Problem 34 Project Euler Solution with Python

Digit factorials

145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.
Find the sum of all numbers which are equal to the sum of the factorial of their digits.

Note: as 1! = 1 and 2! = 2 are not sums they are not included.

The question might seem simple as it is just asking the factorials of the digit and to check if the digit factorials is same as the number. But there are some hurdles one will have to come across to solve this problem.

The first problem will be to find the limit up to which the iterations are to be calculated.

To find the upper bound you will have to take a look at Factorion Wikipedia.

Now to split each and every digit in the number you will have to take a look at: Split an integer to digits to check ISBN - Stack Overflow

First I have imported the factorial function from math module.

As we knew that we will have to find the factorial of the digits form 0-9, instead of calculating it every time in the iteration, I stored the pre-calculated values so that they can used later.

I have used the above Stack overflow answer to split the digits to find the factorial sum.

Started iteration and calculated the value.

That's it. No big algorithm involved.

Program

	#http://radiusofcircle.blogspot.com

	#factorial function for calculating factorial
	from math import factorial

	#time module for calculating time of execution
	import time

	#time at the start of program execution
	start = time.time()

	#factorials of numbers from 0-9
	f = [factorial(0),
	factorial(1),
	factorial(2),
	factorial(3),
	factorial(4),
	factorial(5),
	factorial(6),
	factorial(7),
	factorial(8),
	factorial(9)]

	#sum of factorial of the digits
	def factorial_digits(n):
	s = 0
	while n:
	s += f[n%10]
	n //= 10
	return s

	#variable to save the value of solution
	solution = 0

	#for loop to loop till 1854721
	for i in xrange(10,1854721):
	if factorial_digits(i) == i:
	solution += i

	#printing the solution
	print solution

	#Time at the end of program execution
	end = time.time()

	#total time of execution
	print end - start

view raw pep34.py hosted with ❤ by GitHub

As you can see the program is very basic and simple.

File: pep34.py ------------------------- #http://radiusofcircle.blogspot.com #factorial function for calculating factorial from math import factorial #time module for calculating time of execution import time #time at the start of program execution start = time.time() #factorials of numbers from 0-9 f = [factorial(0), factorial(1), factorial(2), factorial(3), factorial(4), factorial(5), factorial(6), factorial(7), factorial(8), factorial(9)] #sum of factorial of the digits def factorial_digits(n): s = 0 while n: s += f[n%10] n //= 10 return s #variable to save the value of solution solution = 0 #for loop to loop till 1854721 for i in xrange(10,1854721): if factorial_digits(i) == i: solution += i #printing the solution print solution #Time at the end of program execution end = time.time() #total time of execution print end - start

Instead of using the factorial_digits function I have also tried with list comprehension inside the for loop which turned out to be time taking. fd = [f(int(x)) for x in str(i)], I used this code and it ran for about 6 seconds extra. So I have eliminated that part with the basic function.

As always you can download the source code form Github Gist pep34.py.

Output

Summary

This problem was relatively easy but optimizing the performance of code was bit challenging. With the help of internet community I made it happen and I am satisfied with the result. The code is running under 2 seconds for a large number of iterations.

If you have any doubt or didn't understand anything then you can comment in the comment box below and I will be glad to help you.

Please do comment if you have found any typo, or have a better program or have a different program or have any suggestions. I will be glad to view each of them.

You can also contact me.

Thank you. Have a nice day😃.

RADIUS OF CIRCLE

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