## Project Euler: Problem 58 – Spiral primes

#### Problem 58:

Starting with 1 and spiralling anticlockwise in the following way, a square spiral with side length 7 is formed.

```37 36 35 34 33 32 31
38 17 16 15 14 13 30
39 18  5  4  3 12 29
40 19  6  1  2 11 28
41 20  7  8  9 10 27
42 21 22 23 24 25 26
43 44 45 46 47 48 49```

It is interesting to note that the odd squares lie along the bottom right diagonal, but what is more interesting is that 8 out of the 13 numbers lying along both diagonals are prime; that is, a ratio of 8/13 ~ 62%.

If one complete new layer is wrapped around the spiral above, a square spiral with side length 9 will be formed. If this process is continued, what is the side length of the square spiral for which the ratio of primes along both diagonals first falls below 10%?

#### Idea:

Even though I already have a spiral generating method used in Problem 28, I don’t need to generate anything: the diagonals are easily recognized by a pattern of numbers equally spaced apart per layer. Determine which numbers are on a diagonal, check to see if it’s prime, and keep track of the total numbers on the diagonal. While the ratio of primes along the diagonals to the total number of numbers along the diagonal is greater than 10%, keep adding a new layer.

```int answer = -1;

boolean LESS_THAN_10_PERCENT = false;
double numberOfPrimes = 0;
int totalNumbers = 1;
int increment = 0;
int start = 1;

while(!LESS_THAN_10_PERCENT) {
increment += 2;
for(int i = 0; i < 4; i++) {
start += increment;
totalNumbers++;
if(EulerUtils.isProbablePrime(start)) {
numberOfPrimes++;
}
}
double percentOfPrimes = numberOfPrimes/totalNumbers;

if(percentOfPrimes < 0.1) {
System.out.println("Calculated spiral of width: "+(increment+1)+
" and ratio of: "+percentOfPrimes);
LESS_THAN_10_PERCENT = true;
}
}```

## Project Euler: Problem 28 – Number spiral diagonals

#### Problem 28:

Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:

```21 22 23 24 25
20  7  8  9 10
19  6  1  2 11
18  5  4  3 12
17 16 15 14 13```

It can be verified that the sum of the numbers on the diagonals is 101.

What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?

#### Idea:

Given the performance limits (or in this case, lack thereof) of modern computers, I figured it would be trivial for a modern computer to make a two-dimensional array of 1001×1001. I decided it would be easiest to just recreate the spiral and add the diagonals.

```long answer = -1;

int[][] spiral = new int;
int number = spiral.length*spiral.length;

for(int level = 0; level < = spiral.length/2; level++) {
//top row : reverse
for(int i = spiral.length-level-1; i >= level; i--) {
spiral[level][i] = number;
number--;
}
//left column: down
for(int i = level+1; i < spiral.length-level; i++) {
spiral[i][level] = number;
number--;
}
//bottom row: forward
for(int i = level+1; i < spiral.length-level; i++) {
spiral[spiral.length-1-level][i] = number;
number--;
}
//right column: up
for(int i = spiral.length-1-level-1; i > level; i--) {
spiral[i][spiral.length-1-level] = number;
number--;
}
}