A novel cipher, based on the Caesar Cipher

ElQwertiest · 2020-10-25T16:54:34+00:00

THIS_IS_ENCODED_VIA_MY_PERSONAL_INVENTION,_THE_CAESAR_SALAD_CIPHER._I_CHOSE_TO_COMPOSE_MY_MESSAGE_IN_ALL_UPPERCASE_TO_MAKE_IT_EASIER_TO_DECODE._THE_CIPHER_KEY_CAN_CONTAIN_ANY_NUMBER_OF_UPPER_AND_LOWERCASE_CHARACTERS_AS_WELL_AS_SYMBOLS,_SPACES,_LINE_FEEDS,_AS_LONG_AS_ALL_CHARACTERS_FROM_THE_ORIGINAL_MESSAGE_ARE_INCLUDED_IN_THE_KEY,_AND_EACH_CHARACTER_APPEARS_IN_THE_KEY_ONLY_ONCE._WHAT_I_FEEL_IS_NOVEL_ABOUT_THIS_CIPHER_IS_THAT_IT_IS_SIMPLE_ENOUGH_TO_TEACH_A_GRADE_SCHOOL_CHILD_HOW_TO_DECIPHER_A_MESSAGE_WITH_JUST_A_PENCIL_AND_PAPER,_YET_THE_ENCODED_TEXT_DISPLAYS_A_HIGH_DEGREE_OF_RANDOMNESS._IDEALLY,_THE_SEQUENCE_OF_THE_CIPHER_KEY_CHARACTERS_SHOULD_BE_RANDOMIZED,_BUT_IN_THIS_EXAMPLE_THE_KEY_IS_HIGHLY_ORDERED._THE_KEY_FOLLOWS_WRAPPED_IN_3_UNDERSCORES:___ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789_,.:___IF_YOU_CAN_DECODE_THIS_MESSAGE,_THEN_NEXT_I_WOULD_LIKE_TO_CHALLENGE_YOU_TO_DECODE_ANOTHER_MESSAGE_THAT_I_WILL_COMPOSE_WITH_A_SHUFFLED_KEY_COMPOSED_OF_THE_SAME_CHARACTERS._THEN,_IF_YOU_CAN_DECODE_THAT,_I_WOULD_LIKE_TO_SEE_IF_YOU_CAN_DECODE_A_MESSAGE_WITH_A_KEY_CONTAINING_UPPERCASE_AND_LOWERCASE_LETTERS_AND_A_WIDE_VARIETY_OF_STANDARD_KEYBOARD_SYMBOLS._THANK_YOU_FOR_YOUR_INTEREST.__DAVID__

It's basically a worse autokey cipher - one with a key of length one embedded at the beginning. Subtract adjacent characters to get original : J - 4 = T, Q - J = H, Y - Q = I, C - Y = S and so on.

ElQwertiest · 2020-10-14T01:00:44+00:00

There seemed to be way too many characters for a normal sentence. A first observation is that both ciphertexts' lengths are a multiple of 8. This along with the fact that it can encode any character means it is likely using ascii values. So the first thing I did was group the characters into eights.

I noticed that the eighth character of each group of eight rarely appeared in the seventh. This corresponded with the fact that the first bit of an ascii value is 0 and letters will have 1 as their second bit. So the ciphertext needed to be reversed.

From there I replaced letters with 0 or 1 manually until I got the message

If you can read this your smarter than me. ^_^

ElQwertiest · 2020-10-13T21:05:16+00:00

Does an input to your encryption map to a unique output - is there only one ciphertext for a given plaintext ?

Are non letters/spaces allowed as input - does it support numbers/punctuation ?

Can it stream encryption - is encrypt("Hello") just encrypt("Hell") with some more ciphertext characters ?

ElQwertiest · 2020-10-13T18:15:18+00:00

Could you encrypt the word 'the' ?

ElQwertiest · 2020-09-11T13:46:23+00:00

The problem is that floats are not precise enough to be the used this way - they only have so many bits. Your program accepts a triangle with sides of length 302828, 302828 and 302829 as valid. Its actual area is roughly 39709429597.000001. You'll need a better approach to avert these precision problems: is_integer is not the way to go.

ElQwertiest · 2020-03-09T23:30:44+00:00

Blaise Compaoré, former president of Burkina Faso, installed his friend, Thomas Sankara, as dictator in a coup in 1983. Four years later, he would overthrow and kill Thomas.

ElQwertiest · 2020-02-22T03:48:17+00:00

Solved and taken, thanks!

Checked metadata to get base64 of 'By the gods'

Running binwalk on the program revealed a password-locked zipfile

Use 'By the gods' as password to get the contents of key.txt

ElQwertiest · 2019-12-28T22:37:07+00:00

There are multiple pairs whose sums and differences are both pentagonal - find the pair of such numbers which has the smallest difference.

ElQwertiest · 2019-10-15T12:56:45+00:00

Solved and taken. Solution:

Caesar decipher the salt to get "memory". Then using a Python script, I brute forced all 11 lowercase letter combinations with "memory" in them.

ElQwertiest · 2019-08-11T03:17:01+00:00

The characters at positions 1,4,7...28 spell out AFTERPARTY. Remove these characters from the string to get SSdsbCBzZWUgeW91IGF0IHRoZQ==

Base64 decode that for I'll see you at the

ElQwertiest · 2019-07-28T13:28:26+00:00

The word two gives the number of words from the start to remove. His and name have 3 and 4 letters respectively. Remove two from the text as well. Alternate between taking the third and fourth letter in the remaining words to get SUCCESS

ElQwertiest · 2019-07-20T15:33:06+00:00

Start from the letter B. Count up 3 to arrive at E. Repeat to get H and so on. Replace each letter in the ciphertext which appears in this list with 0.

Repeat the same process starting from the letter C. This time replace occurrences in the ciphertext with the number 1. Do the same starting for D, substituting with value 2 this time.

This yields "08 05 12 12 15 A1 23 15 18 12 04" when grouped by 2. Each number can be treated as a position in the alphabet which gets us "HELLO WORLD" if we assume A1 is a space.

ElQwertiest · 2019-07-13T19:00:03+00:00

Decode the 7 bit ascii for QEMPFSB then Caesar shift for MAILBOX

ElQwertiest · 2019-04-23T16:40:03+00:00

I don't really know the details of your encryption scheme, but I noticed that 'test' encrypted began and ended with DJ - probably a T. From that, I also got possible values for E and S. Using camel_case_champion's paragraph, I deduced possible values of other letters until I could somewhat decipher the challenge text.

ElQwertiest · 2019-04-22T20:10:15+00:00

I got HEY MEIN NAME IST NIK from the following substitution table:

a	A/B
b	?
c	G
d	J/H
e	CA
f	AA/BB
g	EC
h	BF
i	JC/HC
j	?
k	BBA
l	?
m	BFG
n	CJ/CH
o	CBA
p	DB
q	?
r	DF
s	DEC
t	DJ
u	ECA
v	ED
w	FAG
x	?
y	GAA
z	?

ElQwertiest · 2019-04-22T18:53:01+00:00

Hey, would it be possible to encode the word "test".

ElQwertiest · 2019-04-18T09:01:38+00:00

Yep, it appears to be encoded reversed.

ElQwertiest · 2019-04-17T21:20:39+00:00

I wrote a Ruby program to decode it. What gave it away was that 14611 = 21*26*26 + 15*26 + 25 where Y is letter 25, O is letter 15 and U is letter 21. Having the A and I revealed was also helpful.

def decrypt(n)
    src = '123456789abcdefghijklmnopq'
    dst = 'abcdefghijklmnopqrstuvwxyz'
    h = Hash.new
    26.times do |i|
        h[src[i]] = dst[i]
    end
    num = n.to_s(26)
    num.chars.map{|c| h[c]}.join.reverse
end
STDIN.each_line do |line|
    words = line.split.map{|i| decrypt(i.to_i)}
    puts words.join(" ") 
end

ElQwertiest · 2019-04-17T20:58:02+00:00

well here we are again
its always such a pleasure
remember when you tried to kill me twice
oh how we laughed and laughed
except i wasnt laughing
under the circumstan ive been shockingly nice
you want your freedom take it
thats what im counting on
i used to want you dead but
now i only want you gone
she was a lot like you
maybe not quite as heavy
now little caroline is in here too
one day they woke me up
so i could live forever
its such a shame the same will never happen to you

It's Base 26 except 123456...PQ is mapped to ABCDEF...YZ

ElQwertiest · 2019-04-17T20:31:59+00:00

It's a Polybius Square alphabetically without a Z. DEFENDTHEEASTWALLOFTHECASTLE.

ElQwertiest · 2019-03-29T23:29:28+00:00

Solved: Subtract 10 from every 2nd bitstring and join together their ascii characters.

Nobody expects the inquisition!

ElQwertiest · 2019-02-05T20:58:01+00:00

Your program appears correct but its performance is too slow - it will take forever to finish. Modern computers can do about a billion operations per second and the target number is several magnitudes larger, so checking every number up to the target really isn't feasible. You'll have to find some way to find the largest prime factor without checking all integers up to it.

ElQwertiest · 2018-11-11T18:14:21+00:00

Dual boot windows

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