Check if a numer is a power of 2 in ruby

~~The~~ One imo not-so-obvious solution is:

# 0 -> false, 1->true, 2->true, 3->false ...
n.to_s(2).scan(/1/).size == 1

And why it works is an exercise for the reader 😉

Update:
A crazy benchmark comparing different versions of power of 2 algorithms by reto
The fastest, still readable versions is:

n & (n – 1) == 0
... followed by ...
Math.log(n) / Math.log(2) % 1 == 0

5 thoughts on “Check if a numer is a power of 2 in ruby”

Reto says:

2010-03-08 at 8:26:23

arggh 🙂

Math.log(n)/Math.log(2) % 1 == 0

=> does not convert a number to a binary string representation
=> does not compile a regexp
=> does not create a array of 1’s
=> does not count them 🙂

probably something with bit arithmetic is still faster than the log 🙂

Reply
Reto says:

2010-03-08 at 8:29:10

or what about

n.to_s(2).match(/^10*$/)

🙂

Reply
eno says:

2010-03-21 at 23:44:56

I have something even faster (and are hellish proud of it):

def ispow2(n)
n & (n – 1) == 0
end

runs twice as fast as anything 😉

Reply
grosser says:

2010-03-22 at 6:06:33

Wow, actually readable and fast, its not very intuitive, but makes sense when you think about it 🙂

Reply
Gary says:

2013-05-07 at 21:10:07

(0..n).map {|i|n[i]}.inject(:+) == 1

That’s a somewhat obscure and fairly inefficient way to do it.

Reply

Leave a Reply Cancel reply

Enter your comment here...

Fill in your details below or click an icon to log in:

Email (required) (Address never made public)

Name (required)

Website

You are commenting using your WordPress.com account. ( Log Out / Change )

You are commenting using your Facebook account. ( Log Out / Change )

Connecting to %s

%d bloggers like this: