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George Duke wrote on Mon, Nov 22, 2010 06:16 PM UTC:
Those four would be the briefest ones. So, the next problemist's dilemma is better stated, after the 4 obvious cases, what is the next speediest Help-checkmate by one of the other 26? They should have asked this first before the entire century's thousands of Helpmates. (War and proliferation could have been prevented.) Notice that pristinely from the array here, there is no concern of so-called duals, in fact the more the better; to be determined is just how many moves minimum. This is new ground, not duplicative of the Orthodox restrictive subject matter. Yet Helpmate as term still usefully applies for the close similarity to fulfill a condition, checkmate. Here is a solution minimizing Knight-g8, and Black help-mated by White helpmates in six: 1 g4, N g8-f6 2 B f1-g2, N f6-h5 3 d3, N h5-f4 (Notice unlike check White can decline capture.) 4 N b1-d2, a6 (''sort of two waiting moves'' by Black this and next) 5 N d2-f1, b6 6 B c1-d2, N f4-g2. Checkmate. Because of the waiting moves, many other 6-move sequences are correct. The Black Knight at g8 has a mate on move 6 unassisted if things go well. For continuation, back to the Rook-a1 or -a8, which are two different cases. Can any Rook achieve mate, unassisted by own in the final checkmate, but helpmated by opposite side in <=9 moves?