2

The puzzle (this may have a known answer) is to construct a position such that:

  • If Black makes move A, it's mate in 1 for white.
  • If Black makes move B, it's mate in 2 for white.
  • ...
  • If Black makes move X, it's mate in N for white.

For instance, in this position

6k1/7p/5K1P/8/8/4Q3/8/8 b - - 0 1
  • If 1..Kh8 2. Qe8#
  • If 1..Kf8 2. Kh8 Qe8#

So this position has N=2.

What's the highest possible N? (All of the 1-N moves must exist)

Extra details:

  • It doesn't count as having mate-in-X if white also has a shorter forced mate.
  • It is acceptable for white to have several different mate-in-X for a given black response.
  • It is acceptable for black to have additional moves available that don't lead to any particular mate-in-X.
  • It is acceptable for black to have two different moves that both forced mate in the same number of moves.

(It might be considered "nicer" to have solutions that don't rely on the last three points though).

5
  • From the standpoint of a problem composer like me, your question should be more precise: Should White, on a certain black defense, have a mate in N, a mate in N with no shorter mates (I strongly guess this is what you mean) or, in problem spirit, a single mate in N? I'm fairly sure your puzzle was already pondered by problemists (especially the N=4 case after the four moves of a black pawn on the 7th or 2nd rank) and ask the experts. May 18, 2022 at 7:54
  • And should the number of all possible first black moves be also N? May 18, 2022 at 8:01
  • @HaukeReddmann the second one - no shorter mates, but multiple ways of mating in N is fine. Also it's ok if black has additional moves that don't lead to unique mate lengths. May 18, 2022 at 23:17
  • @HaukeReddmann I've updated the question to be more specific. May 19, 2022 at 0:56
  • THX for the clarification. OK, in that case my N=4 demo below is completely valid. My own constructions, and possibly that of the experts, will be found there: matplus.net/… May 19, 2022 at 6:38

2 Answers 2

2

A N=4 demo (board from Black view), not necessarily the stuff that you wanted, depending on your demands.

[FEN "8/6B1/8/8/B2Rp3/8/3p1n1n/K1k5 b - - 0 1"]

If Black plays...

~: 1.Rc4#

d1B: 1.Bh6+ e3 2.Bxe3#

d1Q: 1.Bh6+ e3 2.Bxe3+ Qd2 3.Bxd2#

d1N: 1.Bf8 (threat Ba3) Nd3 2.Bh6+ e3 3.Rxd3 Nd1~/~ 4.Bxe3/Rxd1#

Observe this isn't perfect from my view: ~ is not exact, since Black could play Nd1 with a #2. (I'd prefer ~ is a single move with a #1.) In the #3 line Black might shorten it with Qd2. Also, even worse, Black can play d1R, leading to a dual Bxd2/Rc4 in the mating move. In the #4 variant, the line splits again. It shouldn't since it obscures the theme. But at least all White moves are unique.

Thus, again, please clarify the exact demands. My own maximum demands on the task would be

X%: 1.A#!

Y%: 1.B! Y2! 2.B2#!

Z%: 1.C! Z2! 2.C2! Z3! 3.C3#!

and so on, where "%" means "all of the moves, but choice possible" and "!" only black move/only White move that mates in the required number of moves. In contrast, my minimum demand would be:

X: 1.A#!

Y: 1.B! Y2 2.B2#!

Z: 1.C! Z2 2.C2! Z3 3.C3#!

i.e., only that some forced unique lines with n=1,2,...,N exist.

EDIT:

[FEN "8/k1K5/p1R5/8/8/P7/8/8 b - - 0 1"]

"Superexact" N=2 example. All moves are forced (Black's as they are the only legal ones, White's as they are the only fulfilling the stipulation, also no #2 exists in the #1 line!): 0...Ka8 1.Rxa6#, 0...a5 1.a4 Ka8 2.Ra6#.

1
  • Yeah, this works, but I suspect the maximum N is much higher. I fiddled around a bit and made a position with N=6 except it was missing a mate-in-3. May 19, 2022 at 13:07
0

(delete - whereas it's a mate in 1 to N, it is a consecutive one, not a split on Black's 1st move! Sorry!)

1
  • Are you able to clarify how that link is an answer to this question? I'm having trouble understanding what I'm looking at. May 21, 2022 at 8:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.