S2T2

Jim Randell 8:27 pm on 7 February 2019 Permalink Reply
Tags: by: Nick MacKinnon ( 52 )   

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  Jim Randell 11:38 am on 8 February 2019 Permalink | Reply

  I think this puzzle could have been worded more clearly.

  I took the first part to mean we are looking for situations where each team has exactly the same number of points as the number of goals scored by that team. And that the “best game finished 5-0” means that that game had the most goals scored altogether (i.e. the other games had no more than 4 goals scored in total). I took the fact that all matches had different scores to mean that you couldn’t have (for example) one game with a score of 2-0 and another with a score of 0-2.

  This program uses the [[ Football() ]] helper class from the enigma.py library. It looks for possible games for a team that can give the same number of points as the number of goals scored by that team, and then chooses outcomes for the four teams that satisfy the remaining conditions of the puzzle. It runs in 340ms.

  Run: [ @replit ]

  from itertools import (product, combinations)
  from enigma import (defaultdict, Football, ordered, printf)
  
  # scoring system
  football = Football(games='wdl', points=dict(w=3, d=1))
  
  # possible scores (a 5-0 and then no more than 4 goals in any match)
  scores = dict()
  scores['w'] = set([(1, 0), (2, 0), (3, 0), (4, 0), (5, 0), (2, 1), (3, 1)])
  scores['d'] = set([(0, 0), (1, 1), (2, 2)])
  scores['l'] = set((x, y) for (y, x) in scores['w'])
  
  # record possible games by points (= goals for)
  ss = defaultdict(list)
  
  # consider outcomes for three matches for team T against X, Y, Z
  for (TX, TY, TZ) in football.games(repeat=3):
  
    # make the table for team T
    T = football.table([TX, TY, TZ], [0, 0, 0])
  
    # make possible score lines that give "goals for" = "points"
    for (sTX, sTY, sTZ) in product(scores[TX], scores[TY], scores[TZ]):
  
      (fT, aT) = football.goals([sTX, sTY, sTZ], [0, 0, 0])
  
      if not (fT == T.points): continue
  
      ss[fT].append((sTX, sTY, sTZ))
  
  # choose possible points (= goals for) for A, B, C, D
  for (pA, pB, pC, pD) in combinations(sorted(ss.keys(), reverse=1), 4):
  
    # but points = goals for, and the total number of goals is 15
    if not (pA + pB + pC + pD == 15): continue
  
    # choose matches for B
    for (BA, BC, BD) in ss[pB]:
  
      # B vs D has less than 3 goals scored
      if not (sum(BD) < 3): continue
  
      # choose matches for A
      for (AB, AC, AD) in ss[pA]:
        if not (AB == BA[::-1]): continue
  
        # choose matches for C
        for (CA, CB, CD) in ss[pC]:
          if not (CA == AC[::-1] and CB == BC[::-1]): continue
  
          # check matches for D
          if not ((AD[::-1], BD[::-1], CD[::-1]) in ss[pD]): continue
  
          # all games have different scores
          s = set(ordered(*x) for x in (AB, AC, AD, BC, BD, CD))
          if not (len(s) == 6): continue
  
          # there is a 5-0 game
          if not ((0, 5) in s): continue
  
          printf("AB={AB} AC={AC} AD={AD} BC={BC} BD={BD} CD={CD}, A={pA} B={pB} C={pC} D={pD}")
  

  Solution: The scores in B’s games are: B vs A = 1-2; B vs C = 2-2; B vs D = 1-0.

  It turns out that the fact that there is a 5-0 game means that there are only 10 goals to distribute between the remaining 5 matches, and there are no further solutions if games with more than 4 goals scored are considered, so it is enough to know that there is a 5-0 score.

  If we allow “mirror” scores (e.g. one game with a score of 2-0 and another with a score of 0-2), then there are no further solutions either. (Although if we allow repeated scores then there are additional solutions).

  If we relax the conditions that the number of points is exactly the same as the number of goals scored then there are multiple solutions, even if they must be within 1 of each other.

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