# Prime numbers

## Project Euler 47: Find the first four consecutive integers to have four distinct primes factors.

One of the big topics in number theory – Prime factorization – is once again the topic of a Project Euler problem. Problem 47 reads

The first two consecutive numbers to have two distinct prime factors are:

14 = 2 x 7
15 = 3 x 5

The first three consecutive numbers to have three distinct prime factors are:

644 = 2² x 7 x 23
645 = 3 x 5 x 43
646 = 2 x 17 x 19.

Find the first four consecutive integers to have four distinct primes factors. What is the first of these numbers?

Posted by Kristian in Project Euler, 19 comments

## Project Euler 46: What is the smallest odd composite that cannot be written as the sum of a prime and twice a square?

Problem 46 of Project Euler reads

It was proposed by Christian Goldbach that every odd composite number can be written as the sum of a prime and twice a square.

9 = 7 + 2×12
15 = 7 + 2×22
21 = 3 + 2×32
25 = 7 + 2×32
27 = 19 + 2×22
33 = 31 + 2×12

It turns out that the conjecture was false.

What is the smallest odd composite that cannot be written as the sum of a prime and twice a square?

Posted by Kristian in Project Euler, 19 comments

## Project Euler 41: What is the largest n-digit pandigital prime that exists?

This time we mix two old topics together and form a new question. This time Project Euler has mixed pandigital numbers and primes and Problem 41 asks us to find the largest such number. The problem description reads

We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime.

What is the largest n-digit pandigital prime that exists?

I will start solving it by brute force and as we shall see the approach is possible but very inefficient, so after that I will show you how to speed the process up by using a small property of numbers that I was taught in 4th grade or so – And no, I am not particularly gifted, the trick is just really simple. Continue reading →

Posted by Kristian in Project Euler, 13 comments

## Project Euler 35: How many circular primes are there below one million?

In problem 35 of Project Euler we are back to primes., this time it is circular primes that we are focusing on. Before looking more at circular primes lets look at the problem description which reads:

The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.

There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.

How many circular primes are there below one million?

Posted by Kristian in Project Euler, 9 comments

## Project Euler 27: Find a quadratic formula that produces the maximum number of primes for consecutive values of n

We are back to saying something about primes in Problem 27 of Project Euler. So you may already now want to go back to the solution to Problem 10 for an efficient implementation of Sieve of Eratosthenes since we are likely to need that. The problem reads

Euler published the remarkable quadratic formula:

n² + n + 41

It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39. However, when n = 40, 402 + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when n = 41, 41² + 41 + 41 is clearly divisible by 41.

Using computers, the incredible formula  n² – 79n + 1601 was discovered, which produces 80 primes for the consecutive values n = 0 to 79. The product of the coefficients, -79 and 1601, is -126479.

n² + an + b, where |a| < 1000 and |b| < 1000

where |n| is the modulus/absolute value of n
e.g. |11| = 11 and |-4| = 4

Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n = 0.

For me this one reeks of brute force, since it is obvious that we can run through all possible values of a and b. Yes that is 4.000.000 solutions we need to check, but it should be possible. Continue reading →

Posted by Kristian in Project Euler, 37 comments

## Sum of all Primes below 2000000 – Problem 10

This blog post is all about the solution to Problem 10 of Project Euler. Just like Problem 7 the problem is all about primes.  And the solution strategy I posted for problem 7 would be valid for this problem as well.

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

Find the sum of all the primes below two million.

There is nothing particular tricky about this question, and since there isn’t a formula for finding all primes, we will have to brute force a solution. However, the brute force solution can be more or less elegant. In the source code I have made available, you can find the solution approach from problem 7, adopted to this problem. I wont spent any more time on that method, but instead introduce you to Sieve of Eratosthenes.

## Sieve of Eratosthenes

Sieve of Eratosthenes was as the name implies invented by Eratosthenes who was a Greek Mathematician living around 200 BC.

The algorithm needs to have an upper limit for the primes to find. Lets call this limit N

The algorithm works as follows.

1. Create a list l of consecutive integers {2,3,…,N}.
2. Select p as the first prime number in the list, p=2.
3. Remove all multiples of p from the l.
4. set p equal to the next integer in l which has not been removed.
5. Repeat steps 3 and 4 until p2 > N, all the remaining numbers in the list are primes

It is a pretty simple algorithm, and the description of it on Wikipedia has a really nice graphical illustration, which I decided I couldn’t do better my self. So go and have a look at it. The algorithm finds a prime, and then marks all multiples of that primes.  The first new number will always be a prime, since all numbers which are not will have been removed.

It can be pretty straight forward to implement the algorithm, and the challenge is definitely to optimize the implementation both execution and memory wise.

I found an implementation two implementation which are similar over at Stack Overflow and digitalBush which seemed very promising. I have then further optimized it a bit. The optimized code is shown here

```public int[] ESieve(int upperLimit) {
int sieveBound = (int)(upperLimit - 1) / 2;
int upperSqrt = ((int)Math.Sqrt(upperLimit) - 1) / 2;

BitArray PrimeBits = new BitArray(sieveBound + 1, true);

for (int i = 1; i <= upperSqrt; i++) {
if (PrimeBits.Get(i)) {
for (int j = i * 2 * (i + 1); j <= sieveBound; j += 2 * i + 1) {
PrimeBits.Set(j, false);
}
}
}

List numbers = new List((int)(upperLimit / (Math.Log(upperLimit) - 1.08366)));

for (int i = 1; i <= sieveBound; i++) {
if (PrimeBits.Get(i)) {
}
}

return numbers.ToArray();
}
```

Once we have a list of all the primes we need the rest of the code is a trivial for loop summing up the array. You can check the source code for that bit. In the following sections I will touch on different aspects of the code.

### Data representation

It uses a BitArray to store all the numbers. It is a enumerable type which uses one bit per boolean. Using a BitArray means the algorithm will limit the memory usage by a factor 32 compared to an array of booleans according this discussion. However, it will lower the operational performance. We need an array to hold 2.000.000 numbers, which means a difference of 250kB vs. 8MB.

I played around with it a bit, and didn’t see much of a performance difference for a small set of primes. For a large set of primes I noticed that the BitArray was slightly faster. This is likely due to better cache optimization, since the the BitArray is easier to store in the CPU cache, and thus increasing the performance.

### Eliminating even numbers

Over at digitalBush he optimizes his code by skipping the even numbers in the loops. I have chosen a bit of another approach, to avoid allocating memory for it. It just takes a bit of keeping track of my indexes.

Basically I want to  start with three and then for a counter i = {1,2,….,N/2} represent every odd number p = {3,5,7,….,N}. That can be done as p = 2i+1. And that is what I have done in the code. It makes the code a bit more complex, but saves half the memory, and thus it can treat even larger sets.

Furthermore we start our inner loop at p2 = (2i+1)(2i+1) = 4i2 + 4i + 1, which will have the index 2i(i+1), which is where we start the search of the inner loop. By increasing p=2i+1 indexes in every iteration of the inner loop, I am jumping 2p, and thus only taking the odd multiples of p. Since multiplying with an even number will give an even result, and therefore not a prime.

## Sieve of Atkin

Another Sieve method to generate primes is the Sieve of Atkin. It should be faster than the Sieve of Eratosthenes. I have made a reference implementation of it, but I can’t wrap my head around how to optimize it, so I have never been able to optimize it to the same degree as the Sieve of Eratosthenes. However, I have included the reference implementation in the source code, so you can play around with it if you like.

## Wrapping up

This post took a bit of a different approach. I have worked hard to really optimize the code. Since I believe that the Sieve method for finding primes will be used later. So making a reusable bit of fast code, will make later problems easier to solve.

I took three 3 different approaches and tried to optimize them.  The result of the three are

```Prime sum of all primes below 2000000 is 142913828922
Solution took 203,125 ms using Trial Division
Prime sum of all primes below 2000000 is 142913828922
Solution took 31,25 ms using Sieve of Eratosthenes
Prime sum of all primes below 2000000 is 142913828922
Solution took 31,25 ms using Sieve of Atkin
```

The difference between the two sieve methods are not really noticeable for such small numbers, but if we start calculating all primes below one billion, the differences in execution time will show. And as usual you can download the source code.

If you can come up with further optimizations of the sieve methods, I would appreciate you to leave a comment.  I am always eager to learn more and speeding things up further.

Posted by Kristian in Project Euler, 31 comments

## Project Euler – Problem 7

Now we reached Problem 7 of Project Euler which is about prime numbers. The problem reads

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?

We could solve this by brute force checking all the numbers, or we could reuse part of the solution to problem 5, where we generated a list of primes using trial division with already found prime numbers. I haven’t found a way to solve this without finding the 10000 primes before finding the answer.

Wikipedia comes with a great article about prime numbers, which also refers to several methods for checking if a number is a prime. We will take a bit of simpler approach though. But dive into the article, it is pretty interesting.

Posted by Kristian in Project Euler, 9 comments

## Project Euler – Problem 3

I am sorry, I haven’t posted anything for a while. I have been busy moving, and is currently without an Internet connection. However I couldn’t keep away any more.

Problem 3 in Project Euler reads:

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

What is the largest prime factor of the number 600851475143 ?

I used two different approaches for this, and lets get right to them. Continue reading →

Posted by Kristian in Project Euler, 27 comments