Competition Problem 125c
by "Bedouin" (a very long time ago)
South is on lead at no-trumps. North-South to force East-West to win eight tricks, the latter doing their utmost to resist.
Successful solvers: In the update of September 6th, 2015, I expressed a belief that only the intended solution was sound and that Wing-Kai Hon and Sebastian Nowacki, who gave that solution, were the only successful solvers. Since then I have had some correspondence with Jean-Marc Bihl that is causing me to change my mind yet again! He has shown me how the problem can after all be solved by North playing the ♦J at trick one, on either a diamond from South or a club, and in the latter case North could also discard the ♦K. So there are at least four solutions, Sotuh leading either a club or a diamond and North playing either the ♦J or the ♦K in each case!
Well, I'm still giving this problem the maximum DR but now I must add Jean-Marc Bihl's name to the list of solvers, so he, Wing-Kai Hon, and Sebastian Nowacki all get 8 MPs and 2 SPs. But please watch this space as I can never be sure we've got to the bottom of this tricky problem! Tables
East is in danger of being forced to win six club tricks plus two in hearts. South starts with the ♦9, to remove East’s safe exit card.
A. If West plays low the trick is ducked to East’s ♦10. East does best to cash two club tricks, on which North’s ♦KJ are jettisoned, followed by the ♥QJ. South follows with the ♥4 and ♥2 as North underplays with the ♥9 and ♥8. If East cashes a third club before exiting in hearts, South discards the ♦8, North the ♠9. In either case South’s carefully preserved ♥7 wins the trick and West can be put on play in diamonds, starting with the ♦7 if West still has the ♦Q.
B. So West does better to rise with the ♦Q, in which case North wins with the ♦K! The ♥9 follows and East must win and cash two clubs before getting off lead in hearts by playing the ♥J and exiting with the ♥3. Again North’s diamonds go on the clubs and again South’s ♥7 is preserved, leaving this position with East to lead the ♥3 and North-South needing to lose four of the last five tricks:
West is caught in a see-saw squeeze!
1. A diamond discard lets North follow with the ♥5 and discard the ♠9 when South’s ♦2 puts West on play.
2. A spade discard lets North overtake with the ♥A and exit in spades.
In both cases West wins the last four tricks.
How Playing the ♦J at Trick One Succeeds (?)
South starts with the ♦9 as above but when West plays the ♦Q North plays the ♦J. If West now leads a club play follows the intended solution and "declarer" makes an undertrick! If West instead leads a diamond, South can win with the ♦8 and exit in clubs. East must take two club tricks, North discarding the ♦K and ♠9. Three rounds of hearts now give a seesaw squeeze much like that in the intended solution: if West discards a diamond North overtakes and leads a spade; otherwise South wins and leads the ♦2. But what if West leads a spade? Then North with the wins ♠9 and leads the ♥9, won by East with the ♥J. East must now cash two clubs as before and North again discards diamonds.
A. If East discarded a club on the ♠9, we have this position:
A heart now would allow North to win and play a spade, giving West the remainder; so East must cash another club, giving this:
but North still threatens to win a heart and put West in with a spade, so East must try another club, but North simply discards the ♥A.
B. If East discarded the ♥Q on the ♠9, we have this position:
A heart now would allow North to win and play a spade, giving West the remainder; so East must cash another club, but North discards the ♥A and we have:
If East leads the ♥3, South wins and exits to West with the ♦2; otherwise, North's ♥5 and South's ♥7 both go away and East takes the last two tricks.
How a Club at Trick One Also Succeeds (?)
First, assume North discards the ♠9. For a while I thought this one worked too but Sebastian Nowacki has convinced me it doesn't. It seems that East can defeat the "contract" by playing the ♦10 to West's ♦Q (optionally preceded by the ♥Q, which holds, but assume this is not played).
A. If North wins with the ♦K and leads a middle heart, East wins, cashes a club on which North can discard the ♦J, then plays the ♥Q and ♥J, which hold, and exits on a heart. North's remaining ♦6 means that North-South must win one more trick after the heart.
B. If North plays the ♦J, West cashes two spades, East discarding clubs, and leads a second diamond. South overtakes North's ♦6 to put East in with a club, on which North discards the ♦K. East plays the ♥Q and ♥J, which hold, and exits on the ♥3. South hopes to win with the ♥7 and lose the last trick to West's ♦3, but West discards it!
But what if North discards the ♦K at trick one, instead of the ♠9? East returns the ♦10 to West's ♦Q and North's ♦J and West leads a spade. North wins this with the ♠9, South discarding a high diamond. If East discards a club on this trick it appears that the "contract" succeeds. But what if East discards a high heart? Previously I had claimed that this succeeded for East-West, but now I have to change my mind again. Here is the position after East has won the first heart and cashed a second club:
A heart now would allow North to win and play a spade, giving West the remainder; so East must cash two more clubs, West discarding a spade and a heart. But on the second of these North can discard the ♥A and if East plays yet another club North can jettison the ♥5 so that East's ♥3 will win.
Moreover: F.Y. Sing suggests that North could discard the ♦J instead of the ♦K at trick one. Now a diamond from East fails as North can win and lead a middle heart, whereas if East cashes a second club North can discard the ♦K.
See the solution to Competition Problem #4 for the recommended tabular format if you prefer not to write in English prose.
Hugh Darwen, 2015