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Given the following text description, write Python code to implement the functionality described below step by step
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Description:
 # 0. warming up
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Step1:  1. What about making trans?
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Step2:  2. ocr
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Step3:  3. re
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Python Code:
pow(2, 38)
Explanation: # 0. warming up
URL: http://www.pythonchallenge.com/pc/def/0.html
Flag: http://www.pythonchallenge.com/pc/def/274877906944.html
Python docs
 * pow()
End of explanation
from string import ascii_lowercase as al
# Build Caesar table using a shift of 2
table = str.maketrans(al, al[2:] + al[:2])
hint = "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."
print("Hint: \"%s\"\n" % hint.translate(table))
cypher_flag = 'map'
flag = cypher_flag.translate(table)
print(f'Flag: \"{flag}\"')
Explanation: 1. What about making trans?
URL: http://www.pythonchallenge.com/pc/def/map.html
Flag: http://www.pythonchallenge.com/pc/def/ocr.html
Python docs
* str.maketrans()
* str.translate()
* printf-style string formating
* Literal string interpolation
* Lists slices (and other operations)
Other resources
* Caesar cipher
End of explanation
import requests
import re
webpage = requests.get('http://www.pythonchallenge.com/pc/def/ocr.html').text
# Filter comments
comments = re.findall(r'<!--([^>]*)-->', webpage)
cypher_flag = comments[1]  # Get second comment
# Guessed a "rare" character in a mess was something readable, so anything that is alphabetic
regex = re.findall(r'[a-z]', cypher_flag)
lc = ''.join([x for x in cypher_flag if x.isalpha()])
print('Using regex:', regex)
print('Using isalpha:', lc)
Explanation: 2. ocr
URL: http://www.pythonchallenge.com/pc/def/ocr.html
Flag: http://www.pythonchallenge.com/pc/def/equality.html
Python docs
* requests
* re.findall()
* str.isaplha()
* str.join()
* List comprehensions
End of explanation
import requests
import re
# Load website and comment
webpage = requests.get('http://www.pythonchallenge.com/pc/def/equality.html').text
cypher_flag = re.findall(r'<!--([^>]*)-->', webpage)[0]
# One letter surrounded by EXACTLY three big bodyguards on each of its side = 3 upper, 1 lower, 3 upper (EXACTLY)
matches = re.findall(r'[a-z][A-Z]{3}[a-z][A-Z]{3}[a-z]', cypher_flag)
''.join([x[4] for x in matches])
Explanation: 3. re
URL: http://www.pythonchallenge.com/pc/def/equality.html
Flag: http://www.pythonchallenge.com/pc/def/linkedlist.html
End of explanation
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