Step by Step solution about How to resolve the algorithm Executable library step by step in the Python programming language

The provided Python code defines several functions for exploring the "Hailstone sequence" for various input numbers. Here's a breakdown of the code:

1. hailstone(n) Function:

   • This function calculates the Hailstone sequence for a given input number n.
   • It starts with the input number and applies the following rules iteratively:
     • If n is odd, it multiplies by 3 and adds 1 (n = 3 * n + 1).
     • If n is even, it divides by 2 (n = n // 2).
   • The function continues this process until n reaches 1.
   • It returns a list containing the sequence of numbers generated during this process.

2. Main Execution Block (if __name__ == '__main__'):

   • It tests the hailstone(n) function for n = 27.
   • The expected result is a list of 112 numbers, with the first four elements being [27, 82, 41, 124] and the last four elements being [8, 4, 2, 1].
   • It then prints the maximum length of the Hailstone sequence found for any number between 1 and 99,999.

3. function_length_frequency(func, hrange) Function:

   • This function calculates the frequency distribution of the lengths of the Hailstone sequences for a given function func applied to a range of numbers hrange.
   • It uses the Counter class from the collections module to count the occurrences of each sequence length.
   • It returns the most common sequence length and its frequency.

4. Second Main Execution Block (if __name__ == '__main__' inside a module):

   • This block uses an imported module named executable_hailstone_library that contains an optimized implementation of the Hailstone function.
   • It calculates the most common length and frequency of the Hailstone sequences for numbers between 1 and 100,000 using the function_length_frequency function.
   • It prints the results, indicating the most common length and frequency of the Hailstone sequences in that range.

Overall, this code explores the Hailstone sequence for different input numbers and provides insights into the length distribution of these sequences.

def hailstone(n):
    seq = [n]
    while n>1:
        n = 3*n + 1 if n & 1 else n//2
        seq.append(n)
    return seq

if __name__ == '__main__':
    h = hailstone(27)
    assert len(h)==112 and h[:4]==[27, 82, 41, 124] and h[-4:]==[8, 4, 2, 1]
    print("Maximum length %i was found for hailstone(%i) for numbers <100,000" %
          max((len(hailstone(i)), i) for i in range(1,100000)))


from collections import Counter

def function_length_frequency(func, hrange):
    return Counter(len(func(n)) for n in hrange).most_common()

if __name__ == '__main__':
    from executable_hailstone_library import hailstone

    upto = 100000
    hlen, freq = function_length_frequency(hailstone, range(1, upto))[0]
    print("The length of hailstone sequence that is most common for\n"
          "hailstone(n) where 1<=n<%i, is %i. It occurs %i times."
          % (upto, hlen, freq))