Hardest/Impossible Composition Challenge?

@Demonolith
Almost. The rule still exists, and is applied in almost every game. But it was a bit unclear, so in 1972 it was amended. Which destroyed the beautiful solution to the puzzle.

@Panagrellus So if I have never heard of this rule, I cannot solve the puzzle? I have not studied the evolution of the rules in chess. I am just playing with what I was taught.

@Demonolith
Don't overthink it, it's a bit silly.

I'll give you another hint:

This is the pre-1972 rule you have to apply:

Castling:
“The king is transferred from its original square, two squares toward the rook; then that rook toward which the king has moved is transferred over the king to the square immediately adjacent to the king.”

Requirements: Neither the king nor the rook has moved previously, and the king passes through no square guarded by Black.

So it was possible to caste with any rook on the board even vertically?

@Demonolith
Exactly, if the rook hasn't moved (such as after being born at a promotion).... Well, at least in theory it was possible, if you take a very legalistic view on chess rules. I think no one ever dared to try this in a real game. I can imagine that the opponent either would slam you in the face or get a heart attack when you attempted to castle vertically in a tournament game.

@Panagrellus It is a nice solution to the challenge, they should have kept this rule for nice endgames!

I can't resist to post this study, after I have spent too much time trying to solve this impossible riddle.

lichess.org/study/lPkz9EEY

One of the jesters in this study told me that the task which was described in #1 is a mission impossible.

I hope you will not punish him for telling you the truth.

@AyrtonTwigg

Well i found something, but the challenge author took the time to write "in regular chess"... so i guess it doesn't count.


Edit : found an underpromotion in bishop and rook in antichess. Maybe it's possible in other variants ( impossible in racing kings, probably imposs in horde too, same for crazyhouse. maybe possible in atomic and koth but i don't see the point of underpromotion in these variants, and maybe possible in 3-checks too)

Edit #2 : I uploaded a position in the 3rd chapter on the study above ; i think it solves the puzzle as it does not involve "avoiding stalemate" as post #1 says (the goal of the underpromotion is in fact to provoke stalemate). Correct me if i missed something.

@ayan2007
@AyrtonTwigg
@Bishop1964
@classicalMpk
@dboing
@Demonolith
@KingPuzzles
@MoistChess
@Panagrellus
@Sarg0n

There is a simple proof of the impossibility of an example in the challenge.

1. A queen combines the power of a rook with the power of a bishop. (Definition, Basic Rule).

2. No advantage obtained by a weaker piece could not be obtained by substituting a stronger piece for the weaker piece in the same position, if the stronger piece includes all the powers of the weaker piece, and if the extra power does not cause a stalemate. (Tautology).*

3. No advantage using a rook or bishop cannot be obtained by substituting a queen for the weaker piece if the extra power does not cause a stalemate (From 1, 2, and 3).

Therefore

4. No promotion of a pawn to Bishop or Rook can be more advantageous than promotion to a queen, if the extra power does not cause a stalemate. (From 3 and definition of promotion, Basic Rule).

*Note: while a knight has less power than a queen, a queen does not include all the powers of a knight. So not every substitution of a knight by a queen will be at least equal in advantage, but every substitution of a bishop or rook by a queen will preserve at least equal advantage, barring stalemates.

@nayf well look at the 3rd chapter of the study in post #28... h8=B is the best move (every other move loses) and the purpose of the underpromotion is not to AVOID stalemate, thus the position meets the requirements of the problem, and is a solution.