Competition Problem 134a
by Vincent Labbé
South to make three no-trumps. West leads the ♠10.
Successful solvers: Steve Bloom, Ian Budden, Abby Chiu, Radu Mihai, Sebastian Nowacki, F.Y. Sing, Andries van der Vegt, Wim van der Zijden Tables
Best defence is for East to win with the ♠A and return the suit to South’s ♠K. A club to the ♣A is followed by the ♥4! (Trap, below, explains why this might be necessary). If East rises with the ♥A, North has enough entries in hearts to set up the diamonds, whatever East returns. So East ducks and the ♥J wins. South cashes the ♣K and ♣Q, North discarding a spade and a diamond, then leads the ♦2. West must play an honour to prevent the diamonds from being established by losing one to the ♦Q, and East must drop the ♦Q under North’s ♦A for the same reason. North now leads the ♥K. If East ducks, West is thrown in immediately with a spade; otherwise, East can make the ♥A and ♣J but they force spade discards from West and when North gets back in on the ♥Q West is again thrown in with a spade. In either case West eventually has to lead a diamond away from ♦J8 into North’s ♦K9.
Trap: If at trick four, after making the ♣A, North cashes a top diamond, East drops the ♦Q, West the ♦8. If North now plays on hearts from the top, East ducks twice. North can now try throwing West in on the third round of diamonds, only to find that West had played the ♣J at trick three! Now East can overtake West’s club return (otherwise, North discards a heart and South ducks so that West has to concede three of the last four tricks to either North or South).
See the solution to Competition Problem #4 for the recommended tabular format if you prefer not to write in English prose.
Hugh Darwen, 2015