Illegal clusters

29 Sep. 2022 | by Peter Wong

In retro-analytical problems, the solver has to consider the past of a given position and deduce what occurred in the preceding play. A broad range of tasks are possible, such as determining the legality of a castling move, uncovering the last moves played, or reconstructing the whole proof game. Another form of retro is the illegal cluster, invented by T. R. Dawson in the 1930s. In this type of construction task, you are shown a position along with a list of pieces that must be added to it. The goal is to create an illegal position, but one that becomes legal upon the removal of any one unit (except a king). Legal/illegal positions are defined as those that can/cannot be reached via a sequence of legal moves from the opening array. Illegal cluster problems have a sort of built-in elegance to them. Every piece in the solution position must somehow contribute to its illegality, a delicately balanced state that tips over into a legal one if any unit disappears.

Thomas R. Dawson
The Problemist Fairy Chess Supplement 1933

Illegal cluster
Add 7 BPs
(b) Add 8 BPs

Solution to (a)

Solution to (b)

Our first example, by Dawson himself, demonstrates two different kinds of illegality. For part (a), how do you add seven black pawns to the position to make it illegal, while the subsequent removal of any one of these pawns, or the queen, would render it legal? The solution exploits how the queen is checking White, by surrounding the piece with the pawns, such that it had no possible way of delivering the check. This impossible check situation is made legal upon the removal of any pawn, or the queen itself. In part (b), when eight black pawns are to be added, the same checking configuration cannot be used since wherever the eighth pawn is placed, its removal would still leave the position illegal. Instead, all eight pawns are put on their original squares, meaning the queen couldn’t have escaped from the top rank. Removing any pawn will legalise the position by enabling such an escape (or even by allowing the queen to be promoted). Here the illegality stems from the general features of the position, rather than from the lack of specific retraction moves as seen in the first part.

Jorge Lois
feenschach 1976, 4th Commendation

Illegal cluster
Add WQ, 2 BPs
(b) Add WR, 2 BPs
(c) Add WB, 2 BPs
(d) Add WS, 2 BPs
(e) Add WP, 2 BPs

Solution to (a)

Solution to (b)

Solution to (c)

Solution to (d)

Solution to (e)

The next problem is a classic with five homogeneous parts. In each of the settings, you add a different white unit plus two black pawns to create the illegal cluster. All five solutions employ the white unit to give an impossible check, and it’s remarkable how exactly three black pawns (including the one in the diagram) can serve to constrain that unit regardless of its mobility type. Incidentally, the white king’s placement ensures that the configurations of parts (c) and (e) can’t be reflected left-to-right, as the alternatives would leave a checking pawn on e4.

Bernd Schwarzkopf
feenschach 1987

Illegal cluster
Add 5 WPs
(b) BBh5

Solution to (a)

Solution to (b)

Both parts of this problem require adding five white pawns, and the twin (b) is produced by replacing the h5-pawn with a black bishop. So in (a) six white pawns in total and the black bishop will partake in the illegality. Here the solver needs to know a property about the pattern of six white pawns standing on the same file at the edge of the board. Such an arrangement is possible, but it implies that these pawns have captured fifteen units, leaving Black with the bare king. Thus if the five pawns are placed on the h-file along with their sibling, an illegal position arises since Black still has two pieces remaining. Naturally, the capture count becomes attainable when any pawn or the bishop is removed.

Part (b) is easier as a second light-squared bishop provides a big hint. We want to add the pawns on the second rank in such a way that would prevent a promoted light-squared bishop from ever escaping to f5/h5. It’s tempting to put four pawns on a2, c2, e2, and g2, but that doesn’t work because there’s no viable square for the fifth pawn. The solution is similar to this try, the only difference being that two pawns occupy b2 and b3 instead of the one on a2. The position is still illegal because the b3-pawn comes from a2 and a promoted bishop on b1 couldn’t have left the corner. But if, say, the b2-pawn is gone, then the b3-pawn could have originated from b2, meaning the promoted bishop was able to escape before the b2-b3 move was made.

Wolfgang Dittmann
The Problemist 1976

Illegal cluster
Add WK, BB, BP
(b) Rc2 to d8

Solution to (a)

Solution to (b)

Our final example returns to the idea of impossible checks, but with more complicated retraction play. Since we are adding a lone white king plus a black bishop and pawn, the check clearly will be given by one of the four black units. In (a), a rook check seems unlikely because it’d be difficult to involve the distant h7-pawn in generating the illegality, but that pawn is well-placed to constrict a bishop put on g8 to give an illegal check along the a2-g8 diagonal. Where exactly do we add the king and pawn so that removing this pawn or the c2-rook would make the bishop check legal? The solution is to put the king on b3 and the pawn on c3. Now removing the c3-pawn will allow the discovered check, …Rc4-c2+. If the rook disappears, the bishop check becomes legal by means of an en passant capture, i.e. retract …dxc3 e.p.+ c2-c4 and further, …d5-d4+.

For the twin (b), the rook is shifted to d8, suspiciously making a castled position. Suppose we place the white king on either g8 or h8, then a pawn on d7 would prevent the rook from delivering a normal check (not a d7-bishop which could have discovered check). The purpose of the h7-pawn must be to preclude a king move from h7, so that in the final position, if we try to retract the rook check by uncastling, the resulting position (with the black king on e8) would still be illegal, because the white king has no possible last move. This plan also requires the white king to be unable to retract to g7 (and elsewhere), a job that’s handled by the black bishop. If the king is added on g8 and the bishop on g7, that fails because the position is legal – last move could be …Bf8-g7+. Closer is to guard g7 with the bishop on a1 instead, but this is also legal, because we can retract, e.g. …0-0-0+ Kg7-g8 and further, …a1=B+. Finally, the solution is to place the bishop on h8. Here retracting …0-0-0+ Kg7-g8 is illegal as the bishop could not have executed the corner check. Removing the bishop or the h7-pawn will legalise the uncastling check, while removing the d7-pawn will permit a simple rook check that’s preceded by a white king move from f7/f8. This gem of an illegal cluster brings about both en passant and castling play in a very light setting.


4 Oct. 2022

Illegal cluster problems are notoriously prone to be cooked, and it turns out that the instructive Dawson problem above also demonstrates this unfortunate tendency! Thanks to Andrew Buchanan who points out that both parts of this problem are unsound; his clever alternative solutions are diagrammed below.

Cook to (a)

Cook to (b)

In (a), where seven pawns are added, the position is illegal for the following reason. Black’s last move was the double-check …exd4+, a capture that accounts for one of White’s fifteen missing units. The pawn structure resulting from this retraction – doubled pawns on the e-file and five on the h-file – requires another fifteen captures, and the total of sixteen is one too many to be feasible. Now if the queen is removed, the last move could be …cxd4/d5-d4+, and in either case this queen-side pawn didn’t have to take a detour to the e-file, and the total number of captures needed for the pawn structure is only fourteen. Removing any one of the pawns, thus reducing their numbers on the same file (e- or h-), will likewise bring down the necessary captures to below fifteen.

The eight-pawn position for (b) is nearly symmetrical to (a) and the argument for its illegality is virtually the same. We only need to note that if the a7-pawn is removed, one of the b-pawns could have started from the a-file and again the capture count becomes manageable. Both Andrew and I have attempted to correct this illegal cluster problem (such as by placing the black queen and white king in other checking formations) with no luck.

Help retractors

2 Feb. 2024