S2T2

Jim Randell 9:22 am on 16 February 2021 Permalink Reply
Tags: by: B W M Young ( 10 )   

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  Jim Randell 9:23 am on 16 February 2021 Permalink | Reply

  If the First Sea Lord and the Second Sea Lord both predict taking on the same position, I am assuming that the position cannot be that of the First or Second Sea Lord, as then one of them would have made a prediction that was impossible to fulfil (because no-one keeps their position).

  This Python program runs in 53ms.

  Run: [ @repl.it ]

  from enigma import subsets, peek, map2str, printf
  
  # positions
  pos = (1, 2, 3, 4, 5)
  
  # check <n> makes statement <s> is valid in map <m>
  check = lambda m, n, s: (m[n] < n) == bool(s)
  
  # find an index that maps to x
  find = lambda m, x: peek(k for (k, v) in m.items() if v == x)
  
  # consider derangements
  for s in subsets(pos, size=len, select="D"):
    m = dict(zip(pos, s))
  
    # check the statements:
  
    # 5: "two will make a direct exchange"
    if not check(m, 5, (any(m[v] == k for (k, v) in m.items()))): continue
  
    # 4: "3 -> 2"
    if not check(m, 4, (m[3] == 2)): continue
  
    # 3: "a man who makes a true prediction (so is promoted) will take
    # over my job (= 3)"
    if not check(m, 3, (find(m, 3) > 3)): continue
  
    # 1: "1 -> x", 2: "2 -> x"
    for x in pos:
      if x == 1 or x == 2: continue
      if not(check(m, 1, (m[1] == x)) and check(m, 2, (m[2] == x))): continue
  
      printf("x={x}; {m}", m=map2str(m, arr="->"))
  

  Solution: The First and Second Sea Lords both predicted they would take on the role of Third Sea Lord.

  There are two possible maps:

  1→4; 2→5; 3→2; 4→1; 5→3
  1→5; 2→4; 3→2; 4→3; 5→1

  In both scenarios the First and Second Sea Lords are demoted (so made false predictions), and the Third, Fourth and Fifth Sea Lords are promoted (and so made true predictions).

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• 

  Frits 10:54 am on 19 February 2021 Permalink | Reply

  from enigma import SubstitutedExpression
  
  def twoswap(a, b):
    z = [tuple(sorted((x, y))) for (x, y) in zip(a, b)]
    return len(set(z)) < len(z)
  
  # the alphametic puzzle
  p = SubstitutedExpression(
    [
    # (1, 2, 3, 4, 5) --> (V, W, X, Y, Z) 
    
    "sorted([V, W, X, Y, Z]) == [1, 2, 3, 4, 5]",
    
    # 5: "two will make a direct exchange
    "twoswap([1, 2, 3, 4, 5], [V, W, X, Y, Z])",
    
    # 4: "3 -> 2"
    "(X == 2 and Y < 4) or (Y == 5 and X != 2)",
     
    # 3: "a man who makes a true prediction (so is promoted) will take
    # over my job (= 3)"
    "[V, W, X, Y, Z].index(3) > 2 if X < 3 else [V, W, X, Y, Z].index(3) < 3",
    ],
    digits=range(1, 6),
    answer="{3, 4, 5}.difference({V, W})",
    env=dict(twoswap=twoswap),
    # derangement
    # 1: "1 -> x", 2: "2 -> x",  x > 2
    # we only know both statements have to be false
    d2i={1 : "VW", 2 : "VW", 3 : "X", 4 : "Y", 5 : "Z"},
    distinct="VWXYZ",
    verbose=0)   
  
  # print answers 
  answs = set([list(y)[0] for _, y in p.solve()])
  
  for ans in answs:
    print(f"post = {ans}")
  

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  • 

    Frits 11:15 am on 19 February 2021 Permalink | Reply

    @Jim, do you agree with V > 2 and W > 2 for 1: “1 -> x”, 2: “2 -> x”?
    We only know both statements to be false.

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    • 

      Jim Randell 11:39 am on 19 February 2021 Permalink | Reply

      @Frits: I don’t see how to determine the required answer from the output of your program.

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• 

  John Crabtree 4:37 pm on 19 February 2021 Permalink | Reply

  Here are a couple of ways to tackle it manually.
  The 5th SL must be correct. There are three cases to consider:
  3rd wrong, and so the 4th is wrong: 4 -> 5, 3 -> 4, 1/2 -> 3 with no possible exchange
  3rd right, 4th wrong: 3 -> 1, 4 -> 5, 5 -> 3, with no possible exchange
  3rd right, 4th right: 3 -> 2, 4/5 -> 3, 2 -> 4/5 (to allow an exchange), 5/4 -> 1, 1 -> 5/4

  1st and 2nd become 4th and 5th or vice versa. Assuming that a SL could not predict his current position, the 1st and 2nd SLs predicted that they would be 3rd Sea Lord.

  Using Boolean algebra, where 12 means 1 moves to 2 is true,
  5th SL: the statement is true
  4th SL: (41+43).32 + 45.(31+34) = 1
  3rd SL: (31+32).(43+53) + (34+35).(13+23) = 1
  Multiplying the two and removing impossible terms gives:
  41.32.53 + 43.32 + 45.31.53 + 45.34.(13+23) = 1
  These are the same four possibilities shown in the first solution. The last two do not allow an exchange. From the 5th SL:
  14.41.32.53 + 15.51.43.32 = 1, ie
  14.41.32.53.25 + 15.51.43.32.24 = 1, with the rest as before.

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