ALL DE FR ES IT

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Miscellaneous Problems

Solution

A. Frolkin and A. Kornilov
Rex Multiplex 6, Apr. 1983

[88888/2AAAB2/BBBBBBA1/A1CDEAA1]

17. Associate a different chess unit to each letter. a) Diagram b) Ag1 -> b1

Here C and E are the only non-adjacent single-occurrence letters, hence they must be the 2 Kings. Now B must be a wh. Pawn because any other unit would give illegal checks to one of the Kings. Then D cannot be a Queen or Rook (illegal checks) or a Bishop (behind wh. Pawns?). It must be a Knight, white or black.

Because we have 7 wPs, only one wh. promotion is possible, so that the 7 A's cannot be wh. units. They cannot be bl. Pawns (on the 1st row?) or Bishops (on a1?). They cannot be bl. Queens because this would require six bl. promotions and need too many captures.

(a): If we assume the A's are bl. Rooks, they give an impossible check to the wh. King. So that A is a bl. Knight and then D is a wh. Knight. The wh. King must be C and then E is the bl. King.

Here is the deciphered position:

[88888/2nnnP2/PPPPPPn1/n1KNknn1]

The bPd7, e7, f7, g7 & h7 used 6 captures to promote to g1 (or h1). This consumed all six missing wh. units (excepting wBc1 captured at home, but including wPh2 who had to leave its h-file somehow). So that the last move was -1. Nd3+ without capture on d3, preceded by -1 ... Kb1-c1 again without any capture.

(b): with Ab1 the previous solution does not work because White's last move -1 ... Kb1-c1 is not possible. So that A can only be a bl. Rook with wh. King on e1 and Rxf1+ as last move. Then because of this extra capture by Black, the Nd1 cannot be white as in (a). Then the deciphered position is:

[88888/2rrrP2/PPPPPPr1/rrknKr2]