Flips the given canonical coordinates, when cond_right is true. Returns coords unchanged otherwise. The returned coordinates represent those of a left hand.翻译

该函数根据条件 `cond_right` 是否为真来翻转给定的规范化坐标（canonical coordinates）。如果 `cond_right` 为真，则将坐标进行翻转，返回左手的坐标。如果 `cond_right` 为假，则直接返回给定的坐标，不进行翻转。

在计算机图形学和计算机视觉中，规范化坐标通常是指将坐标系的原点和方向规范化为特定的形式，以便进行后续的计算和处理。在三维空间中，左右手系（left-handed and right-handed coordinate systems）是两种常见的坐标系。在左手系中，当拇指指向坐标系的正方向时，其他四根手指的方向为正方向；而在右手系中，当拇指指向坐标系的正方向时，其他四根手指的方向为负方向。当处理不同坐标系之间的数据时，可能需要将数据进行转换，以便在不同坐标系之间进行比较和计算。

相关问题

Coin is heads with probability 1/3 and tails with probability 2/3. The expectation of independent flips until first heads is _____.

The expected number of flips until the first heads is given by the formula:

E(X) = 1/p

where p is the probability of getting heads on any given flip.

In this case, the probability of getting heads on any given flip is 1/3. Therefore, the expected number of flips until the first heads is:

E(X) = 1/(1/3) = 3

So the answer is 3.

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

Here's the code for the pixel_flip function and check_adjacent_for_one helper function:

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.
    """
    if budget == 0:
        # Base case: no more pixels can be flipped
        results.append(lst)
        return
    if i >= len(lst):
        # Base case: all pixels have been checked
        return
    if lst[i] == 0 and check_adjacent_for_one(orig_lst, i):
        # Flip the current pixel and recurse with a lower budget
        new_lst = lst.copy()
        new_lst[i] = 1
        pixel_flip(new_lst, orig_lst, budget - 1, results, i + 1)
    # Recurse without flipping the current pixel
    pixel_flip(lst, orig_lst, budget, results, i + 1)

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.
    """
    row_size = int(len(flat_image)**0.5)
    row = flat_pixel // row_size
    col = flat_pixel % row_size
    if row > 0 and flat_image[(row-1)*row_size+col] == 1:
        return True
    if row < row_size-1 and flat_image[(row+1)*row_size+col] == 1:
        return True
    if col > 0 and flat_image[row*row_size+(col-1)] == 1:
        return True
    if col < row_size-1 and flat_image[row*row_size+(col+1)] == 1:
        return True
    return False

The pixel_flip function takes in the current list being processed, orig_lst which is the original flattened image, budget representing the number of pixels that can still be flipped, results which is initially an empty list of resulting flattened images with flipped pixels, and i representing the index of the pixel being processed.

The function 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. It first checks if the budget is 0 or if all pixels have been checked, and returns accordingly. If the current pixel is 0 and has an adjacent pixel with the value of 1, it flips the current pixel and recurses with a lower budget. Otherwise, it recurses without flipping the current pixel.

The check_adjacent_for_one helper function takes in the flattened image and the index of the pixel in question, and checks if the pixel has an adjacent pixel with the value of 1. It calculates the row and column of the pixel using integer division and modulus, and checks if the adjacent pixels in the vertical and horizontal directions have the value of 1. If any of the adjacent pixels have the value of 1, it returns True, otherwise it returns False.