S2T2

Jim Randell 8:57 am on 17 July 2026 Permalink Reply
Tags: by: Bob Walker ( 12 )   

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  Jim Randell 8:58 am on 17 July 2026 Permalink | Reply

  Presumably the puzzle is asking how many possible combinations remain after the initial 4 attempts.

  This Python program runs in 64ms. (Internal runtime is 820µs).

  from enigma import (subsets, printf)
  
  # candidates already tried
  tries = [
    (6, 1, 2, 5, 3, 4),
    (4, 6, 5, 3, 2, 1),
    (2, 3, 4, 1, 6, 5),
    (3, 2, 6, 5, 4, 1),
  ]
  
  # start with all orderings
  ss = list(subsets((1, 2, 3, 4, 5, 6), size=6, select='P'))
  
  # count the number of correct positions
  def count(xs, ys):
    return sum(1 for (x, y) in zip(xs, ys) if x == y)
  
  # more than one of the digits in each candidate was in the correct position
  for ts in tries:
    ss = list(ns for ns in ss if 1 < count(ns, ts) < 6)
  
  # output solution
  printf("remaining candidates = {n}", n=len(ss))
  printf("-> {ss}")
  

  Solution: There are 3 combinations remaining.

  The remaining candidates are:

  2 6 4 5 3 1
  4 3 2 5 6 1
  6 3 4 5 2 1

  In each case, each of the initial 4 attempts was correct for exactly 2 digits.

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• 

  Ruud 4:17 pm on 17 July 2026 Permalink | Reply

  import istr
  
  
  def matches(n0, n1):
      return sum(i0 == i1 for i0, i1 in zip(n0, n1))
  
  
  tries = istr(["612534", "465321", "234165", "326541"])
  for n0 in istr.permutations(range(1, 7), join=True):
      if all(2 <= matches(n0, n1) <= 5 for n1 in tries):
          print(n0)
  

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