# Brute-force attacks practical limit

My question is about practical limit for brute force attacks. As I know 3DES with 56 bits key length can be broken via brute force. I also heard the same news about 64 bit key length (correct me if I am wrong).

My question is about the minimal length of the key that can be considered as a non breakable by classical computers (not quantum ones). I have 90 bits in my mind but unfortunately don't remember the reasons for the number. I.e. it's impossible to perform $2^{90}$ operations on a classical computer in a reasonable time. Is it correct? Could somebody provide me explanations for the number (90) or provide another answer.

• The current hash rate of Bitcoin network is 2^73 hashes per second, which is approximately 2^89 hashes per day, can it be assumed that such effort for brute forcing an AES key is also possible? – khan Sep 12 '17 at 7:47
• @Resa I think that it can be considered as a possible effort for a brute forcing an AES key. As result the suggested number $90$ is incorrect. But I know that 128 bits length is considered as safe (if we don't take into consideration quantum computers). I.e. the requested number is between 90 and 128 – Ivan Sep 12 '17 at 7:55
• @Raza: you're off by a thousand; bitcoin is currently 8e18 ~ 2^63 /sec. If an AES-128 trial costs about the same as a bitcoin hash, then the resources used for bitcoin (thus demonstrated to be feasible) would brute-force AES-128 in about 500 billion years on average, except our Earth and Sun won't last nearly that long. – dave_thompson_085 Sep 13 '17 at 0:12
• yeah you are right bitcoin hashrate is ~ 2^63/sec. I calculated it wrongly and took Tera as 2^50 instead of 2^40 which led the calculation to a difference of 2^10. – khan Sep 13 '17 at 10:14