Here reciprocal change is effortlessly executed in Meredith. Using bK flight squares is a common, but striking way to pull of reciprocal change. We have relevant set play: 1...Kxf5 2.Rxe3 1...Ke4 2.Rxd5 The pendulum type key threatens mate on e6 and now the mates are swapped after the bK moves: 1.Qb6! (>2.Qe6) 1...Kxf5 2.Rxd5 1...Ke4 2.Rxe3 In order to pull off reciprocal change one needs two effects after the key: the new mates must work and the old mates must fail. In this case obviously the wQ plays a crucial role. Before the key she guards the squares d5 and g5 and she gives these up for guard on e3 and e6. |

In this problem we have a similar set up with an indirect battery aimed at the bK flights. Moreover, each of these flights is set with a dual: 1...Kd4 2.Rd3/Rc5 1...Ke5 2.Rd3/Rc5 The try and key separate the duals and reciprocally change them. Here the mechanism is uses the guard on the squares f5 and c4 and the opening of black lines to swap the mates. 1.Sge4? (>2.Qd6) 1...Kd4 2.Rd3 1...Ke5 2.Rc5 1.Sde4! (>2.Qd6) 1...Kd4 2.Rc5 1...Ke5+ 2.Rd3 |

Here we have a doubling of the theme. A common mechanism for reciprocal effects is the forming of batteries: should the rear piece play first forming the battery or should the front piece play and clear the way for the rear piece. In this amazing problem we see this strategy to perfect. We see two possible Q+R batteries to be formed one with the wQ to f1 and another with wR capturing e6. However, we can move the front piece for two nice tries. 1.Re3 X? (>2.Rc3) 1...Sxa4 2.Qf1 A 1...Sd7 2.Qxe6 B 1...Qh8! 1.Re5 Y? (>2.Rxc5) 1...Sxa4 2.Qxe6 B 1...Sd7 2.Qf1 A 1...Qf2! 1.Qf1 A? (>2.Re4) 1...Sxa4 2.Re3 X 1...Sd7 2.Re5 Y 1...Sd3! 1.Rxe6! (>2.Re4) 1...Sxa4 2.Re5 Y 1...Sd7 2.Re3 X 1...Sxe6 2.Qxe6 1...Qd4 2.Rxd4 |

Here we have a twinning mechanism to achieve reciprocal change. The reciprocally changed mates come from two different knights playing to the same square. a) 1.Qb5 (>2.Qb3) 1...Se2 2.Sfe5 1...Se4 2.Sce5 b) 1.Qa4 (>2.Qb3) 1...Se2 2.Sce5 1...Se4 2.Sfe5 |

We have a wonderful prize winning problem. Notice the wR,wS, and wB on the e-file. This is what is known as a half-battery. Guarding e6 (in particular if any move of wSe7 or wBe6) will threaten 2.Sf7. However, something must be done about the defense 1...Sg5. Let's try moving the wQ first. 1.Qa2? (>2.Sf7) 1...Qxe7 2.Qd5 1...Qxe6 2.Qxe6 1...Sg5 2.Qh2 but 1...Bxe3 This almost works but the wQ abandoned her access to e3. Now it is obvious that we should move the wBe6 or wSe7. Let's try the wB which should go to f5 (why?) 1.Bf5? (>2.Sf7) 1...Qxe7 2.Qc5 (2.Qd4?) 1...Qe6 2.Qd4 1...Sg5 2.f4 (notice the wB now guards e4) 1...Bxe3 2.Qxe3 but 1...Qd5! 1.Sg8! 1...Qe7 2.Qd4 1...Qxe6 2.Qc5 (2.Qd4?) 1...Sg5 2.Bxf6 1...Bxe3 2.Qxe3 1...Qxd7 2.Bxd7 Thus we see the half-battery theme, reciprocal change, and a 3x3 Zagoruiko (3 different mates after the defenses 1...Qe7,Qe6,Sg5). Wow! |

We end with my reciprocal change problem which is obtained by means of twinning. The twins have the same key, which can be considered a flaw, but here it might be a feature. Part a) is a usual Novotny whereas part b) is what I call a reversed Novotny. The reciprocal change happens by pinning and line opening, i.e., if the bQ is removed from the board then after the captures of the key pawn on d4 both mates Bb3 and Qxe5 would work. However, in part a) she blocks the wRd1's line so only the standard Novotny mates work, where as in part b) the moves of the bBc3 and bRf4 open the bQ's lines so the other mates are the ones that work. One thing that was really important to me was working in the defense 1...Qe3 2.Sxf4 in part a) so that the wRd1 was utilized. a) 1.d4! (>2.Bb3 A/Qxe5 B) 1...Bxd4 2.Bb3 A 1...Rxd4 2.Qxe5 B 1...Qe3 2.Sxf4 1...Kxe6 2.Qf7 b) 1.d4! (>2.Bb3 A/Qxe5 B) 1...Bxd4 2.Qxe5 B 1...Rxd4 2.Bb3 A |