So I have two identical files representing a matrix :
4x3
1 2 3
1 2 3
1 2 3
1 2 3
I have to sum the two matrix , but I have to use processe in which each process is responsible for calculating each column of the matrix.
I have this :
int addmx(const char *file1, const char *file2){
FILE *f1 = fopen(file1,"r");
if (f1 == NULL){
perror("Error in opening file 1");
return -1;
}
FILE *f2 = fopen (file2, "r");
if (f2 == NULL){
perror("Error in opening file 2");
return -1;
}
struct stat sb;
if (fstat(fileno(f1),&sb) == -1){
perror("Couldn't get file size.\n");
}
printf("The file size is %ld\n",sb.st_size);
struct stat sb2;
if (fstat(fileno(f2),&sb2) == -1){
perror("Couldn't get file size.\n");
}
printf("The file size is %ld\n",sb2.st_size);
char *file_in_memory = mmap(NULL,sb.st_size ,PROT_READ ,MAP_SHARED,fileno(f1),0);
char *file_in_memory2 = mmap(file_in_memory,sb2.st_size,PROT_READ,MAP_SHARED,fileno(f2),0);
char *file_in_memory3 = mmap(file_in_memory2,sb.st_size ,PROT_WRITE,MAP_SHARED,0,0);
for (int i = 0; i < sb.st_size;i++){
printf("%c",file_in_memory[i]);
}
printf("\n");
for (int i = 0; i < sb2.st_size;i++){
printf("%c",file_in_memory2[i]);
}
int counter = 1;
char ch1;
do {
ch1 = fgetc(f1);
if (ch1 == '\n'){
break;
}
else counter++;
} while(1);
unsigned int row = atoi(&file_in_memory[0]) ;
unsigned int col = atoi(&file_in_memory[2]) ;
//file_in_memory3[0] = row + '0'; file_in_memory3[1] = 'x'; file_in_memory3[2] = col + '0';
pid_t pid_matrix[col];
for (size_t i = 1; i <= col;i++) {
pid_matrix[col] = fork();
if (pid_matrix[col] == -1) {
//erro
}
else if (pid_matrix[col] == 0) {
//child process
for (int j = 1; j <= row;j++){
int new_col = atoi(&file_in_memory[counter + (j-1)*(col*2) + ((i-1)*2)] )+ atoi(&file_in_memory2[counter + (j-1)*(col*2) + ((i-1)*2)]);
file_in_memory3[counter+ (j-1)*(col*2) + ((i-1)*2)] = new_col + '0';
printf("%d\n",new_col);
}
printf("Child process with pid %d for column %d\n", getpid(),i);
exit(EXIT_SUCCESS);
}
else {
//parent process
printf("Created a new process with pid %d\n", pid_matrix[col]);
}
}
for ( int i = 0; i < sb.st_size;i++){
printf("%c",file_in_memory3[i]);
}
munmap(file_in_memory,sb.st_size);
munmap(file_in_memory2,sb2.st_size);
close(f1);
close(f2);
return 0;
}
However, the child processes are not executing ( because it is not printing "Child process with pid ... ") but when I remove the line of code :
"file_in_memory3[counter + (j-1)(col2) + ((i-1)*2)] = new_col + '0';"
the child processes actually execute...
Related
Node * create(Node * root, int I, int J, string str)
{
if (I == J) { root -> data =str[I]; root -> left=NULL; root -> right=NULL; }
int i = 0, j = 0, k = 0, l = 0;
//// to store the data of root
string val;
for (i = I; i < J; i++) {
if (str[i] == '(') break;
val.push_back(str[i]);
}
root -> data = stoi(val);
stack < char > st;
for (j = i; j < J; j++) {
if (str[j] == '(') st.push(str[j]);
else if (str[j] == ')') st.pop();
if (st.empty()) break;
}
for (l = j + 1; l < J; l++) {
if (str[l] == '(') st.push(str[l]);
else if (str[l] == ')') st.pop();
if (st.empty()) break;
}
k = j + 1;
if (j - i == 2) root -> left -> data = str[i + 1];
else
root -> left = create(root -> left, i + 1, j - 1, str);
if (l == k) root -> right=NULL;
else if (l - k == 2) root -> right -> data = str[k + 1];
else
root -> right = create(root -> right, k + 1, l - 1, str);
return root;
}
Node * treeFromString(string str){
Node * p = create(p, 0, str.size() - 1, str);
return p;
}
Here I have initialized variables i , j , k , l to track the left child and right child bracket in the string. I , J are the range of the node for a particular activation record of recursion.
I assume you parsed some expression, by your code I pre-implemented it.
I just implement build a tree for following expression:
expression: |<symbol>| <integer> '(' <expression> ')' '('<expression>')'
symbol : any C++ `char` single byte character.
integer : any C++ valid int type value.
#include <iostream>
#include <string>
#include <string_view> // C++17 std::string_view
#include <charconv> // C++17 std::from_chars.
#include <cassert>
//simulate Node class
struct Node
{
Node *left, *right;
int data;
};
//1. use string_view instead of triple I, J, std::string.
//2. First argument Node* root - needn't. It should be local variable.
Node * create(std::string_view str)
{
assert(!str.empty());
if (str.size() == 1)
{
Node* root = new Node; //-> YOU should allocate a memory for root.
root -> data =str[0];
root -> left=nullptr; // use nullptr instead of NULL for c++11 or later.
root -> right=nullptr;
return root; // exit needed!
}
Node* root = new Node;
root->left = nullptr;
root->right = nullptr;
root->data = 0;
//// to store the data of root
//2. THERE NEED take an integer until first '(' symbol.
{
std::size_t i = 0;
while (i < str.size() && str[i] != '(' )
++i;
// str[0 .. i) - interval is an integer.
int val = 0;
(void)std::from_chars(str.data(), str.data() + i, val); // FOR simplifity don't check validness of conversation.
str.remove_prefix(i);
root->data = val;
}
//3. SKIP balanced '(' and ')'
/*stack < char > st;
for (j = i; j < J; j++) {
if (str[j] == '(') st.push(str[j]);
else if (str[j] == ')') st.pop();
if (st.empty()) break;
}
* */
/** we can implement it another way */
assert(!str.empty() && str[0] == '(' );
std::string_view within_bracket;
{
int balanced_brackets = 0;
std::size_t i = 0;
while (i < str.size())
{
if (str[i] == '(') ++ balanced_brackets;
else if (str[i] == ')' ) --balanced_brackets;
i++;
if (balanced_brackets == 0)
break;
}
assert (i > 0 && str[ i - 1 ] == ')' );
// 0 1 2 3
// str[0..i) - is '(' ... ')' symbols.
within_bracket = std::string_view(str.data() + 1, i - 2);
str.remove_prefix(i);
}
/****4. THIS second balanced bracket check */
std::string_view second_within_bracket;
/*
for (l = j + 1; l < J; l++) {
if (str[l] == '(') st.push(str[l]);
else if (str[l] == ')') st.pop();
if (st.empty()) break;
}
k = j + 1;
*/
assert(!str.empty() && str[0] == '(' );
// ========== second balanced brackets check ==========
{
std::size_t i = 0;
int balanced_brackets = 0;
while (i < str.size())
{
if (str[i] == '(') ++ balanced_brackets;
else if (str[i] == ')' ) --balanced_brackets;
i++;
if (balanced_brackets == 0)
break;
}
// 0 1 2 3
// str[0..i) - is '(' ... ')' symbols.
second_within_bracket = std::string_view(str.data() + 1, i - 2);
str.remove_prefix(i);
}
//================================
/*
if (j - i == 2) root -> left -> data = str[i + 1];
else
root -> left = create(i + 1, j - 1, str);
if (l == k) root -> right=NULL;
else if (l - k == 2) root -> right -> data = str[k + 1];
else
root -> right = create(root -> right, k + 1, l - 1, str);
*/
root->left = create(within_bracket);
root->right = create(second_within_bracket);
return root;
}
Node * treeFromString(std::string_view str){
Node * p = create(str);
return p;
}
void printTree(Node* root, int level = 0)
{
if (root==nullptr) return;
for (int i= 0; i < level; ++i) std::cout << "--";
std::cout << " data = " << root->data << std::endl;
printTree(root->left, level + 1);
printTree(root->right, level + 1);
}
int main(){
std::string str = "12(8(7)(5))(9(3)(2(1)(8)))";
Node * expr = treeFromString(str);
printTree(expr);
}
godbold output:
Program returned: 0
data = 12
-- data = 8
---- data = 55
---- data = 53
-- data = 9
---- data = 51
---- data = 2
------ data = 49
------ data = 56
This answer is a bit different, it assumes you are loading a tree of integer values from a string, that could be loaded from a file. Next time you ask a question about code, could you please explain a little bit what the code does? Guessing does take some time and some effort.
I've reused the main() and PrintTree() functions from Khurshid Normuradov's answer. I hope he won't mind.
I took the liberty to add modern c++ coding techniques because this question is tagged c++17, so you're getting an example in c++17.
#include <algorithm>
#include <iostream>
#include <memory>
#include <string>
#include <string_view>
struct Node {
std::unique_ptr<Node> left = nullptr;
std::unique_ptr<Node> right = nullptr;
int value = 0;
};
std::unique_ptr<Node> treeFromString(std::string_view str) {
std::cout << "treeFromString(\"" << str << "\")\n";
if (str.empty()) return {};
auto node = std::make_unique<Node>();
// extract an int
auto pos = str.find_first_not_of("0123456789");
auto val = str.substr(0, pos);
// optional: std::stoi() would throw anyway in this case
if (val.empty())
throw std::runtime_error("invalid value in expression");
node->value = std::stoi(std::string{val});
if (val.length() == str.length()) return node;
str = str.substr(val.length());
// Both left/right parsing are similar and use this
// common subroutine.
// expects parens delimited string as input.
// param str in: str string to parse from
// out: whatever's left to parse
// returns string content within parens, parens not included.
auto extract_parens_contents = [](std::string_view& str) {
// right here would be the perfect place to insert code to skip
// whitespace if you ever needed to do that.
// find parens extent
int parens = 0;
auto parens_end =
std::find_if(str.begin(), str.end(), [&parens](char c) {
parens += (c == '(') - (c == ')');
return (parens == 0);
});
if (parens_end == str.end())
throw std::runtime_error("unbalanced parens in expression");
// extract result
auto result = std::string_view(
str.begin() + 1, std::distance(str.begin() + 1, parens_end));
// remove spent bytes from input stream
str = std::string_view(
parens_end + 1,
str.length() - std::distance(str.begin(), parens_end + 1));
return result;
};
node->left = treeFromString(extract_parens_contents(str));
node->right = treeFromString(extract_parens_contents(str));
return node;
}
// special thanks to user Khurshid Normuradov, who originally wrote the two functions below.
// it would be difficult to writing something that would be any better for the
// intended purpose.
void printTree(Node* root, int level = 0) {
if (root == nullptr) return;
for (int i = 0; i < level; ++i) std::cout << "--";
std::cout << " data = " << root->value << std::endl;
printTree(root->left.get(), level + 1);
printTree(root->right.get(), level + 1);
}
int main() {
std::string str = "12(8(7)(5))(9(3)(2(1)(8)))";
auto expr = treeFromString(str);
printTree(expr.get());
}
You can play with the code here: https://godbolt.org/z/ch3zv5KTT
I'm trying to create a IWICBitmap from an EXR file (error checks removed).
#pragma pack(push,1)
struct fl
{
float r, g, b, a;
};
#pragma pack(pop)
HRESULT Open(const char* f,IWICBitmap** d)
{
exr_context_initializer_t ctxtinit = EXR_DEFAULT_CONTEXT_INITIALIZER;
exr_context_t myfile = {};
exr_result_t rv = exr_start_read(&myfile, f, &ctxtinit);
int part_index = 0;
const exr_attr_chlist_t* chl = 0;
exr_get_channels(myfile, part_index, &chl);
int32_t ck = 0;
rv = exr_get_chunk_count(myfile, part_index, &ck);
int32_t sl = 0;
rv = exr_get_scanlines_per_chunk(myfile, part_index, &sl);
int y = 0;
int wi = 0;
int he = 0;
std::vector<fl> data; // put here the floats
exr_decode_pipeline_t dec = {};
for (int32_t cuk = 0; cuk < ck; cuk++)
{
exr_chunk_info_t ch = {};
exr_read_scanline_chunk_info(myfile, part_index, y, &ch);
wi = ch.width;
he += ch.height;
y += sl;
bool first = 0;
if (dec.decompress_fn == 0)
{
rv = exr_decoding_initialize(myfile, part_index, &ch, &dec);
rv = exr_decoding_choose_default_routines(myfile, part_index, &dec);
first = 1;
}
if (!first)
rv = exr_decoding_update(myfile, part_index,&ch,&dec);
rv = exr_decoding_run(myfile, part_index, &dec);
int NumPixels = (wi * ch.height);
auto BytesPerPixel = ch.unpacked_size / NumPixels;
if (true)
{
// RGB(A)
if (chl->entries[0].pixel_type == EXR_PIXEL_HALF)
{
if (BytesPerPixel == chl->num_channels * 2)
{
auto ds = data.size();
data.resize(ds + NumPixels);
auto p = data.data() + ds;
char* x = (char*)dec.unpacked_buffer;
for (int j = 0; j < NumPixels; j++)
{
uint16_t* u = (uint16_t*)x;
p->a = 1.0f;
for (int jH = 0; jH < chl->num_channels; jH++)
{
half ha(Imath_3_2::half::FromBits,*u);
ha.setBits(*u);
if (strcmp(chl->entries[jH].name.str, "R") == 0) p->r = ha.operator float();
if (strcmp(chl->entries[jH].name.str, "G") == 0) p->g = ha.operator float();
if (strcmp(chl->entries[jH].name.str, "B") == 0) p->b = ha.operator float();
if (strcmp(chl->entries[jH].name.str, "A") == 0) p->a = ha.operator float();
u++;
}
x += BytesPerPixel;
p++;
}
}
else
break;
}
if (chl->entries[0].pixel_type == EXR_PIXEL_FLOAT)
{
// code removed for simplicity, I guess the same issue happens here unless it's a problem of the half-float
}
}
}
rv = exr_decoding_destroy(myfile, &dec);
exr_finish(&myfile);
CComPtr<IWICImagingFactory2> wbfact = 0;
CoCreateInstance(CLSID_WICImagingFactory2, 0, CLSCTX_INPROC_SERVER,
__uuidof(IWICImagingFactory2), (void**)&wbfact);
return wbfact->CreateBitmapFromMemory(wi, he, GUID_WICPixelFormat128bppPRGBAFloat, wi * 16,(UINT)data.size()*16, (BYTE*)data.data(), d);
}
What am I doing wrong? The pixel number I'm reading is correct (in this image 800x800).
My result:
Photoshop:
Is there a problem with the half-float? I'm just using the OpenEXR's IMath implementation.
I have an array which is constituted of only 0s and 1s. Task is to find index of a 0, replacing which with a 1 results in the longest possible sequence of ones for the given array.
Solution has to work within O(n) time and O(1) space.
Eg:
Array - 011101101001
Answer - 4 ( that produces 011111101001)
My Approach gives me a result better than O(n2) but times out on long string inputs.
int findIndex(int[] a){
int maxlength = 0; int maxIndex= -1;
int n=a.length;
int i=0;
while(true){
if( a[i] == 0 ){
int leftLenght=0;
int j=i-1;
//finding count of 1s to left of this zero
while(j>=0){
if(a[j]!=1){
break;
}
leftLenght++;
j--;
}
int rightLenght=0;
j=i+1;
// finding count of 1s to right of this zero
while(j<n){
if(a[j]!=1){
break;
}
rightLenght++;
j++;
}
if(maxlength < leftLenght+rightLenght + 1){
maxlength = leftLenght+rightLenght + 1;
maxIndex = i;
}
}
if(i == n-1){
break;
}
i++;
}
return maxIndex;
}
The approach is simple, you just need to maintain two numbers while iterating through the array, the current count of the continuous block of one, and the last continuous block of one, which separated by zero.
Note: this solution assumes that there will be at least one zero in the array, otherwise, it will return -1
int cal(int[]data){
int last = 0;
int cur = 0;
int max = 0;
int start = -1;
int index = -1;
for(int i = 0; i < data.length; i++){
if(data[i] == 0){
if(max < 1 + last + cur){
max = 1 + last + cur;
if(start != -1){
index = start;
}else{
index = i;
}
}
last = cur;
start = i;
cur = 0;
}else{
cur++;
}
}
if(cur != 0 && start != -1){
if(max < 1 + last + cur){
return start;
}
}
return index;
}
O(n) time, O(1) space
Live demo: https://ideone.com/1hjS25
I believe the problem can we solved by just maintaining a variable which stores the last trails of 1's that we saw before reaching a '0'.
int last_trail = 0;
int cur_trail = 0;
int last_seen = -1;
int ans = 0, maxVal = 0;
for(int i = 0; i < a.size(); i++) {
if(a[i] == '0') {
if(cur_trail + last_trail + 1 > maxVal) {
maxVal = cur_trail + last_trail + 1;
ans = last_seen;
}
last_trail = cur_trail;
cur_trail = 0;
last_seen = i;
} else {
cur_trail++;
}
}
if(cur_trail + last_trail + 1 > maxVal && last_seen > -1) {
maxVal = cur_trail + last_trail + 1;
ans = last_seen;
}
This can be solved by a technique that is known as two pointers. Most two-pointers use O(1) space and O(n) time.
Code : https://www.ideone.com/N8bznU
#include <iostream>
#include <string>
using namespace std;
int findOptimal(string &s) {
s += '0'; // add a sentinel 0
int best_zero = -1;
int prev_zero = -1;
int zeros_in_interval = 0;
int start = 0;
int best_answer = -1;
for(int i = 0; i < (int)s.length(); ++i) {
if(s[i] == '1') continue;
else if(s[i] == '0' and zeros_in_interval == 0) {
zeros_in_interval++;
prev_zero = i;
}
else if(s[i] == '0' and zeros_in_interval == 1) {
int curr_answer = i - start; // [start, i) only contains one 0
cout << "tried this : [" << s.substr(start, i - start) << "]\n";
if(curr_answer > best_answer) {
best_answer = curr_answer;
best_zero = prev_zero;
}
start = prev_zero + 1;
prev_zero = i;
}
}
cout << "Answer = " << best_zero << endl;
return best_zero;
}
int main() {
string input = "011101101001";
findOptimal(input);
return 0;
}
This is an implementation in C++. The output looks like this:
tried this : [0111]
tried this : [111011]
tried this : [1101]
tried this : [10]
tried this : [01]
Answer = 4
Example:
Given “25525511135”
Output : [“255.255.11.135”, “255.255.111.35”]. (sorted order)
Kindly let me know if we could do a depth first search over here ?(that's the only thing striking me )
Why is it important to have an 'optimal' approach for answering this?
There are not many permutations so the simple approach of checking every combination that fits into the IP format and then filtering out those that have out of range numbers will easily work.
It's unlikely to be a bottle neck for whatever this is part of.
You probably want a dynamic programming algorithm for the general case (something like
http://www.geeksforgeeks.org/dynamic-programming-set-32-word-break-problem/).
Instead of testing whether prefixes can be segmented into words in the dictionary, you'd be testing to see whether the prefixes are prefixes of some valid IPv4 address.
Brutal DFS is acceptable in this problem:
class Solution{
private:
vector<string> ans;
int len;
string cur, rec, str;
bool IsOk(string s) {
if(s[0] == '0' && s.size() > 1) return false;
int sum = 0;
for(int i = 0; i < s.size(); i ++) {
if(s[i] == '.') return false;
sum = sum * 10 + s[i] - '0';
}
if(sum >= 0 && sum <= 255) return true;
return false;
}
void dfs(int x, int cnt) {
if(x == len) {
if(str.size() != len + 4) return ;
string tmp(str);
tmp.erase(tmp.size() - 1, 1);
if(cnt == 4) ans.push_back(tmp);
return ;
}
if(cnt > 4 || str.size() > len + 4) return ;
string tmp = cur;
cur += rec[x];
if(!IsOk(cur)) {
cur = tmp;
return ;
}
dfs(x + 1, cnt);
string tmp2 = cur + '.';
str += tmp2;
cur = "";
dfs(x + 1, cnt + 1);
str.erase(str.size() - tmp2.size(), tmp2.size());
cur = tmp;
}
public:
vector<string> restoreIpAddresses(string s) {
this->len = s.size();
this->rec = s;
cur = str = "";
ans.clear();
dfs(0, 0);
return ans;
}
};
Here is a recursive solution on JavaScript. The result is not sorted.
// Task from https://www.geeksforgeeks.org/program-generate-possible-valid-ip-addresses-given-string/
// Given a string containing only digits, restore it by returning all possible valid IP address combinations.
//
// Example:
// Input : 25525511135
// Output : [“255.255.11.135”, “255.255.111.35”]
//
(function () {
function getValidIP(str) {
const result = [];
const length = str.length;
check(0, 0, '');
function check(start, level, previous){
let i = 0;
let num;
if (level === 3) {
num = str.substring(start);
if (num && num < 256) {
result.push(`${previous}.${num}`);
}
return;
}
num = str.substring(start, start + 1);
if (num == 0) {
check(start + 1, level + 1, level === 0 ? `${num}`: `${previous}.${num}`);
} else {
while (num.length < 4 && num < 256 && start + i + 1 < length) {
check(start + i + 1, level + 1, level === 0 ? `${num}`: `${previous}.${num}`);
i++;
num = str.substring(start, start + i + 1);
}
}
}
return result;
}
console.log('12345:')
console.time('1-1');
console.log(getValidIP('12345'));
console.timeEnd('1-1');
console.log('1234:')
console.time('1-2');
console.log(getValidIP('1234'));
console.timeEnd('1-2');
console.log('2555011135:')
console.time('1-3');
console.log(getValidIP('2555011135'));
console.timeEnd('1-3');
console.log('222011135:')
console.time('1-4');
console.log(getValidIP('222011135'));
console.timeEnd('1-4');
})();
I am using the code sample for stereo matching that came with openCV source code, here:
/*
* stereo_match.cpp
* calibration
*
* Created by Victor Eruhimov on 1/18/10.
* Copyright 2010 Argus Corp. All rights reserved.
*
*/
#include "opencv2/calib3d/calib3d.hpp"
#include "opencv2/imgproc/imgproc.hpp"
#include "opencv2/highgui/highgui.hpp"
#include "opencv2/contrib/contrib.hpp"
#include <stdio.h>
using namespace cv;
static void print_help()
{
printf("\nDemo stereo matching converting L and R images into disparity and point clouds\n");
printf("\nUsage: stereo_match <left_image> <right_image> [--algorithm=bm|sgbm|hh|var] [--blocksize=<block_size>]\n"
"[--max-disparity=<max_disparity>] [--scale=scale_factor>] [-i <intrinsic_filename>] [-e <extrinsic_filename>]\n"
"[--no-display] [-o <disparity_image>] [-p <point_cloud_file>]\n");
}
static void saveXYZ(const char* filename, const Mat& mat)
{
const double max_z = 1.0e4;
FILE* fp = fopen(filename, "wt");
for(int y = 0; y < mat.rows; y++)
{
for(int x = 0; x < mat.cols; x++)
{
Vec3f point = mat.at<Vec3f>(y, x);
if(fabs(point[2] - max_z) < FLT_EPSILON || fabs(point[2]) > max_z) continue;
fprintf(fp, "%f %f %f\n", point[0], point[1], point[2]);
}
}
fclose(fp);
}
int main(int argc, char** argv)
{
const char* algorithm_opt = "--algorithm=";
const char* maxdisp_opt = "--max-disparity=";
const char* blocksize_opt = "--blocksize=";
const char* nodisplay_opt = "--no-display=";
const char* scale_opt = "--scale=";
if(argc < 3)
{
print_help();
return 0;
}
const char* img1_filename = 0;
const char* img2_filename = 0;
const char* intrinsic_filename = 0;
const char* extrinsic_filename = 0;
const char* disparity_filename = 0;
const char* point_cloud_filename = 0;
enum { STEREO_BM=0, STEREO_SGBM=1, STEREO_HH=2, STEREO_VAR=3 };
int alg = STEREO_SGBM;
int SADWindowSize = 0, numberOfDisparities = 0;
bool no_display = false;
float scale = 1.f;
StereoBM bm;
StereoSGBM sgbm;
StereoVar var;
for( int i = 1; i < argc; i++ )
{
if( argv[i][0] != '-' )
{
if( !img1_filename )
img1_filename = argv[i];
else
img2_filename = argv[i];
}
else if( strncmp(argv[i], algorithm_opt, strlen(algorithm_opt)) == 0 )
{
char* _alg = argv[i] + strlen(algorithm_opt);
alg = strcmp(_alg, "bm") == 0 ? STEREO_BM :
strcmp(_alg, "sgbm") == 0 ? STEREO_SGBM :
strcmp(_alg, "hh") == 0 ? STEREO_HH :
strcmp(_alg, "var") == 0 ? STEREO_VAR : -1;
if( alg < 0 )
{
printf("Command-line parameter error: Unknown stereo algorithm\n\n");
print_help();
return -1;
}
}
else if( strncmp(argv[i], maxdisp_opt, strlen(maxdisp_opt)) == 0 )
{
if( sscanf( argv[i] + strlen(maxdisp_opt), "%d", &numberOfDisparities ) != 1 ||
numberOfDisparities < 1 || numberOfDisparities % 16 != 0 )
{
printf("Command-line parameter error: The max disparity (--maxdisparity=<...>) must be a positive integer divisible by 16\n");
print_help();
return -1;
}
}
else if( strncmp(argv[i], blocksize_opt, strlen(blocksize_opt)) == 0 )
{
if( sscanf( argv[i] + strlen(blocksize_opt), "%d", &SADWindowSize ) != 1 ||
SADWindowSize < 1 || SADWindowSize % 2 != 1 )
{
printf("Command-line parameter error: The block size (--blocksize=<...>) must be a positive odd number\n");
return -1;
}
}
else if( strncmp(argv[i], scale_opt, strlen(scale_opt)) == 0 )
{
if( sscanf( argv[i] + strlen(scale_opt), "%f", &scale ) != 1 || scale < 0 )
{
printf("Command-line parameter error: The scale factor (--scale=<...>) must be a positive floating-point number\n");
return -1;
}
}
else if( strcmp(argv[i], nodisplay_opt) == 0 )
no_display = true;
else if( strcmp(argv[i], "-i" ) == 0 )
intrinsic_filename = argv[++i];
else if( strcmp(argv[i], "-e" ) == 0 )
extrinsic_filename = argv[++i];
else if( strcmp(argv[i], "-o" ) == 0 )
disparity_filename = argv[++i];
else if( strcmp(argv[i], "-p" ) == 0 )
point_cloud_filename = argv[++i];
else
{
printf("Command-line parameter error: unknown option %s\n", argv[i]);
return -1;
}
}
if( !img1_filename || !img2_filename )
{
printf("Command-line parameter error: both left and right images must be specified\n");
return -1;
}
if( (intrinsic_filename != 0) ^ (extrinsic_filename != 0) )
{
printf("Command-line parameter error: either both intrinsic and extrinsic parameters must be specified, or none of them (when the stereo pair is already rectified)\n");
return -1;
}
if( extrinsic_filename == 0 && point_cloud_filename )
{
printf("Command-line parameter error: extrinsic and intrinsic parameters must be specified to compute the point cloud\n");
return -1;
}
int color_mode = alg == STEREO_BM ? 0 : -1;
Mat img1 = imread(img1_filename, color_mode);
Mat img2 = imread(img2_filename, color_mode);
if( scale != 1.f )
{
Mat temp1, temp2;
int method = scale < 1 ? INTER_AREA : INTER_CUBIC;
resize(img1, temp1, Size(), scale, scale, method);
img1 = temp1;
resize(img2, temp2, Size(), scale, scale, method);
img2 = temp2;
}
Size img_size = img1.size();
Rect roi1, roi2;
Mat Q;
if( intrinsic_filename )
{
// reading intrinsic parameters
FileStorage fs(intrinsic_filename, CV_STORAGE_READ);
if(!fs.isOpened())
{
printf("Failed to open file %s\n", intrinsic_filename);
return -1;
}
Mat M1, D1, M2, D2;
fs["M1"] >> M1;
fs["D1"] >> D1;
fs["M2"] >> M2;
fs["D2"] >> D2;
M1 *= scale;
M2 *= scale;
fs.open(extrinsic_filename, CV_STORAGE_READ);
if(!fs.isOpened())
{
printf("Failed to open file %s\n", extrinsic_filename);
return -1;
}
Mat R, T, R1, P1, R2, P2;
fs["R"] >> R;
fs["T"] >> T;
stereoRectify( M1, D1, M2, D2, img_size, R, T, R1, R2, P1, P2, Q, CALIB_ZERO_DISPARITY, -1, img_size, &roi1, &roi2 );
Mat map11, map12, map21, map22;
initUndistortRectifyMap(M1, D1, R1, P1, img_size, CV_16SC2, map11, map12);
initUndistortRectifyMap(M2, D2, R2, P2, img_size, CV_16SC2, map21, map22);
Mat img1r, img2r;
remap(img1, img1r, map11, map12, INTER_LINEAR);
remap(img2, img2r, map21, map22, INTER_LINEAR);
img1 = img1r;
img2 = img2r;
}
numberOfDisparities = numberOfDisparities > 0 ? numberOfDisparities : ((img_size.width/8) + 15) & -16;
bm.state->roi1 = roi1;
bm.state->roi2 = roi2;
bm.state->preFilterCap = 31;
bm.state->SADWindowSize = SADWindowSize > 0 ? SADWindowSize : 9;
bm.state->minDisparity = 0;
bm.state->numberOfDisparities = numberOfDisparities;
bm.state->textureThreshold = 10;
bm.state->uniquenessRatio = 15;
bm.state->speckleWindowSize = 100;
bm.state->speckleRange = 32;
bm.state->disp12MaxDiff = 1;
sgbm.preFilterCap = 63;
sgbm.SADWindowSize = SADWindowSize > 0 ? SADWindowSize : 3;
int cn = img1.channels();
sgbm.P1 = 8*cn*sgbm.SADWindowSize*sgbm.SADWindowSize;
sgbm.P2 = 32*cn*sgbm.SADWindowSize*sgbm.SADWindowSize;
sgbm.minDisparity = 0;
sgbm.numberOfDisparities = numberOfDisparities;
sgbm.uniquenessRatio = 10;
sgbm.speckleWindowSize = bm.state->speckleWindowSize;
sgbm.speckleRange = bm.state->speckleRange;
sgbm.disp12MaxDiff = 1;
sgbm.fullDP = alg == STEREO_HH;
var.levels = 3; // ignored with USE_AUTO_PARAMS
var.pyrScale = 0.5; // ignored with USE_AUTO_PARAMS
var.nIt = 25;
var.minDisp = -numberOfDisparities;
var.maxDisp = 0;
var.poly_n = 3;
var.poly_sigma = 0.0;
var.fi = 15.0f;
var.lambda = 0.03f;
var.penalization = var.PENALIZATION_TICHONOV; // ignored with USE_AUTO_PARAMS
var.cycle = var.CYCLE_V; // ignored with USE_AUTO_PARAMS
var.flags = var.USE_SMART_ID | var.USE_AUTO_PARAMS | var.USE_INITIAL_DISPARITY | var.USE_MEDIAN_FILTERING ;
Mat disp, disp8;
//Mat img1p, img2p, dispp;
//copyMakeBorder(img1, img1p, 0, 0, numberOfDisparities, 0, IPL_BORDER_REPLICATE);
//copyMakeBorder(img2, img2p, 0, 0, numberOfDisparities, 0, IPL_BORDER_REPLICATE);
int64 t = getTickCount();
if( alg == STEREO_BM )
bm(img1, img2, disp);
else if( alg == STEREO_VAR ) {
var(img1, img2, disp);
}
else if( alg == STEREO_SGBM || alg == STEREO_HH )
sgbm(img1, img2, disp);
t = getTickCount() - t;
printf("Time elapsed: %fms\n", t*1000/getTickFrequency());
//disp = dispp.colRange(numberOfDisparities, img1p.cols);
if( alg != STEREO_VAR )
disp.convertTo(disp8, CV_8U, 255/(numberOfDisparities*16.));
else
disp.convertTo(disp8, CV_8U);
if( !no_display )
{
namedWindow("left", 1);
imshow("left", img1);
namedWindow("right", 1);
imshow("right", img2);
namedWindow("disparity", 0);
imshow("disparity", disp8);
printf("press any key to continue...");
fflush(stdout);
waitKey();
printf("\n");
}
if(disparity_filename)
imwrite(disparity_filename, disp8);
if(point_cloud_filename)
{
printf("storing the point cloud...");
fflush(stdout);
Mat xyz;
reprojectImageTo3D(disp, xyz, Q, true);
saveXYZ(point_cloud_filename, xyz);
printf("\n");
}
return 0;
}
And I have tried all of the algorithms on the following set of images:
http://imageshack.com/a/img607/4641/utam.jpg (left)
http://imageshack.com/a/img62/5939/2hkc.jpg (right)
Here is my result:
http://imageshack.com/a/img856/4274/1n50.jpg
And here is the desired result:
http://i.stack.imgur.com/W9PBr.jpg
What could be the problem?
change parameters! For example, say bm, my experience is SADWindowSize and minDisparity really matters a lot! so you can add trackbars for each parameter and play with it until you get desired result. I've tried the same pair of images and it works fine. So all you need to do it try, try and try parameters.
I know it sounds crazy, but I had similar output when I was accidentally passing the same image as both the left and right input to the SGBMStereo algorithm. Just taking a shot in the dark...