S2T2

Jim Randell 9:01 am on 11 February 2021 Permalink Reply
Tags: by: Eric Emmet ( 10 )   

• 

  Jim Randell 9:02 am on 11 February 2021 Permalink | Reply

  (See also: Puzzle 81).

  This Python program uses the [[ Football ]] helper class from the enigma.py library. It runs in 198ms.

  from enigma import (Football, subsets, cproduct, printf)
  
  # scoring system
  football = Football(points=dict(w=2, d=1))
  
  # possible scores in the matches (no match has more than 7 goals scored)
  scores = {
    'w': [
      (1, 0), (2, 0), (3, 0), (4, 0), (5, 0), (6, 0), (7, 0),
      (2, 1), (3, 1), (4, 1), (5, 1), (6, 1),
      (3, 2), (4, 2), (5, 2),
      (4, 3),
    ],
    'd': [(0, 0), (1, 1), (2, 2), (3, 3)],
    'x': [None],
  }
  scores['l'] = list((b, a) for (a, b) in scores['w'])
  
  # choose outcomes for each of the matches
  for (AB, AC, AD, BC, BD, CD) in subsets('wdlx', size=6, select="M"):
    # calculate the points
    A = football.table([AB, AC, AD], [0, 0, 0])
    B = football.table([AB, BC, BD], [1, 0, 0])
    C = football.table([AC, BC, CD], [1, 1, 0])
    D = football.table([AD, BD, CD], [1, 1, 1])
  
    # each team has played at least one game
    if not all(x.played > 0 for x in [A, B, C, D]): continue
  
    # count the discrepancies in the points
    dp = abs(A.points - 3) + abs(B.points - 5) + C.points
    if dp > 1: continue
  
    # choose the scores in C's games
    for (sAC, sBC, sCD) in cproduct([scores[AC], scores[BC], scores[CD]]):
      (fC, aC) = football.goals([sAC, sBC, sCD], [1, 1, 0])
      dC = aC
      if dp + dC > 1: continue
  
      # choose the scores in A's games
      for (sAB, sAD) in cproduct([scores[AB], scores[AD]]):
        (fA, aA) = football.goals([sAB, sAC, sAD], [0, 0, 0])
        dA = abs(aA - 5)
        if dp + dA + dC > 1: continue
  
        # remaining game
        for sBD in scores[BD]:
          (fB, aB) = football.goals([sAB, sBC, sBD], [1, 0, 0])
          dB = abs(aB - 6)
          if dp + dA + dB + dC > 1: continue
  
          (fD, aD) = football.goals([sAD, sBD, sCD], [1, 1, 1])
          dD = abs(aD - 7)
          if dp + dA + dB + dC + dD != 1: continue
  
          printf("A=({aA} {A.points}) B=({aB} {B.points}) C=({aC} {C.points}) D=({aD} {D.points}) / \\")
          printf("AB={AB}:{sAB} AC={AC}:{sAC} AD={AD}:{sAD} BC={BC}:{sBC} BD={BD}:{sBD} CD={CD}:{sCD}")
  

  Solution: The scores in the played matches are: A vs B = 3-3; A vs D = 3-2; B vs C = 1-0; B vs D = 4-3.

  The A vs C and C vs D matches are not yet played.

  The incorrect value is C’s “goals against”. It should be 1, not 0.

  It is clear that the incorrect value must be one of C’s, as if they have played at least one game they cannot have both 0 points and 0 goals against.

  LikeLike