Steve P. - 7 months ago 31

Java Question

I just finished Project Euler problem 9 (warning **spoilers**):

`A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,`

a^2 + b^2 = c^2

For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.

There exists exactly one Pythagorean triplet for which a + b + c = 1000.

Find the product abc.

Here's my solution:

`public static int specPyth(int num)`

{

for (int a = 1; a < num; a++)

for (int b = 2; b < a; b++)

{

if (a*a +b*b == (num-a-b)*(num-a-b))

return a*b*(num-a-b); //ans = 31875000

}

return -1;

}

I can't help but think that there's a solution that involves only one loop. Anyone have ideas? I'd prefer answers using only one loop, but anything that's more efficient than what I currently have would be nice.

Answer

```
if a + b +c = 1000
```

then

```
a + b + sqroot(a² + b²) = 1000
-> (a² + b²) = (1000 - a - b)²
-> a² + b² = 1000000 - 2000*(a+b) + a² + 2*a*b + b²
-> 0 = 1000000 - 2000*(a+b) + 2*a*b
-> ... (easy basic maths)
-> a = (500000 - 1000*b) / (1000 - b)
```

Then you try every b until you find one that makes a natural number out of a.

```
public static int specPyth(int num)
{
double a;
for (int b = 1; b < num/2; b++)
{
a=(num*num/2 - num*b)/(num - b);
if (a%1 == 0)
return (int) (a*b*(num-a-b));
}
return -1;
}
```

EDIT: b can't be higher than 499, because c>b and (b+c) would then be higher than 1000.

Source (Stackoverflow)