## Problem no. 20, 2003

### The problem

My good friend and former team mate Göran Ofsén recently showed me this puzzle. I couldn't solve it. Maybe you can...

 A Q 9 8 7 6 5 A K 4 3 2 A K J 10 4 3 Q 7 6 J 10 9 8 5 Q J 10 9 8 K Q J 10 9 7 6 5 2 4 3 2 A K 7 6 5 8 4 3 2

It is notrump and the opening lead is the ten of spades. As you can see, 11 tricks is the limit for North-South. Now the question is: How many tricks can they take against perfect defense, if they are allowed to swap one card from East to West – and how should the play go then?

### Solution

If North and South are allowed to make the swap in the most favorable way for them, they can take all thirteen tricks, believe it or not. This is how the deal looks after the swap:

 A Q 9 8 7 6 5 A K 4 3 2 A K J 10 4 3 Q 7 6 J 10 9 8 5 Q J 10 9 8 K Q J 10 9 7 6 5 2 4 3 2 A K 7 6 5 8 4 3 2

The moved cards are the ten of spades and the nine of clubs. That means North will be the declarer (or that East leads out of turn, South accepting it...), but in the premises there was no mentioning of South being the declarer.
In the new setting, the opening lead of the spade ten blows a trick, and since West now is the only one to guard the minor suits, North can squeeze him. It looks like South will be squeezed ahead of West, but that isn't so. A criss-cross squeeze comes to North's rescue.
After seven spade tricks and the ace of hearts, these cards remain:

 – K 4 3 2 A – – – J 10 Q J 10 – K Q 7 6 5 – – A K 7 8 4

On the king of hearts, South discards the seven of diamonds – but West has no card to spare. If he gets rid of a diamond, North cashes ace-king of diamonds and has the ace of clubs as an entry to the long diamond; and if he pitches a club, North cashes the ace of clubs. Then, South's hand is high.

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