Velocity Reviews > Ruby > Hamming Distance

# Hamming Distance

Martin Hansen
Guest
Posts: n/a

 03-07-2011
What is the fastest way to calculate the Hamming Distance between two
equally long strings in Ruby? A have looked at the amatch gem, but I was
wondering what can be done with Ruby only?

Here is a very interesting discussion on the subject:

http://www.perlmonks.org/?node_id=500235

Note the elegant and extremely fast Perl solution:

sub hd {
return (\$_[0] ^ \$_[1]) =~ tr/\001-\255//;
}

Cheers,

Martin

--
Posted via http://www.ruby-forum.com/.

Kirk Haines
Guest
Posts: n/a

 03-07-2011
On Mon, Mar 7, 2011 at 8:47 AM, Martin Hansen <(E-Mail Removed)> wrote:
> What is the fastest way to calculate the Hamming Distance between two
> equally long strings in Ruby? A have looked at the amatch gem, but I was
> wondering what can be done with Ruby only?
>
> Here is a very interesting discussion on the subject:
>
> http://www.perlmonks.org/?node_id=3D500235
>
>
> Note the elegant and extremely fast Perl solution:
>
> sub hd {
> =A0 =A0return (\$_[0] ^ \$_[1]) =3D~ tr/\001-\255//;
> }

There are a bunch of ways. There isn't a built in xor of strings in
Ruby, but it's easy to define one. Once you have that, you can do it
pretty similarly to the Perl example:

def hd(l, r)
l.xor(r).tr("\x00",'').length
end

You can implement xor of two strings using pure ruby, but to get it
fast, you'll probably want to utilize an extension. Here's one
approach using a technique that, I believe, was in the mailing list
several years ago:

require "narray"

class String
def xor(other)
if other.empty?
self
else
left =3D self
right =3D other

if left.length < right.length
n,r =3D right.length.divmod(left.length)
left =3D left * n + left[0, r]
elsif right.length < left.length
n,r =3D left.length.divmod(right.length)
right =3D right * n + right[0, r]
end

left_na =3D NArray.to_na(left, "byte")
right_na =3D NArray.to_na(right, "byte")

(left_na ^ right_na).to_s
end
end

def hamming_distance(other)
self.xor(other).tr("\x00",'').length
end
end

Kirk Haines
Software Engineer
Engine Yard

Robert Klemme
Guest
Posts: n/a

 03-07-2011
On 07.03.2011 18:11, Kirk Haines wrote:
> On Mon, Mar 7, 2011 at 8:47 AM, Martin Hansen<(E-Mail Removed)> wrote:
>> What is the fastest way to calculate the Hamming Distance between two
>> equally long strings in Ruby? A have looked at the amatch gem, but I was
>> wondering what can be done with Ruby only?
>>
>> Here is a very interesting discussion on the subject:
>>
>> http://www.perlmonks.org/?node_id=500235
>>
>>
>> Note the elegant and extremely fast Perl solution:
>>
>> sub hd {
>> return (\$_[0] ^ \$_[1]) =~ tr/\001-\255//;
>> }

>
> There are a bunch of ways. There isn't a built in xor of strings in
> Ruby, but it's easy to define one. Once you have that, you can do it
> pretty similarly to the Perl example:
>
> def hd(l, r)
> l.xor(r).tr("\x00",'').length
> end
>
> You can implement xor of two strings using pure ruby, but to get it
> fast, you'll probably want to utilize an extension. Here's one
> approach using a technique that, I believe, was in the mailing list
> several years ago:
>
>
> require "narray"
>
> class String
> def xor(other)
> if other.empty?
> self
> else
> left = self
> right = other
>
> if left.length< right.length
> n,r = right.length.divmod(left.length)
> left = left * n + left[0, r]
> elsif right.length< left.length
> n,r = left.length.divmod(right.length)
> right = right * n + right[0, r]
> end
>
> left_na = NArray.to_na(left, "byte")
> right_na = NArray.to_na(right, "byte")
>
> (left_na ^ right_na).to_s
> end
> end
>
> def hamming_distance(other)
> self.xor(other).tr("\x00",'').length
> end
> end

I'll throw in two other solutions for whoever wants to do a benchmark.

class String
def hd_1(s)
c = 0

each_codepoint.zip(s.each_codepoint) do |l, r|
c += 1 unless l == r
end

c
end

def hd_2(s)
each_codepoint.zip(s.each_codepoint).select {|l, r| l != r}.length
end
end

Cheers

robert

--
remember.guy do |as, often| as.you_can - without end
http://blog.rubybestpractices.com/

Kirk Haines
Guest
Posts: n/a

 03-07-2011
On Mon, Mar 7, 2011 at 10:11 AM, Kirk Haines <(E-Mail Removed)> wrote:

> You can implement xor of two strings using pure ruby, but to get it
> fast, you'll probably want to utilize an extension. =A0Here's one

Actually, I should revise that statement. To get it really fast with
MRI, you'll probably want to utilize an extension. I am guessing, as
I haven't benchmarked anything, but one may well be able to get much
faster pure ruby speeds with jruby or rubinius. When in doubt,
benchmark it!

Kirk Haines

botp
Guest
Posts: n/a

 03-08-2011
On Tue, Mar 8, 2011 at 5:49 AM, Kirk Haines <(E-Mail Removed)> wrote:
> On Mon, Mar 7, 2011 at 10:11 AM, Kirk Haines <(E-Mail Removed)> wrote:
>> You can implement xor of two strings using pure ruby, but to get it
>> fast, you'll probably want to utilize an extension. =A0Here's one

> Actually, I should revise that statement. =A0To get it really fast with
> MRI, you'll probably want to utilize an extension. =A0I am guessing, as
> I haven't benchmarked anything, but one may well be able to get much
> faster pure ruby speeds with jruby or rubinius. =A0When in doubt,
> benchmark it!

am staying w your modest narray implem

\$ ruby test_hamming_distance.rb
Rehearsal ------------------------------------------------
kirk narray 0.140000 0.030000 0.170000 ( 0.160551)
robert zip 1 22.570000 0.010000 22.580000 ( 22.589685)
robert zip 2 22.870000 0.020000 22.890000 ( 22.922881)
-------------------------------------- total: 45.640000sec

user system total real
kirk narray 0.050000 0.030000 0.080000 ( 0.08034
robert zip 1 22.490000 0.090000 22.580000 ( 22.805407)
robert zip 2 22.900000 0.040000 22.940000 ( 22.963609)

wow, your narray scheme is immune to length of string...

thanks kirk and robert
best regards -botp