Project Euler Problem 12 - Highly Divisible Triangular Number - Python Solution Optimization

Question

Project Euler Problem 12 asks (paraphrased):

Considering triangular numbers T_n = 1 + 2 + 3 + ... + n, what is the first T_n with over 500 divisors? (For example, T₇ = 28 has six divisors: 1, 2, 4, 7, 14, 28.)

I have written the following solution. The code is calculates the correct answer, but the run time is appalling (6m9s on a 2.4GHz i7!!!)...

I know Python is definitely not the fastest of languages, but I'm clearly missing a trick here.

Could anyone highlight any optimizations that could be made to bring the execution time down to something sensible?

from math import sqrt, floor
class findDivisors:
 def isPrime(self, num):
 for i in range(2, int(floor(sqrt(num)) + 1)):
 if not num % i:
 return False
 return True
 def numDivisors(self, num):
 factors = {}
 numFactors = 2
 currNum = num
 currFactor = 2
 while not self.isPrime(currNum):
 if not currNum % currFactor:
 if currFactor not in factors:
 factors[currFactor] = 1
 else:
 factors[currFactor] += 1
 currNum = currNum / currFactor
 else:
 currFactor += 1
 if currNum not in factors:
 factors[currNum] = 1
 else:
 factors[currNum] += 1
 total = 1
 for key in factors:
 total *= factors[key] + 1
 return total
 def firstWithNDivisors(self, noDivisors):
 triGen = genTriangular()
 candidate = triGen.next()
 while self.numDivisors(candidate) < noDivisors:
 candidate = triGen.next()
 return candidate
def genTriangular():
 next = 1
 current = 0
 while True:
 current = next + current
 next += 1
 yield current
if __name__ == '__main__':
 fd = findDivisors()
 print fd.firstWithNDivisors(5)

Addendum: I have since made to revisions of the code The first is the alteration made as suggested by nabla; check for currentNum being> 1 instead of prime. This gives an execution time of 2.9s

I have since made a further improvement by precomputing the first 10k primes for quick lookup. This improves runtime to 1.6s and is shown below. Does anyone have any suggestions for further optimisations to try and break below the 1s mark?

from math import sqrt, floor
from time import time
class findDivisors2:
 def __init__(self):
 self.primes = self.initPrimes(10000)
 def initPrimes(self, num):
 candidate = 3
 primes = [2]
 while len(primes) < num:
 if self.isPrime(candidate):
 primes.append(candidate)
 candidate += 1
 return primes
 def isPrime(self, num):
 for i in range(2, int(floor(sqrt(num)) + 1)):
 if not num % i:
 return False
 return True
 def primeGen(self):
 index = 0
 while index < len(self.primes):
 yield self.primes[index]
 index += 1
 def factorize(self, num):
 factors = {}
 curr = num
 pg = self.primeGen()
 i = pg.next()
 while curr > 1:
 if not curr % i:
 self.addfactor(factors, i)
 curr = curr / i
 else:
 i = pg.next()
 return factors
 def addfactor(self, factor, n):
 if n not in factor:
 factor[n] = 1
 else:
 factor[n] += 1
 def firstWithNDivisors(self, num):
 gt = genTriangular()
 candidate = gt.next()
 noDivisors = -1
 while noDivisors < num:
 candidate = gt.next()
 factors = self.factorize(candidate)
 total = 1
 for key in factors:
 total *= factors[key] + 1
 noDivisors = total
 return candidate
def genTriangular():
 next = 1
 current = 0
 while True:
 current = next + current
 next += 1
 yield current
if __name__ == '__main__':
 fd2 = findDivisors2()
 start = time()
 res2 = fd2.firstWithNDivisors(500)
 fin = time()
 t2 = fin - start
 print "Took", t2, "s", res

• Joel Cornett: Is prime is checking every number from 2 to sqrt(n). Half of those are even numbers.
• rcbevans: Can't believe I missed that, good spot (y)

Answer 1

Here is one possible improvement:

Don't check whether currNum is prime, instead simply search factors until currNum becomes 1. Checking for prime takes O(sqrt(n)) time, but the rest of the loop checking for factors takes approximately O(1) per loop iteration, so you can save those O(sqrt(n)) here:

 while currNum > 1:
 if not currNum % currFactor:
 if currFactor not in factors:
 factors[currFactor] = 1
 else:
 factors[currFactor] += 1
 currNum = currNum / currFactor
 else:
 currFactor += 1
 total = 1
 for key in factors:
 total *= factors[key] + 1

Result and timing for firstWithNDivisors(300) with the original code:

 2162160 2.21085190773

and without checking for prime:

 2162160 0.0829100608826

In fact with this improvement alone the firstWithNDivisors(501) only takes 2.3 sec on my machine.

There are very likely much more efficient algorithms to do this. One can probably make use of the fact that the n-th triangle number is n*(n+1)/2.

• rcbevans: Thanks for a good suggestion, this made a world of difference! I managed to enhance the run-time further by precomputing primes. This has run time on my machine down to 1.6s. Any further suggestions?
• user33895: Precomputing primes would have been my next suggestion, also you can calculate the number of divisors for n and n+1 separately because they are coprime, then you can save one the result of n+1 as n for the next run, effectively reducing runtime by a factor of 2. But there are surly much better algorithmic approaches that I haven't looked into.