Can you use repeating numbers like $\pi$, and $e$, as the Key to OTP ciphers?

Question

Since $\pi$ and other numbers repeat could you use them as a OTP key. Ex: A 5 character message is sent and the first 5 digits of $\pi$ are used. After that the next $X$ number of digits are used and so on. From what I understand of Cryptography (it's my hyper fixation for the week) and potential 3rd party readers would have to know the number of messages sent and their length to try and decode the messages. If I understand it, it would be near impossible to brute force and would require digital forensics (data remanence can be an issue). Obviously, the community knows more than me, any insight would be much appreciated.

Answer 1

An OTP is completely broken if you use a key that can be predicted. As such, $\pi$ would be a terrible choice. The key needs to be unpredictable, nonrepeating, and completely random. $\pi$ satisfies the nonrepeating aspect, and (looks like it) satisfies the randomness attribute, but is predictable since I can simply search online for 1 million digits of $\pi$.

To clarify, any number people have actually heard of is a bad choice. $\pi$, $e$, the coefficient of rolling friction across a carpet, are all bad choices. You need to use a number that has never occurred before, and never will; aka randomly generated by you, not taken from a mathematical constant.

Answer 2

In modern cryptography, we work with the Kerckhoffs's Principles, in which we consider everything public but the key.

In your encryption scheme you consider the secret as the number of messages and the message length. Those are not secret for a constant observer of your system and hiding the length of a message is not an easy and most of the time we consider that the an observer has this knowledge.

The digits/bits of $\pi$ are not secret. Everybody can calculate any base16 digits of $\pi$ without calculating the previous ones by Bailey–Borwein–Plouffe formula, $$ \pi = \sum_{i=0}^{\infty} \frac1{16^i}\left( \frac{4}{8i+1}-\frac{2}{8i+4}-\frac{1}{8i+5}-\frac{1}{8i+6}\right),$$ this produces the digit in base16. So you and your attacker in a race to calculate the digits. It is like there is no secret at all.

In an actual protocol you need to send the position, too. This is where such a system will easily collapse.

In modern cryptography, we achieve similar to this with stream ciphers. The key and the nonce points a position on the stream of the cipher. The key is secret and the nonce must be used only once with the same key.

The attacker either can try to bruteforce the cryptosystem or try to cryptanalyze it. We prevent the bruteforce by increasing the key size over 128-bit. To countermeasure against the cryptanalyze is requires knowledge and experience.

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And a side not, the $\pi$ sequence is not even good a random stream it is predictable. A nice property of the sequence of $\pi$ is that contains full of pattern if you run enough (this is called normal ^*). This doesn't mean that it is periodic like LFSR, otherwise it won't be a transcendental number.

An OTP key must be truly random and, ofcource, unpredictable.

$\pi$, $e$, and other constants (transcendental) can have similar issues with $\pi$

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_{^* Normal: A number is called "normal" with respect to a given base if, when the number is expressed in that base, the asymptotic frequencies of occurrence of each distinct string of $k$ digits are equal, and this applies to every positive integer $k$. $\pi$ and $e$ are believed to be normal}

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• This Math.SE answers to Does $\pi$ contain all possible number combinations? have good answers about the possibility of the strings.