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|reply to norwegian |
Re: Can anyone crack pigeon's wartime code?
said by norwegian:Think of it as representing each plaintext message letter by a number (for its alphabet position), then add a random digit to it (a digit copied from the corresponding position in the key string). Their sum would represent the alphabetic positional value of the coded letter to be sent. eg: a plaintext "b" (=2) in the first character position of the message would get added to, say, an "e" (=5) or even the actual digit "5" in the first character position of the key string... the sum is 7, which corresponds in the alphabet to "g" which is what gets sent. If the sum exceeds 26 (the max number of letters in the alphabet, the count simply starts over so that a 29 corresponds to a 3 (or a "c"). The recipient simply reverses the process, subtracting his identical key's character value for the first key string position from the value of the sent-message first-character position to recover the correct plaintext message value.
So I gather using a random cipher stops the patterns of the English language from forming?
Such as 'E' is the most common letter.
'Q' has to have a 'U' follow it, etc.
Little subtleties like that to give away the form of the message? ...
By using truly random key strings from the one-time pad, the message sent is randomized beyond unwanted recovery, provided that the key is only ever used once and that the message is kept short enough. To assure shortness, the key page is limited in length. Each message sent uses a different one-time pad, so a recoverable pattern never emerges. The "pad" term derives from the original practice of providing key strings (one to a page) in a pad of pages, tearing off, using and then destroying them one at a time as messages are sent.
One time pads are a pain to use unless employing computers/machines to assist - but those create undesirable trails, so one-times are usually reserved for very critical or crisis messaging.
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