# Enigmatic Code

Programming Enigma Puzzles

## Puzzle 33: Football and addition (Letters for digits)

From New Scientist #1084, 5th January 1978 [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.

(2 points are given for a win and 1 point to each side in a drawn match).

Find the scores in the football matches, and write out the addition sum with numbers substituted for letters.

This completes the archive of puzzles from 1978. There is now a full archive of puzzles from New Scientist from January 1978 to September 1988, and also from May 1999 up to the final Enigma puzzle in December 2013. Together with the Tantalizer puzzles from Mar 1976 to May 1977, there are a total of more than 1350 puzzles available on the site.

There are 31 puzzles remaining to post from the Puzzle series, all from 1977, which fill the gap after the final Tantalizer puzzle.

[puzzle33]

### One response to “Puzzle 33: Football and addition (Letters for digits)”

1. Jim Randell 19 December 2018 at 7:46 am

This Python program uses the [[ Football() ]] and [[ SubstitutedSum() ]] classes from the enigma.py library. It runs in 170ms.

Run: [ @repl.it ]

```from enigma import Football, SubstitutedSum, irange, update

# the scoring system
football = Football(points=dict(w=2, d=1))

# the teams
(A, B, C) = (0, 1, 2)

# the alphametic sum
p = SubstitutedSum(['rg', 'ph'], 'hm')

# solve the sum (to get: g h m p r)
for d in p.solve():

# solve the table (to get: d)
for (ms, d2) in football.substituted_table(dict(played='dg?', d='??m', points='?r?'), d=d):

# choose a value for t
for t in set(irange(0, 9)).difference(d2.values()):
d3 = update(d2, [('t', t)])

# determine the scores (using teams A, C)
for ss in football.substituted_table_goals('r?g', 't?p', ms, teams=[A, C], d=d3):

# check B.goals_for = h
(fB, aB) = football.goals([ss[(A, B)], ss[(B, C)]], [1, 0])
if not(fB == d3['h']): continue

# output solutions
football.output_matches(ms, ss, teams="ABC", d=d3)
```

Solution: The scores in the played matches are: A vs B = 4-5; B vs C = 3-2. The A vs C match is not yet played. The sum is: 42 + 38 = 80.

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