Minimum password length for maximum entropy

Question

Assuming a SHA 256 hash and a completely random password using extended ASCII encoding, is there a certain length after which additional characters do not offer an increase in entropy, and if so, what is it?

Thanks.

Answer 1

SHA-256 has 256 bits, obviously. The minimum character length of UTF-8 is one byte, i.e. 8 bit Thus, any password longer than 256/8 = 32 characters is guaranteed , most likely, it will come across a shorter one.

Is that what you meant?

Answer 2

A hash does not increase entropy; it simply, so to speak, distills it. Since SHA256 outputs 256 bits of output, if you provide it with a password that is completely unpredictable (i.e., each input bit represents one bit of entropy), then more than 256 input bits are more or less lost.

However, than from a truly random source, it is very difficult to get an input that has one entropy bit for each input bit. For a typical English text, Shannon's testing showed about one bit of entropy per character.

Answer 3

I came to about the same conclusion as the others, but with a different justification.

Generally speaking, a prototype (brute force) attack on SHA-256 requires 2 ^ 256 ratings regardless of the password length. In other words, a password hash of thousands of characters will still take an average of 2 ^ 256 duplication attempts. 2 ^ 256 is about 1.2 × 10 ~ 77. However, a very short password, where the number of possibilities is less than 2 ^ 256, is even easier to break.

The threshold is transmitted when the number of possibilities is greater than 2 ^ 256.

If you use ISO 8859-1, which has 191 characters, there are 191 ^ n possible random passwords of length n, where n is the length of the password. 191 ^ 33 is about 1.9 × 10 ~ 75 and 191 ^ 34 is about 3.6 × 10 77, so the threshold would be 33 characters .

If you used plain ASCII with 128 characters, there would be 128 possible possible random passwords of length n, where n is the length of the password. 128 ^ 36 is about 7.2 x 10 ^ 75 and 128 ^ 37 is about 9.3 x 10 77, so the threshold will be 36 characters .

Some of the other answers seem to imply that the threshold always has 32 characters. However, if my logic is correct, the threshold changes, depending on the number of characters that you have in the character set .

Actually, suppose you used only the characters az and 0-9, you continue to add the strength of the password until your password is 49 characters long! (36 ^ 49 is about 1.8 × 10 ~ 76)

I hope this answer gives you a mathematical basis for answering the question.

As a side note, if a birthday (collision) attack were possible on the SHA-256, theoretically only 2 ^ 128 ratings (average) would be required, which is about 3.4 × 10 38. In this case, the threshold value for ISO 8859-1 will have a value of 16 characters (191 ^ 16 is about 3.1 × 10 ~ 36). Fortunately, such an attack has not yet been publicly demonstrated.

Please view SHA-2 Wikipedia articles, prefix attacks and birthday attacks .

Answer 4

I do not think there is an “effective” limit. A password of any length will be effective if it is effectively created (ordinary rules, no words, mixed numbers, letters, cases and characters). It is better to force the user to follow these rules, and not limit the length. But the minimum length should be superimposed, like 8-10 characters, to save users from themselves.