S2T2

Jim Randell 7:48 am on 14 February 2019 Permalink Reply
Tags: by: Philip Spencer   

• 

  Jim Randell 9:25 am on 14 February 2019 Permalink | Reply

  Here’s a programmed solution in Python that uses the [[ Football() ]] helper class from the enigma.py library. It runs in 108ms.

  Run: [ @replit ]

  # there are four remaining matches: AC, BC, AX, BX
  
  from enigma import (Football, irange, cproduct, subsets, printf)
  
  # the scoring system
  football = Football(games='wdl', points=dict(w=2, d=1))
  
  # possible scores (no team scored more than 5)
  scores = dict()
  scores['w'] = set((x, y) for x in irange(1, 5) for y in irange(0, x - 1))
  scores['d'] = set((x, x) for x in irange(0, 5))
  scores['l'] = set((y, x) for (x, y) in scores['w'])
  
  # possible outcomes in the remaining 4 games
  for (AC, BC, AX, BX) in football.games(repeat=4):
  
    # table for A, B, C in these four games
    A = football.table([AC, AX], [0, 0])
    B = football.table([BC, BX], [0, 0])
    C = football.table([AC, BC], [1, 1])
  
    # A and B got the same number of additional points, and C got 2 more
    if not (A.points == B.points == C.points - 2): continue
  
    # choose scores for the 4 matches
    for (sAC, sBC, sAX, sBX) in cproduct(scores[x] for x in (AC, BC, AX, BX)):
  
      # in total 10 goals are scored in the 4 matches
      if not (sum(x + y for (x, y) in (sAC, sBC, sAX, sBX)) == 10): continue
  
      # goals for/against A, B, C
      (fA, aA) = football.goals([sAC, sAX], [0, 0])
      (fB, aB) = football.goals([sBC, sBX], [0, 0])
      (fC, aC) = football.goals([sAC, sBC], [1, 1])
  
      # A failed to score, C scored 5 goals
      if not (fA == 0 and fC == 5): continue
  
      # C worsened their goal average: (70 + fC) / (35 + aC) < 70 / 35
      if not (fC < 2 * aC): continue
  
      # A, B started off with 70 goals for and a, b goals against
      # their goal average is better than 70/35, so b < a < 35
      for (b, a) in subsets(irange(1, 34), size=2):
  
        # and we ended with A having the best goal average, then C, then B
        if not ((70 + fA) * (35 + aC) > (70 + fC) * (a + aA)): continue
        if not ((70 + fC) * (b + aB) > (70 + fB) * (35 + aC)): continue
  
        # output solution
        printf("AC={AC}:{sAC} BC={BC}:{sBC} AX={AX}:{sAX} BX={BX}:{sBX} / a={a} b={b}")
        printf("A={A} f={fA} a={aA}", fA=70 + fA, aA=a + aA)
        printf("B={B} f={fB} a={aB}", fB=70 + fB, aB=b + aB)
        printf("C={C} f={fC} a={aC}", fC=70 + fC, aC=35 + aC)
        printf()
  

  Solution: A: for = 70, against = 35; B: for = 73, against = 37; C: for = 75, against = 38.

  There are two scenarios. Assuming 2 points for a win and 1 for a draw, we start with:

  A: points = 32; goals for = 70; goals against = 33; avg = 70/33 = 2.121
  B: points = 32; goals for = 70; goals against = 32; avg = 70/32 = 2.188
  C: points = 30; goals for = 70; goals against = 35; avg = 70/35 = 2.000

  Then the remaining four matches are played. The outcomes are one of:

  (1) A vs C = 0-0; B vs C = 3-5; A vs X = 0-2; B vs X = 0-0
  (2) A vs C = 0-2; B vs C = 3-3; A vs X = 0-0; B vs X = 0-2

  Each of these scenarios has 10 goals scored, none of them by A and 5 of them by C.

  The final table then becomes:

  A: points = 33; goals for = 70; goals against = 35; avg = 70/35 = 2.000
  B: points = 33; goals for = 73; goals against = 37; avg = 73/37 = 1.973
  C: points = 33; goals for = 75; goals against = 38; avg = 75/38 = 1.974

  LikeLike