In 1985, Swedish player Anders Brunzell, East in the diagram below, was defending 3NT on this layout (hands rotated):
♠ K 9 4
♥ A K 7 3
♦ 10 6 3
♣ J 6 5
♠ 7 ♠ Q 10 6 2
♥ 10 9 8 6 5 ♥ J 2
♦ 9 8 2 ♦ A J 5
♣ 10 9 4 3 ♣ A K 7 2
♠ A J 8 5 3
♥ Q 4
♦ K Q 7 4
♣ Q 8
West led the ♥10 won in dummy with the ♥K. Declarer now led a spade to his ♠J. Next he cashed the ♥Q, West following with the 9, showing he had four left. Declarer continued with the ♦K and West again followed with a 9, this time showing an odd number. Declarer had shown five spades in the bidding, so this was now a double dummy problem for Brunzell. How could he defeat 3NT? What would you do after looking at all four hands?
Brunzell could see that declarer hoped to take five spade tricks (if they broke), three hearts and a diamond. If East ducked the ♦K, declarer would find out that spades didn't split, and play on diamonds, successfully. Therefore, he won his ♦A.
A spade or a diamond return would be fatal. What about a club? If Brunzell played ♣K, ♣A and another, his low club would be set up. But then what would he discard on the ♥A? He would be forced to discard his club winner, and now declarer could concede a spade and make his contract.
Brunzell exited with the ♣2, taken in dummy with the jack. Declarer cashed the ♠K, then the ♥A. On the heart ace, Brunzell discarded the ♣A!
Now, South could not avoid defeat. He threw East in with a spade, but the communications were intact between the East and West hands and declarer went two down. Yes, Brunzell won the prize for that year.
Do you see how declarer can counter this defense? When Brunzell shifts to a low club, he has to win with the queen, not an easy play to find (and wrong on most layouts).
Here is the deal in the BBO Handviewer: