I'm attempting to run some fairly deep recursive code in R and it keeps giving me this error:
Error: C stack usage is too close to the limit
My output from CStack_info()
is:
Cstack_info()
size current direction eval_depth
67108864 8120 1 2
I have plenty of memory on my machine, I'm just trying to figure out how I can increase the CStack for R.
EDIT: Someone asked for a reproducible example. Here's some basic sample code that causes the problem. Running f(1,1) a few times you'll get the error. Note that I've already set --max-ppsize = 500000 and options(expressions=500000) so if you don't set those you might get an error about one of those two things instead. As you can see, the recursion can go pretty deep here and I've got no idea how to get it to work consistently. Thanks.
f <- function(root=1,lambda=1) {
x <- c(0,1);
prob <- c(1/(lambda+1),lambda/(lambda+1));
repeat {
if(root == 0) {
break;
}
else {
child <- sample(x,2,replace=TRUE,prob);
if(child[1] == 0 && child[2] == 0) {
break;
}
if(child[1] == 1) {
child[1] <- f(root=child[1],lambda);
}
if(child[2] == 1 && child[1] == 0) {
child[2] <- f(root=child[2],lambda);
}
}
if(child[1] == 0 && child[2] == 0) {
break;
}
if(child[1] == 1 || child[2] == 1) {
root <- sample(x,1,replace=TRUE,prob);
}
}
return(root)
}
I suspect that, regardless of stack limit, you'll end up with recursions that are too deep. For instance, with lambda = Inf, f(1) leads to an immediate recursion, indefinitely. The depth of the recursion seems to be a random walk, with some probability r of going deeper, 1 - r of finishing the current recursion. By the time you've hit the stack limit, you've made a large number of steps 'deeper'. This implies that r > 1 / 2, and the very large majority of time you'll just continue to recurse.
Also, it seems like it is almost possible to derive an analytic or at least numerical solution even in the face of infinite recursion. One can define p as the probability that f(1) == 1, write implicit expressions for the 'child' states after a single iteration, and equate these with p, and solve. p can then be used as the chance of success in a single draw from a binomial distribution.