Here is a chess "rebus" problem. In this position, white and black circles correspond to White and Black pieces respectively. Each letter corresponds to exactly one piece type (king, queen, rook, knight, bishop or pawn). The task is to find letter-piece correspondence.

enter image description here

I know that the answer is:

A = rook; B = king; C = bishop; D = queen; E = knight.

Can you a list of moves which leads to such a position?

2 Answers 2


Highly non-optimal, but here's one line that reproduces the position. The basic idea is to not capture the existing pieces, but capture pawns. A single pawn capture can free 3 pawns to promote. Here's a sample game.

   [FEN ""]

   1.h4 a6 2.h5 g5 3.hxg6 Nf6 4.g7 h5 5.g8=R h4 6.g4 h3 7.g5 h2 8.Nf3 Nd5 9.
   g6 Nb6 10.g7 Rh4 11.Rg1 h1=R 12.Rh8 Nd5 13.g8=R Rh5 14.R1g5 R1h4 15.Rh6 
   Nb6 16.Rg8g6 f5 17.e4 fxe4 18.Nd4 e3 19.Bh3 e5 20.Ke2 e4 21.Qh1 Nd5 22.Kf1
   e2+ 23.Kg2 e1=R 24.f4 e3 25.f5 Bb4 26.f6 e2 27.f7+ Ke7 28.f8=R Rf1 29.Nb3 
   e1=R 30.Rf4 Re6 31.Rf3 Rf6 32.Rgg3 Re1 33.R6g4 Ree6 34.Nd4 Kf7 35.Ne2 Bf8 
   36.a4 Bg7 37.a5 b5 38.axb6 Nc6 39.b7 a5 40.b8=R a4 41.Raa3 Ra6 42.Rae3 a3 
   43.b4 a2 44.b5 a1=R 45.Ra8 R6a5 46.b6 Nce7 47.b7 Nc6 48.b8=R Nb6 49.Kf2 d5
   50.Bb2 d4 51.c4 dxc3 52.d4 c2 53.d5 Qe7 54.d6 c1=R 55.d7 Nb4 56.d8=R c5 
   57.Rg2 c4 58.R4g3 c3 59.Na3 c2 60.Nb5 Rd1 61.Ned4 c1=R 62.Nc2 Nd3+ 63.Ke2 
   Ne1 64.Nc7 Ra6 65.Ne8 Na4 66.Rhg6 Bb7 67.R6g4 Rff5 68.Rb4 Rh7 69.Ree4 
   Rh4h6 70.Qg1 Rhf6 71.Qd4 Ra3 72.Qc4 Rad3 73.Qa2 Rdd6 74.Qa1 Qc7 75.Rff2 
   Qc3 76.Ree3 Qa3 77.Rbc8 Qa2 78.Rab8 Qb1 79.Reb3 Rd1d5 80.Ra3 Rb5 81.Qa2 
   Rdc6 82.Rdd3 Rg6 83.Qa1 Bc3 84.Qa2 Rgf6 85.Qa1 Rfd5 86.Qa2 Red6 87.Qa1 
   Rff5 88.Re4 Rdf6 89.Rb4 Rcc5 90.Rbb3 Rdd6 91.Qa2 Be4 92.Qa1 Rf4 93.Qa2 Bg6
   94.Qa1 Rf4f5 95.Qa2 Rcc6 96.Qa1 Nc5 97.Qa2 Nb7 98.Rd8 Rbb6 99.Qa1 Rc7 100.
   Rde3 Rde6 101.Rd2 Rec6 102.Rbd8 Re7 103.R8d3 Rd7 104.Ref3 Rdd6 105.Nb4 Bd4
   106.Qa2 Rc1c5 107.Qa1 Rcd5 108.Nc2 Bc3 109.Qa2 Ke7 110.Qa1 Ke6 
  • 1
    +1 The key is that one pawn capture not only frees itself and the pawn behind it, but the pawn opposite it in its original file.
    – Daniel
    Commented Jun 20, 2014 at 5:03

Each side has 14 units and has thus made two captures. Thus A is not pawn: if so, White's pawn structure entails cxd3, exf3, hxg3. As Wes & Daniel pointed out, the way to let as many pawns as possible promote with as few captures as possible is for some pawns to capture pawns and then promote. For example White axPb lets White's a- & b-pawns and Black's a-pawn promote. Two such captures by White and two by Black let each side promote 6 pawns. But with each side making only two captures, no more promotions are possible. So each side has 2 original & 6 promoted A's, so A is not queen, but rook, bishop or knight. Moreover, every pawn either promoted or was captured, so no letter represents pawn.

Which letter means king? Not A, C or E: White has 2 or more of each. Nor D, for then the Ds check each other. Therefore B.

E is neither queen nor rook: if so, both kings would be in check. Nor is E pawn. Therefore bishop or knight. If E is bishop, White has promoted a pawn to bishop on a white square. But it has already been proved that there were 6 White promotions (to A) & that only 6 are possible. So E is knight.

Queen is C or D. But there have been no promotions except to A, so queen is not C but D.

A is not knight and so is rook or bishop. Suppose that A is bishop. Then there are 7 promoted bishops on white squares (3 white and 4 black) and 5 on black squares (3 white and 2 black). Each PxP allows 3 promotions all on the same colour square. So the total number of promotions on each colour square is a multiple of 3. So the supposition is false. So A is not bishop but rook.

So C is bishop.

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