Why python implementation of miller-rabin faster than ruby by a lot? - performance

Question

For one of my classes I recently came across both a ruby and a python implementations of using the miller-rabin algorithm to identify the number of primes between 20 and 29000. I am curious why, even though they are seemingly the same implementation, the python code runs so much faster. I have read that python was typically faster than ruby but is this much of a speed difference to be expected?
miller_rabin.rb
def miller_rabin(m,k)
t = (m-1)/2;
s = 1;
while(t%2==0)
t/=2
s+=1
end
for r in (0...k)
b = 0
b = rand(m) while b==0
prime = false
y = (b**t) % m
if(y ==1)
prime = true
end
for i in (0...s)
if y == (m-1)
prime = true
break
else
y = (y*y) % m
end
end
if not prime
return false
end
end
return true
end
count = 0
for j in (20..29000)
if(j%2==1 and miller_rabin(j,2))
count+=1
end
end
puts count
miller_rabin.py:
import math
import random
def miller_rabin(m, k):
s=1
t = (m-1)/2
while t%2 == 0:
t /= 2
s += 1
for r in range(0,k):
rand_num = random.randint(1,m-1)
y = pow(rand_num, t, m)
prime = False
if (y == 1):
prime = True
for i in range(0,s):
if (y == m-1):
prime = True
break
else:
y = (y*y)%m
if not prime:
return False
return True
count = 0
for j in range(20,29001):
if j%2==1 and miller_rabin(j,2):
count+=1
print count
When I measure the execution time of each using Measure-Command in Windows Powershell, I get the following:
Python 2.7:
Ticks: 4874403
Total Milliseconds: 487.4403
Ruby 1.9.3:
Ticks: 682232430
Total Milliseconds: 68223.243
I would appreciate any insight anyone can give me into why their is such a huge difference

Answer 1

In ruby you are using (a ** b) % c to calculate the modulo of exponentiation. In Python, you are using the much more efficient three-element pow call whose docstring explicitly states:
With three arguments, equivalent to (x**y) % z, but may be more
efficient (e.g. for longs).
Whether you want to count the lack of such built-in operator against ruby is a matter of opinion. On the one hand, if ruby doesn't provide one, you might say that it's that much slower. On the other hand, you're not really testing the same thing algorithmically, so some would say that the comparison is not fair.
A quick googling reveals that there are implementations of modulo exponentiation for ruby.

Answer 2

I think these profile results should answer your question:
%self total self wait child calls name
96.81 43.05 43.05 0.00 0.00 17651 Fixnum#**
1.98 0.88 0.88 0.00 0.00 17584 Bignum#%
0.22 44.43 0.10 0.00 44.33 14490 Object#miller_rabin
0.11 0.05 0.05 0.00 0.00 32142 <Class::Range>#allocate
0.11 0.06 0.05 0.00 0.02 17658 Kernel#rand
0.08 44.47 0.04 0.00 44.43 32142 *Range#each
0.04 0.02 0.02 0.00 0.00 17658 Kernel#respond_to_missing?
0.00 44.47 0.00 0.00 44.47 1 Kernel#load
0.00 44.47 0.00 0.00 44.47 2 Global#[No method]
0.00 0.00 0.00 0.00 0.00 2 IO#write
0.00 0.00 0.00 0.00 0.00 1 Kernel#puts
0.00 0.00 0.00 0.00 0.00 1 IO#puts
0.00 0.00 0.00 0.00 0.00 2 IO#set_encoding
0.00 0.00 0.00 0.00 0.00 1 Fixnum#to_s
0.00 0.00 0.00 0.00 0.00 1 Module#method_added
Looks like Ruby's ** operator is slow as compared to Python.
It looks like (b**t) is often too big to fix in a Fixnum, so you are using Bignum (or arbitrary-precision) arithmetic, which is much slower.