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Why do we say shift cipher is perfectly secure when it is easy to break it (source)? Let's say I have a plaintext. "Australia is a big country"; I encrypt it using a shift of 2; That ciphertext can be broken using brute force search over 26 numbers.

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    $\begingroup$ Could you please provide a link to a claim where it says that a shift cipher is perfectly secure? I think you're mistaking the simple caesar cipher with the secure one-time pad. $\endgroup$ – AleksanderRas Jan 28 '19 at 15:47
  • $\begingroup$ it is given in the book "Cryptography theory and practice" by douglas stinson .it is here."theswissbay.ch/pdf/Gentoomen%20Library/Cryptography/…" page no -75, theorem 2.3. $\endgroup$ – Manoharsinh Rana Jan 28 '19 at 15:54
  • $\begingroup$ It appears that you are confused about the message-space of the "ShiftCipher" discussed in the source. It is stated quite clearly that the message is an element of $\mathbb{Z}_{26}$. "Australia is a big country" is in no way an element of $\mathbb{Z}_{26}$. $\endgroup$ – Maeher Jan 28 '19 at 16:28
  • $\begingroup$ because of space? if i remove spaces from the message then also I can break it. I will have "Australiaisabigcountry". which is easy to understand. $\endgroup$ – Manoharsinh Rana Jan 28 '19 at 16:30
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    $\begingroup$ No, $\mathbb{Z}_{26}$ are the integers $\{0,\dots,25\}$. Those can be thought of as an encoding of the 26 letters in the English alphabet if you like. But a message can consist of only a single letter. $\endgroup$ – Maeher Jan 28 '19 at 16:43
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What you're referring to with "using a shift of 2" is in fact called a Caesar cipher and is not secure. You correctly pointed out, that it can easily be broken if you brute force the 26 possibilities.

But what the author is referring to is a so-called one-time pad (as described on page 78). It has the property of information-theoretic security.

The caesar-cipher can easily be brute forced to the original message, but on the other hand, OTP can be decrypted to any message of the same length. If we don't know any information about the message space then all keys are probable.

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  • $\begingroup$ I am talking about theorem 2.3 on page no 75.which states that Caesar cipher is perfectrly secret. $\endgroup$ – Manoharsinh Rana Jan 28 '19 at 16:17
  • $\begingroup$ Suppose the 26 keys in the Shift Cipher are used with equal probability 1/26. Then for any plaintext probability distribution, the Shift Cipher has perfect secrecy. That's exactly what a one-time pad does. For a caesar-cipher all letters are shifted by the same key, i.e. 2. $\endgroup$ – AleksanderRas Jan 28 '19 at 16:23
  • $\begingroup$ if I use a random key for each plain text, then also I can break each time. $\endgroup$ – Manoharsinh Rana Jan 28 '19 at 16:29
  • $\begingroup$ If you use it for the whole plain-text then yes (Caesar cipher). If you use for each letter in the plain text a random key then it's impossible, have a look at this example. $\endgroup$ – AleksanderRas Jan 28 '19 at 16:33
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    $\begingroup$ I could say like: ... hand, OTP can be decrypted to any message of the same length. If we don't know any information about the message space then all keys are probable. $\endgroup$ – kelalaka Jan 28 '19 at 18:09

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