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Pete Tamburro on

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We will finish our little lesson on interference themes with a dandy double version—appropriately so, because H. Johner beat P. Johner with this mate in three back in 19005.


  1. Be4 (1. h3+ Kg3 2. Be4 leads to the same thing) 1... Bxe4 (1... Rxe4 2. h3+ Kg3 3. Rf3#) (1...c3 2. h3+ Kg3 3. Be1# [or 3. Rf3#]) 2. h3+ Kg3 3. Be1#

 


Send questions and comments to PTamburro@aol.com.

 

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No more castling problems, but I was intrigued with G.F. Anderson’s work with the April 15th puzzle. Here’s one he did with C.C. Palmer. A mate in two.

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If you did Friday’s puzzle, you have a hint on today’s solution. But don’t forget to do your income tax first! It’s a mate in four.

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Many of you will recognize this Sam Loyd classic. I’m having it do an encore bow so as to give you a hint as to next week’s puzzles.

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A dazzling display by a minister during the Major Open of the British Championship in 1929.


 

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