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Consolidate binary exponentiation files (#10742)
* Consolidate binary exponentiation files
* updating DIRECTORY.md
* Fix typos in doctests
* Add suggestions from code review
* Fix timeit benchmarks
---------
Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
This commit is contained in:
Tianyi Zheng
GitHub
parent
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commit
06edc0eea0
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@ -578,9 +578,7 @@
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* [Bailey Borwein Plouffe](maths/bailey_borwein_plouffe.py)
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* [Base Neg2 Conversion](maths/base_neg2_conversion.py)
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* [Basic Maths](maths/basic_maths.py)
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* [Binary Exp Mod](maths/binary_exp_mod.py)
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* [Binary Exponentiation](maths/binary_exponentiation.py)
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* [Binary Exponentiation 2](maths/binary_exponentiation_2.py)
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* [Binary Multiplication](maths/binary_multiplication.py)
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* [Binomial Coefficient](maths/binomial_coefficient.py)
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* [Binomial Distribution](maths/binomial_distribution.py)
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@ -1,28 +0,0 @@
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def bin_exp_mod(a: int, n: int, b: int) -> int:
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"""
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>>> bin_exp_mod(3, 4, 5)
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1
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>>> bin_exp_mod(7, 13, 10)
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7
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"""
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# mod b
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assert b != 0, "This cannot accept modulo that is == 0"
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if n == 0:
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return 1
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if n % 2 == 1:
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return (bin_exp_mod(a, n - 1, b) * a) % b
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r = bin_exp_mod(a, n // 2, b)
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return (r * r) % b
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if __name__ == "__main__":
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try:
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BASE = int(input("Enter Base : ").strip())
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POWER = int(input("Enter Power : ").strip())
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MODULO = int(input("Enter Modulo : ").strip())
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except ValueError:
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print("Invalid literal for integer")
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print(bin_exp_mod(BASE, POWER, MODULO))
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@ -1,48 +1,196 @@
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"""Binary Exponentiation."""
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# Author : Junth Basnet
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# Time Complexity : O(logn)
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def binary_exponentiation(a: int, n: int) -> int:
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"""
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Compute a number raised by some quantity
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>>> binary_exponentiation(-1, 3)
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-1
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>>> binary_exponentiation(-1, 4)
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1
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>>> binary_exponentiation(2, 2)
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4
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>>> binary_exponentiation(3, 5)
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Binary Exponentiation
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This is a method to find a^b in O(log b) time complexity and is one of the most commonly
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used methods of exponentiation. The method is also useful for modular exponentiation,
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when the solution to (a^b) % c is required.
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To calculate a^b:
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- If b is even, then a^b = (a * a)^(b / 2)
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- If b is odd, then a^b = a * a^(b - 1)
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Repeat until b = 1 or b = 0
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For modular exponentiation, we use the fact that (a * b) % c = ((a % c) * (b % c)) % c
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"""
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def binary_exp_recursive(base: float, exponent: int) -> float:
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"""
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Computes a^b recursively, where a is the base and b is the exponent
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>>> binary_exp_recursive(3, 5)
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243
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>>> binary_exponentiation(10, 3)
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1000
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>>> binary_exponentiation(5e3, 1)
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5000.0
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>>> binary_exponentiation(-5e3, 1)
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-5000.0
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>>> binary_exp_recursive(11, 13)
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34522712143931
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>>> binary_exp_recursive(-1, 3)
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-1
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>>> binary_exp_recursive(0, 5)
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0
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>>> binary_exp_recursive(3, 1)
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3
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>>> binary_exp_recursive(3, 0)
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1
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>>> binary_exp_recursive(1.5, 4)
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5.0625
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>>> binary_exp_recursive(3, -1)
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Traceback (most recent call last):
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...
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ValueError: Exponent must be a non-negative integer
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"""
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if n == 0:
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if exponent < 0:
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raise ValueError("Exponent must be a non-negative integer")
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if exponent == 0:
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return 1
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elif n % 2 == 1:
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return binary_exponentiation(a, n - 1) * a
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if exponent % 2 == 1:
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return binary_exp_recursive(base, exponent - 1) * base
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else:
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b = binary_exponentiation(a, n // 2)
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b = binary_exp_recursive(base, exponent // 2)
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return b * b
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def binary_exp_iterative(base: float, exponent: int) -> float:
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"""
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Computes a^b iteratively, where a is the base and b is the exponent
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>>> binary_exp_iterative(3, 5)
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243
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>>> binary_exp_iterative(11, 13)
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34522712143931
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>>> binary_exp_iterative(-1, 3)
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-1
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>>> binary_exp_iterative(0, 5)
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0
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>>> binary_exp_iterative(3, 1)
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3
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>>> binary_exp_iterative(3, 0)
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1
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>>> binary_exp_iterative(1.5, 4)
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5.0625
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>>> binary_exp_iterative(3, -1)
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Traceback (most recent call last):
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...
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ValueError: Exponent must be a non-negative integer
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"""
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if exponent < 0:
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raise ValueError("Exponent must be a non-negative integer")
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res: int | float = 1
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while exponent > 0:
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if exponent & 1:
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res *= base
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base *= base
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exponent >>= 1
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return res
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def binary_exp_mod_recursive(base: float, exponent: int, modulus: int) -> float:
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"""
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Computes a^b % c recursively, where a is the base, b is the exponent, and c is the
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modulus
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>>> binary_exp_mod_recursive(3, 4, 5)
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1
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>>> binary_exp_mod_recursive(11, 13, 7)
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4
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>>> binary_exp_mod_recursive(1.5, 4, 3)
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2.0625
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>>> binary_exp_mod_recursive(7, -1, 10)
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Traceback (most recent call last):
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...
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ValueError: Exponent must be a non-negative integer
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>>> binary_exp_mod_recursive(7, 13, 0)
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Traceback (most recent call last):
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...
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ValueError: Modulus must be a positive integer
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"""
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if exponent < 0:
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raise ValueError("Exponent must be a non-negative integer")
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if modulus <= 0:
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raise ValueError("Modulus must be a positive integer")
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if exponent == 0:
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return 1
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if exponent % 2 == 1:
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return (binary_exp_mod_recursive(base, exponent - 1, modulus) * base) % modulus
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r = binary_exp_mod_recursive(base, exponent // 2, modulus)
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return (r * r) % modulus
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def binary_exp_mod_iterative(base: float, exponent: int, modulus: int) -> float:
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"""
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Computes a^b % c iteratively, where a is the base, b is the exponent, and c is the
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modulus
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>>> binary_exp_mod_iterative(3, 4, 5)
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1
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>>> binary_exp_mod_iterative(11, 13, 7)
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4
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>>> binary_exp_mod_iterative(1.5, 4, 3)
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2.0625
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>>> binary_exp_mod_iterative(7, -1, 10)
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Traceback (most recent call last):
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...
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ValueError: Exponent must be a non-negative integer
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>>> binary_exp_mod_iterative(7, 13, 0)
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Traceback (most recent call last):
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...
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ValueError: Modulus must be a positive integer
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"""
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if exponent < 0:
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raise ValueError("Exponent must be a non-negative integer")
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if modulus <= 0:
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raise ValueError("Modulus must be a positive integer")
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res: int | float = 1
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while exponent > 0:
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if exponent & 1:
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res = ((res % modulus) * (base % modulus)) % modulus
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base *= base
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exponent >>= 1
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return res
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if __name__ == "__main__":
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import doctest
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from timeit import timeit
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doctest.testmod()
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a = 1269380576
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b = 374
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c = 34
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try:
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BASE = int(float(input("Enter Base : ").strip()))
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POWER = int(input("Enter Power : ").strip())
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except ValueError:
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print("Invalid literal for integer")
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RESULT = binary_exponentiation(BASE, POWER)
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print(f"{BASE}^({POWER}) : {RESULT}")
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runs = 100_000
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print(
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timeit(
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f"binary_exp_recursive({a}, {b})",
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setup="from __main__ import binary_exp_recursive",
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number=runs,
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)
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)
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print(
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timeit(
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f"binary_exp_iterative({a}, {b})",
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setup="from __main__ import binary_exp_iterative",
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number=runs,
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)
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)
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print(
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timeit(
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f"binary_exp_mod_recursive({a}, {b}, {c})",
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setup="from __main__ import binary_exp_mod_recursive",
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number=runs,
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)
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)
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print(
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timeit(
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f"binary_exp_mod_iterative({a}, {b}, {c})",
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setup="from __main__ import binary_exp_mod_iterative",
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number=runs,
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)
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)
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@ -1,61 +0,0 @@
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"""
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Binary Exponentiation
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This is a method to find a^b in O(log b) time complexity
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This is one of the most commonly used methods of exponentiation
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It's also useful when the solution to (a^b) % c is required because a, b, c may be
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over the computer's calculation limits
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Let's say you need to calculate a ^ b
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- RULE 1 : a ^ b = (a*a) ^ (b/2) ---- example : 4 ^ 4 = (4*4) ^ (4/2) = 16 ^ 2
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- RULE 2 : IF b is odd, then a ^ b = a * (a ^ (b - 1)), where b - 1 is even
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Once b is even, repeat the process until b = 1 or b = 0, because a^1 = a and a^0 = 1
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For modular exponentiation, we use the fact that (a*b) % c = ((a%c) * (b%c)) % c
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Now apply RULE 1 or 2 as required
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@author chinmoy159
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"""
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def b_expo(a: int, b: int) -> int:
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"""
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>>> b_expo(2, 10)
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1024
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>>> b_expo(9, 0)
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1
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>>> b_expo(0, 12)
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0
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>>> b_expo(4, 12)
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16777216
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"""
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res = 1
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while b > 0:
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if b & 1:
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res *= a
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a *= a
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b >>= 1
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return res
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def b_expo_mod(a: int, b: int, c: int) -> int:
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"""
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>>> b_expo_mod(2, 10, 1000000007)
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1024
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>>> b_expo_mod(11, 13, 19)
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11
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>>> b_expo_mod(0, 19, 20)
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0
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>>> b_expo_mod(15, 5, 4)
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3
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"""
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res = 1
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while b > 0:
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if b & 1:
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res = ((res % c) * (a % c)) % c
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a *= a
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b >>= 1
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return res
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