Question

How to calculate the number of attempts necessary to break a password that uses secure hash functions?

Example: How many attempts would it take to break an 8 character (only lower case) long password stored as a 32 bit long hash value?

26^8 or would it be more like 2^26 * 8 * 2^32 attempts?

Answer 1

So to understand this let us build from smaller number of characters

Take 1 character for example .number of attempts necessary?

So obviously answer is 26 as there are 26 different lower case alphabets.

Now take 2 characters.number of attempts necessary?

In this case if the first character is 'a' then second character cab be any one out of 'a' to 'z' So 26 combinations with 'a' as first character

Similarly if first character is 'b' then swcond character be again any one out of 26 character so 26 combinations with 'b' as first character

This is true for all 26 characters

So 26 is added 26 times

that is 26 * 26

Anothere way to understand is first character has 26 options. second option has 26 options and they are not dependent on each other

so 26*26

So for 3 characters

Ans is 26*26*26

For 4 characters ans is 26*26*26*26 = 26^4

So for n character ans is 26^n

Heres something general answer for you

Let there be p options for each character instead of 26 and length of password is 'n'

So ans will be p^n

Please comment for any further assistance