Competition Problem 106b
each of the layouts (a) and (b) below your task is the same:
Successful solvers: Jean-Marc Bihl, Steve Bloom, Ian Budden, Leigh Matheson, Radu Mihai, Sebastian Nowacki, F.Y. Sing, Dick Yuen, Wim van der Zijden. Updated tables.
1. West must lead the ♣K! If this is allowed to hold, West must switch to spades to break up the threatened black suit squeeze. If instead South wins with the ♣A, the best chance is to play a diamond to the ♦A followed by four heart tricks. However, there is no squeeze so long as West discards nothing but diamonds on the hearts and then a club on the ♦K.
2. Assume West leads a spade. Then North wins and cashes the ♦A (key play) on which South plays the ♦2. Four rounds of hearts follow and West again does best to discard diamonds, coming down to four clubs and three spades. North discards spades. Now South leads the ♦K.
(i) If West discards a spade, North plays the ♦3. Now a diamond to the ♦Q is followed by ♠Q and another spade to force a club lead from West. The ♣Q becomes not only a winner but also an entry to North’s good spade.
(ii) If West discards a club, North drops the ♦Q. South leads a low club and regardless of whether West ducks or wins with the ♣K, South’s heart loser goes on the ♠Q and the long club can be established with the ♦4 as entry.
The only difference in layout (b) is that West must start with a low club! On the lead of a spade or the ♣K, the diamond at trick two comes from South, of course. On a spade lead the play is then exactly as in (ii) above. On the ♣K lead North can win the first diamond with either the ♦A or the ♦Q. Then, because South, in the ending, still has the ♠A and North the ♣Q, it doesn’t matter which hand wins the second diamond trick, just so long as South leads the ♦K and North plays high or low depending on West’s discard.
See the solution to Competition Problem #4 for the recommended tabular format if you prefer not to write in English prose.
© Hugh Darwen, 2014
Date last modified: 11 March, 2017