S \in \binom{E_1}{3}$, where $S$ is composed of the edges $e_1, e_2, e_3$ and $f_2^{-1}(S)$ is set of all $3$-edge subsets of the set $\{(e_1, 0), (e_1, 1), (e_2, 0), (e_2, 1), (e_3, 0), (e_3, 1)\}$.
|
||||
|
||||
\begin{itemize}
|
||||
\item $3$-matching ($\threedis$)
|
||||
|
|
@ -278,7 +261,7 @@ Consider the $S = \threedis$ pattern. Note that edges in $f_2^{-1}(S)$ are {\em
|
|||
\begin{itemize}
|
||||
\item Disjoint Two-Path ($\twopathdis$)
|
||||
\end{itemize}
|
||||
For $S = \twopathdis$ edges $e_2, e_3$ form a $2$-path with $e_3$ being disjoint. This means that $(e_2, 0), (e_2, 1), (e_3, 0), (e_3, 1)$ form a $4$-path while $(e_1, 0), (e_1, 1)$ is its own disjoint $2$-path. We can only pick $(e_1, 0)$ or $(e_1, 1)$ from $f_2^{-1}$, and then we need to pick a $2$-matching from the mapping of the $e_1, e_2$ under $f_2^{-1}.$ Note that a four path allows there to be 3 possible 2 matchings, specifically, $\pbrace{(e_2, 0), (e_3, 0)}, \pbrace{(e_2, 0), (e_3, 1)}, \pbrace{(e_2, 1), (e_3, 1)}$. Since these two selections can be made independently, there are $2 \cdot 3 = 6$ choices for $3$-matchings in $f_2^{-1}$.
|
||||
For $S = \twopathdis$ edges $e_2, e_3$ form a $2$-path with $e_3$ being disjoint. This means that $(e_2, 0), (e_2, 1), (e_3, 0), (e_3, 1)$ form a $4$-path while $(e_1, 0), (e_1, 1)$ is its own disjoint $2$-path. We can only pick $(e_1, 0)$ or $(e_1, 1)$ from $f_2^{-1}(S)$, and then we need to pick a $2$-matching from the mapping of the $e_1, e_2$ under $f_2^{-1}(S).$ Note that a four path allows there to be 3 possible 2 matchings, specifically, $\pbrace{(e_2, 0), (e_3, 0)}, \pbrace{(e_2, 0), (e_3, 1)}, \pbrace{(e_2, 1), (e_3, 1)}$. Since these two selections can be made independently, there are $2 \cdot 3 = 6$ choices for $3$-matchings in $f_2^{-1}(S)$.
|
||||
|
||||
\begin{itemize}
|
||||
\item $3$-star ($\oneint$)
|
||||
|
|
@ -288,12 +271,12 @@ When $S = \oneint$, in $f_2^{-1}$, the inner edges $(e_i, 1)$ are all connected,
|
|||
\begin{itemize}
|
||||
\item $3$-path ($\threepath$)
|
||||
\end{itemize}
|
||||
When $S =\threepath$ it is the case that all edges beginning with $e_1$ and ending with $e_3$ are successively connected. This translates to a $6$-path in the edges of $f_2^{-1}$, where all edges from $(e_0, 0),(e_0,1),(e_2,0),(e_2, 1)$ are successively connected. For a $3$-matching to exist, there must be at least one edge separating edges picked from a sequence. There are four such possibilities: $\pbrace{(e_1, 0), (e_2, 0), (e_3, 0)}, \pbrace{(e_1, 0), (e_2, 0), (e_3, 1)}, \pbrace{(e_1, 0), (e_2, 1), (e_3, 1)},$\newline $\pbrace{(e_1, 1), (e_2, 1), (e_3, 1)}$ . Thus, there are four possible 3-matchings in $f_2^{-1}(S)$.
|
||||
When $S =\threepath$ it is the case that all edges beginning with $e_1$ and ending with $e_3$ are successively connected. This translates to a $6$-path in the edges of $f_2^{-1}(S)$, where all edges from $(e_1, 0),\ldots,(e_3, 1)$ are successively connected. For a $3$-matching to exist, there must be at least one edge separating edges picked from a sequence. There are four such possibilities: $\pbrace{(e_1, 0), (e_2, 0), (e_3, 0)}, \pbrace{(e_1, 0), (e_2, 0), (e_3, 1)}, \pbrace{(e_1, 0), (e_2, 1), (e_3, 1)},$\newline $\pbrace{(e_1, 1), (e_2, 1), (e_3, 1)}$ . Thus, there are four possible 3-matchings in $f_2^{-1}(S)$.
|
||||
|
||||
\begin{itemize}
|
||||
\item Triangle ($\tri$)
|
||||
\end{itemize}
|
||||
For $S = \tri$, note that it is the case that the edges in $f_2^{-1}$ are connected in a successive manner, but this time in a cycle, such that $(e_1, 0)$ and $(e_3, 1)$ are also connected. While this is similar to the discussion of the six path above, one must use caution not to consider the first and last edges as disjoint, since they are connected. This rules out both $(e_1, 0), (e_2, 0), (e_3, 1)$ and $(e_1, 0), (e_2, 1), (e_3, 1)$ leaving us with two remaining edge combinations that produce a 3 matching.
|
||||
For $S = \tri$, note that it is the case that the edges in $f_2^{-1}(S)$ are connected in a successive manner, but this time in a cycle, such that $(e_1, 0)$ and $(e_3, 1)$ are also connected. While this is similar to the discussion of the six path above, one must use caution not to consider the first and last edges as disjoint, since they are connected. This rules out both subsets of $(e_1, 0), (e_2, 0), (e_3, 1)$ and $(e_1, 0), (e_2, 1), (e_3, 1)$ leaving us with $2$ remaining edge combinations that produce a 3 matching.
|
||||
|
||||
\begin{itemize}
|
||||
\item $2$-matching ($\twodis$), $2$-path ($\twopath$), $1$ edge ($\ed$)
|
||||
|
|
@ -328,7 +311,7 @@ When $S = \ed$, $f_3^{-1}(S)$ gives one result, $(e_1, 0), (e_1, 1), (e_1, 2)$,
|
|||
\begin{itemize}
|
||||
\item $2$-path ($\twopath$)
|
||||
\end{itemize}
|
||||
Fix then $S = \twopath$ and now we have $f_3^{-1}(S)$ yielding all $3$-edged subsets of $S'$. All edges in $S'$ form a $6$-path, and as stated above in \cref{lem:3m-G2}, this leads to $4$ $3$-matchings in $f_3^{-1}(S)$.
|
||||
Fix then $S = \twopath$ and now we have $f_3^{-1}(S)$ yielding all $3$-edged subsets of $S'$. All edges in $S'$ form a $6$-path, and as discussed in \cref{lem:3m-G2}, this leads to $4$ $3$-matchings in $f_3^{-1}(S)$.
|
||||
|
||||
\begin{itemize}
|
||||
\item $2$-matching ($\twodis$)
|
||||
|
|
@ -338,7 +321,7 @@ For $S = \twodis$, then all subsets in the output of $f_3^{-1}(S)$ are predicate
|
|||
\begin{itemize}
|
||||
\item Triangle ($\tri$)
|
||||
\end{itemize}
|
||||
Now, we consider the 3-edge subgraphs of $\graph{1}$, starting with a $S = \tri$. As discussed in \cref{lem:3m-G2} for the case of $\tri$, $f_3^{-1}(S)$ subsets are conditioned on the fact that all the edges in $S'$ are a cyclic sequence, and we must be careful not to pair $(e_1, 0)$ with $(e_3, 2)$ in a $3$-matching. For any $s \in f_3^{-1}(S)$, $s$ is a $3$-matching when we have that for the edges $(e_1, b_1), (e_2, b_2), (e_3, b_3)$ where $b_1, b_2, b_3 \in \{0, 1, 2\}$, such that, for all $i \in [3]$ it is the case that if $b_i = 2$ then $b_{i + 1 \mod{4}} = 0$. Iterating through all possible combinations producing 3-matchings, i.e. $\pbrace{(e_1, 0), (e_2, 0), (e_3, 0)}, \pbrace{(e_1, 0), (e_2, 0), (e_3, 1)}, \pbrace{(e_1, 0), (e_2, 1), (e_3, 0)},\pbrace{(e_1, 0), (e_2, 1), (e_3, 1)}, \pbrace{(e_1, 0), (e_2, 2), (e_3, 1)}, \pbrace{(e_1, 1), (e_2, 0), (e_3, 0)}$\newline$\ldots, \pbrace{(e_1, 1), (e_2, 1), (e_3, 2)} \pbrace{(e_1, 1), (e_2, 2), (e_3, 1)}, \pbrace{(e_1, 1), (e_2, 2), (e_3, 2)}, \pbrace{(e_1, 2), (e_2, 1), (e_3, 0)},\ldots, \pbrace{(e_1, 2), (e_2, 2), (e_3, 1)}, \pbrace{(e_1, 2), (e_2, 2), (e_3, 2)}$, giving a total of 18 3-matchings in $f_3^{-1}(S)$.
|
||||
Now, we consider the 3-edge subgraphs of $\graph{1}$, starting with $S = \tri$. As discussed in \cref{lem:3m-G2} for the case of $\tri$, $f_3^{-1}(S)$ subsets are conditioned on the fact that all the edges in $S'$ are a cyclic sequence, and we must be careful not to pair $(e_1, 0)$ with $(e_3, 2)$ in a $3$-matching. For any $s \in f_3^{-1}(S)$, $s$ is a $3$-matching when we have that for the edges $(e_1, b_1), (e_2, b_2), (e_3, b_3)$ where $b_1, b_2, b_3 \in \{0, 1, 2\}$, such that, for all $i \in [3]$ it is the case that if $b_i = 2$ then $b_{i + 1 \mod{4}} = 0$. Iterating through all possible combinations producing 3-matchings, i.e. $\pbrace{(e_1, 0), (e_2, 0), (e_3, 0)}, \pbrace{(e_1, 0), (e_2, 0), (e_3, 1)}, \pbrace{(e_1, 0), (e_2, 1), (e_3, 0)},\pbrace{(e_1, 0), (e_2, 1), (e_3, 1)}, \pbrace{(e_1, 0), (e_2, 2), (e_3, 1)}, \pbrace{(e_1, 1), (e_2, 0), (e_3, 0)}$\newline$\ldots, \pbrace{(e_1, 1), (e_2, 1), (e_3, 2)} \pbrace{(e_1, 1), (e_2, 2), (e_3, 1)}, \pbrace{(e_1, 1), (e_2, 2), (e_3, 2)}, \pbrace{(e_1, 2), (e_2, 1), (e_3, 0)},\ldots, \pbrace{(e_1, 2), (e_2, 2), (e_3, 1)}, \pbrace{(e_1, 2), (e_2, 2), (e_3, 2)}$, giving a total of 18 3-matchings in $f_3^{-1}(S)$.
|
||||
|
||||
\begin{itemize}
|
||||
\item $3$-path ($\threepath$)
|
||||
|
|
@ -348,12 +331,12 @@ Consider when $S = \threepath$ and $f_3^{-1}(S)$ has the constraint that all edg
|
|||
\begin{itemize}
|
||||
\item Disjoint Two-Path ($\twopathdis$)
|
||||
\end{itemize}
|
||||
Assume $S = \twopathdis$, then $f_3^{-1}$ has successive connectivity from $(e_1, 0)$ through $(e_1, 2)$, and successive connectivity from $(e_2, 0)$ through $(e_3, 2)$. It is the case that the edges in $S'$ form a 6-path with a disjoint 3-path. There exist $8$ distinct two matchings (with at least one $(e_2,\cdot)$ and at least one $(e_3,\cdot)$ edge) in the $6$-path $(e_2, 0),\ldots, (e_3, 2)$ of the form $\pbrace{(e_2, 0), (e_3, 0)},\ldots, \pbrace{(e_2, 1), (e_3, 2)}, \pbrace{(e_2, 2), (e_3, 1)}, \pbrace{(e_2, 2), (e_3, 2)}$. These matchings can be paired independently with either of the $3$ remaining edges of $(e_1, b)$, for a total of $8 \cdot 3 = 24$ 3-matchings in $f_3^{-1}(S)$.
|
||||
Assume $S = \twopathdis$, then $f_3^{-1}(S)$ has successive connectivity from $(e_1, 0)$ through $(e_1, 2)$, and successive connectivity from $(e_2, 0)$ through $(e_3, 2)$. It is the case that the edges in $S'$ form a 6-path with a disjoint 3-path. There exist $8$ distinct two matchings (with at least one $(e_2,\cdot)$ and at least one $(e_3,\cdot)$ edge) in the $6$-path $(e_2, 0),\ldots, (e_3, 2)$ of the form $\pbrace{(e_2, 0), (e_3, 0)},\ldots, \pbrace{(e_2, 1), (e_3, 2)}, \pbrace{(e_2, 2), (e_3, 1)}, \pbrace{(e_2, 2), (e_3, 2)}$. These matchings can be paired independently with either of the $3$ remaining edges of $(e_1, b)$, for a total of $8 \cdot 3 = 24$ 3-matchings in $f_3^{-1}(S)$.
|
||||
|
||||
\begin{itemize}
|
||||
\item $3$-star ($\oneint$)
|
||||
\end{itemize}
|
||||
Given $S = \oneint$, the subsets of $f_3^{-1}(S)$ are restricted such that the outer edges $(e_i, 0)$ are disjoint from another, the middle edges $(e_i, 1)$ are also disjoint to each other, and only the inner edges $(e_i, 2)$ intersect with one another at exactly one common endpoint. To be precise, any outer edge $(e_i, 0)$ is disjoint to every middle edge $(e_j, 1)$ for $i \neq j$. As discussed previously, at most one inner edge may appear in a $3$-matching. For arbitrary inner edge $(e_i, 2)$, we have $4$ combinations of the middle and outer edges of $e_j, e_k$, where $i \neq j \neq k$. These choices are independent and we have $4 \cdot 3 = 12$ 3-matchings. We are not done yet, as we need to consider the middle and outer edge combinations. Notice that for each $e_i$, we have $2$ choices, i.e. a middle or outer edge, contributing $2^3 = 8$ additional $3$-matchings, for a total of $8 + 12 = 20$ $3$-matchings in $f_3^{-1}(S)$.
|
||||
Given $S = \oneint$, the subsets of $f_3^{-1}(S)$ are restricted such that the outer edges $(e_i, 0)$ are disjoint from another, the middle edges $(e_i, 1)$ are also disjoint to each other, and only the inner edges $(e_i, 2)$ intersect with one another at exactly one common endpoint. To be precise, any outer edge $(e_i, 0)$ is disjoint to every middle edge $(e_j, 1)$ for $i \neq j$. As previously mentioned in \cref{lem:3m-G2}, at most one inner edge may appear in a $3$-matching. For arbitrary inner edge $(e_i, 2)$, we have $4$ combinations of the middle and outer edges of $e_j, e_k$, where $i \neq j \neq k$. These choices are independent and we have $4 \cdot 3 = 12$ 3-matchings. We are not done yet, as we need to consider the middle and outer edge combinations. Notice that for each $e_i$, we have $2$ choices, i.e. a middle or outer edge, contributing $2^3 = 8$ additional $3$-matchings, for a total of $8 + 12 = 20$ $3$-matchings in $f_3^{-1}(S)$.
|
||||
|
||||
\begin{itemize}
|
||||
\item $3$-matching ($\threedis$)
|
||||
|
|
@ -402,11 +385,11 @@ The number of triangles in $\graph{k}$ for $k \geq 2$ will always be $0$ for the
|
|||
\end{proof}
|
||||
\qed
|
||||
|
||||
\AH{\LARGE \bf New material starts here.}
|
||||
\AH{\LARGE \bf Newest material starts here.}
|
||||
\subsection{Developing a Linear System}
|
||||
In \cref{lem:qE3-exp} is the identity for $\rpoly(\prob,\ldots, \prob)$ when $\poly(\wElem_1,\ldots, \wElem_N) = q_E(\wElem_1,\ldots, \wElem_\numTup)^3$. (This lemma still needs a proof, but for now we will pretend the proof is there.)
|
||||
|
||||
All of \cref{eq:1e}, \cref{eq:2p}, \cref{eq:2m}, \cref{eq:3s}, \cref{eq:2pd-3d} show that we can compute their respective edge patterns in $O(m)$ time. Rearrange $\rpoly$ with all linear time computations on one side, leaving only the hard computations,
|
||||
All of \cref{eq:1e}, \cref{eq:2p}, \cref{eq:2m}, \cref{eq:3s}, \cref{eq:2pd-3d} show that we can compute the respective edge patterns in $O(m)$ time. Rearrange $\rpoly$ with all linear time computations on one side, leaving only the hard computations,
|
||||
\begin{align}
|
||||
&\rpoly(\prob,\ldots, \prob) = \numocc{G}{\ed}\prob^2 + 6\numocc{G}{\twopath}\prob^3 + 6\numocc{G}{\twodis}\prob^4 + 6\numocc{G}{\tri}\prob^3 + 6\numocc{G}{\oneint}\prob^4 + 6\numocc{G}{\threepath}\prob^4 + 6\numocc{G}{\twopathdis}\prob^5 + 6\numocc{G}{\threedis}\prob^6\nonumber\\
|
||||
&\rpoly(\prob,\ldots, \prob) - \numocc{G}{\ed}\prob^2 - 6\numocc{G}{\twopath}\prob^3 - 6\numocc{G}{\twodis}\prob^4 - 6\numocc{G}{\oneint}\prob^4 = 6\numocc{G}{\tri}\prob^3 + 6\numocc{G}{\threepath}\prob^4 + 6\numocc{G}{\twopathdis}\prob^5 + 6\numocc{G}{\threedis}\prob^6\label{eq:LS-rearrange}\\
|
||||
|
|
@ -414,7 +397,7 @@ All of \cref{eq:1e}, \cref{eq:2p}, \cref{eq:2m}, \cref{eq:3s}, \cref{eq:2pd-3d}
|
|||
&\frac{\rpoly(\prob,\ldots, \prob)}{6\prob^3} - \frac{\numocc{G}{\ed}}{6\prob} - \numocc{G}{\twopath} - \numocc{G}{\twodis}\prob - \numocc{G}{\oneint}\prob - \big(\numocc{G}{\twopathdis} + \numocc{G}{\threedis}\big)\prob^2 = \numocc{G}{\tri} + \numocc{G}{\threepath}\prob - \numocc{G}{\threedis}\left(\prob^2 - \prob^3\right)\label{eq:LS-subtract}
|
||||
\end{align}
|
||||
|
||||
\cref{eq:LS-rearrange} is the result of simply subtracting from both sides terms that are not wanted on the RHS. Reducing all terms by the common factor of $6\prob^3$ gives \cref{eq:LS-reduce}. The final equation, \cref{eq:LS-subtract}, is the result of subtracting the term $\left(\numocc{G}{\twopathdis} + \numocc{G}{\threedis}\right)\prob^2$ from both sides. Equation \ref{eq:LS-subtract} holds for $\rpoly$ over any graph $G$
|
||||
\cref{eq:LS-rearrange} is the result of simply subtracting from both sides terms that have $O(m)$ complexity. Reducing all terms by the common factor of $6\prob^3$ gives \cref{eq:LS-reduce}. The final equation, \cref{eq:LS-subtract}, is the result of subtracting the term $\left(\numocc{G}{\twopathdis} + \numocc{G}{\threedis}\right)\prob^2$ from both sides. Equation \ref{eq:LS-subtract} holds for $\rpoly$ over any graph $G$
|
||||
|
||||
As previously outlined, assume graph $\graph{1}$ to be an arbitrary graph, with $\graph{2}, \graph{3}$ constructed from $\graph{1}$ as defined in \cref{def:Gk}.
|
||||
|
||||
|
|
@ -432,27 +415,27 @@ Replace the next term with the identity of \cref{lem:3p-G2} and the last term wi
|
|||
\end{equation*}
|
||||
Rearrange terms into groups of those patterns that can be computed in $O(m)$ and those that are 'hard' to compute,
|
||||
\begin{equation*}
|
||||
\linsys{2} = -\pbrace{2 \cdot \numocc{\graph{1}}{\tri} + 4 \cdot \numocc{\graph{1}}{\threepath} + \left(8 \cdot \numocc{\graph{1}}{\threedis} + 6 \cdot \numocc{\graph{1}}{\twopathdis}\right)}\left(\prob^2 - \prob^3\right) + \pbrace{2 \cdot \numocc{\graph{1}}{\twopath}\prob^2 - 4 \cdot \numocc{\graph{1}}{\oneint}\left(\prob^2 - \prob^3\right)}.
|
||||
\linsys{2} = -\pbrace{2 \cdot \numocc{\graph{1}}{\tri} + 4 \cdot \numocc{\graph{1}}{\threepath} + \left(8 \cdot \numocc{\graph{1}}{\threedis} + 6 \cdot \numocc{\graph{1}}{\twopathdis}\right)}\left(\prob^2 - \prob^3\right) + \pbrace{2 \cdot \numocc{\graph{1}}{\twopath}\prob - 4 \cdot \numocc{\graph{1}}{\oneint}\left(\prob^2 - \prob^3\right)}.
|
||||
\end{equation*}
|
||||
|
||||
Let $\lintime{2}$ represent all the terms we wish to remove from $\linsys{2}$. Then, the following are true.
|
||||
\begin{align}
|
||||
\lintime{2} &= -6\cdot\left(\numocc{\graph{1}}{\threedis} + \numocc{\graph{1}}{\twopathdis}\right)\left(\prob^2 - \prob^3\right) - 4 \cdot \numocc{\graph{1}}{\oneint}\left(\prob^2 - \prob^3\right) + 2\cdot \numocc{\graph{1}}{\twopath}\prob^2\nonumber\\
|
||||
\linsys{2'} = \linsys{2} -\lintime{2} &= -\pbrace{2 \cdot \numocc{\graph{1}}{\tri} + 4 \cdot \numocc{\graph{1}}{\threepath} + 2 \cdot \numocc{\graph{1}}{\threepath}}\left(\prob^2 - \prob^3\right)\label{eq:LS-G2'}\\
|
||||
\lintime{2} &= -6\cdot\left(\numocc{\graph{1}}{\threedis} + \numocc{\graph{1}}{\twopathdis}\right)\left(\prob^2 - \prob^3\right) - 4 \cdot \numocc{\graph{1}}{\oneint}\left(\prob^2 - \prob^3\right) + 2\cdot \numocc{\graph{1}}{\twopath}\prob\nonumber\\
|
||||
\linsys{2'} = \linsys{2} -\lintime{2} &= -\pbrace{2 \cdot \numocc{\graph{1}}{\tri} + 4 \cdot \numocc{\graph{1}}{\threepath} + 2 \cdot \numocc{\graph{1}}{\threedis}}\left(\prob^2 - \prob^3\right)\label{eq:LS-G2'}\\
|
||||
\aug{2'} = \aug{2} - \lintime{2} &= \frac{\rpoly(\prob,\ldots, \prob)}{6\prob^3} - \frac{\numocc{\graph{2}}{\ed}}{6\prob} - \numocc{\graph{2}}{\twopath} - \numocc{\graph{2}}{\twodis}\prob - \numocc{\graph{2}}{\oneint}\prob - \big(\numocc{\graph{2}}{\twopathdis} + \numocc{\graph{2}}{\threedis}\big)\prob^2 +\nonumber \\
|
||||
&6\cdot\left(\numocc{\graph{1}}{\threedis} + \numocc{\graph{1}}{\twopathdis}\right)\left(\prob^2 - \prob^3\right) + 4 \cdot \numocc{\graph{1}}{\oneint}\left(\prob^2 - \prob^3\right) - 2\cdot \numocc{\graph{1}}{\twopath}\prob^2\nonumber\\
|
||||
&6\cdot\left(\numocc{\graph{1}}{\threedis} + \numocc{\graph{1}}{\twopathdis}\right)\left(\prob^2 - \prob^3\right) + 4 \cdot \numocc{\graph{1}}{\oneint}\left(\prob^2 - \prob^3\right) - 2\cdot \numocc{\graph{1}}{\twopath}\prob\nonumber\\
|
||||
\implies \aug{2'} &= \linsys{2'}\nonumber
|
||||
\end{align}
|
||||
|
||||
We now have a linear equation in terms of $\graph{1}$ for $\graph{2}$ where $\linsys{2'}$ is the linear combination and $\aug{2}$ is the augmented side of the matrix, or the constant value, since all of the terms in $\aug{2}$ can be solved in $O(m)$ time.
|
||||
We now have a linear equation in terms of $\graph{1}$ for $\graph{2}$ where $\linsys{2'}$ is the linear combination and $\aug{2}$ is the augmented side of the matrix, or the constant value, as all of the terms in $\aug{2}$ can be solved in $O(m)$ time.
|
||||
|
||||
\subsubsection{$\graph{3}$}
|
||||
|
||||
Following the same line of reasoning for $\graph{3}$, using \cref{lem:3m-G3}, \cref{lem:3p-G3}, and \cref{lem:tri}, substitute the identities into $\linsys{3}$,
|
||||
Following the same reasoning for $\graph{3}$, using \cref{lem:3m-G3}, \cref{lem:3p-G3}, and \cref{lem:tri}, substitute the identities into $\linsys{3}$,
|
||||
\begin{align}
|
||||
\linsys{3} =& \pbrace{\numocc{\graph{1}}{\ed} + 2 \cdot \numocc{\graph{1}}{\twopath}}\prob - \left\{4 \cdot \numocc{\graph{1}}{\twopath} + 6 \cdot \numocc{\graph{1}}{\twodis} + 18 \cdot \numocc{\graph{1}}{\tri} + 21 \cdot \numocc{\graph{1}}{\threepath} + 24 \cdot \numocc{\graph{1}}{\twopathdis} +\right.\nonumber\\
|
||||
&\left.20 \cdot \numocc{\graph{1}}{\oneint} + 27 \cdot \numocc{\graph{1}}{\threedis}\right\}\left(\prob^2 - \prob^3\right)\label{eq:LS-G3-sub}\\
|
||||
=&+ \pbrace{ -18\numocc{\graph{1}}{\tri} - 21 \cdot \numocc{\graph{1}}{\threepath} - 24 \cdot \numocc{\graph{1}}{\twopathdis} - 27 \cdot \numocc{\graph{1}}{\threedis}} + \left\{\pbrace{-20 \cdot \numocc{\graph{1}}{\oneint} - 4\cdot \numocc{\graph{1}}{\twopath} - 6 \cdot \numocc{\graph{1}}{\twodis}}\left(\prob^2 - \prob^3\right) +\right.\nonumber\\
|
||||
=&\pbrace{ -18\numocc{\graph{1}}{\tri} - 21 \cdot \numocc{\graph{1}}{\threepath} - 24 \cdot \numocc{\graph{1}}{\twopathdis} - 27 \cdot \numocc{\graph{1}}{\threedis}}\left(\prob^2 - \prob^3\right) + \left\{\pbrace{-20 \cdot \numocc{\graph{1}}{\oneint} - 4\cdot \numocc{\graph{1}}{\twopath} - 6 \cdot \numocc{\graph{1}}{\twodis}}\left(\prob^2 - \prob^3\right) +\right.\nonumber\\
|
||||
&\left.\numocc{\graph{1}}{\ed}\prob + 2 \cdot \numocc{\graph{1}}{\twopath}\prob\right\}. \label{eq:LS-G3-rearrange}
|
||||
\end{align}
|
||||
|
||||
|
|
@ -460,10 +443,10 @@ Equation \ref{eq:LS-G3-sub} follows from simple substitution of all lemma identi
|
|||
|
||||
Collecting terms we would like to send to the other side, the following equations are true,
|
||||
\begin{align}
|
||||
\lintime{3} =& \pbrace{- 24 \cdot \left(\numocc{\graph{1}}{\twopathdis} + \numocc{\graph{1}}{\threedis}\right) - 20 \cdot \numocc{\graph{1}}{\oneint} - 4\cdot \numocc{\graph{1}}{\twopath} - 6 \cdot \numocc{\graph{1}}{\twodis}}\left(\prob^2 - \prob^3\right) + \pbrace{\numocc{\graph{1}}{\ed}\prob + 2 \cdot \numocc{\graph{1}}{\twopath}}\prob\nonumber\\
|
||||
\lintime{3} =& \pbrace{- 24 \cdot \left(\numocc{\graph{1}}{\twopathdis} + \numocc{\graph{1}}{\threedis}\right) - 20 \cdot \numocc{\graph{1}}{\oneint} - 4\cdot \numocc{\graph{1}}{\twopath} - 6 \cdot \numocc{\graph{1}}{\twodis}}\left(\prob^2 - \prob^3\right) + \pbrace{\numocc{\graph{1}}{\ed} + 2 \cdot \numocc{\graph{1}}{\twopath}}\prob\nonumber\\
|
||||
\linsys{3'} =& \linsys{3} - \lintime{3} = \pbrace{- 18 \cdot \numocc{\graph{1}}{\tri} - 21 \cdot \numocc{\graph{1}}{\threepath} - 3 \cdot \numocc{\graph{1}}{\threedis}}\left(p^2 - p^3\right)\label{eq:LS-G3'} \\
|
||||
\aug{3'} =& \aug{3} - \lintime{3} = \frac{\rpoly(\prob,\ldots, \prob)}{6\prob^3} - \frac{\numocc{\graph{3}}{\ed}}{6\prob} - \numocc{\graph{3}}{\twopath} - \numocc{\graph{3}}{\twodis}\prob - \numocc{\graph{3}}{\oneint}\prob - \big(\numocc{\graph{3}}{\twopathdis} + \numocc{\graph{3}}{\threedis}\big)\prob^2 + \nonumber\\
|
||||
& \pbrace{24 \cdot \left(\numocc{\graph{1}}{\twopathdis} + \numocc{\graph{1}}{\threedis}\right) + 20 \cdot \numocc{\graph{1}}{\oneint} + 4\cdot \numocc{\graph{1}}{\twopath} + 6 \cdot \numocc{\graph{1}}{\twodis}}\left(\prob^2 - \prob^3\right) - \pbrace{\numocc{\graph{1}}{\ed}\prob + 2 \cdot \numocc{\graph{1}}{\twopath}}\prob\nonumber\\
|
||||
& \pbrace{24 \cdot \left(\numocc{\graph{1}}{\twopathdis} + \numocc{\graph{1}}{\threedis}\right) + 20 \cdot \numocc{\graph{1}}{\oneint} + 4\cdot \numocc{\graph{1}}{\twopath} + 6 \cdot \numocc{\graph{1}}{\twodis}}\left(\prob^2 - \prob^3\right) - \pbrace{\numocc{\graph{1}}{\ed} + 2 \cdot \numocc{\graph{1}}{\twopath}}\prob\nonumber\\
|
||||
&\implies \aug{3'} = \linsys{3'}\nonumber
|
||||
\end{align}
|
||||
|
||||
|
|
@ -507,12 +490,12 @@ Compute each RHS term starting with the left and working to the right,
|
|||
\end{equation}
|
||||
The middle term then is
|
||||
\begin{equation}
|
||||
-\prob\left(-2(\prob^2 - \prob^3)\cdot -3(\prob^2 - \prob^3)\right) - \left(-18(\prob^2 - \prob^3) \cdot -2(\prob^2 - \prob^3)\right) =
|
||||
-\prob\left(\left(-2(\prob^2 - \prob^3)\cdot -3(\prob^2 - \prob^3)\right) - \left(-18(\prob^2 - \prob^3) \cdot -2(\prob^2 - \prob^3)\right)\right) =
|
||||
-\prob\left(6\cdot (\prob^2 - \prob^3)^2 - 36(\prob^2 - \prob^3)^2\right) = -\prob\left(-30(\prob^2 - \prob^3)^2\right).\label{eq:det-2}
|
||||
\end{equation}
|
||||
Finally, the rightmost term,
|
||||
\begin{equation}
|
||||
\left(\prob^2 - \prob^3\right) \left(-2(\prob^2 - \prob^3)\cdot -21(\prob^2 - \prob^3)\right) - \left(-18(\prob^2 - \prob^3) \cdot -4(\prob^2 - \prob^3)\right) = \left(\prob^2 - \prob^3\right)\left(42\cdot (\prob^2 - \prob^3)^2 - 72(\prob^2 - \prob^3)^2\right) = \left(\prob^2 - \prob^3\right)\left(-30(\prob^2 - \prob^3)^2\right).\label{eq:det-3}
|
||||
\left(\prob^2 - \prob^3\right) \left(\left(-2(\prob^2 - \prob^3)\cdot -21(\prob^2 - \prob^3)\right) - \left(-18(\prob^2 - \prob^3) \cdot -4(\prob^2 - \prob^3)\right)\right) = \left(\prob^2 - \prob^3\right)\left(42\cdot (\prob^2 - \prob^3)^2 - 72(\prob^2 - \prob^3)^2\right) = \left(\prob^2 - \prob^3\right)\left(-30(\prob^2 - \prob^3)^2\right).\label{eq:det-3}
|
||||
\end{equation}
|
||||
|
||||
Putting \cref{eq:det-1}, \cref{eq:det-2}, \cref{eq:det-3} together, we have,
|
||||
|
|
@ -524,11 +507,13 @@ Expanding out $\left(\prob^2 - \prob^3\right)^2$ gives
|
|||
\end{equation*}
|
||||
Further algebraic manipulations result in
|
||||
\begin{align}
|
||||
\dtrm{\mtrix{\rpoly}} &= \left(30\prob^4\right)\left(-1 + 2\prob - \prob^2\right) -\prob\left(-1\left(1 - 2\prob + \prob^2\right)\right) + \left(\prob^2 - \prob^3\right)\left(-1\left(1 - 2\prob + \prob^2\right)\right)\label{eq:det-factor}\\
|
||||
\dtrm{\mtrix{\rpoly}} &= \left(30\prob^4\right)\left(\left(-1 + 2\prob - \prob^2\right) -\prob\left(-1\left(1 - 2\prob + \prob^2\right)\right) + \left(\prob^2 - \prob^3\right)\left(-1\left(1 - 2\prob + \prob^2\right)\right)\right)\label{eq:det-factor}\\
|
||||
&=\left(30\prob^4\right)\left(\left(-1 + 2\prob - \prob^2\right) + \left(\prob - 2\prob^2 + \prob^3\right) + \left( - \prob^2 + 2\prob^3 - \prob^4 + \prob^3 - 2\prob^4 + \prob^5\right)\right)\label{eq:det-mult}\\
|
||||
&= \left(30\prob^4\right)\left(\prob^5 - 3\prob^4 + 4\prob^3 - 4\prob^2 + 3\prob - 1\right)\label{eq:det-combine}.
|
||||
\end{align}
|
||||
|
||||
\cref{eq:det-factor} results from factoring out common terms. \cref{eq:det-mult} is the case when multiplying terms in the right hand factor. We arrive at \cref{eq:det-combine} through a simple rearranging and combining of like terms.
|
||||
|
||||
The roots for \cref{eq:det-combine} are $p = 0, p = 1$, and $p = i$. Thus, we have proved the lemma for fixed $p \in (0, 1)$.
|
||||
The roots for \cref{eq:det-combine} are $p = 0, p = 1$, and $p = i$. Thus, we have proved the lemma for fixed $p \in (0, 1)$.
|
||||
\end{proof}
|
||||
\qed
|
||||
Loading…
Reference in a new issue