Sunday Times Teaser 2567 – A Close Thing

by Neil MacKinnon

Published: 4 December 2011 (link)

Our football league consists of five teams, each playing each other once, earning three points for a win, one for a draw. Last season was close. At the end, all five teams had tied on points and on “goal difference”. So the championship was decided on “goals scored”, which, for the five teams, consisted of five consecutive numbers. The scores in the games were all different, including a no-score draw, and no team scored more than five goals in any game.

How many goals did the league champions score in total?

One Comment Leave one →

Brian Gladman permalink

A solution using my football solver class


from football import Football

def find_scores(f, d0, Eg):

  for d1 in f.matches(d0, 'E', Eg, Eg):
    Et = f.table(d1, 'E')

    for d2 in f.matches(d1, 'D', Eg + 1, Eg + 1):
      Dt = f.table(d2, 'D')

      m = f.results(d2, ordered=True)[1]
      if Dt <= Et or len(m) != len(set(m)) or any(x[1] > 5 for x in m):
        continue

      for d3 in f.matches(d2, 'C', Eg + 2, Eg + 2):
        Ct = f.table(d3, 'C')
                
        m = f.results(d3, ordered=True)[1]
        if Ct <= Dt or len(m) != len(set(m)) or any(x[1] > 5 for x in m):
          continue

        for d4 in f.matches(d3, 'B', Eg + 3, Eg + 3):
          Bt = f.table(d4, 'B')

          m = f.results(d4, ordered=True)[1]
          if Bt <= Ct or len(m) != len(set(m)) or any(x[1] > 5 for x in m):
            continue

          for d5 in f.matches(d4, 'A', Eg + 4, Eg + 4):
            At = f.table(d5, 'A')
                      
            m = f.results(d5, ordered=True)[1]
            if At <= Bt or len(m) != len(set(m)) or any(x[1] > 5 for x in m) or (0, 0) not in m:
              continue
            
            yield d5, (At, Bt, Ct, Dt, Et)

f = Football(teams='ABCDE', outcomes='ldw', sep='')

# The 10 games give a minimum and maximum total scores of 20 and 30
# points. Since all scores are equal, this total is a multiple of 5
# and cannot be 20 or 30 so it must be 25. Hence each team has five
# points, which is 1 win, 2 draws and a loss.

for d1 in f.games(dict(), 'A'):
  At = f.table(d1, 'A')
  if At.pts != 5:
    continue

  for d2 in f.games(d1, 'B'):
    Bt = f.table(d2, 'B')
    if Bt != At:
      continue

    for d3 in f.games(d2, 'C'):
      Ct = f.table(d3, 'C')
      if Ct != At:
        continue

      for d4 in f.games(d3, 'D'):
        Dt = f.table(d4, 'D')
        if Dt != At:
          continue

        for d5 in f.games(d4, 'E'):
          Et = f.table(d5, 'E')
          if Et != At:
            continue

          for Eg in range(3, 8):
            
            for dr, ft in find_scores(f, d5, Eg):

              print(f"The Champions scored {ft[0].gf} goals\n")
              print(f.results(dr, 'A')[1], ft[0])
              print(f.results(dr, 'B')[1], ft[1])
              print(f.results(dr, 'C')[1], ft[2])
              print(f.results(dr, 'D')[1], ft[3])
              print(f.results(dr, 'E')[1], ft[4])
              print()
              exit()

from football import Football

def find_scores(f, d0, Eg):

for d1 in f.matches(d0, 'E', Eg, Eg):

Et = f.table(d1, 'E')

for d2 in f.matches(d1, 'D', Eg + 1, Eg + 1):

Dt = f.table(d2, 'D')

m = f.results(d2, ordered=True)[1]

if Dt <= Et or len(m) != len(set(m)) or any(x[1] > 5 for x in m):

continue

for d3 in f.matches(d2, 'C', Eg + 2, Eg + 2):

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

m = f.results(d3, ordered=True)[1]

if Ct <= Dt or len(m) != len(set(m)) or any(x[1] > 5 for x in m):

continue

for d4 in f.matches(d3, 'B', Eg + 3, Eg + 3):

Bt = f.table(d4, 'B')

m = f.results(d4, ordered=True)[1]

if Bt <= Ct or len(m) != len(set(m)) or any(x[1] > 5 for x in m):

continue

for d5 in f.matches(d4, 'A', Eg + 4, Eg + 4):

At = f.table(d5, 'A')

m = f.results(d5, ordered=True)[1]

if At <= Bt or len(m) != len(set(m)) or any(x[1] > 5 for x in m) or (0, 0) not in m:

continue

yield d5, (At, Bt, Ct, Dt, Et)

f = Football(teams='ABCDE', outcomes='ldw', sep='')

# The 10 games give a minimum and maximum total scores of 20 and 30

# points. Since all scores are equal, this total is a multiple of 5

# and cannot be 20 or 30 so it must be 25. Hence each team has five

# points, which is 1 win, 2 draws and a loss.

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

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

if At.pts != 5:

continue

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

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

if Bt != At:

continue

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

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

if Ct != At:

continue

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

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

if Dt != At:

continue

for d5 in f.games(d4, 'E'):

Et = f.table(d5, 'E')

if Et != At:

continue

for Eg in range(3, 8):

for dr, ft in find_scores(f, d5, Eg):

print(f"The Champions scored {ft[0].gf} goals\n")

print(f.results(dr, 'A')[1], ft[0])

print(f.results(dr, 'B')[1], ft[1])

print(f.results(dr, 'C')[1], ft[2])

print(f.results(dr, 'D')[1], ft[3])

print(f.results(dr, 'E')[1], ft[4])

print()

exit()

Sunday Times Teaser 2567 – A Close Thing

by Neil MacKinnon

Published: 4 December 2011 (link)

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