Competition Problem 120a
by Steve Bloom
South is in five
Successful solvers: Jean-Marc Bihl, Ian Budden, Abby Chiu, Leigh Matheson, Sebastian Nowacki, Wim van der Zijden. Those who failed mostly missed declarer's opportunity in clubs and the resulting need for West's key play in that suit at trick two. Tables
Yes, so long as West leads the ♦J! Assume North plays the ♦10 and South the ♦Q. Then South tries the ♣2, but West rises with the ♣10 to deny North two entries in that suit. Still hopeful, declarer runs the clubs, East discarding a spade and three hearts, and then throws West in on a spade. Now we have the following ending, with West to lead.
A. If West cashes the club, the diamond winners will then squeeze East.
B. If West leads a low diamond, North plays the ♦7 and East is caught in a seesaw (or entry-shifting) squeeze, South playing the ♦8 if East comes down to three spades, the ♦A if East comes down to two hearts. In either case a trick is conceded to East in the unguarded suit, with the remaining diamond as entry to the established winner(s).
C. However, if West leads the ♦9, Eastís choice of discard depends on Northís play: a heart if North plays the ♦A, otherwise a spade. The seesaw squeeze has been foiled and the contract defeated. The situation is effectively the same if North plays the ♦K at trick one and South overtakes.
If West leads a lower diamond, say the ♦9, North can win with the ♦10 and then play the ♦7 at trick seven. The third round can now be won by either the ♦A or the ♦K and so the seesaw squeeze takes effect.
If West plays a low club at trick 2, North wins with the ♣9 and immediately gives up a spade. South scores in some order the ♦A, a spade and lower club honour, then crosses to North on a diamond to lead a low club towards the ♣AQ. East is again caught in the seesaw squeeze, this time with clubs as the entry-shifting suit.
See the solution to Competition Problem #4 for the recommended tabular format if you prefer not to write in English prose.
Hugh Darwen, 2015