As I understand it, the Sprague-Grundy theorem states that, under the operation of combining games in parallel, every impartial game can be reduced to a game of Nim. I see that the definition itself specifies this parallel combination operator explicitly: two impartial games G and G' are assigned the same nimber iff for all games H, G + H and G' + H have the same outcome (here, + denotes the parallel combination operator)---this definition is from Wikipedia.
However, what about other methods of combining games? For instance, we combine two games A and B such that they are still played in parallel, but as soon as a player has no moves left in A, they instantly lose, regardless of the position of B. In general, can we still treat games A and B as their corresponding nimbers? If so, why? If not, what's a counterexample?
EDIT (which I added in a comment below):
More precisely, my question is:
Let $$$n(G)$$$ be the nimber associated with game G, according to the Sprague-Grundy theorem. Moreover, let $$$f(G, H)$$$ be a function of any two games that outputs another game. Then, I would like to know: for all possible $$$f$$$, does $$$n(f(G, H)) = f(n(G), n(H))$$$?
Actually, Sprague-Grundy theorem applies specifically to disjunctive sums where players choose one game to play per turn and all games must be played to completion. In your variant where a player loses immediately upon having no moves in one subgame the structure don't fit this model completely.This is more like a conjunctive game with early termination and in such setting Grundy numbers don't behave additively. Suppose an example like two nim heaps where the first empty heap causes a loss then the XOR trick fails there. So NO, we can't reduce such games to nim directly they require separate analysis.
Yes, the XOR trick fails, but we can still analyze the resulting Grundy number via mex. For example, if we have two piles with $$$i$$$ and $$$j$$$ stones respectively, and losing the first game also means losing the second, I believe this becomes equivalent to a single pile with $$$(i - 1) \oplus j + 1$$$ stones.
More precisely, my question is: if SG theorem reduces game $$$G$$$ to nimber $$$*i$$$, and game $$$H$$$ to nimber $$$*j$$$, will every method $$$f$$$ of combining $$$G$$$ and $$$H$$$ into $$$f(G, H)$$$ be equivalent to $$$f(*i, *j)$$$?
But according to me in general not every way of combining two games preserves nimber behavior. The sprague grundy theorem only works when the games are combined using the standard disjunctive sum where players choose one subgame to play on each turn and all subgames are played independently. If you change the rules like making the player lose as soon as one subgame ends then you're creating a different kind of combination that doesn't follow the same algebra as nim. So combining games using your custom rule won’t necessarily behave the same way as combining their nimbers. Your approach of using the mex function to analyze the outcome is fine but it creates a new way of assigning values to positions based on your specific rules. These values may not follow the same rules as in normal nim, so you can’t always treat them like standard nimbers.
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