S2T2

Jim Randell 11:19 am on 26 December 2025 Permalink Reply
Tags: by: C Higgins ( 35 ), by: H Bradley ( 35 )   

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  Jim Randell 11:20 am on 26 December 2025 Permalink | Reply

  This Python program runs in 70ms. (Internal runtime is 1.7ms).

  from enigma import (irange, inf, subsets, div, printf)
  
  # choose male (= 0) and female (= 1) values (longest serving -> shortest serving)
  for vs in subsets([0, 1], size=6, select='M'):
    # find possible (i, j) values for Mr S and Ms J
    ijs = list((i, j) for (i, j) in subsets(irange(6), size=2) if j >= i + 3 and vs[i] == 0 and vs[j] == 1)
    if not ijs: continue
  
    # consider increasing values for the male voucher
    for x in irange(1, inf):
      # calculate the years of service for the longest serving employee (>= 6)
      n = 500 + sum(i * (x + v) for (i, v) in enumerate(vs))
      d = 6 * x + sum(vs)
      if n < 6 * d: break
      y = div(n, d)
      if y is None: continue
  
      # calculate actual gift amounts
      gs = list((y - i) * (x + v) for (i, v) in enumerate(vs))
  
      # find shared values for Mr S and Ms J
      for (i, j) in ijs:
        if gs[i] == gs[j]:
          # output solution
          printf("vs={vs}, y={y}, x={x}, gs={gs}, i={i}, j={j}")
  

  Solution: The common value is £80.

  The red vouchers (male) are worth £4 each. The green vouchers (female) are worth £5 each.

  The amounts given to the employees are:

  21 years, female → £105
  20 years, male → £80 (Mr Smith)
  19 years, female → £95
  18 years, male → £72
  17 years, male → £68
  16 years, female → £80 (Ms Jones)
  total = £500

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• 

  Ruud 1:17 pm on 26 December 2025 Permalink | Reply

  Extreme brute force:

  import peek
  import itertools
  import istr
  
  for n in range(1, 30):
      for bonus in range(100):
          for is_womans in itertools.product((False, True), repeat=6):
              if sum(k * (bonus + is_woman) for k, is_woman in enumerate(is_womans, n)) == 500:
                  for jones, smith in itertools.product(range(6), repeat=2):
                      if not is_womans[jones] and is_womans[smith] and jones > smith + 2 and (jones + n) * bonus == (smith + n) * (bonus + 1):
                          peek((jones + n) * bonus, (smith + n) * (bonus + 1))
  

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