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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.