S2T2

Jim Randell 9:03 am on 19 September 2021 Permalink Reply
Tags: by: Richard Beetham ( 2 )   

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  Jim Randell 9:03 am on 19 September 2021 Permalink | Reply

  B and J were equal after the first 9 holes (i.e. 4.5 holes each). And each won 4 of the next 8 holes, so that the match was decided on the 18th hole.

  If the stake on the 10th hole was k, then it increases in powers of 2, i.e. k, 2k, 4k, 8k, 16k, …, 256k.

  In order to come out on top B must have won the 256k hole (as the remaining holes sum to 255k), and so B won the match.

  In binary, B’s stake multiplier has 5 bits and is between 100001111 (= 271) and 111110000 (= 496), and J’s multiplier is (111111111 − B).

  In order for B to win 11s + 9d = 141d the stake on the 10th hole is 141 / (B − J), and I assume the stake is some sensible amount (i.e. a multiple of 1/4 d).

  This Python program runs in 53ms.

  from enigma import (bit_permutations, div, irange, printf)
  
  # choose B's stake multiplier
  for B in bit_permutations(0b100001111):
    J = 0b111111111 - B
    k = div(564, B - J)
    if k:
      hs = list(h + 10 for h in irange(0, 8) if B & (1 << h))
      printf("holes = {hs}; stake @ 10 = {k:g}d", k=0.25 * k)
    if B == 0b111110000: break
  

  Solution: Brown won the match. Brown won the 10th, 11th, 12th, 14th, 18th holes. The stake on the 10th hole was 3d.

  The holes Brown won give: (1 + 2 + 4 + 16 + 256)k = 279k

  And the holes Brown lost give: (8 + 32 + 64 + 128)k = 232k

  The difference is: 279k − 232k = 47k, so: k = 141d / 47 = 3d

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  Hugh Casement 10:14 am on 19 September 2021 Permalink | Reply

  To bring it up to date a bit, he won £2.35
  — though when we compare (for example) the cost of sending a letter then and now I feel an equivalent initial stake would be quite a bit more than 5p.

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  Frits 5:52 pm on 4 October 2021 Permalink | Reply

    
  from enigma import SubstitutedExpression, div
  
  # the alphametic puzzle 
  p = SubstitutedExpression(
    [
     # Brown won 4 holes on holes 10-17
     "sum([A, B, C, D, E, F, G, H]) == 4",
     # Brown won the last hole
     "I = 1",
     # Brown's profit (141 pence) is a multiple of the stake
     "div(141, sum([2 ** i if x else -1 * 2 ** i \
          for i, x in enumerate([A, B, C, D, E, F, G, H, I])]))",
    ],
    base=2,
    answer="(A, B, C, D, E, F, G, H, I), \
           sum([2 ** i if x else -1 * 2 ** i \
               for i, x in enumerate([A, B, C, D, E, F, G, H, I])])",
    verbose=0,
    distinct="",
  )
  
  # print answer
  for (_, ans) in p.solve():
    print(f"Brown won the match and the holes "
          f"{[i + 10 for i, x in enumerate(ans[0]) if x]}"
          f" with side-stake of {141 // ans[1]} pence")
  

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