Propagation of the correct eq(10).
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%root main.tex
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\section{$\rpoly$ cancellations due to $\bi$ constraint}
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\paragraph{Problem Definition}
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Since $\bi$ has the constraint that all tuples from the same block are mutually exclusive from one another, it is the case that there exist query polynomials $\poly$ such that $\rpoly$ will cancel out monomials that violate this condition. Let us assume that we have the following $\poly = \poly_1 \cdot \poly_2$, where $\poly_1 = \sum_{i = 1}^\numvar \tup^{1_i}$ and $\poly_2 = \sum{j = 1}^\numvar \tup^{2_j}$, and $\tup^{a_i}$ is a monomial as defined in ~\cref{def:monomial}, i.e., every term in $\tup^{a_i}$ is a single variable factor of the monomial as opposed to allowing product of sums. Note that each $\tup^{a_i}$ has at most a degree of $k$ and that each of its variables are associated with a particular block $\block$. We can assume WLOG that each monomial $\tup^{a_i}$ has at most one variable from each block since any $\tup^{a_i}$ having non-identitcal variables from the same $\block$ can easily be pruned in a $O(\numvar)$ scan.
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\paragraph{High Level Description of Solution}
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We claim that we can compute the number of cancelations in $O(\numvar\cdot \log{\numvar})$ time.
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Before digging into the details of computing the exact number of cancellations of $\rpoly$, we describe the high-level details of the solution.
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main.tex
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main.tex
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@ -160,6 +160,7 @@ sensitive=true
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\input{ra-to-poly}
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\input{poly-form}
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\input{approx_alg}
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\input{bi_cancellation}
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@ -132,8 +132,8 @@ For any graph $G$, the following formulas compute $\numocc{G}{H}$ for their resp
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\end{align}
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A quick argument to why \cref{eq:2m} is true. Note that for edge $(i, j)$ connecting arbitrary vertices $i$ and $j$, finding all other edges in $G$ disjoin to $(i, j)$ is equivalent to finding all edges that are not connected to either vertex $i$ or $j$. The number of such edges is $m - d_i - d_j + 1$, where we add $1$ since edge $(i, j)$ is removed twice when subtracting both $d_i$ and $d_j$. Since the summation is iterating over all edges, division by $2$ eliminates the double counting.
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\AH{The formula ~\cref{eq:2pd-3d} doesn't seem to work. The argument below I think is incorrect. This is because of triple counting the 3-matchings.}
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\textcolor{gray}{Equation ~\ref{eq:2pd-3d} is true for similar reasons. For edge $(i, j)$, it is necessary to find two additional edges, disjoint or connected. As in ~\cref{eq:2m}, once the number of edges disjoint to $(i, j)$ have been computed, then we only need to consider all possible combinations of two edges from the set of disjoint edges, since it doesn't matter if the two edges are connected or not. The sum over all such edge combinations is precisely then $\numocc{G}{\twopathdis} + \numocc{G}{\threedis}$.}
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\AH{The formula ~\cref{eq:2pd-3d} has been fixed to reflect the triple counting of 3-matchings. Notice the factor of 3 on the right term (3-matchings) in the LHS. 110220}
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Equation ~\ref{eq:2pd-3d} is true for similar reasons. For edge $(i, j)$, it is necessary to find two additional edges, disjoint or connected. As in ~\cref{eq:2m}, once the number of edges disjoint to $(i, j)$ have been computed, then we only need to consider all possible combinations of two edges from the set of disjoint edges, since it doesn't matter if the two edges are connected or not. Note, the factor $3$ of $\threedis$ is necessary to account for the triple counting of $3$-matchings. It is also the case that, since the two path in $\twopathdis$ is connected, that there will be no double counting by the fact that the summation automatically 'disconnects' the current edge, meaning that a two matching at the current edge will not be counted. The sum over all such edge combinations is precisely then $\numocc{G}{\twopathdis} + 3\numocc{G}{\threedis}$.
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Now consider the query $q_E(X_1,\ldots, X_\numvar) = \sum\limits_{(i, j) \in E} X_i \cdot X_j$. For the following discussion, set $\poly_{G}(\vct{X}) = \left(q_E(X_1,\ldots, X_\numvar)\right)^3$.
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@ -174,16 +174,14 @@ All of \cref{eq:1e}, \cref{eq:2p}, \cref{eq:2m}, \cref{eq:3s}, \cref{eq:2pd-3d}
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&\rpoly_{G}(\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\nonumber\\
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&\rpoly_{G}(\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}\\
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&\frac{\rpoly_{G}(\prob,\ldots, \prob)}{6\prob^3} - \frac{\numocc{G}{\ed}}{6\prob} - \numocc{G}{\twopath} - \numocc{G}{\twodis}\prob - \numocc{G}{\oneint}\prob = \numocc{G}{\tri} + \numocc{G}{\threepath}\prob + \numocc{G}{\twopathdis}\prob^2 + \numocc{G}{\threedis}\prob^3\label{eq:LS-reduce}\\
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&\frac{\rpoly_{G}(\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}
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&\frac{\rpoly_{G}(\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} + 3\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}
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\end{align}
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\cref{eq:LS-rearrange} is the result of simply subtracting from both sides terms that have $O(\numedge)$ complexity. Dividing all terms by the common factor of $6\prob^3$ gives \cref{eq:LS-reduce}. Equation ~\ref{eq:LS-subtract}, is the result of subtracting the term $\left(\numocc{G}{\twopathdis} + \numocc{G}{\threedis}\right)\prob^2$ from both sides.
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\cref{eq:LS-rearrange} is the result of simply subtracting from both sides terms that have $O(\numedge)$ complexity. Dividing all terms by the common factor of $6\prob^3$ gives \cref{eq:LS-reduce}. Equation ~\ref{eq:LS-subtract}, is the result of subtracting the $O(\numedge)$ computable term $\left(\numocc{G}{\twopathdis} + 3\numocc{G}{\threedis}\right)\prob^2$ from both sides.
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However, one additional step is needed on the left hand side. We previously have shown that $\numocc{G}{\twopathdis} + 3\numocc{G}{\threedis}$ can be computed in $O(\numedge)$ time. Subtracting $\numocc{G}{\threedis}\prob^2$ from both sides gives
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\begin{equation}
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\frac{\rpoly_{G}(\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} + 3\numocc{G}{\threedis}\big)\prob^2 = \numocc{G}{\tri} + \numocc{G}{\threepath}\prob - \numocc{G}{\threedis}\left(3\prob^2 - \prob^3\right)
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\end{equation}
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%\begin{equation}
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%\frac{\rpoly_{G}(\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} + 3\numocc{G}{\threedis}\big)\prob^2 = \numocc{G}{\tri} + \numocc{G}{\threepath}\prob - \numocc{G}{\threedis}\left(3\prob^2 - \prob^3\right)
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%\end{equation}
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The implication in \cref{claim:four-two} follows by the above and \cref{lem:qE3-exp}.
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@ -211,7 +209,7 @@ If we can compute $\rpoly_{G}(\vct{X})$ in $T(\numedge)$ time for $O(1)$ distinc
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\begin{Theorem}\label{lem:const-p}
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If we can compute $\rpoly_{G}(\vct{X})$ in T(\numedge) time for $X_1 =\cdots= X_\numvar = \prob$, then we can count the number of triangles, 3-paths, and 3-matchings in $G$ in $T(\numedge) + O(\numedge)$ time.
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\end{Theorem}
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\AH{Theorem ~\ref{lem:const-p} is different from ~\cref{lem:gen-p} in that $\prob$ is constant, while for ~\cref{lem:gen-p} there are $O(1)$ \textit{distinct} values of $\prob$ in $\vct{p}$.}
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\AH{Theorem ~\ref{lem:const-p} is different from ~\cref{lem:gen-p} is that ~\cref{lem:const-p} has exactly one $\vct{p}$, while for ~\cref{lem:gen-p} there are $O(1)$ \textit{distinct} $\vct{p}_i$, where as explicitly stated in both statements for distinct $\vct{p}$, all values $\prob_j \in \vct{p}$ are equal to one another.}
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@ -297,7 +295,7 @@ Using the identities of lemmas [\ref{lem:3m-G2}, \ref{lem:3m-G3}, \ref{lem:3p-G2
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\AH{I didn't think of a more appropriate name for $\vct{b}$, so I have just stuck with what Atri called it on chat.}
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Using \cref{def:Gk} we construct graphs $\graph{2}$ and $\graph{3}$ from arbitrary graph $\graph{1}$.
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We then show that for any of the patterns $\threedis, \threepath, \tri$ which are all known to be hard to compute, we can use linear combinations in terms of $\graph{1}$ from Lemmas \ref{lem:3m-G2}, \ref{lem:3m-G3}, \ref{lem:3p-G2}, \ref{lem:3p-G3}, \ref{lem:tri} to compute $\numocc{\graph{i}}{S}$, where $i$ in $\{2, 3\}$ and $S \in \{\threedis, \threepath, \tri\}$. Then, using \cref{lem:qE3-exp} and \cref{lem:lin-sys}, we can combine all three linear combinations into a linear system, solving for $\numocc{\graph{1}}{S}$.
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We then show that for any of the patterns $\threedis, \threepath, \tri$ which are all known to be hard to compute, we can use linear combinations in terms of $\graph{1}$ from Lemmas \ref{lem:3m-G2}, \ref{lem:3m-G3}, \ref{lem:3p-G2}, \ref{lem:3p-G3}, \ref{lem:tri} to compute $\numocc{\graph{i}}{S}$, where $i$ in $\{2, 3\}$ and $S \in \{\threedis, \threepath, \tri\}$. Then, using ~\cref{claim:four-two}, \cref{lem:qE3-exp} and \cref{lem:lin-sys}, we can combine all three linear combinations into a linear system, solving for $\numocc{\graph{1}}{S}$.
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%$%^&*(
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@ -516,43 +514,43 @@ The number of triangles in $\graph{k}$ for $k \geq 2$ will always be $0$ for the
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\subsection{Developing a Linear System}
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\AH{The results for the bug in ~\cref{eq:2pd-3d} need to be propogated in this section.}
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\AH{The changes in ~\cref{eq:2pd-3d} have been propagated 110220. Barring any errors, everything should be updated and correct.}
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\begin{proof}[Proof of Lemma \ref{lem:lin-sys}]
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Our goal is to build a linear system $M \cdot (x~y~z)^T = \vct{b}$, such that, assuming an indexing starting at $1$, each $i^{th}$ row in $M$ corresponds to the RHS of ~\cref{eq:LS-subtract} for $\graph{i}$ \textit{in} terms of $\graph{1}$. The vector $\vct{b}$ analogously has the terms computable in $O(\numedge)$ time for each $\graph{i}$ at its corresponing $i^{th}$ entry. Lemma ~\ref{lem:qE3-exp} gives the identity for $\rpoly_{G}(\prob,\ldots, \prob)$ when $\poly_{G}(\vct{X}) = q_E(X_1,\ldots, X_\numvar)^3$, and using
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%Let us maintain a vector $\vct{b}$ to hold the entries for the terms that are computable in $O(\numedge)$ time, for each of $\graph{1}, \graph{2},$ and $\graph{3}$. From
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~\cref{eq:LS-subtract}, $\vct{b}[1] = \frac{\rpoly_{G}(\prob,\ldots, \prob)}{6\prob^3} - \frac{\numocc{G}{\ed}}{6\prob} - \numocc{G}{\twopath} - \numocc{G}{\twodis}\prob - \numocc{G}{\oneint}\prob - \big(\numocc{G}{\twopathdis} + \numocc{G}{\threedis}\big)\prob^2$.
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~\cref{eq:LS-subtract}, $\vct{b}[1] = \frac{\rpoly_{G}(\prob,\ldots, \prob)}{6\prob^3} - \frac{\numocc{G}{\ed}}{6\prob} - \numocc{G}{\twopath} - \numocc{G}{\twodis}\prob - \numocc{G}{\oneint}\prob - \big(\numocc{G}{\twopathdis} + 3\numocc{G}{\threedis}\big)\prob^2$.
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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}.
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\subsubsection{$\graph{2}$}
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Let us call the linear equation for graph $\graph{2}$ $\linsys{2}$. Using the hard to compute terms of the RHS in ~\cref{eq:LS-subtract}, define $\linsys{2} = \numocc{\graph{2}}{\tri} + \numocc{\graph{2}}{\threepath}\prob - \numocc{\graph{2}}{\threedis}\left(\prob^2 - \prob^3\right)$. By \cref{claim:four-two} we can compute $\linsys{2}$ in $O(T(\numedge) + \numedge)$ time with $\numedge = |E_2|$, and more generally, $\numedge = |E_k|$ for a graph $\graph{k}$.
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Let us call the linear equation for graph $\graph{2}$ $\linsys{2}$. Using the hard to compute terms of the RHS in ~\cref{eq:LS-subtract}, define $\linsys{2} = \numocc{\graph{2}}{\tri} + \numocc{\graph{2}}{\threepath}\prob - \numocc{\graph{2}}{\threedis}\left(3\prob^2 - \prob^3\right)$. By \cref{claim:four-two} we can compute $\linsys{2}$ in $O(T(\numedge) + \numedge)$ time with $\numedge = |E_2|$, and more generally, $\numedge = |E_k|$ for a graph $\graph{k}$.
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Using \cref{lem:3m-G2}, \cref{lem:3p-G2}, and \cref{lem:tri}, construct now a linear equation for $\graph{2}$ in \textit{terms} of $\graph{1}$ for $\tri$, $\threepath$, and $\threedis$.
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By \cref{lem:tri}, the first term of $\linsys{2}$ is $0$, and then $\linsys{2} = \numocc{\graph{2}}{\threepath}\prob - \numocc{\graph{2}}{\threedis}\left(\prob^2 - \prob^3\right)$.
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By \cref{lem:tri}, the first term of $\linsys{2}$ is $0$, and then $\linsys{2} = \numocc{\graph{2}}{\threepath}\prob - \numocc{\graph{2}}{\threedis}\left(3\prob^2 - \prob^3\right)$.
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Replace the next term with the identity of \cref{lem:3p-G2} and the last term with the identity of \cref{lem:3m-G2},
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\begin{equation*}
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\linsys{2} = 2 \cdot \numocc{\graph{1}}{\twopath}\prob - \pbrace{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}}\left(\prob^2 - \prob^3\right).
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\linsys{2} = 2 \cdot \numocc{\graph{1}}{\twopath}\prob - \pbrace{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}}\left(3\prob^2 - \prob^3\right).
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\end{equation*}
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Rearrange terms into groups of those patterns that are 'hard' to compute and those that can be computed in $O(\numedge)$,
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\begin{equation*}
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\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)}.
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\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(3\prob^2 - \prob^3\right) + \pbrace{2 \cdot \numocc{\graph{1}}{\twopath}\prob - 4 \cdot \numocc{\graph{1}}{\oneint}\left(3\prob^2 - \prob^3\right)}.
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\end{equation*}
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Note that there are terms computable in $O(\numedge)$ time which can be subtracted from $\linsys{2}$ and added to the other side of \cref{eq:LS-subtract}, i.e., $\vct{b}[2]$. This leaves us with
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\begin{equation}
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\linsys{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'}\\
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\linsys{2} = -\pbrace{2 \cdot \numocc{\graph{1}}{\tri} + 4 \cdot \numocc{\graph{1}}{\threepath} + 2 \cdot \numocc{\graph{1}}{\threedis}}\left(3\prob^2 - \prob^3\right)\label{eq:LS-G2'}\\
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\end{equation}
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and
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\begin{align*}
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\vct{b}[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 +\\
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&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
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\vct{b}[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} + 3\numocc{\graph{2}}{\threedis}\big)\prob^2 +\\
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&6\cdot\left(\numocc{\graph{1}}{\threedis} + \numocc{\graph{1}}{\twopathdis}\right)\left(3\prob^2 - \prob^3\right) + 4 \cdot \numocc{\graph{1}}{\oneint}\left(3\prob^2 - \prob^3\right) - 2\cdot \numocc{\graph{1}}{\twopath}\prob
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\end{align*}
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We now have a linear equation in terms of $\graph{1}$ for $\graph{2}$.
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@ -562,28 +560,28 @@ We now have a linear equation in terms of $\graph{1}$ for $\graph{2}$.
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Following the same reasoning for $\graph{3}$, using \cref{lem:3m-G3}, \cref{lem:3p-G3}, and \cref{lem:tri}, substitute the identities into $\linsys{3}$,
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\begin{align}
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\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\\
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&\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}\\
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=&\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\\
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&\left.+ \numocc{\graph{1}}{\ed}\prob + 2 \cdot \numocc{\graph{1}}{\twopath}\prob\right\}. \label{eq:LS-G3-rearrange}
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&\left.20 \cdot \numocc{\graph{1}}{\oneint} + 27 \cdot \numocc{\graph{1}}{\threedis}\right\}\left(3\prob^2 - \prob^3\right)\label{eq:LS-G3-sub}\\
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=&\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(3\prob^2 - \prob^3\right) \nonumber\\
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&+ \left\{\pbrace{-20 \cdot \numocc{\graph{1}}{\oneint} - 4\cdot \numocc{\graph{1}}{\twopath} - 6 \cdot \numocc{\graph{1}}{\twodis}}\left(3\prob^2 - \prob^3\right)+ \numocc{\graph{1}}{\ed}\prob + 2 \cdot \numocc{\graph{1}}{\twopath}\prob\right\}. \label{eq:LS-G3-rearrange}
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\end{align}
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Equation \ref{eq:LS-G3-sub} follows from simple substitution of all lemma identities. We then get \cref{eq:LS-G3-rearrange} by simply rearranging the operands into a formation that is more organized for our purposes.
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Removing terms to the other side of \cref{eq:LS-subtract}, we get
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\begin{equation}
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\linsys{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'}
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\linsys{3} = \pbrace{- 18 \cdot \numocc{\graph{1}}{\tri} - 21 \cdot \numocc{\graph{1}}{\threepath} - 3 \cdot \numocc{\graph{1}}{\threedis}}\left(3p^2 - p^3\right)\label{eq:LS-G3'}
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\end{equation}
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and
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\begin{align*}
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\vct{b}[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 + \\
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& \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
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\vct{b}[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} + 3\numocc{\graph{3}}{\threedis}\big)\prob^2 + \\
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& \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(3\prob^2 - \prob^3\right) - \pbrace{\numocc{\graph{1}}{\ed} + 2 \cdot \numocc{\graph{1}}{\twopath}}\prob
|
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\end{align*}
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We now have a linear system consisting of three linear combinations, for $\graph{1}, \graph{2}, \graph{3}$ in terms of $\graph{1}$. To make it easier, use the following variable representations: $x = \numocc{\graph{1}}{\tri}, y = \numocc{\graph{1}}{\threepath}, z = \numocc{\graph{1}}{\threedis}$. Using $\linsys{2}$ and $\linsys{3}$, the following matrix is obtained,
|
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\[ \mtrix{\rpoly} = \begin{pmatrix}
|
||||
1 & \prob & -(\prob^2 - \prob^3)\\
|
||||
-2(\prob^2 - \prob^3) & -4(\prob^2 - \prob^3) & -2(\prob^2 - \prob^3)\\
|
||||
-18(\prob^2 - \prob^3) & -21(\prob^2 - \prob^3) & -3(\prob^2 - \prob^3)
|
||||
1 & \prob & -(3\prob^2 - \prob^3)\\
|
||||
-2(3\prob^2 - \prob^3) & -4(3\prob^2 - \prob^3) & -2(3\prob^2 - \prob^3)\\
|
||||
-18(3\prob^2 - \prob^3) & -21(3\prob^2 - \prob^3) & -3(3\prob^2 - \prob^3)
|
||||
\end{pmatrix},\]
|
||||
and the following linear equation
|
||||
\begin{equation}
|
||||
|
|
@ -592,28 +590,30 @@ and the following linear equation
|
|||
\AR{
|
||||
Also the top right entry should be $-(p^2-p^3)$-- the negative sign is missing. This changes the rest of the calculations and has to be propagated. If my calculations are correct the final polynomial should be $-30p^2(1-p)^2(1-p-p^2+p^3)$. This still has no root in $(0,1)$}
|
||||
|
||||
\AH{While propagating changes in ~\cref{eq:2pd-3d}, I noticed and corrected some errors, most notably, that for pulling out the \textbf{$a^2$} factor as described next, I hadn't squared it. That has been addressed. 110220}
|
||||
|
||||
Now we seek to show that all rows of the system are indeed independent.
|
||||
|
||||
The method of minors can be used to compute the determinant, $\dtrm{\mtrix{\rpoly}}$, giving
|
||||
The method of minors can be used to compute the determinant, $\dtrm{\mtrix{\rpoly}}$.
|
||||
|
||||
We make use of the fact that for a matrix with entries $ab, ac, ad,$ and $ae$, the determinant is $a^2be - a^2cd = a^2(be - cd)$.
|
||||
We also make use of the fact that for a matrix with entries $ab, ac, ad,$ and $ae$, the determinant is $a^2be - a^2cd = a^2(be - cd)$.
|
||||
\begin{equation*}
|
||||
\begin{vmatrix}
|
||||
1 & \prob & -(\prob^2 - \prob^3)\\
|
||||
-2(\prob^2 - \prob^3) & -4(\prob^2 - \prob^3) & -2(\prob^2 - \prob^3)\\
|
||||
-18(\prob^2 - \prob^3) & -21(\prob^2 - \prob^3) & -3(\prob^2 - \prob^3)
|
||||
1 & \prob & -(3\prob^2 - \prob^3)\\
|
||||
-2(3\prob^2 - \prob^3) & -4(3\prob^2 - \prob^3) & -2(3\prob^2 - \prob^3)\\
|
||||
-18(3\prob^2 - \prob^3) & -21(3\prob^2 - \prob^3) & -3(3\prob^2 - \prob^3)
|
||||
\end{vmatrix}
|
||||
= (\prob^2 - \prob^3) \cdot
|
||||
= (3\prob^2 - \prob^3)^2 \cdot
|
||||
\begin{vmatrix}
|
||||
-4 & -2\\
|
||||
-21 & -3
|
||||
\end{vmatrix}
|
||||
~ - ~ \prob(\prob^2 - \prob^3)~ \cdot
|
||||
~ - ~ \prob(3\prob^2 - \prob^3)^2~ \cdot
|
||||
\begin{vmatrix}
|
||||
-2 & -2\\
|
||||
-18 & -3
|
||||
\end{vmatrix}
|
||||
- ~(\prob^2 - \prob^3)^2~ \cdot
|
||||
- ~(3\prob^2 - \prob^3)^3~ \cdot
|
||||
\begin{vmatrix}
|
||||
-2 & -4\\
|
||||
-18 & -21
|
||||
|
|
@ -622,23 +622,26 @@ We make use of the fact that for a matrix with entries $ab, ac, ad,$ and $ae$, t
|
|||
|
||||
Compute each RHS term starting with the left and working to the right,
|
||||
\begin{equation}
|
||||
(\prob^2 - \prob^3)\cdot \left((-4 \cdot -3) - (-21 \cdot -2)\right) = (\prob^2 - \prob^3)\cdot(12 - 42) = -30(\prob^2 - \prob^3).\label{eq:det-1}
|
||||
(3\prob^2 - \prob^3)^2\cdot \left((-4 \cdot -3) - (-21 \cdot -2)\right) = (3\prob^2 - \prob^3)^2\cdot(12 - 42) = -270(\prob^2 - \prob^3)^2.\label{eq:det-1}
|
||||
\end{equation}
|
||||
The middle term then is
|
||||
\begin{equation}
|
||||
-\prob(\prob^2 - \prob^3) \cdot \left((-2 \cdot -3) - (-18 \cdot -2)\right) = -\prob(\prob^2 - \prob^3) \cdot ( 6 - 36) = 30\prob(\prob^2 - \prob^3).\label{eq:det-2}
|
||||
-\prob(3\prob^2 - \prob^3)^2 \cdot \left((-2 \cdot -3) - (-18 \cdot -2)\right) = -\prob(3\prob^2 - \prob^3)^2 \cdot ( 6 - 36) = 270\prob(\prob^2 - \prob^3)^2.\label{eq:det-2}
|
||||
\end{equation}
|
||||
Finally, the rightmost term,
|
||||
\begin{equation}
|
||||
-\left(\prob^2 - \prob^3\right)^2 \cdot \left((-2 \cdot -21) - (-18 \cdot -4)\right) = -\left(\prob^2 - \prob^3\right)^2 \cdot (42 - 72) = 30\left(\prob^2 - \prob^3\right)^2.\label{eq:det-3}
|
||||
-\left(3\prob^2 - \prob^3\right)^3 \cdot \left((-2 \cdot -21) - (-18 \cdot -4)\right) = -\left(3\prob^2 - \prob^3\right)^3 \cdot (42 - 72) = 810\left(\prob^2 - \prob^3\right)^3.\label{eq:det-3}
|
||||
\end{equation}
|
||||
|
||||
Putting \cref{eq:det-1}, \cref{eq:det-2}, \cref{eq:det-3} together, we have,
|
||||
\begin{equation}
|
||||
\dtrm{\mtrix{\rpoly}} = -30(\prob^2 - \prob^3) + 30\prob(\prob^2 - \prob^3) + 30(\prob^2 - \prob^3)^2 = 30(\prob^2 - \prob^3)\left(-1 + \prob + (\prob^2 - \prob^3)\right).\label{eq:det-final}
|
||||
\dtrm{\mtrix{\rpoly}} = -270(\prob^2 - \prob^3)^2 + 270\prob(\prob^2 - \prob^3)^2 + 810(\prob^2 - \prob^3)^3 = 270(\prob^2 - \prob^3)^2\left(-1 + \prob + 3(\prob^2 - \prob^3)\right).\label{eq:det-final}
|
||||
\end{equation}
|
||||
|
||||
Equation \cref{eq:det-final} has no roots in $(0, 1)$.\AH{I need to understand how lemma ~\ref{lem:lin-sys} follows.}
|
||||
\AH{It appears that the equation below has roots at p = 0, p = 1, and p = 0.5774, \textit{UNLESS} I made a mistake somewhere.}
|
||||
|
||||
%Equation \cref{eq:det-final} has no roots in $(0, 1)$.
|
||||
\AH{I need to understand how lemma ~\ref{lem:lin-sys} follows.}
|
||||
\end{proof}\AH{End proof of Lemma \ref{lem:lin-sys}}
|
||||
|
||||
\qed
|
||||
|
|
|
|||
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Reference in New Issue