S2T2

Jim Randell 4:25 pm on 20 March 2020 Permalink Reply
Tags: by: Richard England ( 9 )   

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  Jim Randell 4:30 pm on 20 March 2020 Permalink | Reply

  We can use the [[ SubstitutedSum() ]] solver from the enigma.py library to solve this puzzle.

  The following run file executes in 126ms.

  Run: [ @replit ]

  #! python3 -m enigma -r
  
  SubstitutedSum
  
  --base="11"
  --digits="1-10"
  
  "THREE + THOUS + ANDTH = TEASER"
  

  Solution: TEASER = 12X523.

  There are two ways to assign values to the letters, as D and O are interchangeable:

  17322 + 17495 + X6817 = 12X523 / A=10 D=8 E=2 H=7 N=6 O=4 R=3 S=5 T=1 U=9
  17322 + 17895 + X6417 = 12X523 / A=10 D=4 E=2 H=7 N=6 O=8 R=3 S=5 T=1 U=9

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  • 

    Jim Randell 12:08 pm on 1 July 2023 Permalink | Reply

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

    The internal runtime of this run file is 495µs.

    Run: [ @replit ]

    #! python3 -m enigma -rr
    
    SubstitutedExpression.split_sum
    
    --base="11"
    
    "{THREE} + {THOUS} + {ANDTH} = {TEASER}"
    
    --invalid="0,ADEHNORSTU"  # no 0 digits
    --code="base_digits('0123456789X')"  # use X for digit 10
    --answer="int2base({TEASER}, 11)"  # output solution in base 11
    

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  GeoffR 1:48 pm on 21 March 2020 Permalink | Reply

  The teaser uses the 10 different letters ([T,H,R,E,O,U,S,A,N,D] in this teaser), which can convieniently represent the digits (1..10), as zero is not required,

  The programme produces a single solution with the third digit as 10 (or X), alltough there are two sets of digits, with values of O and D interchangeable.

  The programme uses the standard method of adding digits in columns, with carry digits to adjacent columns for column digit sums exceeding 11.

  % A Solution in MiniZinc
  include "globals.mzn";
  
  %    T H R E E
  %    T H O U S
  %    A N D T H
  %  -----------
  %  T E A S E R
  %  -----------
  
  % Using digits 1..10 in base 11
  var 1..10:T; var 1..10:H; var 1..10:R; var 1..10:E;
  var 1..10:O; var 1..10:U; var 1..10:S; var 1..10:A;
  var 1..10:N; var 1..10:D;
  
  constraint all_different ([T,H,R,E,O,U,S,A,N,D]);
  
  var 0..2: carry1; var 0..2: carry2; var 0..2: carry3;
  var 0..2: carry4; var 0..2: carry5;
  
  % Adding up digits in columns, starting from right side
  constraint (E + S + H) mod 11 == R 
  /\ carry1 == (E + S + H) div 11;
  
  constraint (E + U + T + carry1) mod 11 == E 
  /\ carry2 == (E + U + T + carry1) div 11;
  
  constraint (R + O + D + carry2) mod 11 == S 
  /\ carry3 == (R + O + D + carry2) div 11;
  
  constraint (H + H + N + carry3) mod 11 == A 
  /\ carry4 = (H + H + N + carry3) div 11;
  
  constraint (T + T + A + carry4) mod 11 == E 
  /\ carry5 == (T + T + A + carry4) div 11;
  
  constraint T == carry5;
  
  solve satisfy;
  
  output ["TEASER = " ++ show(T) ++ " " ++ show(E) ++ " " ++ show(A)
  ++ " " ++ show(S) ++ " " ++ show(E) ++ " " ++ show(R)  
  ++"\n" ++ "10 digits are [T,H,R,E,O,U,S,A,N,D] = " ++ show([T,H,R,E,O,U,S,A,N,D]) ];
  
  
  

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