Brain-Teaser 492: Light on the matches
From The Sunday Times, 1st November 1970 [link]
Six football sides, A, B, C, D, E and F are all to play each other once, with 2 points for a win, 1 point for a draw.
After some of the matches have been played a battered piece of paper is discovered on which has obviously been written particulars of the results so far.
All that can be read is shown below:
What were the details (opponents and scores) of E’s matches? (Example: EvP = 4-3, EvQ = 0-2, etc. E’s score first).
This puzzle is included in the book Sunday Times Brain Teasers (1974).
[teaser492]
John Crabtree and 
S2T2 
Jim Randell 2:26 pm on 8 August 2019 Permalink |
F has 7 points and has played 5 games, but they only have 2 goals for, so cannot have won more than 2 games. Their 7 points must be made by 2w + 3d, which must be 1-0, 1-0, 0-0, 0-0, 0-0, so their goals against is 0. Hence the total number of goals scored is 25, and between them C and E have scored 17 goals.
This Python program chooses values for C and E’s “goals for” values and then uses the [[
Football()]] helper class from the enigma.py library to determine the outcomes and scorelines of each match.It runs in 550ms.
Run: [ @repl.it ]
from enigma import Football, digit_map, irange, int2base, sprintf # scoring system football = Football(points=dict(w=2, d=1)) # labels for the teams (A, B, C, D, E, F) = (0, 1, 2, 3, 4, 5) # digits stand for themselves, and could go up to 17 M = 17 d = digit_map(0, M) # columns of the table (without goals against) (table, gf, ga) = (dict(played="24?3?5", w="1????2", l="??1??0", d='?????3', points="?373?7"), "41{C}1{E}2", "247570") # find possible match outcomes from the table for (ms, _) in football.substituted_table(table, teams=[F, A, B, C, D, E], d=d): # consider possible goals for C and E for (fC, fE) in zip(irange(0, M), irange(M, 0, step=-1)): gf1 = sprintf(gf, C=int2base(fC, M + 1), E=int2base(fE, M + 1)) # find possible scorelines for ss in football.substituted_table_goals(gf1, ga, ms, d=d): # output solution football.output_matches(ms, ss, teams="ABCDEF")Solution: The outcomes of E’s matches are: E vs A = not yet played; E vs B = 0-1; E vs C = 3-5; E vs D = not yet played; E vs F = 0-1.
There is only one scenario that works, which is when C has 14 goals for, and E has 3.
The full scores are:
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John Crabtree 6:24 pm on 8 August 2019 Permalink |
This teaser has a logical solution which leads directly to the answer, with no other paths.
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John Crabtree 2:36 pm on 13 August 2019 Permalink |
F has 7 points, but only scores 2 goals, and so F’s line reads 5 2 0 3 2 0 7
C has 7 points, but loses a game, and so C’s line reads 5 3 1 1 ? 7 7
As A plays 2 games, E must play 3 games, not 5.
The total of the Played column = 22 = the total of the Points column.
A has a minimum of 2 points and so the minimum Points total = 22
And so A has 2 points and so A’s line reads 2 1 1 0 4 2 2
And so E has 0 points and so B’s line reads 3 0 3 0 ? 7 0
B can only draw one game not three, and so B’s line reads 4 1 2 1 1 4 3
D can only draw one game,not three, and so D’s line reads 3 1 1 1 1 5 3
By inspection A v F 0-1, B v F 0-0, C v F 0-0, D v F 0-0, E v F 0-1.
Then A beats C. C beats B, D and E. B loses to D and beats E.
From A’s GF and GA, A v C 4-1.
From D”s GF = 1, , B v D 0-1
From D’s GF and GA, C v D 5-0
From B’s GF = 1, B v E 1-0
From B’s GF and GA, C v B 3-0
From C’s GA and E’s GA, C v E 5 -3
And so E played 3 matches: E v B 0-1, E v C 3-5 and E v F 0-1
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