为什么不通过迷宫解决代码工作 [英] Why Wont Maze solver Code Work

问题描述

很抱歉，格式不正确，代码量很大，我只是一名初学者，不知道如何诊断错误代码量。我哪里错了，它只是返回发现空0,0并退出。我查看了其他代码，并且无法理解给出的答案，需要对此特定代码进行更详细的说明。请帮助

  grid = [[T，，，＃，＃，＃ ]，
 [＃，＃，，，，＃]，
 [＃，，，＃， ，＃]，
 [＃，＃，＃，，，＃]，
 [＃，＃， ，，＃，]，
 [＃，＃，＃，＃，＃，E]] 
 
 def print_grid（）：
 pr_grid =
用于网格中的键：
用于键中的num：
 pr_grid + = str（num）
 pr_grid + = 
 print pr_grid 
 pr_grid =
 
 print pr_grid 
 
 run = True 
 
 def main（x，y ）：
 if grid [x] [y] ==或者grid [x] [y] ==T：
 print在％d％d找到空％（x， y）
 grid [x] [y] =x
 elif grid [x] [y] ==＃：
 print在％d％d找到的墙％ （x，y）
 elif grid [x] [y] ==E：
 print发现在％d％d出口％（x，y）
 
如果y < len（grid）-1：
 main（x，y + 1）
 if y> 0：
 main（x，y  -  1）
如果x < len（grid [x]） -  1：
 main（x + 1，y）
 if x> 0：
 main（x  -  1，y）
 
 
 print_grid（）
 
 main（0，0）
 
 print_grid（）
  

解决方案

比其他答案快，但如果你想返回True / False，如果你已经找到了出口，那么你可以为每个递归调用维护一个布尔值。

（另外，您的 print_grid 方法可以缩短）

  def print_grid（）：
打印\\\
.join（''.join（row）for grid in grid）
  
 
 
 
 
 
无论如何，这里是我对程序的修改。 
  def main（x，y）：
＃检查超出界限
 if y< 0或y> = len（网格）：
返回False 
 if x< 0或x> = len（grid [y]）：
返回False 
 
 if grid [x] [y] ==E：
 printFound exit at％d％d％（x，y）
 return True 
 elif grid [x] [y] ==＃：
 printFound d％d％d ％（x，y）
返回False 
 elif grid [x] [y] ==或者grid [x] [y] ==T：
 printFound empty在％d％d％（x，y）
 grid [x] [y] =x
＃没有返回，我们要继续搜索
 else：＃catch invalid characters 
返回False 
 
找到= False 
＃Bubble-up搜索出口
＃的结果也通过跟踪退出来限制搜索空间被发现
，如果y < len（grid）-1而未找到：
 found = main（x，y + 1）
 if y> 0且未找到：
 found = main（x，y  -  1）
 if x < len（grid [x]） -  1且未找到：
 found = main（x + 1，y）
 if x> 0且未找到：
 found = main（x  -  1，y）
 
找到的返回
  
 
Sorry for the bad formatting and large amount of code, I am only a beginner and do not know how to diagnose the error down to a smaller amount of code. Where am i going wrong it just returns "found empty at 0, 0" and exits. I have viewed other code under this nature and cannot understand the answers given and need more elaboration on this specific code. Please Help
 grid = [["T", " ", " ", "#", "#", "#"],
        ["#", "#", " ", " ", " ", "#"],
        ["#", " ", " ", "#", " ", "#"],
        ["#", "#", "#", " ", " ", "#"],
        ["#", "#", " ", " ", "#", " "],
        ["#", "#", "#", "#", "#", "E"]]

def print_grid():
    pr_grid = ""
    for key in grid:
        for num in key:
            pr_grid += str(num)
            pr_grid += " "
        print pr_grid
        pr_grid = ""

    print pr_grid

run = True

def main(x, y):
    if grid[x][y] == " " or grid[x][y] == "T":
        print "Found empty at %d %d" % (x, y)
        grid[x][y] = "x"
    elif grid[x][y] == "#":
        print "Found wall at %d %d" % (x, y)
    elif grid[x][y] == "E":
        print "Found exit at %d %d" % (x, y)

    if y < len(grid)-1:
        main(x, y + 1)
    if y > 0:
        main(x, y - 1)
    if x < len(grid[x])-1:
        main(x + 1, y)
    if x > 0:
        main(x - 1, y)


print_grid()

main(0, 0)

print_grid()

解决方案
No claims as to whether this is faster than the other answer, but if you want to return True/False if you have found the exit, then you can maintain a boolean for each of the recursive calls. 

(Also, your print_grid method can be shorter)
def print_grid():
    print "\n".join(' '.join(row) for row in grid)




Anyways, here are the modifications that I made to the program. 
def main(x, y):
    # Check going out of bounds
    if y < 0 or y >= len(grid): 
        return False
    if x < 0 or x >= len(grid[y]):
        return False

    if grid[x][y] == "E":
        print "Found exit at %d %d" % (x, y)
        return True
    elif grid[x][y] == "#":
        print "Found wall at %d %d" % (x, y)
        return False
    elif grid[x][y] == " " or grid[x][y] == "T":
        print "Found empty at %d %d" % (x, y)
        grid[x][y] = "x"
        # no return, we want to continue searching
    else: # catch invalid characters
        return False

    found = False
    # "Bubble-up" the results from searching for the exit 
    # Also limit the search space by keeping track if the exit was found
    if y < len(grid)-1 and not found:
        found = main(x, y + 1)
    if y > 0 and not found:
        found = main(x, y - 1)
    if x < len(grid[x])-1 and not found:
        found = main(x + 1, y)
    if x > 0 and not found:
        found = main(x - 1, y)

    return found


                        
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