# My Hand Cipher, Can some one tell if it secure [closed]

In easy steps how it works,

1. Convert txt to numbers . mod 1-26.
2. Generate random numbers (By my other cipher) equal to plain txt.
3. Write random numbers under txt numbers.

txt= LOI - txt number= 12 15 09 - random number = 73 96 14

encryption system, random numbers under txt numbers

12 15 09 --> txt number

73 96 14 --> random number

85 01 13 --> encrypted number

• 1+7=8 , 2+3=5 , 1+9=0 not '10' if the result two digit take the 2nd digit 5+6=1 , and so on.

• If the result two digit '10,11,12,13,...,19' take the second digit '0,1,...,9'

• Last step will convert the numbers to letter and numbers.

8 5 0 11 3 --> HE0KC or 8 5 0 1 1 3 --> HE0AAC

So Is it secure or not>

• Revers the steps to get the plain text, If the encrypted number smaller than the random number that mean we take the second digit 5+6=1 Reverse 1-6=5 because 5+6=11 2nd digit =1.
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 Are your random numbers uniformly distributed between 00 and 99? In that case this part of the system should be secure but inefficient (ciphertext is twice the size of the plaintext). Simple modular addition is better. Essentially an inferior method for combining a keypad with a message, similar to stream ciphers/one-time-pads. – CodesInChaos Nov 29 '12 at 9:57 Your real problem is "By my other cipher", because I'm 99% sure that the other cipher sucks, and obviously you need to generate a new pad for each message. Like with most stream ciphers, pad reuse is absolutely fatal. – CodesInChaos Nov 29 '12 at 9:59 Please have a look at the FAQ. There is a specific part that states that your type of question is offtopic. Thus I will have to close it. Do not let this discourage you, however. We love good, on topic questions here. – mikeazo♦ Nov 29 '12 at 12:21

## closed as off topic by CodesInChaos, Ilmari Karonen, fgrieu, Thomas, mikeazo♦Nov 29 '12 at 12:29

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The description isn't very clear but from what I understand it's a one time pad with a weird encoding of your input : If you take and encoded input you know that every even position is either $0$, $1$ or $2$ and if it is $2$ then the subsequent digit is between $0$ and $5$.