333 lines
32 KiB
TeX
333 lines
32 KiB
TeX
%root: main.tex
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%!TEX root = ./main.tex
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%\onecolumn
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\section{Polynomial Formulation}
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We can think of $\poly(\vct{w})$ as a function whose input are the variables $X_1,\ldots, X_M$ as in $\poly(X_1,\ldots, X_M)$. Denote the sum of products expansion of $\poly(X_1,\ldots, X_\numTup)$ as $\poly(X_1,\ldots, X_\numTup)_{\Sigma}$
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\begin{Definition}\label{def:qtilde}
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Define $\rpoly(X_1,\ldots, X_\numTup)$ as the reduced version of $\poly(X_1,\ldots, X_\numTup)_{\Sigma}$, of the form
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$\rpoly(X_1,\ldots, X_\numTup) = $
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\[\poly(X_1,\ldots, X_\numTup)_{\Sigma} \mod \wbit_1^2-\wbit_1\cdots\mod \wbit_\numTup^2 - \wbit_\numTup.\]
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\end{Definition}
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Intuitively, $\rpoly(\textbf{X})$ is the expanded sum of products form of $\poly(\textbf{X})$ such that if any $X_j$ term has an exponent $e > 1$, it is reduced to $1$, i.e. $X_j^e\mapsto X_j$ for any $e > 1$.
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Alternatively, one can gain intuition for $\rpoly$ by thinking of $\rpoly$ as the resulting sum of product expansion of $\poly$ when $\poly$ is in a factorized form such that none of its terms have an exponent $e > 1$, with an idempotent product operator.
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The usefulness of this reduction will be seen shortly.
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\begin{Lemma}\label{lem:pre-poly-rpoly}
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When $\poly(X_1,\ldots, X_\numTup) = \sum\limits_{\vct{d} \in \{0,\ldots, D\}^\numTup}q_{\vct{d}} \cdot \prod\limits_{\substack{i = 1\\s.t. d_i\geq 1}}^{\numTup}X_i^{d_i}$, we have then that $\rpoly(X_1,\ldots, X_\numTup) = \sum\limits_{\vct{d} \in \{0,\ldots, D\}^\numTup} q_{\vct{d}}\cdot\prod\limits_{\substack{i = 1\\s.t. d_i\geq 1}}^{\numTup}X_i$.
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\end{Lemma}
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\begin{proof}
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Follows by the construction of $\rpoly$ in \cref{def:qtilde}.
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\end{proof}
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\qed
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Note the following fact:
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\begin{Proposition}
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\[\text{For all } (\wbit_1,\ldots, \wbit_\numTup) \in \{0, 1\}^\numTup, \poly(\wbit_1,\ldots, \wbit_\numTup) = \rpoly(\wbit_1,\ldots, \wbit_\numTup).\]
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\end{Proposition}
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\begin{proof}
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Note that any $\poly$ in factorized form is equivalent to its sum of product expansion. For each term in the expanded form, further note that for all $b \in \{0, 1\}$ and all $e \geq 1$, $b^e = 1$.
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\end{proof}
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\qed
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Define all variables $X_i$ in $\poly$ to be independent.
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\begin{Lemma}\label{lem:exp-poly-rpoly}
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The expectation of a possible world in $\poly$ is equal to $\rpoly(\prob_1,\ldots, \prob_\numTup)$.
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\begin{equation*}
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\expct_{\wVec}\pbox{\poly(\wVec)} = \rpoly(\prob_1,\ldots, \prob_\numTup).
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\end{equation*}
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\end{Lemma}
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\begin{proof}
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%Using the fact above, we need to compute \[\sum_{(\wbit_1,\ldots, \wbit_\numTup) \in \{0, 1\}}\rpoly(\wbit_1,\ldots, \wbit_\numTup)\]. We therefore argue that
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%\[\sum_{(\wbit_1,\ldots, \wbit_\numTup) \in \{0, 1\}}\rpoly(\wbit_1,\ldots, \wbit_\numTup) = 2^\numTup \cdot \rpoly(\frac{1}{2},\ldots, \frac{1}{2}).\]
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Let $\poly$ be the generalized polynomial, i.e., the polynomial of $\numTup$ variables with highest degree $= D$: %, in which every possible monomial permutation appears,
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\[\poly(X_1,\ldots, X_\numTup) = \sum_{\vct{d} \in \{0,\ldots, D\}^\numTup}q_{\vct{d}}\cdot \prod_{\substack{i = 1\\s.t. d_i \geq 1}}^\numTup X_i^{d_i}\].
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Then for expectation we have
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\begin{align}
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\expct_{\wVec}\pbox{\poly(\wVec)} &= \sum_{\vct{d} \in \{0,\ldots, D\}^\numTup}q_{\vct{d}}\cdot \expct_{\wVec}\pbox{\prod_{\substack{i = 1\\s.t. d_i \geq 1}}^\numTup w_i^{d_i}}\label{p1-s1}\\
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&= \sum_{\vct{d} \in \{0,\ldots, D\}^\numTup}q_{\vct{d}}\cdot \prod_{\substack{i = 1\\s.t. d_i \geq 1}}^\numTup \expct_{\wVec}\pbox{w_i^{d_i}}\label{p1-s2}\\
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&= \sum_{\vct{d} \in \{0,\ldots, D\}^\numTup}q_{\vct{d}}\cdot \prod_{\substack{i = 1\\s.t. d_i \geq 1}}^\numTup \expct_{\wVec}\pbox{w_i}\label{p1-s3}\\
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&= \sum_{\vct{d} \in \{0,\ldots, D\}^\numTup}q_{\vct{d}}\cdot \prod_{\substack{i = 1\\s.t. d_i \geq 1}}^\numTup \prob_i\label{p1-s4}\\
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&= \rpoly(\prob_1,\ldots, \prob_\numTup)\label{p1-s5}
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\end{align}
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In steps \cref{p1-s1} and \cref{p1-s2}, by linearity of expectation (recall the variables are independent), the expecation can be pushed all the way inside of the product. In \cref{p1-s3}, note that $w_i \in \{0, 1\}$ which further implies that for any exponent $e \geq 1$, $w_i^e = w_i$. Next, in \cref{p1-s4} the expectation of a tuple is indeed its probability.
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%\OK{
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% You don't need to tie this to TI-DBs if you define the variables ($X_i$) to be independent.
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% Annotations
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% Boolean expressions over uncorrelated boolean variables are sufficient to model TI-, BI-, and
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% PC-Tables. This should still hold for arithmetic over the naturals.
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%}
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Finally, observe \cref{p1-s5} by construction in \cref{lem:pre-poly-rpoly}, that $\rpoly(\prob_1,\ldots, \prob_\numTup)$ is exactly the product of probabilities of each variable in each monomial across the entire sum.
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\qed
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\end{proof}
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\begin{Corollary}
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If $\poly$ is given to us in a sum of monomials form, the expectation of $\poly$ ($\ex{\poly}$) can be computed in $O(|\poly|)$, where $|\poly|$ denotes the total number of multiplication/addition operators.
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\end{Corollary}
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\begin{proof}
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Note that \cref{lem:exp-poly-rpoly} shows that $\ex{\poly} = \rpoly(\prob_1,\ldots, \prob_\numTup)$. Therefore, if $\poly$ is already in sum of products form, one only needs to compute $\poly(\prob_1,\ldots, \prob_\numTup)$ ignoring exponent terms (note that such a polynomial is $\rpoly(\prob_1,\ldots, \prob_\numTup)$), which is indeed has $O(|\poly|)$ compututations.\qed
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\end{proof}
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\subsection{When $\poly$ is not in sum of monomials form}
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We would like to argue that in the general case there is no computation of expectation in linear time.
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To this end, consider the following graph $G(V, E)$, where $|E| = m$, $|V| = \numTup$, and $i, j \in [\numTup]$. Consider the query $q_E(X_1,\ldots, X_\numTup) = \sum\limits_{(i, j) \in E} X_i \cdot X_j$.
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\AR{The two lemmas need to be re-written once notation for representing a query is finalized in Section 1.}
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\AH{I think that we are okay with this now. We can use both polynomial and query notation interchangably. Does it matter which we use in the lemmas, i.e. $\poly(\vct{w})$ vs. $\poly(\wElem_1,\ldots, \wElem_N)$. Please let me know.}
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\begin{Lemma}\label{lem:const-p}
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If we can compute $\poly(\wElem_1,\ldots, \wElem_\numTup) = q_E(\wElem_1,\dots, \wElem_\numTup)^3$ in T(m) time for $\wElem_1 =\cdots= \wElem_\numTup = \prob$, then we can count the number of 3-matchings in $G$ in $T(m) + O(m)$ time.
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\end{Lemma}
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\begin{Lemma}\label{lem:gen-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-matchings) in $G$ in O(T(m) + m) time.
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\end{Lemma}
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\begin{Lemma}\label{lem:qE3-exp}
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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,
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\begin{align}
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&\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\\
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&\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}
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\end{align}
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\end{Lemma}
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\AH{\cref{lem:qE3-exp} needs to be proven. I think I might need a gentle nudge on this, I can understand intuitively, but I think there is a combinatorics argument to prove this formally, I'm just a bit unsure.}
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\AH{The warm-up below is fine for now, but will need to be removed for the final draft}
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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 $\numocc{G}{H}$ denote the number of occurrences that $H$ occurs in $G$. So, e.g., $\numocc{G}{\ed}$ is the number of edges ($m$) in $G$.
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\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.}
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\AH{UPDATE: I did a quick google, and it \textit{appears} that there is a bit of a learning curve to implement node/edge symbols in LaTeX. So, maybe, if time is of the essence, we go with another notation.}
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\begin{Claim}
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We can compute $\rpoly(\prob,\ldots, \prob)^2$ in O(m) time.
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\end{Claim}
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\begin{proof}
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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 $\rpoly_2(\prob,\ldots, \prob) = \numocc{G}{\ed} \cdot \prob^2 + 2\cdot \numocc{G}{\twopath}\cdot \prob^3 + 2\cdot \numocc{G}{\twodis}\cdot \prob^4$.
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\begin{enumerate}
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\item First note that
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\begin{align*}
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\poly^2(\wVec) &= \sum_{(i, j) \in E} (\wElem_i\wElem_j)^2 + \sum_{(i, j), (k, \ell) \in E s.t. (i, j) \neq (k, \ell)} \wElem_i\wElem_j\wElem_k\wElem_\ell\\
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&= \sum_{(i, j) \in E} (\wElem_i\wElem_j)^2 + \sum_{\substack{(i, j), (j, \ell) \in E\\s.t. i \neq \ell}}\wElem_i
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\wElem_j^2\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\\
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\end{align*}
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By definition of $\rpoly$,
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\begin{equation*}
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\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}
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\end{equation*}
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Notice that the first term is $\numocc{G}{\ed}\cdot \prob^2$, the second $\numocc{G}{\twopath}\cdot \prob^3$, and the third $\numocc{G}{\twodis}\cdot \prob^4.$
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\item Note that
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\AH{We need the correct formula for two-matchings below.}
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\begin{align*}
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&\numocc{G}{\ed} = m,\\
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&\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}}
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\end{align*}
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\end{enumerate}
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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
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\end{proof}
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\AH{END of the 'warm-up'}
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We are now ready to state the claim we need to prove \cref{lem:const-p} and \cref{lem:gen-p}.
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Let $\poly(\wVec) = q_E(\wVec)^3$.
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\begin{Claim}\label{claim:four-two}
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If one can compute $\rpoly(\prob,\ldots, \prob)$ in time T(m), then we can compute the following in O(T(m) + m):
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\[\numocc{G}{\tri} + \numocc{G}{\threepath} \cdot \prob - \numocc{G}{\threedis}\cdot(\prob^2 - \prob^3).\]
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\end{Claim}
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\begin{proof}
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We have either shown or will show that the following subgraph cardinalities can be computed in $O(m)$ time:
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\[\numocc{G}{\ed}, \numocc{G}{\twopath}, \numocc{G}{\twodis}, \numocc{G}{\oneint}, \numocc{G}{\twopathdis} + \numocc{G}{\threedis}.\]
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By definition we have that
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\[\poly(\wElem_1,\ldots, \wElem_\numTup) = \sum_{\substack{(i_1, j_1),\\ (i_2, j_2),\\ (i_3, j_3) \in E}} \prod_{\ell = 1}^{3}\wElem_{i_\ell}\wElem_{j_\ell}.\]
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Rather than list all the expressions in full detail, let us make some observations regarding the sum. Let $e_1 = (i_1, j_1), e_2 = (i_2, j_2), e_3 = (i_3, j_3)$. Notice that each expression in the sum consists of a triple $(e_1, e_2, e_3)$. There are three forms the triple $(e_1, e_2, e_3)$ can take.
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\underline{case 1:} $e_1 = e_2 = e_3$, where all edges are the same. There are exactly $m$ such triples, each with a $\prob^2$ factor.
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\underline{case 2:} This case occurs when there are two distinct edges of the three. All 6 combinations of two distinct values consist of the same monomial in $\rpoly$, i.e. $(e_1, e_1, e_2)$ is the same as $(e_2, e_1, e_2)$. This case produces the following edge patterns: $\twodis, \twopath$.
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\underline{case 3:} $e_1 \neq e_2 \neq e_3$, i.e., when all edges are distinct. This case consists of the following edge patterns: $\threedis, \twopathdis, \threepath, \oneint, \tri$.
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\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.}
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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.
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\[\numocc{G}{\oneint} = \sum_{u \in V} \binom{d_u}{3}\]
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$\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.
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\[\numocc{G}{\twopathdis} + \numocc{G}{\threedis} = \sum_{(u, v) \in E} \binom{m - d_u - d_v + 1}{2}\] The implication in \cref{claim:four-two} follows by the above and \cref{lem:qE3-exp}.
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\AH{Justify the last sentence.}
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\end{proof}
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\qed
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\begin{proof}[Proof of \cref{lem:gen-p}]
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%\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.}
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\cref{claim:four-two} says that if we know $\rpoly_3(\prob,\ldots, \prob)$, then we can know in O(m) additional time
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\[\numocc{G}{\tri} + \numocc{G}{\threepath} \cdot \prob - \numocc{G}{\threedis}\cdot(\prob^2 - \prob^3).\] 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.}
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\AH{This Vandermonde matrix I need to research.}
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\end{proof}
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\qed
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\AH{Below is only a transcription of the notes. The claims need to be verified and further worked out.}
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\begin{proof}[Proof of \cref{lem:const-p}]
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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.
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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 $\numocc{G_2}{\tri} = 0$, and if we can prove that
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\begin{itemize}
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\item $\numocc{G_2}{\threepath} = 2 \cdot \numocc{G_1}{\twopath}$
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\item $\numocc{G_2}{\threedis} = 8 \cdot \numocc{G_1}{\threedis} + 6 \cdot \numocc{G_1}{\twopathdis} + 4 \cdot \numocc{G_1}{\oneint} + 4 \cdot \numocc{G_1}{\threepath} + 2 \cdot \numocc{G_1}{\tri}$
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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 $\numocc{G}{\threedis}$, a hard problem.
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\end{proof}
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\AH{Proving the above linear combination for 3-matchings in $G_2$ always holds for an arbitrary $G_1$.}
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Consider graph $G_2$, constructed from an arbitrary graph $G_1$. We wish to show that the number of 3-matchings in $G_2$ will always be the linear combination above, regardless of the construction of $G_1$.
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\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 $G_2$ and $G_3$ outside of the proofs below.}
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\subsubsection{$f_k$ and $G_k$}
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\begin{Definition}\label{def:Gk}
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For $k > 1$, let graph $G_k$ be a graph generated from an arbitrary graph $G_1$, by replacing every edge $e$ of $G_1$ 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 $G_1$.
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\end{Definition}
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For any graph $G_k$, we denote its edges to be a pair $(e_i, b)$, such that $b \in \{0,\ldots, k\}$.
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\begin{Definition}\label{def:fk}
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Define $f_k: \binom{E_k}{3} \mapsto \binom{E_1}{\leq3}$.
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\end{Definition}
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The function $f_k$ is a mapping from every $3$-edge shape in $G_k$ to its generating subgraph in $G_1$. The notation $\binom{S}{t}$ is standard and is used to denote the set of subsets in $S$ with exactly $t$ edges. The set of edges in $G_k$ is written as $E_k$.
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\begin{Lemma}
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$f_k$ is a function.
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\end{Lemma}
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Note that $f_k$ is properly defined. For any $S \in \binom{E_k}{3}$, $|f(S)| \leq 3$, since it has to be the case that any subset of $3$ edges in $E_k$ will map to at most 3 edges in $G_1$. All mappings are in the required range. Then, since for any $b \in \{0,\ldots, k\}$ the edge $(e, b) \mapsto e$ is a mapping for which $(e, b)$ maps to no other edge than $e$, and this implies that $f_k$ is a function.
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\subsubsection{Subgraph patterns with 3 edges}
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\begin{itemize}
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\item Triangle ($\tri$)
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\item 3-path ($\threepath$)
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\item 3-star ($\oneint$)--this is the graph that results when all three edges share exactly one common endpoint. The remaining endpoint for each edge is disconnected from any endpoint of the three edges.
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\item Disjoint Two-Path ($\twopathdis$)--this subgraph consists of a two path and a remaining disjoint edge.
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\item 3-matching ($\threedis$)--this subgraph is composed of three disjoint edges.
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\end{itemize}
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\subsection{Three Matchings}
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\begin{Lemma}\label{lem:3m-G_2}
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The number of $3$-matchings in graph $G_2$ is computed by the following identity,
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\[\numocc{G_2}{\threedis} = 8 \cdot \numocc{G_1}{\threedis} + 6 \cdot \numocc{G_1}{\twopathdis} + 4 \cdot \numocc{G_1}{\oneint} + 4 \cdot \numocc{G_1}{\threepath} + 2 \cdot \numocc{G_1}{\tri}.\]
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\end{Lemma}
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\begin{proof}
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Denote $3_{match}$ as the set of 3-matchings in $G_2$. \AR{I do not like the notation $3_{match}$: perhaps $\mathcal{M}_3$ would be better?} Denote $SG$ as the set of subgraphs imposed on $G_1$. \AR{The set of all edge-subgraphs of $G_1$ already has an existing notation-- $2^{E_1}$, i.e. the power set of the set of edges of $G_1$.} Notate each edge in $G_2$ as $(e, b)$ such that $b \in \{0, 1\}$, where $b$ identifies either the first or second edge of the two path that replaced the original $G_1$ edge.
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In $G$, the following possible subgraphs that can be composed of three edges.
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\AH{I think that all of the above can be generalized to 3-paths as well.}
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\AR{Yep, this is why I am advocating for defining $f_k$ more generally above.}
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Given any $S \in \binom{E_1}{\leq3}$, we consider $f^{-1}(S)$, which is the set of all possible edges in $S \times \{0, 1\}$ which $f$ maps to $S$. Then we count the number of $3$-matchings in the $3$-edge subgraphs of $G_2$ in $f^{-1}(S)$.
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Beginning with the leftmost of RHS terms and proceeding to the consecutive rightmost terms, let us show \cref{lem:3m-G_2} to be true.
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\AR{I think it would be good to give an overview of the argument below. Something along these lines. Given any $S\in\binom{E_1}{\le 3}$, we consider $f^{-1}(S)$ (which is the set of all possible subsets of three edges in $S\times \{0,1\}$ that $f$ maps to $S$). Then in $f^{-1}(S)$, we count how many of the corresponding three edge subgraphs in $G_2$ is a 3-matching.}
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Consider the $\threedis$ pattern. For $G_2$, this gives us three disjoint two paths.
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\AH{The comment below is an important comment.}
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\AR{I think your argument seems to implicitly assume that $G_1$ is the subset $S$ and $G_2$ is the corresponding mapping under $f^{-1}$. This is {\bf not} correct. You should present the argument as in the outline above above. I.e. fix an $S\in\binom{E_1}{\le 3}$ in $G_1$ and then consider all possible subgraphs in $G_2$ in $f^{-1}(S)$. {\bf Propagate} this change to the rest of the proof.} One can see that we have a total of two possible choices for each disjoint two path, which yields $2^3 = 8$ possible 3-matchings in $G_2$.
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For the $\twopathdis$, i.e. DT pattern, when we convert to $G_2$, we have a four path and disjoint two path. Let the generating subgraph in $G_1$ be $S = \{e_0, e_1, e_2\}$, and $e_1, e_2$ form the $2$-path. We can only pick $(e_0, 0)$ or $(e_0, 1)$ from $f^{-1}$, and then we need to pick a $2$-matching from the mapping of the $e_1, e_2$ under $f^{-1}.$ The corresponding $4$-path in $G_2$ is $(e_1, 0), (e_1, 1), (e_2, 0), (e_2, 1).$ A four path allows there to be 3 possible 2 matchings, specifically, $\pbrace{(e_1, 0), (e_2, 0)}, \pbrace{(e_1, 0), (e_2, 1)}, \pbrace{(e_1, 1), (e_2, 1)}$. Since these two selections can be made independently, there are $2 \cdot 3 = 6$ choices. Edge $e_0$ cannot produce a $2$-matching, and are done with $\twopathdis$.
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For the $\oneint$, i.e. 3-star pattern, the resulting $G_2$ graph produces 3 two paths, each of which has one and only one endpoint that intersects with one and only one endpoint of the other two paths. Call the intersecting edges of the two paths inner edges. Note that for a valid 3 matching it must be the case that at most one inner edge can be part of the set of disjoint edges. When exactly one inner edge is chosen, there are 3 such possibilities. The remaining possible 3-matching occurs when all 3 outer edges are chosen. Thus, there are $3 + 1 = 4$ 3-matchings for a 3-star subgraph. \AR{Overall this argument if good. But remember to make the change that should follow from the comment in the first case on how to setup the argument.}
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When $S \in G_1$ is $\threepath$ it is the case that $S$ becomes a six path. Let $S = (e_0, e_1, e_2)$ and the six path be $(e_0, 0), (e_0, 1), (e_1, 0), (e_1, 1), (e_2, 0), (e_2, 1)$. The following edge combinations, $\pbrace{(e_1, 0), (e_2, 0), (e_3, 0)}, \pbrace{(e_1, 0), (e_2, 0), (e_3, 1)}, \pbrace{(e_1, 0), (e_2, 1), (e_3, 1)}, \pbrace{(e_1, 1), (e_2, 1), (e_3, 1)}$ produce four possible 3-matchings.
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For $S = \tri$, note that it is the case that the generated subgraph in $G_2$ is a 'triangle of two paths', where each trianglular edge is a two path. 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.
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Consider when $S$ consists of at most $2$ distinct edges. When $S$ has exactly $2$ edges, the generated subgraphs would either be a $4$-path or $2$ disjoint $2$-paths. Neither of these can produce a $3$-matching. For the former, a 3-matching would require at least to alternate edges to ensure disjointedness, which requires $\geq 5$-path. The latter requires that at least one of the disjoint $2$-paths be a $3$-path in order to produce a $2$-matching. When $S$ has $< 2$ edges, it is the case we have a generated $2$-path, which as stated above cannot produce a $3$-matching. Therefore only subgraphs $S$ of $3$ edges need to be considered.
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Observe that all of the arguments above focused solely on the shape/pattern of $S$. In other words, all $S$ of a given shape yield the same number of $3$-matchings, and this is why we get the required identity.
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\AR{I did not make a pass on the rest-- though see a quick comment later on. Please propagate the changes above to the rest of the argument.}
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\end{proof}
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\qed
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\subsubsection{$G_3$}
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\AH{Linear Equation computing 3-matchings in $G_3$ using all 3-edge subgraphs in $G_1$.}
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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. Because of $G_3$ construction, we now need to also account for two paths in $G_1$.
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The linear combination of 3-edge $G_1$ subgraphs to compute the number of 3-matchings in $G_3$ follows is
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\begin{align*}
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\numocc{G_3}{\threedis} = &4\pbrace{\numocc{G_1}{\twopath}} + 6\pbrace{\numocc{G_1}{\twodis}} + 18\pbrace{\numocc{G_1}{\tri}} + 21\pbrace{\numocc{G_1}{\threepath}}\\
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&+ 24\pbrace{\numocc{G_1}{\twopathdis}} + 20\pbrace{\numocc{G_1}{\oneint}} + 27\pbrace{\numocc{G_1}{\threedis}}.
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\end{align*}
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\AH{Justification.}
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Enumerate through the RHS in a similar fashion. Beginning with a two path $\twopath$ in $G_1$, it is the case that in $G_3$ this becomes a six-path. As discussed previously, this yields four three matching subgraphs.
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For subgragh of two disjoint edges, $\twodis$, this becomes two disjoint 3-paths. It is the case in one 3-path, that we have one subgraph of two disjoint edges, where a third disjoint edge can be picked from any of the three edges in the remaining disjoint 3-path. The process can be repeated starting with the alternative 3-path, giving $2 * 3 = 6$ unique 3-matchings.
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Now for the 3-edge subgraphs, starting with a triangle. Note that we are considering 3-matchings $M$ $s.t.$ $f(M) = G'$, where $G'$ is a 3-edge subgraph. In other words, the function $f$ will produce 3 \textit{distinct} edge outputs. This then disalllows double counting of 3-matchings from a 3-edge subgraph using only two of the edges.
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Note that for the case of $G_3$, edges are denoted as $(e_i, b_i)$ where $b \in \{0, 1, 2\}$. When a triangle in $G_1$ is transformed into $G_3$, it becomes a 'triangle' where each leg is a three-path. This is very similar to a 9-path, with the caveat that the first and last edge $(e_1, 0)$ and $(e_3, 2)$ cannot be in the same 3-matching set together. For this $G_3$ it is also the case that for all $i \in \{0, 1, 2\}$ $(e_i, 2)$ and $(e_{i + 1}, 0)$ are neighbors and cannot share a 3-matching. Iterating through all possible combinations producing 3-matchings, i.e. $\pbox{(e_1, 0), (e_2, 0), (e_3, 0)}, \pbox{(e_1, 0), (e_2, 0), (e_3, 1)}, \pbox{(e_1, 0), (e_2, 1), (e_3, 0)},$\newline$\ldots, \pbox{(e_1, 0), (e_2, 2), (e_3, 1)},\ldots, \pbox{(e_1, 2), (e_2, 2), (e_3, 2)}$ gives a total of 18 3-matchings.
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Consider next a 3-path in $G_1$, where the resulting subgraph in $G_3$ is a 9-path. In this case, because the endpoints are disconnected, we have 3 other 3-matchings that couldn't be counted in the case of the Tri subgraph, namely $\pbox{(e_1, 0), (e_2, 0), (e_3, 2)},\pbox{(e_1, 0), (e_2, 1), (e_3, 2)}, \pbox{(e_1, 0), (e_2, 2), (e_3, 2)}$, thus $18$ (from the Tri analysis)$ + 3 = 21$ three-matchings.
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For the $\twopathdis$ subgraph, it is the case that this graph becomes a 6-path with a disjoint 3-path in $G_3$. Starting with the 6-path, there are 8 distinct two matchings in the form of $\pbox{(e_1, 0), (e_2, 0), (e_3, 0)}, \pbox{(e_1, 0), (e_2, 0), (e_3, 1)}, \pbox{(e_1, 0), (e_2, 0), (e_3, 2)}, \ldots \pbox{(e_1, 2), (e_2, 1), (e_3, 0)},\ldots, \pbox{(e_1, 2), (e_2, 2), (e_3, 2)}$. Notice again that the edge pattern $\pbox{(e_1, 2), (e_2, 0)\ldots}$ is forbidden. Each of these 8 2-matchings can be paired with one of the 3 edges in the disjoint 3-path, yielding $8 * 3 = 24$ 3-matchings.%We have to consider 3 possibilities of 3-matchings. First, the 6-path produces 4. Second, it is the case that the 6-path produces 10 two-matchings, which can be paired with any one of the three edges in the disjoint 3-path, producing $10 * 3$ 3-matchings. Third, the disjoint 3-path can produce one 3-matching which can be paired with a third edge from any one of the edges in the 6-path, giving $1 * 6 = 6$ three-matchings, for a total of $4 + 30 + 6 = 40$ three-matchings.
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Given the 3-star subgraph, where 3 distinct edges are connected at one common endpoint, occurring in $G_1$. In $G_3$, this becomes 3 3-paths joined at one common endpoint. Note the non-intersecting outer and middle edges. For each 3-path, there are 2 possible choices to create a 3-matching, yielding $2^3 = 8$ 3-matchings. Finally, consider the joined inner edges, recalling that at most one of them can be in a 3-matching. Pick an arbitrary edge, and there are four possible combinations that result from the middle and outer edges of the other 2 3-paths. This gives $3 * 4 = 12$ more 3-matchings, a total of $8 + 12 = 20$.% 3 distinct 9-paths, with each 9-path intersecting the others at one common and shared endpoint. If we consider the outermost non-intersecting edges along with the middle non-intersecting edges, we have $2 * 2 * 2 = 8$ possible 3-matchings. Considering the inner, intersecting edges, we have the condition that only one can appear at a time in a 3-matching set. When we pick an arbitrary inner edge, we have one of two possibilities, we can pick the outer edge of the same 3-path the inner edge is located on, while picking any of the other 4 remaining edges in the middle and outer edges of the other two 3-paths. This gives $4 * 3$ more unique 3-matchings. The remaining possibility exists in combining the arbitrary inner edge with any of the 4 combinations of the middle and outer edges of the other 3-paths. This yields again $3 * 4 = 12$ unique three-matchings, together which make $8 + 12 + 12 = 32$ three-matchings.
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Given the $\threedis$ subgraph occurring in $G_1$, the resulting graph consists of three disjoint 3-paths in $G_3$. Since only one edge can be used at a time from each 3-path, it results that there are $3^3 = 27$ 3-matchings.%There are two considerations. First, if we pull one edge from each disjoint 3-path, we have three choices from each path, which is $3^3 = 27$ three-matchings. The second consideration is that we can pull a two matching from any of the given disjoint 3-paths, matching it with a third disjoint edge from any of the other edges in the other 2 three-paths, giving $3 * 6 = 18$ more unique 3-matchings for a total of $27 + 18 = 45$ three-matchings.
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\subsection{Three Paths}
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Computing the number of 3-paths in $G_2$ and $G_3$ consists of much simpler linear combinations.
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\subsubsection{$G_2$}
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It can be easily verified that a single edge in $G_1$ becomes a 2-path in $G_2$, and generates $0$ 3-paths. Similarly, it is the case for 2 disjoint edges. When we look at output of $f$ that has 3 edges, we see the same result for the case of 3 disjoint edges in $G_1$. When we have a 2-path and 1 disjoint edge, there is no way to create a 3-path out of all three edges. The same holds for the remaining subgraphs consisting of 3 edges, the 3-star, triangle, and 3-path.\AR{For each pattern here, you should {\em explicitly} argue why they give rise to zero 3-paths. It does not have to be a separate argument for each shape-- you can make a general argument that rules out everything except the 2-path. Whatever is easier.}
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All the above to say that there exists only one subgraph in $G_1$ that will produce 3-paths, namely the 2-path. This subgraph becomes a 4-path in $G_2$, and using both edges (as required by definition of $f$), we have two 3-paths generated: $\pbox{(e_1, 0), (e_1, 1), (e_2, 0)}$ and $\pbox{(e_1, 1), (e_2, 0), (e_2, 1)}$. Thus,
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\[\numocc{G_2}{\threepath} = 2 * \numocc{G_1}{\twopath}.\]
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\subsubsection{$G_3$}
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In a similar fashion, enumerate through the various subgraphs in $G_1$ with $\leq 3$ edges, starting with the smallest. Note, that one edge in $G_1$ generates one 3-path in $G_3$. Moving on to a 2-path, again we see that we have 2 2-paths that consist of both $G_1$ generating edges. For the subgraph of 2 disjoint edges, as in the case of $G_2$, there is no way to make a 3-path out of disjoint edges, and this rolls over into the subgraph consisting of 3 disjoint edges, the subgraph made of a 2 path and disjoint edge, 3-star, triangle, and 3-path. All of these subgraphs provide no way to create a 3-path from all edges $e_1, e_2, e_3$ in $G_1$. The combination is then
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\[\numocc{G_3}{\threepath} = \numocc{G_1}{\ed} + 2 \times \numocc{G_1}{\twopath}.\]
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\subsection{Triangle}
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The number of triangles in both $G_2$ and $G_3$ will always be $0$ for the simple fact that when we replace a single edge with $\geq 2$-path, the possibility of a triangle of single edge sides disappears, since the only way a single edged triangle could exist is if it existed in $G_1$ and then was passed to $G_2$ or $G_3$ without replacing each single edge with $\geq 2$-paths.
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