Here is a clear example. The bK has a flight d5 and capturing the wR leads to 1...Kxd5 2.Qf3 1...Bxd5 2.Sd2. Notice that the wBd4 plays an important role guarding e3 and e4. As soon as the wB moves it threatens Rd4 but relinquishes guard on at least one of these squares.
1.B~ (say Bc5) >2.Rd4 but 1...Bxd5! However, the wB can provide for this mate by cutting into Black's lines. 1.Be5? corrects by blocking the bBb8 guard on f4. However this also blocks the wR's access to e5. 1.Be5!? >2.Rd4 1...Bxd5 2.Qxe4 but 1...Ba7! We can keep going up 1.Bf6? now blocks the bR guard to f5 but obstructs the wSh7. 1.Bf6!!? >2.Rd4 1...Bxd5 2.Bxf5 but 1...Rd8! Finally the square g7 does it by blocking the bRg8's line. 1.Bg7! >2.Rd4 1...Bxd5 2.Sg5 1...Kxd5 2.Qf3 1...Ba7 2.Re5 1...Rd8 2.Sf6 |

Another memorable problem from the great two-move composer. Not the prettiest position but the beauty is in the logic it displays. There are three relevant set mates:
1...Sf6 2.Qxf6 1...Bxc4 2.Re3 1...gxf5 2.Qe3 A random move of the knight on d5 creates a threat of Rd5 but commits the error of unguarding f4. 1.Sd~ (>2.Rd5) but 1...Sf6! Now the correction begins. 1.Sc3!? (>2.Rd5) 1...Sf6 2.Bg4 but 1...Bxc4! (Re3?) The problem was that key piece blocked the masked line of the wRa3. 1.Sf4!? (extra guard on e6) >2.Rd5 1...Sf6 2.Bxg6 but 1...gxf5! (Qe3) The two tries fail by blocking the wR's and wQ's access to e3. Surprisingly the key commits both of these errors! 1.Se3! (extra guard on f5) >2.Rd5 1...Sf6 2.Qxg3 1...Bxc4 2.Sxc4 1...gxf5 2.Qxf5 1...fxe6 2.Bxg6 1...dxe6 2.Ra5 1...Sf4 2.Qf6 |

The above two problems had more of a pure form of White correction in the sense that there was a set mate to the general refutation. Often this requirement is relaxed. Consider the following problem. A random move of the wSe5 creates two threats: 2.Qe5/Qe6. These threats also have two refutations: 1...Re1/Bf1. The specific moves 1.Sd7? and 1.Sf7? hand 1...Re1 but not 1...Bf1!. Instead the wS must cut into the e2-a6 diagonal. 1.Sd3? (2.Qe5 only) 1...Re1 2.Sb4 1...Qxf4 2.Sxf4 1...Be4! (Qc4?) 1.Sc4! (2.Qe6 only) 1...Re1 2.Sxb6 1...Be4 2.Se3 1...Qh6 2.Qxg2 |

The great Italian composer offers a nice example. 1.S~ (2.Qxe7) 1...Re1 2.Be4 1...Qe1 2.g5 1...Se6! 1.Sc5!? (2.Qxe7) 1...Se6 2.Sd7 1...Re1! 1.Sg5!? (2.Qxe7) 1...Se6 2.Sh7 1...Qe1! Finally the flight granting key works and changes both replies to 1...Re1,Qe1. 1.Sf4! (2.Qxe7) 1...Se6 2.Rxg6 1...Re1 2.Sd5 1...Qe1 2.Sh5 |

The master of correction play offers a three fold arrival form of correction. Marjan Kovacevic selects this as one of his favorite British #2 problems. The point is that a dummy piece on e3 creates the threat 2.Sf4. However each arrival of such piece destroys a set mate. Set: 1...Sh5 2.e4 1...Bxd3 2.Rxd3 1...Rxb4 2.Sxb4 However, the problem is more subtle. Each try actually destroys two set mates but changes one and is refuted by the other. 1.e3? (>2.Sf4) 1...Sh5 2.e4 1...Bxd3 2.Qxd3 1...Rxb4! 1...Re3? (>2.Sf4) 1...Bxd3 2.Rxd3 1...Rxb4 2.Bxc6 1...Sh5! 1...Be3? (2.Sf4) 1...Rxb4 2.Sxb4 1...Sh5 2.Qf3 1...Bxd3! Finally, the wQ imposes a pinning threat and the three mates are preserved. 1.Qd1! (>2.Sf4) 1...Sh5 2.e4 1...Bxd3 2.Rxd3 1...Rxb4 2.Sxb4 I guess my only dislike is that there are not more changes after the key, but overall a wonderful problem. |

Here is an outstanding problem that combines White and Black correction. The White correction is departure based while the Black correction is arrival based. Both White correction moves give a flight. Excellent work. Placing the wK must have been a struggle! 1.S~? (>2.Qg5) 1...Re5 2.Qf3 1...Sde5 2.Qxd2 1...Sce5! 1.Sg3? (>2.Qg5) 1...Sce5 2.Se2 1...Sde5! 1.Sd4! (>2.Qg5) 1...Sde5 2.Qe4 1...Sce5 2.Se2 |

Here is another Christopher Reeves masterpiece. This time we have reciprocal White correction perfectly unified with a White and Black Grimshaw unpinning defense. Beautiful work. 1...Bb5 2.Sb6 1...Rb5 2.dxc4 1.Rg~? (>2.Qe6) 1...Bb5 2.Sb6 1...Rb5! (2.dxc4?) 1.Rf4? (>2.Qe6) 1...Rb5 2.dxc4 1...Bb5! (Sb6?) 1.Bg~? (>2.Qf3) 1...Rb5 2.dxc4 1...Bb5! 1.Bf4? (>2.Qf3) 1...Bb5 2.Sb6 1...Rb5! 1.Bh2! |

Finally, here is my offering with a light 14 pieces. I was fortunate on this to work with the expert D. Shire who basically remade one of my earlier problems. We were able to obtain two obstruction tries and an obstruction key with a beautiful Grimshaw changed to a white interference mate. 1...Rxe5 2.Qd3 1...Bxe5 2.Qf3 1...Re6 2.Bxb7 1.Se~ (>2.Qxd6) 1...Rxe5 2.Qd3 1...Bxe5 2.Qf3 1...Re6 2.Bxb7 1...Rg8 2.Qxg8 1...Sxc7! 1.Sf7!? (>2.Qxd6) placing extra guard on d6 1...Rg8! (Qxg8?) 1.Sc6!!? (>2.Qxd6) 1...Sxc7 2.Sb4 1...Re6! (Bxb7?) 1.Sd3! (>2.Qxd6) the key obstructs Qd3 but changes it. 1...Sxc7 2.Sb4 1...Rxe5 2.Sf4 1...Bxe5 2.Qf3 1...Re6 2.Bxb7 1...Rg8 2.Qxg8 |