S2T2

Jim Randell 1:07 pm on 6 June 2019 Permalink Reply
Tags: by: R Postill ( 20 )   

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  Jim Randell 1:09 pm on 6 June 2019 Permalink | Reply

  Although not explicitly mentioned in the puzzle text, presumably we are talking about a cricket match.

  B is one of the first pair to go into bat, and at the 10th dismissal he is “not out”, so he is paired with every other batsman. If there is an A, he would be in the first pair (and the first to be dismissed), and C would be next (and be the second to be dismissed). If there isn’t an A, C would be in the first pair and be the first to be dismissed. So C is either the first or second batsman to be dismissed.

  I tackled this problem in 2 parts. The first part deals with the first three dismissals, where the score is always the product of B’s score and the dismissed batsman’s score (and two of the scores have to be the square of the dismissed batsman’s score – the one that isn’t identifies C).

  The second part deals with the remaining dismissals (4th – 10th) the scores have to be the square of the dismissed batsman’s score.

  When the 10th batsman is dismissed the total score must be a square less than 155, so not more than 12².

  This program runs in 80ms.

  Run: [ @repl.it ]

  from enigma import (irange, div, printf)
  
  # solve the first three dismissals, at each the total score is the
  # product of B's score and the dismissed batsman's score and all,
  # except C, the total is the square of the dismissed batsman's score.
  # return ([(<B's score>, <dismissed score>, <extras>), ...], <C>)
  # t = total so far
  # b = B's current score
  # ds = scores at dismissals
  # x = current extras
  # c = identify C
  def solve1(t=0, b=0, ds=[], x=0, c=None):
    n = len(ds)
    # are we done?
    if n == 3:
      # check C has been allocated
      if c in (0, 1):
        yield (ds, c)
    else:
      # possible extra
      for x2 in irange(x, 1):
        # possible current score for B
        for b2 in irange(b, 71):
          # calculate the score for the dismissed batsman (d)
          #   t2 = t + (b2 - b) + d + (x2 - x) = b2 * d
          # so:
          #   d = (t + b2 + x2 - b - x) / (b2 - 1)
          d = div(t + b2 + x2 - b - x, b2 - 1)
          if d is None or d < 1: continue
          t2 = b2 * d
          # is the total the square of d?
          c2 = c
          if t2 != d * d:
            # this must be C
            if c is not None: continue
            c2 = n
          yield from solve1(t2, b2, ds + [(b2, d, x2)], x2, c2)
  
  # solve the remaining dismissals, at each the total score is the
  # square of the dismissed batman's score
  # t = total score so far
  # b = B's current score
  # ds = scores of B and dismissed batsman
  # x = extra (0 or 1)
  def solve2(t=0, b=0, ds=[], x=0):
    n = len(ds)
    # are we done?
    if n == 10:
      # check an extra has been scored and B's score is 71
      if x == 1 and ds[-1][0] == 71:
        yield ds
    else:
      # possible extra
      for x2 in irange(x, 1):
        # possible scores for next batsman
        for d in irange(1, 12):
          # the new total is...
          t2 = d * d
          if not (t < t2 < 155): continue
          # calculate the current score for B
          b2 = t2 + b + x - t - d - x2
          if b2 < b: continue
          yield from solve2(t2, b2, ds + [(b2, d, x2)], x2)
            
  # solve for the first 3 dismissals
  for (ds1, c) in solve1():
    (b, d, x) = ds1[-1]
    # solve for the remaining dismissals
    for ds in solve2(b * d, b, ds1, x):
      # output the solution
      printf("[ds={ds}]")
      printf("-> C={C} [i={c}]; B[i=6]={B}", C=ds[c][1], B=ds[6][0])
  

  Solution: Collins scored 5. Bond had scored 41 at the 7th dismissal.

  The scores for the batsman (in dismissal order) were:

  2, 5 (Collins), 4, 5, 6, 8, 9, 10, 11, 12, 71 (not out, Bond)

  And the total scores at each dismissal were:

  4, 10 (Collins), 16, 25, 36, 64, 81, 100, 121, 144

  So there is an A (who was dismissed first). The extra was scored between A’s dismissal and C’s dismissal.

  Bond’s scores at each dismissal were:

  2, 2, 4, 8, 13, 33, 41, 50, 60, 71

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