Just got an idea: It's kind of like the "What am I"-challenge, but this time it isn't cars, it's about crypted text. Whoever finds out what the original text was may post another encrypted message - with only one condition: it has to be reversible, so don't use MD5 or something, that wouldn't be fun

let's start of really easy:

i got stuck

n = o though, which leaves you this, i've bolded the "known" bit

Lhc, guog jof eroyyl rofl

im geussing that the end is rofl, if so then r is r, f is f and l is l, which means i need a 3 letter "l" word:

lab
lac
lad
lag
lam
lap
law
lax
lay
lbs
lea
led
leg
lei
lek
leo
let
leu
lev
lex
ley
lib
lid
lie
lim
lip
liq
lit
lob
loc
log
lop
lot
low
lox
lpm
lug
lux
lye

one word. eh?

Quote from herki :

let's start of really easy:

Lhc, gung jnf ernyyl rnfl

Well, since you said it was easy, my first guess was rot13.. so:
Yup, that was really easy

My entry:

Tip: This might seem much harder, but infact - it isn't.

Maybe jnf = the

lhc, guhg the ernhyyl rhel

I think the yy pair is a good place to start too, there are not very much letters that can be doubled in a word... The text is pretty small for a frequency analysis to work, but here it goes:

f(C) = 1
f(E) = 1
f(F) = 2
f(G) = 2
f(H) = 1
f(J) = 1
f(L) = 3
f(N) = 4
f(R) = 2
f(U) = 1
f(Y) = 2

Damn you herki, I really enjoy these crypto things and I have other things to do!!! You want me not to pass my exams or what?

i have a German exam tommorow. I have to memorise 150 word essay, and 1 and a half minutes of answers i havent even written yet!

herki, yeah, i agree, that was really easy

bored tho...

Quote from TheDeppchef :

Well, since you said it was easy, my first guess was rot13.. so:
Yup, that was really easy

My entry:

~45? 5? - .5@ 4->01>

Tip: This might seem much harder, but infact - it isn't.

My thoughts at the moment:
"this is a .i@ ha>01>"

...

wait a minute...
"this is a bit harder"

edit: new challenge, increasing the difficulty

the 13 might be a good hint

That was rot42 btw

Quote from TheDeppchef :

That was rot42 btw

you used all ascii-characters? that's evil, rot-encryption is supposed to be letters only
but every rot-encryption is solvable by replacing letters, the most basic form of encryption

Quote from herki :

you used all ascii-characters? that's evil, rot-encryption is supposed to be letters only
but every rot-encryption is solvable by replacing letters, the most basic form of encryption

Not all of them, rot47 is possible, too

"rot13 again, but every other letter is used"


Hope i didn't make a typo..

Quote from NotAnIllusion :

"rot13 again, but every other letter is used"

Quote:
2124424344 5234423114 521142 431323153215

Hope i didn't make a typo..

use [ code ]-tags next time
[maybe spoiler] GJUVW ZPUMD ZAU VCIENE [/spoiler] ...anything near? or am I completely on the wrong path

Spoiler - click to reveal

I assumed the numbers are pairs based on the 6-system and because the numbers were to big to be letters I deducted 6 of every number and then came up with this

Well... [hint]think matrices [/hint]

Now I get to try out my ASCII encryption program I made:

Quote from TheDeppchef :

My entry:

~45? 5? - .5@ 4->01>

Tip: This might seem much harder, but infact - it isn't.

²his is a bit harder

Not sure why the T didn't come out right. And you didn't "encrypt" the spaces either.

Edit: Ah, googled ROT-13, only encrypts certain characters. Even ROT-47 excludes spaces. Weird.

Here's mine:


Clue - what was a good year for british footy?

Quote from Bob Smith :

Here's mine:

Ž§¯±°b…ª§§µ¥£*§b¨±´b¶ª§b¹«°

Clue - what was a good year for british footy?

let me try out mine (5 min of work )
Lemon Cheescaçe for the win

...something seems not to work out perfectly

edit: NotAnIllusion, don't worry, still at it

edit2:

Spoiler - click to reveal

so is it 11 for A, 12 for B,... you have to leave a letter out then... i wonder which one...

edit3: (god, i'm editing a lot), NotAnIllusion, is it "First World War Scheme"? (I'm pretty sure it is)...the missing letter is "N"?

yes

that could be quite hard for now

[little help]
This table is used
ABCDEFGHIJKLMNOPQRSTUVWXYZ
abcdefghijklmnopqrstuvwxyz
0123456789!"#$%&'()*+,-./°
:;<=>?@[\]^_{|}ÄÖÜäöüߧ~
[/help]

Hmm, don't think
y)Ycj1$xK7A or Ite-O1Gx'71 are correct

Quote from NotAnIllusion :

Hmm, don't think
y)Ycj1$xK7A or Ite-O1Gx'71 are correct

They cerainly aren't ... I'll give you a few hints:

[hint1]The order of the letters is 21436587...[/hint1]

[hint2]There's some kind of progressive addition[/hint2]

With all three hints it should be possible... btw, that's an quite simple encryption I made myself, I'm thinking about making a better one, but I'm quite busy lately

I deal in encryption quite frequently as a programmer, one of the horrible things I do to data - after learning what foiled me most as a hacker - is compress it, along with some fractal encryption and bit scrambling, which is just so outright rude to a decipher that I wont post anything for you to hack, besides, the output is usually binary and wont post on the forums.

Quote from Becky Rose :

I deal in encryption quite frequently as a programmer, one of the horrible things I do to data - after learning what foiled me most as a hacker - is compress it, along with some fractal encryption and bit scrambling, which is just so outright rude to a decipher that I wont post anything for you to hack, besides, the output is usually binary and wont post on the forums.

that would be quite mean, the point of this thread is, that people should be actually able to find out what the encrypted message is
You could actually post hex-code, there are enough free hex-editors which can paste hex-code seperated by spaces (e.g. EC B0 00), but as you said, very few people would actually be able to solve it

btw: couldn't get my figured out, eh? *challenge*

since no one got anything until now
[hint3]First 4 letters: Who' [/hint]