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MatPlus.Net Forum General Most Checkmates In One In A Position Without Promoted Pieces

### Most Checkmates In One In A Position Without Promoted Pieces

(= 13+3 )

Here’s a challenge for you all. Construct a position, with no promoted pieces, with the most mates in one that beats my 42 in the given diagram above that I constructed. I found that solution myself.

It must be legal of course.

For some suggestions, I would try to add in more queen moves or another pawn promotion, if at all workable. I was unable to do either of these myself.

The 42 mates in my position go as follows: Rg2-14, Be5-13, Nc5-7 Ne8-2, Qb5-2, Ph7-1, Pa7-1, Pd3-1. Pe3-1

Good luck!

Just you enter Capital letter F after opening 2 brackets and close the FEN with same 2 brackets....then diagram Visible in the thread...before that preview the diagram for correctness.

Thanks! And fixed!

The pawns each have two mates, because they can promote to Queen or Bishop!

I only count promotion as one move. Even if we make a separate category allowing your idea, it’s still pretty much the same position anyhow.

From Schach und Mathematik (Gik):

(= 12+3 )

Kf7: 6 Rg5: 12 Qd4: 4 Be4: 13 Sh4: 2 Sc3: 2 Pd2: 1 Pf2: 2 Ph2: 1 Total 43

And if promotions are allowed:
(= 13+2 )

Se8: 2 Pa7: 1 Pd7: 1 Ph7: 1 Be5: 13 Qf5: 4 Rc4: 14 Sd3: 7 Pe2: 1
Total 44 (Gik claims 47, so he sees promotion to queen and bishop/rook as separate moves)

This is nice! The funny thing is that you can just add a promotion into the non-promotion position and still get 44.

Is this the book that you sourced from? https://www.amazon.de/Schach-Mathematik-Evgeni-J-Gik/dp/3871449873

Yes, that's the book.

I don't know Gik's book, but the 2nd diagram given by Mr de Heer friday is in the book of Nenad Petrović, Šahovski problem (1949), page 166. The author of this problem is J. C. West, 1880 (!!).

Nothing new under the sun !

You may find it in a version the 02/06/2012 and also read funny things about "mate in 1" here :

https://www.france-echecs.com/article.php?art=20120531233813551

(11) Posted by Rewan Demontay (Real Name: James Malcom) [Sunday, May 5, 2019 21:16]; edited by Rewan Demontay (Real Name: James Malcom) [19-05-05]

Yes. I don't know why, only the 2nd works.

Turns out that Otto Blathy also found 47 but with 4 promoting pawns!

P1325213
Joseph Ney Babson
Brentano's Chess Monthly 1882

(= 14+3 )