Yes, the disambiguating stuff involves the existence of other pieces. Yes, BigDoggProblem got it correct. (The bottom-right part apparently doesn't work as intended, but not a major issue there.)
OK, my bad. I might give it another try later.
EDIT: just to clarify. The order of preference for disambiguation is the usual file, rank, file and rank?
Yes. (I actually had the Wikipedia article "Algebraic notation" opened while composing to make sure of that.)
By the way chessniu's has 0 invalid moves if I count correctly. Will post position later.
So many things!
@BigDogg: Hm. Intention was "white to move" (and indeed should be sound) but the black to move cooks are also interesting. It should be soundly twinnable, but not in the current form. I might look further into it.
@chaotic_iak: I think I can confirm your monstrosity is sound. Everything in BigDogg's solution is determined except the precise nature of the bottom right corner, but it doesn't affect the solution. Reminds me of regular logic puzzles how the positions of the bishops are absolutely determined.
Also, nice solving for mine. The key idea was that fxg5+ and gxh5+ cannot both be legal at the same time - but this assumption cannot be made of fxg6+ and hxg6+! (It happens that my puzzle didn't exploit this asymmetry and the logic held; I'm thinking of how to showcase that effect which should be obvious when I mention it.)
@chessniu: Oooh, looks fun. 0 invalid moves - and that's with both white and black sets simultaneously. (Position is legal with either side to move.)
Black has 8Q 1R, white 7Q 1R. White made 6 promotions, black made 7, and as we all know at least 5 captures total are needed for that to happen, which is perfectly legal.
fxg6+ and hxg6+ clearly can be both valid. But if you meant fxg6+ and gxh6+, I'm still not sure how both can be valid, even if gxh6+ is en passant. Move one file to the left (exf6+ and fxg6+) and I can see that: Bh6 and Re6 checking Ke3 is possible.
Oh right. Wrong files. That's even why I changed it to the h file. (Originally fxe and gxf, but realised the cook).
Qaxh8 Qaxh1 Qbxh8 Qcxh8 Qcxh1 Qdxh1 Qexh1 Qfxa1 Qgxa1 Qhxa1
how many moves must be illegal
So many things!
@BigDogg: Hm. Intention was "white to move" (and indeed should be sound) but the black to move cooks are also interesting. It should be soundly twinnable, but not in the current form. I might look further into it.
@chaotic_iak: I think I can confirm your monstrosity is sound. Everything in BigDogg's solution is determined except the precise nature of the bottom right corner, but it doesn't affect the solution. Reminds me of regular logic puzzles how the positions of the bishops are absolutely determined.
Also, nice solving for mine. The key idea was that fxg5+ and gxh5+ cannot both be legal at the same time - but this assumption cannot be made of fxg6+ and hxg6+! (It happens that my puzzle didn't exploit this asymmetry and the logic held; I'm thinking of how to showcase that effect which should be obvious when I mention it.)
@chessniu: Oooh, looks fun. 0 invalid moves - and that's with both white and black sets simultaneously. (Position is legal with either side to move.)
Black has 8Q 1R, white 7Q 1R. White made 6 promotions, black made 7, and as we all know at least 5 captures total are needed for that to happen, which is perfectly legal.
I thought there would be nowhere for the kings but i was wrong
Bxe4++ Rxd2++ Kf2+ cxd4+ dxc4+ 0-0 R1h3+
#69: Should be 4 invalid moves; I'll see if this is achievable. (Only two queens can attack the same corner square without being on the same file as the target square, so at least one to h8, two to h1, one to a1 are illegal.)
#71: What does ++ mean? Double check, checkmate, or something else?
#69: No matter whether the moves are done by one or two sides, only two of each move to a corner are possible. (Not three because the third requires being on the same file, and there's no move from a file to the same file.) So we already have four invalid moves.
Suppose we can get away with four invalid moves, so each corner has two valid moves. Now, consider the two moves to h8 and to a1. Both of them uses the same a1-h8 diagonal. But the one going to h8 needs to come from a1, b2, or c3, while the one going to a1 needs to come from f6, g7, or h8; they obstruct each other. So this is also not achievable; we need to discard one more move along the diagonal to five invalid moves.
The same applies to a1 and h1; they both use the first rank, but the one coming to h1 must come from e1 or further left while the one coming to a1 must come from f1 or further right. So we also need to discard another move to six invalid moves. This one is easily legal, with valid moves being Qaxh8, Qaxh1, Qcxh1, Qfxa1.
@chaotix_iak -- '++' mean double check.
The problems should state who to play... In all cases, I always assume White is to move (and so I only consider the moves as White's), unless specified otherwise by the problem.
Solution to #71
cxd4+ and dxc4+ cannot be both legal; see my solution to Remellion's problems.
If 0-0 is legal, then wK is on e1, wR on h1, and none on f1 and g1. If Kf2+ is also legal, then it discovers a check along the first rank (no other line), and cannot discover to the right, so bK is on a1/b1/c1. This makes Rxd2++ illegal. So among 0-0, Kf2+, Rxd2++, one is illegal. So now we have two illegal moves. Let's try to do this.
Note that Bxe4++ and R1h3+ must be legal. In both cases, the piece moving is the one giving check (Bxe4++ because it's a double check and there's no way for two checks to be discovered at once, R1h3+ because moving away from h1 cannot discover any check), so Black king must be on a square attacked by both: d3, f3, or h7. (Not h1 since wR occupied h1 before doing R1h3+.)
This means 0-0 and Kf2+ cannot be both legal, since we knew if this is the case then bK is on the first rank. So Rxd2++ is legal, which means a wR on d2 will also give check, so Black king must be on d3. This disallows dxc4 for obvious reasons (d3 is occupied). This also disallows Kf2+ (since there's no way to discover check passing through e1 to d3), so we have the two illegal moves dxc4 and Kf2+. Are the rest possible?
Since cxd4+ is legal, we have a unit on c3 and d4. Now, Bxe4++ can only discover an orthogonal check (either a rank check or a file check). Since c3 and d4 are occupied, Bxe4++ can only discover along e3-h3 or along d2-d1. The latter is impossible; no way for the bishop to get to e4 from d2 or d1, so there's a discovered check along e3-h3. But this means R1h3+ should not be a check (stopped by wB on f3). So we cannot have two illegal moves, but we can have three easily.
Remove Kf2+, dxc4+, R1h3+. The rest (Bxe4++, Rxd2++, cxd4+, 0-0) are possible:
how about fairy chess pieces?
Exf6# exf6# Bxc6 bxc6 Wxf8# Axb8# axb8=Q# e4 Pxg6# Rxb7 Gd2 Gxf8 Wg7# Gxf8=Q# Rhh5 fxg4 Fxg4
E=empress=rook+knight
P=princess=bishop+knight
A=amazon=queen+knight
W=wazir=a rook that can move only 1 square
G=grasshopper=a queen that jumps over exactly 1 piece and lands on the square after it
F=ferz=a bishop that can only move 1 square diagonally
What is the starting position? Or are fairy pieces only obtainable through promotions?
this doesnt have to be legal so the fairy pieces are from promotions
there are only 6 fairy pieces anyway
Wh7+++ 0-0-0 Gd1 Gc1 Bb2 Rxf7 Wxf7 Nxf7+ Wg8+++ for white
c1=Q+ R7f1 Axa1+ Ab6xc8 Fxf2+ Fxd2+ Ac3+ Rb7 for black
4 invalid
@chaotic_iak: From your diagram to #71, -bPd4, +wRh4, wBf3->d5, wRg3->d6. Presto, cxd4+ illegal but R1h3+ legal. cxd4+ and dxc4+ have a NAND relationship not XOR - they can both be illegal at the same time.
@chessniu: WOAH fairies should not exist. #76 is a nightmare I can't quite optimise. #79 has multiple solutions, and I'm trying to envisage a cook but can't quite make it.
R7f1 is invalid no matter what. At least one of 0-0-0 and c1=Q+ is invalid. Wxf7 should be a check if Wh7+++ and Wg8+++ are both valid.
(Fairy units are white Gc8, Gh5, Gb3, Wg7 and black Aa6, Ab6, Ab8, Fe3, all other units are orthodox) This has Wxf7, ...c1=Q+, ...R7f1 and ...Ac3+ illegal. A discovered wazir check is met by black ...K*g8, and Nxf7+ ...Nxf7.
Replace the white Pc2 -> black Pc2 to get 0-0-0 illegal and ...c1=Q+ legal.
Cleverness gets another set of illegalities.
This time it's (white Gc8, Gh5, Wg7, black Aa1, Aa6, Ab6, Ab8, Fe3, Ga2), and 0-0-0, Wxf7, R7f1 and Axa1+ are illegal. ...c1=Q+ is a discovery of the Ga2, ...Fxd2/f2+ are discoveries of Re4.
It would be nice if the problems are set such that at least the invalid moves, if not all the piece locations, are uniquely determined.
2 quickies.
(1 invalid) Ka1+, Ka2+, Kb1+, Qg8+, Qh7+, Qh8+
(1 invalid) Na1+, Na2+, Nb1+, Rg8+, Rh7+, Rh8+
I can't do complex compositions, so I'll stick to what I know: simple, cute, hopefully uncooked. Still thinking about the twinning of the earlier one.
Yes, the disambiguating stuff involves the existence of other pieces. Yes, BigDoggProblem got it correct. (The bottom-right part apparently doesn't work as intended, but not a major issue there.)