python

I have a function I wrote for a project euler that calculates the sum of the digits in a number. This is my first attempt which simply converts each letter to an integer and sums them.

Method 1:

def digitsum(x):
	total=0
	for letter in str(x):
		total+=int(letter)
	return total

I thought that this could be improved using ord, which converts a letter into its decimal ascii number. Numbers ‘0’, ‘1’, ‘2’ … ‘9’ correspond to the ascii values of 48 – 57 and then took the moduli of this with 48 to give the integer value. I later realised that this was completely nonsensical and should have just subtracted 48, but I decided to include it for the purposes of the speed test.

Method 2:

def digitsum2(x):
	total=0
	for letter in str(x):
		total+=ord(letter)%48
	return total 

Method 3:

def digitsum3(x):
	total=0
	for letter in str(x):
		total+=ord(letter)-48
	return total 

Speed Test:
The test uses a long number and one million repetitions for each method.

from time import time

# .. functions go here

# Nice long number to sum
x=981234153134415646571899783156122451653

tic = time()
for i in range(0,1000000):
	digitsum(x)
print time() - tic, 'Seconds elapsed'

tic = time()
for i in range(0,1000000):
	digitsum2(x)
print time() - tic, 'Seconds elapsed'

tic = time()
for i in range(0,1000000):
	digitsum3(x)
print time() - tic, 'Seconds elapsed'

Results:

#Method 1
29.3496568203 Seconds elapsed
#Method 2
12.185685873 Seconds elapsed
#Method 3
9.59367895126 Seconds elapsed

So we can see that the first method is much slower, avoiding the integer conversion by using ord speeds it up the function by ~60% and that using subtraction rather than modulus (a division based operation) saves a further ~20% on top of this.

I’ve been doing a lot of problem solving on Project Euler recently. If your not aware of Project Euler (PE) it is a website with lots of maths / programming based puzzles to solve. Many of the problems involve using or checking prime numbers, in many cases a sieve method would be applicable (see: Sieve of Eratosthenes for a clear explanation / animation).

However sometimes this is not necessary if you only require one prime number, or if memory limitations mean a sieve would be inconceivable. The following is the code I wrote to check if a number is prime or not in python. It tests upwards checking if the number is perfectly divisble, and as it does this it lowers the maximum number need to reach to stop as we know that if a number is not divisible by 2 then it will not be uniquely divisible (a repatition of a divislbe may exist) by anything greater than N/2

def isprime(number):
	if number<=1:
		return 0
	check=2
	maxneeded=number
	while check<maxneeded+1:
		maxneeded=number/check
		if number%check==0:
			return 0
		check+=1
	return 1

I hope that this made sense, and is useful to someone. If anyone has any more efficient methods I would be happy to hear from you.

Check out my profile to see which problems I’ve completed. If you’ve completed any I haven’t please get in contact and give me some tips.

Update:
Mike sent a suggestion that I could speed up the program by ignoring all factors of 2, it could also propbably be sped up by looking at 3,4,5 .. etc until certain point.

def isprime(number):
	if number<=1 or number%2==0:
		return 0
	check=3
	maxneeded=number
	while check<maxneeded+1:
		maxneeded=number/check
		if number%check==0:
			return 0
		check+=2
	return 1

So lets test the speed difference by doing 1,000 checks of the number 982,451,653 (known to be prime from this usefulsite)

Original Code: 9.99917411804 Seconds elapsed
Improved Code: 5.2977039814 Seconds elapsed

That’s approximately a factor of two speed increase, but this makes me think that a combination of the sieve method with this one may lead to an even further increase.