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It turns out the B cannot have lost any matches, so they only have 1 draw and so their “goals for” and “goals against” values must be the same (i.e. 3), and this means C’s “goals for” must be 7. This is enough information to allow the [[

`Football()`

]] helper class from theenigma.pylibrary to determine the match outcomes and scorelines.This Python program runs in 101ms.

Run:[ @repl.it ]Solution:The scores are: A vs C = 0-3, A vs D = 5-2, B vs D = 3-3, C vs D = 4-0. The other matches remain to be played.