Python Grid Labyrinth Problem

Asked 1 weeks ago, Updated 1 weeks ago, 1 views

Python is really a beginner. It hasn't been long since I started.

The study group gave me a question, but I didn't get close yet and asked, so Python gave me a question If you solve it, I'll analyze it one by one.

python

2022-09-20 10:23

2 Answers

I deleted the unproductive comment I just wrote. Instead, I'm going to think about the problem. For your information, it is a difficult problem that takes a long time to actually code.

In theory...

...can solve the problem

So that's a very simple form of theory, and there's a lot more to consider in practice. There is a lot of room for loopholes in the above theory itself. It's like a comprehensive gift set that requires all sorts of applications in the first place. If you ask me to code right now, I think I'll have to use it for a week at my level.

Good luck.


2022-09-20 10:23

I don't usually do other people's homework.No, I don't know because I'm not good enough. I did it because it looked fun.

It's not my homework anyway, so it doesn't matter if I get it wrong, right? LOL

I think I need to use a different module, but I don't know how to use it, so I just roughly did it.

from random import randrange

list_a = [
    ['a', 'b', 't', 't', 't'],
    ['t', 'c', 'd', 'e', 't'],
    ['t', 't', 't', 'f', 't'],
    ['b', 'a', 'h', 'g', 'f'],
    ['c', 'd', 'e', 'f', 'g'],
]



count = 1
move = 0

num1 = randrange(0, 4)
num2 = randrange(0, 4)

print(num1, num2)

a = list_a[num1][num2]

print('reference value', a, order(a))


while 1:
    dict_a = {}
    a = list_a[num1][num2]
    print (f' current location: {num1}, {num2}, {a}')
    for i in list_a:
        print(i)

    try:
        if num1 + 1 > 4:
            raise
        dict_a[ord(list_a[num1 + 1][num2])] = [num1 + 1, num2]
    except:
        pass
    try:
        if num1 - 1 < 0:
            raise
        dict_a[ord(list_a[num1 - 1][num2])] = [num1 - 1, num2]
    except:
        pass
    try:
        if num2 + 1 > 4:
            raise
        dict_a[ord(list_a[num1][num2 + 1])] = [num1, num2 + 1]
    except:
        pass
    try:
        if num2 - 1 < 0:
            raise
        dict_a[ord(list_a[num1][num2 - 1])] = [num1, num2 - 1]
    except:
        pass

    print('dict_a', dict_a)

    if count:
        print ('There is a chance to move')
        r = randrange(0, 1)
        if r:
            print ('Use travel opportunities')
            count -= 1
            b = {i:dict_a[i] for i in dict_a if i < ord(a)}
        else:
            print ('no opportunity to move')
            b = {i:dict_a[i] for i in dict_a if i > ord(a)}
    else:
        print ('no opportunity to move')
        b = {i:dict_a[i] for i in dict_a if i > ord(a)}

    print(b)

    bb = 0
    print (previous location: {num1}, {num2}, {a}')
    if b:
        num_b = randrange(0, len(b))
        print('num_b', num_b)
        for n, i in enumerate(dict_a.keys()):
            c = i
            if n == num_b:
                break
        bb += 1
        print('c', c)
        num1 = dict_a[c][0]
        num2 = dict_a[c][1]
    else:
        print ('no moveable position')
        if count:
            print ('Use travel opportunities')
            count -= 1
            b = {i:dict_a[i] for i in dict_a if i < ord(a)}
            if b:
                num_b = randrange(0, len(b))
                print('num_b', num_b)
                for n, i in enumerate(dict_a.keys()):
                    c = i
                    if n == num_b:
                        break
                bb += 1
                print('c', c)
                num1 = dict_a[c][0]
                num2 = dict_a[c][1]


    if not bb:
        print ("End of Movement")
        break
    move += 1
    print('Number of moves:', move)

    input()
print ('Final number of movements:', move)


2022-09-20 10:23

If you have any answers or tips


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