Optimising loops for pixel processing - optimization

I'm implementing an algorithm (in OpenCV) that iterates over every pixel in an image, and for each pixel calculates block matches with pixels in the neighbourhood in order to evalaute the similarity of these neighbouring pixels. A "naive" implementation with very deep loops is very slow, so I was wondering how I might try to improve the performance. The following is an extract of my current code:
for(nCh=1;nCh<=channels;nCh++) { // Loop over three channels
for(i=0;i<h;i++) { // "vertical" loop
for(j=0;j<w;j++) { // "horizontal" loop
for (si=-sw_height; si<sw_height; si++){ // vertical search window loop
for (sj=-sw_width; sj<sw_width; sj++){ // horizontal search window loop
dist = 0;
for (blki=0; blki<blk_height; blki++){ // block match loop
for (blkj=0; blkj<blk_width; blkj++){ // block match loop
current_pxl = data[(i+blki)*step+(j+blkj)*channels+nCh];
search_pxl = data[(i+blki+si)*step+(j+blkj+sj)*channels+nCh];
dist += pow((current_pxl - search_pxl),2);
}
}
// ... further processing
}
}
}
}
}

You're calling pow in the innermost loop. Don't.
Also you're doing a lot of index calculation in there.
I bet you can move some out that out of the inner loop.
You should be able to get it so your inner loop looks more like this:
for (blkj = 0; blkj < blk_width; blkj++, pc += channels, ps += channels){
int diff = (*pc - *ps);
dist += (diff * diff);
}
And then, you might even want to unroll it a bit.
BTW, a little more whitespace might help :-)

Related

Parallel Dynamic Programming with CUDA

It is my first attempt to implement recursion with CUDA. The goal is to extract all the combinations from a set of chars "12345" using the power of CUDA to parallelize dynamically the task. Here is my kernel:
__device__ char route[31] = { "_________________________"};
__device__ char init[6] = { "12345" };
__global__ void Recursive(int depth) {
// up to depth 6
if (depth == 5) return;
// newroute = route - idx
int x = depth * 6;
printf("%s\n", route);
int o = 0;
int newlen = 0;
for (int i = 0; i<6; ++i)
{
if (i != threadIdx.x)
{
route[i+x-o] = init[i];
newlen++;
}
else
{
o = 1;
}
}
Recursive<<<1,newlen>>>(depth + 1);
}
__global__ void RecursiveCount() {
Recursive <<<1,5>>>(0);
}
The idea is to exclude 1 item (the item corresponding to the threadIdx) in each different thread. In each recursive call, using the variable depth, it works over a different base (variable x) on the route device variable.
I expect the kernel prompts something like:
2345_____________________
1345_____________________
1245_____________________
1234_____________________
2345_345_________________
2345_245_________________
2345_234_________________
2345_345__45_____________
2345_345__35_____________
2345_345__34_____________
..
2345_245__45_____________
..
But it prompts ...
·_____________
·_____________
·_____________
·_____________
·_____________
·2345
·2345
·2345
·2345
...
What I´m doing wrong?
What I´m doing wrong?
I may not articulate every problem with your code, but these items should get you a lot closer.
I recommend providing a complete example. In my view it is basically required by Stack Overflow, see item 1 here, note use of the word "must". Your example is missing any host code, including the original kernel call. It's only a few extra lines of code, why not include it? Sure, in this case, I can deduce what the call must have been, but why not just include it? Anyway, based on the output you indicated, it seems fairly evident the launch configuration of the host launch would have to be <<<1,1>>>.
This doesn't seem to be logical to me:
I expect the kernel prompts something like:
2345_____________________
The very first thing your kernel does is print out the route variable, before making any changes to it, so I would expect _____________________. However we can "fix" this by moving the printout to the end of the kernel.
You may be confused about what a __device__ variable is. It is a global variable, and there is only one copy of it. Therefore, when you modify it in your kernel code, every thread, in every kernel, is attempting to modify the same global variable, at the same time. That cannot possibly have orderly results, in any thread-parallel environment. I chose to "fix" this by making a local copy for each thread to work on.
You have an off-by-1 error, as well as an extent error in this loop:
for (int i = 0; i<6; ++i)
The off-by-1 error is due to the fact that you are iterating over 6 possible items (that is, i can reach a value of 5) but there are only 5 items in your init variable (the 6th item being a null terminator. The correct indexing starts out over 0-4 (with one of those being skipped). On subsequent iteration depths, its necessary to reduce this indexing extent by 1. Note that I've chosen to fix the first error here by increasing the length of init. There are other ways to fix, of course. My method inserts an extra _ between depths in the result.
You assume that at each iteration depth, the correct choice of items is the same, and in the same order, i.e. init. However this is not the case. At each depth, the choices of items must be selected not from the unchanging init variable, but from the choices passed from previous depth. Therefore we need a local, per-thread copy of init also.
A few other comments about CUDA Dynamic Parallelism (CDP). When passing pointers to data from one kernel scope to a child scope, local space pointers cannot be used. Therefore I allocate for the local copy of route from the heap, so it can be passed to child kernels. init can be deduced from route, so we can use an ordinary local variable for myinit.
You're going to quickly hit some dynamic parallelism (and perhaps memory) limits here if you continue this. I believe the total number of kernel launches for this is 5^5, which is 3125 (I'm doing this quickly, I may be mistaken). CDP has a pending launch limit of 2000 kernels by default. We're not hitting this here according to what I see, but you'll run into that sooner or later if you increase the depth or width of this operation. Furthermore, in-kernel allocations from the device heap are by default limited to 8KB. I don't seem to be hitting that limit, but probably I am, so my design should probably be modified to fix that.
Finally, in-kernel printf output is limited to the size of a particular buffer. If this technique is not already hitting that limit, it will soon if you increase the width or depth.
Here is a worked example, attempting to address the various items above. I'm not claiming it is defect free, but I think the output is closer to your expectations. Note that due to character limits on SO answers, I've truncated/excerpted some of the output.
$ cat t1639.cu
#include <stdio.h>
__device__ char route[31] = { "_________________________"};
__device__ char init[7] = { "12345_" };
__global__ void Recursive(int depth, const char *oroute) {
char *nroute = (char *)malloc(31);
char myinit[7];
if (depth == 0) memcpy(myinit, init, 6);
else memcpy(myinit, oroute+(depth-1)*6, 6);
myinit[6] = 0;
if (nroute == NULL) {printf("oops\n"); return;}
memcpy(nroute, oroute, 30);
nroute[30] = 0;
// up to depth 6
if (depth == 5) return;
// newroute = route - idx
int x = depth * 6;
//printf("%s\n", nroute);
int o = 0;
int newlen = 0;
for (int i = 0; i<(6-depth); ++i)
{
if (i != threadIdx.x)
{
nroute[i+x-o] = myinit[i];
newlen++;
}
else
{
o = 1;
}
}
printf("%s\n", nroute);
Recursive<<<1,newlen>>>(depth + 1, nroute);
}
__global__ void RecursiveCount() {
Recursive <<<1,5>>>(0, route);
}
int main(){
RecursiveCount<<<1,1>>>();
cudaDeviceSynchronize();
}
$ nvcc -o t1639 t1639.cu -rdc=true -lcudadevrt -arch=sm_70
$ cuda-memcheck ./t1639
========= CUDA-MEMCHECK
2345_____________________
1345_____________________
1245_____________________
1235_____________________
1234_____________________
2345__345________________
2345__245________________
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...
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1234__234________________
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1234__1234__123___12____1
========= ERROR SUMMARY: 0 errors
$
The answer given by Robert Crovella is correct at the 5th point, the mistake was in the using of init in every recursive call, but I want to clarify something that can be useful for other beginners with CUDA.
I used this variable because when I tried to launch a child kernel passing a local variable I always got the exception: Error: a pointer to local memory cannot be passed to a launch as an argument.
As I´m C# expert developer I´m not used to using pointers (Ref does the low-level-work for that) so I thought there was no way to do it in CUDA/c programming.
As Robert shows in its code it is possible copying the pointer with memalloc for using it as a referable argument.
Here is a kernel simplified as an example of deep recursion.
__device__ char init[6] = { "12345" };
__global__ void Recursive(int depth, const char* route) {
// up to depth 6
if (depth == 5) return;
//declaration for a referable argument (point 6)
char* newroute = (char*)malloc(6);
memcpy(newroute, route, 5);
int o = 0;
int newlen = 0;
for (int i = 0; i < (6 - depth); ++i)
{
if (i != threadIdx.x)
{
newroute[i - o] = route[i];
newlen++;
}
else
{
o = 1;
}
}
printf("%s\n", newroute);
Recursive <<<1, newlen>>>(depth + 1, newroute);
}
__global__ void RecursiveCount() {
Recursive <<<1, 5>>>(0, init);
}
I don't add the main call because I´m using ManagedCUDA for C# but as Robert says it can be figured-out how the call RecursiveCount is.
About ending arrays of char with /0 ... sorry but I don't know exactly what is the benefit; this code works fine without them.

Optimized recalculating all pairs shortest path when removing vertexes dynamically from an undirected graph

I use following dijkstra implementation to calculate all pairs shortest paths in an undirected graph. After calling calculateAllPaths(), dist[i][j] contains shortest path length between i and j (or Integer.MAX_VALUE if no such path available).
The problem is that some vertexes of my graph are removing dynamically and I should recalculate all paths from scratch to update dist matrix. I'm seeking for a solution to optimize update speed by avoiding unnecessary calculations when a vertex removes from my graph. I already search for solution and I now there is some algorithms such as LPA* to do this, but they seem very complicated and I guess a simpler solution may solve my problem.
public static void calculateAllPaths()
{
for(int j=graph.length/2+graph.length%2;j>=0;j--)
{
calculateAllPathsFromSource(j);
}
}
public static void calculateAllPathsFromSource(int s)
{
final boolean visited[] = new boolean[graph.length];
for (int i=0; i<dist.length; i++)
{
if(i == s)
{
continue;
}
//visit next node
int next = -1;
int minDist = Integer.MAX_VALUE;
for (int j=0; j<dist[s].length; j++)
{
if (!visited[j] && dist[s][j] < minDist)
{
next = j;
minDist = dist[s][j];
}
}
if(next == -1)
{
continue;
}
visited[next] = true;
for(int v=0;v<graph.length;v++)
{
if(v == next || graph[next][v] == -1)
{
continue;
}
int md = dist[s][next] + graph[next][v];
if(md < dist[s][v])
{
dist[s][v] = dist[v][s] = md;
}
}
}
}
If you know that vertices are only being removed dynamically, then instead of just storing the best path matrix dist[i][j], you could also store the permutation of each such path. Say, instead of dist[i][j] you make a custom class myBestPathInfo, and the array of an instance of this, say myBestPathInfo[i][j], contain members best distance as well as permutation of the best path. Preferably, the best path permutation is described as an ordered set of some vertex objects, where the latter are of reference type and unique for each vertex (however used in several myBestPathInfo instances). Such objects could include a boolean property isActive (true/false).
Whenever a vertex is removed, you traverse through the best path permutations for each vertex-vertex pair, to make sure no vertex has been deactivated. Finally, only for broken paths (deactivated vertices) do you re-run Dijkstra's algorithm.
Another solution would be to solve the shortest path for all pairs using linear programming (LP) techniques. A removed vertex can be easily implemented as an additional constraint in your program (e.g. flow in <=0 and and flow out of vertex <= 0*), after which the re-solving of the shortest path LP:s can use the previous optimal solution as a feasible basic feasible solution (BFS) in the dual LPs. This property holds since adding a constraint in the primal LP is equivalent to an additional variable in the dual; hence, previously optimal primal BFS will be feasible in dual after additional constraints. (on-the-fly starting on simplex solver for LPs).

A* Pathfinding - how to modify G and H to include rough terrain movement cost?

I have A* pathfinding implemented in my 2D game and it works well on a plain map with obstacles. Now I'm trying to understand how to modify the algorithm, so it counts rough terrain (hills, forest, etc) as 2 moves instead of 1.
With the 1 movement cost, the algorithm uses integers 10 and 14 in the move cost function. Im interested in how to modify these values if one cell actually has a movement cost of 2? will it be 20:17?
Here's how my current algorithm currently computes G and H (adopted from Ray Wenderleich):
// Compute the H score from a position to another (from the current position to the final desired position
- (int)computeHScoreFromCoord:(CGPoint)fromCoord toCoord:(CGPoint)toCoord
{
// Here we use the Manhattan method, which calculates the total number of step moved horizontally and vertically to reach the
// final desired step from the current step, ignoring any obstacles that may be in the way
return abs(toCoord.x - fromCoord.x) + abs(toCoord.y - fromCoord.y);
}
// Compute the cost of moving from a step to an adjecent one
- (int)costToMoveFromStep:(ShortestPathStep *)fromStep toAdjacentStep:(ShortestPathStep *)toStep
{
return ((fromStep.position.x != toStep.position.x)
&& (fromStep.position.y != toStep.position.y))
? 14 : 10;
}
If some of the edges have movement cost 2, you will simply add 2 to the G of the parent node, rather than 1.
As for H: it doesn't need to change. The resulting heuristic will still be admissible/consistent.
I think I got it, with this line the tutorial author checks if the move is 1 square or 2 squares(diagonal) from the move that is currently being considered.
return ((fromStep.position.x != toStep.position.x)
&& (fromStep.position.y != toStep.position.y))
? 14 : 10;
Unfortunately, this is a really simple case and does not really explain what has to be done. Number 10 is used to make calculations easier (10 = 1 move cost), and (14 = 1 diagonal move) is an approximation of sqrt(10*10).
I attempted to introduce terrain cost below, and this requires extra information - I need to know which cell I'm going through to reach the destination. This turned out to be really annoying, and the code below is clearly not my best, but I attempted to spell out what's going on at each step.
If I'm making a diagonal move, I need to know it's move cost AND the move cost of 2 squares that can be used to get there. I can then pick the lowest movement cost among two squares and plug it into the equation of the form:
moveCost = (int)sqrt(lowestMoveCost*lowestMoveCost + (stepNode.moveCost*10) * (stepNode.moveCost*10));
Here's the entire loop that checks adjacent steps and creates new steps out of them with the move cost. It finds tile in my map array and returns it's terrain cost.
NSArray *adjSteps = [self walkableAdjacentTilesCoordForTileCoord:currentStep.position];
for (NSValue *v in adjSteps) {
ShortestPathStep *step = [[ShortestPathStep alloc] initWithPosition:[v CGPointValue]];
// Check if the step isn't already in the closed set
if ([self.spClosedSteps containsObject:step]) {
continue; // Ignore it
}
tileIndex = [MapOfTiles tileIndexForCoordinate:step.position];
DLog(#"point (x%.0f y%.0f):%i",step.position.x,step.position.y,tileIndex);
stepNode = [[MapOfTiles sharedInstance] mapTiles] [tileIndex];
// int moveCost = [self costToMoveFromStep:currentStep toAdjacentStep:step];
//in my case 0,0 is bottom left, y points up x points right
if((currentStep.position.x != step.position.x) && (currentStep.position.y != step.position.y))
{
//move one step away - easy, multiply move cost by 10
moveCost = stepNode.moveCost*10;
}else
{
possibleMove1 = 0;
possibleMove2 = 0;
//we are moving diagonally, figure out in which direction
if(step.position.y > currentStep.position.y)
{
//moving up
possibleMove1 = tileIndex + 1;
if(step.position.x > currentStep.position.x)
{
//moving right and up
possibleMove2 = tileIndex + tileCountTall;
}else
{
//moving left and up
possibleMove2 = tileIndex - tileCountTall;
}
}else
{
//moving down
possibleMove1 = tileIndex - 1;
if(step.position.x > currentStep.position.x)
{
//moving right and down
possibleMove2 = tileIndex + tileCountTall;
}else
{
//moving left and down
possibleMove2 = tileIndex - tileCountTall;
}
}
moveNode1 = nil;
moveNode2 = nil;
CGPoint coordinate1 = [MapOfTiles tileCoordForIndex:possibleMove1];
CGPoint coordinate2 = [MapOfTiles tileCoordForIndex:possibleMove2];
if([adjSteps containsObject:[NSValue valueWithCGPoint:coordinate1]])
{
//we know that possible move to reach destination has been deemed walkable, get it's move cost from the map
moveNode1 = [[MapOfTiles sharedInstance] mapTiles] [possibleMove1];
}
if([adjSteps containsObject:[NSValue valueWithCGPoint:coordinate2]])
{
//we know that the second possible move is walkable
moveNode2 = [[MapOfTiles sharedInstance] mapTiles] [possibleMove2];
}
#warning not sure about this one if the algorithm has to backtrack really far back
//find out which square has the lowest move cost
lowestMoveCost = fminf(moveNode1.moveCost, moveNode2.moveCost) * 10;
moveCost = (int)sqrt(lowestMoveCost*lowestMoveCost + (stepNode.moveCost*10) * (stepNode.moveCost*10));
}
// Compute the cost form the current step to that step
// Check if the step is already in the open list
NSUInteger index = [self.spOpenSteps indexOfObject:step];
if (index == NSNotFound) { // Not on the open list, so add it
// Set the current step as the parent
step.parent = currentStep;
// The G score is equal to the parent G score + the cost to move from the parent to it
step.gScore = currentStep.gScore + moveCost;
// Compute the H score which is the estimated movement cost to move from that step to the desired tile coordinate
step.hScore = [self computeHScoreFromCoord:step.position toCoord:toTileCoord];
// Adding it with the function which is preserving the list ordered by F score
[self insertInOpenSteps:step];
}
else { // Already in the open list
step = (self.spOpenSteps)[index]; // To retrieve the old one (which has its scores already computed ;-)
// Check to see if the G score for that step is lower if we use the current step to get there
if ((currentStep.gScore + moveCost) < step.gScore) {
// The G score is equal to the parent G score + the cost to move from the parent to it
step.gScore = currentStep.gScore + moveCost;
// Because the G Score has changed, the F score may have changed too
// So to keep the open list ordered we have to remove the step, and re-insert it with
// the insert function which is preserving the list ordered by F score
// Now we can removing it from the list without be afraid that it can be released
[self.spOpenSteps removeObjectAtIndex:index];
// Re-insert it with the function which is preserving the list ordered by F score
[self insertInOpenSteps:step];
}
}
}
These types of problems are quite common in, say, chip routing and, yes, gamedev.
Standard approach is to have your graph (in C++ I would say you have Boost "grid graph" or similar structure). If you can afford to have an object each vertex, then the solution is quite easy.
You connect two vertices (neighbors or diagonally adjacent) by an edge, unless there is an obstacle between them. You assign this edge a weight equal to edge length (10 or 14) times terrain cost. Sometimes people prefer not to exclude obstacle edges but assign extremely high weights to them (an advantage is that with such approach you are guaranteed to find at least some path, even when object is stuck at an island).
Then you apply A* algorithm. Your heuristic function (H) can be "pessimistic" (equal to Euclidean distance times the max move cost) or "optimistic" (Euclidean distance times min move cost) or anything in between. Different heuristics will result in slightly different "personalities" of your search but usually do not matter much.

game maker random cave generation

I want to make a cave explorer game in game maker 8.0.
I've made a block object and an generator But I'm stuck. Here is my code for the generator
var r;
r = random_range(0, 1);
repeat(room_width/16) {
repeat(room_height/16) {
if (r == 1) {
instance_create(x, y, obj_block)
}
y += 16;
}
x += 16;
}
now i always get a blank frame
You need to use irandom(1) so you get an integer. You also should put it inside the loop so it generates a new value each time.
In the second statement, you are generating a random real value and storing it in r. What you actually require is choosing one of the two values. I recommend that you use the function choose(...) for this. Here goes the corrected statement:
r = choose(0,1); //Choose either 0 or 1 and store it in r
Also, move the above statement to the inner loop. (Because you want to decide whether you want to place a block at the said (x,y) location at every spot, right?)
Also, I recommend that you substitute sprite_width and sprite_height instead of using the value 16 directly, so that any changes you make to the sprite will adjust the resulting layout of the blocks accordingly.
Here is the code with corrections:
var r;
repeat(room_width/sprite_width) {
repeat(room_height/sprite_height) {
r = choose(0, 1);
if (r == 1)
instance_create(x, y, obj_block);
y += sprite_height;
}
x += sprite_width;
}
That should work. I hope that helps!
Looks like you are only creating a instance if r==1. Shouldn't you create a instance every time?
Variable assignment r = random_range(0, 1); is outside the loop. Therefore performed only once before starting the loop.
random_range(0, 1) returns a random real number between 0 and 1 (not integer!). But you have if (r == 1) - the probability of getting 1 is a very small.
as example:
repeat(room_width/16) {
repeat(room_height/16) {
if (irandom(1)) {
instance_create(x, y, obj_block)
}
y += 16;
}
x += 16;
}
Here's a possible, maybe even better solution:
length = room_width/16;
height = room_height/16;
for(xx = 0; xx < length; xx+=1)
{
for(yy = 0; yy < height; yy+=1)
{
if choose(0, 1) = 1 {
instance_create(xx*16, yy*16, obj_block); }
}
}
if you want random caves, you should probably delete random sections of those blocks,
not just single ones.
For bonus points, you could use a seed value for the random cave generation. You can also have a pathway random generation that will have a guaranteed path to the finish with random openings and fake paths that generate randomly from that path. Then you can fill in the extra spaces with other random pieces.
But in regards to your code, you must redefine the random number each time you are placing a block, which is why all of them are the same. It should be called inside of the loops, and should be an integer instead of a decimal value.
Problem is on the first line, you need to put r = something in the for cycle

Looping with iterator vs temp object gives different result graphically (Libgdx/Java)

I've got a particle "engine" whom I've implementing a Pool system to and I've tested two different ways of rendering every Particle in a list. Please note that the Pooling really doesn't have anything with the problem to do. I just followed a tutorial and tried to use the second method when I noticed that they behaved differently.
The first way:
for (int i = 0; i < particleList.size(); i++) {
Iterator<Particle> it = particleList.iterator();
while (it.hasNext()) {
Particle p = it.next();
if (p.isDead()){
it.remove();
}
p.render(batch, delta);
}
}
Which works just fine. My particles are sharp and they move with the correct speed.
The second way:
Particle p;
for (int i = 0; i < particleList.size(); i++) {
p = particleList.get(i);
p.render(batch, delta);
if (p.isDead()) {
particleList.remove(i);
bulletPool.free(p);
}
}
Which makes all my particles blurry and moving really slow!
The render method for my particles look like this:
public void render(SpriteBatch batch, float delta) {
sprite.setX(sprite.getX() + (dx * speed) * delta * Assets.FPS);
sprite.setY(sprite.getY() + (dy * speed) * delta * Assets.FPS);
ttl--;
sprite.setScale(sprite.getScaleX() - 0.002f);
if (ttl <= 0 || sprite.getScaleX() <= 0)
isDead = true;
sprite.draw(batch);
}
Why do the different rendering methods provide different results?
Thanks in advance
You are mutating (removing elements from) a list while iterating over it. This is a classic way to make a mess.
The Iterator must have code to handle the delete case correctly. But your index-based for loop does not. Specifically when you call particleList.remove(i) the i is now "out of sync" with the content of the list. Consider what happens when you remove the element at index 3: 'i' will increment to 4, but the old element 4 got shuffled down into index 3, so it will get skipped.
I assume you're avoiding the Iterator to avoid memory allocations. So, one way to side-step this issue is to reverse the loop (go from particleList.size() down to 0). Alternatively, you can only increment i for non-dead particles.