# Euler Problem 9 : Special Pythagorean Triple

## Euler Problem 9 Definition

Scatter plot of the legs (a,b) of the first Pythagorean triples with a and b less than 6000. Negative values are included to illustrate the parabolic patterns. By Dearjean13Own work, CC BY-SA 4.0, Link

A Pythagorean triple 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 of a, b and c.

## Brute Force Solution

This solution uses brute force and checks all combinations of a, b and c. To limit the solution space I used the fact that a < b < c, which implies that a < s/3,  and a < b < s/2, where s is the sum of the three sides.

```a <- 0
b <- 0
c <- 0
s <- 1000
found <- FALSE
for (a in 1:floor((s/3))) {
for (b in a:(s/2)) {
c <- s - a - b
if (a^2 + b^2 == c^2) {
found <- TRUE
break
}
}
if (found)
break
}
answer <- a * b * c
```

View this code on GitHub.

## 4 thoughts on “Euler Problem 9 : Special Pythagorean Triple”

1. Pingback: Euler Problem 9 : Special Pythagorean Triple - Use-R!Use-R!

2. That was fun.

You could speed that up slightly by having b loop through (a+1):(s/2) instead of a:(s/2) since b can’t equal a.

I tried a different tack. Since min(a)=1 and max(b)=499, I walked a and b towards each other. That way there’s a max of 498 possibilities instead of 292*497 in yours.

I ran yours and mine 100 times and collected the times.

user system elapsed
19.38 0.14 19.61

Mine:
user system elapsed
0.11 0.10 0.22

Neither of our codes verifies that c is an integer, and if there were a second solution, they would both miss the second.

```# Your code slightly modified
ptm <- proc.time()
for (round in 1:100) {
a <- 0
b <- 0
c <- 0
s <- 1000
found <- FALSE
for (a in 1:floor((s/3))) {
for (b in (a+1):(s/2)) {
c <- s - a - b
if (a^2 + b^2 == c^2) {
found <- TRUE
break
}
}
if (found) break
}
answer <- a * b * c
}
proc.time()-ptm

# Mine
ptm <- proc.time()
a <- 1
b <- 499
repeat{
c <- sqrt(a^2 + b^2)
if (a + b + c > 1000) {b <- b-1}
else if (a + b + c < 1000) {a <- a+1}
else if (a + b + c ==1000) {break}
}
answer <- a * b * c
proc.time()-ptm
```
• Thanks Alan,

Project Euler is indeed a lot of fun. I am aiming to do at least 100 problems.

I knew there had to be faster code, so thanks for contributing.

I toyed with the Euclidean formulas for generating triples but moved to the next problem.

Peter