Something like this?
n<-1000
pythag<-function(n){
for(i in 3:((n-3)/3))
for(j in (i+1):((n-i)/2))
if(i^2+j^2==(n-i-j)^2) return(i*j*(n-i-j))
}
system.time(print(pythag(n)))
Interesting -- seems to improve speed by ~12%. Not sure if that's
because stop() has more overhead, or if functions are innately faster
somehow. Still seems like there should be a way to break out of nested
loops, though...
-b
On Tue, Dec 18, 2012 at 4:50 PM, Duncan Murdoch
<murdoch.dun...@gmail.com> wrote:
> 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
>
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