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The following pictures depict chess positions one move before white checkmated black:

boards

The problem is, I seemed to have completely lost which pieces each letter/number represents. All I know is that the number substitution of the pieces is consistent across the board and that

1-7 include {2 pawns, 1 knight, 1 bishop, 1 rook, 1 queen, and 1 king}

A-F include {1 pawn, 1 knight, 1 bishop, 1 rook, 1 queen, and 1 king}.

Text boards:

-------A
E-------
-----4--
----CB3-
------26
------D-
--------
-------7

---B----
-16--F--
---3-E--
-A--4---
---7----
------D-
--------
--------

-3------
574-----
--------
--------
--------
--------
--------
---C-D--

So... what piece does each character represent?


Took inspiration from the snakes and ladders retrograde analysis puzzle and decided to make a non-retrograde analysis chess puzzle (which doesn't count in the fortnightly topic challenge :( )

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  • 3
    $\begingroup$ Like the puzzle idea! +1 $\endgroup$ – BmyGuest Oct 24 '18 at 7:17
  • $\begingroup$ I came in here expecting to see a puzzle about arranging all of white's pieces such that none of them has a valid move, thus checkmating the white king without using any black pieces. $\endgroup$ – Ian MacDonald Oct 24 '18 at 14:10
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    $\begingroup$ @IanMacDonald What you're describing sounds like stalemate, not checkmate? (Checkmate requires check, which must be delivered by an opposing piece.) $\endgroup$ – mathmandan Oct 24 '18 at 16:03
  • $\begingroup$ @mathmandan Good point! It would also likely require more pieces than are available in a standard chess set. $\endgroup$ – Ian MacDonald Oct 24 '18 at 17:59
  • $\begingroup$ @IanMacDonald I have turned your comments into a puzzle: puzzling.stackexchange.com/questions/74350/… $\endgroup$ – mathmandan Oct 24 '18 at 21:32
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The #3 looks like a

back rank mate, so 3 is the black king. C is missing from #2, so that can't be the king, which means that the white pieces in #3 are D=king and C=queen/rook. For the position to be a mate in 1, 4 needs to be a knight/pawn and 7 a rook/pawn.

The pieces in #1 and #2

can't be attacking the opposing king, so
- B can't be a queen or a rook (#1)
- E can't be a queen or a rook (#2)
- F can't be a knight (#1)
- 2 can't be a queen or a rook (#1)
- 4 can't be a bishop or a queen (#2)
- 6 can't be a bishop or a pawn (#1)
- 7 can't be a knight (#1)

Pawns can't be

on the 1st/8th rank, so A, B and 7 can't be pawns.

In #1 the mating move looks like

Qh6 by A=queen, protected by B=knight.

So the white pieces are

A queen, B knight, C rook, D king, with E and F being a bishop or a pawn.

Substituting the known pieces in #2,

Having F be a pawn would make it possible to mate with f8=Q (or f8=B for extra pizzazz), with all escape squares covered by white pieces.

Now we just give the black pieces some combination which makes the #2 position a checkmate after that move.

1 pawn, 2 bishop, 3 king, 4 pawn, 5 queen, 6 knight, 7 rook.
A queen, B knight, C rook, D king, E bishop, F pawn.

The boards with all pieces:

enter image description here enter image description here enter image description here

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  • 7
    $\begingroup$ I spent the last two hours typing out in paragraph form all of this. You beat me to it though by 7 minutes. Nice work. $\endgroup$ – Arpeggio Oct 24 '18 at 7:00
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    $\begingroup$ Good job! Some logic questions: On your first step, couldn't it be possible for 4 to be a knight? Also, on your second step, white could possibly be in check as long as they removed the threat on the mate, right? $\endgroup$ – Jo. Oct 24 '18 at 7:12
  • $\begingroup$ @Jo. Hmm that's true, good points. I had thought about 4 being a knight, then disregarded it when I thought C was the white king, and forgot to add the possibility back. The second point hadn't occurred to me at all :P $\endgroup$ – jafe Oct 24 '18 at 7:16
  • $\begingroup$ ah .... too late on my end too :p Keeping my answer for those who want to see it in a different manner $\endgroup$ – Kryesec Oct 24 '18 at 7:57
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    $\begingroup$ the only way the position is a mate in 1 is if 4 is a pawn and 7 is a rook. If both are pawn it also works. $\endgroup$ – SJuan76 Oct 24 '18 at 8:18
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Well, it is obvious that:

White has king D, as it is the only figure appearing on all boards.

Now,

Board 3 has only one possible mate - C going up to mate king (3). The rest on the board are clearly impossible as no figure move could mate in one (nearest would be mate of king at (5) if C was one place to the right, but (5) cannot be king anyway as it isn't present on other boards).

Now, what we additionally learn from the checkmate:

5 is not knight, 7 is not knight, bishop or queen = rook or pawn and 4 cannot be bishop, rook or queen = knight or pawn.

That is exhausted for now, moving on.

White's B isn't pawn from board 2, it cannot be on the back rank. It cannot be rook or queen either otherwise king is already in check/mated. So, it is bishop or knight. A and F cannot be knights, and E cannot be rook or queen. Board 1 additionally puts constraints on 2 to be something else than a rook or a queen, while 6 cannot be pawn, queen or bishop and A and 7 cannot be pawns.

To summarize figures up to now (including all resulting eliminations):

1: Q, B, p
2: B, p
3: K
4: P
5: Q, B, p
6: N
7: R
A: Q, R, B
B: B, N
C: Q, R
D: K
E: B, N, P
F: Q, R, B, P.

Now, how to mate on board 1:

There are two possibilities: A=Q, B=N, Qh6. C=Q, B=B, A=R, Qf4.
C=Q/R, B=N, A=Q/R with Ne7 fails as 4 takes the piece at C.
Other possible checkmates violate what we know about pieces.

So, this additionally restricts:

F = B/P, as A and C take the Q and R slots in both valid combinations to mate on first board.

Looking at the board 2 now:

We have A = Q/R, B, N and P in addition to the K that is far enough to be useless. The trick here is that A=Q, as no other combination can attack c3, c4 and d2 at the same time while performing any move to mate the king (Rd5 with F=B and B=N isn't possible as N or R will take the rook). There are some moves that can mate the king, but with A=Q let's rather go back and eliminate the other possibility on board 1.

So, we know:

A=Q, B=N, Qh6 is the mate on board 1. That has no further implication for figures on board 1 as 4 is already p and 6 already N and the rest cannot interfere. Now, what happens with board 2? E and F are B and P, so the poor king cannot move anywhere already. Now, it is time to mate it. With E = P and F = B, Nxb7 delivers the mate. With E = B and F = P, p->Q or B on f8 does the trick.

This already proves all boards are a mate in 1 and the @jafe answer before "Substituting the known pieces in #2," is the unique solution, not just one possible that is easy to see. The rest is unanswerable as there are many piece combinations that are possible and cannot be ruled out even by considering possible previous moves. The final pieces are:

1: Q, B, p
2: B, p
3: K
4: P
5: Q, B, p
6: N
7: R
A: Q
B: N
C: R
D: K
E: B, P
F: B, P.

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  • $\begingroup$ F must be a pawn, and E must be a bishop. If A attacks in board 2, it can only do so in the square threatened by 1. We don't know 1 for sure, but any of its possible roles (p,b,q) leave it open to attack A after it moves, and kill the 1 move mate. Thus, the only possible mate is to move F and promote it. $\endgroup$ – Ethan Oct 24 '18 at 23:27
  • $\begingroup$ @Ethan I never considered A to attack, except to show why it has to be a queen (= it cannot attack as a rook on d5, it will get taken by 6=N or 7=R). What is wrong with E=P, F=B, (B=N) Nxb7 to deliver mate? $\endgroup$ – Zizy Archer Oct 25 '18 at 0:54
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My answer is :

King = 3,D
Queen = 1,A
Rook = 7,C
Bishop = 2,E
Knight = 5,B
Pawn = 4,6,F

But the truth is .....

that there are many possible permutations for black, as I shall explain why.


Taking Board 1/2/3 from left to right,
First resolving for Kings:

For White, only D exist throughout the 3 boards, hence D must be King
For Black, it is a choice between 3 & 4. However in Board 3, there is too much room for 4 to move around, making checkmate impossible. Hence 3 is King.

Next for White:

C in particular:
For board 3, the most logical way to achieve a checkmate is that C is a queen/rook. This also means that:
5 != Knight
7 != Queen/Knight/Bishop
4 != Queen/Rook/Knight/Bishop, conveniently making it a pawn

Moving over to Board 1:

To make it not a 0 move checkmate, B != Queen/Rook

Here we have 2 possibilities:
1. C = Queen
Move C to R5C6 to achieve checkmate. This is possible if A is a rook & B is a pawn/bishop.
Further verifying with board 2, this is not possible to achieve a checkmate as there are too many possibilities for 3 to escape (i.e. to R3C3, R3C5 etc ...)
(I skipped the in-depth analysis for this for now, due to excessive permutations)
2. C = Rook
& Hence A has to be a queen. Move A to R3C8 to checkmate. B has to be a knight
Verifying this with Board 2, A being queen effectively seals off most of the moves for Black King (3). B being a knight seals off R3C3 & R3C5. Only move for BK is to R2C5. E/F could be pawn/bishop. Way to secure checkmate is that if F is a pawn and moves to R1C6 to promote to a queen/bishop.
This leaves E to be a bishop

Further eliminating possibilities for Black:

For the above scenario to work,
from Board 1 to avoid check on king: 7 != knight
2 != Queen/Rook
6 != Queen/Bishop
to avoid blocking of checkmate, 2 !=Knight, 6 != Rook

Summarizing possibilities for Black:

|K|Qn|Rk|Bs|Kn|Pn
1|N|??|??|??|??|??
2|N|NN|NN|??|NN|??
3|Y|NN|NN|NN|NN|NN
4|N|NN|NN|NN|NN|YY
5|N|??|??|??|NN|??
6|N|NN|NN|NN|??|??
7|N|NN|??|NN|NN|??

IMO

The rest of the pieces could be filled randomly, as long as it fulfils the above matrix.

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  • $\begingroup$ In your fourth hidden block, I don't agree with "4 != Queen/Rook/Knight/Bishop" : up to that point, there is no reason 4 wouldn't be a Knight. Also, the standard chess notation for the square you designate as 'R4C3' is 'c4'. $\endgroup$ – Evargalo Oct 24 '18 at 8:26
  • $\begingroup$ 7 cannot be a pawn because it is on the eighth rank in board 1 $\endgroup$ – awesomepi Oct 24 '18 at 17:03
  • $\begingroup$ 6 can't be a pawn because then the white king is in check $\endgroup$ – awesomepi Oct 24 '18 at 17:21

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