The Babson Task – Part 2

In orthodox helpmates, the Babson task occurs when Black and White promote to matching pieces of each type – queens, rooks, bishops, and knights – across four distinct solutions. The demanding scheme means relatively few realisations, and the design constraints are such that examples often share some basic elements. At least three moves are needed to show the theme in a standard helpmate, because for the black queen promotion part, the piece must make two further moves (one orthogonal and one diagonal) to be irreplaceable by a promoted rook or bishop. The motivations for the promotions are usually the same: White employs the new pieces to deliver mate, Black uses them to block the king’s flights. Incorporating extra thematic contents, besides the echoed promotions, is exceedingly rare. Such exceptional helpmates do exist though there may be constructional costs involved, as we shall see.

The Hungarian-American Aurel Karpati (1916-2006) introduced the task in helpmate form during the 1950s. His very first rendition, diagrammed above, has no formal flaws at all. In the four parts, the same black and white pawns promote respectively on the same squares, as called for by the theme. Although shifting the black king in twinning is a bit drastic, such an adjustment is normal for the task and applying it for all four positions may even be viewed as pleasing. 1.b1=S b7 2.Sc3 b8=S 3.Sxb5 Sc6, (b) 1.b1=R b7 2.Rd1 b8=R 3.Rxd5 Re8, (c) 1.b1=Q b7 2.Qxb5 b8=Q 3.Qe8 Qd6, (d) 1.b1=B b7 2.Bxe4 b8=B 3.Bh7 Be5. Later the same composer improved the economy with a setting that contains just eight units: P0508538.

The next demonstration employs twinning that never shifts the black king. That in itself is noteworthy, but what elevates this problem is a bonus unifying effect: Black’s promoted pieces all self-block on the same square, e5. These additional features come at a cost, though, in that the promotions take place on various squares. 1.e1=S f7 2.Sf3 fxe8=S 3.Se5 Sxg7, (b) 1.c1=B f7 2.Bf4 fxe8=B 3.Be5 Bd7, (c) 1.e1=Q fxg7 2.Qa5 g8=Q 3.Qe5 Qxf7, (d) 1.e1=R fxg7 2.Rxe4 g8=R 3.Re5 Rxg6. Except for the queen-promotion phase, each solution ends with a model mate.

The laudable goal to complement the Babson task with four model mates is surprisingly elusive. Most examples involve formal defects, such as a twin position that requires excessive adjustments. I could find only one helpmate that achieves the combination with perfectly acceptable twinning. 1.d1=B d7 2.Bc2 d8=B 3.Bh7 Bf6, (b) 1.d1=Q d7 2.Qd6 d8=Q+ 3.Qf8 Qxf8, (c) 1.d1=S d7 2.Sc3 d8=S 3.Sb5 Sc6, (d) 1.d1=R d7 2.Ra1 d8=R 3.Ra4 Rd5. Compare this with Stepan Tsyrulik’s four-model precursor, P0508635, in which creating the twins (c) and (d) entails double-changes. However, the play in Tsyrulik’s helpmate is so similar to Bakke’s that the latter problem could be seen as largely anticipated!

In helpmates, the multi-solution form is preferred over twinning when feasible, so it’s a natural question as to whether a Babson could stem from a single position with four solutions. The Swedish composer Christer Jonsson answered this in the affirmative with the four-move problem shown. A significant blemish exists, though, in that three of the solutions have repeated first moves, besides the minor flaw of the black king starting in check. Nonetheless, this is an impressive feat of construction for the task, especially in view of the light position utilising only kings and pawns. 1.Kxe4 e6 2.e1=Q e7 3.Qa1 e8=Q+ 4.Qe5 Qxe5, 1.Kxe4 e6 2.e1=B e7 3.Bf2 e8=B 4.Bd4 Bc6, 1.Kxe4 e6 2.e1=S e7 3.Sc2 e8=S 4.Sd4 Sf6, 1.Kc6 e6 2.e1=R e7 3.Rc1 e8=R 4.Rc5 Re6.

The Babson Task – Part 1

22 Dec. 2025