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If we have data of 2 Tb, how to express this in 2 to the power of x? (to the closest one)
The answer needs to be a number in decimal. I feel that it is 40.99 but turns out to be wrong. I'm wondering how to solve this problem. Thanks!
The Metric System prefixes all represent powers of ten, and “tera” means 1012. So two terabytes is 2,000,000,000,000 bytes, not 2,199 billion bytes.
There are binary prefixes for powers of two. These replace the second syllable of the Metric System prefixes with “bi,” for “binary”. A tebibyte is 10244 = 240 bytes, and two tebibytes is 241 bytes = 2,199,023,255,552 bytes.
Given a number x, you can calculate what power of two it is as log2 x, that is, the base-two logarithm of x. If you do not have a base-two logarithm function, you can calculate it as log x / log 2, where “log” is a logarithm function that uses any base. (Dividing by log 2 effectively adjusts the base.)
log2 2,199,000,000,000 is about 40.9999847, and that is near 41. So, if a number of bytes is near 2,199 billion bytes, it is near 241 bytes.
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I also noticed that multiples of 0.0625 such as 1.25 keep giving zero but I was unable to find an explanation
as per CostantinoGrana
"Do you know how IEEE 754 works? The numbers which you say "give 0"
are multiples of a not too negative negative power of 2. So when you
store them in little endian, the 32 less significant bits, that are
all 0, are the first thing you find in memory as int. Your int is 32
bits wide, thus 0."
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What is the best way to count the number of 1's in an 32-bit integer x in C without using for or while loops, and without using constants greater than 0xFF?
What I thought of is shifting x 24 to the right and count how many 1's in the shifted integer and store that in a variable count. And then, shifting x 16 to the right and increment count by the number of 1's in the shifted integer, and so on.
So, any ideas of a better solution?
You can tabulate the number of 1's in all d bits numbers. This takes a table of 2^d entries, each not exceeding the value d (<255).
Now you can cut your number in slices of d bits and lookup the counts for all slices.
A good compromise between space/number of operations is probably with d=4 (8 slices, table size=16).
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Is there a generic way to represent a number containing n bits using n-1 bits; e.g. showing 1001 which has 4 bits using XXX where X = {0|1} with 3 bits. Also, the mapping should be able to retrieve the original binary back without any collision.
This page is the most relevant reference that I have found so far which tries to count the number of bits, but not reduce the number of bits.
Edit: I knew this sounds impossible, but I'm curious if there exists a workaround to do so!
There are 2^n possible values with n bits, and 2^(n-1) with (n-1) bits. So you can't convert from the former to the latter losslessly.
If it were at all possible, you could also recursively represent your n-1 bits using n-2 bits, etc. Everything would be representable with 0 bits :)
You get misleaded by the page you linked, which explains that x &= x-1 removes a 1 in a bit string.
10100
& 10011
= 10000
You will be very clever if you manage to establish a one-to-one mapping between the 16 numbers
0000,0001,0010,0011,0100,0101,0110,0111,1000,1001,1010,1011,1100,1101,1110,1111
and the eight numbers
000,001,010,011,100,101,110,111.
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I am looking for fast check sum algorithm that produce 1 byte checksum.
I checked CRC8 and Adler8, but I do not fully understand the samples.
Also different CRC8 implementations give different results.
In all cases I do not need anything that fancy.
CRC's are based on a type of finite field math, using polynomials with 1 bit coefficients (math modulo 2). An 8 bit CRC is the result of treating data as a very long polynomial dividend with 1 bit coefficients and dividing it by a 9 bit polynomial divisor, which produces an 8 bit remainder. Since 1 bit coefficients are used, add or subtract effectively become exclusive or. You don't really need to understand finite field math to implement a CRC, just use a lookup table or use an algorithm to generate the CRC.
You could just add up all the bytes into a 1 byte sum, and use that as a checksum. The advantage of a CRC is that if bytes are missing or out of order, it has a better chance of detecting that.
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how to find all prime numbers between 1 and 10^9 , i know we can use Sieve_of_Eratosthenes for smaller range, but what when range is too large equivalent to 10^6 ?
Up to 10^9 is not really a big deal. First, only look at odd numbers (because there is only one even prime). Second, use a bit array, so you only need 500 million bits or about 62 Megabyte. Even straightforward code should do that in a few seconds at most.
If you go further, you'd do a sieve for numbers from 1 to 10^9, then from 10^9 + 1 to 2 * 10^9 and so on. Above 10^13 it gets interesting and you need to put a bit more effort into it.