[14+10. Matt in 4 Zügen]
Under chess problem conventions, in a problem stipulating 'Mate in
4', White has the move. Only 1.PxPe.p. will suffice for a mate in four;
the en passant capture can be justified only if Black's last move was
Pf7-f5. This is what we now prove.
Captures generally. In the diagram as presented,
White has 14 units (i.e. 2 were captured by Black) and Black has 10
(6 were captured by White).
White's captures. The White P at a7 could have originated
only on f2. In order to get from f2 to a7 it accounted for 5 of the
6 White captures, each occurring on a dark square. The White Queen's
Rook Pawn (QRP) must have captured from the a file to the b file to
account for the configuration of the Ps at b3 and b4, which also accounts
for the 6th White capture. As the Black Queen's Bishop is missing from
the diagram it must have been captured; that B occupied only light squares,
so it must have been captured by the White QRP on b3. All White captures
were by Ps.
Black's captures. Black's 2 captures were made by
its KRP, which explains the two Black Ps on the f file in the diagram.
The Black KRP must have captured both of the missing White Bishops,
one on a light square and one on a dark square. The only possible route
requires that the Black KRP captured the White KB at g6 and the White
QB at f4 (where that P appears in the diagram).
Black's QRP. As noted, all captures by both sides
were made by Ps. The absent Black QRP must have been captured by the
White KBP (the only other capturing White unit, the QRP, having captured
the Black QB). Either the Black QRP (i) promoted at a1 (after the White
QRP captured the Black QB at b3 and cleared the file), and is still
on the board, or (ii) did not promote and was captured on the a file.
(It could not have swerved to another file as that would entail a capture
by Black in addition to the 2 required for the Black KRP to reach the
f file). The only square on the a file reached by a White P is a7. But
if the Black QRP was captured at a7 by the White KBP, then the square
a7 was continuously occupied by a P, and the Bishop at b8 in the diagram
could never have reached b8. Therefore the Black QRP promoted at a1,
and it is still on the board, in its promoted form (so it must be the
rook on g8)
The last move. The Ps at c7, d7, and g7 never moved.
Neither the B at b8 nor the R at g8 could have made the last move, all
squares from which they could have moved being occupied. As discussed
above, the P at f4 came from g5 when it captured the White QB; thus
it couldn't have made the last move. The P at b6 came from b7, but had
it done so as recently as the last move the Black QB could not have
left its original square (c8) to be captured, as it was, on b3. Similar
reasoning shows that the P at e6 did not come from e7 on the last move,
as then the Black KB could not have departed f8 (ultimately to reach
b8, where it stands in the diagram). The P at e6 could not have come
from f7 on the last move as that would entail too many captures by Black.
The only remaining possibilities are a move by the Black K, or a move
by the P at f5. The Black K could have come from a square on which it
would have been in check, but only if that check could have been delivered
by White on the immediately preceding move. Therefore the Black K could
not have come from h5, h4 or h3 on its last move, as in any of those
cases the White Q could not have delivered the check on White's immediately
preceding move. The Black K never could have occupied f3 or g3, as neither
the White KP nor the White KRP, which deliver checks on those squares
(respectively), has ever moved. Nor could the last Black move have been
the capture by the Black K, while moving from h3 or h4, of a White B
on g4 (which White B could have delivered a discovered check by having
moved from h5 to g4 on White's immediately preceding move), as the 2
Black captures were made by the P at f4.
Therefore the P at f5 made Black's last move. That P could have moved
only from f6 or f7. Assume that it came from f6 (and occupied that square
when White made its last move). This confronts the crux of Loyd's innovation.
In all previous retro problems with an e.p. key move the diagram position
presented a K on the same rank and adjacent to the P that advanced two
squares, in order to negate the possibility that the P had moved only
one square (from a location where, illegally, it would have been moving
while the other side was in check). Loyd's 1894 position pushed the
analysis back another move, by showing that White would have had no
possible immediately preceding move if the Black P had occupied the
Assume that the P at f5 were on f6, and examine the possible last
moves for White. Ps at c2, d2, e2, and h2 have never moved. The P at
a7 came from b6, so it has no last move. The P at g5 may not have swerved
off the g file (which would require too many captures by White), and
its path on the file is blocked, so it didn't move last. The P at b3
could not have come from b2 recently or the White QB would not have
been able to depart from c1 (which it earlier did, having been captured
at g6). Nor could the P at b3 have moved from a2 to capture the Black
QB on the last move, because if it recently occupied the a file the
Black QRP would not have been able, earlier, to advance along that file
and promote on a1(as established above). No square is available from
which any of the White K, N at a1, N at f8, or R at h8 could have moved.
Any square from which the White Q could have come (h3, h4, or h5) would
have involved its moving while delivering a check to Black; clearly
White cannot be on the move if Black is in check. Finally, if Black
had its KBP on f6 when White last moved, the R at g6, which otherwise
could have come from f6, would have no available square from which it
might have moved.
Therefore the Black KBP did not come from f6 on the last move, it
came from f7, and White, immediately before, had moved its R from f6
to g6. Black's last move having been a 2 square P advance, W is free
to capture en passant and mate in 4.
Solution by Tom VOlet