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Competition Problem 148a composed
by
Stefan
Ralescu
South to make six no-trumps against any defence. Successful solvers: Steve Bloom, Ian Budden, Johnson, Leigh Matheson, Radu Mihai, Sebastian Nowacki, A.V. Ramana Rao, Zoran Sibinović,, Rajeswar Tewari, Andries van der Vegt, Dick Yuen, Wim van der Zijden Tables
Solution West does best to lead a club to North. Three diamond winners follow and West does best to throw a spade—otherwise either the hearts or the clubs can be established with an entry (though this needs a second squeeze when West discards a heart, in which case North is entered on a club to cash the ♦J). North is entered on the ♥A and then the ♦J, South throwing a heart, wrings another spade from West. This allows South to overtake North’s ♠K, score the ♥K, then lead the ♠8 to subject West to a seesaw squeeze in this position:
A. If West discards a club North overtakes and has an entry in spades when South jettisons the ♠J on the next club. North can play a low club next or the ♣A and another to establish a club trick. East then has to give North the lead on the ♠7. B. If West discards a heart, North plays the ♠7 and South gives up a heart to establish the suit with ♠J as entry. On the lead of the ♠9 North must play the ♠10, South the ♠J or ♠8. Then, when West discards two spades on the diamonds, South comes to hand on a spade, keeping hold of the best spade, squeezing West without the count. If West discards a heart, South sets up the suit with a spade entry. If West discards a club, declarer cashes the ♥K and then sets up the clubs, discarding that spade so that East has to give North the remainder. Similar endings can be obtained by a variety of plays against other opening leads. See the solution to Competition Problem #4 for the recommended tabular format if you prefer not to write in English prose.
©
Hugh Darwen, 2017 |