S2T2

Jim Randell 2:26 pm on 14 November 2021 Permalink Reply
Tags: by: Quentin Fowler ( 3 )   

• 

  Jim Randell 2:27 pm on 14 November 2021 Permalink | Reply

  See also: Teaser 1908, Teaser 2756, Enigma 1602, Enigma 1278, Enigma 1292.

  It turns out that choosing 2 symbols to be the same, and then checking for other values with the same value as the 2 we chose is much faster than trying all combinations of 4 symbols.

  The following Python program runs in 105ms.

  Run: [ @replit ]

  from enigma import (irange, int2words, union, matrix, subsets, trim, join, map2str, printf)
  
  # we are interested in the values of the words "one" ... "seventeen"
  words = dict((i, int2words(i)) for i in irange(1, 17))
  
  # determine symbols used
  symbols = sorted(union(words.values()))
  n = len(symbols)
  
  # a function for making equations
  eq = matrix.equation(symbols)
  
  eqs = list(eq(w, i) for (i, w) in words.items())
  
  # choose the first k symbols to have the same (positive) value
  k = 2
  for ss in subsets(trim(symbols, tail=4 - k), size=k, fn=list):
    s0 = ss[0]
    eqs_ = list(eq ({s0: -1, s: 1}) for s in ss[1:])
  
    try:
      vs = matrix.linear(eqs + eqs_)
    except ValueError:
      continue
  
    # check the shared value is positive, and used exactly 4 times
    m = dict(zip(symbols, vs))
    v = m[s0]
    sv = sorted(s for s in symbols if m[s] == v)
    if not (v > 0 and len(sv) == 4 and sv[:k] == ss): continue
  
    # output solution
    ans = sum(m[s] for s in 'nought')
    printf("nought={ans} {vs} {sv}",
      vs=map2str(m, sep=" ", enc="[]"),
      sv=join(sv, sep="=", enc="[]"),
    )
  

  Solution: NOUGHT = 23/2 (= 11.5).

  The values are:

  E = I = S = W = 0
  F = R = U = V = 5/2
  G = 15/2
  H = −5
  L = 4
  N = 9/2
  O = −7/2
  T = 11/2
  X = 6

  So the four letters that share a positive value are F, R, U, V.

  (E, I, S, W, also share a value, but this is zero).

  ---

  We immediately see that E = 0. (As 14 = 4 + 10, 16 = 6 + 10, 17 = 7 + 10), and then I = 0 (as 13 = 3 + 10). So we could exclude these from the letters we consider when we choose some of them to be the same, to give a slight speed improvement.

  This observation can also form the start of a manual solution.

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  • 

    Frits 3:33 pm on 15 November 2021 Permalink | Reply

    @Jim, symbols[:n + k – 4] makes more sense to me. fi, now you miss the situation e, v, w and x being the same.

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    • 

      Jim Randell 3:45 pm on 15 November 2021 Permalink | Reply

      @Frits: Yes it should be [:n + k − 4].

      I might add a [[ trim(s, head, tail) ]] function to enigma.py to make it easier to avoid these “off-by-1” errors (which are all too easy to make with Python indices).

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