Problem 808. Hamming Weight - Fast

Created by Richard Zapor

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The Hamming Weight, <a href="http://en.wikipedia.org/wiki/Hamming_weight">wiki Hamming Weight</a>, in its most simple form is the number of ones in the binary representation of a value.The task here is to create a fast Hamming Weight function/method such that processing many 4K x 4K images of 32 bit integer can be evaluated rapidly. Saw this question posed on Stack Overflow.Input: Vector of length N of 32 bit integer values.Output: Vector of number of ones of the binary representationScoring: Time in milliseconds to process a [4096*4096,1] vectorExamples: Input [7 ; 3], output=[3;2]; [16 32], output [1;1]; [0 4294967295] output [0;32]Timing Test vector: uint32(randi(2^32,[4096*4096,1])-1)Minimum vector length/increment: 65536Helpful, possibly, global variables.b1=uint32(1431655765); b2=uint32(858993459); b3=uint32(252645135) b4=uint32(16711935); b5=uint32(65535);Hex: b1=55555555 b2=33333333 b3=0F0F0F0F b4=00FF00FF b5=0000FFFFThe array num_ones is created for values 0-65535 (0:2^16-1).
num_ones(1)=0, num_ones(2)=1, num_ones(3)=1,num_ones(4)=2,...num_ones(65536)=15Due to lack of zero indexing num_ones(value+1) is number of ones for value.Hint: Globals are not good for time performance.Hint: Segmentation appears to provide significant time optimization potential.

The Hamming Weight, <http://en.wikipedia.org/wiki/Hamming_weight wiki Hamming Weight>, in its most simple form is the number of ones in the binary representation of a value.

The task here is to create a fast Hamming Weight function/method such that processing many 4K x 4K images of 32 bit integer can be evaluated rapidly. Saw this question posed on Stack Overflow.

*Input:* Vector of length N of 32 bit integer values.

*Output:* Vector of number of ones of the binary representation

*Scoring:* Time in milliseconds to process a [4096*4096,1] vector

*Examples:* Input [7 ; 3], output=[3;2];  [16 32], output [1;1]; [0 4294967295] output [0;32]

*Timing Test vector:* uint32(randi(2^32,[4096*4096,1])-1)

*Minimum vector length/increment:* 65536

Helpful, possibly, global variables.

b1=uint32(1431655765); b2=uint32(858993459); b3=uint32(252645135) b4=uint32(16711935); b5=uint32(65535);

Hex: b1=55555555 b2=33333333 b3=0F0F0F0F b4=00FF00FF b5=0000FFFF

The array num_ones is created for values 0-65535 (0:2^16-1).
num_ones(1)=0, num_ones(2)=1, num_ones(3)=1,num_ones(4)=2,...num_ones(65536)=15

Due to lack of zero indexing num_ones(value+1) is number of ones for value.

Hint: Globals are not good for time performance.

Hint: Segmentation appears to provide significant time optimization potential.

Solve

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60 Solutions
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Last Solution submitted on Mar 06, 2019

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Solution 1319034

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1 Comment

Shlomo Geva on 29 Oct 2017

1. This solution is much faster on re-invocation than the one without the persistent num_ones variable. Unless of course it is performed on a much larger (than num_ones) array of 32-bit integer.

2. It is essential to have the statement
x= double(x);
The reason for this is that the floor() function has a problem with precision. If can fail with 32-bit integer that are close to 2^32.
For instance, consider this Matlab code and system response:

>> p=uint32(4294946031)

p =

uint32

4294946031

>> floor(p/65536)

ans =

uint32

65536

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bit manipulation bitand bitshift compression floor mod uint32

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Problem 808. Hamming Weight - Fast

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Problem 808. Hamming Weight - Fast

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