# Blog Archives

## Longest Collatz sequence

The following iterative sequence is defined for the set of positive integers:

n n/2 (n is even)
n 3n + 1 (n is odd)

Using the rule above and starting with 13, we generate the following sequence:

13 40 20 10 5 16 8 4 2 1

It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.

Which starting number, under one million, produces the longest chain?

NOTE: Once the chain starts the terms are allowed to go above one million.

Solution:

```
#include <iostream>

int main ()
{
int N           = 1000000;
int CHECKPOINT  = 800000;
long long nTemp = N;
int chainCount  = 1;
int check       = 0;
long long myNum = 0;

for (int i = N; i > CHECKPOINT; --i)
{
while (nTemp != 1)
{
if ((nTemp%2) == 0)
{
nTemp = nTemp/2;
}else
{
nTemp = (nTemp*3) + 1;
}
++chainCount;
}

--N;
nTemp = N;

if (check < chainCount)
{
check = chainCount;
myNum = nTemp;
}
chainCount = 1;
}

std::cout << "The number " << myNum << " has longest chain of " << check << std::endl;
}
```

## Project Euler problem#9 solution in C++

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:

#include <iostream>
using namespace std;

int main()
{
int a;
int b;
int c;
for (a=1; a<=500; a++)
{
for (b=1; b<=500; b++)
{
c=1000-b-a;
if (a*a+b*b-c*c == 0 && a<b )
{
cout<<a<<” “<<b<<” “<<” “<<c<<” “<<a*b*c<<endl;
}
}
}
return 0;
}

## Project Euler Problem# 8 solution in C++

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:

#include <iostream>
#include <cstdlib>

using namespace std;

int main()
{
int maxNum =1;
char a ;
string test ;
int result =1;
int checker = 1;

string numString  = (“7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450”);

for (int i = 0; i < 1000; i++)
{
if (numString[i] == ‘9’)
{
for (int j = 0; j < 5; j++)
{
a = numString[i+j];
maxNum *= atoi(&a);
}
if (result < maxNum)
{
result = maxNum;
}else
{
cout << result << endl;
}
maxNum = 1;
}
}
}

## Project Euler Problem# 6 solution in C++

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.

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

Efficient solution without using any loops. So it will be constant time method:

#include<iostream>

using namespace std;

int main ()
{
int maxNum = 0;
int sum_of_square = 0;
int square_of_sum = 0;

cout << “Enter Maximum number: “;
cin >> maxNum;

sum_of_square = ((maxNum*(maxNum +1))*((2*maxNum)+1))/6;

square_of_sum *= square_of_sum = maxNum* (maxNum+1)/2;

cout << “sum of squares till ” << maxNum << ” is: ” << sum_of_square << endl;
cout << “squares of sum till ” << maxNum << ” is: ” << square_of_sum << endl;
cout << “difference between the sum of the squares the square of the sum: ” << square_of_sum – sum_of_square  << endl;
}

## Project Euler Problem# 5 solution in C++

/**
Project Euler problem# 5:
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?

Below is an efficient solution of above problem using C++. Time complexity in the random access machine model is O(n log log n)
operations, a direct consequence of the fact that the prime harmonic series asymptotically approaches log log n.

*/

#include <iostream>

using namespace std;

//Finding Prime Numbers using the sieve of Eratosthenes, one of a number of prime number sieves,
//is a simple, ancient algorithm for finding all prime numbers up to any given limit.
// it is not an trial and check algorithm but find prime numbers by jumping out.
//The sieve of Eratosthenes is one of the most efficient ways to find all of the smaller primes (below 10 million or so)
int find_prime (int *numArray, int maxNum)
{
int factor = 2; // we will make 2 as the starting point.
numArray = 0; // rule out 1 from our logic to avoid incorrect results.

// loop condition will check, if we are in our maximum number limit.
//maxNum is the number till there we can find the prime numbers.
while ((factor * factor) < maxNum)
{
for (int i = 2; i < maxNum; i++) // we start our itration from number 2 not from 0 or 1 to get correct results.
{
if (numArray[i] != 0) //if a number on current array index is not zero, it is a prime or we haven’t yet checked it.
{
// we are putting zeros on all multiples of prime numbers, one by one.
if (numArray[i] != factor && (numArray[i]% factor) == 0)
{
numArray[i] = 0;
}
}
}
++factor;
}// after the loop, array will have zeros on all non prime locations and prime numbers.
}

int main ()
{
int maxNum = 0;
int result = 1;
int num = 0;

cout << “enter max number: “;
cin >> maxNum;

int *myArray = new int [maxNum];

//we fill up the array till the number we want to find the smallest positive number that is evenly divisible.
for (int i = 0; i < maxNum; i++)
{
myArray[i]= i;
}

// we will get prime numbers till the maxNum by calling below function.
find_prime(myArray, maxNum);

// returning array will have prime numbers and zeros on non prime locations.
// we start our loop from 2 as first two locations are not use full so we reduce the itration of the loop.
for (int i = 2; i < maxNum; i ++)
{
// we only do our calculation on prime numbers.
if ((myArray [i]) != 0)
{
num = myArray[i];
// we take multiples of each prime number till the maxNum. meaning we raise each prime to power of a number.
// which result will remain till maxNum.
while ((num * (myArray[i])) < maxNum)
{
num = num * (myArray[i]);
}
result = num * result;
}
}// after the loop result will have smallest positive number that is evenly divisible by all of the numbers from 1 to maxNum.
cout << “smallest positive number that is evenly divisible by all of the numbers from 1 to ” << maxNum
<< ” is: ” << result <<endl;
delete []myArray;
}

## Project Euler Problem #2 Solution in Java

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.

/**
* @author Khuram Ali
*/
public class Problem2
{
public static void main(String[] args)
{
int term_1 = 1;
int term_2 = 2;
int term_3 = 0;
int sum = 0;
long MaxTerm = 4000000;

while (term_1 <= MaxTerm)
{
System.out.println (term_1);
if ((term_1 %2) == 0){sum += term_1;}

term_3 = term_1 + term_2;
term_1 = term_2;
term_2 = term_3;
}
System.out.println (“the sum of even value numbers is: “);
System.out.println (sum);
}
}