Here is a knight rebus. Each letter represents a different type of piece. Uppercase letters are one color and lowercase letters are the other.

Determine the position. If possible, also determine the previous moves.


n and N are kings and h is pawn. What is the rest?

  • 1
    Are you sure this is chess?
    – SmallChess
    Jun 22 '17 at 12:02
  • I'd say it isn't. At least not with standard rules.
    – Annatar
    Jun 22 '17 at 12:18
  • @Peter What is the source of this problem? Is there a significance to the choice of letters used to encode the pieces?
    – yrodro
    Jun 27 '17 at 14:24
  • @yrodro The source is XX all-Ukrainian tournament among young mathematicians. "Knight" is just a word of six different letters. Jun 28 '17 at 5:26
  • What's I, which side of the board is white, and which types of pieces are white.
    – Tony Ennis
    Oct 12 '17 at 3:41

Assume the board is in it's usual position (the bottom row is the first row) then this is the solution:

[FEN "1B6/8/1k6/R7/1n6/2K5/r6p/q1r1n3 - - - 0 1"]

and Black's last move was b2xc1R++.

I could write an explanation here about I came to this result, but it would essentially be copying @Maxwell's answer. I didn't use it, though.

  • Heh, I even thought about the possibility of multiple queens by promotion, but this is quite an interesting twist. Well done.
    – Annatar
    Jun 22 '17 at 12:53
  • 3
    I assume ++ is a checkmate symbol, but is it really a checkmate? I think white king can run away to b3 or capture b4, or did I misunderstand something? Or is it "double check"?
    – Andrew T.
    Jun 22 '17 at 18:35
  • 2
    @AndrewT. it's double-check. I'd use # for mate.
    – Glorfindel
    Jun 22 '17 at 19:23
  • This can't be the solution. The King must have walked illegally into that double check. "i" is Bishop, "K" is knight. "g" is rook.
    – LocalFluff
    Oct 12 '17 at 10:41
  • 1
    @LocalFluff Black's last move was with a pawn on b2, capturing a White piece on c1 and promoting to Rook. i can't be a Bishop, because a White Bishop on a5 would put the Black King in check as well.
    – Glorfindel
    Oct 12 '17 at 10:49

Yes indeed, n corresponds to king, as we need two kings of a different color to have a legal chess position.

Also, as you give, h = pawn, as otherwise we would have a pawn on the first or last rank, which would be illegal.

Consider i. It cannot be a queen, as both kings would be in check. It cannot be a knight, as the "i" on a2 would give check. There is no legal way where "knight a2 check" is the last move. It cannot be a bishop, as the "I" on a5 would give check. There is no legal way where "bishop a5 check" is the last move. Therefore, i = rook.

We see that the king on c3 is in check by the rook on c1. Therefore, g cannot be a queen or a bishop, because there is no legal way to obtain the resulting position. So, g = knight.

The "K" on b8 should not give check to the king on b6. Therefore, k = bishop and t = queen.

[However, the resulting position seems to be impossible to achieve too... As far as I can see, the puzzle has no solution.]

Edit: the resulting position is indeed possible, see the answer of Glorfindel.

  • You're welcome and thank you for the question.
    – Maxwell86
    Jun 22 '17 at 14:13

This is a basic example of an orthodox retro form which has been taken to a new level over the last year by Ukrainian composer Andrey Frolkin and Canadian Jeff Coakley. I recently acquired a lovely landscape painting by Andrey's wife, Nina, which contains an embedded rebus by Andrey, who I bet authored the problem OP posted!

Although Maxwell's excellent proof relied at a couple of place on the edge of the board preventing retro-movement, this is not actually necessary. Shift all the pieces one square to the right (except for h on h2), and the problem remains sound (and in my opinion slightly superior :D )!

Proof: first count the pieces. K N I G H T Upper 1 1 1 0 0 0 Lower 0 1 2 2 1 1

So N is king and by examining first and last ranks, H is pawn. Observe that whatever I represents, one instance is giving check to the enemy king. So any other checking unit implies a double check. So G is not a diagonal unit (bishop or queen), and K is not an orthogonal unit (rook or queen), since the resulting double checks are impossible. And I is not queen, or both kings would be in check!

So by elimination the queen is represented by T and we do have double check. So I can only be rook, and the position is achieved through promotion. Hence lower case is Black. G must be knight so by elimination K is bishop.

[title "Frolkin? version Laska rebus: solution"]
[fen "2B5/8/2k5/1R6/2n5/3K4/1r5p/1q1r1n2 w - - 0 1"]

and Black's last move was c2xd1=R++

  • So, what was the last move? That the white King walked illegally into that double check? "i" is Bishop, "K" is knight. "g" is rook
    – LocalFluff
    Oct 12 '17 at 10:44
  • @LocalFluff: Thanks for the question. Last move was black pawn on c2 capturing something on d1 and promoting to rook check, also discovering check from queen. The solution would have to be something like this to allow white and black to be distinguished. If I,K&G are as you suggested, then both kings are in check: wKd3 from bQb1 and bKc6 from wBb5.
    – Laska
    Oct 12 '17 at 11:16

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