/*
* Solves for times by processing samples in the active list in parallel.
*/
private void solveParallel(
final ActiveList al,
final float[][][] t,
final int m,
final float[][][] times,
final int[][][] marks) {
int mbmin = 64; // target minimum number of samples per block
int nbmax = 256; // maximum number of blocks
final float[][] dtask = new float[nbmax][];
final ActiveList[] bltask = new ActiveList[nbmax];
while (!al.isEmpty()) {
final int n = al.size(); // number of samples in active (A) list
final int mbmax = max(mbmin, 1 + (n - 1) / nbmax); // max samples per block
final int nb = 1 + (n - 1) / mbmax; // number of blocks <= nbmax
final int mb = 1 + (n - 1) / nb; // evenly distribute samples per block
Parallel.loop(
nb,
new Parallel.LoopInt() { // for all blocks, ...
public void compute(int ib) {
if (bltask[ib] == null) { // if necessary for this block, make ...
dtask[ib] = new float[6]; // work array for tensor coefficients
bltask[ib] = new ActiveList(); // and an empty active list
}
int i = ib * mb; // beginning of block
int j = min(i + mb, n); // beginning of next block (or end)
for (int k = i; k < j; ++k) { // for each sample in block, ...
Sample s = al.get(k); // get k'th sample from A list
solveOne(t, m, times, marks, s, bltask[ib], dtask[ib]); // do sample
}
bltask[ib].setAllAbsent(); // needed when merging B lists below
}
});
// Merge samples from all B lists to a new A list. All samples
// in B lists are currently marked as absent in the A list. As
// samples in B lists are appended to the A list, their absent
// flags are set to false, so that no sample is appended more
// than once to the new A list.
al.clear();
for (int ib = 0; ib < nb; ++ib) {
if (bltask[ib] != null) {
al.appendIfAbsent(bltask[ib]);
bltask[ib].clear();
}
}
}
}
/*
* Solves for times by sequentially processing each sample in active list.
*/
private void solveSerial(
ActiveList al, float[][][] t, int m, float[][][] times, int[][][] marks) {
float[] d = new float[6];
ActiveList bl = new ActiveList();
int ntotal = 0;
while (!al.isEmpty()) {
// al.shuffle(); // demonstrate that solution depends on order
int n = al.size();
ntotal += n;
for (int i = 0; i < n; ++i) {
Sample s = al.get(i);
solveOne(t, m, times, marks, s, bl, d);
}
bl.setAllAbsent();
al.clear();
al.appendIfAbsent(bl);
bl.clear();
}
trace("solveSerial: ntotal=" + ntotal);
trace(" nratio=" + (float) ntotal / (float) (_n1 * _n2 * _n3));
}
/*
* Solves for times by processing samples in the active list in parallel.
*/
private void solveParallelX(
final ActiveList al,
final float[][][] t,
final int m,
final float[][][] times,
final int[][][] marks) {
int nthread = Runtime.getRuntime().availableProcessors();
ExecutorService es = Executors.newFixedThreadPool(nthread);
CompletionService<Void> cs = new ExecutorCompletionService<Void>(es);
ActiveList[] bl = new ActiveList[nthread];
float[][] d = new float[nthread][];
for (int ithread = 0; ithread < nthread; ++ithread) {
bl[ithread] = new ActiveList();
d[ithread] = new float[6];
}
final AtomicInteger ai = new AtomicInteger();
int ntotal = 0;
// int niter = 0;
while (!al.isEmpty()) {
ai.set(0); // initialize the shared block index to zero
final int n = al.size(); // number of samples in active (A) list
ntotal += n;
final int mb = 32; // size of blocks of samples
final int nb = 1 + (n - 1) / mb; // number of blocks of samples
int ntask = min(nb, nthread); // number of tasks (threads to be used)
for (int itask = 0; itask < ntask; ++itask) { // for each task, ...
final ActiveList bltask = bl[itask]; // task-specific B list
final float[] dtask = d[itask]; // task-specific work array
cs.submit(
new Callable<Void>() { // submit new task
public Void call() {
for (int ib = ai.getAndIncrement(); ib < nb; ib = ai.getAndIncrement()) {
int i = ib * mb; // beginning of block
int j = min(i + mb, n); // beginning of next block (or end)
for (int k = i; k < j; ++k) { // for each sample in block, ...
Sample s = al.get(k); // get k'th sample from A list
solveOne(t, m, times, marks, s, bltask, dtask); // process sample
}
}
bltask.setAllAbsent(); // needed when merging B lists below
return null;
}
});
}
try {
for (int itask = 0; itask < ntask; ++itask) cs.take();
} catch (InterruptedException e) {
throw new RuntimeException(e);
}
// Merge samples from all B lists to a new A list. As samples
// are appended, their absent flags are set to false, so that
// each sample is appended no more than once to the new A list.
al.clear();
for (int itask = 0; itask < ntask; ++itask) {
al.appendIfAbsent(bl[itask]);
bl[itask].clear();
}
// ++niter;
}
es.shutdown();
// trace("solveParallel: ntotal="+ntotal);
// trace(" nratio="+(float)ntotal/(float)(_n1*_n2*_n3));
}