采用广度优先搜索(BFS)、深度优先搜索(DFS)和启发式搜索算法(A/A*算法) 编程求解八数码问题（初始状态允许任意），找出一种从初始状态到目标状态 移动步数最少的步骤。

下面是Python代码实现：

from queue import Queue, PriorityQueue

# 目标状态
target = [[1, 2, 3], [4, 5, 6], [7, 8, 0]]
# 移动方向
dx = [0, 1, 0, -1]
dy = [1, 0, -1, 0]


def bfs(start):
    # 初始化
    q = Queue()
    q.put((start, '', 0))
    visited = set()

    while not q.empty():
        node, path, steps = q.get()

        if node == target:
            return path

        # 找到0在九宫格中的位置
        x, y = 0, 0
        for i in range(3):
            for j in range(3):
                if node[i][j] == 0:
                    x, y = i, j

        # 枚举所有可能的移动方向
        for k in range(4):
            nx, ny = x + dx[k], y + dy[k]
            if nx < 0 or nx >= 3 or ny < 0 or ny >= 3:
                continue
            new_node = [[node[i][j] for j in range(3)] for i in range(3)]
            new_node[x][y], new_node[nx][ny] = new_node[nx][ny], new_node[x][y]
            # 判断是否已经访问过
            state = tuple([tuple(row) for row in new_node])
            if state in visited:
                continue
            visited.add(state)
            q.put((new_node, path + str(k), steps + 1))

    return ''


def dfs(node, path, steps, max_depth, visited):
    if node == target:
        return path

    if steps > max_depth:
        return ''

    # 找到0在九宫格中的位置
    x, y = 0, 0
    for i in range(3):
        for j in range(3):
            if node[i][j] == 0:
                x, y = i, j

    # 枚举所有可能的移动方向
    for k in range(4):
        nx, ny = x + dx[k], y + dy[k]
        if nx < 0 or nx >= 3 or ny < 0 or ny >= 3:
            continue
        new_node = [[node[i][j] for j in range(3)] for i in range(3)]
        new_node[x][y], new_node[nx][ny] = new_node[nx][ny], new_node[x][y]
        # 判断是否已经访问过
        state = tuple([tuple(row) for row in new_node])
        if state in visited:
            continue
        visited.add(state)
        res = dfs(new_node, path + str(k), steps + 1, max_depth, visited)
        if res:
            return res

    return ''


def iddfs(start):
    visited = set()
    # 最大搜索深度
    depth = 0
    while True:
        res = dfs(start, '', 0, depth, visited)
        if res:
            return res
        depth += 1


def heuristic(node):
    # 最小曼哈顿距离
    h = 0
    for i in range(3):
        for j in range(3):
            if node[i][j] == 0:
                continue
            row, col = (node[i][j] - 1) // 3, (node[i][j] - 1) % 3
            h += abs(row - i) + abs(col - j)
    return h


def a_star(start):
    # 初始化
    q = PriorityQueue()
    q.put((heuristic(start), start, '', 0))
    visited = set()

    while not q.empty():
        _, node, path, steps = q.get()

        if node == target:
            return path

        # 找到0在九宫格中的位置
        x, y = 0, 0
        for i in range(3):
            for j in range(3):
                if node[i][j] == 0:
                    x, y = i, j

        # 枚举所有可能的移动方向
        for k in range(4):
            nx, ny = x + dx[k], y + dy[k]
            if nx < 0 or nx >= 3 or ny < 0 or ny >= 3:
                continue
            new_node = [[node[i][j] for j in range(3)] for i in range(3)]
            new_node[x][y], new_node[nx][ny] = new_node[nx][ny], new_node[x][y]
            # 判断是否已经访问过
            state = tuple([tuple(row) for row in new_node])
            if state in visited:
                continue
            visited.add(state)
            q.put((heuristic(new_node) + steps + 1, new_node, path + str(k), steps + 1))

    return ''


if __name__ == '__main__':
    # 输入初始状态
    start = []
    for i in range(3):
        line = input().strip().split()
        start.append([int(x) for x in line])
    # 广度优先搜索
    res_bfs = bfs(start)
    print('BFS:', res_bfs)
    # 深度优先搜索
    res_dfs = iddfs(start)
    print('DFS:', res_dfs)
    # 启发式搜索
    res_a_star = a_star(start)
    print('A*:', res_a_star)

在输入初始状态后，分别采用BFS、DFS和A*算法进行求解，输出最短的移动步骤。