Lemma 1 and Lemma 2 of poly write up
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@ -100,10 +100,9 @@ To this end, consider the following graph $G(V, E)$, where $|E| = m$, $|V| = \nu
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If we can compute $\poly(\wElem_1,\ldots, \wElem_\numTup) = q_E(\wElem_1,\ldots, \wElem_\numTup)^3$ in T(m) time for fixed $\prob$, then we can count the number of triangles in $G$ in T(m) + O(m) time.
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\end{Lemma}
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\begin{Lemma}\label{lem:const-p}
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If we can compute $\poly(\wElem_1,\ldots, \wElem_\numTup) = q_E(\wElem_1,\ldots, \wElem_\numTup)^3$ in T(m) time for O(1) distinct values of $\prob$ then we can count the number of triangles (and the number of 3-paths, the number of 3-mathcings) in $G$ in O(T(m) + m) time.
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If we can compute $\poly(\wElem_1,\ldots, \wElem_\numTup) = q_E(\wElem_1,\ldots, \wElem_\numTup)^3$ in T(m) time for O(1) distinct values of $\prob$ then we can count the number of triangles (and the number of 3-paths, the number of 3-matchings) in $G$ in O(T(m) + m) time.
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\end{Lemma}
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\begin{proof}
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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 $G$.
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\begin{Claim}
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@ -163,7 +162,29 @@ $\numocc{\twopathdis} + \numocc{\threedis} = $ the number of occurrences of thre
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\[\numocc{\twopathdis} + \numocc{\threedis} = \sum_{(u, v) \in E} \binom{m - d_u - d_v - 1}{2}\] The implication in \cref{claim:four-two} follows by the above and \cref{claim:four-one}.\qed
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\end{proof}
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\end{Claim}
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\begin{proof}
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\underline{Lemma 2}
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\cref{claim:four-two} of Claim 4 implies that if we know $\rpoly_3(\prob,\ldots, \prob)$, then we can know in O(m) additional time
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\[\numocc{\tri} + \numocc{\threepath} \cdot \prob - \numocc{\threedis}\cdot(\prob^2 - \prob^3).\] 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 equation. In the worst case, without independence, 4 distince values of $\prob$ would suffice...because Atri said so.
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\[\numocc{\tri} + \numocc{\threepath} \cdot \prob - \numocc{\threedis}\cdot(\prob^2 - \prob^3).\] 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 equation. 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.\qed
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\end{proof}
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\begin{proof}
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\underline{Lemma 1}
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The argument for lemma 2 cannot be applied to lemma 1 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.
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The following is an option.
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\begin{enumerate}
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\item Let $G_1$ be an arbitrary graph
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\item Build $G_2$ from $G_1$, where each edge in $G_1$ gets replaced by a 2 path.
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\end{enumerate}
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Then $\numocc{\tri}_2 = 0$, and if we can prove that
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\begin{itemize}
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\item $\numocc{\threepath}_2 = 2 \cdot \numocc{\twopath}_1$
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\item $\numocc{\threedis}_2 = 8 \cdot \numocc{\threedis}_1$
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\end{itemize}
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we solve our problem for $q_E^3$ based on $G_2$ and we can compute $\numocc{\threedis}$, a hard problem.
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\end{proof}
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