Level 1

code

2**38

solution

http://www.pythonchallenge.com/pc/def/274877906944.html

Level 2

code

code = "g fmnc wms bgblr rpylqjyrc gr zw fylb. rfyrq ufyr amknsrcpq ypc dmp. bmgle gr gl zw fylb gq glcddgagclr ylb rfyr'q ufw rfgq rcvr gq qm jmle. sqgle qrpgle.kyicrpylq() gq pcamkkclbcb. lmu ynnjw ml rfc spj."

def solution(code=code):
    crack = ''
    for x in code:
        if (ord(x) > ord('a')) & (ord(x) < ord('y')):
            crack += chr(ord(x)+2)
        elif x == 'y':
            crack += 'a'
        elif x == 'z':
            crack += 'b'
        else:
            crack += x
    return crack

solution()
>>> "i hope you didnt translate it by hand. thats what computers are for. doing it in by hand is inefficient and that's why this text is so long. using string.maketrans() is recommended. now apply on the url."

solution("http://www.pythonchallenge.com/pc/def/map.html")
>>> "jvvr://yyy.ravjqpejcnngpig.eqo/re/fgh/ocr.jvon"

solution

http://www.pythonchallenge.com/pc/def/ocr.html

Level 3

solution

google “page source”
found website to view page source of websites, https://www.view-page-source.com/
found the code to crack with
split the code into a loooong list
pd.DataFrame(list).unique() – noticed there are letters hidden in the code : e q u a l i t y

Level 4

solution

Apparently “little candle” is not the answer.
Checking out the page source again and found the code.
Apparently, it’s requiring regular expression.
I hate regular expression. So I tried ‘a-z’. But disappointingly, it didn’t work.

code

code = "the looong code"
crack =[]
cracker = ''
# the following code shows how much I hate regular expression
for i in range(1, len(code)-6):
    if ord(code[i-1]) >= ord('a'):
        if ord(code[i]) <= ord('Z'):
            if ord(code[i+1]) <= ord('Z'):
                if ord(code[i+2]) <= ord('Z'):
                    if ord(code[i+3]) >= ord('a'):
                        if ord(code[i+4]) <= ord('Z'):
                            if ord(code[i+5]) <= ord('Z'):
                                if ord(code[i+6]) <= ord('Z'):
                                    if ord(code[i+7]) >= ord('a'):
                                        if '\n' not in code[i:i+7]:
                                            crack.append(code[i-1:i+8])
                                            cracker += code[i+3]
                                    continue
                                continue
                            continue
                        continue
                    continue
                continue
            continue
        continue
        
crack,cracker

>>> (['qIQNlQSLi',
  'eOEKiVEYj',
  'aZADnMCZq',
  'bZUTkLYNg',
  'uCNDeHSBj',
  'kOIXdKBFh',
  'dXJVlGZVm',
  'gZAGiLQZx',
  'vCJAsACFl',
  'qKWGtIDCj'],
 'linkedlist')

Level 5

code

import requests
code = '12345'
try:
    for i in range(50):
        link = "http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing="+code
        f = requests.get(link)
        print(f.text)
        code = f.text[-5:]
except:
    print('aaaaaaaaaaaaaaa')

# Eyeball the outliers and correct the codes, until I saw : "peak.html"

Level 6

Wasn’t able to hack this one all by myself. I thought about pickle but wasn’t able to realize that banner.p could actually be an address. And for the code-cracking part, I used some hint from the Internet. It’s kind of a pity though, since I really appreciated the trickiness of this problem once I knew the answer. Should have put more thoughts into it.

solution

View the page source - get “peak hell” - “pickle” and the address, “http://www.pythonchallenge.com/pc/def/banner.p”
Then do the coding part:

code

import pickle
import urllib.request
url = "http://www.pythonchallenge.com/pc/def/banner.p"
with urllib.request.urlopen(url) as f:
    result = pickle.load(f)
    for r in result:
        print(r)

# observe the printed result then do the following: 

for x in result:
    s = ''
    for y in x: 
        s += y[0]*y[1]
    print(s)

>>>"channel"