Category: Programming

Euler 4

Posted on 2010-08-032018-08-20 by Roberto's Chronicles

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

Euler 3

Posted on 2010-08-032018-08-20 by Roberto's Chronicles

Project Euler’s Problem #3 statement is —

The prime factors of 13195 are 5, 7, 13 and 29.

What is the largest prime factor of the number 600851475143 ?

My solution is really ugly and is basically dumb-, brute-force.

#!/usr/bin/python

import math

top_number = 600851475143

def is_divisor(divisor):
        return top_number % divisor == 0

def is_prime(divided):
        divisor = 3
        sqrt_divided = math.sqrt(divided)
        if divided == 1:
                return true
        while divisor <= sqrt_divided:
                if divided == divisor:
                        return True
                elif divided % divisor == 0:
                        return False
                else:
                        divisor += 2
        return True

def main():
        count = 3
        go_to = top_number

        first_list =[]
        while count <= go_to:
                if is_divisor(count):
                        first_list.append(count)
                        go_to = top_number / count
                        first_list.append(go_to)

                count += 2

        second_list = map(is_prime, first_list)
        print "%s" % max(zip(second_list, first_list))[-1]

if __name__ == "__main__":
        main()

Euler 8

Posted on 2010-08-032018-08-20 by Roberto's Chronicles

Project Euler’s Problem #8 statement is —

Find the greatest product of five consecutive digits in the 1000-digit number.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

This is another very straightforward problem. My solution —

number = '73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450'

max_prod = 0
for i in range(1, len(number) - 4):
        prod = reduce(lambda x,y: x*y, [int(s) for s in number[i:i+5]])
        if prod > max_prod:
                max_prod = prod

print max_prod

This ran in 0m0.023s.

Euler 7

Posted on 2010-08-022018-08-20 by Roberto's Chronicles

Project Euler’s Problem #7 statement is —

By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.

What is the 10001st prime number?

It’s a straightforward problem to solve using brute force, but I went with a mixed approach doing some math instead.

#!/usr/bin/python
import math

def is_prime(n):
    return not (n < 2 or any(n % x == 0 for x in xrange(2, int(n ** 0.5) + 1)))

currentMax =0
primes = 1
counter = 3

while (primes < 10001):
    if (is_prime(counter)):
        currentMax = counter
        primes+=1
        counter+=2

print currentMax

I initially used a dumb brute-force which ended up running fast enough. But reading about primes, I came across this property by which if a number cannot be divided by any number smaller than sqrt(n), it is a prime. That required significantly less tests to be performed inside is_prime, but then again, it had to perform a square root calculation for every number we need to test. In the end, the algorithm above performed slightly better (0.02s to be exact.)

Updated: Thanks to Eduardo Habkost for pointing out the fact that I had mistakenly pasted the wrong version of the script. I fixed the text to reflect that.