Much can be deduced from the placement of Black’s units:

The bPs have made at least five moves to reach their current squares.
The bNe4 has made at least two moves (g8–f6 and f6–e4).
The bBh5 has also made at least two moves (c8–g4 and g4–h5).
The bRd4 has likewise made at least two moves (h8–h4 and h4–d4).
The bQe3 has made at least two moves as well (d8–g5 and g5–e3).
The bKf8 must have moved at least once (e8–f8).

That already tallies 5 + 2×4 + 1 = 14 of Black’s 17 moves.

Next, we see that the wK is checkmated by bBg3 — evidently the final move of the solution — which could only have come from e5 or f4, implying at least two prior steps (f8–g7 and g7–e5 or f8–h6 and h6–f4). That completes the count: 14 + 3 = 17, fully accounting for Black’s moves. Moreover, Black’s sequence is tightly constrained by logic. For instance, d6 and Bg4 must precede e6 and Qg5. Similarly, Rh4–Rd4 can only occur after hxg6 has been played, and g7–g5 only becomes possible once the bQ has reached e3 via g5 — and so forth — the order very much dictated by design.

In contrast, White’s pieces yield few clues, much like in Markus Ott’s PG 8.0. The f- and h-pawns have each advanced once; the bNb1 is conspicuously absent, while the rest seemingly untouched. Naturally, the missing white knight must have been captured on g6 — the only explanation for Black’s doubled pawns on the g-file. Yet in the diagrammed position, the knights and rooks alone cannot account for White’s 15 remaining moves, since together they can only contribute an even number. This suggests that another piece — presumably the wK — made a round trip, shedding a tempo to balance the count.

Armed with these findings, we can promptly chart out more than half of the sequence: 1.f3 d6 2.Kf2 Bg4 3.Ke3 e6 4.Ke4 f5+ 5.Kf4 Qg5+ 6.Kg3 Nf6 7.Kf2 Qe3+ 8.Ke1 g5. At this point, however, we realise that sacrificing the b1 knight on g6 requires at least four moves — yet hxg6 (and the ensuing Rh4) cannot wait for that knight to complete them. For example: 9.Nc3 Bg7 10.Nd5 Kf8 11.Nf4, aiming for Ng6 next, leads to a dead end, as Black has no spare moves. Playing 11...Ne4 is premature, as it blocks the fourth rank before Rh4–Rd4 is in place. For the same reason, 11...Bh5 would be equally ill-timed. This means the g1 knight has to be sacrificed on g6 — requiring just three moves — and the b1 knight must later take its place on g1. In other words, we have what might be called a knight switcheroo!

Thus, the rest unfolds as follows: 9.Nh3 Bg7 10.Nf4 Kf8 11.Ng6+ hxg6 12.Nc3 Rh4 13.Ne4 Bh5 14.Nf2 Rd4 15.Nh3 Ne4 16.Ng1 Be5 17.h3 Bg3#.

A masterful blend of Rundlauf and impostor knight: the wK pirouettes through a tempo-losing circuit, while the two knights trade identities — one sacrificing itself on g6, the other returning to g1 as its “impostor.”

Help-Selfmate in 2 (2 Solutions)