Mdthbs

Tempoverlustspiel-A German word that roughly translates to "loss of tempo game." An interesting chess term that I leaned a while ago

With White to move in the below position, how many moves will it take, with both sides helping each other out, for Black to help White lose a tempo , and reach the same position, but with Black to move instead?

This was created by Michael Schreckenbach, and published in Die Schwalbe on 04/2015. I pulled this from the Schwable Chess Problem Database website.

Have fun, and no looking up the solution please!

At risk of being wrong because of a mental mistake:

It's easy to reason that the total amount of moves by White must be even:

The only moves White can make without irrevocably changing the position are 1. Re1, 2. Rd1, 3. Re1... ad infinitum. This means Black needs to recreate the current position of the Black pieces in an odd number of moves, as to allow (even):Rd1.

Given that

Black's only piece that can break parity is the King, since the Bishops are both effectively trapped as they can never capture

our first Black moves are

Bg5-h4-g3-h2-g1, followed by 6... Kh2-g3-h4-g5-h6-h7-g8-f7-e8-d8-c8-b8-a8, 19... Bb8, 20... Ka7-a6-b5-a4-a3-b2-c3 and 27... Kd4 (!)

at which point we can finally lose our tempo. Since White's next move is 28. Rd1,

we can play 28... Ke5, 29. Re1+ Kd5 30. Rd1 Kd4, and since our path from the starting position to Kd4 took us 27 moves, so will the inverse.

for a total of

57 Black moves and 58. Rd1 recreating the initial position?

Well, I'd thought to prove that it was insoluble

Pawns

All pawn positions are locked in place. They can neither move nor be taken.

Black Bishops

The bottom black bishop is locked into place. He can't move at all without taking a pawn. The black bishop on a7 can only move back and forth between there and b8. He always consumes an even number of moves to return to origin. The black bishop on h6 can move to g5, h4, g3, h2, and g1. He always consums an even number of moves to return to origin

White King

The white king is locked into place by the fact that he cannot take a pawn and the fact that he cannot move into check.

Black Knight

The black knight is locked into place. All of his available moves would either involve taking a pawn, or involve putting the white king in check - which would require the white king to take a pawn.

Rooks

- The only white piece that can move is the rook, and he can only move between d1 and e1. If he attempts to leave the 1 row, he puts his king in check from the black rook, and the black rook cannot escape the pin without first moving to d1, which would force white to make a non-reversible move. he requires an even number of turns to return to origin. Thus, the black rook is also locked in place.

Movable pieces

the only movable pieces are two of the three black bishops (who can only burn even numbers of turns), the white rook (who can only burn even numbers of turns) and the Black King (who therefore must return to his starting position in an odd number of turns in order to solve this).

...and this is the point where my proof falls apart and I realize that @Braegh has the right of it because...

The king can sneak by the white pawns at the top if you move the black bishop all the way down first.

Ah, well.