The Babson Task – Part 3

In the first two parts of this series on the Babson task, we saw how the theme – four matching pairs of black and white promotions to the same kind of piece – was accomplished in selfmates and helpmates. Serious attempts by established composers to achieve the task in directmates began in the 1960s, and the extreme difficulty was such that no-one succeeded for decades. Their valiant efforts produced “proto-Babson” positions that did not pass muster as they are illegal or contain promoted force. Then in 1983, an unknown Russian composer named Leonid Yarosh (1957-) caused a sensation by publishing a sound mate-in-4 Babson problem, with a diagram that is legal and has no promoted pieces. This first directmate Babson (P1052449) unfortunately starts with a brutal key-move that captures a black knight; but just months later, Yarosh constructed another setting without the serious flaw. The improved rendition, solved by an excellent key that pushes the thematic pawn, is now considered one of the greatest chess compositions ever.

The black queen is largely tied up preventing Rxf4 mate, and most of the piece’s moves are provided with set white responses. In particular, the line 1…Qxd8+ 2.Kg7 Qf8+ (2…Qb8 3.d8=Q+ Qxd8 4.Rxf4) 3.Kxf8 axb1=Q 4.d8=Q relies on the b1-rook’s placement to avoid stalemate. Therefore, not 1.Re1? (threatening 2.Re4) Qxd8+! 2.Kg7 Qf8+ 3.Kxf8 stalemate.

The key 1.a7! threatens 2.axb8=Q/R/B/S, and since the four threat-moves are individually forced by four black defences, an incidental Fleck effect is shown. After any of the thematic promotions on b1, White cannot continue with 2.Qxb1?, again because of the stalemate tactic starting with 2…Qxd8+.

(1) 1…axb1=Q 2.axb8=Q (threats: 3.Rxf4+/Qxf4+/Qxb3/Qd6+). To handle two strong follow-ups by Black’s promoted queen, White must choose a queen as well. 2…Qxb2 creates flights on d3/c4, which are retaken by 3.Qxb3 (4.Rxf4/Qaxb2), then 3…Qxa1 4.Rxf4, 3…Qc3 4.Qaxc3/Qbxc3. If 2…Qe4 then 3.Rxf4 Qxf4 4.Qxf4 or 3.Qxf4 Qxf4 4.Rxf4.

(2) 1…axb1=R 2.axb8=R (3.Rxf4). The black rook has only one strong follow-up, yet it refutes 2.axb8=Q? when 2…Rxb2! 3.Qxb3 produces stalemate, or 3.Qxf4+ Kd3. White’s new rook is ready to meet 2…Rxb2 with 3.Rxb3, which leaves one flight for the king: 3…Kxc4 4.Qa4. The alternative 2…Re1 merely extends the threat line, 3.Rxf4+ Re4 4.Rxe4.

(3) 1…axb1=B 2.axb8=B (3.Rxf4+/Bxf4/Sd6). Black’s promoted bishop has its own strong defence that defeats 2.axb8=Q?, despite White’s numerous threats – 2…Be4!, freeing the f4-bishop. Then 3.Rxf4 or 3.Qxf4 gives another stalemate, while 3.Qxb3 is thwarted by 3…Bh6+. The new white bishop counters 2…Be4 with 3.Bxf4 B~ 4.Be3/Be5, a dual mate.

(4) 1…axb1=S 2.axb8=S (3.Rxf4). The black knight spoils 2.axb8=Q? with 2…Sxd2!, which gains a flight on c3 and defends b3 directly. Here 3.Qc1 (4.Qxd2) S~ 4.Rxf4/Qbxf4/Qe3, but the correction move 3…Se4 refutes. White forestalls such a defence by promoting to a knight, then 2…Sxd2 3.Qc1 (4.Qxd2) S~ 4.Rxf4/Qe3, and the correction 3…Se4 is exploited as a self-block, 4.Sc6.

The marvellous intricacy of the four main variations is supplemented with by-play that is also quite complicated. Here is a sample of side-lines that involve both precise and dualized white play. 1…Qe5 2.Bxe7 Qd6 (2…Qb8+ 3.axb8=Q, 2…Qc7 3.Bxf6+, 2…Qxf5 3.d8=Q+) 3.Sxd6 Ke5 4.Sd3. 1…Qd6 2.Re1 Qc6 3.Rxf4+ Qe4 4.Rfxe4/Rexe4. 1…Qxa8 2.Rxf4+ Qe4 3.a8=Q axb1=Q 4.Qd5. 1…Qxd8+ 2.Kg7 Qf8+ (2…Qg8+ 3.Kxg8, 2…Qh8+ 3.Kxh8) 3.Kxf8 axb1=Q 4.d8=Q.

In the time since Yarosh’s Babson first appeared, about two dozen directmate realisations of the task have been published by various composers. Among these problems, some are further distinguished by presenting the four thematic variations accurately without white duals. The Yarosh position is flawed in this respect, since the bishop-promotion sequence ends with a dual mate. Also, its queen-promotion sequence branches into many lines of which only one is dual-free (2…Qxb2 3.Qxb3 Qxa1 4.Rxf4). Such a precise sub-variation is sufficient to make the thematic queen-promotion part valid, according to Jeremy Morse in Chess Problems: Tasks and Records. But another authority, Werner Keym, proposes that all sub-variations in a matched-promotion part should be without duals.

Based on Morse’s definition, Yarosh himself created a Babson directmate with four accurate thematic variations, soon after the original masterpiece: P1038119 (both compositions earned the maximum 12-points in the FIDE Album). However, if we take Keym’s stricter approach, only a handful of Babson problems have rendered four totally dual-free thematic lines, and the first was P1290995 by the German composer, Peter Hoffmann. Although that four-mover contains a poor key (capture of a bishop), improved versions of it have appeared and perhaps the best with an acceptable key is diagrammed below.

The black king has flights to g8 and h8, of which is latter is provided: 1…Kh8 2.fxg7+. The key 1.fxg7! removes that flight and overloads the black rook, partly by opening the f-file for the white one. The threats are 2.gxf8=Q/R/S+ and 2.Rxf8. Black’s capturing promotions on d1 and f1 are ignored as weak defences that don’t stop these threats.

(1) 1…e1=Q 2.gxf8=Q (3.Qg7/Qf7/Qh8/Rf7/Sf6/g6). To prepare against a strong black queen’s check on e4, White must promote to a queen, which controls not only the 8th-rank flights but also g7. 2…Qxe4+ cannot be answered by 3.dxe4/Rxe4? causing stalemate. Instead, 3.d4 pins the queen and threatens 4.Rf7/Qxe4, then 3…Qf5 4.Qxf5 or 3…Qxc2 4.Rf7. Not 2.gxf8=R? Qxe4+! 3.d4 Qf5, or 2.Rxf8? Qxe4+! 3.d4 Kxg7.

(2) 1…e1=R 2.gxf8=R (3.Rh8/R4f7). Black’s promoted rook also threatens a check on e4, but here 2.gxf8=Q? fails because after 2…Rxe4+!, not only 3.dxe4/Rxe4 but 3.d4 also results in stalemate. The white rook promotion purposefully leaves g7 unguarded for 2…Rxe4+ 3.d4 Kg7 4.R4f7. Similarly, not 2.Rxf8? Rxe4+! 3.d4 Kxg7.

(3) 1…e1=B 2.gxf8=B. The immobilised black bishop means that 2.gxf8=Q/R? and 2.Rxf8? all stalemate immediately. On the other hand, White’s bishop promotion exploits Black’s lack of mobility, by putting Black in zugzwang. 2…Kg8 3.Bcg7 Kh7 4.Sf6. If 2.gxf8=S+? Kg8, the black king is trapped but all attempts to generate a mating threat, such as 3.Kxb5/d8=Q, again stalemate.

(4) 1…e1=S 2.gxf8=S+. The black knight threatens two consecutive checks, thus 2.gxf8=Q/R? and 2.Rxf8? are foiled by 2…Sxc2+! 3.Kxb5 Sa3+/Sd4+. White checks first with the knight promotion, and after 2…Kg8, 3.Kxb5 not only evades Black’s checks but threatens a mate that the black piece is unable to halt, 3…S~ 4.b4.

When Black defends with the f8-rook, the ensuing by-play is largely dualized. 1…Rxf4 2.d8=Q/R Rxe4+ 3.dxe4 ~ 4.Qh8/g8=Q. 1…Rf6 2.d8=Q Kg7 3.Qxf6+. 1…Ra8 2.Rf8 Rxf8 (2…Ra4+ 3.bxa4) 3.gxf8=Q/R. 1…Rg8 2.Rf7 Rxg7 3.d8=Q/R or 3.Rxg7+.

This first-class Babson problem is taken from 100 Years: Babson Task in the Orthodox Directmate (2025 edition) by Peter Hoffmann with Erik Zierke. It’s a free e-book in German text (and some English) that comprehensively covers the history of the theme, with all known examples shown with full solutions. The 150-page book, which can be downloaded here, also examines Babson variants such as 3/4 matching promotions and cyclic Babsons (note that the table of contents is on the last page).

The so-called “perfect Babson” specifies that in a directmate Babson problem, all full-length variations – the four main lines as well as the subsidiary ones – are without duals. This special task has yet to be realised, and the nearest attempt seems to be P1425088 by Werner Keym (based on a matrix credited to three other composers). A good non-capturing key leads to four accurate thematic variations, with only one side-line bringing duals – tantalisingly close! In his book, Chess Problems Out of the Box, Keym offered a cash prize of 100 euros for the first successful demonstration of a perfect Babson.