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154 lines
4.2 KiB
Python
154 lines
4.2 KiB
Python
"""
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File: time_complexity.py
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Created Time: 2022-11-25
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Author: krahets (krahets@163.com)
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"""
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def constant(n: int) -> int:
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"""Constant order"""
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count = 0
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size = 100000
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for _ in range(size):
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count += 1
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return count
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def linear(n: int) -> int:
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"""Linear order"""
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count = 0
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for _ in range(n):
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count += 1
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return count
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def array_traversal(nums: list[int]) -> int:
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"""Linear order (traversing array)"""
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count = 0
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# Number of iterations is proportional to the array length
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for num in nums:
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count += 1
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return count
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def quadratic(n: int) -> int:
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"""Quadratic order"""
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count = 0
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# Number of iterations is quadratically related to the data size n
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for i in range(n):
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for j in range(n):
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count += 1
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return count
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def bubble_sort(nums: list[int]) -> int:
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"""Quadratic order (bubble sort)"""
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count = 0 # Counter
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# Outer loop: unsorted range is [0, i]
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for i in range(len(nums) - 1, 0, -1):
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# Inner loop: swap the largest element in the unsorted range [0, i] to the rightmost end of that range
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for j in range(i):
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if nums[j] > nums[j + 1]:
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# Swap nums[j] and nums[j + 1]
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tmp: int = nums[j]
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nums[j] = nums[j + 1]
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nums[j + 1] = tmp
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count += 3 # Element swap includes 3 unit operations
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return count
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def exponential(n: int) -> int:
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"""Exponential order (loop implementation)"""
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count = 0
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base = 1
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# Cells divide into two every round, forming sequence 1, 2, 4, 8, ..., 2^(n-1)
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for _ in range(n):
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for _ in range(base):
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count += 1
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base *= 2
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# count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
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return count
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def exp_recur(n: int) -> int:
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"""Exponential order (recursive implementation)"""
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if n == 1:
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return 1
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return exp_recur(n - 1) + exp_recur(n - 1) + 1
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def logarithmic(n: int) -> int:
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"""Logarithmic order (loop implementation)"""
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count = 0
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while n > 1:
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n = n / 2
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count += 1
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return count
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def log_recur(n: int) -> int:
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"""Logarithmic order (recursive implementation)"""
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if n <= 1:
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return 0
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return log_recur(n / 2) + 1
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def linear_log_recur(n: int) -> int:
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"""Linearithmic order"""
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if n <= 1:
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return 1
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# Divide into two, the scale of subproblems is reduced by half
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count = linear_log_recur(n // 2) + linear_log_recur(n // 2)
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# Current subproblem contains n operations
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for _ in range(n):
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count += 1
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return count
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def factorial_recur(n: int) -> int:
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"""Factorial order (recursive implementation)"""
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if n == 0:
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return 1
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count = 0
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# Split from 1 into n
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for _ in range(n):
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count += factorial_recur(n - 1)
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return count
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"""Driver Code"""
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if __name__ == "__main__":
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# You can modify n to run and observe the trend of the number of operations for various complexities
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n = 8
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print("Input data size n =", n)
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count = constant(n)
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print("Number of operations of constant order =", count)
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count = linear(n)
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print("Number of operations of linear order =", count)
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count = array_traversal([0] * n)
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print("Number of operations of linear order (traversing array) =", count)
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count = quadratic(n)
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print("Number of operations of quadratic order =", count)
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nums = [i for i in range(n, 0, -1)] # [n, n-1, ..., 2, 1]
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count = bubble_sort(nums)
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print("Number of operations of quadratic order (bubble sort) =", count)
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count = exponential(n)
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print("Number of operations of exponential order (loop implementation) =", count)
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count = exp_recur(n)
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print("Number of operations of exponential order (recursive implementation) =", count)
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count = logarithmic(n)
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print("Number of operations of logarithmic order (loop implementation) =", count)
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count = log_recur(n)
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print("Number of operations of logarithmic order (recursive implementation) =", count)
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count = linear_log_recur(n)
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print("Number of operations of linearithmic order (recursive implementation) =", count)
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count = factorial_recur(n)
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print("Number of operations of factorial order (recursive implementation) =", count)
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