八数码问题是一个经典的搜索问题，即在九宫格中寻找一种方法以使下面的状态变为上面的状态。我们可以使用 Python 编程求解这个问题。

首先，我们需要定义初始状态和目标状态。这两个状态通常以列表的形式表示，其中数字 0 表示空格。

initial_state = [1, 2, 3, 4, 5, 6, 7, 8, 0]
target_state = [0, 1, 2, 3, 4, 5, 6, 7, 8]

接下来，我们需要定义状态转移函数，即如何从一个状态转移到另一个状态。在八数码问题中，我们可以将空格上下左右移动，从而得到新的状态。

def move_up(state):
    new_state = state.copy()
    position = new_state.index(0)
    if position > 2:
        new_state[position - 3], new_state[position] = new_state[position], new_state[position - 3]
        return new_state
    else:
        return None

def move_down(state):
    new_state = state.copy()
    position = new_state.index(0)
    if position < 6:
        new_state[position + 3], new_state[position] = new_state[position], new_state[position + 3]
        return new_state
    else:
        return None

def move_left(state):
    new_state = state.copy()
    position = new_state.index(0)
    if position % 3 != 0:
        new_state[position - 1], new_state[position] = new_state[position], new_state[position - 1]
        return new_state
    else:
        return None

def move_right(state):
    new_state = state.copy()
    position = new_state.index(0)
    if position % 3 != 2:
        new_state[position + 1], new_state[position] = new_state[position], new_state[position + 1]
        return new_state
    else:
        return None

然后，我们需要使用广度优先搜索算法来搜索解。在搜索过程中，我们需要将初始状态加入队列，并逐个将状态转移后的新状态加入队列。当找到目标状态时，搜索过程结束，输出解路径。

from queue import Queue

def bfs(initial_state, target_state):
    q = Queue()
    q.put(initial_state)
    visited = set()
    visited.add(tuple(initial_state))

    while not q.empty():
        state = q.get()
        if state == target_state:
            return state
        for move in [move_up, move_down, move_left, move_right]:
            new_state = move(state)
            if new_state is not None and tuple(new_state) not in visited:
                q.put(new_state)
                visited.add(tuple(new_state))

    return None

solution = bfs(initial_state, target_state)
print(solution)

最后运行代码，即可求解八数码问题。