为什么这段代码做的是 closed[init[0]][init[1]] 而不是 closed[init[0]][init[0]]？

Question

我正在阅读第一个搜索程序 - 机器人算法的人工智能，我正在阅读它的 python 代码。在这里，我们创建了一个封闭数组来检查单元格一旦展开并且不再展开。我们定义了一个名为 closed 的数组，并将其大小定义为我们的网格。作者说它有两个值 0 和 1。0 表示打开，1 表示关闭，但我看到它只是零。

他把起点0,0用1标记了，直到不去查，但是他把坐标0和1放在这一行closed[init[0]][init[1]] = 1。为什么他把 0 和 1 而不是 0,0?

python代码在这里：

#grid format
# 0 = navigable space
# 1 = occupied space

grid=[[0,0,1,0,0,0],
      [0,0,1,0,0,0],
      [0,0,0,0,1,0],
      [0,0,1,1,1,0],
      [0,0,0,0,1,0]]

init = [0,0]                         
goal = [len(grid)-1,len(grid[0])-1]   


delta=[[-1, 0],      #up
       [ 0,-1],      #left
       [ 1, 0],      #down
       [ 0, 1]]      #right

delta_name = ['^','<','V','>']        #The name of above actions
cost = 1

def search():
    #open list elements are of the type [g,x,y] 
    closed = [[0 for row in range(len(grid[0]))] for col in range(len(grid))]

    #We initialize the starting location as checked
    closed[init[0]][init[1]] = 1
    # we assigned the cordinates and g value
    x = init[0]
    y = init[1]
    g = 0
    #our open list will contain our initial value
    open = [[g,x,y]]


    found = False #flag that is set when search complete
    resign= False #Flag set if we can't find expand

    #print('initial open list:')
    #for i in range(len(open)):
            #print('  ', open[i])
    #print('----')


    while found is False and resign is False:
        #Check if we still have elements in the open list
        if len(open)==0: #If our open list is empty
            resign=True
            print('Fail')
            print('############# Search terminated without success')
        else: 
            #if there is still elements on our list
            #remove node from list
            open.sort()       
            open.reverse()    #reverse the list
            next = open.pop() 
            #print('list item')
            #print('next')

            #Then we assign the three values to x,y and g. Which is our expantion
            x = next[1]
            y = next[2]
            g = next[0]

            #Check if we are done

            if x == goal[0] and y == goal[1]:
                found = True
                print(next) #The three elements above this if
                print('############## Search is success')
            else:
                #expand winning element and add to new open list
                for i in range(len(delta)): 
                    x2 = x+delta[i][0]
                    y2 = y+delta[i][1]
                    #if x2 and y2 falls into the grid
                    if x2 >= 0 and x2 < len(grid) and y2 >=0 and y2 <= len(grid[0])-1:
                        #if x2 and y2 not checked yet and there is not obstacles
                        if closed[x2][y2] == 0 and grid[x2][y2] == 0:
                            g2 = g+cost #we increment the cose
                            open.append([g2,x2,y2])#we add them to our open list
                            #print('append list item')
                            #print([g2,x2,y2])
                            #Then we check them to never expand again
                            closed[x2][y2] = 1

search()

Answer 1

he put the coordinate as 0 and 1 in this line closed[init[0]][init[1]] = 1

closed[init[0]][init[1]] 并不代表 "set the value at coordinates (0,1) to 1"。意思是"using init[0] as the x coordinate and init[1] at the y coordinate, set the value to 1"。 init[0] 为 0，init[1] 为 0，因此 closed[init[0]][init[1]] = 1 将 closed[0][0] 设置为 1。

假设起始坐标是init = [2,5]。将该行更改为 closed[init[2]][init[5]] = 1 是不正确的。这会因 IndexError 而崩溃，因为 init 只有两个元素，所以你只能用 0 或 1 索引它。

为什么这段代码做的是 closed[init[0]][init[1]] 而不是 closed[init[0]][init[0]]？

Why does this code do closed[init[0]][init[1]] instead of closed[init[0]][init[0]]?

python

algorithm

depth-first-search

breadth-first-search