An answer to an old Nimber question

Near to the start of this blog, I posed a question:

if the nimbers for a game are bounded above, does this imply that there is an efficient algorithm, or shortcut, to evaluate impartial games?

I recently realized there is strong evidence against this. It is true that any impartial ruleset where all the nimber values are either *0 or *1 is very easy to evaluate---all options from one position have the opposite value (otherwise there would be *2 states)---so all paths down the game tree have the same parity. All that's needed to create PSPACE-hard instances, however, are states with the value *2. Why is this? Well, we can change the rules for any game with higher values into an equivalently-winnable* game with maximum nimber *2. The plan here is to break down the options from a game when there are more than two. We do this by splitting up those options into separate groups, then preventing the other player from being able to make any choices.

For example, we can transform the game G = {A, B, C} into the game G' = { { { A, B } }, { { C } } }.

Here's a diagram of the game trees, with the nimbers *0, *1, and *2 respectively assigned to A, B, and C.
For positions with more than 3 options, this splitting can be done in a recursive fashion. Thus, games with any range of nimbers can be transformed into a game with maximum nimber of *2.

* game states may not have the same value, but winning strategies from one translate into winning strategies in the other.