Programming

So I recently wrote an ugly solution to Project Euler’s problem #9. It was written in Python and even though every other Project Euler solution I wrote ran in less than one second, this one took over 18 seconds to get to the answer. Obviously it’s my fault for using a purely brute force algorithm.

Anyway, boredom is a great motivator and I just rewrote the exact same algorithm in C, just to see how much faster it would be.

#include <stdio.h>

int is_triplet(int a, int b, int c)
{
        return (((a < b) && (b < c)) && ((a * a + b * b) == (c * c)));
}

int main(void)
{
        for (int a = 0; a < 1000; a++)
                for (int b = a + 1; b < 1000; b++)
                        for (int c = b + 1; c < 1000; c++)
                                if ((a + b + c == 1000) && is_triplet(a, b, c)) {
                                        printf("%d %d %d = %dn", a, b, c,
                                               a * b * c);
                                        return 0;
                                }
}

It’s the exact same algorithm, but instead of 18 seconds, it ran in 0.072s.

Update: And here’s the version suggested by Gustavo Niemeyer:

#include <stdio.h>

int is_triplet(int a, int b, int c)
{
        return ((a * a + b * b) == (c * c));
}

int main(void)
{
        for (int a = 1; a < 1000; a++)
                for (int b = a + 1; b < (1000 - a) / 2; b++) {
                        int c = 1000 - a - b;
                        if (is_triplet(a, b, c)) {
                                printf("%d %d %d = %dn", a, b, c,
                                       a * b * c);
                                return 0;
                        }
                }
}

It now runs in 0.002s :–)

So the other night I was a bit bored and decided to do something to pass the time. I first came across Project Euler a while ago, but had never gone further than problem #1. Boredom is a great motivator and I went through problems #2 thru #9 last night and I decided to post my solutions in search of better ones. Feel free to comment with your suggestions.

Project Euler’s Problem #9 statement is —

A Pythagorean triplet is a set of three natural numbers, $latex a < b < c$, for which,

[latex]a² + b² = c²[/latex]

For example, $latex 3² + 4² = 9 + 16 = 25 = 5²$.

There exists exactly one Pythagorean triplet for which [pmath]a + b + c = 1000[/pmath].

Find the product [pmath]abc[/pmath].

Butt ugly, super slow, brute-force solution:

from sys import exit

def is_triplet(a, b, c):
        if a < b and b < c:
                return a ** 2 + b ** 2 == c ** 2

for a in range(0, 1000):
        for b in range(a + 1, 1000):
                for c in range(b + 1, 1000):
                        if a + b + c == 1000 and is_triplet(a, b, c):
                                print a * b * c
                                exit(0)

I’ll have to rethink this, as it’s really inefficient. As it is, it runs in 18.034s!

Updated: take a look at another take at this problem and algorithm.

Updated 2010/08/04: Gustavo Niemeyer pointed an obvious optimization to the algorithm above (written in C)—the innermost loop is unnecessary. I rewrote it in Python to see the difference:

from sys import exit

def is_triplet(a, b, c):
        return (a ** 2 + b ** 2 == c ** 2)

for a in range(1, 1000):
        for b in range(a + 1, (1000 - a) / 2):
                c = 1000 - a - b
                if is_triplet(a, b, c):
                        print "%d * %d * %d = %d" % (a, b, c, a * b * c)

From 18s to 0.076s. As well, following Eduardo Habkost’s suggestion, I used psyco and execution time went down to 0.023s.

So the other night I was a bit bored and decided to do something to pass the time. I first came across Project Euler a while ago, but had never gone further than problem #1. Boredom is a great motivator and I went through problems #2 thru #9 last night and I decided to post my solutions in search of better ones. Feel free to comment with your suggestions.

Project Euler’s Problem #6 statement is —

The sum of the squares of the first ten natural numbers is,

[pmath]1² + 2² + … + 10² = 385[/pmath]

The square of the sum of the first ten natural numbers is,

pmath² = 55² = 3025[/pmath]

Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is [pmath]3025 – 385 = 2640[/pmath].

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

Can’t be more straightforward than that, really.

sqr_sum = 0
num_sum = 0
for i in range(1,100 + 1):
        num_sum += i
        sqr_sum += i**2

num_sum = num_sum**2
print sqr_sum - num_sum

I’m pretty sure Niemeyer could rewrite this in one line, but whatever. This runs in 0.015s.

So the other night I was a bit bored and decided to do something to pass the time. I first came across Project Euler a while ago, but had never gone further than problem #1. Boredom is a great motivator and I went through problems #2 thru #9 last night and I decided to post my solutions in search of better ones. Feel free to comment with your suggestions.

Project Euler’s Problem #5 statement is —

2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

My first attempt was using brute force, but it didn’t go very well. It took a considerable amount of time, so I decided to think about the problem a bit. In the end, it boils down to the a relation between the lesser common multiplier and the greater common divisor.

def gcd(a, b):
        if b== 0: return a
        return gcd(b, a % b)

def lcm(a, b):
        return a * b / gcd(a, b)

if __name__ == "__main__":
        b = 2
        for i in range(3,20):
                b = lcm(b, i)
        print b

While the original code took several seconds before I killed it, this runs in 0.015s.

So the other night I was a bit bored and decided to do something to pass the time. I first came across Project Euler a while ago, but had never gone further than problem #1. Boredom is a great motivator and I went through problems #2 thru #9 last night and I decided to post my solutions in search of better ones. Feel free to comment with your suggestions.

Project Euler’s Problem #4 statement is —

A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 99.

Find the largest palindrome made from the product of two 3-digit numbers.

And here’s the solution.

def int_to_list(number):
  return map(int, str(number))

def is_palindrome(number):
  local_list = int_to_list(number)
  return local_list == list(reversed(local_list))

pal = 0
if __name__ == "__main__":
  for i in range(100,1000):
        for j in range(100,1000):
                if is_palindrome(i * j):
                        print "%d * %d = %d"%(i, j, i*j)
                        if i * j > pal:
                                pal = i * j

  print pal