S2T2

Jim Randell 11:30 am on 8 February 2024 Permalink Reply
Tags: by: Peter G Chamberlain ( 11 )   

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  Jim Randell 11:31 am on 8 February 2024 Permalink | Reply

  Here is a solution using the [[ SubstitutedExpression ]] solver from the enigma.py library (although the method of constructing alphametic words is different from the usual alphametic rules).

  It runs in 194ms. (Internal runtime of the generated program is 78ms).

  Run: [ @replit ]

  #! python3 -m enigma -rr
  
  SubstitutedExpression
  
  --base=26
  
  --code="nconc = lambda *ss: int(concat(ss))"
  --macro="@TWO = nconc(T, W, O)"
  --macro="@FOUR = nconc(F, O, U, R)"
  
  "sq(@TWO) == @FOUR"
  
  --answer="@FOUR"
  
  # T, W, O are chosen from: 2, 3, 5, 7, 22, 23, 25
  --invalid="0|1|4|6|8-21|24,TWO"
  
  # [optional] speeds things up a bit
  # FOUR is a square ...
  "is_square(@FOUR)"
  # ... so R must be the final digits of a square
  --invalid="2-3|7-8|10-15|17-20|22-23,R"
  

  Solution: FOUR = 105625.

  We have:

  TWO = 3:2:5 = 325
  FOUR = 10:5:6:25 = 105625 = 325^2

  ---

  And the solution to the TWO × TWO = THREE puzzle is:

  % python3 enigma.py SubstitutedExpression "TWO * TWO = THREE"
  (TWO * TWO = THREE)
  (138 * 138 = 19044) / E=4 H=9 O=8 R=0 T=1 W=3
  [1 solution]
  

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  GeoffR 4:04 pm on 8 February 2024 Permalink | Reply

  from enigma import all_different
  from math import isqrt
  
  def is_sq(x):
     return isqrt(x) ** 2 == x
  
  tup_TWO = (2, 3, 5, 7, 22, 23, 25)
   
  for F in range(1, 26):
    for O in tup_TWO:
      if O == F:continue
      for U in range(26):
        if U in (F, O):continue
        for R in (4, 9, 25): # max value of R = 25
          if R in (U, O, F):continue
          FOUR = int(str(F) + str(O) + str(U) + str(R))
          if not is_sq(FOUR):continue
          for T in tup_TWO:
            for W in tup_TWO:
              if not all_different(T, W, O, F, U, R):continue
              TWO = int(str(T) + str(W) + str(O))
              if TWO * TWO == FOUR:
                print(f"TWO = {TWO}, FOUR = {FOUR}.")
                
  # TWO = 325, FOUR = 105625.
  
  

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  GeoffR 8:55 pm on 8 February 2024 Permalink | Reply

  TWO × TWO = THREE is much easier than TWO × TWO = FOUR, as the former has only digits 0..9 to consider.

  from itertools import permutations
  
  for p1 in permutations('1234567890', 3):
     T, W, O = p1
     if T == '0':continue
     TWO = int(T + W + O)
     q1 = set('01234560789').difference([T, W, O])
     for p2 in permutations(q1, 3):
        R, H, E = p2
        THREE = int(T + H + R + E + E)
        if TWO * TWO == THREE:
           print(f"Sum: {TWO} * {TWO} = {THREE}.")
  
  # Sum: 138 * 138 = 19044.
  

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