S2T2

Jim Randell 9:48 am on 9 September 2021 Permalink Reply
Tags: by: J H Perryman ( 5 )   

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  Jim Randell 9:49 am on 9 September 2021 Permalink | Reply

  Using P for “POW!” and W for “WHAM!”:

  If we interpret “interposed” to mean that P and W are inserted between 2 existing symbols, then we find a unique solution.

  The following Python program runs in 51ms.

  Run: [ @replit ]

  from enigma import (subsets, irange, base2int, int2base, join, printf)
  
  # categorise a string
  # 'new' -> entirely composed of new symbols
  # 'old' -> entirely composed of old symbols
  # 'mix' -> a mixture of new and old symbols
  def categorise(s):
    s = set(s)
    assert bool(s)
    if s.issubset('PW'): return 'new'
    if not s.intersection('PW'): return 'old'
    return 'mix'
  
  # make possible base 12 symbol sets
  for (w, p) in subsets(irange(1, 9), size=2):
    syms = list("0123456789")
    syms.insert(p, 'P')
    syms.insert(w, 'W')
  
    # convert between strings and numbers
    s2n = lambda s: base2int(s, base=12, digits=syms)
    n2s = lambda n: int2base(n, base=12, digits=syms)
  
    # calculate: A = '35' * '64', B = '54' * '66'
    A = n2s(s2n('35') * s2n('64'))
    B = n2s(s2n('54') * s2n('66'))
    # one is entirely composed of new symbols and one entirely composed of old
    if set(map(categorise, (A, B))) == {'old', 'new'}:    
      printf("digits 0-11 = {syms}", syms=join(syms))
      printf("  35 * 64 = {A}; 54 * 66 = {B}")
  

  Solution: The two sums are: 35 × 64 = 2445 and 54 × 66 = ↓↑↑↓.

  The digits from 0 to 11 are represented by:

  0 1 2 3 W 4 5 P 6 7 8 9

  So the sums are:

  35 × 64 = 2445 [Inertial base 12]
  36 × 85 = 2556 [standard base 12]
  42 × 101 = 4242 [standard decimal]

  54 × 66 = WPPW [Inertial base 12]
  65 × 88 = 4774 [standard base 12]
  77 × 104 = 8008 [standard decimal]

  However, if we allow W and P to be inserted anywhere (but still with W before P), we find an additional solution using the following digits:

  W 0 1 2 3 P 4 5 6 7 8 9

  And the sums are:

  35 × 64 = 2194 [Inertial base 12]
  47 × 86 = 32B6 [standard base 12]
  55 × 102 = 5610 [standard decimal]

  54 × 66 = PPWW [Inertial base 12]
  76 × 88 = 5500 [standard base 12]
  90 × 104 = 9360 [standard decimal]

  We can see this additional solution by using the following call to [[ subsets() ]] in line 15:

  subsets(irange(0, 10), size=2, select='R')
  

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