S2T2

Jim Randell 4:44 pm on 16 October 2020 Permalink Reply
Tags: by: John Owen ( 20 )   

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  Jim Randell 5:03 pm on 16 October 2020 Permalink | Reply

  This Python program runs in 46ms.

  Run: [ @repl.it ]

  from enigma import irange, div, peek, printf
  
  # calculate PS
  PS = lambda p, t: div(100 * p, t)
  
  # generate scenarios with p balls potted (at the end)
  def generate(p):
    # consider total number of attempts (at the end)
    for t in irange(p + 1, 100):
      # final PS (ps4)
      ps4 = PS(p, t)
      if ps4 is None: continue
      # last shot was a miss, and the PS before it (ps3)
      ps3 = PS(p, t - 1)
      if ps3 is None: continue
      # before that 2 lots of k balls were potted
      for k in irange(2, (p - 1) // 2):
        ps2 = PS(p - k, t - 1 - k)
        if ps2 is None: continue
        ps1 = PS(p - 2 * k, t - 1 - 2 * k)
        if ps1 is None or not(ps1 < ps2): continue
        # return (t, k, PS) values
        yield (t, k, (ps1, ps2, ps3, ps4))
  
  # consider number of balls potted (at the end)
  for p in irange(5, 99):
    # accumulate ps values
    s = set(ps for (t, k, ps) in generate(p))
    # is there only one?
    if len(s) == 1:
      # output solution
      printf("balls potted = {p} -> PS = {ps}", ps=peek(s))
  

  Solution: You potted 13 balls.

  This corresponds to the following scenario:

  13 balls potted (2 runs of 5), PS = (3/15 = 20%; 8/20 = 40%; 13/25 = 52%; 13/26 = 50%)

  The other possible scenarios are:

  12 balls potted (2 runs of 5), PS = (2/5 = 40%; 7/10 = 70%; 12/15 = 80%; 12/16 = 75%)
  12 balls potted (2 runs of 4), PS = (4/16 = 25%; 8/20 = 40%; 12/24 = 50%; 12/25 = 48%)

  57 balls potted (2 runs of 15), PS = (27/45 = 60%; 42/60 = 70%; 57/75 = 76%; 57/76 = 75%)
  57 balls potted (2 runs of 25), PS = (7/28 = 25%; 32/50 = 64%; 57/75 = 76%; 57/76 = 75%)

  It seems plausible that these could correspond to a snooker match (it is possible to pot 10 reds + 10 colours + 6 colours = 26 balls in one frame, and we know at least 2 frames are involved). Although if just one of them did not, then the other scenario would provide a further solution.

  The program ensures that the PS values we are considering are non-zero, but allowing the initial PS to be zero gives a further solution:

  18 balls potted (2 runs of 9), PS = (0/6 = 0%; 9/15 = 60%; 18/24 = 75%; 18/25 = 72%)

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  Frits 11:06 am on 17 October 2020 Permalink | Reply

  @Jim, I think you should also consider p=4 as ps1 might be zero and zero also is a whole number .

  Also if t would be (p + 1) then ps1, ps2 and ps3 would be 100 and ps2 would not be higher than ps1.
  If you start t from (p+ 2) you don’t have to code the ps1 < ps2 check as it will always be the case.

  Of course there always is a balance between efficiency and seeing right away that the code solves the requirements.

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  • 

    Jim Randell 11:50 am on 17 October 2020 Permalink | Reply

    @Frits: Unfortunately the term “whole number” isn’t precise. It can be used to mean the integers, the non-negative integers, or just the positive integers.

    For this puzzle I took it to be the positive integers (which gets around a possible problem when considering PS values of 0), so I didn’t have to consider PS values of zero. Which is probably what the setter intended, as I’m sure the puzzle is meant to have a unique solution.

    I also wanted to make the [[ ps1 < ps2 ]] condition explicit in the code (as without it there would be further solutions – which can be seen by removing the test).

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  Frits 1:28 pm on 17 October 2020 Permalink | Reply

  Assuming whole numbers are in the range (1, …, 100) and with the “improved PS” check (which is not needed if we use “AB < CD").

  If we include zero as a whole number there are 2 solutions.

  from enigma import  SubstitutedExpression, multiset
  
  r = multiset()
  
  p = SubstitutedExpression([
      # we start with AB balls potted from CD attempts
      "AB <= CD",
      "CD > 0",
      # "I then potted several balls (XY > 1) in a row"
      # Assume CD + 2 * XY + 1 is roughly less than 100
      "XY > 1",
      "XY < (99 - CD) // 2",
       # the 4 pot successes being whole numbers (meaning 1..100 ???)
      "div(100 * AB, CD) > 0",
      "div(100 * (AB + XY), CD + XY) > 0",
      "div(100 * (AB + 2 * XY), CD + 2 * XY) > 0",
      "div(100 * (AB + 2 * XY), CD + 2 * XY + 1) > 0",
      # improved PS
      "div(100 * (AB + XY), CD + XY) > div(100 * AB, CD)",
      ],
      verbose=0,
      d2i={},
      answer="AB + 2 * XY, AB, CD, XY, " 
             "(100 * AB / CD, 100 * (AB + XY) / (CD + XY),"
             " 100 * (AB + 2 * XY) / (CD + 2 * XY),"
             " 100 * (AB + 2 * XY) / (CD + 2 * XY + 1))",
      distinct="",
  )
  
  # Solve and print answers
  
  print("potted   start  several     pot success" )
  for (_, ans) in p.solve():
    print(f" {ans[0]:>2}     {str(ans[1:3]):<9}  {str(ans[3]):<3}  {ans[4]}")
    r.add(ans[0])
  
  for (k, v) in r.items():
    if v == 1:
      print(f"\n{k} balls potted")
  

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