S2T2

Jim Randell 8:14 am on 24 October 2025 Permalink Reply
Tags: by: Graham Smithers ( 82 )   

• 

  Jim Randell 8:14 am on 24 October 2025 Permalink | Reply

  Here is a solution using the [[ SubstitutedExpression ]] solver from the enigma.py library, that is a direct interpretation of the puzzle statement.

  The following run file executes in 227ms. (Internal runtime of the generated code is 59ms).

  #! python3 -m enigma -rr
  
  SubstitutedExpression
  
  "is_square(FOUR)"
  "is_cube(EIGHT)"
  "is_prime(ONE + TEN)"
  
  --answer="THREE"
  

  Solution: THREE = 42911.

  We have:

  FOUR = 7569 = sq(87)
  EIGHT = 13824 = cb(24)
  ONE + TEN = 501 + 410 = 911 (prime)

  LikeLike

  • 

    Ruud 7:48 am on 25 October 2025 Permalink | Reply

    import peek
    import istr
    
    for x2 in istr.range(32, 100):
        for x3 in istr.range(22, 47):
            for n in istr.range(10):
                f, o, u, r = [*x2**2]
                e, i, g, h, t = [*x3**3]
                if ((o | n | e) + (t | e | n)).is_prime() and ((f | o | u | r | e | i | g | h | t | n).all_distinct()):
                    peek(t | h | r | e | e)
    

    LikeLike

• 

  GeoffR 8:58 pm on 25 October 2025 Permalink | Reply

  
  % A Solution in MiniZinc
  include "globals.mzn";
  
  var 0..9: U; var 0..9: I; var 0..9: R;
  var 0..9: G; var 0..9: H; var 0..9: N;
  var 1..9: O; var 1..9: F; var 1..9: E; var 1..9: T;
  
  constraint all_different([U,I,R,G,H,N,O,F,E,T]);
  var 100..999:ONE == 100*O +10*N + E;
  var 100..999:TEN == 100*T +10*E + N;
  var 1000..9999: FOUR == 1000*F + 100*O + 10*U + R;
  var 10000..99999: EIGHT == 10000*E + 1000*I + 100*G + 10*H + T;
  var 10000..99999: THREE == 10000*T + 1000*H + 100*R + 11*E;
  
  set of int: sq4 = {n*n | n in 32..99}; 
  set of int: cb5 = {n*n*n | n in 22..46}; 
  
  predicate is_prime(var int: x) = 
  x > 1 /\ forall(i in 2..1 + ceil(sqrt(int2float(ub(x))))) 
  ((i < x) -> (x mod i > 0));
  
  constraint FOUR in sq4 /\ EIGHT in cb5 /\ is_prime(ONE + TEN);
  
  solve satisfy;
  
  output ["THREE = " ++ show(THREE)
  ++ "\n" ++ "FOUR = " ++ show(FOUR)
  ++ "\n" ++ "EIGHT = " ++ show(EIGHT)
  ++ "\n" ++ "ONE = " ++ show(ONE) 
  ++ "\n" ++ "TEN = " ++ show(TEN)];
  
  % THREE = 42911
  % FOUR = 7569
  % EIGHT = 13824
  % ONE = 501
  % TEN = 410
  % ----------
  % ==========
  
  
  

  LikeLike

• 

  ruudvanderham 3:04 pm on 26 October 2025 Permalink | Reply

  With the latest version of istr, we can decompose and compose to/from one letter variables, resulting in cleaner code:

  import istr
  import peek
  
  for x2 in istr.range(32, 100):
      for x3 in istr.range(22, 47):
          for n in istr.range(10):
              (x2**2).decompose("four")
              (x3**3).decompose("eight")
              if (istr.compose("one") + istr.compose("ten")).is_prime() and istr.compose("foureightn").all_distinct():
                  peek(istr.compose("three"))
  

  LikeLike