Enigmatic Code
Programming Enigma Puzzles
Enigma 235: Double trouble
Posted by on 5 November 2014
From New Scientist #1381, 27th October 1983 [link]
In the following football table and addition sum letters have been substituted for digits (from 0 to 9). The same letter stands for the same digit wherever it appears and different letters stand for different digits.
The three teams are eventually going to play each other once — or perhaps they have already done so.
(Two points are given for a win and one point to each side in a drawn match).
Find the scores in the football matches and write the addition sum out with numbers substituted for letters.
Note: I now have a phone line installed at my new house, but I’m waiting for ADSL to be activated on it, so I only have sporadic access to the internet at the moment. The current estimate is that I should have an internet connection next week.
[enigma235]
Enigmatic Code 
This Python program uses the [[
Football()]] helper class from the enigma.py library. It runs in 33ms.from enigma import irange, Football, printf digits = set(irange(0, 9)) football = Football(points={ 'w': 2, 'd': 1 }) # consider possible match outcomes for (AB, AC, BC) in football.games(repeat=3): # generate the table A = football.table([AB, AC], [0, 0]) B = football.table([AB, BC], [1, 0]) C = football.table([AC, BC], [1, 1]) # values for n, y, z (n, y, z) = (C.l, B.d, C.points) # determine p from the sum: 2(10n + p) = 10p + z (p, r) = divmod(20 * n - z, 8) if r > 0: continue ds1 = digits.difference([n, y, z, p]) if len(ds1) != 6: continue # total number of goals against = y + n + p # and this is the same as the total of the goals for column # = x + q + q, where x = Goals for A g = y + n + p for q in irange(0, g // 2): if q not in ds1: continue x = g - 2 * q # generate possible scorelines for A for (sAB, sAC) in football.scores([AB, AC], [0, 0], x, y): # and the remaining match for B for (sBC,) in football.scores([BC], [0], q, n, [sAB], [1]): # and check C if football.goals([sAC, sBC], [1, 1]) != (q, p): continue printf("AB={AB}:{sAB} AC={AC}:{sAC} BC={BC}:{sBC}") printf("A={A}") printf("B={B}") printf("C={C}") printf("n={n} y={y} z={z} p={p} q={q} x={x}")Solution: A vs. B was a 0-0 draw; A vs. C was a 2-1 win for A; B vs. C was a 3-2 win for B. The addition sum is 25 + 25 = 50.