about loops

Hi,

One approach could be a while loop as follows. Note that your circle
should have radius 0.5 if I understood the problem correctly.


N <- 5

npoints <- 0
ntests <- 0
points.in.circle <- matrix(NA, ncol=2, nrow=N)

while (npoints < N) {

  test.point <- runif(2, -0.5, 0.5) # generate new point in 2D
  if ( sqrt(test.point %*% test.point) <= 0.5){ # test its distance to
the origin
    npoints <- npoints + 1
    points.in.circle[npoints, ] <- test.point
  }
  ntests <- ntests + 1
}

print(paste(npoints, "out of", ntests, 'where in the circle') )

HTH,

baptiste


2010/1/21 Wolfgang?Amadeus <fdfcz123 at 163.com>:

Hello !
?I have a quick question about loops.
?For example, ?I have a 1 * 1 square and its inscribed circle tangent i, whose radius, of course, is also 1.
?The loop here is as the following:
?generate n points, say 5, in the square randomly repeatedly until we have five in total in the circle, then we stop, otherwise we continue.
?I do not know how !
Help me Please ~
Thank you very much for your time.
?Yours
Wolfgang Amadeus


? ? ? ?[[alternative HTML version deleted]]

______________________________________________
R-help at r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Hello !
I have a quick question about loops.
For example,  I have a 1 * 1 square and its inscribed circle tangent  
i, whose radius, of course, is also 1.
The loop here is as the following:
generate n points, say 5, in the square randomly repeatedly until we  
have five in total in the circle, then we stop, otherwise we continue.
I do not know how !
Help me Please ~
Thank you very much for your time.
Yours
Wolfgang Amadeus

What part don't you know how to do? Sounds like homework (especially  
with the piteous appeals from a pseudonym), so we expect more evidence  
of effort on your part.

The condition for being "outside", assuming this box's lower left-hand  
corner is (0,0) is

(x-0.5)^2 + (y-0.5)^2 > 1

David Winsemius, MD
Heritage Laboratories
West Hartford, CT