Project Euler on Rohit Jha

0x000d - Project Euler 011

Wed, 09 Nov 2011 13:07:59 +0000

Problem:

In the 20 x 20 grid below, four numbers along a diagonal line have been marked.

08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00 81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65 52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91 22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80 24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50 32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70 67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21 24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72 21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95 78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92 16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57 86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58 19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40 04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66 88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69 04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36 20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16 20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54 01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48

0x000c - Project Euler 010

Wed, 02 Nov 2011 12:08:12 +0000

Problem:

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.

Find the sum of all the primes below two million.

Solution:

1. Brute-force approach:


class PE010_brute
{
 static boolean isPrime(long n)
 {
 if(n == 2 || n == 3) return true;
	if(n < 2 || (n % 2 == 0)) return false;
 if(n < 9) return true;
 if(n % 3 == 0) return false;

 int r = (int)Math.sqrt(n), f = 5;

 while(f <= r) {
 if ((n % f == 0) || (n % (f + 2) == 0))
 return false;
 f += 6;
 }
 return true;
 }
}

2. Efficient approach (using the Sieve of Eratosthenes) :

0x000a - Project Euler 009

Tue, 01 Nov 2011 12:57:54 +0000

Problem:

A Pythagorean triplet is a set of three natural numbers, a < b < c, for which, a2 + b2 = c2

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

There exists exactly one Pythagorean triplet for which a + b + c = 1000. Find the product abc.

Solution:

1. Brute-force approach:


s = 1000
for a in range(3, s) :
 for b in range(a+1, s) :
 c = s-a-b
 if c**2 == a**2 + b**2:
 print a*b*c
 exit

2. Efficient approach:

0x0009 - Project Euler 008

Mon, 31 Oct 2011 23:01:41 +0000

Problem:

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

Solution:


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

max = 0
for i in range(0, len(s) - 4) :
 p = 1
 for j in range(i, i+5) :
 p *= int(s[j])
 if p > max:
 max = p

print max

Analysis:

The 1000-digit number is stored as a string in variable s. The backslash ‘\’ at the end of a line implies that the following physical line is actually in the same logical line of code.

0x0008 - Project Euler 007

Mon, 31 Oct 2011 19:31:18 +0000

Problem:

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 10,001st prime number?

Solution:

1. Brute-force approach:


from math import sqrt
def is_prime(n) :
 for i in range(2, int(sqrt(n)+1)) :
 if n % i == 0:
 return False
 return True

limit, count, num = 10001, 1, 1
while count < limit:
 num += 2
 if is_prime(num) : count += 1
print num

2. Efficient approach:


from math import sqrt
def is_prime(n) :
 if n == 2 or n == 3: return True
 if n < 2 or n % 2 == 0: return False
 if n % 3 == 0: return False
 if n < 9: return True

 r, f = int(sqrt(n)), 5
 while f <= r:
 if n % f == 0 or n % (f + 2) == 0: return False
 f += 6
 return True

limit, count, num = 10001, 1, 1
while count < limit:
 num += 2
 if is_prime(num) : count += 1
print num

Analysis:

0x0006 - Project Euler 006

Sat, 29 Oct 2011 14:56:48 +0000

Problem:

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

12 + 22 + … + 102 = 385

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

(1 + 2 + … + 10)2 = 552 = 3025

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

0x0005 - Project Euler 005

Sat, 29 Oct 2011 12:59:35 +0000

Problem:

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?

Solution:

1. Brute-Force Approach:


def is_divisible_20(n) :
 for i in range(2, 20) :
 if n%i != 0: return False
 return True

i = 20
while True:
 if is_divisible_20(i) :
 print i
 break
 i += 20

2. Efficient Approach:


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

 while b != 0:
 a, b = b, a%b
 return a

result = 2
for i in range(2,21) :
 result = i / gcd(result,i) * result
print result

Analysis:

0x0004 - Project Euler 004

Fri, 28 Oct 2011 22:57:56 +0000

Problem:

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

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

Solution:

1. Brute Force Approach:


def is_palindrome(n) :
 return str(n) == str(n)[::-1]

largest = 0
for a in range(100, 1000) :
 for b in range(100, 1000) :
 c = a*b
 if is_palindrome(c) and c > largest:
 largest = c
print largest

2. Efficient Approach:


largest = 0
for i in range(999,99,-2) :
 for j in range(i,99,-2) :
 x = i*j
 if x%11==0:
 y = str(x)
 if x > largest and y == y[::-1]:
 largest = x
print largest

Analysis:

The brute-force approach would be to calculate product of pairs of three digit numbers and then find the largest palindrome from them.

0x0003 - Project Euler 003

Fri, 28 Oct 2011 21:06:30 +0000

Problem:

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

What is the largest prime factor of the number 600851475143 ?

Solution:

A neat little trick for Linux/UNIX users in Python is:


from commands import *
print int(getoutput("factor 600851475143").split()[::-1][0])

But a general program could be:


p, n = 2, 600851475143

while (p*p <= n) :
 if (n % p == 0) :
 n /= p
 if p == 2: p += 1
 else: p += 2

print n

Analysis:

I have solved this problem in two ways using Python 2.7 in Ubuntu 10.10. In the first code, initially I’ve imported all methods from the ‘commands’ module in Python, using which I can run Operating System(OS) commands and get the result from within my Python program. The next line is where the “cheat” lies.

0x0002 - Project Euler 002 – More efficient

Sun, 25 Sep 2011 18:42:33 +0000

Problem:

Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …

By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

Solutions:

1. Python Solution:


limit = 4000000
x, y, sum = 3, 5, 2

while sum < limit:
 sum += (x + y)
 x, y = x + 2*y, 2*x + 3*y

print(sum)

2. Java Solution:

0x0001 - Project Euler 002

Wed, 07 Sep 2011 19:18:01 +0000

Problem:

Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …

By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

Solutions:

1. Python solution:


from math import sqrt

limit, current, sum = 4000000, 2, 0
phi3 = ((1+sqrt(5))/2)**3

while sum < limit:
 sum += current
 current = round(phi3*current)

print (sum)

2. Java solution:

0x0000 - Project Euler 001

Sat, 03 Sep 2011 17:14:31 +0000

Problem:

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

Solutions: 1. Python solution:


def sum_n(n, d) :
 n = int(n/d)
 return int(d * n * (n + 1) / 2)

print (sum_n(999, 3) + sum_n(999, 5) - sum_n(999, 15))

2. Java solution:


class PE001
{
	public static int sum_n(int n,int d)
	{
		n = (int)(n/d);
		return (int)(d * n * (n + 1) / 2);
	}

	public static void main(String args[])
	{
		System.out.println(sum_n(999, 3) + sum_n(999, 5) - sum_n(999, 15));
	}
}

3. Perl solution: