Are variants that are wins or losses easier to solve than ones that are draws?

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acgusta2

I noticed that sometimes in the case of a forced mate there is just one line of best moves.

For instance in this case the line shown is the only line, in which every move is a best move as if on move 1 black was to move the king to the g file that would blunder a mate in 1 and any other moves white could play would also lead to a slower checkmate.

In this position however there are a lot of different move sequences that would involve both players playing best moves, in terms of maintaining the book draw, so it looks like this position should take a larger search tree to solve by brute force than the previous position.

So what I wonder is whether in general variants that are forced wins might be easier to solve, using brute force ,than variants that are draws with best play of similar complexity.

pds314

It's hard to know because chess isn't solved. We don't know if chess, when played perfectly, is a draw. In fact, we don't even know it's not a win for black. And most non-trivial variants have had exponentially less work put into solving them than chess. Indeed, chess isn't solved even for some very strong positions. e.g. I have never seen any proof that giving white 7 Queens at the start and a row of pawns in front against the normal starting position is winning for white. +50? Sure, but not Mate in N moves. There is actually no absolute proof that white has a winning strategy even with a ton of queens.

So you can see the issue here. Any chess variant with a ton of pieces that is remotely balanced and doesn't have obvious winning strategies (like in chess with the knights replaced with 2,5 leapers where white has mate in 1 from the starting position. Perhaps the only situation where a 2,5 leaper is actually useful).