如何使用 python 函数从列表中生成多个列表

Question

I am trying to make a 8 puzzle problem solver using different algorithms, such as BFS,DFS, A* etc. using python. For those who are not familiar with the problem, 8 puzzle problem is a game consisting of 3 rows and 3 columns. You can move the empty tile only horizontally or vertically, 0 represents the empty tile. It looks like this (I couldn't add the images due to my accounts reputation.):

https://miro.medium.com/max/679/1*yekmcvT48y6mB8dIcK967Q.png

initial_state = [0,1,3,4,2,5,7,8,6]
goal_state = [1,2,3,4,5,6,7,8,0]
    
def find_zero(state):
       global loc_of_zero
       loc_of_zero = (state.index(0))


def swap_positions(list, pos1, pos2):
       first = list.pop(pos1)
       second = list.pop(pos2-1)

       list.insert(pos1,second)
       list.insert(pos2,first)
       return list

 def find_new_nodes(state):
      if loc_of_zero == 0:
         right = swap_positions(initial_state,0,1)
         left = swap_positions(initial_state,0,3)
         return(right,left)




find_zero(initial_state)
print(find_new_nodes(initial_state))   

The problem I have is this, I want the function "find_new_nodes(state)" return 2 different lists, so I can choose the most promising node, depending on the algorithm) and so on. But the output of my code consists of two identical lists.

This is my output: ([4, 0, 3, 1, 2, 5, 7, 8, 6], [4, 0, 3, 1, 2, 5, 7, 8, 6])

What can I do to make it return 2 different lists? My goal is to return all possible moves depending on where the 0 is, using the find_new_nodes function. Apologies if this is an easy question, This is my first time making a project this complicated.

Answer 1

The problem is that swap_positions obtains a reference to the global initial_state and not a clone of it. So both calls to swap_positions mutate the same array. A solution would be to clone the array on the first call: right = swap_positions(initial_state[:],0,1)

probably a better solution for swap_positions would also be:

# please do not name variables same as builtin names
def swap_positions(lis, pos1, pos2):
       # create a new tuple of both elements and destruct it directly
       lis[pos1], lis[pos2] = lis[pos2], lis[pos1]
       return lis

see also here

Answer 2

You don't really have "two identical list", you only have one list object that you're returning twice. To avoid modifying the original list and also two work with different lists, you should pass copies around.

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

def find_zero(state):
    global loc_of_zero
    loc_of_zero = (state.index(0))


def swap_positions(states, pos1, pos2):
    first = states.pop(pos1)
    second = states.pop(pos2-1)

    states.insert(pos1,second)
    states.insert(pos2,first)
    return states

def find_new_nodes(states):
    if loc_of_zero == 0:
        right = swap_positions(states.copy(),0,1) # pass around a copy
        left = swap_positions(states.copy(),0,3) # pass around a copy
        return(right,left)

find_zero(initial_state)
print(find_new_nodes(initial_state))

Side note 1: I have renamed your vairable list to states , otherwise it would shadow the built in list function

Side note 2: find_new_nodes did not work with the parameter, instead it used the global list. I changed that, too.

Side note 3: There are different ways to create a copy of your (shallow) list. I think list.copy() is the most verbose one. You could also use the copy module, use [:] or something else.

Output:

([1, 0, 3, 4, 2, 5, 7, 8, 6], [4, 1, 3, 0, 2, 5, 7, 8, 6])

Answer 3

Ok, first of all, some thoughts...

1. Try to not use "list" as a variable, it's a Python identifier for "list" type. It seems that you are redefining the term.

2. Usually, it's a bad idea to use global vars such as loc_of_zero.

About your problem:

I believe that the problem is that you are getting a lot of references of the same variable. Try to avoid it. One idea:

from copy import deepcopy
def swap_positions(list0, pos1, pos2): 
    list1 = deepcopy(list0) 
    first = list1.pop(pos1) 
    second = list1.pop(pos2-1) 

    list1.insert(pos1,second) 
    list1.insert(pos2,first) 
    return list1