S2T2

Jim Randell 5:08 pm on 13 June 2024 Permalink Reply
Tags: by: Graham Smithers ( 82 )   

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  Jim Randell 5:08 pm on 13 June 2024 Permalink | Reply

  Here is a solution using the [[ SubstitutedExpression ]] solver from the enigma.py library.

  It runs in 77ms. (Internal runtime of the generated program is 230µs).

  Run: [ @replit ]

  #! python3 -m enigma -rr
  
  # suppose the number is: ABCDEFGHI
  #                  pos = 123456789
  
  SubstitutedExpression
  
  --digits="1-9"
  
  # even positions are the sum of the adjacent digits
  "A + C = B"
  "C + E = D"
  "E + G = F"
  "G + I = H"
  
  # solution is divisible by A, I and 11
  "ABCDEFGHI % A = 0"
  "ABCDEFGHI % I = 0"
  "ABCDEFGHI % 11 = 0"
  
  --answer="ABCDEFGHI"
  

  Solution: Sam’s number was: 143968275.

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  • 

    Ruud 7:49 pm on 15 June 2024 Permalink | Reply

    No a priori knowledge, just brute force:

    from istr import istr
    
    for i in istr.concat(istr.permutations(istr.digits("1-"), 9)):
        if (
            i[1] == i[0] + i[2]
            and i[3] == i[2] + i[4]
            and i[5] == i[4] + i[6]
            and i[7] == i[6] + i[8]
            and i.divisible_by(11)
            and i.divisible_by(i[0])
            and i.divisible_by(i[8])
        ):
            print(i)
    
    

    Note that this requires the latest-soon to be published- version of istr.

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• 

  GeoffR 6:56 pm on 13 June 2024 Permalink | Reply

  
  % A Solution in MiniZinc
  include "globals.mzn";
  
  var 1..9:a; var 1..9:b; var 1..9:c;
  var 1..9:d; var 1..9:e; var 1..9:f;
  var 1..9:g; var 1..9:h; var 1..9:i;
  
  constraint all_different([a,b,c,d,e,f,g,h,i]);
  
  var 123456789..987654321:N == a*pow(10,8) + b*pow(10,7) + c*pow(10,6) 
  + d*pow(10,5) + e*pow(10,4) + f*pow(10,3) + g*pow(10,2) + h*pow(10,1) + i;
   
  constraint N mod a == 0;
  constraint N mod i == 0;
  constraint N mod 11 == 0;
  
  constraint a + c == b /\ c + e == d /\ e + g == f /\ g + i ==h;
  
  solve satisfy;
   
  output [ "His number was " ++ show(N)];
  
  % His number was 143968275
  % ----------
  % ==========
  % Finished in 461 msec - Chuffed Solver
  
  

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• 

  Frits 7:01 pm on 13 June 2024 Permalink | Reply

       
  #! python3 -m enigma -r
   
  # suppose the number is: ABCDEFGHI
  #                  pos = 123456789
   
  SubstitutedExpression
   
  --digits="1-9"
  
  # divisibility of 11 rule, abs(sum odd digits - sum even digits) is 0 or a 
  # multiple of 11, sum odd: A + C + E + G + I, sum even: A + 2.(C + G + E) + I
   
  "(11 if C + E < 11 else 22) - (C + E) = G"
   
  # even positions are the sum of the adjacent digits
  "A + C = B"
  "C + E = D"
  "E + G = F"
  "G + I = H"
   
  # solution is divisible by A and I
  "ABCDEFGHI % A = 0"
  "ABCDEFGHI % I = 0"
   
  --answer="ABCDEFGHI"
  

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  • 

    Frits 9:23 pm on 13 June 2024 Permalink | Reply

    You only need to know three digits (like the 1st, 3rd and 5th digits).

    from functools import reduce
    
    # convert digit sequences to number
    d2n = lambda s: reduce(lambda x,  y: 10 * x + y, s)
    
    dgts8 = set(range(1, 9))
    dgts9 = dgts8 | {9}
    
    for C in dgts8:
      for E in dgts8 - {C}:
        # 11 divisibility rule
        G = (11 if C + E < 11 else 22) - (C + E)
        D, F = C + E, E + G
        # different digits
        if len(r1 := dgts9 - {C, E, G, D, F}) != 4: continue
        for A in r1:
          B = A + C
          if len(r2 := sorted(r1 - {A, B})) != 2: continue
          H, I = r2[1], r2[0]
          if H != G + I: continue
          
          ABCDEFGHI = d2n([A, B, C, D, E, F, G, H, I])
          if ABCDEFGHI % A: continue
          if ABCDEFGHI % I: continue
          print("answer:", ABCDEFGHI)     
    

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• 

  Ruud 6:44 am on 16 June 2024 Permalink | Reply

  This version does not require the divisible_by method:

  from istr import istr
  
  for i in istr.concat(istr.permutations(istr.digits("1-"), 9)):
      if i[1] == i[0] + i[2] and i[3] == i[2] + i[4] and i[5] == i[4] + i[6] and i[7] == i[6] + i[8] and i % 11 == 0 and i % i[0] == 0 and i % i[8] == 0:
          print(i)
  

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