Whyatt Medal winner: Dennis K. Hale

The Whyatt Medal is presented once every four years by the Australian Chess Federation for outstanding achievements in the field of chess problem composition. The 2019 selection for this prestigious lifetime award has been announced and the winner is Dennis Hale, to whom we offer our hearty congratulations. A worthy recipient, Dennis composes mostly retro-analytical problems in an intricate style. He favours the twinning device – not commonly seen in traditional retros – to maximise the contents of a problem. His marvellous compositions have appeared in many national publications as well as in the top problem journal, The Problemist.

The notice for this award, published in the ACF Newsletter of 31/1/2020, includes the following citation by Nigel Nettheim: “Dennis has been a prominent Australian specialist in retrograde-analysis chess problems for many years. He has especially studied the question of the legality of a given position. Biographical information and samples of his output can be seen here [link to Dennis’s page on this site]. His ‘magnum opus’ was a stupendous achievement: a 63-position series of twinned retrograde-analysis problems dedicated to W. A. Whyatt, one position for each year of Whyatt’s life. Those problems were published in the book W. A. Whyatt’s Chess Problems by Bob Meadley (1979); the Whyatt medal is, therefore, all the more fitting for Dennis.”

Diagrammed above is the amazing dedicatory problem mentioned. The whole composition is far too elaborate to quote in full here, but I shall examine the first two of its 63 parts. Is the diagram position legal? The black rook is checking White and it must have just captured a knight (i.e. retract …Re1xSf1+) as any other captive on f1 would leave White with no possible last move. That white knight can then retract to e3, uncapturing a piece on f1 to avoid an illegal check by the black rook, and the captive must be a black knight since a bishop couldn’t have reached f1 and a rook or a queen would have no way of unchecking White from that square.

Now we consider how the position after that pair of retractions (containing two white knights and three black ones) could have arisen. The white knight on h1 must have moved to that square from g3 before the latter is blocked by the white pawn, so by the time White opened the h-file with hxg3, the knight was already obstructing h1 and that would have prevented the original h1-rook from ever leaving the first rank via the h-file. Neither of the original white rooks could have escaped via the c-file either because of the trapped c1-bishop, and that implies the rook on h2 is promoted. By the same token, the h1-knight and c1-bishop would also have barred either original black rook from ever reaching the first rank, meaning the f1-rook in the diagram is promoted as well. Thus, including the uncaptured third black knight, three promotions have apparently occurred; are they all possible, given the multiple captures the pawn treks would entail? The answer is yes – for instance: White’s c-pawn made one capture on the b-file before promoting to a rook; then Black’s c-pawn marched down to c2 and promoted to a knight by capturing the original a1-rook on b1, and Black’s d-pawn captured the queen on the c-file before promoting to a rook on d1 by capturing the original h1-rook. Therefore the diagram position is legal.

For part 2, the twinning change is to add a black pawn on c7. In this position, the immediate retractions …Re1xSf1+ Se3xSf1 are still necessary to avoid situations where either side is stuck with no possible last move. Then the argument for how three promotions are needed applies again, but could they have actually occurred when Black’s original c-pawn is still on c7? The answer is no, because a minimum of four captures are now required for the black pawn treks, e.g. bPxcxb1=S and dPxcxd1=R (or gPxhxg1=R), when only three missing white units are available for these captures – the queen and two original rooks (the original f1-bishop never moved). Therefore the position with a black pawn on c7 is illegal.

Here the solver’s task is to determine the unique last move leading to the diagram. A preliminary analysis shows that the white rook on h7 is promoted (since an original rook couldn’t have passed through Black’s king-side pawns and trapped f8-bishop), and that requires White’s original f-pawn to have made at least one capture, fxg8=R. This capture plus one by the doubled b-pawns account for two of Black’s three missing units – queen, rook, and pawn. White is missing six units – queen, two original rooks, two knights, and the c1-bishop. Now the black king is in check by the f1-bishop, which couldn’t have just moved, so it was a discovered check. There are five potential ways White could have delivered this check: (1) e2-e4+, (2) Ke2-e1+ with a possible capture on e1, (3) axb6 e.p.+, (4) cxb6 e.p.+, and (5) dxe4+. The last two options can be ruled out quickly because the resulting pawn structures (involving doubled c-pawns and d-pawns respectively) entail more white captures than missing black units.

Option (1), retracting the white pawn to e2, implies that the f1-bishop had never moved and hence Black’s g1-rook (analogous to the h7-rook) is promoted. Such a promotion requires Black to have made a total of six pawn captures, e.g. bPxcxdxexfxg1=R and fPxg5, but White’s six missing units include the c1-bishop which was captured by something else at home. That means Black couldn’t have made such a rook promotion and option (1) is false. For option (2), if the white king has just moved from e2 without capturing, the piece would have been in an impossible check by the g4-bishop. However, if White retracts Ke2xSe1+, the uncaptured black knight could have made the discovered check, …Sf3-e1+. Could Black have promoted a pawn to arrive at this position with three black knights? Yes, Black’s b-pawn, for example, could have knighted by making just one capture on the a-file. That leaves Black with two missing units (queen and original h8-rook), which are sufficient to account for White’s doubled b-pawns and promotion capture, fxg8=R. Thus option (2) is possible, but can we discount option (3)? Suppose White has just played axb6 e.p.+; Black must then retract …b7-b5 and this leads to another position with the black king in check by the f1-bishop. At this point, however, the uncheck Ke2xSe1+ no longer works because the e.p. uncapture has left all eight black pawns on the board and so none could have promoted to a third knight. Hence option (3) fails to resolve legally. Only option (2) remains, so White’s last move in the diagram was Ke2xSe1+.

In part (b), the black pawn on g5 is shifted to f6, and one effect of this change is that the white f-pawn couldn’t have marched through the black one to promote with a single capture, fxg8=R. Rather, White must have played fPxgxh7 before promoting on h8, and these two captures plus one by the b6-pawn account for all three missing black units. Consequently, there are no spare black units that could have been captured by any other white pieces, including the king; thus in this position, the option (2) of retracting Ke2xSe1+ is not viable anymore. But another effect of the twinning is that Black no longer has doubled g-pawns, meaning there is an extra spare white unit that could have been captured elsewhere. This leads us back to option (1), retracting e2-e4+, which locks the f1-bishop so that only a promoted black rook could have arrived on g1. Now such a promotion is indeed possible, because the five missing white units (disregarding the c1-bishop captured at home) are enough to permit a sequence like bPxcxdxexfxg1=R. We can exclude the remaining options for reasons similar to those raised before, e.g. if White retracts axb6 e.p.+, that forces Black to take back …b7-b5, which creates another position with the f1-bishop giving check; but here White cannot uncheck with e2-e4+ because that would imply the g1-rook is promoted while eight black pawns are present. Therefore option (1) is correct in position (b) and the last move was e2-e4+. A terrific piece of work!