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The problem I am facing is the following:
I have a function based on the BFS search algorithm that I use in a NxM grid, the mission of this function is to return the following Direction from a set of possible Directions = {Up, Down, Left , Right} (No diagonal moves!)to which a player has to move, so that in each "round / frame" where there is a type of item of a game (For example, in this specific case, a bazooka) is closer to the item.
To address the problem, I have created a Map class made of vector <vector <Cell> > where vector is from the standard library and Cell is what the grid is made of and has some consulting methods on what is in one of the NxM cells (if there is a building, an enemy, a Bazooka, etc.)
So, for implementing a solution for this, I made a struct TrackingBFS to reconstruct the path of the BFS search:
struct TrackingBFS {
pair <int,int> anterior;
bool visited;
};
And this is the BFS search implementation:
//Pre: origen is a valid position on the grid where the player is
//Post:Returns a pair of bool and a direction to the closest bazooka. If we have access to a bazooka, then we will return a pair (true,Dir) where Dir is a direction to take to be closer to the bazooka else a pair (false, Dir) where dir is just the same direction as origen.
pair <bool,Dir> direction_to_closest_gun (const Pos& origen) {
//R = board_rows() C = board_cols()
//m = mapa
//sr,sc = origin positions
int sr = origen.i;
int sc = origen.j;
//Variables para mantener tracking de los pasos cogidos
queue <int> rq; //Queue of x-row coordinates
queue <int> cq; //Queue of y-row coordinates
int move_count = 0; //Number of steps
int nodes_left_in_layer = 1; //How many nodes we need to de-queue before doing a step
int nodes_in_next_layer = 0; //How many nodes we add to the expansio of the BFS so we can upgrade nodes_left_in_layer in the next iteration
bool arma_mejor_encontrada = false;
//Visited is a MxN matrix of TrackingBFS that for every i,j stores the following information:
//If we visited the cell at position visited [i][j], then the TrackingBFS struct will be visited = true and last_node = (k,l) where k and l are the components of the LAST cell on the grid we visited in the BFS search.
//Else if we didn't visited the cell at [i][j], the TrackingBFS struct will be visited = true and last_node (-1,-1).
TrackingBFS aux;
aux.last_node = make_pair(-1,-1);
aux.visited = false;
//We create a Matrix of TrackingBFS named visited of NxM filled with the unvisited cells
vector < vector <TrackingBFS> > visited (board_rows(), vector <TrackingBFS> (board_cols(),aux));
//--------------------------------SOLVE---------------------------------
rq.push(sr);
cq.push(sc);
visited[sr][sc].visited = true;
visited[sr][sc].last_node = make_pair(sr,sc);
int xfound;
int yfound;
while (!rq.empty()) {
int r = rq.front();
int c = cq.front();
rq.pop();
cq.pop();
if (mapa[r][c].weapon == Bazooka) {
arma_mejor_encontrada = true;
xfound = r;
yfound = c;
break;
}
//Explore neighbours
Pos p (r,c);
for (Dir d : dirs) {
Pos searching = p + d;
int rr = searching.i;
int cc = searching.j;
//If the position we are searching is out of range or it's been visited before or there is a obstacle then continue
if (!pos_ok(searching) or visited[rr][cc].visited or mapa[rr][cc].type == Building or mapa[rr][cc].resistance != -1 or mapa[rr][cc].id != -1) {
//NADA
}
//Else we add the visited node to the queue, and fill the information on visited[rr][cc] with the node we are coming and mark it as visited
else {
rq.push(rr);
cq.push(cc);
visited[rr][cc].visited = true;
visited[rr][cc].last_node = make_pair (r,c);
nodes_in_next_layer++;
}
}
nodes_left_in_layer--;
if (nodes_left_in_layer == 0) {
nodes_left_in_layer = nodes_in_next_layer;
nodes_in_next_layer = 0;
move_count++;
}
}
//If we found the Bazooka
if (arma_mejor_encontrada) {
//Return the pair (true,next direction of the player at position (sr,sc) to approach the bazooka)
return make_pair(arma_mejor_encontrada, reconstruir_camino(visited,xfound,yfound,sr,sc));
}
else {
//If there is no bazooka we return (false,Up (this second component is meaningless))
return make_pair(arma_mejor_encontrada, Up);
}
}
The reconstruir_camino (recosntruct_path in english) implementation:
//This function is made to reconstruct the path from where we found de bazooka (x,y) to where our player is (ox,oy), whe are only interested in the next position of because this is run each frame, so we are constantly refreshing the state of the map.
Dir reconstruir_camino(const vector < vector <TrackingBFS> >& visited, const int& x, const int& y, const int& ox, const int& oy) {
//In v we save the pair of coordinates of our path, this was done only for debuging and in debug_matriz_tracking(visited) is also only for debuging
vector <pair<int,int>> path;
debug_matriz_tracking(visited);
//
int i = visited[x][y].last_node.first;
int j = visited[x][y].last_node.second;
while (not (i == ox and j == oy)) { //While the next node is not iqual as the original node we started de search (The one where our player is)
path.push_back(make_pair(i,j)); //Just for debuging
i = visited[i][j].last_node.first;
j = visited[i][j].last_node.second;
}
//So now in path we have the start and end nodes of every edge on the path
int siguiente_x = path.back().first;
int siguiente_y = path.back().second;
debug_camino(path);
return direccion_contiguos(ox,oy,siguiente_x,siguiente_y);
}
And direccion_contiguos (contiguous direction / relative direction in english) implementation:
//Returns the RELATIVE direction to go if we want to go from (ox, oy) to (dx, dy) being these two contiguous positions, that is, (dx, dy) is in Up, Down, Left or Right with respect to (ox, oy) IT WORKS FINE, NOTHING WRONG WITH THIS DON'T TOUCH
Dir direccion_contiguos (const int& ox, const int& oy, const int& dx, const int& dy) {
Pos o (ox,oy);
Pos d (dx,dy);
if (o + Up == d) {
return Up;
}
else if (o + Down == d){
return Down;
}
else if (o + Left == d) {
return Left;
}
else if (o + Right == d) {
return Right;
}
return Down;
}
So now in visited, we have the information to reconstruct the path, in fact I debuged it (it's kinda messy i know, sorry), in a visual way so this is what I got for a Player in origen = (7,10) and bazooka at position = (4,11):
[Imgur link of the Visual representation of the Matrix for reconstructing the path from origin to bazooka][1]
To read this image, at the top and left there are the coordinates of every cell of the visited matrix, the ones with green font, are the ones that have been visited, and they store THE NEXT cell/vertex of the path, and the ones with (-1,-1) are the ones that have not been visited by the BFS algorithm and thus they don't have any previous node and are in white.
So, NICE! It seems to work, at least the visited matrix.
My problem is when I debug the vector of edges/directions of the graph/grid, this is what I used in the example of the image:
void debug_camino(const vector <pair<int,int>>& v) {
cerr << "----------------------CAMINO-------------------- DEBUG_CAMINO" << endl;
for (long unsigned int i = 0; i < v.size(); ++i) {
cerr << "(" << v[i].first << "," << v[i].second << "),";
}
cerr << endl;
}
And when I executed the program, this is the path that I got with debug_camino():
If you see the image attached you can see that that's almost the path but not quite yet.
(4,12),(4,13),(4,14),(3,15),(3,16),(4,16),(5,16),(6,16),(7,15),(7,14),(7,13),(7,12),(7,11)
These ones bolded are not real (even valid because they are diagonal moves) reconstructions of the path and I don't really know WHY this is happening, but it's provoking my player to not following right path, and I want to fix the error and I'm kinda desperate because I don't really know where the error is and I've been trying for days :( ! I hope somebody can help me with this. Thanks for reading all this and sorry if the code is in some parts in Spanish or if it's not all that readable.
[1]: https://i.stack.imgur.com/vZ2Go.png
Okay, I actually managed to fix this error, i was overwriting the i variable so that was causing the error.
//This function is made to reconstruct the path from where we found de bazooka (x,y) to where our player is (ox,oy), whe are only interested in the next position of because this is run each frame, so we are constantly refreshing the state of the map.
Dir reconstruir_camino(const vector < vector <TrackingBFS> >& visited, const int& x, const int& y, const int& ox, const int& oy) {
//In v we save the pair of coordinates of our path, this was done only for debuging and in debug_matriz_tracking(visited) is also only for debuging
vector <pair<int,int>> path;
debug_matriz_tracking(visited);
//
int i = visited[x][y].last_node.first;
int j = visited[x][y].last_node.second;
while (not (i == ox and j == oy)) { //While the next node is not iqual as the original node we started de search (The one where our player is)
path.push_back(make_pair(i,j)); //Just for debuging
**i** = visited[i][j].last_node.first;
j = visited[**i**][j].last_node.second;
}
//So now in path we have the start and end nodes of every edge on the path
int siguiente_x = path.back().first;
int siguiente_y = path.back().second;
debug_camino(path);
return direccion_contiguos(ox,oy,siguiente_x,siguiente_y);
}
It's already fixed
Background
For a computer vision assignment I've been given the task of implementing RANSAC to fit a plane to a given set of points and filter that input list of points by the consensus model using Eigenvalue Decomposition.
I have spent days trying to tweak my code to achieve correct plane filtering behavior on an input set of test data. All you algorithm junkies, this one's for you.
My implementation uses a vector of a ROS data structure (Point32) as inputs, but this is transparent to the problem at hand.
What I've done
When I test for expected plane filtering behavior (correct elimination of outliers >95-99% of the time), I see in my implementation that I only eliminate outliers and extract the main plane of a test point cloud ~30-40% of the time. Other times, I filter a plane that ~somewhat~ fits the expected model, but leaves a lot of obvious outliers inside the consensus model. The fact that this works at all suggests that I'm doing some things right, and some things wrong.
I've tweaked my constants (distance threshold, max iterations, estimated % points fit) to London and back, and I only see small differences in the consensus model.
Implementation (long)
const float RANSAC_ESTIMATED_FIT_POINTS = .80f; // % points estimated to fit the model
const size_t RANSAC_MAX_ITER = 500; // max RANSAC iterations
const size_t RANDOM_MAX_TRIES = 100; // max RANSAC random point tries per iteration
const float RANSAC_THRESHOLD = 0.0000001f; // threshold to determine what constitutes a close point to a plane
/*
Helper to randomly select an item from a STL container, from stackoverflow.
*/
template <typename I>
I random_element(I begin, I end)
{
const unsigned long n = std::distance(begin, end);
const unsigned long divisor = ((long)RAND_MAX + 1) / n;
unsigned long k;
do { k = std::rand() / divisor; } while (k >= n);
std::advance(begin, k);
return begin;
}
bool run_RANSAC(const std::vector<Point32> all_points,
Vector3f *out_p0, Vector3f *out_n,
std::vector<Point32> *out_inlier_points)
{
for (size_t iterations = 0; iterations < RANSAC_MAX_ITER; iterations ++)
{
Point32 p1,p2,p3;
Vector3f v1;
Vector3f v2;
Vector3f n_hat; // keep track of the current plane model
Vector3f P0;
std::vector<Point32> points_agree; // list of points that agree with model within
bool found = false;
// try RANDOM_MAX_TRIES times to get random 3 points
for (size_t tries = 0; tries < RANDOM_MAX_TRIES; tries ++) // try to get unique random points 100 times
{
// get 3 random points
p1 = *random_element(all_points.begin(), all_points.end());
p2 = *random_element(all_points.begin(), all_points.end());
p3 = *random_element(all_points.begin(), all_points.end());
v1 = Vector3f (p2.x - p1.x,
p2.y - p1.y,
p2.z - p1.z ); //Vector P1P2
v2 = Vector3f (p3.x - p1.x,
p3.y - p1.y,
p3.z - p1.z); //Vector P1P3
if (std::abs(v1.dot(v2)) != 1.f) // dot product != 1 means we've found 3 nonlinear points
{
found = true;
break;
}
} // end try random element loop
if (!found) // could not find 3 random nonlinear points in 100 tries, go to next iteration
{
ROS_ERROR("run_RANSAC(): Could not find 3 random nonlinear points in %ld tries, going on to iteration %ld", RANDOM_MAX_TRIES, iterations + 1);
continue;
}
// nonlinear random points exist past here
// fit a plane to p1, p2, p3
Vector3f n = v1.cross(v2); // calculate normal of plane
n_hat = n / n.norm();
P0 = Vector3f(p1.x, p1.y, p1.z);
// at some point, the original p0, p1, p2 will be iterated over and added to agreed points
// loop over all points, find points that are inliers to plane
for (std::vector<Point32>::const_iterator it = all_points.begin();
it != all_points.end(); it++)
{
Vector3f M (it->x - P0.x(),
it->y - P0.y(),
it->z - P0.z()); // M = (P - P0)
float d = M.dot(n_hat); // calculate distance
if (d <= RANSAC_THRESHOLD)
{ // add to inlier points list
points_agree.push_back(*it);
}
} // end points loop
ROS_DEBUG("run_RANSAC() POINTS AGREED: %li=%f, RANSAC_ESTIMATED_FIT_POINTS: %f", points_agree.size(),
(float) points_agree.size() / all_points.size(), RANSAC_ESTIMATED_FIT_POINTS);
if (((float) points_agree.size()) / all_points.size() > RANSAC_ESTIMATED_FIT_POINTS)
{ // if points agree / total points > estimated % points fitting
// fit to points_agree.size() points
size_t n = points_agree.size();
Vector3f sum(0.0f, 0.0f, 0.0f);
for (std::vector<Point32>::iterator iter = points_agree.begin();
iter != points_agree.end(); iter++)
{
sum += Vector3f(iter->x, iter->y, iter->z);
}
Vector3f centroid = sum / n; // calculate centroid
Eigen::MatrixXf M(points_agree.size(), 3);
for (size_t row = 0; row < points_agree.size(); row++)
{ // build distance vector matrix
Vector3f point(points_agree[row].x,
points_agree[row].y,
points_agree[row].z);
for (size_t col = 0; col < 3; col ++)
{
M(row, col) = point(col) - centroid(col);
}
}
Matrix3f covariance_matrix = M.transpose() * M;
Eigen::EigenSolver<Matrix3f> eigen_solver;
eigen_solver.compute(covariance_matrix);
Vector3f eigen_values = eigen_solver.eigenvalues().real();
Matrix3f eigen_vectors = eigen_solver.eigenvectors().real();
// find eigenvalue that is closest to 0
size_t idx;
// find minimum eigenvalue, get index
float closest_eval = eigen_values.cwiseAbs().minCoeff(&idx);
// find corresponding eigenvector
Vector3f closest_evec = eigen_vectors.col(idx);
std::stringstream logstr;
logstr << "Closest eigenvalue : " << closest_eval << std::endl <<
"Corresponding eigenvector : " << std::endl << closest_evec << std::endl <<
"Centroid : " << std::endl << centroid;
ROS_DEBUG("run_RANSAC(): %s", logstr.str().c_str());
Vector3f all_fitted_n_hat = closest_evec / closest_evec.norm();
// invoke copy constructors for outbound
*out_n = Vector3f(all_fitted_n_hat);
*out_p0 = Vector3f(centroid);
*out_inlier_points = std::vector<Point32>(points_agree);
ROS_DEBUG("run_RANSAC():: Success, total_size: %li, inlier_size: %li, %% agreement %f",
all_points.size(), out_inlier_points->size(), (float) out_inlier_points->size() / all_points.size());
return true;
}
} // end iterations loop
return false;
}
Pseudocode from wikipedia for reference:
Given:
data – a set of observed data points
model – a model that can be fitted to data points
n – minimum number of data points required to fit the model
k – maximum number of iterations allowed in the algorithm
t – threshold value to determine when a data point fits a model
d – number of close data points required to assert that a model fits well to data
Return:
bestfit – model parameters which best fit the data (or nul if no good model is found)
iterations = 0
bestfit = nul
besterr = something really large
while iterations < k {
maybeinliers = n randomly selected values from data
maybemodel = model parameters fitted to maybeinliers
alsoinliers = empty set
for every point in data not in maybeinliers {
if point fits maybemodel with an error smaller than t
add point to alsoinliers
}
if the number of elements in alsoinliers is > d {
% this implies that we may have found a good model
% now test how good it is
bettermodel = model parameters fitted to all points in maybeinliers and alsoinliers
thiserr = a measure of how well model fits these points
if thiserr < besterr {
bestfit = bettermodel
besterr = thiserr
}
}
increment iterations
}
return bestfit
The only difference between my implementation and the wikipedia pseudocode is the following:
thiserr = a measure of how well model fits these points
if thiserr < besterr {
bestfit = bettermodel
besterr = thiserr
}
My guess is that I need to do something related to comparing the (closest_eval) with some sentinel value for the expected minimum eigenvalue corresponding to a normal for planes that tend to fit the model. However this was not covered in class and I have no idea where to start figuring out what's wrong.
Heh, it's funny how thinking about how to present the problem to others can actually solve the problem I'm having.
Solved by simply implementing this with a std::numeric_limits::max() starting best fit eigenvalue. This is because the best fit plane extracted on any n-th iteration of RANSAC is not guaranteed to be THE best fit plane and may have a huge error in consensus amongst each constituent point, so I need to converge on that for each iteration. Woops.
thiserr = a measure of how well model fits these points
if thiserr < besterr {
bestfit = bettermodel
besterr = thiserr
}
I am trying to get the left polygonal chain given a set of consecutive points. (NOTE: edges are non-intersecting.)
Image 1. Sample polygon and its bound.
What I did was:
Get the minY, maxY and minX. (Bound.)
Find the point that contains minY (or maxY) then save it as the first point.
Save any points until point with minY or maxY is found while checking for point with minX.
If the same Y is found first, save it as the new first point and repeat from #3.
If other Y is found first and the saved points has minX, this is the chain. Otherwise, save as the new first point and repeat from #3.
Image 2. The left chain of points.
But using this steps might give wrong result for some polygon, like this:
Since one point is (minX, maxY), either of the side will be returned.
EDIT:
With the idea of the left-bottom- and left-top-most points, here is the current code that I am using:
Get the min (left-bottom-most) and max (left-top-most) point.
std::vector<Coord> ret;
size_t i = 0;
Coord minCoord = poly[i];
Coord maxCoord = poly[i];
size_t minIdx = -1;
size_t maxIdx = -1;
size_t cnt = poly.size();
i++;
for (; i < cnt; i++)
{
Coord c = poly[i];
if (c.y < minCoord.y) // new bottom
{
minCoord = c;
minIdx = i;
}
else if (c.y == minCoord.y) // same bottom
{
if (c.x < minCoord.x) // left most
{
minCoord = c;
minIdx = i;
}
}
if (c.y > maxCoord.y) // new top
{
maxCoord = c;
maxIdx = i;
}
else if (c.y == maxCoord.y) // same top
{
if (c.x < maxCoord.x) // left most
{
maxCoord = c;
maxIdx = i;
}
}
}
Get the points connected to the max point.
i = maxIdx;
Coord mid = poly[i];
Coord ray1 = poly[(i + cnt - 1) % cnt];
Coord ray2 = poly[(i + 1) % cnt];
Get which has smallest angle. This will be the path we will follow.
double rad1 = Pts2Rad(mid, ray1);
double rad2 = Pts2Rad(mid, ray2);
int step = 1;
if (rad1 < rad2)
step = cnt - 1;
Save the points.
while (i != minIdx)
{
ret.push_back(poly[i]);
i = (i + step) % cnt;
}
ret.push_back(poly[minIdx]);
To be specific, I am assuming that no vertex is duplicated and define the "left chain" as the sequence of vertices from the original polygon loop that goes from the leftmost vertex in the top side of the bounding box, to the leftmost vertex in the bottom side of the bounding box. [In case the top and bottom sides coincide, these two vertices also coincide; I leave it to you what to return in this case.]
To obtain these, you can scan all vertices and keep the left-topmost so far and left-bottommost so far. Then compare to the next vertex. If above the left-topmost, becomes the new lef-topmost. If at the same level and to the left, becomes the new left-topmost. Similarly for the left-bottommost.
Given N boxes. How can i find the tallest tower made with them in the given order ? (Given order means that the first box must be at the base of the tower and so on). All boxes must be used to make a valid tower.
It is possible to rotate the box on any axis in a way that any of its 6 faces gets parallel to the ground, however the perimeter of such face must be completely restrained inside the perimeter of the superior face of the box below it. In the case of the first box it is possible to choose any face, because the ground is big enough.
To solve this problem i've tried the following:
- Firstly the code generates the rotations for each rectangle (just a permutation of the dimensions)
- secondly constructing a dynamic programming solution for each box and each possible rotation
- finally search for the highest tower made (in the dp table)
But my algorithm is taking wrong answer in unknown test cases. What is wrong with it ? Dynamic programming is the best approach to solve this problem ?
Here is my code:
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cstdlib>
#include <cstring>
struct rectangle{
int coords[3];
rectangle(){ coords[0] = coords[1] = coords[2] = 0; }
rectangle(int a, int b, int c){coords[0] = a; coords[1] = b; coords[2] = c; }
};
bool canStack(rectangle ¤t_rectangle, rectangle &last_rectangle){
for (int i = 0; i < 2; ++i)
if(current_rectangle.coords[i] > last_rectangle.coords[i])
return false;
return true;
}
//six is the number of rotations for each rectangle
int dp(std::vector< std::vector<rectangle> > &v){
int memoization[6][v.size()];
memset(memoization, -1, sizeof(memoization));
//all rotations of the first rectangle can be used
for (int i = 0; i < 6; ++i) {
memoization[i][0] = v[0][i].coords[2];
}
//for each rectangle
for (int i = 1; i < v.size(); ++i) {
//for each possible permutation of the current rectangle
for (int j = 0; j < 6; ++j) {
//for each permutation of the previous rectangle
for (int k = 0; k < 6; ++k) {
rectangle &prev = v[i - 1][k];
rectangle &curr = v[i][j];
//is possible to put the current rectangle with the previous rectangle ?
if( canStack(curr, prev) ) {
memoization[j][i] = std::max(memoization[j][i], curr.coords[2] + memoization[k][i-1]);
}
}
}
}
//what is the best solution ?
int ret = -1;
for (int i = 0; i < 6; ++i) {
ret = std::max(memoization[i][v.size()-1], ret);
}
return ret;
}
int main ( void ) {
int n;
scanf("%d", &n);
std::vector< std::vector<rectangle> > v(n);
for (int i = 0; i < n; ++i) {
rectangle r;
scanf("%d %d %d", &r.coords[0], &r.coords[1], &r.coords[2]);
//generate all rotations with the given rectangle (all combinations of the coordinates)
for (int j = 0; j < 3; ++j)
for (int k = 0; k < 3; ++k)
if(j != k) //micro optimization disease
for (int l = 0; l < 3; ++l)
if(l != j && l != k)
v[i].push_back( rectangle(r.coords[j], r.coords[k], r.coords[l]) );
}
printf("%d\n", dp(v));
}
Input Description
A test case starts with an integer N, representing the number of boxes (1 ≤ N ≤ 10^5).
Following there will be N rows, each containing three integers, A, B and C, representing the dimensions of the boxes (1 ≤ A, B, C ≤ 10^4).
Output Description
Print one row containing one integer, representing the maximum height of the stack if it’s possible to pile all the N boxes, or -1 otherwise.
Sample Input
2
5 2 2
1 3 4
Sample Output
6
Sample image for the given input and output.
Usually you're given the test case that made you fail. Otherwise, finding the problem is a lot harder.
You can always approach it from a different angle! I'm going to leave out the boring parts that are easily replicated.
struct Box { unsigned int dim[3]; };
Box will store the dimensions of each... box. When it comes time to read the dimensions, it needs to be sorted so that dim[0] >= dim[1] >= dim[2].
The idea is to loop and read the next box each iteration. It then compares the second largest dimension of the new box with the second largest dimension of the last box, and same with the third largest. If in either case the newer box is larger, it adjusts the older box to compare the first largest and third largest dimension. If that fails too, then the first and second largest. This way, it always prefers using a larger dimension as the vertical one.
If it had to rotate a box, it goes to the next box down and checks that the rotation doesn't need to be adjusted there too. It continues until there are no more boxes or it didn't need to rotate the next box. If at any time, all three rotations for a box failed to make it large enough, it stops because there is no solution.
Once all the boxes are in place, it just sums up each one's vertical dimension.
int main()
{
unsigned int size; //num boxes
std::cin >> size;
std::vector<Box> boxes(size); //all boxes
std::vector<unsigned char> pos(size, 0); //index of vertical dimension
//gets the index of dimension that isn't vertical
//largest indicates if it should pick the larger or smaller one
auto get = [](unsigned char x, bool largest) { if (largest) return x == 0 ? 1 : 0; return x == 2 ? 1 : 2; };
//check will compare the dimensions of two boxes and return true if the smaller one is under the larger one
auto check = [&boxes, &pos, &get](unsigned int x, bool largest) { return boxes[x - 1].dim[get(pos[x - 1], largest)] < boxes[x].dim[get(pos[x], largest)]; };
unsigned int x = 0, y; //indexing variables
unsigned char change; //detects box rotation change
bool fail = false; //if it cannot be solved
for (x = 0; x < size && !fail; ++x)
{
//read in the next three dimensions
//make sure dim[0] >= dim[1] >= dim[2]
//simple enough to write
//mine was too ugly and I didn't want to be embarrassed
y = x;
while (y && !fail) //when y == 0, no more boxes to check
{
change = pos[y - 1];
while (check(y, true) || check(y, false)) //while invalid rotation
{
if (++pos[y - 1] == 3) //rotate, when pos == 3, no solution
{
fail = true;
break;
}
}
if (change != pos[y - 1]) //if rotated box
--y;
else
break;
}
}
if (fail)
{
std::cout << -1;
}
else
{
unsigned long long max = 0;
for (x = 0; x < size; ++x)
max += boxes[x].dim[pos[x]];
std::cout << max;
}
return 0;
}
It works for the test cases I've written, but given that I don't know what caused yours to fail, I can't tell you what mine does differently (assuming it also doesn't fail your test conditions).
If you are allowed, this problem might benefit from a tree data structure.
First, define the three possible cases of block:
1) Cube - there is only one possible option for orientation, since every orientation results in the same height (applied toward total height) and the same footprint (applied to the restriction that the footprint of each block is completely contained by the block below it).
2) Square Rectangle - there are three possible orientations for this rectangle with two equal dimensions (for examples, a 4x4x1 or a 4x4x7 would both fit this).
3) All Different Dimensions - there are six possible orientations for this shape, where each side is different from the rest.
For the first box, choose how many orientations its shape allows, and create corresponding nodes at the first level (a root node with zero height will allow using simple binary trees, rather than requiring a more complicated type of tree that allows multiple elements within each node). Then, for each orientation, choose how many orientations the next box allows but only create nodes for those that are valid for the given orientation of the current box. If no orientations are possible given the orientation of the current box, remove that entire unique branch of orientations (the first parent node with multiple valid orientations will have one orientation removed by this pruning, but that parent node and all of its ancestors will be preserved otherwise).
By doing this, you can check for sets of boxes that have no solution by checking whether there are any elements below the root node, since an empty tree indicates that all possible orientations have been pruned away by invalid combinations.
If the tree is not empty, then just walk the tree to find the highest sum of heights within each branch of the tree, recursively up the tree to the root - the sum value is your maximum height, such as the following pseudocode:
std::size_t maximum_height() const{
if(leftnode == nullptr || rightnode == nullptr)
return this_node_box_height;
else{
auto leftheight = leftnode->maximum_height() + this_node_box_height;
auto rightheight = rightnode->maximum_height() + this_node_box_height;
if(leftheight >= rightheight)
return leftheight;
else
return rightheight;
}
}
The benefits of using a tree data structure are
1) You will greatly reduce the number of possible combinations you have to store and check, because in a tree, the invalid orientations will be eliminated at the earliest possible point - for example, using your 2x2x5 first box, with three possible orientations (as a Square Rectangle), only two orientations are possible because there is no possible way to orient it on its 2x2 end and still fit the 4x3x1 block on it. If on average only two orientations are possible for each block, you will need a much smaller number of nodes than if you compute every possible orientation and then filter them as a second step.
2) Detecting sets of blocks where there is no solution is much easier, because the data structure will only contain valid combinations.
3) Working with the finished tree will be much easier - for example, to find the sequence of orientations of the highest, rather than just the actual height, you could pass an empty std::vector to a modified highest() implementation, and let it append the actual orientation of each highest node as it walks the tree, in addition to returning the height.
I've been working on this for several weeks but have been unable to get my algorithm working properly and i'm at my wits end. Here's an illustration of what i have achieved:
If everything was working i would expect a perfect circle/oval at the end.
My sample points (in white) are recalculated every time a new control point (in yellow) is added. At 4 control points everything looks perfect, again as i add a 5th on top of the 1st things look alright, but then on the 6th it starts to go off too the side and on the 7th it jumps up to the origin!
Below I'll post my code, where calculateWeightForPointI contains the actual algorithm. And for reference- here is the information i'm trying to follow. I'd be so greatful if someone could take a look for me.
void updateCurve(const std::vector<glm::vec3>& controls, std::vector<glm::vec3>& samples)
{
int subCurveOrder = 4; // = k = I want to break my curve into to cubics
// De boor 1st attempt
if(controls.size() >= subCurveOrder)
{
createKnotVector(subCurveOrder, controls.size());
samples.clear();
for(int steps=0; steps<=20; steps++)
{
// use steps to get a 0-1 range value for progression along the curve
// then get that value into the range [k-1, n+1]
// k-1 = subCurveOrder-1
// n+1 = always the number of total control points
float t = ( steps / 20.0f ) * ( controls.size() - (subCurveOrder-1) ) + subCurveOrder-1;
glm::vec3 newPoint(0,0,0);
for(int i=1; i <= controls.size(); i++)
{
float weightForControl = calculateWeightForPointI(i, subCurveOrder, controls.size(), t);
newPoint += weightForControl * controls.at(i-1);
}
samples.push_back(newPoint);
}
}
}
//i = the weight we're looking for, i should go from 1 to n+1, where n+1 is equal to the total number of control points.
//k = curve order = power/degree +1. eg, to break whole curve into cubics use a curve order of 4
//cps = number of total control points
//t = current step/interp value
float calculateWeightForPointI( int i, int k, int cps, float t )
{
//test if we've reached the bottom of the recursive call
if( k == 1 )
{
if( t >= knot(i) && t < knot(i+1) )
return 1;
else
return 0;
}
float numeratorA = ( t - knot(i) );
float denominatorA = ( knot(i + k-1) - knot(i) );
float numeratorB = ( knot(i + k) - t );
float denominatorB = ( knot(i + k) - knot(i + 1) );
float subweightA = 0;
float subweightB = 0;
if( denominatorA != 0 )
subweightA = numeratorA / denominatorA * calculateWeightForPointI(i, k-1, cps, t);
if( denominatorB != 0 )
subweightB = numeratorB / denominatorB * calculateWeightForPointI(i+1, k-1, cps, t);
return subweightA + subweightB;
}
//returns the knot value at the passed in index
//if i = 1 and we want Xi then we have to remember to index with i-1
float knot(int indexForKnot)
{
// When getting the index for the knot function i remember to subtract 1 from i because of the difference caused by us counting from i=1 to n+1 and indexing a vector from 0
return knotVector.at(indexForKnot-1);
}
//calculate the whole knot vector
void createKnotVector(int curveOrderK, int numControlPoints)
{
int knotSize = curveOrderK + numControlPoints;
for(int count = 0; count < knotSize; count++)
{
knotVector.push_back(count);
}
}
Your algorithm seems to work for any inputs I tried it on. Your problem might be a that a control point is not where it is supposed to be, or that they haven't been initialized properly. It looks like there are two control-points, half the height below the bottom left corner.