R help - Dec 2012 - Breaking out of multiple loops

Hey all,

I'm currently working through the problems at Project Euler -- this
question came up while working on Problem 9
(http://projecteuler.net/problem=9):

"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."

Not too hard:

n=1000
for(i in 1:floor(n/3))
  for(j in (i+1):floor(n/2-i/2))
    if(i^2+j^2==(n-i-j)^2) {print(i*j*(n-i-j)); stop()}

I could just let the for loops finish looping after it finds the
answer, and it would still run in under a second, but the goal of
Project Euler is sort of to see how efficiently (and quickly) you can
solve these problems, so in that spirit I would like to break out of
the for loops early once the answer is found -- hence the call to
stop(). However, this seems "improper", as it throws up an error. Is
there a way to exit out of both for loops with a call to "break" or
similar that would not throw errors (or is it fine the way I've coded
it)? (I realize I could put an "if(i^2+j^2==(n-i-j)^2) break"
statement in the outer loop, but again that's inefficient, as it's
checking that conditional hundreds of times.)

So is there a way to "cleanly" break out of multiple loops?

Thanks,
-bryan

On 12-12-18 1:02 PM, McCloskey, Bryan wrote:
> Hey all,
>
> I'm currently working through the problems at Project Euler -- this
> question came up while working on Problem 9
> (http://projecteuler.net/problem=9):
>
> "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."
>
> Not too hard:
>
> n=1000
> for(i in 1:floor(n/3))
>    for(j in (i+1):floor(n/2-i/2))
>      if(i^2+j^2==(n-i-j)^2) {print(i*j*(n-i-j)); stop()}
>
> I could just let the for loops finish looping after it finds the
> answer, and it would still run in under a second, but the goal of
> Project Euler is sort of to see how efficiently (and quickly) you can
> solve these problems, so in that spirit I would like to break out of
> the for loops early once the answer is found -- hence the call to
> stop(). However, this seems "improper", as it throws up an error.
Is
> there a way to exit out of both for loops with a call to "break"
or
> similar that would not throw errors (or is it fine the way I've coded
> it)? (I realize I could put an "if(i^2+j^2==(n-i-j)^2) break"
> statement in the outer loop, but again that's inefficient, as it's
> checking that conditional hundreds of times.)
>
> So is there a way to "cleanly" break out of multiple loops?
Put them in a function, and return from the function.

Duncan Murdoch