Example
In this simple example we use the English alphabet of 26 characters from a-z. Thus we can not encrypt numbers, commas, spaces and other symbols. The random numbers in the keystream then have to be at least between 0 and 25.
To encrypt we add the keystream numbers to the plaintext. And to decrypt we subtract the same keystream numbers from the ciphertext to get the plaintext.
If a ciphertext number becomes larger than 25 we wrap it to a value between 0-25. Thus 26 becomes 0 and 27 becomes 1 and so on. (Such wrapping is called modular arithmetic.)
Here the plaintext message "attack at dawn" is combined by addition with the keystream "kjcngmlhylyu" and produces the ciphertext "kcvniwlabluh".
| Plaintext | a | t | t | a | c | k | a | t | d | a | w | n |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Plaintext as numbers | 0 | 19 | 19 | 0 | 2 | 10 | 0 | 19 | 3 | 0 | 22 | 13 |
| Keystream | k | j | c | n | g | m | l | h | y | l | y | u |
| Keystream as numbers | 10 | 9 | 2 | 13 | 6 | 12 | 11 | 7 | 24 | 11 | 24 | 20 |
| Ciphertext as numbers | 10 | 28 | 21 | 13 | 8 | 22 | 11 | 26 | 27 | 11 | 46 | 33 |
| Ciphertext as numbers wrapped to 0-25 |
10 | 2 | 21 | 13 | 8 | 22 | 11 | 0 | 1 | 11 | 20 | 7 |
| Ciphertext as text | k | c | v | n | i | w | l | a | b | l | u | h |
Read more about this topic: Keystream
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