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Programming Enigma Puzzles

3 May 2015

Posted by on **From New Scientist #2392, 26th April 2003** [link]

Albion, Borough, City, Rangers and United have played another tournament in which each team played each of the other teams once. Two matches took place on each of five successive Saturdays, each of the five teams having one Saturday without a match.

Three points were awarded for a win and one for a draw. Albion had a clear points lead after the first Saturday, likewise Borough after the second Saturday, City after the third and Rangers after the fourth; and United were joint leaders after the final Saturday.

Knowing this you would be able to deduce with certainty the results of both the matches played on the final Saturday if I told you which matches were played on that day.

Which matches were played on the final Saturday and what was the result for each one?

[enigma1236]

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We don’t need to bother with the actual scores in the matches, so here’s a solution that doesn’t use the

Football()helper class fromenigma.py. Instead of coding it as a bunch of nested loops I used a recursive routine, although even that gets a little involved.This Python 3 program runs in 115ms.

Solution:The matches played on the final Saturday were Borough vs. United, which was a win for United, and City vs. Rangers, which was a draw.There is only one possible sequence of matches that leads to a unique solution:

Week 1: A vs. R, A wins; B vs. C, match drawn. Points: A=3, B=1, C=1, R=0, U=0.

Week 2: A vs. B, B wins; C vs. U, match drawn. Points: B=4, A=3, C=2, U=1, R=0.

Week 3: A vs. C, C wins; R vs. U, R wins. Points: C=5, B=4, A=3, R=3, U=1.

Week 4: A vs. U, U wins; B vs. R, R wins. Points: R=6, C=5, B=4, U=4, A=3.

Week 5: B vs. U, U wins; C vs. R, match drawn. Points: R=7, U=7, C=6, B=4, A=3.

There is a unique solution but in the sequence week four matches can be interchanged- A v R, R wins and B v U, U wins.

I’m wrong. They can’t be interchanged because each team plays each other once.