# Capture cycles in proof games

### 24Apr. 2022 | by Peter Wong

Cyclic play as a problem’s theme comes in myriad forms, but basically it indicates a pattern where at least three elements of some sort get rotated from ABC to BCA. In previous Walkabouts, we have seen demonstrations of such in directmates (e.g. cyclic shift of mates) and helpmates (e.g. cyclic change of functions, second problem). In proof games, the scope for such rotational effects is far more limited; one unusual example, an idea I originated some thirty years ago, is the capture cycle. In this theme, at least three units of a player are captured in a certain order in one solution, but in another phase, their order of capture is cyclically shifted. While this idea sounds simple enough (and the problems themselves are fairly short and easy to solve), implementing it is another story, as multi-phase proof games are not straightforward to construct. Recently I decided to revisit the theme and tried to extend the pattern as much as possible. Shown below are the resulting problems along with their precursors; they represent all instances of this cyclic effect, as far as I know.

Peter Wong
feenschach 1993, Hon. Mention

Proof game in 10, 2 solutions

In the first diagram, Black’s move sequence is largely fixed and the main question is how does White place the four missing pieces – the queen, c1-bishop, and two knights – into the paths of the developing black units. Black does have a single spare or “invisible” move that must be allocated to the b8-knight, to facilitate its own capture. The first solution goes 1.Sc3 Sf6 2.Se4 Sd5 3.Sf6 exf6 [A] 4.d3 Bd6 5.Bf4 Bxf4 [B] 6.Sf3 d6 7.Se5 Sd7 8.Sxd7 Bxd7 [C] 9.c4 Qc8 10.Qa4 Bxa4 [D]. The thematic captures are marked to highlight the order in which the four pieces disappear: A = b1-knight, B = bishop, C = g1-knight, and D = queen. The second solution is 1.d3 Sf6 2.Bg5 Sd5 3.Bf6 [B] exf6 4.Sh3 Bd6 5.Sf4 Bxf4 [C] 6.c4 d6 7.Qa4+ Sd7 8.Qxd7+ Bxd7 [D] 9.Sc3 Qc8 10.Sa4 Bxa4 [A], which sees a cyclic shift of the capture order to BCDA. The repetition of Black’s play in the two phases seems to be a weakness, though that actually helps to create another formal cycle, viz. pairs of white pieces take turns to occupy the same square (AB-f6, BC-f4, CD-d7, DA-a4).

Soon after this problem was published, effecting a 4x2 capture cycle, I also produced a 5x2 rendition of the theme for Phénix. The latter game is similar in depicting unchanged black play and a supplementary white cycle, but alas it was cooked and incurable. Here’s a link to the problem on the PDB database.

Peter Wong
Original

Proof game in 8
Twin (b) Rh2 to a2

In retackling the theme, I sought to maximise the number of thematic captures in the two phases. Using a single piece to make these captures seems conducive to the task, and it also leads to less repetitive play but no additional cycle. Experimenting with different pieces as the capturer, I found the rook to be the most suitable, as it’s flexibly strong but less prone to cooks than a queen. Still, it took a surprising amount of effort to concoct another 5x2 cycle with this scheme. When that was done, I aimed for 6x2 and nearly gave up on it before arriving at the twin setting above. Part (a) is solved by 1.a4 h5 2.a5 Rh6 3.a6 Rxa6 [A] 4.b4 Rxa1 [B] 5.Ba3 Rxa3 [C] 6.f3 Rxf3 [D] 7.h3 Rxh3 [E] 8.Rh2 Rxh2 [F]. One aspect I had to abandon is starting the second-phase captures with piece B, which would have been clearer for the theme. Instead the new sequence begins with the removal of piece C: (b) 1.b4 h5 2.Bb2 Rh6 3.Bf6 Rxf6 [C] 4.f4 Rxf4 [D] 5.h4 Rxh4 [E] 6.Rh3 Rxh3 [F] 7.a3 Rxa3 [A] 8.Ra2 Rxa2 [B]. All six white units change the squares where they are captured.

Peter Wong
The Problemist 1992, 4th Hon. Mention

Proof game in 7
Twin (b) Qd3 to b3, (c) Qd3 to h4

The next problem was the earliest to realise the theme, and it remains a personal favourite. The triplet of positions employs the black queen to capture three white pieces (A = queen, B = knight, C = bishop) to attain a full 3x3 cycle: ABC, BCA, and CAB. 1.c3 Sh6 2.Qa4 Sf5 3.Qxd7+ Qxd7 [A] 4.Sf3 Kd8 5.Sd4 Qxd4 [B] 6.e3 Sd7 7.Bd3 Qxd3 [C], (b) 1.Sf3 Sh6 2.Se5 Sf5 3.Sxd7 Qxd7 [B] 4.e3 Kd8 5.Bb5 Qxb5 [C] 6.c3 Sd7 7.Qb3 Qxb3 [A], and (c) 1.e3 Sh6 2.Bb5 Sf5 3.Bxd7+ Qxd7 [C] 4.c3 Kd8 5.Qa4 Qxa4 [A] 6.Sf3 Sd7 7.Sh4 Qxh4 [B]. Once again, there’s a complete change of capture squares for each missing white piece.

Peter Wong
Phénix 2020

Proof game in 5
Progressive twin (b) Re4 to d6, (c) & Rd6 to g5, (d) & Pa5 to a4, (e) & Rg5 to d5, (f) & Pa4 to a5, Pe2 to e3

When three elements are arranged in specific orders, there are six possible permutations: ABC, ACB, BAC, BCA, CAB, and CBA. Our last example attains this maximum number of permutations in its capture cycle, involving White’s missing bishop (A), pawn (B), and queen (C). Progressive twinning (signified by ‘&’ in the stipulation) is utilised here, and it differs from standard twinning in that each specified adjustment is applied to the starting position of the previous part, not to the diagram. (The last twin (f) regrettably requires a double-shift of two units.) The solutions are: 1.d4 a5 2.Bf4 Ra6 3.Bd6 Rxd6 [A] 4.Qd3 Rxd4 [B] 5.Qe4 Rxe4 [C], (b) 1.d4 a5 2.d5 Ra6 3.d6 Rxd6 [B] 4.Bh6 Rxh6 [A] 5.Qd6 Rxd6 [C], (c) 1.d4 a5 2.d5 Ra6 3.d6 Rxd6 [B] 4.Qd5 Rxd5 [C] 5.Bg5 Rxg5 [A], (d) 1.d4 a5 2.Qd2 a4 3.Qa5 Rxa5 [C] 4.d5 Rxd5 [B] 5.Bg5 Rxg5 [A], (e) 1.d4 a5 2.Qd2 a4 3.Qa5 Rxa5 [C] 4.Bg5 Rxg5 [A] 5.d5 Rxd5 [B], and (f) 1.d4 a5 2.Bh6 Ra6 3.e3 Rxh6 [A] 4.Qh5 Rxh5 [C] 5.d5 Rxd5 [B]. The cost of this maximum task is apparent, namely a few repetitive parts where some white units don’t vary their capture squares. On the other hand, six thematically related phases in a proof game may well be a first.

29 Apr. 2021