Mystery #7

The Original Switch Problem

In the previous mystery, we challenged detectives with the general form of "the switch problem." That research thread began, some time during the late sixties, with a specific construction problem of Ken Lebensold's. Bart Bramley has improved the formulation several times; here, we present an updated form of the original problem. In the realm of puzzledom, this problem has several interesting aspects. Perhaps foremost, it represents an attempt to pose a full-deal construction problem with only one given hand, a dramatic advance from the usual set-ups in this genre. Also, it comes with two strands, one of which allows for creative activity unfettered by the constraint of believing that there is only one solution, already discovered by the proposer. The other thread provides a full mystery of its own, a chase of the elusive dream of uniqueness.

South, declarer in a six-notrump contract that cannot be defeated by any defense, holds,

♠ A J 6 4  ♥ Q 8  ♦ A J 6 3 2  ♣ K 5

The contract can be defeated by best defense if any one of these interchanges is made: the four and three of spades, the five and four of hearts, the five and four of diamonds, the four and three of clubs.

Mystery 7-A: Find one or more deals that satisfy these conditions. (Essentially different solutions may exist.)

A "power-reducing exchange" (PRE) is an interchange of two cards that reduces the number of tricks that declarer could take by direct leads of the suit, presuming adequate entries. For example, in this suit layout,

	NORTH ♠ 9 8
WEST ♠ 7 6 5		EAST ♠ J 10 3 2
	SOUTH ♠ A K Q 4

the exchange of the four and three of spades is a PRE (it reduces declarer's four top tricks to three). But, in this layout,

	NORTH ♠ 8 7
WEST ♠ 6 5		EAST ♠ J 10 9 3 2
	SOUTH ♠ A K Q 4

the exchange of the four and three of spades is not a PRE (declarer would still have only three top tricks, although the exchange prevents him from creating a spade tenace).

Mystery 7-B: Find one or more deals that satisfy the conditions in such a way that none of the four exchanges is a PRE. (There is hope that the solution may be unique. Can you prove it? Or can you find two different solutions?)

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