Winning strategy of an "edge or isolated vertex" deletion game Is this perfect information game played on graphs know/studied?


Given a graph $G= (V,E)$, two players alternate picking an edge or an isolated node. If the player picks an edge $e = (u,v)$ the two nodes $u$ and $v$ are deleted along with their incident edges. If the player picks an isolated node, the node is deleted. The first player unable to move loses the game.


What is the complexity of finding the winner?

Any references to similar games?
I post an update as a self-answer only to keep it distinct from the question (which is still open).

As shown in the comments (thanks to Tsuyoshi Ito) the problem is polynomial-time solvable for paths:

$Win(P_n) = 1$ iif $(n \bmod 34) \in \{3,7,23,27\}$

Starting from 0, the (calculated) sequence of the nim values is periodic:

0,1,1,0,2,1,3,0,1,1,3,2,2,3,4,1,5,3,2,2,3,1,1,0,3,1,2,0,1,1,4,4,2,6,
4,1,1,0,2,1,3,0,1,1,3,2,2,3,4,4,5,7,2,2,3,1,1,0,3,1,2,0,1,1,4,4,3,6,
4,1,1,0,2,1,3,0,1,1,3,2,2,3,4,4,5,7,2,2,3,1,1,0,3,1,2,0,1,1,4,4,3,6,
...
the subsequence rseq of length 34:
4,1,1,0,2,1,3,0,1,1,3,2,2,3,4,4,5,7,2,2,3,1,1,0,3,1,2,0,1,1,4,4,3,6
is repeated


I didn't work on a rigorous mathematical proof, but the idea is:

suppose that we want to calculate element $Win(P_n), n = k*34 + x \; (k\geq 4, 0 \leq x &lt; 34)$, then the first move (pick an edge) can split the path in $\lceil n / 2 \rceil$  different ways (n-2,0),(n-3,1),(n-4,2),...). The new nim value is equal to:

$mex \{ P_{n-2}+P_0, P_{n-3}+P_1, ..., P_{\lceil n / 2 \rceil}+P_{n-\lceil n / 2 \rceil}\}$

The first 34 elements of the set are produced by the first non repeating sequence (0,1,1,0,...) (nim)summed with the elements of the repeating sequence in reverse order starting from element $(34-2-x) \bmod 34$.

For example: for $x = 0$:

     0,1,1,0,2,1,3,0,1,1,3,2,2,3,4,1,5,3,2,2,3,1,1,0,3,1,2,0,1,1,4,4,2,6 +
     3,4,4,1,1,0,2,1,3,0,1,1,3,2,2,7,5,4,4,3,2,2,3,1,1,0,3,1,2,0,1,1,4,6 =
mex{ 3,5,5,1,3,1,1,1,2,1,2,3,1,1,6,6,0,7,6,1,1,3,2,1,2,1,1,1,3,1,5,5,6,0 } = 4


For x=0..33 the resulting mex sequence is equal to the repeating sequence:

4,1,1,0,2,1,3,0,1,1,3,2,2,3,4,4,5,7,2,2,3,1,1,0,3,1,2,0,1,1,4,4,3,6


The remaining elements of the set are calulated only on the repeating sequence(s): $rseq[j \bmod 34] + rseq[(34-2-x-j) \bmod 34]$ (for $j \geq 34$ the pairs are repeated, so they don't alter the mex result). The resulting mex sequence for x=0..33 is:

4,1,1,0,2,1,3,0,1,1,3,2,2,3,4,4,4,7,2,2,3,1,1,0,3,1,2,0,1,1,4,4,3,4,


Which is equal to the repeating sequence except for $x=16$ and $x=33$; but the values are lower than the corresponding mex on the non-repeating sequence, so: 

$mex \{ P_{n-2}+P_0, P_{n-3}+P_1, ..., P_{\lceil n / 2 \rceil}+P_{n-\lceil n / 2 \rceil}\}$ = $mex \{ P_{n-2}+P_0, P_{n-3}+P_1, ..., P_{n-2-33}+P_{33}\}$

and for $(k\geq 4, 0 \leq x &lt; 34)$, $Win(P_{k*34 + x}) = Win(P_{34+x}) = Win(P_x)\; $