Hardest Mate in 4 of All Time

I apologize if this has been posted before, but I was just so enthusiastic about the following puzzle (which is not mine by the way) to the point I couldn't help but share it with you guys.

Believe it or not, there is a mate in 4 in this position.

The fun thing about it is that not even Stockfish could help you come up with the correct solution, and as such I highly doubt any of you will be able to figure it out ( and hence why I called it the hardest mate in 4 of all time).

I will give hints at a later stage.

Now enjoy!

I found a mate in 5

Could you post the solution please?

I will give you a clue when enough people have tried.

And finally, in case nobody can find the solution even after the clue's been posted, I will then reveal the solution :)

Please, remember to post your solutions/what you have found out so far :)

how the white pawn reached b6?

Mate in 2:

b7 Kd8

b8 Promote to Rook #

Pretty easy, it came from e2 then c5. Black c5 pawn comes from e7. There are a lot of pieces missing so you have more than enough captures available. I can post a proof game if you want.

Except black can castle. If you can't disprove that a side can castle and if the puzzle description doesn't say otherwise, you have to assume that side can castle.

This solution is incorrect.

The real solution will be totally mind-blowing :)

The real solution will be totally mind-blowing :)

In that case, the solution must be:

1. Qc4! bxc4 2. O-O!! any 3. b7 any 4. b8Q# (or 4. b8R#)

The hard part (which I haven't worked out yet) is showing that if White can castle, then Black can't.

Not sure I can do it in 4, but Qh3 looks like a quick win.  If black doesn't castle, the b-pawn promotes.  If he does castle, Qxd7 should do the trick.

I can see mate in 5

1)Nf5  0-0 2)Qe2 Kh8(if Re8 then Qxe8+#) 3)Qe7 Rg8 4)Rxh6+ gxh6 5)Qh7#

White is missing R+B+B+P while Black has made two pawn captures.

If we assume that White can castle, then White's R and black-squared B never got out, so Blacks pawn(s) captured the white-square B and P.

The captures must have occured on b6/b5 and c6 (not d6 and c5, because it's a white-squared B). For this to happen, the White pawn must first have been promoted.

We can already see 5 pawn captures by White, and Black is missing only 7 men (6 pieces and 1 P), so there are only two spare captures available to promote the White pawn. This means that the promotion must have occurred either on f8 (from f7) or on h8.

Either of these possibilities means that Black can't castle.

Good stuff.

I assume that this type of analysis is exactly why Stockfish cannot solve the problem.

The laws of aesthetics outrank the laws of energy. Any hard problem that is not easy to solve, is ugly. This one takes no more than 5 seconds and that speaks in its favor.

caveatcanis #13 2 hrs ago
White is missing R+B+B+P while Black has made two pawn captures.
If we assume that White can castle, then White's R and black-squared B never got out, so Blacks pawn(s) captured the white-square B and P.
The captures must have occured on b6/b5 and c6 (not d6 and c5, because it's a white-squared B). For this to happen, the White pawn must first have been promoted.
We can already see 5 pawn captures by White, and Black is missing only 7 men (6 pieces and 1 P), so there are only two spare captures available to promote the White pawn. This means that the promotion must have occurred either on f8 (from f7) or on h8.
Either of these possibilities means that Black can't castle.

Which means that my "Mate in 2" solution is correct.

@Bawker I suspect that it can be proven that either white can castle or black can castle, but not both.

I will post the solution soon enough!