JAGS, dinterval

Edit: I posed this question to Martyn Plummer and got a very useful response. I'll post it once I figure out the details.

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One of the hazards of using something like JAGS, with complicated innards written by someone else, is that you don't have much of a chance to the problems coming ahead.  I happen to have a continuous process which is observed only in discrete units and it happens to mimic the dinterval distribution in JAGS.  Unfortunately I recently had to change the number of breaks used by dinterval by an order of magnitude and it turns out that it bogs down fairly quickly.  Here's an example:


cat('
model{
    for ( i in 1:N) {
        x[i] ~ dunif(0,10^K)
        o[i] ~ dinterval(x[i],breaks)
    }
}
', file = './dintervalProblem.bugs'
)

library(rjags)
mod <- list()
samp <- list()
timings <- list()
k <- c(3,4,5)
for ( i in 1:length(k) ) {
    N <- 100
    K <- k[i]
    x <- runif(N,0,10^K)
    o <- ceiling(x)
    breaks <- seq(0,10^K,1)

    timings[[i]] <- system.time(
        {
            mod[[i]] <- jags.model(
                file = './dintervalProblem.bugs',
                data = list(
                    N = N,
                    K = K,
                    o = o,
                    breaks = breaks
                ),
                inits = list(
                    o = o,
                    x = x
                ),
                n.chains = 1,
                n.adapt = 1000,
                quiet = FALSE
            )

            samp[[i]] <- jags.samples(
                model = mod1,
                variable.names = c('x','o'),
                n.iter = 1000,
                thin = 1
            )
        }
    )

}

My digging in the JAGS internals hasn't turned up anything yet which could be responsible for this behavior.  The value of the dinterval node is calculated by:


static unsigned int value(vector const &par, unsigned int ncut)
{
    double t = T(par);
    for (unsigned int i = 0; i < ncut; ++i) {
  if (t <= CUTPOINTS(par)[i])
      return i;
    }
    return ncut;
}

It's not like there's a separate node for every breakpoint... though the number of nodes generated in the graphs does go up with K.

Suggestions?  (Especially all those folks coming here from Andrew Gelman's blog...)