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Hello,

I made a program wich can encrypt data. It is pretty simple to crack if you know the technique, but I'd like to know how difficult it is to crack when you don't know it.

I'd really appreciate it if someone tries to crack this:

111111111111111111111111\0000000000\111111\00000000000000000\11111\00000\11111\11111111111\

It are 8 letters.

Good luck and thank you very much!!

-Luc

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lucb1e,

You have posted a very small piece of ciphertext, too small to do any cryptanalysis on. Just about any cipher would be very hard to crack with only a small amount of text.

Matt_S

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the answer is 42.. don't forget your towel!

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youdidit

I believe that's right, I could be wrong, though.

Angoth

Edit: 1's are moving one character to the right in the alphabet. 0's are moving one character to the left.

Start with A. Move 24 to the right. Y

Start with Y. Move 10 to the left. O

Start with O. Move 6 to the right. U

Start with U. Move 17 to the left. D

Start with D. Move 5 to the right. I

Start with I. Move 5 to the left. D

Start with D. Move 5 to the right. I

Start with I. Move 10 to the right. T

The more text you encrypt with your method, the more times someone will be hit over the head with the realization that you'll never get more than 25 1's or 0's per character. That's a pretty big clue. But, honestly, everyone starts with something like this. It's just moving around the alphabet. It's the differential that is the clue, especially the longer that you extend it.

The only problem I have is that it's such a short piece of ciphertext that I have arrived at one *possible* solution. The phrase does indicate that I got it right, though.

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Sorry for my late reply, but you are right Angoth

I already got a new version of it which does make it a little more tricky, it now doesn't limit to 25 0's or 1's. But thanks for the tip

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Angoth is right, the cipher text gives away alot of information about both the key and plain text. key being ether 0 or 1 plus the length and plain text being first element. He is also right about the whole 26 thing. one way of making it more empowered and easier to work on is converting each element so its the same length... and also hiding the way the rotation works by hiding it in your key that must be used, instead of including it in the cipher text with the 0 and 1. converting the elements so there the same length is as easy as finding how much info you can jam into a bit. since your using only letters we will assume a starts at 0 and z is 25. since there are 25 elements you can find out how much data is actually stored (called entropy) by LOG(n)/LOG(2) n being the number of elements. so we find that 25 elements can be stored in some 4.6bits or 5 bits. starting from 00000 for a to 11001 to z. now for the key... an easy way of making your algo key based is making it as long as your plain text (message) and using ether 0 or 1 for shift left and right. this also saves memory by making each message exactly 5 bits long assuming that each letter appears equally in a message(usually not the case but for simplicity, and that if this algo is stronger then data should appear random) then your cipher text size is reduced by almost 50% even if you use bytes as storage!

so your message

111111111111111111111111\0000000000\111111\00000000000000000\11111\00000\11111\11111111111\ becomes

11000 01010 00110 10001 00101 00101 00101 01010

and the key is : 10101011

hope that helps

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