hammingweight

Hamming weight for a list of integers in Matlab [duplicate]

一世执手 提交于 2019-12-08 06:16:53
问题 This question already has answers here : Calculating Hamming weight efficiently in matlab (9 answers) Closed 6 years ago . Quite a simple problem: I have a list of integers, e.g., a = [7 8] Now I want to have a seperate list, that contains the Hamming Weight (that is the number of 1 bits in the binary represenation) for each of the integers in the list. That means the result for the integer list above should look as follows: res = [3 1] Anyone an idea how I could do this quickly? 回答1: This is

calculate hamming distance and weight in sqlite

你离开我真会死。 提交于 2019-12-08 03:29:56
问题 Is there a good way to calculate hamming distance and weight in sqlite? It supports bit-wise operators but I want to order results based on hamming weight, and there is no support for bitcount in sqlite. To be more elaborate, let's say I have those rows: 1011 1000 1100 0011 and given the 1st row (1011) I would like to get as a result last row (0011) which has the most 1s if you AND them. In my case the numbers will be about 650 digits long, and I have about 3500 rows. I've found this solution

Hamming weight for a list of integers in Matlab [duplicate]

情到浓时终转凉″ 提交于 2019-12-07 15:19:25
This question already has answers here : Calculating Hamming weight efficiently in matlab (9 answers) Closed 6 years ago . Quite a simple problem: I have a list of integers, e.g., a = [7 8] Now I want to have a seperate list, that contains the Hamming Weight (that is the number of 1 bits in the binary represenation) for each of the integers in the list. That means the result for the integer list above should look as follows: res = [3 1] Anyone an idea how I could do this quickly? This is a little hacky, but it works: res = sum( dec2bin(a).' == '1' ); It converts a to binary representation,

Find next number with specific hamming weight

只谈情不闲聊 提交于 2019-12-02 03:39:54
问题 Given a certain integer x , I wish to compute the next higher integer y which has a certain hamming weight w . Mind that the hamming weight of x does not have to be w as well. So, for example x = 10 (1010) and w = 4 the result should be y = 15 (1111). Obviously, I could achieve this by just incrementing x but that would be a very slow solution for high numbers. Can I achieve this by the means of bit shifts somehow? 回答1: There are three cases: the Hamming weight (a.k.a. bitwise population

Find next number with specific hamming weight

 ̄綄美尐妖づ 提交于 2019-12-02 03:12:22
Given a certain integer x , I wish to compute the next higher integer y which has a certain hamming weight w . Mind that the hamming weight of x does not have to be w as well. So, for example x = 10 (1010) and w = 4 the result should be y = 15 (1111). Obviously, I could achieve this by just incrementing x but that would be a very slow solution for high numbers. Can I achieve this by the means of bit shifts somehow? There are three cases: the Hamming weight (a.k.a. bitwise population count) is reduced, unchanged, or increased. Increased Hamming weight Set enough successive low-order 0-bits to

How does this algorithm to count the number of set bits in a 32-bit integer work?

感情迁移 提交于 2019-11-30 10:24:47
问题 int SWAR(unsigned int i) { i = i - ((i >> 1) & 0x55555555); i = (i & 0x33333333) + ((i >> 2) & 0x33333333); return (((i + (i >> 4)) & 0x0F0F0F0F) * 0x01010101) >> 24; } I have seen this code that counts the number of bits equals to 1 in 32-bit integer, and I noticed that its performance is better than __builtin_popcount but I can't understand the way it works. Can someone give a detailed explanation of how this code works? 回答1: OK, let's go through the code line by line: Line 1: i = i - ((i >

C code to count the number of '1' bits in an unsigned char

妖精的绣舞 提交于 2019-11-29 10:08:58
I need C code to return the number of 1's in an unsigned char in C. I need an explanation as to why it works if it's not obvious. I've found a lot of code for a 32-bit number but not much for an unsigned char. The same code will work for an unsigned char. Loop over all bits testing them. See this . const unsigned char oneBits[] = {0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4}; unsigned char CountOnes(unsigned char x) { unsigned char results; results = oneBits[x&0x0f]; results += oneBits[x>>4]; return results } Have an array that knows the number of bits for 0 through 15. Add the results for each nibble.

Hamming weight/population count in T-SQL

我们两清 提交于 2019-11-28 14:16:03
I'm looking for a fast way to calculate the hamming weight/population count/"the number of 1 bits" of a BINARY(1024) field. MySQL has a BIT_COUNT function that does something like that. I couldn't find a similar function in T-SQL? Or would you suggest storing the binary data in a field of another type? If you don't know what I'm talking about, here's a Wikipedia article about the hamming weight . You could use a helper table with precalculated Hamming weights for small numbers, like bytes, then split the value accordingly, join to the helper table and get the sum of partial Hamming weights as

Calculating Hamming weight efficiently in matlab

倖福魔咒の 提交于 2019-11-28 08:25:39
Given a MATLAB uint32 to be interpreted as a bit string, what is an efficient and concise way of counting how many nonzero bits are in the string? I have a working, naive approach which loops over the bits, but that's too slow for my needs. (A C++ implementation using std::bitset count() runs almost instantly). I've found a pretty nice page listing various bit counting techniques, but I'm hoping there is an easy MATLAB-esque way. http://graphics.stanford.edu/~seander/bithacks.html#CountBitsSetNaive Update #1 Just implemented the Brian Kernighan algorithm as follows: w = 0; while ( bits > 0 )

Hamming weight based indexing

孤街醉人 提交于 2019-11-28 07:42:33
Assume we have a integer of bitsize n=4; The problem I am describing is how you would go about indexing a number to an array position based on the Hamming weight and its value knowing the bitsize . E.g. An array with 16 elements for bitsize 4 would/could look like this: |0|1|2|4|8|3|5|6|9|10|12|7|11|13|14|15| Where elements are grouped by their Hamming weight(necessary) and sorted based on size(not necessary). Sorting is not necessary as long as you can take e.g. 3(0011) do some operations and get back index 5, 5(0101) -> 6 etc. All combinations of n bits will be present and there will be no