Santa needs a new password that meets the "security" requirements. The puzzle is very similar to day 5, finding naughty strings.
New Python features
zip is a function that allows to iterate on two or more sequences at once. Usually it's used over two or more lists, dicts, tuples etc. Example:
>>> countries=["Austria", "United Kingdom", "Denmark", "Sweden", "Norway", "Poland", "Germany"]
>>> areacodes=[43, 44, 45, 46, 47, 48, 49]
>>> for country, code in zip(countries,areacodes):
... print(f"Area code for {country} is {code}")
...
Area code for Austria is 43
Area code for United Kingdom is 44
Area code for Denmark is 45
Area code for Sweden is 46
Area code for Norway is 47
Area code for Poland is 48
Area code for Germany is 49
Here I used it like this: for ch1, ch2, ch3 in zip (text, text[1:], text[2:]) which means it combines 3 sequences:
some string, the same string starting from the second characted and the same string starting from the 3rd character.
On the first iteration, it returns characters 0, 1, 2; on the second iteration 1, 2, 3 etc. Note that the sequences
have different length, second and third one are 1 and 2 characters shorter from the first one. It doesn't matter for
the zip function, it just ends when the shortest sequence ends.
ord returns the Unicode (or ASCII, it's the same in this case) code of a character, chr is an opposite function that returns character for the given code. In C and C++ I could just do arithmetic operations directly on the characters, suprisingly Python typing is sometimes more strict, you can't just treat character as a number without explicitly converting it first.
Code
#!/usr/bin/python3
import re
FORBIDDEN=['i', 'o', 'l']
rx1=re.compile(r"(\w)\1.*(\w)\2")
password="cqjxjnds"
def increment(text: str) ->str:
char_array=list(text)
for i in range(7,-1,-1):
char_array[i]=chr(ord( char_array[i] ) + 1)
if char_array[i]>'z':
char_array[i]='a'
else:
break
return "".join(char_array)
def rule1(text: str) -> bool: # 3 characters in a sequence
for ch1, ch2, ch3 in zip (text, text[1:], text[2:]):
if (ord(ch3)-ord(ch2)==1) and (ord(ch2)-ord(ch1)==1):
return True
return False
def rule2(text: str) -> bool: # does not contain forbidden characters
for char in text:
if char in FORBIDDEN:
return False
return True
def rule3(text: str) -> bool: # contains at least 2 different pairs
if pairsearch := rx1.search(text):
pair1, pair2= pairsearch.groups()
if pair1 != pair2: # and they are different
return True
return False
print ("Old password: ", password)
while(True):
password=increment(password)
if rule1(password) and rule2(password) and rule3(password):
break
print("Part 1: ", password)
while(True):
password=increment(password)
if rule1(password) and rule2(password) and rule3(password):
break
print("Part 2: ", password)