Solutions for Quantum Logic Chess Problems

These are terrific problems - full of concepts, tactics and humour.

Here are some solutions mostly provided by Indian enthusiast, S.Mukhuty! Thanks so much for sending them!



En passants play quite a humorous role in this cute little problem. In the first part, mate comes with a cross-check 1.Bxd5+ e4 2.fxe3ep+ Nf3# while in part (b) mate comes with an en passant capture 1.Bf7 Ng4+ 2.e5 fxe6ep#
Quite impressive how both the sides get to make an en passant capture in the two phases of play.

Part (a) is simple 1.Kf4+ e4 2.dxe3ep+ Nd3#
But (b) is considerably harder, there is no way you can prove that the last move was e2-e4 and hence 1.dxe3(e.p) isn't admissible.
The solution is 1.Qf6 cxd4 2.Be6 dxc5 3.Kxe5 Nd3#

The solution for part (a) is 1.f6 Kxc5 2.Kf7 Nxh6+ 3.Ke6 Nf5 4.Bf7 Ng7# but the problem would have been a trivial h#2 if black could castle long: 1.0-0-0 Kb6 2.Bb8 Nxe7#.
But why can't black castle? Well, white's pawn structure on the Q-side suggests 4 captures by white pawns, accounting for all of black's missing pieces. Evidently, bRh8 must have come out of the pawn chain after bPg7×h6 happened, implying bNg8 was on f6 at some point checking bK. Hence, bK must have moved!
In part (b) of the problem, with bPf7-f6 the bR gets some room to escape the pawn chain via f7 & g7. Therefore, in this case the wN could have gone to g8 simply via h6 without disturbing the king and the bBe5 could be a promotee. Here indeed, the trivial solution works 1.0-0-0 Kb6 2.Bb8 Nxe7#

For part (a) 1.Ke8+ g4 2.hxg3+ Kg2 3.Bd7 Rh8# works. For part (b), with bKd7 moved to e8 and wPg2 moved to g4, neither can we prove that white's last move was the double-step pawn push g2-g4 nor can we disprove black's right to castle long. Therefore, the solution here is 1.Bd7 Rb2 2.0-0-0 Ne7#
Note: Unlike {K} no 'A Posteriori' rule is not required here.

(a) 1. Qg4+ f5 2. gxf6ep#
(b) In the position after the first two moves, we can no longer be sure that the e.p. is legal. So there is no #1, and instead the following #2: 1. Qc4+ d5 2. cxd6ep#
(c) Again, in the position after the first two moves, we have "forgotten"; whether e.p. is legal. So the #1 is gone. But another #2 beckons: 1. Qe2 (returning to the start square) then 1...d/f4 2. Qc/g4#.


The first question that pops up is whether white can go 1.0-0-0 with the idea of 2.Rf1+ followed by mate. Also note that even 1.d4, threatening 2.Rxe5#, is an option. If 1...e5 then after 2.dxe6 (e.p) either 3.Re5# or 3.Rf3# follows. But which of these two moves actually work or are they both correct?
Well, let's analyse: Firstly, the wBc1 never left it's home square so wBa5 is a promotee. Next, wPe2 (now on d5) & wPf2 (now on g5) made one capture each. Also, the wPa2 had to make a capture before promoting on b8, this captured unit must have been nothing else but the bPf7 which itself got promoted on f1 prior to that.
Thus, wKe1 must have come under a check with a bP on f2 and therefore had to move, implying that 1.0-0-0 is illegal. Hence, 1.d4! is the only key and the problem isn't a cook!
In part (b) however, with wPd2 on d4 & bPe7 on e5, 1.0-0-0 is possible (because now wBa5 don't have to be a promotee!) and is actually our required key.

The first part is basic, 1. Ke1 is the key move that threatens 2.Rd8#. Black's only way to thwart this is 1...Ke8 but after 2.Bb6, 3.Rd8# is unavoidable.
Now for part (b) with white and black kings moved to e1 and e8 respectively we have to analyse the provisions of castling for both sides. It is clear from the position that white made five pawn captures on dark squares (a3,c3,b4,a5,g3) and black made three pawn captures cxb, dxc, cxb accounting for all of white's missing units. Now, with regards to the missing wPe2 & bPe7 there are two possibilities:
(i) wPe2 made two captures to be captured on the b-file, in which case bPe7 had to promote to e1 to be captured later by white's pawns. Clearly 0-0 for white isn't legal in this case.
or (ii) bPe7 did not promote, in which case it couldn't have been captured by white implying that wPe2 promoted and contributed to one of the three captures by black pawns. And clearly in this case 0-0 isn't possible for black.
Thus, this is a scenario of mutually exclusive castling rights.
Hence, the key for part (b) is 1.0-0! with the idea of 2.Rfe1 & 3.Rd8#. Note that black can't defend with 1....0-0 anymore after white has castled. If 1.Bb6? (Threatening 2.Rd8#), then 1...0-0! is legal and refutes mate in 3.

Part (a) is easy 1...Rxh1+ 2. Kg5 f4# and the (b) part uses the so called 'A posteriori' rule. With bK under check it was clearly white who moved last. We have only two legal retracts namely, Pf2-f4+ & Pf3-f4+. If the white's last move was Pf3-f4+ then black's last move could only have been bKh4-g5 (king moving out of a check) prior to which white must have moved the wRh1. Thus, white can have any provision of castling only if the last move was Pf2-f3+ (and en passant capture immediately next is legal) because this leaves room for the bRe3 to be retracted on f3 etc. Now, in this scenario the 'A Posteriori' convention states that you can take the P en passant, provided that you validate this capture by actually castling later in the "game". Then the castling move is the a posteriori justification that the en passant capture was indeed legal. Hence, exploiting this rule, our required solution is 1. gxf3 (e.p) 0-0 2. Rxe2 Nh1 3. Rg2#.

S.Mukhuty also comments:
Probably {H} is a cook. Both 1.Rd7 & 1.Re8+ here leads to mate in 5. I am yet to solve {A} & {B} and problem {G} has already been solved by you.
I found this article very entertaining, it serves as a very good introductory text for anyone who wants to learn the conventions regarding castling and en passant in retroanalysis.
The idea of so called "Amnesia" that you have written about is fascinating. Of course, generally one would expect a h#n to convert into a h#(n-1) after the first move of the solution has been played out. But the problems from {C} to {K} goes on to show how the knowledge (or ignorance) of the game history can seemingly muddle up this "obvious" arithmetic. OTB players hardly realize the tricky retro-analytic nature of en passant and castling when they play chess.

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