S2T2

Jim Randell 10:28 am on 22 May 2022 Permalink Reply
Tags: by: R Postill ( 20 )   

• 

  Jim Randell 10:29 am on 22 May 2022 Permalink | Reply

  I made a MiniZinc model to solve this puzzle:

  %#! minizinc -a --solver chuffed --output-mode json
  
  % results for A
  % x[<year>, <event>] = 0 (loss) | 1 (draw) | 2 (win)
  % <year> = 1 (1957) .. 5 (1961)
  % <event> = 1 .. 5 (C, F, H, S, A)
  array [1..5, 1..5] of var 0..2: x;
  
  % points for each year
  function var int: ptsA(var int: i) = 3 * (x[i, 1] + x[i, 2]) + 2 * (x[i, 3] + x[i, 4] + x[i, 5]);
  function var int: ptsB(var int: i) = 24 - ptsA(i);
  
  % cumulative points
  function var int: cumA(var int: y) = sum (i in 1..y) (ptsA(i));
  function var int: cumB(var int: y) = sum (i in 1..y) (ptsB(i));
  
  % A won the cup 3 times, B won the cup twice
  constraint sum (i in 1..5) (ptsA(i) > ptsB(i)) = 3;
  constraint sum (i in 1..5) (ptsB(i) > ptsA(i)) = 2;
  
  % final cumulative total is equal
  constraint cumA(5) = cumB(5);
  
  % winning margin decreased each year ...
  constraint forall (i in 1..4) (abs(ptsA(i) - ptsB(i)) > abs(ptsA(i + 1) - ptsB(i + 1)));
  % ... from 12 points in 1957 ...
  constraint abs(ptsA(1) - ptsB(1)) = 12;
  % ... to 2 points in 1961
  constraint abs(ptsA(5) - ptsB(5)) = 2;
  
  % each sport has exactly one tie
  constraint forall (j in 1..5) (sum (i in 1..5) (x[i, j] = 1) = 1);
  
  % H was drawn in 1959
  constraint x[3, 3] = 1;
  
  % B last won C in 1958
  constraint x[2, 1] = 0;
  constraint forall (i in 3..5) (x[i, 1] > 0);
  
  % B won the cup in 1958
  constraint ptsA(2) < ptsB(2);
  
  % A have never won the swimming
  constraint forall (i in 1..5) (x[i, 4] < 2);
  
  % team A have a better record at event A
  constraint sum (i in 1..5) (x[i, 5]) > 5;
  
  % team A won event A in 1957
  constraint x[1, 5] = 2;
  
  solve satisfy;
  

  And a Python wrapper to format the output (using the minizinc.py library):

  from enigma import join, printf
  from minizinc import MiniZinc
  
  for [x] in MiniZinc("teaser87.mzn").solve(result='x', use_shebang=1):
    printf("      C F H S A | pA pB | cA cB")
    cA = cB = 0
    for (y, vs) in enumerate(x, start=1957):
      # points for A, B
      pA = sum(x * y for (x, y) in zip(vs, [3, 3, 2, 2, 2]))
      pB = 24 - pA
      # cumulative totals
      cA += pA
      cB += pB
      # output the table row
      r = join(("BdA"[x] for x in vs), sep=" ")
      printf("{y}: {r} | {pA:2d} {pB:2d} | {cA:2d} {cB:2d}")
    printf()
  

  Solution: The full results are (d = drawn):

  1957: cricket = A, football = A, hockey = B, swimming = d, athletics = A; cup = A (18 – 6)
  1958: cricket = B, football = d, hockey = B, swimming = B, athletics = A; cup = B (17 – 7)
  1959: cricket = A, football = B, hockey = d, swimming = B, athletics = B; cup = B (16 – 8)
  1960: cricket = A, football = A, hockey = B, swimming = B, athletics = d; cup = A (14 – 10)
  1961: cricket = d, football = A, hockey = B, swimming = B, athletics = A; cup = A (13 – 11)

  At the end of the 5 years each team has a cumulative total of 60 points each.

  The winning margin for each year is: 12, 10, 8, 4, 2.

  LikeLike