S2T2

Jim Randell 10:35 am on 28 March 2024 Permalink Reply
Tags: by: Andrew Skidmore ( 85 )   

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  Jim Randell 10:36 am on 28 March 2024 Permalink | Reply

  At least one of the numbers is odd, so at least one of E or R must be odd.

  We can check that one of the numbers is the sum of the other two by adding all of them together, and then this total must be twice one of the original numbers:

  SIMPLE + EASTER + TEASER ∈ { 2×SIMPLE, 2×EASTER, 2×TEASER }

  So the LHS must be an even number, and if both E and R were odd, the LHS would be an odd number, so only one of them can be odd. And if E is odd then the LHS is also odd, hence E must be even and R must be odd.

  We can solve the puzzle using this expression directly with the [[ SubstitutedExpression ]] solver from the enigma.py library.

  The following run file executes in 383ms. (Internal runtime of the generated program is 201ms).

  Run: [ @replit ]

  #! python3 -m enigma -rr
  
  SubstitutedExpression
  
  "SIMPLE + EASTER + TEASER in { 2 * SIMPLE, 2 * EASTER, 2 * TEASER }"
  
  # E is even, R is odd
  "E % 2 = 0"
  "R % 2 = 1"
  
  --answer="EASTER"
  

  Solution: EASTER = 278521.

  And the sum is:

  EASTER + TEASER = SIMPLE → 278521 + 527821 = 806342

  There is a further solution to this alphametic sum where all the terms are even:

  EASTER + TEASER = SIMPLE → 417342 + 341742 = 759084

  ---

  For a faster program we can consider the three possible sums (which can be solved using the [[ SubstitutedExpression.split_sum() ]] solver.

  The following Python program runs in 58ms. (Internal runtime is 8.6ms).

  Run: [ @replit ]

  from enigma import (SubstitutedExpression, delete, sprintf, join, printf)
  
  # words involved in the alphametic
  words = ['SIMPLE', 'EASTER', 'TEASER']
  
  # try each word as the result
  for (i, result) in enumerate(words):
    expr = sprintf("{terms} = {result}", terms=join(delete(words, [i]), sep=" + "))
    # construct the alphametic puzzle
    p = SubstitutedExpression.split_sum(
      expr,
      extra=["(E % 2 != 0) or (R % 2 != 0)"],  # at least one is odd
    ).solver()
    # solve the puzzle
    for s in p.solve(verbose=0):
      # output solution
      printf("{expr} / {s}", s=p.substitute(s, expr))
  

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• 

  GeoffR 1:47 pm on 28 March 2024 Permalink | Reply

  % A Solution in MiniZinc
  include "globals.mzn";
  
  var 1..9:S; var 0..9:I; var 0..9:M; var 0..9:P; var 0..9:L;
  var 1..9:E; var 0..9:A; var 1..9:T; var 1..9:R;
  
  constraint all_different ([S, I, M, P, L, E, A, T, R]);
  
  % At least one of three numbers is odd.
  constraint sum([E mod 2 == 1, R mod 2 == 1]) > 0;
  
  var 100000..999999:SIMPLE == 100000*S + 10000*I + 1000*M + 100*P + 10*L + E;
  var 100000..999999:EASTER == 100000*E + 10000*A + 1000*S + 100*T + 10*E + R;
  var 100000..999999:TEASER == 100000*T + 10000*E + 1000*A + 100*S + 10*E + R;
  
  %  One of the three numbers is the sum of the other two
  constraint SIMPLE == EASTER + TEASER
  \/ EASTER == SIMPLE + TEASER
  \/ TEASER == SIMPLE + EASTER;
  
  solve satisfy;
  
  output[ "EASTER = " ++ show(EASTER) 
  ++ "\n" ++ "SIMPLE = " ++ show(SIMPLE) 
  ++ "\n" ++ "TEASER = " ++ show(TEASER) ];
  
  % EASTER = 278521
  % SIMPLE = 806342
  % TEASER = 527821
  % ----------
  % ==========
  % So SIMPLE = EASTER + TEASER
  

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• 

  Frits 9:13 pm on 28 March 2024 Permalink | Reply

    
  from itertools import permutations
  
  # EASTER
  # TEASER     only valid sum ot the three sums (otherwise E = 0)
  # ------ +
  # SIMPLE
  
  # A can't be zero as it would also lead to M = 0
  for R, E, S, T in permutations(range(1, 10), 4):
    # basic checks (R must be odd)
    if not ((2 * R) % 10 == E and R % 2 and E + T in {S - 1, S}): continue
  
    if (L := (2 * E + (R > 5)) % 10) in {R, E, S, T}: continue
    if (P := (S + T + (E > 4)) % 10) in {R, E, S, T, L}: continue
    
    for A in set(range(1, 10)).difference({R, E, S, T, L, P}):
      EASTER = int(''.join(str(x) for x in [E, A, S, T, E, R]))
      TEASER = int(''.join(str(x) for x in [T, E, A, S, E, R]))
      
      if (SIMPLE := EASTER + TEASER) // 100000 != S: continue
      IM = (SIMPLE // 1000) % 100
      if any(int(x) in {R, E, S, T, L, P, A} for x in str(IM)): continue
      print("answer:", EASTER)   
  

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