S2T2

Jim Randell 8:51 am on 22 July 2019 Permalink Reply
Tags: by: Andrew Skidmore ( 85 )   

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  Jim Randell 8:52 am on 22 July 2019 Permalink | Reply

  EASTER has 6 digits, (EGG)² has 5 or 6 digits, but whichever is larger BOX is the difference between them.

  The alphametic expression:

  abs(EASTER − (EGG)²) = BOX

  only has one solution (when we don’t use the digit 0):

  % python -m enigma SubstitutedExpression --digits="1-9" "abs(EASTER - sq(EGG)) = BOX"
  (abs(EASTER - sq(EGG)) = BOX)
  (abs(954891 - sq(977)) = 362) / A=5 B=3 E=9 G=7 O=6 R=1 S=4 T=8 X=2
  [1 solution]
  

  So this gives us our solution (BOX=362), even without needing to know which of EASTER or (EGG)² is larger. (In fact EASTER > (EGG)²).

  But here is a full solution, that solves the two possible sums as alphametic problems, and looks for situations where there is a single solution for BOX. It runs in 122ms.

  Run: [ @replit ]

  from enigma import (SubstitutedExpression, irange, printf)
  
  # look for solutions to the alphametic sum that give a unique value for ans
  def check(expr, answer):
    p = SubstitutedExpression(expr, digits=irange(1, 9), answer=answer, verbose=0)
    r = set(x for (_, x) in p.solve())
    if len(r) == 1:
      x = r.pop()
      printf("{answer}={x} [{expr}]")
  
  # there are two possible largest values
  check("sq(EGG) + BOX = EASTER", "BOX") # if EASTER is largest
  check("sq(EGG) - BOX = EASTER", "BOX") # if EGG^2 is largest
  

  Solution: BOX = 362.

  In fact this program runs faster than the single alphametic expression above, as each alphametic expression only requires considering the possibilities for 5 digits, rather than 6 with the single sum.

  If (non-leading) 0 digits are allowed, then there are three further solutions to the (EGG)² + BOX = EASTER sum (but still none for (EGG)² − BOX = EASTER).

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  GeoffR 8:52 am on 30 July 2019 Permalink | Reply

  % A Solution in MiniZinc
  include "globals.mzn";
  
  var 1..9:E;   var 1..9:A;   var 1..9:S;   var 1..9:T;
  var 1..9:R;   var 1..9:G;   var 1..9:B;   var 1..9:O;
  var 1..9:X;
  
  constraint all_different ([E, A, S, T, R, G, B, O, X]);
  
  var 100..999:EGG = 100*E + 11*G;
  var 100..999:BOX = 100*B + 10*O + X;
  var 100000..999999: EASTER = 100000*E + 10000*A + 1000*S + 100*T + 10*E + R;
  
  % Two of the numbers add to make the third
  constraint (EASTER - EGG*EGG == BOX) \/ (EGG*EGG - EASTER == BOX);
  
  solve satisfy;
  
  output [ "EASTER = " ++ show(EASTER) 
  ++ ", EGG = " ++ show(EGG)  ++ ", BOX = " ++ show(BOX) ];
  
  % EASTER = 954891, EGG = 977, BOX = 362
  % ----------
  % Finished in 232msec
  
  

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