Completely forced checkmates

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blueemu
paper_llama wrote:

I still think white can force a mate with my sequence of moves, just maybe not as fast or something.

That turns out NOT to be the case.

Your proposed line "loses the opposition", so to speak.

paper_llama
blueemu wrote:
paper_llama wrote:

1.Rd8 Bd5

2.Be4 Bb7

3.Rf8

Take it.

Ok, now I see.

1.Rd8 Bd5

2.Be4 Bb7

3.Rf8 Bxe4

4.Rf3

And of course black doesn't have to capture the rook... hard to get my brain into losing mode heh.

paper_llama

So the rook can't be on the same file as the bishop and can't go "behind" it (until Rh8 wins).

If it goes on same file or less, then black can capture and there is no forced mate. When the rook is in front (or on h8) then capturing the bishop results in checkmate. That's the initial logic of the puzzle.

blueemu

Think of it as an "opposition" or "triangulation" problem. The question of White or Black to move is critical. EVERY POSITION that is winning for White is only drawing instead if the wrong side is on the move.

paper_llama
blueemu wrote:

Think of it as an "opposition" or "triangulation" problem. The question of White or Black to move is critical. EVERY POSITION that is winning for White is only drawing instead if the wrong side is on the move.

Sure, like corresponding squares. Using your numbers I could solve the puzzle, but I didn't understand it until I found out why the incorrect moves don't work.

Splunker33

I have been wondering if this is possible. Thanks

MaestroDelAjedrez2025

I'm pretty it's possible