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斐波那契数列

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"""
Calculates the Fibonacci sequence using iteration, recursion, memoization,
and a simplified form of Binet's formula

NOTE 1: the iterative, recursive, memoization functions are more accurate than
the Binet's formula function because the Binet formula function  uses floats

NOTE 2: the Binet's formula function is much more limited in the size of inputs
that it can handle due to the size limitations of Python floats

See benchmark numbers in __main__ for performance comparisons/
https://en.wikipedia.org/wiki/Fibonacci_number for more information
"""

import functools
from collections.abc import Iterator
from math import sqrt
from time import time


def time_func(func, *args, **kwargs):
    """
    Times the execution of a function with parameters
    """
    start = time()
    output = func(*args, **kwargs)
    end = time()
    if int(end - start) > 0:
        print(f"{func.__name__} runtime: {(end - start):0.4f} s")
    else:
        print(f"{func.__name__} runtime: {(end - start) * 1000:0.4f} ms")
    return output


def fib_iterative_yield(n: int) -> Iterator[int]:
    """
    Calculates the first n (1-indexed) Fibonacci numbers using iteration with yield
    >>> list(fib_iterative_yield(0))
    [0]
    >>> tuple(fib_iterative_yield(1))
    (0, 1)
    >>> tuple(fib_iterative_yield(5))
    (0, 1, 1, 2, 3, 5)
    >>> tuple(fib_iterative_yield(10))
    (0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55)
    >>> tuple(fib_iterative_yield(-1))
    Traceback (most recent call last):
        ...
    ValueError: n is negative
    """
    if n < 0:
        raise ValueError("n is negative")
    a, b = 0, 1
    yield a
    for _ in range(n):
        yield b
        a, b = b, a + b


def fib_iterative(n: int) -> list[int]:
    """
    Calculates the first n (0-indexed) Fibonacci numbers using iteration
    >>> fib_iterative(0)
    [0]
    >>> fib_iterative(1)
    [0, 1]
    >>> fib_iterative(5)
    [0, 1, 1, 2, 3, 5]
    >>> fib_iterative(10)
    [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55]
    >>> fib_iterative(-1)
    Traceback (most recent call last):
        ...
    ValueError: n is negative
    """
    if n < 0:
        raise ValueError("n is negative")
    if n == 0:
        return [0]
    fib = [0, 1]
    for _ in range(n - 1):
        fib.append(fib[-1] + fib[-2])
    return fib


def fib_recursive(n: int) -> list[int]:
    """
    Calculates the first n (0-indexed) Fibonacci numbers using recursion
    >>> fib_iterative(0)
    [0]
    >>> fib_iterative(1)
    [0, 1]
    >>> fib_iterative(5)
    [0, 1, 1, 2, 3, 5]
    >>> fib_iterative(10)
    [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55]
    >>> fib_iterative(-1)
    Traceback (most recent call last):
        ...
    ValueError: n is negative
    """

    def fib_recursive_term(i: int) -> int:
        """
        Calculates the i-th (0-indexed) Fibonacci number using recursion
        >>> fib_recursive_term(0)
        0
        >>> fib_recursive_term(1)
        1
        >>> fib_recursive_term(5)
        5
        >>> fib_recursive_term(10)
        55
        >>> fib_recursive_term(-1)
        Traceback (most recent call last):
            ...
        Exception: n is negative
        """
        if i < 0:
            raise ValueError("n is negative")
        if i < 2:
            return i
        return fib_recursive_term(i - 1) + fib_recursive_term(i - 2)

    if n < 0:
        raise ValueError("n is negative")
    return [fib_recursive_term(i) for i in range(n + 1)]


def fib_recursive_cached(n: int) -> list[int]:
    """
    Calculates the first n (0-indexed) Fibonacci numbers using recursion
    >>> fib_iterative(0)
    [0]
    >>> fib_iterative(1)
    [0, 1]
    >>> fib_iterative(5)
    [0, 1, 1, 2, 3, 5]
    >>> fib_iterative(10)
    [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55]
    >>> fib_iterative(-1)
    Traceback (most recent call last):
        ...
    ValueError: n is negative
    """

    @functools.cache
    def fib_recursive_term(i: int) -> int:
        """
        Calculates the i-th (0-indexed) Fibonacci number using recursion
        """
        if i < 0:
            raise ValueError("n is negative")
        if i < 2:
            return i
        return fib_recursive_term(i - 1) + fib_recursive_term(i - 2)

    if n < 0:
        raise ValueError("n is negative")
    return [fib_recursive_term(i) for i in range(n + 1)]


def fib_memoization(n: int) -> list[int]:
    """
    Calculates the first n (0-indexed) Fibonacci numbers using memoization
    >>> fib_memoization(0)
    [0]
    >>> fib_memoization(1)
    [0, 1]
    >>> fib_memoization(5)
    [0, 1, 1, 2, 3, 5]
    >>> fib_memoization(10)
    [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55]
    >>> fib_iterative(-1)
    Traceback (most recent call last):
        ...
    ValueError: n is negative
    """
    if n < 0:
        raise ValueError("n is negative")
    # Cache must be outside recursuive function
    # other it will reset every time it calls itself.
    cache: dict[int, int] = {0: 0, 1: 1, 2: 1}  # Prefilled cache

    def rec_fn_memoized(num: int) -> int:
        if num in cache:
            return cache[num]

        value = rec_fn_memoized(num - 1) + rec_fn_memoized(num - 2)
        cache[num] = value
        return value

    return [rec_fn_memoized(i) for i in range(n + 1)]


def fib_binet(n: int) -> list[int]:
    """
    Calculates the first n (0-indexed) Fibonacci numbers using a simplified form
    of Binet's formula:
    https://en.m.wikipedia.org/wiki/Fibonacci_number#Computation_by_rounding

    NOTE 1: this function diverges from fib_iterative at around n = 71, likely
    due to compounding floating-point arithmetic errors

    NOTE 2: this function doesn't accept n >= 1475 because it overflows
    thereafter due to the size limitations of Python floats
    >>> fib_binet(0)
    [0]
    >>> fib_binet(1)
    [0, 1]
    >>> fib_binet(5)
    [0, 1, 1, 2, 3, 5]
    >>> fib_binet(10)
    [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55]
    >>> fib_binet(-1)
    Traceback (most recent call last):
        ...
    ValueError: n is negative
    >>> fib_binet(1475)
    Traceback (most recent call last):
        ...
    ValueError: n is too large
    """
    if n < 0:
        raise ValueError("n is negative")
    if n >= 1475:
        raise ValueError("n is too large")
    sqrt_5 = sqrt(5)
    phi = (1 + sqrt_5) / 2
    return [round(phi**i / sqrt_5) for i in range(n + 1)]


if __name__ == "__main__":
    from doctest import testmod

    testmod()
    # Time on an M1 MacBook Pro -- Fastest to slowest
    num = 30
    time_func(fib_iterative_yield, num)  # 0.0012 ms
    time_func(fib_iterative, num)  # 0.0031 ms
    time_func(fib_binet, num)  # 0.0062 ms
    time_func(fib_memoization, num)  # 0.0100 ms
    time_func(fib_recursive_cached, num)  # 0.0153 ms
    time_func(fib_recursive, num)  # 257.0910 ms
关于此算法

在数学中,斐波那契数列(通常用 F(n) 表示)形成一个序列,称为斐波那契序列,其中每个数字都是前两个数字的和,从 0 和 1 开始。序列如下所示

[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...]

应用

查找此序列的第 N 个成员在许多应用中都很有用

  • 最近,斐波那契数列和黄金分割率在许多科学领域的研究中引起了极大的兴趣,

包括高能物理学、量子力学、密码学和编码。

步骤

  1. 准备基础矩阵
  2. 计算此矩阵的幂
  3. 从矩阵中获取相应的值

示例

查找斐波那契数列的第 8 个成员

步骤 0

| F(n+1)  F(n)  |
| F(n)    F(n-1)|

步骤 1

Calculate matrix^1
| 1 1 |
| 1 0 |

步骤 2

Calculate matrix^2
| 2 1 |
| 1 1 |

步骤 3

Calculate matrix^4
| 5 3 |
| 3 2 |

步骤 4

Calculate matrix^8
| 34 21 |
| 21 13 |

步骤 5

F(8)=21

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