S2T2

Jim Randell 10:23 am on 17 March 2022 Permalink Reply
Tags: by: Ian Kay ( 5 )   

• 

  Jim Randell 10:24 am on 17 March 2022 Permalink | Reply

  This Python program uses the [[ SubstitutedExpression ]] solver, from the enigma.py library, to find possible sums, and then the [[ filter_unique ]] function to find those that give a unique result if we know the number of even numbers involved.

  It runs in 58ms.

  Run: [ @replit ]

  from enigma import (SubstitutedExpression, filter_unique, unpack, multiset, printf)
  
  # find viable alphametic sums
  p = SubstitutedExpression.split_sum(
    "ABC + DEF = GHIJ",
    extra=["A < D"], # remove duplicates
    answer="(ABC, DEF, GHIJ)",
  )
  
  # if we knew how many of the numbers were even, we could work out the result
  rs = filter_unique(
    p.answers(verbose=0),
    (lambda ns: sum(n % 2 == 0 for n in ns)),
    unpack(lambda x, y, z: z),
  )
  
  # output possible sums
  m = multiset()
  for (x, y, z) in rs.unique:
    printf("[{x} + {y} = {z}]")
    m.add(z)
  
  # output solution
  for (v, n) in m.most_common():
    printf("result = {v} [{n} sums]")
  

  Solution: The result of the sum is 1098.

  There are 4 possible sums:

  356 + 742 = 1098
  346 + 752 = 1098
  352 + 746 = 1098
  342 + 756 = 1098

  (And another 4 where the summands are swapped around).

  LikeLike

• 

  Hugh+Casement 8:58 am on 18 March 2022 Permalink | Reply

  Either all three numbers are even, or only one of them is.

  If just one of the three-digit numbers is even, there are 24 possible sets,
  with sum 1035, 1053, 1305, 1503, or 1089.
  If only the four-digit number is even, there are 20 possible sets,
  with sum 1026, 1062, 1206, 1602, or 1098.

  Only if both three-digit numbers are even do we have a unique sum.

  LikeLike

• 

  GeoffR 2:59 pm on 18 March 2022 Permalink | Reply

  A MiniZinc program does indeed confirm that two even numbers are required to sum to a unique four digit number.

  % A Solution in MiniZinc
  include "globals.mzn";
   
  var 1..9:A; var 0..9:B; var 0..9:C;
  var 1..9:D; var 0..9:E; var 0..9:F;
  var 1..9:G; var 0..9:H; var 0..9:I; var 0..9:J;
  
  constraint all_different ([A,B,C,D,E,F,G,H,I,J]);
  
  var 100..999:ABC = 100*A + 10*B + C;
  var 100..999:DEF = 100*D + 10*E + F;
  var 1000..9999:GHIJ = 1000*G + 100*H + 10*I + J;
  constraint ABC > DEF;
  
  % confirm two even 3-digit numbers are needed 
  % to sum to a unique 4-digit number 
  constraint ABC mod 2 == 0 /\ DEF mod 2 == 0;
  constraint ABC + DEF == GHIJ;
  
  solve satisfy;
  output ["Sum is " ++ show(ABC) ++ " + " ++ show(DEF) 
  ++ " = " ++ show(GHIJ) ];
  
  % Sum is 756 + 342 = 1098
  % ----------
  % Sum is 752 + 346 = 1098
  % ----------
  % Sum is 746 + 352 = 1098
  % ----------
  % Sum is 742 + 356 = 1098
  % ----------
  % ==========
  
  
  
  

  LikeLike