Write a function pair_numbers that takes in two lists of numbers with the same length and returns a list of tuples. Example: pair_numbers([1],[1]) = [(1,1)] pair_numbers([1,2],[3,4]) = [(1,3),(1,4),(2,3),(2,4)] pair_numbers([],[]) = []

时间: 2023-12-28 13:04:51 浏览: 28
Certainly! Here is the implementation of the `pair_numbers` function: ```python def pair_numbers(list1, list2): pairs = [] for num1 in list1: for num2 in list2: pairs.append((num1, num2)) return pairs ``` The `pair_numbers` function takes in two lists of numbers, `list1` and `list2`, with the same length. It creates an empty list called `pairs` to store the resulting tuples. It then iterates over each element in `list1` and `list2` using nested loops. For each combination of elements, it creates a tuple `(num1, num2)` and appends it to the `pairs` list. Finally, it returns the `pairs` list containing all the tuples. Here are a few examples of using the `pair_numbers` function: ```python print(pair_numbers([1], [1])) # Output: [(1, 1)] print(pair_numbers([1, 2], [3, 4])) # Output: [(1, 3), (1, 4), (2, 3), (2, 4)] print(pair_numbers([], [])) # Output: [] ``` In the first example, it pairs the single elements `[1]` and `[1]`, resulting in `[(1, 1)]`. In the second example, it pairs each element from the first list with each element from the second list, resulting in `[(1, 3), (1, 4), (2, 3), (2, 4)]`. In the third example, as both lists are empty, it returns an empty list `[]`.

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Create a function pixel_flip(lst, orig_lst, budget, results, i=0) that uses recursion to generate all possible new unique images from the input orig_lst, following these rules: • The input lst is the current list being processed. Initially, this will be the same as orig_lst which is the original flattened image. • The input budget represents the number of pixels that can still be flipped. When the budget reaches 0, no more pixels can be flipped. • The input results is a list of resulting flattened images with flipped pixels. Initially, this will be an empty list. • The input i represents the index of the pixel being processed, by default set to 0, which is used to drive the recursive function towards its base case (i.e., initially starting from i=0). At termination of the function, the argument results should contain all possibilities of the input orig_lst by only flipping pixels from 0 to 1 under both the budget and the adjacency constraints. fill code at #TODO def pixel_flip(lst: list[int], orig_lst: list[int], budget: int, results: list, i: int = 0) -> None: """ Uses recursion to generate all possibilities of flipped arrays where a pixel was a 0 and there was an adjacent pixel with the value of 1. :param lst: 1D list of integers representing a flattened image . :param orig_lst: 1D list of integers representing the original flattened image. :param budget: Integer representing the number of pixels that can be flipped . :param results: List of 1D lists of integers representing all possibilities of flipped arrays, initially empty. :param i: Integer representing the index of the pixel in question. :return: None. """ #TODO def check_adjacent_for_one(flat_image: list[int], flat_pixel: int) -> bool: """ Checks if a pixel has an adjacent pixel with the value of 1. :param flat_image: 1D list of integers representing a flattened image . :param flat_pixel: Integer representing the index of the pixel in question. :return: Boolean. """ #TODO

Create a function pixel_flip(lst, orig_lst, budget, results, i=0) that uses recursion to generate all possible new unique images from the input orig_lst, following these rules: • The input lst is the current list being processed. Initially, this will be the same as orig_lst which is the original flattened image. • The input budget represents the number of pixels that can still be flipped. When the budget reaches 0, no more pixels can be flipped. • The input results is a list of resulting flattened images with flipped pixels. Initially, this will be an empty list. • The input i represents the index of the pixel being processed, by default set to 0, which is used to drive the recursive function towards its base case (i.e., initially starting from i=0). At termination of the function, the argument results should contain all possibilities of the input orig_lst by only flipping pixels from 0 to 1 under both the budget and the adjacency constraints. fill code at #TODO def pixel_flip(lst: list[int], orig_lst: list[int], budget: int, results: list, i: int = 0) -> None: """ Uses recursion to generate all possibilities of flipped arrays where a pixel was a 0 and there was an adjacent pixel with the value of 1. :param lst: 1D list of integers representing a flattened image . :param orig_lst: 1D list of integers representing the original flattened image. :param budget: Integer representing the number of pixels that can be flipped . :param results: List of 1D lists of integers representing all possibilities of flipped arrays, initially empty. :param i: Integer representing the index of the pixel in question. :return: None. """ #TODO

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