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618f9ca885
* Add Proth number to maths * Add test for 0 and more informative output * Fixing test failure issue - unused variable * Update proth_number.py Co-authored-by: Christian Clauss <cclauss@me.com>
78 lines
1.9 KiB
Python
78 lines
1.9 KiB
Python
"""
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Calculate the nth Proth number
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Source:
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https://handwiki.org/wiki/Proth_number
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"""
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import math
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def proth(number: int) -> int:
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"""
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:param number: nth number to calculate in the sequence
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:return: the nth number in Proth number
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Note: indexing starts at 1 i.e. proth(1) gives the first Proth number of 3
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>>> proth(6)
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25
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>>> proth(0)
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Traceback (most recent call last):
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...
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ValueError: Input value of [number=0] must be > 0
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>>> proth(-1)
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Traceback (most recent call last):
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...
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ValueError: Input value of [number=-1] must be > 0
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>>> proth(6.0)
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Traceback (most recent call last):
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...
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TypeError: Input value of [number=6.0] must be an integer
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"""
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if not isinstance(number, int):
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raise TypeError(f"Input value of [number={number}] must be an integer")
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if number < 1:
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raise ValueError(f"Input value of [number={number}] must be > 0")
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elif number == 1:
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return 3
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elif number == 2:
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return 5
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else:
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block_index = number // 3
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"""
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+1 for binary starting at 0 i.e. 2^0, 2^1, etc.
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+1 to start the sequence at the 3rd Proth number
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Hence, we have a +2 in the below statement
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"""
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block_index = math.log(block_index, 2) + 2
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block_index = int(block_index)
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proth_list = [3, 5]
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proth_index = 2
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increment = 3
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for block in range(1, block_index):
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for move in range(increment):
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proth_list.append(2 ** (block + 1) + proth_list[proth_index - 1])
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proth_index += 1
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increment *= 2
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return proth_list[number - 1]
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if __name__ == "__main__":
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for number in range(11):
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value = 0
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try:
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value = proth(number)
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except ValueError:
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print(f"ValueError: there is no {number}th Proth number")
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continue
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print(f"The {number}th Proth number: {value}")
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