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Oct 18, 2022 at 17:09 comment added Vepir @StevenStadnicki Well it is an open problem, but I agree, thanks for the comment.
Oct 18, 2022 at 17:02 history edited Vepir CC BY-SA 4.0
further clarification
Oct 18, 2022 at 16:53 comment added Vepir @PeterTaylor I've checked again, it seems your definition is more common when talking about simple graphs, and mine when talking about multigraphs, but I've seen it used vice versa. Maybe it is best to just specify $V,E$.
Oct 18, 2022 at 16:46 comment added Peter Taylor The standard I'm used to from the literature is $G = (V, E)$ where $E$ is a set of unordered or ordered pairs according to whether the graph is undirected or directed.
Oct 18, 2022 at 16:34 comment added Steven Stadnicki You might want 'NP complete' — there are after all no problems known to be in NP and not in P...
Oct 18, 2022 at 15:42 history edited Vepir CC BY-SA 4.0
clarifications
Oct 18, 2022 at 15:37 comment added Vepir @PeterTaylor Ah yes sorry, I meant to ask that (In NP and not in P).
Oct 18, 2022 at 15:35 comment added Vepir @PeterTaylor It is the incidence function, mapping every edge to an unordered pair of vertices (that is, an edge is associated with two distinct vertices). Is that not the standard way to define graphs? Using sets of edges and vertices, and a function that relates them.
Oct 18, 2022 at 15:31 history edited Vepir CC BY-SA 4.0
added 8 characters in body; edited title
Oct 18, 2022 at 14:05 comment added Peter Taylor Also, the question itself is trivial: the sequence of $f_k$ is a certificate which can be verified in polynomial time. Perhaps the question is intended to be whether the problem is in P?
Oct 18, 2022 at 14:02 comment added Peter Taylor What is $\phi$? I thought that it was a vertex weighting function and that $f(v_1)$ was a typo for $\phi(v_1)$ until I looked at the linked examples and realised that $f(v_1)$ is a typo for $f_k(v_1)$.
Oct 18, 2022 at 13:06 comment added Vepir @AlekseiKulikov You misunderstood the rules. For example, a star graph is not solvable in leaf vertices if it has $3$ or more leaves. The $f$ frogs must jump exactly over $f$ edges, and can't jump to empty vertices (with no frogs).
Oct 18, 2022 at 12:59 comment added Aleksei Kulikov I am slightly confused and maybe misunderstood the problem, but isn’t the frog game winnable for any vertex of any tree (and a fortiori any connected graph by passing to a spanning tree): just put a root at this vertex and start moving frogs from the bottom level up. All this is doable in polynomial (maybe even linear) time.
Oct 18, 2022 at 12:27 history asked Vepir CC BY-SA 4.0