S2T2

Jim Randell 9:28 am on 7 September 2023 Permalink Reply
Tags: by: Andrew Skidmore ( 85 ), flawed ( 53 )   

• 

  Jim Randell 9:29 am on 7 September 2023 Permalink | Reply

  With the extra condition that “FIVE is greater than FOUR” we find the published answer.

  So maybe the setter intended to say: “… I have found that FIVE − FOUR is a positive number, whose square is ONE” (or: “the square root of ONE is equal to FIVE – FOUR“).

  This Python program runs in 147ms.

  Run: [ @replit ]

  from enigma import (SubstitutedExpression, group, unpack, item, peek, seq2str, printf)
  
  # the alphametic problem (with an additional condition)
  p = SubstitutedExpression(
    ["sq(FIVE - FOUR) = ONE", "FIVE > FOUR"],
    answer="(THREE, FOUR, FIVE)",
  )
  
  # collect the answers, and record values of FIVE (= item(2))
  # by (THREE divisible by 3, FOUR divisible by 4).
  fn = unpack(lambda n3, n4, n5: (int(n3 % 3 == 0), int(n4 % 4 == 0)))
  d = group(p.answers(verbose=0), by=fn, f=item(2), fn=set)
  
  # output the collections, and find the solution
  for (k, vs) in d.items():
    if len(vs) == 1:
      printf("(d3, d4) = {k} -> FIVE = {v} [*** SOLUTION ***]", v=peek(vs))
    else:
      printf("(d3, d4) = {k} -> FIVE = {vs}", vs=seq2str(vs, sort=1, enc="{}"))
  

  Solution: FIVE = 5901.

  If we are told THREE is not divisible by 3, and FOUR is divisible by 4, then the only possible assignments are:

  FIVE = 5901, FOUR = 5872, ONE = 841, THREE = 36211
  FIVE = 5901, FOUR = 5872, ONE = 841, THREE = 63211

  Any of the other of the divisibility possibilities lead to multiple candidate values for FIVE, so this gives the answer.

  ---

  However if we do not require FIVE > FOUR, then we find many further possible solutions to the alphametic problem:

  FIVE = 1496, FOUR = 1520, ONE = 576, THREE = 38066
  FIVE = 1496, FOUR = 1520, ONE = 576, THREE = 83066

  FIVE = 3791, FOUR = 3820, ONE = 841, THREE = 56011
  FIVE = 3791, FOUR = 3820, ONE = 841, THREE = 65011

  FIVE = 5791, FOUR = 5820, ONE = 841, THREE = 36011
  FIVE = 5791, FOUR = 5820, ONE = 841, THREE = 63011

  FIVE = 6791, FOUR = 6820, ONE = 841, THREE = 35011
  FIVE = 6791, FOUR = 6820, ONE = 841, THREE = 53011

  FIVE = 8496, FOUR = 8520, ONE = 576, THREE = 13066
  FIVE = 8496, FOUR = 8520, ONE = 576, THREE = 31066

  And so we cannot work out the value of FIVE.

  We can see these candidates by removing the [[ "FIVE > FOUR" ]] on line 5 of the program.

  LikeLike

  • 

    Frits 12:32 pm on 8 September 2023 Permalink | Reply

    The reported run time 147ms was probably with PyPy (I get 423ms with CPython).

    With this program CPython performs better than PyPY.

    from enigma import (SubstitutedExpression, group, unpack, item, peek, seq2str, printf)
    
    # invalid digit / symbol assignments
    d2i = dict()
    for d in range(0, 10):
      vs = set()
      if d == 0: vs.update('OFT')
      # if E is odd then R is even and if E is even then R is even as well
      if d % 2: vs.update('R')
      # a number that ends in 2, 3, 7 or 8 is not a perfect square
      if d in {2, 3, 7, 8}: vs.update('E')
      d2i[d] = vs
     
    # the alphametic problem (with an additional condition)
    p = SubstitutedExpression(
      [ # FIVE > FOUR
        "I > O", 
        "sq(IVE - OUR) = ONE",],
      answer="(FOUR, FIVE, THREE)",
      #answer="(FOUR, FIVE, \
      #         R + 2*E + sum(i for i in range(10) if i not in {F,I,V,E,O,U,R,N}))",
      d2i=d2i,
      verbose=0
    )
     
    # collect the answers, and record values of FIVE (= item(1))
    # by (THREE divisible by 3, FOUR divisible by 4).
    fn = unpack(lambda n4, n5, n3: (int(n3 % 3 == 0), int(n4 % 4 == 0)))
    d = group(p.answers(), by=fn, f=item(1), fn=set)
     
    # output the collections, and find the solution
    for (k, vs) in d.items():
      if len(vs) == 1:
        printf("(d3, d4) = {k} -> FIVE = {v} [*** SOLUTION ***]", v=peek(vs))
      else:
        printf("(d3, d4) = {k} -> FIVE = {vs}", vs=seq2str(vs, sort=1, enc="{}"))   
    

    LikeLike

    • 

      Jim Randell 12:13 pm on 9 September 2023 Permalink | Reply

      @Frits: Yes, 147ms was with PyPy 7.3.12. With CPython 3.12 I get an overall runtime of 255ms.

      Although, if you are solving the rescued problem you can use the expression [[ "is_square(ONE) + FOUR = FIVE" ]] which is a bit faster, and it is even faster to use [[ "is_square(ONE) + OUR = IVE" ]] (93ms with CPython 3.12).

      LikeLike