POLYA023: Relation on graphs from games (proposed by Stephen Locke, 11/2/01).
The Game G: Two players select a graph G and then take alternate turns.
Set G_0=G, and for move k, the player selects a simple path P_k from G_{k-1}
and sets G_k = G_{k-1} - Edges(P_k). [Note that only the edges of
the path are deleted, not the vertices.] The player who takes the last
edge wins. Let G \sim H for two disjoint graphs G and H, if
G \cup H is a loss
for the first player. It is obvious that \sim is reflexive and symmetric.
Prove that \sim is also transitive (and, therefore, an equivalence relation).
Discussion. [LZ, 1/5/02]: Suppose
F~G and and G~K and the first player deletes any P_1 from F to get F_1.
Then the 2nd player can choose P_2 as follows:
Keep removing paths from G as long as one (and only one
in fact) of the (F_1 \cup G_k) and (G_k \cup K) is still a winning graph.
Then the winning path for this winning graph must be in F_1 or K. This
path is P_2.
It's then easy to verify this is a winning strategy for
the 2nd payer.
[SCL, 1/8/02]: When I proposed this problem, I
was thinking of it as a potential problem for a Discrete Mathematics type
of class, so you're correct that it is easy.
My own solution is only a little longer, and that because
I chose to be more wordy. My real question is whether or not anybody has
seen this, or a similar result about games. There is a concept called
Grundy labeling, but that doesn't quite seem to fit this situation.
By the way, the game was mentioned by Harary in a talk
at LSU (my son attended, but I didn't), in which Harary asked for
a wining strategy for the game when one starts with a caterpillar. A caterpillar
is a tree (a connected acyclic graph) such that removing all of the endvertices
simultaneously results in (at most) a path. (The resultant graph might
be called the "spine".) I catalogued the winning situations for spines
with at most four vertices, and one could easily continue to five, six,
etc. A general description of the winning moves, a la Nim, wasn't apparent
to me.
(I don't know that one should credit Harary, though,
since the game is so similar to Hackenbush and its variants. See Conway's
books. On the other hand, if Conway didn't list it, maybe one can
say it is new.)
Bibliography.
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