Competition Problem 95b
by Paolo Treossi
South can make six no-trumps against any lead, but which one forces precise play by declarer and how is the contract made against that lead?
Successful solvers: Steve Bloom, Ian Budden, Kukuh Indrayana, F.Y. Sing, Wim van der Zijden
There's a very interesting footnote to this problem, provided by F.Y. Sing. See below.
Only the ♦K forces precise play. Against a heart lead North can play from any suit except diamonds at trick 2. Against a low diamond lead, South can similarly win and play from any suit except diamonds. A lead of the ♣Q allows South to win or duck at trick 1. A spade or low club lead forces declarer to play all the hearts, squeezing West, but after the first two it doesn’t matter which hand wins the next three.
Against the lead of the ♦K, North wins and crosses to the ♠K for the heart finesse. Three rounds of hearts, South preserving the ♥2, force West to discard a diamond. South comes to hand on the ♦Q to lead the ♥2 in this position:
West is caught in a classic seesaw squeeze (also known as an entry-shifting squeeze). A spade discard allows North to win with the ♥3 and set up a spade trick with ♥9 as entry. A club discard allows North to win with the ♥9 and duck a club to West (South winning and continuing the suit if East rises with the ♣J), setting up the clubs with the ♥4 as entry.
F.Y. Sing points out that North's ♣9 is not needed and can be exchanged with East's ♣3. Furthermore, the contract now becomes very interesting on a spade lead. Declarer must immediately play five rounds of hearts, ending in North. If West discards any black card, that will allow the twelfth trick to be established in that suit. But to keep all those black cards means West will have to discard all three diamonds, making winners out of South's ♦Q and ♦J. West discards a spade on the first of those winners, but has to discard a club on the second as a second spade discard lets North overtake and cash spade winners. Now whichever defender wins the third round of clubs has to let North's remaining two aces win the last two tricks! It seems that South's diamonds constitute a kind of extended clash menace, the like of which I don't think I've seen before. Moreover, the ending includes a seesaw element and a stepping-stone element as well. Does anybody know of an antecedent for this ending?
Further addition to Footnote:
F.Y. Sing points out that when West keeps three spades, two diamonds and two clubs, declarer's play isn't quite as straightforward as I suggested. If East keeps one spade and three of each minor suit, then North cashes the ♠A and South plays three rounds of clubs, winning the last three tricks with two diamonds and a club. But if East keeps two spades and two diamonds, then South must cash the top clubs and lead a diamond. West covers to block the suit but South comes to hand on a diamond and throws East in with a club, this being the stepping-stone to North's ♠A.
I should further add that the composer, Paolo Treossi, has expressed his appreciation of F.Y. Sing's observations.
See the solution to Competition Problem #4 for the recommended tabular format if you prefer not to write in English prose.
© Hugh Darwen, 2012
Date last modified: 03 June, 2019