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Programming Enigma Puzzles

2 September 2015

Posted by on **From New Scientist #1455, 9th May 1985** [link]

The team of seven has remained unchanged since 1981. The Chancellor described the annual team photographs to me as follows:

1981: Fred, Graham, Hermann, Jack; and men from Trinity, Unity and Varsity.

1982: Ken, Fred, Graham; and men from Unity, Varsity, Westminster and Sanctity.

1983: Graham, Hermann, Jack, Ken; and men from Varsity, Sanctity and Trinity.

1984: Jack, Ken, Fred; and men from Westminster, Unity, Trinity and Sanctity.“Can you match up the names and colleges?”

“Surely”, I said, “there is some duplication, isn’t there?”

“Yes. There are three men of the same name. But there are not three men from one college”.

I found I still couldn’t do a complete matching, but I could say with certainty that ______ college had two men in the team.

Which college, and which two names?

[enigma307]

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This Python program examines all assignments of names to colleges. It runs in 1.4s.

Solution:It is certain that Westminster College has two men in the team. They are Graham and Jack.There are two possible scenarios:

Each possibility has three men with the same name (three Freds in the first case and three Kens in the second one), and two colleges with two men (Westminster and Unity in the first case, Sanctity and Westminster in the second).

I found this quite an intricate Enigma to code. Some preliminary analysis can identify the colleges attended by the 3 men with the same name:

If K is the triple name, then 1981 implies 3 K’s attend T,U,V

If J is the triple name, then 1982 implies 3 J’s attend 3 of U,V,W,S

1981 implies 2 J’s attend 2 of T,U,V, therefore U,V.

1983 implies 2 J’s attend 2 of V,S,T, therefore V,S

3 J’s must attend U,V,S

If H is the triple name, 1982 implies 3 H’s attend 3 of U,V,W,S

1981 implies 2 H’s attend 2 of T,U,V, therefore U,V

1983 implies 2 H’s attend 2 of S,T,V, therefore S,V

3 H’s attend S,U,V

If F is the triple name, 1983 implies 3 F’s attend S,T,V

If G is the triple name, 1984 implies 3 G’s attend 3 of S,T,U,W

1981 implies 2 G’s attend 2 of T,U,V, therefore T,U

1983 implies 2 G’s attend 2 of V,S,T, therefore S,T

3 G’s must attend S,T,U

So, some code: