I hear you're good with computers and programming

I hear you're good with computers and programming.
Surely you know how basic encryption works, user?

23, 63, 13, 62, 1, 58, 40, 41
53, 127, 116, 57, 57, 58, 42, 42
26, 14, 15, 12, 13, 16, 22, 19

The values above are originally from three strings, each eight characters long.
They've been transformed by a keyphrase of the same length, through bitwise XOR.

It's essentially a one-time pad, but here the keyphrase has been used three times - which allows for cryptoanalysis.

What are the three original strings, and what's the password?

Stop asking others to do your homework faggot

Stop asking others to do your homework faggot

Stop asking others to do your homework faggot

Stop asking others to do your homework faggot

Stop asking others to do your homework faggot

Stop asking others to do your homework faggot

>Stop asking others to do your homework faggot

C-C-C-COMBO BREAKER

The content of the strings aren't exactly the kind of stuff you would get in school.
But if you can't solve it, feel free to meme.

hello /r/Sup Forums in advance

Stop asking others to do your homework faggot

>hello /r/Sup Forums in advance
fuck you leftist plebbit

no i just implement the encryption through libraries

How would i go about solving something like this? I converted the decimals to binary, but i have no idea how you would decypher the XOR passphrase.

>Surely you know how basic encryption works, user?
Anyone, from the most clueless amateur to the best cryptographer, can create an algorithm that he himself can't break. It's not even hard. What is hard is creating an algorithm that no one else can break, even after years of analysis.

k^a = ka
ka^kb = a^b
assume all symbols in a,b,c,k are in range 0-9 a-z A-Z
you get ka, kb, kc and from this you can get a^b, a^c, b^c and see how the values on each bit differ, from legal range then get small range of valid options

for example on position 2 you get a={r,t,v,x}, b={2,4,6,8}, c={C,E,G,I}, k={M,K,I,G}
do this for each position and you will still get thousands valid options, bruteforce it with eyes and maybe some will create valid words

Extremely difficult. It's basically a vigenere cipher, which means you have to figure out the length the keyword to even start performing a symbol frequency analysis. Figuring out the lenght of the keyword from the ciphertext this small is very very hard.

Stop asking others to do your homework faggot

Leave and never return

>sit there for an indeterminant amount of time guessing random key phrases because I haven't given enough text for proper frequency analysis
No.

Stop asking others to do your homework, faggot.

Stop asking others to do your homework faggot

>good with computers and programming
>asking about cryptoanalysis

>you have to figure out the length the keyword to even start performing a symbol frequency analysis

>The values above are originally from three strings, each eight characters long.
>They've been transformed by a keyphrase of the same length

keyphrase length = 8

Install Gentoo

OP so I have about 12**8 valid solutions and finding some meaningful strings will eat too much time not to ask:
are there any symbols beside 0-9A-Za-z?
are the strings meaningful?
is the key meaningful?

K: BFAJLYAY
A: UyLtMcip
B: w95sucks
C: XHNFAIWJ

probably not the original solution but numbers check

Happy to see someone trying it out.

>are there any symbols beside 0-9A-Za-z?
Almost exclusively letters, big and small.

>are the strings meaningful?
Most of the strings are actual words.

>is the key meaningful?
Yes, it has meaning because it creates ciphertext. :^)
But its composition is quite random.

More info soon.

>Almost exclusively letters, big and small.
the second character in the second string doesn't seem to be a letter.
is it suppose to be a number?

Glad to see at least 2 people taking a shot at it. Variables:
>plaintext p(1,2,3); keyphrase k; ciphertext c(1,2,3)

P L A I N T E X T 1 = p1
K E Y P H R A S E = k
23, 63, 13, 62, 1, 58, 40, 41 = c1

P L A I N T E X T 2 = p2
K E Y P H R A S E = k
53, 127, 116, 57, 57, 58, 42, 42 = c2

P L A I N T E X T 3 = p3
K E Y P H R A S E = k
26, 14, 15, 12, 13, 16, 22, 19 = c3

>"^" is used to describe XOR operations below

c1 = p1^k; c2 = p2^k; c3 = p3^k

c1^c2 = (p1^k)^(p2^k)
= (p1^p2)^(k^k)
= (p1^p2)^0
= p1^p2

c1^c3 = (p1^k)^(p3^k)
= (p1^p3)^(k^k)
= (p1^p3)^0
= p1^p3

c2^c3 = (p2^k)^(p3^k)
= (p2^p3)^(k^k)
= (p2^p3)^0
= p3^p3

So what this /means/ is that:

>c1^c2
23, 63, 13, 62, 1, 58, 40, 41 = c1
53, 127, 116, 57, 57, 58, 42, 42 = c2
34, 64, 121, 7, 56, 0, 2, 3 = p1^p2

>c1^c3
23, 63, 13, 62, 1, 58, 40, 41 = c1
26, 14, 15, 12, 13, 16, 22, 19 = c3
13, 49, 2, 50, 12, 42, 62, 58 = p1^p3

>c2^c3
53, 127, 116, 57, 57, 58, 42, 42 = c2
26, 14, 15, 12, 13, 16, 22, 19 = c3
47, 113, 123, 53, 52, 42, 60, 57 = p2^p3

could you tell me the two numbers in p2? that would help a lot

00

then the key should be:
>?OD?????

4 unknowns from 3 inputs

OP is a faggot

>OP is a faggot
Kek, that's a great guess.

In fact:
>p1 = OpIsHomo

just slighly over 30k options left
K: XODM????
A: OpIs????
B: m00t????
C: BAKA????

k
K: XODMIUEF
A: OpIsHomo
B: m00tpool
C: BAKADESU

kek, checks out

a = {23, 63, 13, 62, 1, 58, 40, 41};
b = {53, 127, 116, 57, 57, 58, 42, 42};
c = {26, 14, 15, 12, 13, 16, 22, 19};
k = "XODMIUEF" // ToCharacterCode;
BitXor[#, k] & /@ {a, b, c} // FromCharacterCode
>{"OpIsHomo", "m00tpool", "BAKADESU"}

Impressive, very nice
what did you use to automate the guesswork?

it's good to see people being able to actively do computations on the provided ciphertext

but that's basic boolean algebra, anyone with a 3-digit IQ should be able to do it

just a small program, for each position find a set of chars from charset (0-9A-Za-z) that xor all ciphertexts to the charset
then I noticed that p2 has those 2 numbers and that sometimes key and p3 can either be both either upper or lower, so I picked those dictated by positions with numbers for consistency
after you said numers are 00 there were few guesses (m00t, r00t, d00t), but m00t gave OpIs and BAKA on other plaintexts
30240 possibilities, someone suggested that you are faggot and that was it

Go away.

You are the greater among us, truly I can learn much from you. May I become your apprentice that you would teach me and thus I would become greater.

great to hear how you did it.
I hope you had fun

are you fucking retarded?

t. brainlet

Did you mean to reply to somebody?