S2T2

Jim Randell 4:25 pm on 28 January 2022 Permalink Reply
Tags: by: Angela Newing ( 38 )   

• 

  Jim Randell 4:47 pm on 28 January 2022 Permalink | Reply

  Each boy loses a hole at one end of the course, so we can just consider course with 8 holes, as we are only interested in groups of consecutive wins.

  This Python program runs in 47ms.

  Run: [ @replit ]

  from enigma import (irange, subsets, intersect, printf)
  
  # consider a course with 8 holes
  holes = irange(1, 8)
  
  # calculate the winnings for a nephew who wins holes <ws>
  def winnings(ws):
    (t, h) = (0, 0)
    for n in holes:
      if n in ws:
        h += 1
        t += h
      else:
        h = 0
    return t
  
  # choose 5 to 8 extra holes for J and K
  X = dict()
  for n in irange(5, 8):
    X[n] = set(winnings(ws) for ws in subsets(holes, size=n))
  
  # consider number of holes won by J
  for j in irange(5, 8):
    k = 13 - j
    # look for equal amounts
    for w in intersect([X[j], X[k]]):
      printf("I pays {t} [J wins {j}; K wins {k}]", t=w + w)
  

  Solution: Ian paid £32 to the boys.

  Ian lost to one of the boys on 5 consecutive holes and 1 other, paying (1 + 2 + 3 + 4 + 5) + 1 = £16.

  And to the other boy on 4 consecutive holes and another 3 consecutive holes, playing (1 + 2 + 3 + 4) + (1 + 2 + 3) = £16.

  So, in total Ian pays £32 for the 13 holes he lost.

  From 8 holes there are 6 ways that runs of 5 and 1 can be made:

  (1 2 3 4 5) (7)
  (1 2 3 4 5) (8)
  (2 3 4 5 6) (8)
  (1) (3 4 5 6 7)
  (1) (4 5 6 7 8)
  (2) (2 5 6 7 8)

  And there are 2 ways that runs of 4 and 3 can be made:

  (1 2 3 4) (6 7 8)
  (1 2 3) (5 6 7 8)

  So if John wins 5+1 and Ken wins 4+3 there are 12 possibilities.

  And if John wins 4+3 and Ken wins 5+1 there are also 12 possibilities.

  Giving 24 possible ways to achieve the required games.

  ---

  Manually, we see that the boys won 13 holes between them, so the possible splits are (8, 5) or (7, 6).

  Winning 8 holes would give a prize of T(8) which must be more than the winnings from 5 holes. So the splits are (7, 6).

  Possible consecutive runs for 7 (and corresponding prizes) are:

  (7) → T(7) = 28
  (1, 6) → T(1) + T(6) = 22
  (2, 5) → T(2) + T(5) = 18
  (3, 4) → T(3) + T(4) = 16

  Possible runs for 6 (and corresponding prizes) are:

  (6) → T(6) = 21
  (1, 5) → T(1) + T(5) = 16
  (2, 4) → T(2) + T(4) = 13
  (3, 3) → T(3) + T(3) = 12
  (1, 1, 4) → T(1) + T(1) + T(4) = 12
  (1, 2, 3) → T(1) + T(2) + T(3) = 10
  (2, 2, 2) → T(2) + T(2) + T(2) = 9

  The only overlap is for a prize of £16, with consecutive runs of (3, 4) and (1, 5).

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  • 

    Jim Randell 11:04 am on 29 January 2022 Permalink | Reply

    Here’s an alternative approach, which instead of generating the numbers of the holes won, finds possible runs of wins among the 8 holes.

    It is slightly faster than my original solution (above).

    Run: [ @replit ]

    from enigma import (irange, decompose, tri, intersect, printf)
    
    # generate possible runs of <n> wins for <k> holes
    def generate(n, k):
      # split n into j parts
      for j in irange(1, k + 1 - n):
        yield from decompose(n, j, sep=0)
    
    # consider winning runs totalling 5-8 holes on an 8 hole course
    X = dict()
    for n in irange(5, 8):
      X[n] = set(sum(tri(x) for x in ws) for ws in generate(n, 8))
    
    # consider number of holes won by J
    for j in irange(5, 8):
      k = 13 - j
      # look for equal winnings
      for w in intersect([X[j], X[k]]):
        printf("I pays {t} [J wins {j}; K wins {k}]", t=w + w)
    

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