I am very new to cryptography but can someone explain what exactly does 0...0 and 1...1 mean in general term. Is it that 1st and last bit can be 0,0 and `1,1 respectively and in-between bits can be anything $(0,1)$?

In case it helps… I stumbled upon this while trying to solve the following multiple choice question:

Let F be a block cipher with n-bit block length. Consider the message authentication code for 2n-bit messages defined by Mack(m1,m2)=Fk(m1⊕m2). Which of the following gives a valid attack on this scheme?

  • Obtain tag t on message m,0…0 (with m≠0…0), and then output the tag t on the message 0…0,0…0.
  • Obtain tag t on message m1,m2 (with m1≠m2), and then output the tag t on the message m2,m1.
  • Obtain tag t on message m,0…,0, and then output the tag t⊕(1…1) on the message m,1…1.
  • Obtain tag t on message m,m, and then output the tag 0…0 on the message 0…0,m.

Please note that I do not want an answer to the multiple choice question. I merely want to know how to interpret the notation 0...0 and 1...1.


1 Answer 1


What exactly does $0...0$ and $1...1$ mean usually?

This simply means a (more or less) long string of $0$s or $1$s or more clearly $000000...000000$ and $111111...111111$.

Related notiational notes, you may have to use soon:

Sometimes the notation $0^n$ and $1^n$ is also used for these strings with exactly $n$ zeroes and ones.

Even more generally it's usually expected that people see the pattern, namely that nothing changes (or else there would be a note about this), for example $\lambda_0,...,\lambda_n$ is expected to be filled with $\lambda_i$ with $i\in \mathbb N_0, 0\leq i\leq n$ or relatedly $(\alpha,...,\alpha)\in \mathbb R^n, \alpha \in \mathbb R$ would be the same principle, that you'd have to fill the blanks with as many $\alpha$s as needed.

One last note: If you'd want to indicate that you want a leading string $0$ and a trailing string $0$ and stuff you don't care about in the middle you'd usually notate it as $n\in\{0,1\}^*$ and $0||n||0$ which means "place an arbitrary string of arbitrary length between the zeroes".

  • 1
    $\begingroup$ Before anybody asks: No, there's no special reason why I used examples from standard math / linear algebra in the fourth paragraph. $\endgroup$
    – SEJPM
    May 21, 2016 at 18:56

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