Instead of representing states explicitly, we may represent a world using variables: A state is described by a set of variables (i.e., features, attributes). We use capital letters, e.g., X, to represent variables. Each variable X can take a value from its domain D. A factored representation consists of a set of variables X1, X2, . . . , Xk where each variable Xi is associated with a domain Di A candidate solution (or assignment) of a set of variables is a function that maps each variable in the set with a value in its domain. Let Ω be the set of all candidate solutions.理解分析这些内容

这段话讨论了一种描述状态的方式，即使用变量来表示一个世界。每个状态由一组变量描述，每个变量都可以从其域中取值。这个表示方法称为分解表示，由一组变量X1，X2，...，Xk组成，其中每个变量Xi都与域Di相关联。候选解决方案是将变量集中每个变量映射到其域中的值的函数。所有候选解决方案的集合称为Ω。这种表示方法可以用于描述各种问题，如搜索、规划等。

相关问题

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.

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

Here is the code implementation for the function pixel_flip():

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.
    """
    # Base case: when we reach the end of the list or budget is zero
    if i == len(lst) or budget == 0:
        if lst not in results:
            results.append(lst)
        return

    # If the current pixel is 0 and there is an adjacent 1
    if lst[i] == 0 and check_adjacent(lst, i):
        # Flip the current pixel to 1
        new_lst = lst.copy()
        new_lst[i] = 1

        # Recursively call the function with the new list and reduced budget
        pixel_flip(new_lst, orig_lst, budget - 1, results, i + 1)

    # Call the function again with the original list and without flipping the current pixel
    pixel_flip(lst, orig_lst, budget, results, i + 1)


def check_adjacent(lst: list[int], i: int) -> bool:
    """
    Checks if a pixel has an adjacent 1 in the original flattened image.

    :param lst: 1D list of integers representing a flattened image.
    :param i: Integer representing the index of the pixel in question.
    :return: True if the pixel has an adjacent 1, False otherwise.
    """
    n_rows = int(len(lst) ** 0.5)

    # Check left
    if i % n_rows != 0 and lst[i - 1] == 1:
        return True

    # Check right
    if (i + 1) % n_rows != 0 and lst[i + 1] == 1:
        return True

    # Check up
    if i >= n_rows and lst[i - n_rows] == 1:
        return True

    # Check down
    if i < len(lst) - n_rows and lst[i + n_rows] == 1:
        return True

    return False

The function `pixel_flip` takes in the list `lst`, representing the current list being processed, the original flattened image `orig_lst`, the budget of pixels that can still be flipped, the list `results` to store all possible flipped images, and the index `i` of the pixel being processed.

In each recursive call, the function first checks if the base case has been reached, which is when we have processed all pixels or the budget is zero. If so, the current `lst` is appended to `results` if it is not already in there.

If the base case has not been reached, the function then checks if the current pixel is 0 and there is an adjacent 1 in the original flattened image. If so, the function creates a new list where the current pixel is flipped to 1, and calls itself recursively with this new list and reduced budget.

Finally, the function calls itself recursively again with the original `lst` and without flipping the current pixel.

The helper function `check_adjacent` checks if a pixel has an adjacent 1 in the original flattened image.

Note that this implementation assumes that the input `lst` and `orig_lst` are both 1D lists representing a flattened image with equal length and square shape. Additionally, the input `results` should be initially an empty list.