# Can white checkmate in 1 move?

How can this position be reached (via a valid chess game) so that white can checkmate in 1 move?

S.Nichto,2024

• a) the minimum number of moves in the proof game?

• b) the minimum number of moves by queens in this case?

Good luck!

• Right question: If white's turn to move (Qb4#), which was the last black's move?
– z100
Commented Jul 16 at 17:27
• This is indeed a very correct question. Commented Jul 16 at 17:31
• @ЭлсиРинген I have edited your question to make it more explicit. I hope my change is okay with you. Commented Jul 16 at 17:34
• I'm not picky. Do whatever is more convenient for you. Commented Jul 16 at 17:37
• @ЭлсиРинген Your question is currently subject to a close vote. I edited your question hoping that it wouldn’t get closed. Commented Jul 16 at 17:49

Can it be White's turn?

Yes, it can be white's turn.

Reasoning backwards:
The last moves must be

Bb6,Ka5b5,Ba5,Kxa5
Thus, white must promote to bishop on a black square.
To make backwards) progress, the king cannot have come from B6, so white must have captured something which will turn out to be the e pawn.
White has 4 and (considering bxc6) black has 2 unaccounted captures left.

Regarding promotions of pawns,

White can only promote on d8 or f8 if black's g pawn gets into play.
Since only the rooks (and the out of position a-pawn) are available, gxf2 is unavailable since the rook cannot get to f2 before g2g3 is played.
Also a promotion leaves the a-pawn out of position (bxc6 will then not be available)
conclusion: This is impossible. So White must promote on b8!

After promotion to bishop on b8 promote one can only leaf though a7, and all this must happen before the knight is moved back with Nc6b8, which must be before bxc6.
Due to this order:
- This requires bxa7 axb8B
- The white's white bishop cannot get to a8. This must also be a promotion, which requires an additional 2 captures.
The latter means a black pawn must be utilized on a or b, i.e. be promoted. This accounts for all captures, which means whites last capture must be a pawn on its starting rank.
Since that capture must be with bishop on b6 that capture must be on e3.

All this is possible, as in the following game:

1. a4 e6 2. b4 a6 3. b5 Qe7 4. b6 Ra7 5. bxa7 Qb4 6. axb8=B Bd6 7. Ba7 Nh6 8. Be3 Qb6 9. a5 Nf5 10. Ra4 Nd4 11. axb6 Nc6 12. Rc4 Nb8 13. Rc6 bxc6 14. b7 Ke7 15. bxc8=B Ke8 16. Bb7 Ke7 17. Ba8 Kd8 18. Ba3 e5 19. Qc1 Re8 20. Qb2 Re6 21. Kd1 Rf6 22. Kc1 Rf4 23. Qb3 Rb4 24. Kb2 Rb7 25. Nf3 Ra7 26. Qb7 g5 27. g3 g4 28. Bh3 gxh3 29. Rg1 Ke7 30. Rg2 hxg2 31. Nc3 g1=R 32. Ng5 Ra1 33. Bb4 Ra4 34. Bbc5 Rg4 35. Nh3 Rg8 36. Bb6 Ke6 37. Ne4 Kd5 38. Bg5 Rc8 39. Bd8 Kc4 40. Ka2 Ba3 41. Nhg5 Bc1 42. Ne6 Kb5 43. Nc3+ Kc4 44. Na4 Kb5 45. Nb2 Kb4 46. Kb1 Kb5 47. Ka1 Kb4 48. f3 Kb5 49. Ka2 e4 50. Ka3 e3 51. Bxe3+ Ka5 52. Bb6+ Kb5 53. Ba5+ Kxa5

• A brilliant analysis. Now you can try to find a shorter trial game. It takes about 40 moves. Commented Jul 18 at 8:04
• Ahh, very nice solution! I kept wondering what I was missing here ... Commented Jul 18 at 8:48
• Tim Seifert_Yes, I tried to hint to you that you have found everything you need for a solution. You just didn't pay attention that the mandatory capture of the figure to stop the wK from staggering back and forth can be used to good effect. There, out of about 30 possible captures, only one is suitable. Commented Jul 18 at 14:56
• Such pawn captures, although simple in themselves, are always the last thing that comes to mind. I know it by myself. Commented Jul 18 at 15:03
• Capturing the e3 pawn is the key moment of the decision. The minimum short answer in the retro entry is: 1. ... Kb5xBa5 1. Bb6-a5+ Ka5-b5 2. Be3-b6+ Kb5-a5 3. Bb6xPe3+ e4-e3 Commented Jul 19 at 10:51

Edit: This answer is incorrect. I have pointed out my mistake in the last paragraph.

I am a bit confused by the problem statement, so perhaps I made a mistake somewhere (my reasoning is complicated enough that it seems quite possible to me).

But assuming that my reasoning is valid

it cannot be White's turn right now.

If White is to move, what was Black's last move? The only possibility is

Kb5xBa5.

(No other black piece has the physical freedom to have made a move and every other previous square for the bK is clearly impossible. The check in the previous position must have been discovered and only a bishop can have uncovered it.)

Now, I claim that in order to free up the position, a black piece must have been captured in the recent past, either by the knight on b2 or by a dark squared bishop. (This is a bit fiddly to see and the part I am least confident about.) Assuming that this was not the case, the previous move by White must have been Bb6-a5+. Before that, the bK either came from c4 or a5 (the double check on c5 is impossible). In the former case, the wN must have just captured on b2. (Otherwise, the only possible explanation of the check from bB is via b2xc1=B, which requires altogether more captures from Black's e (or g) pawn than are available.)

Hence, bK must have come from a5 again, which places him in check from the wB on b6. Now, we may have many (sort of) repetitions in which the bK alternates between a5 and b5, always escaping check from a bishop and the (discovered) queen. Since each of these checks was just given, White has no time to free up some other black piece. Therefore, the only way to give Black a move other than the King swinging back and forth is to uncapture something. (Perhaps there is a cleaner way to reach this conclusion.)

As a consequence, White has only four captures available to promote the second dark squared bishop, which I believe to be impossible: Where did the promotion occur? There are two essential cases to consider. A promotion on d8 or f8 requires all available four captures by the b-pawn. In particular, this pawn must have eaten the bPg (or something it replaced later). This pawn can have made at most one capture however (as White's missing a-pawn cannot have left its file). This is only possible, if bPg7 captured on f2 to promote on f1. But this is impossible, as the piece consumed must have been a wR, which cannot have reached f2 at this point (because wBf1 and wPg2 were still at home).

Thus, the wP must have promoted on b8. But this is also tricky - in order to exit into the open, a6 must have already been played. In order for the bN to reach b8 afterwards, this means that bPb7 must have still been at home during this promotion. But this in turn implies that White's light squared bishop was permanently locked out of the cage, so that wBa8 is also promoted. Between them, White's a- and b-pawns must then have captured four times in order to both promote (although the exact distribution is not determined). But this now leads to essentially the same contradiction as before: Black's e- and g-pawns must both have promoted (Edit: Here lies the mistake - one of them may be the piece taken by the bishop at the end of the game!) while having only one capture available each. And still, none of the missing white pieces can have reached f2 to allow the entry of bPg7 while Bf1 and g2 were still at home.

As this rules out the last possible case, the starting assumption must have been wrong - it is not White to move.

(As I mentioned at the start, this seems confusing to me given the problem statement. But I don't know where my analysis went wrong, if it did at all.)

• It's not clear to me why black's pawns only have two captures available to them to promote (in your last paragraph of analysis specifically). White is missing two rooks and a bishop (assuming the bishop on a8 was a promotion) which would be enough. Was there another black capture that I missed somewhere? Commented Jul 17 at 15:38
• @msh210 I think it's because white pawns don't go downwards. Commented Jul 17 at 15:39
• @GoblinGuide Black's b-Pawn also captured, which leaves only two others. Commented Jul 17 at 15:43
• Oh right, I spent so much time reading your answer that I forgot all about that one. Commented Jul 17 at 15:44
• In order for bK to start riding the bus on the a5-b5 route, he bought a ticket, which revived the driver of the Ba5 bus. To stop carrying bK on the bus, the driver of the Ba5-b5-g1 bus must drop off any black stowaway at any black stop. Or bK has to jump off the bus himself on c4. However, there he will "get under a horse", from which a black rider on b2 will fall out of surprise. bK will run away under this noise. Commented Jul 17 at 23:35