S2T2

Jim Randell 9:16 am on 14 February 2021 Permalink Reply
Tags: by: C. W. Mitford   

• 

  Jim Randell 9:18 am on 14 February 2021 Permalink | Reply

  I used the [[ SubstitutedExpression ]] solver from the enigma.py library to solve this puzzle.

  The following run file executes in 333ms.

  (Using the standard Python interpreter we need to provide the additional [[ --denest ]] parameter to work around the limit on statically nested blocks. This is not required if using the PyPy interpreter).

  Run: [ @replit ]

  #! python3 -m enigma -rr
  
  #  A B C D E F G
  #  H I J K L M N
  #  P # Q R S # T
  #  U V W # X Y Z
  #  a b c d e f g
  #  h i # # # j k
  #  m n p q r s t
  
  SubstitutedExpression
  
  --distinct=""
  --reorder=0
  
  # 1a. = ABCD "Square of a multiple of 11"
  "is_square({ABCD}) % 11 = 0"
  
  # 11a. = QR "If added to 4d. it gives a square"
  # 4d. = DKR "See 11 across"
  "is_square({DKR} + {QR})"
  
  # 8a. = HIJ "A multiple of 2d."
  # 2d. = BI "See 8 across"
  "{HIJ} % {BI} = 0"
  
  # 3d. = CJQWc "A cube"
  "is_cube({CJQWc})"
  
  # 6d. = FM "The sum of the digits of 3d."
  "dsum({CJQWc}) = {FM}"
  
  # 7d. = GNTZ "The square of (6d. reversed)"
  "{MF} ** 2 = {GNTZ}"
  
  # 5a. = EFG "A fourth power"
  "is_power({EFG}, 4)"
  
  # 12a. = UVW "Three times a prime number"
  "is_prime(div({UVW}, 3))"
  
  # 1d. = AHPU "The product of two consecutive numbers (neither a prime)"
  --code="
  def check_1d(n):
    k = isqrt(n)
    return n == k * (k + 1) and not(is_prime(k) or is_prime(k + 1))
  "
  "check_1d({AHPU})"
  
  # 13d. = Vbin "The product of 2 consecutive even numbers"
  --code="
  def check_13d(n):
    k = isqrt(div(n, 4))
    return n == 4 * k * (k + 1)
  "
  "check_13d({Vbin})"
  
  # 16a. = ab "A prime number"
  "is_prime({ab})"
  
  # 20a. = hi "A multiple of 12"
  "{hi} % 12 = 0"
  
  # 16d. = ahm "A square"
  "is_square({ahm})"
  
  # 17a. = cde "A cube, which is also the sum of three cubes"
  --code="
  def check_17a(n, k=3):
    if k == 1:
      return is_cube(n)
    else:
      for x in powers(1, inf, 3):
        if k * x > n: break
        if check_17a(n - x, k - 1): return True
      return False
  "
  "is_cube({cde}) and check_17a({cde})"
  
  # 10d. = LSXe "The product of 2 consecutive odd numbers"
  "is_square(div({LSXe} + 1, 4))"
  
  # 14a. = XYZ "A multiple of 11"
  "{XYZ} % 11 = 0"
  # 9a. = KLMN "A cube reversed"
  "is_cube({NMLK})"
  
  # 21a. = jk "A prime number"
  "is_prime({jk})"
  
  # 15d. = Yfjs "A cube"
  "is_cube({Yfjs})"
  
  # 19d. = gkt "An even multiple of (18a. + 21a.)"
  "div({gkt}, {fg} + {jk}) % 2 = 0"
  
  # 18a. = fg "An even number"
  "{fg} % 2 = 0"
  
  # 22a. = npqrs "The square of a palindrome"
  "is_palindrome(nsplit(is_square({npqrs})))"
  
  # [optional]
  --template="{ABCD} {EFG} | {HIJ} {KLMN} | {P} {QR} {S} {T} | {UVW} {XYZ} | {ab} {cde} {fg} | {hi} {jk} | {m} {npqrs} {t}"
  --solution=""
  --header=""
  

  Solution: The completed grid looks like this:

  

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