# NIM with multiple winning final positions

I've been looking at a variant of NIM.

You can skip this bit where I'll describe NIM as usually described:

There's a starting position with some number of piles of counters and two players alternate taking any number of counters from any single pile. In "normal play" the player who takes all of the last non-empty pile wins. In "misère play" the player forced to take the last pile loses. That pile will always be of size one, since otherwise a perfect player would take all but one counter from the last pile and force the other player to take the lone pile of one.

A "P-position" in a game is one where the previous player has a forced win, and the person playing next from that position will lose against a player executing the strategy without error.

The variation I'm exploring is to broaden the winning conditions into "the first player to reach any one of a set of positions." In other words, there's a specified set of "seed" P-positions.

The classical normal game seed set for three piles can be recast as {[0,0,0]}; the winning player is the one to reach the state where all the piles are empty. The classical misère game seed set is {[1,0,0],[0,1,0],[0,0,1]}. Plays that make no win possible are not legal: So for example, if {[1,4,2],[3,5,3]} were the two target winning positions, a move from [17,21,5] to [17,3,5] would be illegal.

Both classical normal and classical misère NIM have perfect strategies that are fairly compact to describe, but looking at other sets of seed P-positions, that seems to be a rarity.

Given a game with some set of seed P-positions, what is that set of all P-positions for that game?