Competition Problem 110a
by F.Y. Sing
South to make four spades. West leads the ♠Q.
Successful solvers: Jean-Marc Bihl, Ian Budden, Bülent İyidoğan, Radu Mihai, Sebastian Nowacki, Andries van der Vegt,Wim van der Zijden Tables
Promotions: Congratulations to Sebastian Nowacki on becoming a Master Problemist, and to Radu Mihai on achieving the norm required to become an Expert Problemist.
North wins and leads the ♠6.
A. If East plays low, South wins as cheaply as possible and leads a low diamond. West wins, North playing any diamond, and returns a club to East’s ♣A, on which South must drop an honour. South wins the red suit return and leads a low spade to ♠J and ♠K. East returns the other red suit but South wins and plays the spade winners to squeeze West in three suits. The winner resulting from West’s discard will then squeeze that player again, so long as South plays high again in clubs if West bares the ♣Q.
If East plays hearts twice in the above sequence, South ruffs the second, draws the last trump, crosses to the ♣K and cashes the ♥Q to squeeze West in the minor suits.
B. If East wins at trick 2, best defence is to cash the ♣A. This time South must play the ♣7. East leads another spade to North’s ♠J and then come the ♥A, ♠10, and ♣J, followed by another club if the ♣J holds. South ruffs a heart when North gets the lead, cashes the third club if it is still held, and is now down to four diamonds opposite ♦1082 and ♥Q, West holding ♦QJ65. A low diamond lead forces an entry to North on the ♦10, allowing the ♥Q to score if needed.
See the solution to Competition Problem #4 for the recommended tabular format if you prefer not to write in English prose.
Hugh Darwen, 2014