S2T2

Jim Randell 9:40 am on 13 April 2021 Permalink Reply
Tags: by: R. J. Pocock ( 2 )   

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  Jim Randell 9:41 am on 13 April 2021 Permalink | Reply

  This seems to be the earliest “football table” Teaser.

  We are told that the sum of Scotland and Ireland’s “goals against” values is 20, which means the total of the “goals against” column must be 38. And so the sum of Scotland and Ireland’s “goals for” values must be 38 − (16 + 8) = 14.

  The following Python program uses the [[ Football() ]] helper class from the enigma.py library. It runs in 1.22s.

  from itertools import product
  from enigma import (Football, ordered, chunk, subsets, irange, multiset, join, printf)
  
  # scoring system
  football = Football(games="wdl", points=dict(w=2, d=1))
  
  # identify matches with the same scoreline
  keys = lambda ss: list(ordered(*s) for s in ss)
  
  # check a sequence of scores, all different and in ordered pairs
  check = lambda ss, ps: len(set(ss)) == len(ss) and all(x > y for (x, y) in chunk(ps, 2))
  
  # margin
  margin = lambda ss: ss[0] - ss[1]
  
  # record scores in the S vs I matches
  rs = multiset()
  
  # scorelines for E (who have won all their matches)
  (ew1, ew2, es1, es2, ei1, ei2) = mes = 'w' * 6
  for ssE in football.scores(mes, [0] * 6, 16, 3):
    # check scorelines
    ss0 = keys(ssE)
    if not check(ss0, ss0): continue
  
    # E wins E vs W by same margin, same as exactly one of the E vs I
    (EW1, EW2, ES1, ES2, EI1, EI2) = ssE
    d = margin(EW1)
    if not (d == margin(EW2) and (margin(EI1), margin(EI2)).count(d) == 1): continue
  
    # W have 2 draws and 2 losses remaining
    for mws in subsets("ddll", size=len, select="mP"):
      for ssW in football.scores(mws, [0] * 4, 8, 15, [EW1, EW2], [1, 1]):
        ss1 = keys(ssW)
        if not check(ss0 + ss1, ss1): continue
  
        # calculate current goals for/against S and I (so far)
        (WS1, WS2, WI1, WI2) = ssW
        (fS, aS) = football.goals([ES1, ES2, WS1, WS2], [1, 1, 1, 1])
        (fI, aI) = football.goals([EI1, EI2, WI1, WI2], [1, 1, 1, 1])
        # goals against S and I sum to 20 (and goals for S and I sum to 14)
        (ga, gf) = (20 - aS - aI, 14 - fS - fI)
        if ga < 0 or gf < 0: continue
  
        # choose outcomes for S vs I matches
        (ws1, ws2, wi1, wi2) = mws
        for (si1, si2) in football.games(repeat=2):
          S = football.table([es1, es2, ws1, ws2, si1, si2], [1, 1, 1, 1, 0, 0])
          I = football.table([ei1, ei2, wi1, wi2, si1, si2], [1, 1, 1, 1, 1, 1])
          if not (12 >= S.points >= I.points >= 2): continue
  
          # chose remaining "for" and "against" goals for S
          for (x, y) in product(irange(0, gf), irange(0, ga)):
            # look for scorelines
            for (SI1, SI2) in football.scores([si1, si2], [0, 0], x, y):
              ss2 = keys([SI1, SI2])
              if not check(ss0 + ss1 + ss2, ss2): continue
              # and check goals "for"/"against" I
              (x_, y_) = football.goals([SI1, SI2], [1, 1])
              if not (x + x_ == gf and y + y_ == ga): continue
              # check S is second and I is third
              if not (S.points > I.points or fS - aS + x - y > fI - aI + x_ - y_): continue
              if not (I.points > 2 or fI - aI + x_ - y_ > -7): continue
              printf("[EW = {EW1} {EW2}; ES = {ES1} {ES2}; EI = {EI1} {EI2}; WS = {WS1} {WS2}; WI = {WI1} {WI2}; SI = {SI1} {SI2}]")
              rs.add((SI1, SI2))
  
  # output solution
  for (k, v) in rs.most_common():
    printf("S vs I = {k} [{v} solutions]", k=join(k, sep=", "))
  

  Solution: The scores in the Scotland vs. Ireland matches were 4-0 and 0-0.

  There are many scenarios which lead to this solution. One of them is:

  E vs W = 3-2, 2-1
  E vs S = 3-0, 2-0
  E vs I = 5-0, 1-0
  W vs S = 2-2, 1-3
  W vs I = 1-4, 1-1
  S vs I = 4-0, 0-0

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  John Crabtree 4:28 pm on 16 April 2021 Permalink | Reply

  There were 38 goals scored, ie the minimum possible, and so the match scores were 0-0; 1-0; 2-0, 1-1; 3-0, 2-1; 4-0, 3-1, 2-2; 5-0, 4-1, 3-2.
  Wales scored 8 goals, with scores of 2-3 vE, 1-2 vE, 1-1, 2-2, 1-3 and 1-4.
  And so the scores in England’s other games were 1-0 vI, 2-0, 3-0 and 5-0.
  The other two matches were between Ireland and Scotland with scores of 0-0 and 4-0.

  It is interesting to know when these puzzles started. Some of the later ones were much more convoluted

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