I am making a Correlogram for an image. For each pixel, a correlogram finds the pixels of same color within a certain range of distance, d. Correlogram is a 2D matrix i.e. correlogram[color][distance]. The calculation of a Correlogram is somewhat similar to that of a Histogram.
My Code: I am posting some major part of the code in which all the calculations are going on. Rest of the code (which i didn't post) is used to fulfill other condtions and therefore is not necessary.
Problem: In my final correlogram[][] , some values are "nan". I have checked the code but i am not able to find where is the problem in my calculation/syntax.
int ColorBins = 180;
int DistanceRange = 5;
double calcCorrelogram(Mat hsvImage)
{
double correlogram[ColorBins][DistanceRange];
int pixelNum[ColorBins]; //Used to count the number of pixels of same color
Mat hsvPlanes[3];
split(hsvImage, hsvPlanes);
for(int pi=0; pi<hsvImage.rows; pi++)
{
for(int pj=0; pj<hsvImage.cols; pj++)
{
int pixelColor = (int)hsvPlanes[0].at<uchar>(pi,pj);
pixelNum[pixelColor]++;
for(int d=1; d<=DistanceRange; d++)
{
int sameColorNum=0; //* number of pixels with same color in the d-distance boundary */
int totalBoundaryNum=0; //* total number of pixels in the d-distance boundary */
for(int i= pi-d, j= pj-d; j<=pj+d; j++)
{
if(i<0)
break;
if(j<0 || j>=hsvImage.cols)
continue;
int neighbourColor = (int)hsvPlanes[0].at<uchar>(i,j);
if(pixelColor == neighbourColor)
{
sameColorNum++;
}
totalBoundaryNum++;
correlogram[pixelColor][d-1] = correlogram[pixelColor][d-1] + (double)sameColorNum / (double)totalBoundaryNum;
}
}
}
for(int c=0; c<ColorBins; c++)
{
for(int d=0; d<DistanceRange; d++)
{
if(pixelNum[c] != 0)
correlogram[c][d] = correlogram[c][d] / (double)pixelNum[c];
}
}
}
NaNs are generally created when you divide zero by zero or multiply zero by infinity. One easy way to check for abnormal numbers like NaN and infinity is to multiply by zero and check if the result is zero:
bool is_valid_double(double x)
{
return x*0.0==0.0;
}
This will return false if x is either NaN or infinity.
Then you can sprinkle your code with assertions to help find where things are going wrong:
assert(is_valid_double(correlogram[c][d]));
Once you get a crash due to an assertion failure, you can use the debugger to look at the state of the program to help determine what is going on.
Related
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'm trying to use C++ to recreate the spectrogram function used by Matlab. The function uses a Short Time Fourier Transform (STFT). I found some C++ code here that performs a STFT. The code seems to work perfectly for all frequencies but I only want a few. I found this post for a similar question with the following answer:
Just take the inner product of your data with a complex exponential at
the frequency of interest. If g is your data, then just substitute for
f the value of the frequency you want (e.g., 1, 3, 10, ...)
Having no background in mathematics, I can't figure out how to do this. The inner product part seems simple enough from the Wikipedia page but I have absolutely no idea what he means by (with regard to the formula for a DFT)
a complex exponential at frequency of interest
Could someone explain how I might be able to do this? My data structure after the STFT is a matrix filled with complex numbers. I just don't know how to extract my desired frequencies.
Relevant function, where window is Hamming, and vector of desired frequencies isn't yet an input because I don't know what to do with them:
Matrix<complex<double>> ShortTimeFourierTransform::Calculate(const vector<double> &signal,
const vector<double> &window, int windowSize, int hopSize)
{
int signalLength = signal.size();
int nOverlap = hopSize;
int cols = (signal.size() - nOverlap) / (windowSize - nOverlap);
Matrix<complex<double>> results(window.size(), cols);
int chunkPosition = 0;
int readIndex;
// Should we stop reading in chunks?
bool shouldStop = false;
int numChunksCompleted = 0;
int i;
// Process each chunk of the signal
while (chunkPosition < signalLength && !shouldStop)
{
// Copy the chunk into our buffer
for (i = 0; i < windowSize; i++)
{
readIndex = chunkPosition + i;
if (readIndex < signalLength)
{
// Note the windowing!
data[i][0] = signal[readIndex] * window[i];
data[i][1] = 0.0;
}
else
{
// we have read beyond the signal, so zero-pad it!
data[i][0] = 0.0;
data[i][1] = 0.0;
shouldStop = true;
}
}
// Perform the FFT on our chunk
fftw_execute(plan_forward);
// Copy the first (windowSize/2 + 1) data points into your spectrogram.
// We do this because the FFT output is mirrored about the nyquist
// frequency, so the second half of the data is redundant. This is how
// Matlab's spectrogram routine works.
for (i = 0; i < windowSize / 2 + 1; i++)
{
double real = fft_result[i][0];
double imaginary = fft_result[i][1];
results(i, numChunksCompleted) = complex<double>(real, imaginary);
}
chunkPosition += hopSize;
numChunksCompleted++;
} // Excuse the formatting, the while ends here.
return results;
}
Look up the Goertzel algorithm or filter for example code that uses the computational equivalent of an inner product against a complex exponential to measure the presence or magnitude of a specific stationary sinusoidal frequency in a signal. Performance or resolution will depend on the length of the filter and your signal.
Basically i have to figure out whether the label on on object is straight. I have an edge image of the object. I would like to calculate the distance between the 2 edges on either side in a single row.
My algorithm involves iterating through a row until a white pixel is found. Then calculating the number of black pixels until the next white is found. However when i run the code the answer is always zero.
Code:
for(int i = 0; i < img.cols; i++)
{
int num = nms_result.at<int>(i,100);
//cout <<num<<endl;
if(num > 0) {
stage2 = true;
}
if (stage2 ==true)
counter4++;
{
int num2 = nms_result.at<int>(i,100);
;
if ((num2 < 1) && (counter4 >=1 )) {
counter2++;
}
else counter4 = 0;
}
}
I have tried a lot of things but none seem to work.
Problem number 1: If I'm reading your code right, 'num' and 'num2' are always the same, since they're in the same loop.
Problem number 2: What's the output here? a little hard to tell with your formatting. Consider using some indentations with your nested ifs.
I am trying to create my personal Blob Detection algorithm
As far as I know I first must create different Gaussian Kernels with different sigmas (which I am doing using Mat kernel= getGaussianKernel(x,y);) Then get the Laplacian of that kernel and then filter the Image with that so I create my scalespace. Now I need to find the Local Maximas in each result Image of the scalespace. But I cannot seem to find a proper way to do so.... my Code so far is
vector <Point> GetLocalMaxima(const cv::Mat Src,int MatchingSize, int Threshold)
{
vector <Point> vMaxLoc(0);
if ((MatchingSize % 2 == 0) ) // MatchingSize has 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;
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;
}
which I found in this thread here, which it supposedly returns a vector of points where the maximas are. it does return a vector of points but all the x and y coordinates of each point are always -17891602... What to do???
Please if you are to lead me in something else other than correcting my code be informative because I know nothing about opencv. I am just learning
The problem here is that your LocMax point is declared inside the inner loop and never initialized, so it's returning garbage data every time. If you look back at the StackOverflow question you linked, you'll see that their similar variable Point maxLoc(0,0) is declared at the top and constructed to point at the middle of the search window. It only needs to be initialized once. Subsequent loop iterations will replace the value with the minMaxLoc function result.
In summary, remove this line in your inner loop:
Point LocMax; // delete this
And add a slightly altered version near the top:
vector <Point> vMaxLoc(0); // This was your original first line
Point LocMax(0,0); // your new second line
That should get you started anyway.
I found it guys. The problem was my threshold was too high. I do not understand why it gave me negative points instead of zero points but lowering the threshold worked
I wrote this simple code which reads a length from the Sharp infrared sensor, end presents the average meter in cm (unit) by serial.
When write this code for the Arduino Mega board, the Arduino starts a blinking LED (pin 13) and the program does nothing. Where is the bug in this code?
#include <QueueList.h>
const int ANALOG_SHARP = 0; //Set pin data from sharp.
QueueList <float> queuea;
float cm;
float qu1;
float qu2;
float qu3;
float qu4;
float qu5;
void setup() {
Serial.begin(9600);
}
void loop() {
cm = read_gp2d12_range(ANALOG_SHARP); //Convert to cm (unit).
queuea.push(cm); //Add item to queue, when I add only this line Arduino crash.
if ( 5 <= queuea.peek()) {
Serial.println(average());
}
}
float read_gp2d12_range(byte pin) { //Function converting to cm (unit).
int tmp;
tmp = analogRead(pin);
if (tmp < 3)
return -1; // Invalid value.
return (6787.0 /((float)tmp - 3.0)) - 4.0;
}
float average() { //Calculate average length
qu1 += queuea.pop();
qu2 += queuea.pop();
qu3 += queuea.pop();
qu4 += queuea.pop();
qu5 += queuea.pop();
float aver = ((qu1+qu2+qu3+qu4+qu5)/5);
return aver;
}
I agree with the peek() -> count() error listed by vhallac. But I'll also point out that you should consider averaging by powers of 2 unless there is a strong case to do otherwise.
The reason is that on microcontrollers, division is slow. By averaging over a power of 2 (2,4,8,16,etc.) you can simply calculate the sum and then bitshift it.
To calculate the average of 2: (v1 + v2) >> 1
To calculate the average of 4: (v1 + v2 + v3 + v4) >> 2
To calculate the average of n values (where n is a power of 2) just right bitshift the sum right by [log2(n)].
As long as the datatype for your sum variable is big enough and won't overflow, this is much easier and much faster.
Note: this won't work for floats in general. In fact, microcontrollers aren't optimized for floats. You should consider converting from int (what I'm assuming you're ADC is reading) to float at the end after the averaging rather than before.
By converting from int to float and then averaging floats you are losing more precision than averaging ints than converting the int to a float.
Other:
You're using the += operator without initializing the variables (qu1, qu2, etc.) -- it's good practice to initialize them if you're going to use += but it looks as if = would work fine.
For floats, I'd have written the average function as:
float average(QueueList<float> & q, int n)
{
float sum = 0;
for(int i=0; i<n; i++)
{
sum += q.pop();
}
return (sum / (float) n);
}
And called it: average(queuea, 5);
You could use this to average any number of sensor readings and later use the same code to later average floats in a completely different QueueList. Passing the number of readings to average as a parameter will really come in handy in the case that you need to tweak it.
TL;DR:
Here's how I would have done it:
#include <QueueList.h>
const int ANALOG_SHARP=0; // set pin data from sharp
const int AvgPower = 2; // 1 for 2 readings, 2 for 4 readings, 3 for 8, etc.
const int AvgCount = pow(2,AvgPow);
QueueList <int> SensorReadings;
void setup(){
Serial.begin(9600);
}
void loop()
{
int reading = analogRead(ANALOG_SHARP);
SensorReadings.push(reading);
if(SensorReadings.count() > AvgCount)
{
int avg = average2(SensorReadings, AvgPower);
Serial.println(gpd12_to_cm(avg));
}
}
float gp2d12_to_cm(int reading)
{
if(reading <= 3){ return -1; }
return((6787.0 /((float)reading - 3.0)) - 4.0);
}
int average2(QueueList<int> & q, int AvgPower)
{
int AvgCount = pow(2, AvgPower);
long sum = 0;
for(int i=0; i<AvgCount; i++)
{
sum += q.pop();
}
return (sum >> AvgPower);
}
You are using queuea.peek() to obtain the count. This will only return the last element in queue. You should use queuea.count() instead.
Also you might consider changing the condition tmp < 3 to tmp <= 3. If tmp is 3, you divide by zero.
Great improvement jedwards, however the first question I have is why use queuelist instead of an int array.
As an example I would do the following:
int average(int analog_reading)
{
#define NUM_OF_AVG 5
static int readings[NUM_OF_AVG];
static int next_position;
static int sum;
if (++next_position >= NUM_OF_AVG)
{
next_position=0;
}
reading[next_position]=analog_reading;
for(int i=0; i<NUM_OF_AVG; i++)
{
sum += reading[i];
}
average = sum/NUM_OF_AVG
}
Now I compute a new rolling average with every reading and it eliminates all the issues related to dynamic memory allocation (memory fragmentation, no available memory, memory leaks) in a embedded device.
I appreciate and understand the use of shifting for a division by 2,4 or 8, however I would stay away from that technique for two reasons.
I think readability and maintainability of the source code is more important then saving a little bit of time with a shift instead of a divide unless you can test and verify the divide is a bottleneck.
Second, I believe most current optimizing compilers will do a shift if possible, I know GCC does.
I will leave refactoring out the for loop for the next guy.