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While developing a brute force based cracking algorithm for unix password I met with this line: "salt is a two-character string chosen from the set [a-zA-Z0-9./]. This string is used to perturb the algorithm in one of 4096 different ways."
I did not get 4096 ways?? 2^12 but how??
The set [a-zA-Z0-9./] is made up of 64 characters. Selecting two characters each from the full 64 character set gives 64 * 64 possibilities, or 4096.
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I want to understand the physical significance of n raise to some decimal power.
Like when i say 2^5. I understand that it means 2 multiplied 5 times. But how do i analyse 2^0.1.
Please suggest.
2^0.1 is the tenth root of 2. For rational powers, x^(p/q)=(x^p)^(1/q) is a combination of powers and roots.
For general real numbers,
x^y = exp(log(x)*y).
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I am trying to find out how many combinations are possible for 000 through to FFF under hex format? thanks
There are 16 choices for each of the 3 digits, so:
16*16*16 = 4096
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I have a random number sequence (say 6 bytes)
I now want to generate a shorter sequence from the original sequence (say 3 bytes)
What is the best way of acheiving this so that the randomness of the original sequence is preserved.
Lets say I run a SHA-1 hash code on the original sequence and then grab some bytes from the hashed output. Does the randomness decrease, increase, or stay the same.
The basic question is - Does a Hash Code of a random number produce less random, more random or same randomness.
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I try to calculate, based on given parameters, integer overflow.
for example, if I have an integer than is <= 200, but when I insert it to an unsigned int, it will be > 200. What is the actual arithmetic process for that?
Operations on fixed size integers are usually made modulo 2m, where m is the number of bits (nowadays usually 32 or 64).
This means that a multiple of 2m is added or subtracted from the result to keep it in the range for the type, be it unsigned (0, 2m-1) or signed (-2m-1, 2m-1-1).
You might be interested in the Mathematical foundations of computer integers.
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I am learning binary division following the step-by-step procedure. I am stuck at Step One Repeat One in the remainder section. To get the remainder, the formula is Remainder = Remainder - Divisor.
How do we get to remainder: 11100110? I just don't understand how we have 1110 on the left half of the register. Any guides or help would greatly be appreciated!
Thanks:
The example is 10/5 using an 8 bit ALU.
I have attached a picture of the example for reference