I have been studying Shannon's proof of Vernams (1917) unbreakable cipher. Can His proof be applied to positional ciphers?

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    $\begingroup$ Schneier's law: "Anyone, from the most clueless amateur to the best cryptographer, can create an algorithm that he himself can't break." Your scheme is not perfectly secure. $\endgroup$ – Maarten Bodewes Jul 11 '19 at 10:01
  • $\begingroup$ The one of many schemes shown in the video, or positional ciphers as a category? $\endgroup$ – Jon Hutton Jul 11 '19 at 18:49
  • $\begingroup$ I was just stating this because you seem fully convinced that you've created a secure cipher, while there is only you that is stating this - and there seems to be an issue even categorizing the algorithm. Besides that, there are easy to see problems, like a plaintext that is just consisting of the letter A, which will very surely not result in a ciphertext that is indistinguishable from random. $\endgroup$ – Maarten Bodewes Jul 11 '19 at 20:18
  • $\begingroup$ I am quite confident it is secure. It uses a one or more one time pad keys, The keys can change in any random order with triggers in the plaintext. The keys themselves can contain digraph, trigraps...etc. The actual algorythim to decode does not decode based on the number but based on the position that number represents in the plaintext. The type of cipher is a new category not found in any of the ones referenced here www.cryptogram.org/resource-area/cipher-types/. It is easily decoded. I do not understand what you meant by "like a plaintext that is just consisting of the letter A, ......" $\endgroup$ – Jon Hutton Jul 12 '19 at 1:02
  • $\begingroup$ I understand that you don't understand. You've just created a very simple cipher and claim it to be unbreakable without any scientific proof or rigor. And I fully expect you to be defending it at all cost, against all reason. $\endgroup$ – Maarten Bodewes Jul 12 '19 at 1:11

Can his proof be applied to positional ciphers.

Shannon's proof requires that there be at least as many possible keys as possible plaintexts; unless your positional cipher has that property, then no, it cannot.

(I'm not sure what you're referring to as a 'position cipher' - would that be a transposition cipher?)

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  • $\begingroup$ I'm finding that there are examples in certain new methods that simply do not fit his proof. An example would be a positional cipher (new) and running key cipher. The latter can be successfully attacked (Wikipedia). The positional cipher is similar except the key is made up of the positions of the letters when you combine the key with the plain text, so they are always in a different place. The key length is a total of the key plus the plain text. The cipher text depending on how the key is constructed can be shorter or longer than the plain text. $\endgroup$ – Jon Hutton Jul 6 '19 at 3:06
  • $\begingroup$ "The key length is a total of the key plus the plain text" - the plaintext cannot be counted as key (and most certainly not in Shannon's proof) $\endgroup$ – poncho Jul 6 '19 at 12:58
  • $\begingroup$ So if pr[C=c|M=m]=1/n the number of keys = the number of messages then a running key cipher meets this yet can be attacked. So does the running key cipher meet this or not...what am i missing $\endgroup$ – Jon Hutton Jul 6 '19 at 21:51
  • $\begingroup$ @JonHutton: what is a "running key cipher"? $\endgroup$ – poncho Jul 7 '19 at 2:29
  • $\begingroup$ A running key cipher uses a phrase, or text from a book truncated to the length of the plaintext. $\endgroup$ – Jon Hutton Jul 7 '19 at 13:34

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