Related
Given a set of n rectangles as {(L1, W1), (L2, W2), …….., (Ln, Wn)}, Li and Wi indicate the length and width of rectangle i, respectively. We say that rectangle i fits in rectangle j if Li< Lj and Wi< Wj.
I need help designing a O( nˆ2 ) dynamic programming algorithm that will find the maximum length of a sequence of rectangles that fit into each other.
I made an attempt, but it is not working in some cases, which are as follows:
LR(i, j)= { 2+ LR(i-1, j-1) if (Li< Lj and Wi< Wj) or (Lj< Li and Wj< Wi)
Max ( LR(i+1, j), LR (I, j-1) otherwise }
Could you please help me improve my solution or find better one?
With DP you can do it as follows:
Sort the rectangles by decreasing width, and sort ties by decreasing height
For each index in the array of rectangles, determine the best solution if that rectangle is the first one taken (the outermost containing rectangle), and store it for later lookoup
Use recursion to determine the greatest number of rectangles that can fit when the current rectangle is taken (1) or not taken (2). Take the greatest result of both.
Here is an implementation in JavaScript which you can run here:
function maxNumberOfFittingRectangles(rectangles) {
// Storage of optimal, intermediate results (DP principle),
// keyed by the index of the first rectangle taken
let memo = new Map;
// Take a copy of rectangles, and sort it in order of decreasing width,
// and if there are ties: by decreasing height
rectangles = [...rectangles].sort( (a, b) => (b.width - a.width)
|| (b.height - a.height) );
function recurse(maxHeight, startIndex) {
for (let i = startIndex; i < rectangles.length; i++) {
if (rectangles[i].height <= maxHeight ) { // Can fit
// Try a solution that includes rectangles[i]
// - Only recurse when we did not do this before
if (!(memo.has(i))) memo.set(i, recurse(rectangles[i].height, i+1)+1);
// Also try a solution that excludes rectangles[i], and
// return best of both possibilities:
return Math.max(memo.get(i), recurse(maxHeight, i+1));
}
}
return 0; // recursion's base case
}
let result = recurse(Infinity, 0);
// Display some information for understanding the solution:
for (let i = 0; i < rectangles.length; i++) {
console.log(JSON.stringify(rectangles[i]),
'if taken as first: solution = ', memo.get(i));
}
return result;
}
// Sample data
let rectangles = [
{ width: 10, height: 8 },
{ width: 6, height: 12 },
{ width: 4, height: 9 },
{ width: 9, height: 9 },
{ width: 2, height: 9 },
{ width: 11, height: 4 },
{ width: 9, height: 5 },
{ width: 8, height: 11 },
{ width: 6, height: 6 },
{ width: 5, height: 8 },
{ width: 2, height: 7 },
{ width: 3, height: 5 },
{ width: 12, height: 7 },
];
let result = maxNumberOfFittingRectangles(rectangles);
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }
Get the actual rectangles
The above will give you the maximised count, but not which rectangles you would need to choose to achieve that count. You can change the algorithm slightly, by building a linked list, where you keep not only the maximised count of rectangles that can be picked after a given rectangle (in the sorted order), but also which one would be the next to pick.
Here it is:
function maxNumberOfFittingRectangles(rectangles) {
let memo = new Map;
rectangles = [...rectangles].sort( (a, b) => (b.width - a.width)
|| (b.height - a.height) );
function recurse(maxHeight, startIndex) {
for (let i = startIndex; i < rectangles.length; i++) {
if (rectangles[i].height <= maxHeight ) { // Can fit
if (!(memo.has(i))) memo.set(i, recurse(rectangles[i].height, i+1));
let result = recurse(maxHeight, i+1);
if (memo.get(i).size < result.size) return result;
return { next: i, size: memo.get(i).size + 1 };
}
}
return { next: null, size: 0 }; // recursion's base case
}
let result = recurse(Infinity, 0);
// Turn linked list into array of rectangles:
let arr = [];
while (result.next !== null) {
arr.push(rectangles[result.next]);
result = memo.get(result.next);
}
return arr;
}
// Sample data
let rectangles = [
{ width: 10, height: 8 },
{ width: 6, height: 12 },
{ width: 4, height: 9 },
{ width: 9, height: 9 },
{ width: 2, height: 9 },
{ width: 11, height: 4 },
{ width: 9, height: 5 },
{ width: 8, height: 11 },
{ width: 6, height: 6 },
{ width: 5, height: 8 },
{ width: 2, height: 7 },
{ width: 3, height: 5 },
{ width: 12, height: 7 },
];
let result = maxNumberOfFittingRectangles(rectangles);
console.log(JSON.stringify(result));
.as-console-wrapper { max-height: 100% !important; top: 0; }
If you really want to design a DP solution, here is one you can consider: we can apply the fact that a rectangle with smaller area can't fit into the one with larger area:
Sort the array of rectangles based on their areas (L x W) in descending order (rectangle with the largest area is at index 0).
Starting with the first largest rectangle (let's call head rectangle), we then find the lengths of all sequences this head rectangle has, and get the max length from those lengths (keep reading)
a. The head rectangle may have many sequences. To find the sequences, take the smaller rectangles that can fit into head rectangle, call each of them next_head and recursively 1) take each second rectangle next_head as a head 2) find all sub-sequences this next_head has. As the array is already sorted, we just need to check the rectangles after next_head.
b. The base case: for the last item of one sequence, it will have no subsequence. Then, return 1 because the length of this subsequence is 1 (the head/last item itself counted as 1). For other items not the last in one sequence, they may have more than one sub-sequences. Compare the lengths those sub-sequences return and take the largest value, call it max_length. Then, return max_length + 1 (max_length is its sub-sequence's maximum length; +1 is the head itself)
Now do Step 2 for other rectangles, getting the maximum length each rectangle has. You may want to skip the rectangles that are already counted in another sequence.
Java code below:
class Rectangle {
double L;
double W;
double area;
}
public int maxSequences(Rectangle[] rects) {
// Sort the array based on area. You can do this by yourself
sortBasedOnArea(rects);
int max = 0;
// Find the maximum length of a sequence of each rectangle:
// starting from biggest rectangle to smallest rectangle
// You can improve this for loop by skipping the rectangles that are already counted in sequence before.
for (int i = 0; i < rects.length; i++) {
// For rectangle at index i,
// temp_max is the maximum length of sequence
// the rectangle can make with other smaller rectangles
// Simply put, temp_max is the max number of smaller rectangles that can fit into rect[i]
int temp_max = maxSequencesHelper(rects, i);
if (temp_max > max)
max = temp_max;
}
return max;
}
public int maxSequencesHelper(Rectangle[] rects, int current_head) {
// Head rectangle
Rectangle head = rects[current_head];
// Max of sub-sequence rectangles, excluding the head
int max = 0;
// Loop through smaller rectangles with area < head
for (int i = current_head + 1; i < rects.length; i++) {
Rectangle next_rect = rects[i];
if (isFit(head, next_rect)) {
// If this rectangle can fit into our head,
// Recursively call this function with this rectangle as the head
int next_head_index = i; // This just improves code readability
int temp_max = maxSequencesHelper(rects, next_head_index);
if (temp_max > max)
max = temp_max;
}
}
if (max == 0)
// max = 0 when the if (isFit) in the for loop never runs
// Therefore, this rectangle is the last item of this sequence
return 1;
else
return max + 1;
}
public boolean isFit(Rectangle big, Rectangle small) {
return small.L < big.L && small.W < big.W;
}
Please note that DP may not be the best method, and again my code above isn't the most efficient! One thing you can improve that DP method is to store the solution max length of each sequence of rectangles, but the basic idea of DP for this problem is captured above.
If there is anything unclear, feel free to add comments below.
Hope this helps!
I'm not sure about the DP algorithm here but here's what can be done in O(n^2):
For your set of rectangles create a Directed Acyclic Graph where vertices correspond to rectangles and an edge goes from A to B if (corresponding) rectangle B fits into A.
Find the longest path in the graph you've constructed - http://www.mathcs.emory.edu/~cheung/Courses/171/Syllabus/11-Graph/Docs/longest-path-in-dag.pdf
That path will correspond to the longest sequence you're searching for.
Given is a puzzle game with nine square cards.
On each of the cards there are 4 pictures at top, right, bottom and left.
Each picture on a card depicts either the front part or the rear part of an animal (a crocodile). Each picture has one of 5 colors.
Goal: to lay out the nine cards in a 3x3 grid in such a way that all "inner" (complete) crocodiles are properly combined with adjacent cards, i.e. have a front and rear end as well as matching colors.
To get a visual grip on the problem, here is a picture of the puzzle:
I found the depicted solution by hand.
Even though the puzzle looks simple at first glance, there is an extremely big number of combinations given that you can rotate each piece in 4 different ways.
The problem is now that I'd like to have an algorithm generating all possible 3x3 layouts in order to check all possible solutions (if there are any others). Preferably in Processing/Java.
Thoughts so far:
My approach would be to represent each of the 9 pieces by an array of 4 integer numbers, representing the 4 rotational states of a piece. Then generate all possible permutations of these 9 pieces, picking 1 of the 4 rotation-states from a piece array. A function isValidSolution() could then check a solution for violation of the constraints (color matching and front-rear matching).
Any ideas on how to implement this?
It is possible to find all the solutions, trying not to explore all the unsuccessful paths of the search tree. The C++ code below, not highly optimized, finds a total of 2 solutions (that turn out to be the same unique solution because there is a duplicated tile, right answer?) almost instantaneously with my computer.
The trick here to avoid exploring all the possibilities is to call to function isValidSolution() while we are still placing the tiles (the function handles empty tiles). Also, to speed up the process, I follow a given order placing the tiles, starting in the middle, then the cross around it at left, right, top and bottom, and then the corners top-left, top-right, bottom-left and bottom-right. Probably other combinations give quicker executions.
It is of course possible to optimize this because of the special pattern distribution in this puzzle (the pattern with the letters only accepts one possible match), but that's beyond the scope of my answer.
#include<iostream>
// possible pattern pairs (head, body)
#define PINK 1
#define YELLOW 2
#define BLUE 3
#define GREEN 4
#define LACOSTE 5
typedef int8_t pattern_t; // a pattern is a possible color, positive for head, and negative for body
typedef struct {
pattern_t p[4]; // four patterns per piece: top, right, bottom, left
} piece_t;
unsigned long long int solutionsCounter = 0;
piece_t emptyPiece = {.p = {0, 0, 0, 0} };
piece_t board[3][3] = {
{ emptyPiece, emptyPiece, emptyPiece},
{ emptyPiece, emptyPiece, emptyPiece},
{ emptyPiece, emptyPiece, emptyPiece},
};
inline bool isEmpty(const piece_t& piece) {
bool result = (piece.p[0] == 0);
return result;
}
// check current solution
bool isValidSolution() {
int i, j;
for (i = 0; i < 2; i++) {
for (j = 0; j < 3; j++) {
if (!isEmpty(board[i][j]) && !isEmpty(board[i+1][j]) && (board[i][j].p[1] != -board[i+1][j].p[3])) {
return false;
}
}
}
for (i = 0; i < 3; i++) {
for (j = 0; j < 2; j++) {
if (!isEmpty(board[i][j]) && !isEmpty(board[i][j+1]) && (board[i][j].p[2] != -board[i][j+1].p[0])) {
return false;
}
}
}
return true;
}
// rotate piece
void rotatePiece(piece_t& piece) {
pattern_t paux = piece.p[0];
piece.p[0] = piece.p[1];
piece.p[1] = piece.p[2];
piece.p[2] = piece.p[3];
piece.p[3] = paux;
}
void printSolution() {
printf("Solution:\n");
for (int i = 0; i < 3; i++) {
for (int j = 0; j < 3; j++) {
printf("\t %2i ", (int) board[j][i].p[0]);
}
printf("\n");
for (int j = 0; j < 3; j++) {
printf("\t%2i %2i", (int) board[j][i].p[3], (int) board[j][i].p[1]);
}
printf("\n");
for (int j = 0; j < 3; j++) {
printf("\t %2i ", (int) board[j][i].p[2]);
}
printf("\n");
}
printf("\n");
}
bool usedPiece[9] = { false, false, false, false, false, false, false, false, false };
int colocationOrder[9] = { 4, 3, 5, 1, 7, 0, 2, 6, 8 };
void putNextPiece(piece_t pieces[9], int pieceNumber) {
if (pieceNumber == 9) {
if (isValidSolution()) {
solutionsCounter++;
printSolution();
}
} else {
int nextPosition = colocationOrder[pieceNumber];
int maxRotations = (pieceNumber == 0) ? 1 : 4; // avoids rotation symmetries.
for (int pieceIndex = 0; pieceIndex < 9; pieceIndex++) {
if (!usedPiece[pieceIndex]) {
usedPiece[pieceIndex] = true;
for (int rotationIndex = 0; rotationIndex < maxRotations; rotationIndex++) {
((piece_t*) board)[nextPosition] = pieces[pieceIndex];
if (isValidSolution()) {
putNextPiece(pieces, pieceNumber + 1);
}
rotatePiece(pieces[pieceIndex]);
}
usedPiece[pieceIndex] = false;
((piece_t*) board)[nextPosition] = emptyPiece;
}
}
}
}
int main() {
// register all the pieces (already solved, scramble!)
piece_t pieces[9] = {
{.p = { -YELLOW, -BLUE, +GREEN, +PINK} },
{.p = { -YELLOW, -GREEN, +PINK, +BLUE} },
{.p = { -BLUE, -YELLOW, +PINK, +GREEN }},
{.p = { -GREEN, -BLUE, +PINK, +YELLOW }},
{.p = { -PINK, -LACOSTE, +GREEN, +BLUE }},
{.p = { -PINK, -BLUE, +GREEN, +LACOSTE }},
{.p = { -PINK, -BLUE, +PINK, +YELLOW }},
{.p = { -GREEN, -YELLOW, +GREEN, +BLUE }},
{.p = { -GREEN, -BLUE, +PINK, +YELLOW }}
};
putNextPiece(pieces, 0);
printf("found %llu solutions\n", solutionsCounter);
return 0;
}
There are only 9 pieces, and thus each potential solution is representable by a small structure (say a 3x3 array of pieces, each piece with it's rotation), so the exact description of the pieces isn't too important.
Trying all the possible permutations is wasteful (to abuse LaTeX here, to place the 9 pieces on the grid can be done in $9!$ orders, as each one can be in 4 different orientations this gives a total of $9! \cdot 4^9 = 95126814720 \approx 10^{11}$, a bit too much to check them all). What you'd do by hand is to place a piece, say at the upper left side, and try to complete the square by fitting matching pieces into the rest. So you'd never consider any combinations where the first and second pieces don't match, cutting the search down considerably. This kind of idea is called backtracking. For it you need a description of the partial solution (the 3x3 grid with the filled in pieces and blank places, and the pieces not yet used; a specific order in which to fill the grid), a way of moving forward (place next piece if it fits, skip that one if it doesn't) and backwards (can't find any fits, undo last move and try the next possibility).
Obviously you have to design a way to find out if a potential match exists (given the filled in neighbors, try all orientations of a piece in it's asigned place). For such a small problem this probably isn't performance critical, but if you'd try to solve, say 100x100 the case is different...
I want to parallelize an OpenMP raytracing algorithm that contains two for loops.
Is there anything more I can do than just setting omp_set_num_threads(omp_get_max_threads()) and putting #pragma omp parallel for in front of the first for loop?
So far I've reached a 2.13-times faster algorithm.
Code:
start = omp_get_wtime();
#pragma omp parallel for
for (int i = 0; i < (viewport.xvmax - viewport.xvmin); i++)
{
for (int j = 0; j < (viewport.yvmax - viewport.yvmin); j++)
{
int intersection_object = -1; // none
int reflected_intersection_object = -1; // none
double current_lambda = 0x7fefffffffffffff; // maximum positive double
double current_reflected_lambda = 0x7fefffffffffffff; // maximum positive double
RAY ray, shadow_ray, reflected_ray;
PIXEL pixel;
SPHERE_INTERSECTION intersection, current_intersection, shadow_ray_intersection, reflected_ray_intersection, current_reflected_intersection;
double red, green, blue;
double theta, reflected_theta;
bool bShadow = false;
pixel.i = i;
pixel.j = j;
// 1. compute ray:
compute_ray(&ray, &view_point, &viewport, &pixel, &camera_frame, focal_distance);
// 2. check if ray hits an object:
for (int k = 0; k < NSPHERES; k++)
{
if (sphere_intersection(&ray, &sphere[k], &intersection))
{
// there is an intersection between ray and object
// 1. Izracunanaj normalu...
intersection_normal(&sphere[k], &intersection, &ray);
// 2. ako je lambda presjecista manji od trenutacnog:
if (intersection.lambda_in < current_lambda)
{
current_lambda = intersection.lambda_in;
intersection_object = k;
copy_intersection_struct(¤t_intersection, &intersection);
}
// izracunaj current lambda current_lambda =
// oznaci koji je trenutacni object : intersection_object =
// kopiraj strukturu presjeka : copy_intersection_struct();
}
}
// Compute the color of the pixel:
if (intersection_object > -1)
{
compute_shadow_ray(&shadow_ray, &intersection, &light);
theta = dotproduct(&(shadow_ray.direction), &(intersection.normal));
for (int l = 0; l<NSPHERES; l++)
{
if (l != intersection_object)
{
if (sphere_intersection(&shadow_ray, &sphere[l], &shadow_ray_intersection) && (theta>0.0))
bShadow = true;
}
}
if (bShadow)
{ // if in shadow, add only ambiental light to the surface color
red = shadow(sphere[intersection_object].ka_rgb[CRED], ambi_light_intensity);
green = shadow(sphere[intersection_object].ka_rgb[CGREEN], ambi_light_intensity);
blue = shadow(sphere[intersection_object].ka_rgb[CBLUE], ambi_light_intensity);
}
else
{
// the intersection is not in shadow:
red = blinnphong_shading(¤t_intersection, &light, &view_point,
sphere[intersection_object].kd_rgb[CRED], sphere[intersection_object].ks_rgb[CRED], sphere[intersection_object].ka_rgb[CRED], sphere[intersection_object].shininess,
light_intensity, ambi_light_intensity);
green = blinnphong_shading(¤t_intersection, &light, &view_point,
sphere[intersection_object].kd_rgb[CGREEN], sphere[intersection_object].ks_rgb[CGREEN], sphere[intersection_object].ka_rgb[CGREEN], sphere[intersection_object].shininess,
light_intensity, ambi_light_intensity);
blue = blinnphong_shading(¤t_intersection, &light, &view_point,
sphere[intersection_object].kd_rgb[CBLUE], sphere[intersection_object].ks_rgb[CBLUE], sphere[intersection_object].ka_rgb[CBLUE], sphere[intersection_object].shininess,
light_intensity, ambi_light_intensity);
}
tabelaPixlov[i][j].red = red;
tabelaPixlov[i][j].green = green;
tabelaPixlov[i][j].blue = blue;
glColor3f(tabelaPixlov[i][j].red, tabelaPixlov[i][j].green, tabelaPixlov[i][j].blue);
intersection_object = -1;
bShadow = false;
}
else
{
// draw the pixel with the background color
tabelaPixlov[i][j].red = 0;
tabelaPixlov[i][j].green = 0;
tabelaPixlov[i][j].blue = 0;
intersection_object = -1;
bShadow = false;
}
current_lambda = 0x7fefffffffffffff;
current_reflected_lambda = 0x7fefffffffffffff;
}
}
//glFlush();
stop = omp_get_wtime();
for (int i = 0; i < (viewport.xvmax - viewport.xvmin); i++)
{
for (int j = 0; j < (viewport.yvmax - viewport.yvmin); j++)
{
glColor3f(tabelaPixlov[i][j].red, tabelaPixlov[i][j].green, tabelaPixlov[i][j].blue);
glBegin(GL_POINTS);
glVertex2i(i, j);
glEnd();
}
}
printf("%f\n št niti:%d\n", stop - start, omp_get_max_threads());
glutSwapBuffers();
}
With ray tracing you should use schedule(dynamic). Besides that I would suggest fusing the loop
#pragma omp parallel for schedule(dynamic) {
for(int n=0; n<((viewport.xvmax - viewport.xvmin)*(viewport.yvmax - viewport.yvmin); n++) {
int i = n/(viewport.yvmax - viewport.yvmin);
int j = n%(viewport.yvmax - viewport.yvmin)
//...
}
Also, why are you setting the number of threads? Just use the default which should be set to the number of logical cores. If you have Hyper Threading ray tracing is one of the algorithms that will benefit from Hyper Threading so you don't want to set the number of threads to the number of physical cores.
In addition to using MIMD with OpenMP I would suggest looking into using SIMD for ray tracing. See Ingo Wald's PhD thesis for an example on how to do this http://www.sci.utah.edu/~wald/PhD/. Basically you shoot four (eight) rays in one SSE (AVX) register and then go down the ray tree for each ray in parallel. However, if one ray finishes you hold it and wait until all four are finished (this is similar to what is done on the GPU). There have been many papers written since which have more advanced tricks based on this idea.
I am looking for a fast polygon triangulation algorithm that can triangulate not very complex 2D concave polygons (without holes) into triangle strips ready to be sent to OpenGL ES for drawing using GL_TRIANGLE_STRIP.
I am aware of some algorithms but I couldn't find one that will fit my needs:
http://www.flipcode.com/archives/Efficient_Polygon_Triangulation.shtml
this algorithm works ok but the problem is it returns simple triangles which you can't draw with GL_TRIANGLE_STRIP, you need to use GL_TRIANGLES which isn't very efficient on a large number of vertices.
http://code.google.com/p/iphone-glu/
it doesn't have any example associated and I couldn't find anyone that has successfully used it on iOS with OpenGL ES 2.0
I don't know what it returns and it seems like it also calls the corresponding OpenGL commands which I don't want - I only need the triangles back
it leaks memory
The platform I am developing for is: iOS, OpenGL ES 2.0, cocos2d 2.0.
Can anyone help me with such an algorithm? Or any other advice will be greatly appreciated.
In 2D and without holes, this is fairly easy. First, you need to break your polygon to one or more monotone polygons.
The monotone polygons are pretty simple to turn into tristrips, just sort the values by y, find the top-most and bottom-most vertex, and then you have lists of vertices to the right and to the left (because vertices come in some defined, say clockwise, order). Then you start with top-most vertex and add vertices from the left and from the right sides in alternating manner.
This technique will work for any 2D polygons without self-intersecting edges, which includes some cases of polygons with holes (the holes must be correctly wound though).
You can try and play with this code:
glMatrixMode(GL_PROJECTION);
glLoadIdentity();
glMatrixMode(GL_MODELVIEW);
glLoadIdentity();
glTranslatef(-.5f, -.5f, 0);
std::vector<Vector2f> my_polygon;
my_polygon.push_back(Vector2f(-0.300475f, 0.862924f));
my_polygon.push_back(Vector2f(0.302850f, 1.265013f));
my_polygon.push_back(Vector2f(0.811164f, 1.437337f));
my_polygon.push_back(Vector2f(1.001188f, 1.071802f));
my_polygon.push_back(Vector2f(0.692399f, 0.936031f));
my_polygon.push_back(Vector2f(0.934679f, 0.622715f));
my_polygon.push_back(Vector2f(0.644893f, 0.408616f));
my_polygon.push_back(Vector2f(0.592637f, 0.753264f));
my_polygon.push_back(Vector2f(0.269596f, 0.278068f));
my_polygon.push_back(Vector2f(0.996437f, -0.092689f));
my_polygon.push_back(Vector2f(0.735154f, -0.338120f));
my_polygon.push_back(Vector2f(0.112827f, 0.079634f));
my_polygon.push_back(Vector2f(-0.167458f, 0.330287f));
my_polygon.push_back(Vector2f(0.008314f, 0.664491f));
my_polygon.push_back(Vector2f(0.393112f, 1.040470f));
// from wiki (http://en.wikipedia.org/wiki/File:Polygon-to-monotone.png)
glEnable(GL_POINT_SMOOTH);
glEnable(GL_LINE_SMOOTH);
glEnable(GL_BLEND);
glBlendFunc(GL_SRC_ALPHA, GL_ONE_MINUS_SRC_ALPHA);
glLineWidth(6);
glColor3f(1, 1, 1);
glBegin(GL_LINE_LOOP);
for(size_t i = 0, n = my_polygon.size(); i < n; ++ i)
glVertex2f(my_polygon[i].x, my_polygon[i].y);
glEnd();
glPointSize(6);
glBegin(GL_POINTS);
for(size_t i = 0, n = my_polygon.size(); i < n; ++ i)
glVertex2f(my_polygon[i].x, my_polygon[i].y);
glEnd();
// draw the original polygon
std::vector<int> working_set;
for(size_t i = 0, n = my_polygon.size(); i < n; ++ i)
working_set.push_back(i);
_ASSERTE(working_set.size() == my_polygon.size());
// add vertex indices to the list (could be done using iota)
std::list<std::vector<int> > monotone_poly_list;
// list of monotone polygons (the output)
glPointSize(14);
glLineWidth(4);
// prepare to draw split points and slice lines
for(;;) {
std::vector<int> sorted_vertex_list;
{
for(size_t i = 0, n = working_set.size(); i < n; ++ i)
sorted_vertex_list.push_back(i);
_ASSERTE(working_set.size() == working_set.size());
// add vertex indices to the list (could be done using iota)
for(;;) {
bool b_change = false;
for(size_t i = 1, n = sorted_vertex_list.size(); i < n; ++ i) {
int a = sorted_vertex_list[i - 1];
int b = sorted_vertex_list[i];
if(my_polygon[working_set[a]].y < my_polygon[working_set[b]].y) {
std::swap(sorted_vertex_list[i - 1], sorted_vertex_list[i]);
b_change = true;
}
}
if(!b_change)
break;
}
// sort vertex indices by the y coordinate
// (note this is using bubblesort to maintain portability
// but it should be done using a better sorting method)
}
// build sorted vertex list
bool b_change = false;
for(size_t i = 0, n = sorted_vertex_list.size(), m = working_set.size(); i < n; ++ i) {
int n_ith = sorted_vertex_list[i];
Vector2f ith = my_polygon[working_set[n_ith]];
Vector2f prev = my_polygon[working_set[(n_ith + m - 1) % m]];
Vector2f next = my_polygon[working_set[(n_ith + 1) % m]];
// get point in the list, and get it's neighbours
// (neighbours are not in sorted list ordering
// but in the original polygon order)
float sidePrev = sign(ith.y - prev.y);
float sideNext = sign(ith.y - next.y);
// calculate which side they lie on
// (sign function gives -1 for negative and 1 for positive argument)
if(sidePrev * sideNext >= 0) { // if they are both on the same side
glColor3f(1, 0, 0);
glBegin(GL_POINTS);
glVertex2f(ith.x, ith.y);
glEnd();
// marks points whose neighbours are both on the same side (split points)
int n_next = -1;
if(sidePrev + sideNext > 0) {
if(i > 0)
n_next = sorted_vertex_list[i - 1];
// get the next vertex above it
} else {
if(i + 1 < n)
n_next = sorted_vertex_list[i + 1];
// get the next vertex below it
}
// this is kind of simplistic, one needs to check if
// a line between n_ith and n_next doesn't exit the polygon
// (but that doesn't happen in the example)
if(n_next != -1) {
glColor3f(0, 1, 0);
glBegin(GL_POINTS);
glVertex2f(my_polygon[working_set[n_next]].x, my_polygon[working_set[n_next]].y);
glEnd();
glBegin(GL_LINES);
glVertex2f(ith.x, ith.y);
glVertex2f(my_polygon[working_set[n_next]].x, my_polygon[working_set[n_next]].y);
glEnd();
std::vector<int> poly, remove_list;
int n_last = n_ith;
if(n_last > n_next)
std::swap(n_last, n_next);
int idx = n_next;
poly.push_back(working_set[idx]); // add n_next
for(idx = (idx + 1) % m; idx != n_last; idx = (idx + 1) % m) {
poly.push_back(working_set[idx]);
// add it to the polygon
remove_list.push_back(idx);
// mark this vertex to be erased from the working set
}
poly.push_back(working_set[idx]); // add n_ith
// build a new monotone polygon by cutting the original one
std::sort(remove_list.begin(), remove_list.end());
for(size_t i = remove_list.size(); i > 0; -- i) {
int n_which = remove_list[i - 1];
working_set.erase(working_set.begin() + n_which);
}
// sort indices of vertices to be removed and remove them
// from the working set (have to do it in reverse order)
monotone_poly_list.push_back(poly);
// add it to the list
b_change = true;
break;
// the polygon was sliced, restart the algorithm, regenerate sorted list and slice again
}
}
}
if(!b_change)
break;
// no moves
}
if(!working_set.empty())
monotone_poly_list.push_back(working_set);
// use the remaining vertices (which the algorithm was unable to slice) as the last polygon
std::list<std::vector<int> >::const_iterator p_mono_poly = monotone_poly_list.begin();
for(; p_mono_poly != monotone_poly_list.end(); ++ p_mono_poly) {
const std::vector<int> &r_mono_poly = *p_mono_poly;
glLineWidth(2);
glColor3f(0, 0, 1);
glBegin(GL_LINE_LOOP);
for(size_t i = 0, n = r_mono_poly.size(); i < n; ++ i)
glVertex2f(my_polygon[r_mono_poly[i]].x, my_polygon[r_mono_poly[i]].y);
glEnd();
glPointSize(2);
glBegin(GL_POINTS);
for(size_t i = 0, n = r_mono_poly.size(); i < n; ++ i)
glVertex2f(my_polygon[r_mono_poly[i]].x, my_polygon[r_mono_poly[i]].y);
glEnd();
// draw the sliced part of the polygon
int n_top = 0;
for(size_t i = 0, n = r_mono_poly.size(); i < n; ++ i) {
if(my_polygon[r_mono_poly[i]].y < my_polygon[r_mono_poly[n_top]].y)
n_top = i;
}
// find the top-most point
glLineWidth(1);
glColor3f(0, 1, 0);
glBegin(GL_LINE_STRIP);
glVertex2f(my_polygon[r_mono_poly[n_top]].x, my_polygon[r_mono_poly[n_top]].y);
for(size_t i = 1, n = r_mono_poly.size(); i <= n; ++ i) {
int n_which = (n_top + ((i & 1)? n - i / 2 : i / 2)) % n;
glVertex2f(my_polygon[r_mono_poly[n_which]].x, my_polygon[r_mono_poly[n_which]].y);
}
glEnd();
// draw as if triangle strip
}
This code is not optimal, but it should be easy to understand. At the beginnig, a concave polygon is created. Then a "working set" of vertices is created. On that working set, a permutation is calculated which sorts the vertices by their y coordinate. That permutation is then looped through, looking for split points. Once a split point is found, a new monotone polygon is created. Then, the vertices used in the new polygon are removed from the working set and the whole process repeats. Finally, the working set contains the last polygon which could not be split. At the end, the monotone polygons are rendered, along with triangle strip ordering. It is a little bit messy, but I'm sure you will figure it out (this is C++ code, just put it inside a GLUT window and see what it does).
Hope this helps ...
You can extract tesselation algorithm from OpenGL Sample Implementation, like described in this post http://choruscode.blogspot.de/2013/03/extracting-tesselation-from-opengl.html , it also has an example.
Does anybody know how to find the local maxima in a grayscale IPL_DEPTH_8U image using OpenCV? HarrisCorner mentions something like that but I'm actually not interested in corners ...
Thanks!
A pixel is considered a local maximum if it is equal to the maximum value in a 'local' neighborhood. The function below captures this property in two lines of code.
To deal with pixels on 'plateaus' (value equal to their neighborhood) one can use the local minimum property, since plateaus pixels are equal to their local minimum. The rest of the code filters out those pixels.
void non_maxima_suppression(const cv::Mat& image, cv::Mat& mask, bool remove_plateaus) {
// find pixels that are equal to the local neighborhood not maximum (including 'plateaus')
cv::dilate(image, mask, cv::Mat());
cv::compare(image, mask, mask, cv::CMP_GE);
// optionally filter out pixels that are equal to the local minimum ('plateaus')
if (remove_plateaus) {
cv::Mat non_plateau_mask;
cv::erode(image, non_plateau_mask, cv::Mat());
cv::compare(image, non_plateau_mask, non_plateau_mask, cv::CMP_GT);
cv::bitwise_and(mask, non_plateau_mask, mask);
}
}
Here's a simple trick. The idea is to dilate with a kernel that contains a hole in the center. After the dilate operation, each pixel is replaced with the maximum of it's neighbors (using a 5 by 5 neighborhood in this example), excluding the original pixel.
Mat1b kernelLM(Size(5, 5), 1u);
kernelLM.at<uchar>(2, 2) = 0u;
Mat imageLM;
dilate(image, imageLM, kernelLM);
Mat1b localMaxima = (image > imageLM);
Actually after I posted the code above I wrote a better and very very faster one ..
The code above suffers even for a 640x480 picture..
I optimized it and now it is very very fast even for 1600x1200 pic.
Here is the code :
void localMaxima(cv::Mat src,cv::Mat &dst,int squareSize)
{
if (squareSize==0)
{
dst = src.clone();
return;
}
Mat m0;
dst = src.clone();
Point maxLoc(0,0);
//1.Be sure to have at least 3x3 for at least looking at 1 pixel close neighbours
// Also the window must be <odd>x<odd>
SANITYCHECK(squareSize,3,1);
int sqrCenter = (squareSize-1)/2;
//2.Create the localWindow mask to get things done faster
// When we find a local maxima we will multiply the subwindow with this MASK
// So that we will not search for those 0 values again and again
Mat localWindowMask = Mat::zeros(Size(squareSize,squareSize),CV_8U);//boolean
localWindowMask.at<unsigned char>(sqrCenter,sqrCenter)=1;
//3.Find the threshold value to threshold the image
//this function here returns the peak of histogram of picture
//the picture is a thresholded picture it will have a lot of zero values in it
//so that the second boolean variable says :
// (boolean) ? "return peak even if it is at 0" : "return peak discarding 0"
int thrshld = maxUsedValInHistogramData(dst,false);
threshold(dst,m0,thrshld,1,THRESH_BINARY);
//4.Now delete all thresholded values from picture
dst = dst.mul(m0);
//put the src in the middle of the big array
for (int row=sqrCenter;row<dst.size().height-sqrCenter;row++)
for (int col=sqrCenter;col<dst.size().width-sqrCenter;col++)
{
//1.if the value is zero it can not be a local maxima
if (dst.at<unsigned char>(row,col)==0)
continue;
//2.the value at (row,col) is not 0 so it can be a local maxima point
m0 = dst.colRange(col-sqrCenter,col+sqrCenter+1).rowRange(row-sqrCenter,row+sqrCenter+1);
minMaxLoc(m0,NULL,NULL,NULL,&maxLoc);
//if the maximum location of this subWindow is at center
//it means we found the local maxima
//so we should delete the surrounding values which lies in the subWindow area
//hence we will not try to find if a point is at localMaxima when already found a neighbour was
if ((maxLoc.x==sqrCenter)&&(maxLoc.y==sqrCenter))
{
m0 = m0.mul(localWindowMask);
//we can skip the values that we already made 0 by the above function
col+=sqrCenter;
}
}
}
The following listing is a function similar to Matlab's "imregionalmax". It looks for at most nLocMax local maxima above threshold, where the found local maxima are at least minDistBtwLocMax pixels apart. It returns the actual number of local maxima found. Notice that it uses OpenCV's minMaxLoc to find global maxima. It is "opencv-self-contained" except for the (easy to implement) function vdist, which computes the (euclidian) distance between points (r,c) and (row,col).
input is one-channel CV_32F matrix, and locations is nLocMax (rows) by 2 (columns) CV_32S matrix.
int imregionalmax(Mat input, int nLocMax, float threshold, float minDistBtwLocMax, Mat locations)
{
Mat scratch = input.clone();
int nFoundLocMax = 0;
for (int i = 0; i < nLocMax; i++) {
Point location;
double maxVal;
minMaxLoc(scratch, NULL, &maxVal, NULL, &location);
if (maxVal > threshold) {
nFoundLocMax += 1;
int row = location.y;
int col = location.x;
locations.at<int>(i,0) = row;
locations.at<int>(i,1) = col;
int r0 = (row-minDistBtwLocMax > -1 ? row-minDistBtwLocMax : 0);
int r1 = (row+minDistBtwLocMax < scratch.rows ? row+minDistBtwLocMax : scratch.rows-1);
int c0 = (col-minDistBtwLocMax > -1 ? col-minDistBtwLocMax : 0);
int c1 = (col+minDistBtwLocMax < scratch.cols ? col+minDistBtwLocMax : scratch.cols-1);
for (int r = r0; r <= r1; r++) {
for (int c = c0; c <= c1; c++) {
if (vdist(Point2DMake(r, c),Point2DMake(row, col)) <= minDistBtwLocMax) {
scratch.at<float>(r,c) = 0.0;
}
}
}
} else {
break;
}
}
return nFoundLocMax;
}
The first question to answer would be what is "local" in your opinion. The answer may well be a square window (say 3x3 or 5x5) or circular window of a certain radius. You can then scan over the entire image with the window centered at each pixel and pick the highest value in the window.
See this for how to access pixel values in OpenCV.
This is very fast method. It stored founded maxima in a vector of
Points.
vector <Point> GetLocalMaxima(const cv::Mat Src,int MatchingSize, int Threshold, int GaussKernel )
{
vector <Point> vMaxLoc(0);
if ((MatchingSize % 2 == 0) || (GaussKernel % 2 == 0)) // MatchingSize and GaussKernel have to be "odd" and > 0
{
return vMaxLoc;
}
vMaxLoc.reserve(100); // Reserve place for fast access
Mat ProcessImg = Src.clone();
int W = Src.cols;
int H = Src.rows;
int SearchWidth = W - MatchingSize;
int SearchHeight = H - MatchingSize;
int MatchingSquareCenter = MatchingSize/2;
if(GaussKernel > 1) // If You need a smoothing
{
GaussianBlur(ProcessImg,ProcessImg,Size(GaussKernel,GaussKernel),0,0,4);
}
uchar* pProcess = (uchar *) ProcessImg.data; // The pointer to image Data
int Shift = MatchingSquareCenter * ( W + 1);
int k = 0;
for(int y=0; y < SearchHeight; ++y)
{
int m = k + Shift;
for(int x=0;x < SearchWidth ; ++x)
{
if (pProcess[m++] >= Threshold)
{
Point LocMax;
Mat mROI(ProcessImg, Rect(x,y,MatchingSize,MatchingSize));
minMaxLoc(mROI,NULL,NULL,NULL,&LocMax);
if (LocMax.x == MatchingSquareCenter && LocMax.y == MatchingSquareCenter)
{
vMaxLoc.push_back(Point( x+LocMax.x,y + LocMax.y ));
// imshow("W1",mROI);cvWaitKey(0); //For gebug
}
}
}
k += W;
}
return vMaxLoc;
}
Found a simple solution.
In this example, if you are trying to find 2 results of a matchTemplate function with a minimum distance from each other.
cv::Mat result;
matchTemplate(search, target, result, CV_TM_SQDIFF_NORMED);
float score1;
cv::Point displacement1 = MinMax(result, score1);
cv::circle(result, cv::Point(displacement1.x+result.cols/2 , displacement1.y+result.rows/2), 10, cv::Scalar(0), CV_FILLED, 8, 0);
float score2;
cv::Point displacement2 = MinMax(result, score2);
where
cv::Point MinMax(cv::Mat &result, float &score)
{
double minVal, maxVal;
cv::Point minLoc, maxLoc, matchLoc;
minMaxLoc(result, &minVal, &maxVal, &minLoc, &maxLoc, cv::Mat());
matchLoc.x = minLoc.x - result.cols/2;
matchLoc.y = minLoc.y - result.rows/2;
return minVal;
}
The process is:
Find global Minimum using minMaxLoc
Draw a filled white circle around global minimum using min distance between minima as radius
Find another minimum
The the scores can be compared to each other to determine, for example, the certainty of the match,
To find more than just the global minimum and maximum try using this function from skimage:
http://scikit-image.org/docs/dev/api/skimage.feature.html#skimage.feature.peak_local_max
You can parameterize the minimum distance between peaks, too. And more. To find minima, use negated values (take care of the array type though, 255-image could do the trick).
You can go over each pixel and test if it is a local maxima. Here is how I would do it.
The input is assumed to be type CV_32FC1
#include <vector>//std::vector
#include <algorithm>//std::sort
#include "opencv2/imgproc/imgproc.hpp"
#include "opencv2/core/core.hpp"
//structure for maximal values including position
struct SRegionalMaxPoint
{
SRegionalMaxPoint():
values(-FLT_MAX),
row(-1),
col(-1)
{}
float values;
int row;
int col;
//ascending order
bool operator()(const SRegionalMaxPoint& a, const SRegionalMaxPoint& b)
{
return a.values < b.values;
}
};
//checks if pixel is local max
bool isRegionalMax(const float* im_ptr, const int& cols )
{
float center = *im_ptr;
bool is_regional_max = true;
im_ptr -= (cols + 1);
for (int ii = 0; ii < 3; ++ii, im_ptr+= (cols-3))
{
for (int jj = 0; jj < 3; ++jj, im_ptr++)
{
if (ii != 1 || jj != 1)
{
is_regional_max &= (center > *im_ptr);
}
}
}
return is_regional_max;
}
void imregionalmax(
const cv::Mat& input,
std::vector<SRegionalMaxPoint>& buffer)
{
//find local max - top maxima
static const int margin = 1;
const int rows = input.rows;
const int cols = input.cols;
for (int i = margin; i < rows - margin; ++i)
{
const float* im_ptr = input.ptr<float>(i, margin);
for (int j = margin; j < cols - margin; ++j, im_ptr++)
{
//Check if pixel is local maximum
if ( isRegionalMax(im_ptr, cols ) )
{
cv::Rect roi = cv::Rect(j - margin, i - margin, 3, 3);
cv::Mat subMat = input(roi);
float val = *im_ptr;
//replace smallest value in buffer
if ( val > buffer[0].values )
{
buffer[0].values = val;
buffer[0].row = i;
buffer[0].col = j;
std::sort(buffer.begin(), buffer.end(), SRegionalMaxPoint());
}
}
}
}
}
For testing the code you can try this:
cv::Mat temp = cv::Mat::zeros(15, 15, CV_32FC1);
temp.at<float>(7, 7) = 1;
temp.at<float>(3, 5) = 6;
temp.at<float>(8, 10) = 4;
temp.at<float>(11, 13) = 7;
temp.at<float>(10, 3) = 8;
temp.at<float>(7, 13) = 3;
vector<SRegionalMaxPoint> buffer_(5);
imregionalmax(temp, buffer_);
cv::Mat debug;
cv::cvtColor(temp, debug, cv::COLOR_GRAY2BGR);
for (auto it = buffer_.begin(); it != buffer_.end(); ++it)
{
circle(debug, cv::Point(it->col, it->row), 1, cv::Scalar(0, 255, 0));
}
This solution does not take plateaus into account so it is not exactly the same as matlab's imregionalmax()
I think you want to use the
MinMaxLoc(arr, mask=NULL)-> (minVal, maxVal, minLoc, maxLoc)
Finds global minimum and maximum in array or subarray
function on you image