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Jim Randell 9:17 am on 13 January 2022 Permalink Reply
Tags: by: Des MacHale ( 11 )   

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  Jim Randell 9:17 am on 13 January 2022 Permalink | Reply

  If n3, n4, n5 are the numbers of par 3, 4, 5 holes respectively we have:

  n3 + n4 + n5 = 18
  3 × n3 + 4 × n4 + 5 × n5 = 70

  We can treat the problem as a “money changing” puzzle, i.e. we want to make an amount of 70 using 18 coins of denominations 3, 4, 5.

  The following Python program uses the [[ express() ]] function from the enigma.py library. It runs in 50ms.

  Run: [ @replit ]

  from enigma import (express, primes, irange, subsets, tuples, printf)
  
  # primes up to 18
  primes = set(primes.between(1, 18))
  
  # odd numbers up to 18
  odds = set(irange(1, 18, step=2))
  
  for (n3, n4, n5) in express(70, [3, 4, 5], min_q=1):
    if not (n3 + n4 + n5 == 18): continue
  
    # choose par 5 holes (prime)
    for p5s in subsets(primes, size=n5):
  
      # choose par 4 holes (odd)
      for p4s in subsets(odds.difference(p5s), size=n4):
  
        # the rest are par 3
        par = [3] * 18
        for k in p5s: par[k - 1] = 5
        for k in p4s: par[k - 1] = 4
  
        # no consecutive sequence of 9 holes sums to 35
        if any(sum(t) == 35 for t in tuples(par, 9)): continue
  
        printf("p4s={p4s} p5s={p5s}; {par}")
  

  Solution: The par 4 holes are: 1, 9.

  The par 5 holes are: 2, 3, 5, 7, 11, 13, 17.

  ---

  We can use a more sophisticated “money changing” algorithm by using:

  from enigma import Denominations
  ...
  
  for (n3, n4, n5) in Denominations(3, 4, 5).express(70, min_q=1):
    ...
  

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