Sunday Times Teaser 2893 – Win Some, Lose Some

by Victor Bryant

Published March 4 2018 (link)

The six teams in our local league play each other once in the season.

Last season no two teams tied on points.

Above is part of the table from the end of that season, with the teams in alphabetical order.

However, digits have been consistently replaced by letters.

Please send in the number of LEMONS.

3 Comments Leave one →

Brian Gladman permalink

Here is my solution using my football solver class:


from functools import reduce
from football import Football

f = Football('ABCDEF', outcomes='ldw', scoring={'w':3, 'd':1}, sep='')

# save solutions here
sol = set()
# consider possible points scored in each of A's games
for d1 in f.games(dict(), 'A'):
  
  # compile a table of results and check that they give
  # valid values for A's results in their letter form
  At = f.table(d1, 'A')
  W, O, N = At.p, At.w, At.l
  s3 = {N, O, W}
  if len(s3) != 3 or not all(0 <= x < 10 for x in s3):
    continue

  # repeat the process for B's remaining games
  for d2 in f.games(d1, 'B'):
    
    Bt = f.table(d2, 'B')
    S, o, M = Bt.w, Bt.l, Bt.d
    s5 = s3.union([M, S])
    if (Bt.pts == At.pts or o != O or len(s5) != 5 
              or not all(0 <= x < 10 for x in s5)):
      continue
    
    # ... and C's
    for d3 in f.games(d2, 'C'):
      
      Ct = f.table(d3, 'C')
      L, o = Ct.l, Ct.d
      s6 = s5.union([L])
      if (Ct.pts in {At.pts, Bt.pts} or o != O or len(s6) != 6 
                or not all(0 <= x < 10 for x in s6)):
        continue

      # now complete the remaining games for D, E and F
      for d4 in f.games(d3, 'D'):
        for d5 in f.games(d4, 'E'):
          for d6 in f.games(d5, 'F'):

            Dt = f.table(d6, 'D')
            Et = f.table(d6, 'E')
            Ft = f.table(d6, 'F')
          
            # check F's letter symbol results are valid
            if Ft.d != S or Ft.pts in s6 or not 0 <= Ft.pts < 10:
              continue
          
            pts, E = {x.pts for x in (At, Bt, Ct, Dt, Et, Ft)}, Ft.pts
            # the are no ties on points
            if len(pts) == 6:
              lemons = reduce(lambda x, y: 10 * x + y, (L, E, M, O, N, S))
              # consider possible values for T
              for T in set(range(10)).difference(s6.union([E])):
            
                # find match results for C that are compatible with their loss,
                # draw and win results and their goals for and goals against
                for d7 in f.matches(d6, 'C', S, T):
                   
                  # find match results for F that are compatible with their loss, 
                  # draw, win results and their goals for and goals against
                  for d8 in f.matches(d7, 'F', O, M):
                    sol.add(lemons)

print('LEMONS = {}'.format(*sol))

from functools import reduce

from football import Football

f = Football('ABCDEF', outcomes='ldw', scoring={'w':3, 'd':1}, sep='')

# save solutions here

sol = set()

# consider possible points scored in each of A's games

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

# compile a table of results and check that they give

# valid values for A's results in their letter form

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

W, O, N = At.p, At.w, At.l

s3 = {N, O, W}

if len(s3) != 3 or not all(0 <= x < 10 for x in s3):

continue

# repeat the process for B's remaining games

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

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

S, o, M = Bt.w, Bt.l, Bt.d

s5 = s3.union([M, S])

if (Bt.pts == At.pts or o != O or len(s5) != 5

or not all(0 <= x < 10 for x in s5)):

continue

# ... and C's

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

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

L, o = Ct.l, Ct.d

s6 = s5.union([L])

if (Ct.pts in {At.pts, Bt.pts} or o != O or len(s6) != 6

or not all(0 <= x < 10 for x in s6)):

continue

# now complete the remaining games for D, E and F

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

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

for d6 in f.games(d5, 'F'):

Dt = f.table(d6, 'D')

Et = f.table(d6, 'E')

Ft = f.table(d6, 'F')

# check F's letter symbol results are valid

if Ft.d != S or Ft.pts in s6 or not 0 <= Ft.pts < 10:

continue

pts, E = {x.pts for x in (At, Bt, Ct, Dt, Et, Ft)}, Ft.pts

# the are no ties on points

if len(pts) == 6:

lemons = reduce(lambda x, y: 10 * x + y, (L, E, M, O, N, S))

# consider possible values for T

for T in set(range(10)).difference(s6.union([E])):

# find match results for C that are compatible with their loss,

# draw and win results and their goals for and goals against

for d7 in f.matches(d6, 'C', S, T):

# find match results for F that are compatible with their loss,

# draw, win results and their goals for and goals against

for d8 in f.matches(d7, 'F', O, M):

sol.add(lemons)

print('LEMONS = {}'.format(*sol))

Arthur Vause permalink


from itertools import permutations
W = 5
for O, N, S, M, L in permutations(range(5)):
  if S + O + M == 5 and O + N <= 5 and L + O <= 5:

    WC = 5 - L - O # wins by City
    if WC <= S:    # City must have scored at least as many goals as wins

      # calculate points for Albion, Borough, City
      PA = 3 * O + (5 - O - N)
      PB = 3 * S + M
      PC = 3 * (5 - L - O) + O

      # E is the number of points scored by Friday
      if len(set((PA, PB, PC))) == 3:
        for E in set(range(10))-set(( W, O, N, S, M, L, PA, PB, PC)):
       
          # Friday must have no more draws than points
          # Friday's points for wins must be a multiple of 3,
          # and Friday's number of wins + draws is at most 5
          if E >= S and (E - S) % 3 == 0 and (E - S) // 3 + S <= 5:
            WF = (E - S) // 3 # wins by Friday
            if WF <= O: # Friday must have no more wins than goals scored
              print "LEMONS =", L, E, M, O, N, S
              print "\n\tP W L D   F A Pts"
              print "Albion\t{} {} {} {}   ? ? {}".format(W, O, N, 5 - O - N, PA)
              print "Borough\t5 {} {} {}   {} ? {}".format(S, O, M, E, PB)
              print "City  \t5 {} {} {}   {} ? {}".format(WC, L, O, S, PC)
              print "Friday\t5 {} {} {}   {} {} {}".format(WF, 5 - WF - S, S, O, M, E)

from itertools import permutations

W = 5

for O, N, S, M, L in permutations(range(5)):

if S + O + M == 5 and O + N <= 5 and L + O <= 5:

WC = 5 - L - O # wins by City

if WC <= S: # City must have scored at least as many goals as wins

# calculate points for Albion, Borough, City

PA = 3 * O + (5 - O - N)

PB = 3 * S + M

PC = 3 * (5 - L - O) + O

# E is the number of points scored by Friday

if len(set((PA, PB, PC))) == 3:

for E in set(range(10))-set(( W, O, N, S, M, L, PA, PB, PC)):

# Friday must have no more draws than points

# Friday's points for wins must be a multiple of 3,

# and Friday's number of wins + draws is at most 5

if E >= S and (E - S) % 3 == 0 and (E - S) // 3 + S <= 5:

WF = (E - S) // 3 # wins by Friday

if WF <= O: # Friday must have no more wins than goals scored

print "LEMONS =", L, E, M, O, N, S

print "\n\tP W L D F A Pts"

print "Albion\t{} {} {} {} ? ? {}".format(W, O, N, 5 - O - N, PA)

print "Borough\t5 {} {} {} {} ? {}".format(S, O, M, E, PB)

print "City \t5 {} {} {} {} ? {}".format(WC, L, O, S, PC)

print "Friday\t5 {} {} {} {} {} {}".format(WF, 5 - WF - S, S, O, M, E)

Jim Randell permalink

I derived a set of constraints from the table and then use the general alphametic solver (SubstitutedExpression()) from the enigma.py library [ https://www.magwag.plus.com/jim/enigma.html ].

Here’s a run file that can be executed directly.


#!/usr/bin/env python -m enigma -r

SubstitutedExpression

--answer="LEMONS"

# each team has played 5 matches
"W = 5"

# won + lost + drawn = 5
"O + N <= 5"
"S + O + M = 5"
"L + O <= 5"
"S <= 5"

# goals for >= won, goals against >= lost
"E >= S"
"S >= 5 - L - O"
"T >= L"
"O >= (E - S) // 3" "(E - S) % 3 = 0"

# points (= 3*won + drawn) are all different
"all_different(3 * O + (5 - O - N), 3 * S + M, 3 * (5 - L - O) + O, E)"

#!/usr/bin/env python -m enigma -r

SubstitutedExpression

--answer="LEMONS"

# each team has played 5 matches

"W = 5"

# won + lost + drawn = 5

"O + N <= 5"

"S + O + M = 5"

"L + O <= 5"

"S <= 5"

# goals for >= won, goals against >= lost

"E >= S"

"S >= 5 - L - O"

"T >= L"

"O >= (E - S) // 3" "(E - S) % 3 = 0"

# points (= 3*won + drawn) are all different

"all_different(3 * O + (5 - O - N), 3 * S + M, 3 * (5 - L - O) + O, E)"

The code can be executed on repl.it. [ https://repl.it/@jim_r/teaser2893 ]

We could provide additional constraints (W=5; L,M,N,O,S are 0,1,2,3,4 (in some order)) but id doesn’t make much difference to the run time.

Sunday Times Teaser 2893 – Win Some, Lose Some

by Victor Bryant

Published March 4 2018 (link)

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