S2T2

Jim Randell 9:01 am on 2 November 2021 Permalink Reply
Tags: by: W H St John-Brooks ( 2 )   

• 

  Jim Randell 9:02 am on 2 November 2021 Permalink | Reply

  This Python program runs in 60ms.

  Run: [ @replit ]

  from enigma import irange, nsplit, printf
  
  # consider the 5-digit number
  for n in irange(10000, 99999):
  
    # consider Bill's statements (can't both have the same truth value)
    if ((n + 798) % 693 == 0) == ((n + 987) % 658 == 0): continue
  
    # and Anne's
    (a, b, c, d, e) = nsplit(n)
    if ((n + 969) % 714 == 0) == (3 * (c + e) == a + b + c + d + e): continue
  
    # and Carol's
    if (b + d == 3 * a) == ((n + 988) % 741 == 0): continue
  
    printf("{n}")
  

  Solution: The number was: 32571.

  Anne’s statements are: False, True.

  Bill’s statements are: False, True.

  Carol’s statements are: True, False.

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  • 

    Jim Randell 10:27 am on 3 November 2021 Permalink | Reply

    Or, checking fewer candidates:

    from enigma import irange, nsplit, printf
    
    # exactly one of Bill's statements must be true
    b1 = set(irange(10290, 99999, step=693))
    b2 = set(irange(10199, 99999, step=658))
    for n in b1.symmetric_difference(b2):
      (a, b, c, d, e) = nsplit(n)
      if (
        # Anne
        ((n + 969) % 714 == 0) != (3 * (c + e) == a + b + c + d + e) and
        # Carol
        (b + d == 3 * a) != ((n + 988) % 741 == 0)
      ):
        printf("{n}")
    

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• 

  GeoffR 8:52 am on 13 January 2022 Permalink | Reply

  
  % A Solution in MiniZinc
  include "globals.mzn";
  
  % the five digits - assuming no digit is zero
  var 1..9:a; var 1..9:b; var 1..9:c; var 1..9:d;  var 1..9:e;
  
  % the five digit number
  var 10000..99999:num = 10000*a + 1000*b + 100*c + 10*d + e;
  
  % Anne - one of two statements is true
  constraint ((num + 969) mod 714 == 0) 
  \/ (3 *(c + e) = (a  + b + c + d + e));
  
  % Bill - one of two statements is true
  constraint ((num + 798) mod 693 == 0) 
  \/ ((num + 987) mod 658 == 0);
  
  % Carol - one of two statements is true
  constraint ((b + d) == 3 * a) 
  \/ ((num  + 988) mod 741 == 0);
  
  solve satisfy;
  output ["Number is " ++  show(num) ];
  
  % Number is 32571
  % ----------
  % ==========
  
  
  

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  • 

    Jim Randell 12:51 pm on 13 January 2022 Permalink | Reply

    @GeoffR: This model checks that at least one of the two statements for A, B, C is true. Rather than exactly one.

    But changing \/ to xor (or !=) will do the job.

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• 

  GeoffR 2:31 pm on 13 January 2022 Permalink | Reply

  @ Jim: Ok, I see what you mean. This programme confirms it gets the same answer.

  
  % A Solution in MiniZinc
  include "globals.mzn";
   
  % the five digits - assuming no digit is zero
  var 1..9:a; var 1..9:b; var 1..9:c; var 1..9:d;  var 1..9:e;
   
  % the five digit number
  var 10000..99999:num = 10000*a + 1000*b + 100*c + 10*d + e;
   
  % Anne - one of two statements is true
  constraint ((num + 969) mod 714 == 0) 
  != (3 *(c + e) = (a  + b + c + d + e));
   
  % Bill - one of two statements is true
  constraint ((num + 798) mod 693 == 0) 
  != ((num + 987) mod 658 == 0);
   
  % Carol - one of two statements is true
  constraint ((b + d) == 3 * a) 
  != ((num  + 988) mod 741 == 0);
   
  solve satisfy;
  output ["Number is " ++  show(num) ];
   
  % Number is 32571
  % ----------
  % ==========
  
  

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