S2T2

Jim Randell 9:34 am on 3 February 2026 Permalink Reply
Tags: by: Nick MacKinnon ( 53 ), flawed ( 54 )   

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  Jim Randell 9:35 am on 3 February 2026 Permalink | Reply

  There are 5 teams in the league, so each team plays 4 matches. For a team to “lose most of the time”, they would have to lose at least 3 of their 4 matches. And for “most teams to lose most of the time”, there have to be at least 3 teams that lost at least 3 of their matches.

  So, of the 10 matches in the tournament, at least 9 of them must be lost/won, i.e. there can be at most 1 drawn match.

  All the scores in the matches must be different, so in increasing numbers of total goals scored in lost/won match we have the possible following scorelines:

  1: 1-0
  2: 2-0
  3: 3-0, 2-1
  4: 4-0, 3-1
  5: 5-0, 4-1, 3-2

  At this point, we have identified 9 matches, and the total number of goals scored in them is 32. So all these must take place, and the remaining match must be a 0-0 draw. And we have identified the scores in all of the 10 matches.

  This Python program allocates the scores to the matches. It runs in 20s (using PyPy).

  from enigma import (Football, subsets, diff, cproduct, ordered, is_consecutive, printf)
  
  # scoring system
  football = Football(games="wdl", points=dict(w=3, d=1))
  
  # scores in the 10 matches; all different, and 32 goals scored in total
  scores = [(1, 0), (2, 0), (3, 0), (2, 1), (4, 0), (3, 1), (5, 0), (4, 1), (3, 2), (0, 0)]
  assert sum(x + y for (x, y) in scores) == 32
  
  # allocate <k> scores from <scs>
  def allocate(k, scs):
    for ss in subsets(scs, size=k, select='P'):
      rs = diff(scs, ss)
      for xys in cproduct({(x, y), (y, x)} for (x, y) in ss):
        yield (xys, rs)
  
  # calculate the table and goal difference from a sequence of scores
  def table(ss, ts):
    T = football.table(football.outcomes(ss), ts)
    (f, a) = football.goals(ss, ts)
    return (T, f - a)
  
  # X did better than Y
  def better(ptsX, ptsY, gdX, gdY):
    return ptsX > ptsY or (ptsX == ptsY and gdX > gdY)
  
  # choose the scores in D's matches
  for ((AD, BD, CD, DE), scs1) in allocate(4, scores):
    # D's goals in each match must be consecutive (in some order)
    if not is_consecutive(ordered(AD[1], BD[1], CD[1], DE[0])): continue
    # calculate the table for D
    (D, gdD) = table([AD, BD, CD, DE], [1, 1, 1, 0])
  
    # choose the scores in C's matches
    for ((AC, BC, CE), scs2) in allocate(3, scs1):
      # C scored 0 goals in the A v C match
      if not (AC[1] == 0): continue
      # calculate the table for C
      (C, gdC) = table([AC, BC, CD, CE], [1, 1, 0, 0])
      # C did better than D
      if not better(C.points, D.points, gdC, gdD): continue
  
      # choose the scores in B's matches
      for ((AB, BE), scs3) in allocate(2, scs2):
        # calculate the table for B
        (B, gdB) = table([AB, BC, BD, BE], [1, 0, 0, 0])
        # B did better than C
        if not better(B.points, C.points, gdB, gdC): continue
  
        # allocate the remaining match
        for ([AE], _) in allocate(1, scs3):
          # calculate the table for A
          (A, gdA) = table([AB, AC, AD, AE], [0, 0, 0, 0])
          # A did better than B
          if not better(A.points, B.points, gdA, gdB): continue
          # A's goal difference is 4 times B's
          if not (gdA == 4 * gdB): continue
  
          # calculate the table for E
          (E, gdE) = table([AE, BE, CE, DE], [1, 1, 1, 1])
          # D is higher than E
          if not better(D.points, E.points, gdD, gdE): continue
  
          # at least 3 teams lost at least 3 matches
          if not sum(X.l >= 3 for X in [A, B, C, D, E]) >= 3: continue
  
          # output scenario
          printf("AB={AB} AC={AC} AD={AD} AE={AE} BC={BC} BD={BD} BE={BE} CD={CD} CE={CE} DE={DE}")
          printf("A={A} gd={gdA}")
          printf("B={B} gd={gdB}")
          printf("C={C} gd={gdC}")
          printf("D={D} gd={gdD}")
          printf("E={E} gd={gdE}")
          printf()
  

  Solution: There are two possible answers:

  DvA = 0-3; DvB = 2-3; DvC = 1-4; DvE = 3-1
  DvA = 1-4; DvB = 2-3; DvC = 0-3; DvE = 3-1

  Only the first of these was given in the published solution.

  These correspond to the following 4 scenarios:

  AvB = 0-0; AvC = 4-0; AvD = 3-0; AvE = 5-0; BvC = 2-1; BvD = 3-2; BvE = 1-0; CvD = 4-1; CvE = 0-2; DvE = 3-1
  AvB = 0-0; AvC = 4-0; AvD = 3-0; AvE = 5-0; BvC = 1-0; BvD = 3-2; BvE = 2-1; CvD = 4-1; CvE = 0-2; DvE = 3-1
  AvB = 0-0; AvC = 4-0; AvD = 4-1; AvE = 5-0; BvC = 2-1; BvD = 3-2; BvE = 1-0; CvD = 3-0; CvE = 0-2; DvE = 3-1
  AvB = 0-0; AvC = 4-0; AvD = 4-1; AvE = 5-0; BvC = 1-0; BvD = 3-2; BvE = 2-1; CvD = 3-0; CvE = 0-2; DvE = 3-1

  i.e. the order of {AvD, CvD} = {3-0, 4-1} and {BvC, BvE} = {2-1, 1-0} is not determined.

  But in each case the final table is:

  1st: A → 3w1d = 10 points; goal diff = 12
  2nd: B → 3w1d = 10 points; goal diff = 3
  3rd: C → 1w3l = 3 points; goal diff = −4
  4th: D → 1w3l = 3 points; goal diff = −5
  5th: E → 1w3l = 3 points; goal diff = −6

  With teams C, D, E (i.e. “most teams”) having lost 3 of their matches (i.e. “lost most of the time”).

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