New Scientist Enigma 495 – Lack of Details

by Eric Emmet

From Issue #1647, 14th January 1989

Four football teams are to play each other once. After some of the matches had been played a document giving a few details of the matches played, won, lost and so on was found. This time I am glad to say that, although it was rather a mess, all the figures given were correct. Here it is:

Team	Played	Lost	Drawn	Goals For	Goals Against	Points
A			0	5	2
B		1			2
C				6	7	4
D	2		1	3

Two points are given for a win and one point to each side in a drawn match.

Find the score in each match.

One Comment Leave one →

Brian Gladman permalink

Using my Football class (see here)


from football import Football

f = Football('ABCD', outcomes='uldw', scoring=dict(w=2,d=1), sep='')

for d1 in f.games(dict(), 'A'):
  At = f.table(d1, 'A')
  if At.d:
    continue
  for d2 in f.games(d1, 'B'):
    Bt = f.table(d2, 'B')
    if Bt.l != 1 or Bt.pts != 1:
      continue
    for d3 in f.games(d2, 'C'):
      Ct = f.table(d3, 'C')
      if Ct.pts != 4:
        continue
      for d4 in f.games(d3, 'D'):
        Dt = f.table(d4, 'D')
        if Dt.p != 2 or Dt.d != 1:
          continue
        
        for d5 in f.matches(d4, 'A', 5, 2):
          for d6 in f.matches(d5, 'C', 6, 7):
            for d7 in f.matches(d6, 'B', (0, 5), 2):
              for d8 in f.matches(d7, 'D', 3, (0, 5)):
                nm, rs, gs = f.results(d8)
                print(', '.join(f"{n} {s}" for n, s in zip(nm, rs)))

from football import Football

f = Football('ABCD', outcomes='uldw', scoring=dict(w=2,d=1), sep='')

for d1 in f.games(dict(), 'A'):

At = f.table(d1, 'A')

if At.d:

continue

for d2 in f.games(d1, 'B'):

Bt = f.table(d2, 'B')

if Bt.l != 1 or Bt.pts != 1:

continue

for d3 in f.games(d2, 'C'):

Ct = f.table(d3, 'C')

if Ct.pts != 4:

continue

for d4 in f.games(d3, 'D'):

Dt = f.table(d4, 'D')

if Dt.p != 2 or Dt.d != 1:

continue

for d5 in f.matches(d4, 'A', 5, 2):

for d6 in f.matches(d5, 'C', 6, 7):

for d7 in f.matches(d6, 'B', (0, 5), 2):

for d8 in f.matches(d7, 'D', 3, (0, 5)):

nm, rs, gs = f.results(d8)

print(', '.join(f"{n} {s}" for n, s in zip(nm, rs)))

New Scientist Enigma 495 – Lack of Details

by Eric Emmet

From Issue #1647, 14th January 1989

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