🔒 String permutations

[cyan shell]

I have a string like james.robert and I want to get all posibilities of capitalized letters like James.robert or JAmes.robert also jameS.robert and so on.

[sinful oxide]

@cyan shell

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[cyan shell]

my_str = "james.robert"
for i in range(len(my_str)):
    if my_str[i] != ".":
        print(my_str[:i] + chr(ord(my_str[i])-32) + my_str[i+1:]) 
``` I tried doing for a single iteration, but rn I cannot think of how to make all the possible combinations

[ocean lance]

I would skip ord and do something more like my_str.upper()[i] or my_str.lower()[i] to get the different cases

You could leverage itertools.product here

[edgy harbor]

You actually want the permutations or do you just want to calculate how many there are?

[cyan shell]

yeah I can also do .capitalize, that's not where I am having troubles tho

[cyan shell]

I want to print all possible strings

[eager urchin]

i'd use itertools.product too; count the letters that can be capital and generate lists of true/false for capitals

[cyan shell]

I see, I thought of that but I am not sure if I know how to use it properly

[cyan shell]

all letters can be capitalized except the point

[eager urchin]

yes, you'd of course not count that because it can't be capital

[rugged tundra]

turn each character into a list of both possible capitalizations (or the only possible capitalization, for characters like .). feed all these lists to product.

note that there'll be like 2048 capitalizations.

[cyan shell]

yeah, thats what I hoping for, the more the better

Is it possible to get the code? I am not trying to achieve a big project with this.

[eager urchin]

whatever you do, i reckon to account for the dot not changing you'd need to generate this new string without special characters and then construct the proper version

[crude quartz]

can I ask why you're trying to achieve this?

[cyan shell]

yeah, the dot is not a big deal, I could insert it after every 6 characters in each reconstructed string

[cyan shell]

Well it might get misinterpreted

[crude quartz]

meaning?

[cyan shell]

You are familiar with CTF challenges right?

[crude quartz]

Nope?

But also if it's for a challenge, then carry on

[cyan shell]

CTF means capture the flag

[crude quartz]

just making sure you weren't coming up with some sort of roundabout solution for something

Ahh yeah I'm familiar with CTF, but not sure how it relates to programming here

[polar pilot]

Are you brute forcing a password?

[cyan shell]

oh well there is a challenge where I have an admin panel and I need to bypass it in order to gain access

[cyan shell]

more or less yeah

this is the challenge - as proof

and I need to guess a 1-4 digits pin in order to gain control, but after 3 attempts the user accounts get locked out, but the server thinks that james.robert is different than james.Robert but at the same time consideres it a valid username, which exists in their database

[ocean lance]

If this is an open/active competition, then asking for help here might constitute an unfair advantage.

[cyan shell]

Ah no, it is not on going.

The challenge was posted 1 month ago

[ocean lance]

Now I see why you said the more the better

[cyan shell]

Yep

[ocean lance]

I was just noticed the "terminate in an hour" thing

[cyan shell]

so is there a chance that you could guide me more forwardly to get that script? I did not user itertools.product much

[ocean lance]

Yes, take a look at what @rugged tundra said

And we can help guide you from there

[cyan shell]

Yeah, its been up for a couple of days, more guys try to do it and I kinda found the solution but donno how to automate it

[cyan shell]

alright, I have to go and eat right now but I will try to be quick

[cyan shell]

Wouldn't this alter the format of the initial string? Like wouldn't I get Jjames? Or unordered usernames?

Like asmej

[rugged tundra]

product's outputs take exactly one item from each iterable and don't reorder them.

[cyan shell]

I see, I will try as soon as I finish eating, thanks

[cyan shell]

This has froze my pc for over 10 minutes

I should work with shorter strings

[cyan shell]

I tried with a comma too, I am not sure I understand how the params are utilised by the function

cuz I get an output like this and it changed its order

[rugged tundra]

that'd be because you're doing repeat=12

[cyan shell]

yeah cuz the total length of the string was 12, I shortened the string and the repeat

[rugged tundra]

you're calculating all 12-length strings consisting of 24 possible characters at each position

that's not what you want - you want to, for each position, provide a list of up-to-2 possible characters.

so make a list consisting of a list for each character (e.g. for the first character, "j", the corresponding list would be ["j", "J"], ["."] for the dot), and unpack it into product.

[cyan shell]

oh, it takes a list of lists?

[rugged tundra]

no, but it does take any number of iterables

!e

import itertools
x = ["aA", "bB"]
print(list(itertools.product(*x))) # equivalent to itertools.product(x[0], x[1])

[sinful oxide]

@rugged tundra :white_check_mark: Your 3.12 eval job has completed with return code 0.

[('a', 'b'), ('a', 'B'), ('A', 'b'), ('A', 'B')]

[cyan shell]

would a list comp work here?

my_list = [[a,b] for a,b in zip(list(my_str), list(my_str.capitalize())]

to get my list of lists?

[cyan shell]

so its a list of strings, not a list of lists

[rugged tundra]

I used the former because it's simpler to type, but it doesn't matter for product - just needs to be iterable

[rugged tundra]

sure, that looks valid

i'd have written it as [[x.lower(), x.upper()] for x in my_str]

[cyan shell]

yeah it gave me an error but I fixed it

I managed to reach my goal, thank you

[cyan shell]

oh makes sense, thank you

[cyan shell]

!close

[sinful oxide]

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