what is the longest tablebase mate with special properties?

[Uri Blass]

Examples:

[d]2k5/8/2K5/2R5/2p5/8/8/8 w - - 0 1

This position is a mate in 3 with a single path for white for the shortest mate.
complexity of it is 2 because there are 2 ways for black to respond in the first move.

This is not actually a single path for White. If Black responds to 1.Rxc4 with 1... Kb8, White has two ways to respond (Ra4, Kb6) with equal DTM.

This is fixable, though: just shift all the pieces to the d- or e-file.

Uri, did you mean a single move for the first move only, or a single move for all of whites moves?
Both are interesting questions!

Andrew

I meant for all moves for the winner when the moves of the loser change nothing and the comment that it is not actually a single path but easy to fix is correct.

[Uri Blass]

[d]k7/P3p3/1q2Pp2/K1p1pP2/1pPpPp2/1P1P1P2/B7/b7 w - - 2 1

This is probably mate in 11 with a single path for the winner.
The loser has a lot of option and did not calculate complexity for it.

It is not a single path to win except the first move.

[jp]

[d]r6R/k7/p2K4/P7/8/8/7P/8 w - - 0 1

This is mate in 9, with a single path:

1.Rxa8+ Kxa8 2. Kc7 Ka7 3. h3! Ka8 4. h4 Ka7 5. h5 Ka8 6. h6 Ka7 7. h7 Ka8 8. h8=Q+ Ka7 9. Qb8#

(1... Kb7 allows mate in 6, so the complexity is only 1.)

Strictly, though, the tablebases don't show at the start that 1.Rxa8+ is the shortest mate, because for 7 pieces we have only DTZ. DTMs are shown only after the exchange.

1 Rh7+ ? (mate in 3)

Oh, Andrew & RIchard are right. I got fooled by the (DTZ) tablebase.

Here's the safe but somewhat ugly fix:

[d]q6R/k7/p2K4/P7/8/8/7P/8 w - - 0 1
Mate in 9 with a single path, as previously claimed.

[Note: I see Uri's fix on the bottom of the previous page too.]

[jp]

trying to fix it to single path but it is not good enough because after 1.Rxa8+ Kb7 it is a shorter mate
so it is not a single path based on my definition.
[d]rR6/k7/p2K4/P7/8/8/7P/8 w - - 0 1

It's true that 1... Kb7? allows the shorter mate e.g. 2. Rh8 (or Rg8, etc.) Ka7 3. Rh7+ Kb8 4. Kc6 Ka8 5. Kb6 Kb8 6. Rh8#...

But surely it is single path (or the definition should allow it to be)?
Because 1... Kb7 is not Black's best move. I'd want the definition to mean that the defender puts up the stiffest opposition.

I was assuming the same (best defensive moves) for the complexity definition too, i.e. that there are multiple paths only if the defender's multiple moves all have the same DTM. Maybe we need extra definitions.

[jp]

Extra terminology:

Let me distinguish different complexities. Loser complexity is the complexity of the loser's moves. Winner complexity is the complexity of the winner's moves.

Then in Uri's composition

[d]k7/P3p3/1q2Pp2/K1p1pP2/1pPpPp2/1P1P1P2/B7/b7 w - - 2 1
(#11 with a single path for the winner)

the loser complexity is high, but the winner complexity is only 1.

In the position

[d]3k4/8/3K4/3R4/3p4/8/8/8 w - - 0 1

loser complexity = winner complexity = 2.


In general

Winner complexity must be less than or equal to loser complexity (because of the requirement of a single path).

If (loser complexity > 1) & (winner complexity = 1),
then the loser's moves are irrelevant to the winner -- he plays the same moves regardless.

[jp]

It seems problemists call single-path problems "dual-free", so I'll switch to that terminology.

What I'd like to see is dual-free mates with high winner complexity (at least greater than 1, but greater than 2 would be better still).

[jp]

[d]3k4/8/3K4/3R4/3p3r/6P1/8/8 w - - 0 1

Mate in 6, dual-free (with best play only),
winner complexity = 2.

1. gxh4 Ke8 2. Ke6 Kf8 3. Kf6 Kg8 4. Rxd4
Any deviation so far by White (Black) allows slower (faster) mate.
But now Black has a choice:
a) 4. ... Kh7 5. Rd8 Kh6 6. Rh8#
b) 4. ... Kh8 5. Kg6 Kg8 6. Rd8#.