S2T2

Jim Randell 8:47 am on 4 March 2021 Permalink Reply
Tags: by: Susan Bricket ( 17 )   

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  Jim Randell 8:48 am on 4 March 2021 Permalink | Reply

  This Python program runs in 47ms.

  Run: [ @repl.it ]

  from enigma import divf, irange, printf
  
  # percentage a/b to the nearest whole number
  percent = lambda a, b: divf(200 * a + b, 2 * b)
  
  # total number of people expected (including aunt)
  for n in irange(2, 100):
    # consider number of vegetarians
    for v in irange(1, n):
      # percentage vegetarians (to nearest whole number)
      p = percent(v, n)
      if p > 9: break
  
      # and if aunt doesn't come, is percentage the same?
      if percent(v - 1, n - 1) == p:
        # output solution
        printf("{n} total; {v} veg [{p}% veg]")
  

  Solution: If the aunt comes there will be 95 people in total. 9 of them vegetarian.

  In this case the percentage of vegetarians is 9/95 ≈ 9.47%, which rounds down to 9%.

  But if the aunt doesn’t come both the number of guests and the number of vegetarians is reduced by 1, giving a percentage of 8/94 ≈ 8.51%, which rounds up to 9%.

  ---

  Note that Python 3’s built-in [[ round() ]] function might not do what you expect:

  % python3.9
  >>> round(11.5)
  12
  >>> round(12.5)
  12
  >>> round(1.15, 1)
  1.1
  >>> round(0.115, 2)
  0.12
  

  The decimal module allows more control over rounding behaviour.

  To keep things predictable, in my program I avoided float approximations by keeping the calculations in the integer domain, and I used the common “round half up” rule.

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