I don’t believe your comment is true in any meaningful sense. Can you explain what you mean?
Details: It’s easy to prove that the first player wins in Hex without the swap rule, but it’s even easier to prove the second wins in any (deterministic, …) game with the swap rule. Neither proof is constructive, and so neither provides an efficient program.
Interpreting your statement differently, it’s easy to write a program that plays any (deterministic, …) game optimally. Just explore the full game tree! The program won’t terminate for a while, however, and this interpretation makes no distinction between the versions with and without the swap rule.
I’ve added Hex and it comes in at 377 bytes gzipped (without the swap rule), supporting your impression that it’s simpler.
I’d like to point out that without the swap rule it’s also very easy to write a program that plays perfectly.
I don’t believe your comment is true in any meaningful sense. Can you explain what you mean?
Details: It’s easy to prove that the first player wins in Hex without the swap rule, but it’s even easier to prove the second wins in any (deterministic, …) game with the swap rule. Neither proof is constructive, and so neither provides an efficient program.
Interpreting your statement differently, it’s easy to write a program that plays any (deterministic, …) game optimally. Just explore the full game tree! The program won’t terminate for a while, however, and this interpretation makes no distinction between the versions with and without the swap rule.
Oops. I was apparently confusing hex with bridge-it.
How well does this map to the human experience of the game? Do two experienced players need the swap rule for the game to remain interesting?
It depends on the skill difference and the size of the board, on smaller boards the advantage is probably pretty large: Discussion on LittleGolem