A lot of the music I collect these days is coming from musicians who are looking for alternate ways of connecting with their audience. I really respect Amanda Palmer for what she has to say here, and giving some really intriguing glimpses at what a post-big-label musical landscape might look like.

# Counting 1 bits in log time

To inaugurate my latest attempt at a blog, I want to talk about something cool I’ve read about lately. I’ve received this question a couple times at interviews, and I think it was just to demonstrate that I could work with bitwise values.

For instance, take the number 57:

`57 = 00111001`

It’s easy enough to see that 57 has 4 bits set to 1. How do you check that in code? My answer has always been to loop through the number of bits (in this case 8, but usually 32), shifting to the right each time and then count how often the lowest bit comes up as a 1.

value = 57 count = 0 for i in xrange(8): count += (value >> i) & 1

Effectively what I’m doing here is just looping through the 8 bits, rotating the value to the right and stripping off the least significant bit and adding it to the total (if it’s 1, then it adds to the total).

Since I’m using python for the example, I can abbreviate it a bit like this:

count1s = lambda x: sum( (x >> i) & 1 for i in xrange(32) ) print count1s(57) > 4

That’s more compact, certainly, but it’s still linear time with respect to the number of bits that you are examining, and that’s where today’s coolness comes into play. It turns out there’s a *log* time algorithm for doing this.

Because it can get a bit long when I show all my work long-hand, let me jump straight to the code that will give the results for an 8 bit integer in 3 steps (and could be extended to do 32 bits in 5 steps, and so on):

x = 57 y = (x & 0x55) + ((x & 0xaa) >> 1) y = (y & 0x33) + ((y & 0xcc) >> 2) y = (y & 0x0f) + ((y & 0xf0) >> 4) print y > 4

Follow through the link if you want the nitty gritty.

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