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The matrix tree theorem for weighted graphs

Seeing this question left me wondering, is it possible to modify the matrix so one can compute the following sum: $$ P'(G) = \sum_{T\subseteq G}{m'(T)} $$ where $m'(T) = \prod_{e \in (E-T)}{\omega_e}$. Is this possible?

Daniele Tampieri
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juan manuel rodriguez
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  • We have $P'(G) = (\prod_{e\in E}\omega_e)^{#T} P(G)\mid_{\omega_e \to \omega_e^{-1}}$, so your $P'(G)$ is very easy to obtain from $P(G)$. – Sam Hopkins Feb 19 '24 at 14:27

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