S2T2

Jim Randell 9:38 pm on 31 July 2021 Permalink Reply
Tags: by: Oliver Tailby   

• 

  Jim Randell 10:01 pm on 31 July 2021 Permalink | Reply

  This Python program constructs the possible triangular tiles for increasing ages, and keeps count of the number that satisfy the given conditions. It runs in 73ms.

  from enigma import (irange, subsets, uniq, printf)
  
  # count triangles with ...
  evens = 0  # ... at least one even number
  odds = 0  # ... at least one odd number
  mul4s = 0  # ... at least one multiple of 4
  
  # canonical form (by rotation)
  def canonical(a, b, c):
    return min((a, b, c), (b, c, a), (c, a, b))
  
  # consider triangles with largest number n
  for n in irange(1, 29):
    # consider rotationally distinct triangles
    for tri in uniq(canonical(n, a, b) for (a, b) in subsets(irange(1, n), size=2, select="M")):
      # increase the counts
      if any(x % 2 == 0 for x in tri): evens += 1
      if any(x % 2 == 1 for x in tri): odds += 1
      if any(x % 4 == 0 for x in tri): mul4s += 1
  
    # output solution
    if evens % 2 == 0 and odds % 2 == 1 and mul4s % 4 == 0:
      printf("n={n} [evens={evens} odds={odds} mul4s={mul4s}]")
  

  Solution: Your age is 17.

  The total number of triangular tiles for any particular age is given by OEIS A006527 [link]:

  a(n) = n(n² + 2) / 3

  ---

  There is the possibility of a solution at age = 1. In this case there is only 1 tile in the set — (1, 1, 1). But the setter might be considering that the numbers of tiles in each category is non-zero (also 1 is a little young to be setting such a puzzle). In any case, we are told that there are two (1, 2, 3) tiles, so we know the age must be at least 3.

  In general we find solutions to the puzzle at ages of the form (16k + 1).

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  • 

    Frits 1:23 pm on 2 August 2021 Permalink | Reply

    Similar.

     
    # consider rotationally distinct triangles containing number n
    def triangles(n):
      for x in range(n, 0, -1):
        for y in range(x, 0, -1): 
          yield [n, x, y]
      for x in range(n - 2, 0, -1):    
        for y in range(n - 1, x, -1): 
          yield [n, x, y]
     
    odds, evens, mul4s = 0, 0, 0
    
    print("age evens  odds mlt4s")
    for age in range(1, 30):
      for tri in triangles(age):
        
        if any(x % 2 == 0 for x in tri): evens += 1
        if any(x % 2 for x in tri): odds += 1
        if any(x % 4 == 0 for x in tri): mul4s += 1
      
      if mul4s % 4 == 0 and evens % 2 == 0 and odds % 2:
        print(f"{age:>3} {evens:>5} {odds:>5} {mul4s:>5}") 
    

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