Algorithms on Rohit Jha

0x0013 - Prime Factorization

Sun, 26 Feb 2012 20:56:01 +0000

In the post “Sieve of Eratosthenes“, I had posted a Java program for the prime sieve. Translating it to Python, I have:


from math import sqrt

def prime_sieve(limit) :
 crosslimit = int(sqrt(limit))
 sieve = [False]*limit

 for n in range(4, limit, 2) :
 sieve[n] = True
 for n in range(3, crosslimit+1, 2) :
 if not sieve[n]:
 for m in range(n*n, limit, 2*n) :
 sieve[m] = True

 num_primes = 0
 primes = []
 for n in range(2, limit) :
 if not sieve[n]:
 num_primes += 1

 for n in range(2, limit) :
 if not sieve[n]:
 primes.append(n)

 return primes

The list of primes generated by this sieve could then be used for factorizing a given number. For calculating the factors, I use the Trial Division algorithm, which isn’t the fastest, but easily implemented and not probabilistic. The algorithm basically tests if a given number (integer) n is divisible by any integer between 1 and n. Since only the prime factors of n are needed, we can use the list of primes generated from the sieve as a base set for the calculation, instead of the set of integers. Furthermore, the trial factors need go no further than because, if n is divisible by some number p, then n = p×q and if q were smaller than p, n would have earlier been detected as being divisible by q or a prime factor of q.

0x0012 - Fibonacci sequence – Part 2

Tue, 03 Jan 2012 18:50:27 +0000

In this post we’ll compare the various methods of generating Fibonacci sequence terms and implementing the code to recognize Fibonacci terms and to determine index of these terms. These have been discussed mathematically in the previous post.

Generating Fibonacci Terms

Method 1 – Iterative method

Using the basic concept of the Fibonacci sequence, that each term is the sum of the previous two terms, the following function in Python generates the nth term:

0x0011 - Fibonacci sequence

Mon, 21 Nov 2011 19:06:33 +0000

The Fibonacci sequence in mathematics is the following sequence of numbers:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233 …

Representation

By definition, the first two terms are 0 and 1, and each of the terms following them are the sum of two previous terms. So, if we say Fn is the nth term in the Fibonacci sequence, then we can express the sequence recursively as:

0x0010 - Searching Algorithms – Part 3

Fri, 18 Nov 2011 18:14:55 +0000

Uniform Binary Search

I came across the uniform binary search algorithm in The Art Of Computer Programming – Volume 3: Sorting and Searching as an optimization of the binary search, invented by the author, Donald Knuth. It works quite well for architectures on which a table lookup is generally faster than an addition and a shift, and also when many searches will be performed on the same array, or on several arrays of the same length.

0x000f - Searching Algorithms – Part 2

Thu, 17 Nov 2011 16:10:43 +0000

Binary Search

The binary search algorithm, a variation of the Dichotomic search, finds the position of a specified “key” within a sorted list/array using the divide and conquer approach.

At each stage, the algorithm compares the input key value with the key value of the middle element of the array. If the keys match, then a matching element has been found so its index, or position, is returned. Otherwise, if the sought key is less than the middle element’s key, then the algorithm repeats its action on the sub-array to the left of the middle element or, if the input key is greater, on the sub-array to the right. If the remaining array to be searched is reduced to zero, then the key cannot be found in the array and a special indication is returned, saying that the key was not found.

0x000e - Searching Algorithms – Part 1

Thu, 10 Nov 2011 08:21:00 +0000

This is the first part of some searching algorithms that I’m implementing. The algorithms are:

Linear Search
Binary Search
Uniform Binary Search
Fibonacci Search
Jump Search
Interpolation Search
Ternary Search

In this part, Linear Search is discussed.

Linear Search

The linear/sequential search technique is the simplest of all searching algorithms and it involves checking every element, one at a time in sequence, until the desired one is found. This type of search is easy to implement, but practical when the list has few elements or the list is unsorted. The best case is when the first element is the one to be found. The worst case is when the element is not found even after searching through the entire list. Thus, both the worst-case and expected cost is asymptotically O(n). Below are some variations of the Linear Search algorithm:

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) :

0x000b - Sieve of Eratosthenes

Tue, 01 Nov 2011 20:19:54 +0000

The Sieve of Eratosthenes, named after the Greek mathematician Eratosthenes, is one of many known prime number sieves. The algorithm is a fairly good one for finding prime numbers from a list of numbers, say, less than 10 million.

Suppose we need to find all prime numbers below a natural number n > 1. Then, the algorithm is given as:

Create a list of consecutive integers from 2 to n: (2, 3, 4, …, n).
Initially, let p equal 2, the first prime number.
Starting from p, count up in increments of p, as long as the value is less than √n and mark each of these numbers greater than p itself in the list. These numbers will be 2p, 3p, 4p, etc.; note that some of them may have already been marked.
Find the first number greater than p in the list that is not marked; let p now equal this number (which is the next prime).
If there were no more unmarked numbers in the list, stop. Otherwise, repeat from step 3.

When, the above algorithm terminates, all unmarked numbers are prime.

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:

0x0007 - Primality Testing

Sat, 29 Oct 2011 22:58:18 +0000

I’ve been doing some programs on Primality tests, and so far, have implemented some naive algorithms. Though not efficient, these are useful for quick implementations. Here I post some of them:

Method 1:

The simplest primality test is as follows: Given an input number n, check whether any integer m from 2 to n − 1 evenly divides n. If n is divisible by any m then n is composite, otherwise it is prime. The implementation in Python 2.7 in the form of a function is:

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: