Repeated square root operations in a given range of numbers












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I was curious to find the solution for the question



'Given a range of numbers, return the maximum no of repeated squares root operations within that range'.



For instance, given the range of numbers are (2,20), the perfect squares are [4,9,16] within that range. Now the maximum no of square root operations is 2. Reason behind



 sqrt(16) -> sqrt(4) -> 2. 


I have come with a solution, however, it takes a lot of time to return the results. If you can provide a better version of this code, please let me know.



import math
# For each perfect square, calculate the depth of square root operations
# applied against it.
def square(n):
if n == 2 or n == 1:
return 0
sol = (int(n**0.5))**2
if sol != n : return 0
return 1 + square(int(n**0.5))


A = int(input())
B = int(input())

# Returns the perfect squares within that range
squares = [x for x in range(A,B) if (int(x**0.5))**2 == x]

print(max([square(i) for i in squares]))









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$endgroup$

















    0












    $begingroup$


    I was curious to find the solution for the question



    'Given a range of numbers, return the maximum no of repeated squares root operations within that range'.



    For instance, given the range of numbers are (2,20), the perfect squares are [4,9,16] within that range. Now the maximum no of square root operations is 2. Reason behind



     sqrt(16) -> sqrt(4) -> 2. 


    I have come with a solution, however, it takes a lot of time to return the results. If you can provide a better version of this code, please let me know.



    import math
    # For each perfect square, calculate the depth of square root operations
    # applied against it.
    def square(n):
    if n == 2 or n == 1:
    return 0
    sol = (int(n**0.5))**2
    if sol != n : return 0
    return 1 + square(int(n**0.5))


    A = int(input())
    B = int(input())

    # Returns the perfect squares within that range
    squares = [x for x in range(A,B) if (int(x**0.5))**2 == x]

    print(max([square(i) for i in squares]))









    share|improve this question







    New contributor




    s326280 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      0












      0








      0





      $begingroup$


      I was curious to find the solution for the question



      'Given a range of numbers, return the maximum no of repeated squares root operations within that range'.



      For instance, given the range of numbers are (2,20), the perfect squares are [4,9,16] within that range. Now the maximum no of square root operations is 2. Reason behind



       sqrt(16) -> sqrt(4) -> 2. 


      I have come with a solution, however, it takes a lot of time to return the results. If you can provide a better version of this code, please let me know.



      import math
      # For each perfect square, calculate the depth of square root operations
      # applied against it.
      def square(n):
      if n == 2 or n == 1:
      return 0
      sol = (int(n**0.5))**2
      if sol != n : return 0
      return 1 + square(int(n**0.5))


      A = int(input())
      B = int(input())

      # Returns the perfect squares within that range
      squares = [x for x in range(A,B) if (int(x**0.5))**2 == x]

      print(max([square(i) for i in squares]))









      share|improve this question







      New contributor




      s326280 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I was curious to find the solution for the question



      'Given a range of numbers, return the maximum no of repeated squares root operations within that range'.



      For instance, given the range of numbers are (2,20), the perfect squares are [4,9,16] within that range. Now the maximum no of square root operations is 2. Reason behind



       sqrt(16) -> sqrt(4) -> 2. 


      I have come with a solution, however, it takes a lot of time to return the results. If you can provide a better version of this code, please let me know.



      import math
      # For each perfect square, calculate the depth of square root operations
      # applied against it.
      def square(n):
      if n == 2 or n == 1:
      return 0
      sol = (int(n**0.5))**2
      if sol != n : return 0
      return 1 + square(int(n**0.5))


      A = int(input())
      B = int(input())

      # Returns the perfect squares within that range
      squares = [x for x in range(A,B) if (int(x**0.5))**2 == x]

      print(max([square(i) for i in squares]))






      python performance algorithm complexity






      share|improve this question







      New contributor




      s326280 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question







      New contributor




      s326280 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question






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      s326280 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









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      s326280 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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