Elegant way the find the Vertices of a Cube - c++

Nearly every OpenGL tutorial lets you implement drawing a cube. Therefore the vertices of the cube are needed. In the example code I saw a long list defining every vertex. But I would like to compute the vertices of a cube rather that using a overlong list of precomputed coordinates.
A cube is made of eight vertices and twelve triangles. Vertices are defined by x, y, and z. Triangles are defined each by the indexes of three vertices.
Is there an elegant way to compute the vertices and the element indexes of a cube?

When i was "porting" the csg.js project to Java I've found some cute code which generated cube with selected center point and radius. (I know it's JS, but anyway)
// Construct an axis-aligned solid cuboid. Optional parameters are `center` and
// `radius`, which default to `[0, 0, 0]` and `[1, 1, 1]`. The radius can be
// specified using a single number or a list of three numbers, one for each axis.
//
// Example code:
//
// var cube = CSG.cube({
// center: [0, 0, 0],
// radius: 1
// });
CSG.cube = function(options) {
options = options || {};
var c = new CSG.Vector(options.center || [0, 0, 0]);
var r = !options.radius ? [1, 1, 1] : options.radius.length ?
options.radius : [options.radius, options.radius, options.radius];
return CSG.fromPolygons([
[[0, 4, 6, 2], [-1, 0, 0]],
[[1, 3, 7, 5], [+1, 0, 0]],
[[0, 1, 5, 4], [0, -1, 0]],
[[2, 6, 7, 3], [0, +1, 0]],
[[0, 2, 3, 1], [0, 0, -1]],
[[4, 5, 7, 6], [0, 0, +1]]
].map(function(info) {
return new CSG.Polygon(info[0].map(function(i) {
var pos = new CSG.Vector(
c.x + r[0] * (2 * !!(i & 1) - 1),
c.y + r[1] * (2 * !!(i & 2) - 1),
c.z + r[2] * (2 * !!(i & 4) - 1)
);
return new CSG.Vertex(pos, new CSG.Vector(info[1]));
}));
}));
};

I solved this problem with this piece code (C#):
public CubeShape(Coord3 startPos, int size) {
int l = size / 2;
verts = new Coord3[8];
for (int i = 0; i < 8; i++) {
verts[i] = new Coord3(
(i & 4) != 0 ? l : -l,
(i & 2) != 0 ? l : -l,
(i & 1) != 0 ? l : -l) + startPos;
}
tris = new Tris[12];
int vertCount = 0;
void AddVert(int one, int two, int three) =>
tris[vertCount++] = new Tris(verts[one], verts[two], verts[three]);
for (int i = 0; i < 3; i++) {
int v1 = 1 << i;
int v2 = v1 == 4 ? 1 : v1 << 1;
AddVert(0, v1, v2);
AddVert(v1 + v2, v2, v1);
AddVert(7, 7 - v2, 7 - v1);
AddVert(7 - (v1 + v2), 7 - v1, 7 - v2);
}
}
If you want to understand more of what is going on, you can check out the github page I wrote that explains it.

Related

Create an orientated Quad in Eigen C++ library

I'm a newbie that is driving nuts with a really simple problem :\ How to define a quad that has its center at position x=10 y=11 z=12 with side 5 and it is facing at camera.
My naive implementation:
Eigen::Vector3f quadPosition(10, 11, 12);
Eigen::Vector3f camPos(10, 5, 12);
float sideLength = 5;
Eigen::Affine3f transform = Eigen::Translation3f(position) * Eigen::AngleAxisf(3.141595, (camPos - quadPosition).normalized());
Eigen::Vector3f v0 = transform * Eigen::Vector3f(-sideLength / 2, -sideLength / 2, 0);
Eigen::Vector3f v1 = transform * Eigen::Vector3f(sideLength / 2, -sideLength / 2, 0);
Eigen::Vector3f v2 = transform * Eigen::Vector3f(sideLength / 2, sideLength / 2, 0);
Eigen::Vector3f v3 = transform * Eigen::Vector3f(-sideLength / 2, sideLength / 2, 0);
Eigen::MatrixXd verts;
verts.resize(4, 3);
verts(0, 0) = v0(0);
verts(0, 1) = v0(1);
verts(0, 2) = v0(2);
verts(1, 0) = v1(0);
verts(1, 1) = v1(1);
verts(1, 2) = v1(2);
verts(2, 0) = v2(0);
verts(2, 1) = v2(1);
verts(2, 2) = v2(2);
verts(3, 0) = v3(0);
verts(3, 1) = v3(1);
verts(3, 2) = v3(2);
Eigen::MatrixXi faces;
faces.resize(2, 3);
faces(0, 0) = 0;
faces(0, 1) = 1;
faces(0, 2) = 2;
faces(1, 0) = 0;
faces(1, 1) = 2;
faces(1, 2) = 3;
Any idea? Thank you in advance!
One common approach is to define the quad in terms of two opposed corners like: [0,0,0] and [1,1,1]. Now you know that anything in between 0 and 1 in X, Y and Z axis belong to the box ... as long as there is no rotation. Otherwise, you could also add it.

Evenly distribute values into array

I have a fixed size boolean array of size 8. The default value of all elements in the array is false. There will be a number of truth values to fill between 1-8.
I want to distribute the truth values as far away from one another as possible. I also wish to be able to randomize the configuration. In this scenario the array wraps around so position 7 is "next to" position 0 in the array.
here are some examples for fill values. I didn't include all possibilities, but hopefully it gets my point across.
1: [1, 0, 0, 0, 0, 0, 0, 0] or [0, 1, 0, 0, 0, 0, 0, 0]
2: [1, 0, 0, 0, 1, 0, 0, 0] or [0, 1, 0, 0, 0, 1, 0, 0]
3: [1, 0, 0, 1, 0, 0, 1, 0] or [0, 1, 0, 0, 1, 0, 0, 1]
4: [1, 0, 1, 0, 1, 0, 1, 0] or [0, 1, 0, 1, 0, 1, 0, 1]
5: [1, 1, 0, 1, 1, 0, 1, 0]
6: [1, 1, 0, 1, 1, 1, 0, 1]
7: [1, 1, 1, 1, 1, 1, 1, 0]
8: [1, 1, 1, 1, 1, 1, 1, 1]
The closest solution I have come up with so far hasn't quite produced the results I'm looking for...
I seek to write it in c++ but here is a little pseudo-code of my algorithm so far...
not quite working out how I wanted
truths = randBetween(1, 8)
values = [0,0,0,0,0,0,0,0]
startPosition = randBetween(0, 7) //starting index
distance = 4
for(i = 0; i < truths; i++) {
pos = i + startPosition + (i * distance)
values[pos % 8] = 1
}
this is an example output from my current code. those marked with a star are incorrect.
[0, 0, 0, 0, 1, 0, 0, 0]
[0, 1, 0, 0, 1, 0, 0, 0]*
[0, 1, 0, 0, 1, 0, 1, 0]
[0, 1, 0, 1, 1, 0, 1, 0]*
[1, 1, 0, 1, 1, 0, 1, 0]
[1, 1, 0, 1, 1, 1, 1, 0]*
[1, 1, 1, 1, 1, 1, 1, 0]
[1, 1, 1, 1, 1, 1, 1, 1]
I'm looking for a simple way to distribute the truth values evenly throughout the array without having to code for special cases.
Check this out:
#include <cassert>
#include <vector>
#include <iostream>
#include <iomanip>
/**
* Generate an even spaced pattern of ones
* #param arr destination vector of ints
* #param onescnt the requested number of ones
*/
static inline
void gen(std::vector<int>& arr, size_t onescnt) {
const size_t len = arr.size();
const size_t zeroscnt = len - onescnt;
size_t ones = 1;
size_t zeros = 1;
for (size_t i = 0; i < len; ++i) {
if (ones * zeroscnt < zeros * onescnt) {
ones++;
arr[i] = 1;
} else {
zeros++;
arr[i] = 0;
}
}
}
static inline
size_t count(const std::vector<int>& arr, int el) {
size_t cnt = 0;
for (size_t i = 0; i < arr.size(); ++i) {
cnt += arr[i] == el;
}
return cnt;
}
static inline
void gen_print(size_t len, size_t onescnt) {
std::vector<int> arr(len);
gen(arr, onescnt);
std::cout << "gen_printf(" << std::setw(2) << len << ", " << std::setw(2) << onescnt << ") = {";
for (size_t i = 0; i < len; ++i) {
std::cout << arr[i] << ",";
}
std::cout << "}\n";
assert(count(arr, 1) == onescnt);
}
int main() {
for (int i = 0; i <= 8; ++i) {
gen_print(8, i);
}
for (int i = 0; i <= 30; ++i) {
gen_print(30, i);
}
return 0;
}
Generates:
gen_printf( 8, 0) = {0,0,0,0,0,0,0,0,}
gen_printf( 8, 1) = {0,0,0,0,0,0,0,1,}
gen_printf( 8, 2) = {0,0,0,1,0,0,0,1,}
gen_printf( 8, 3) = {0,1,0,0,1,0,0,1,}
gen_printf( 8, 4) = {0,1,0,1,0,1,0,1,}
gen_printf( 8, 5) = {1,0,1,1,0,1,0,1,}
gen_printf( 8, 6) = {1,1,0,1,1,1,0,1,}
gen_printf( 8, 7) = {1,1,1,1,1,1,0,1,}
gen_printf( 8, 8) = {1,1,1,1,1,1,1,1,}
gen_printf(30, 0) = {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,}
gen_printf(30, 1) = {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,}
gen_printf(30, 2) = {0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,}
gen_printf(30, 3) = {0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,}
gen_printf(30, 4) = {0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,}
gen_printf(30, 5) = {0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,}
gen_printf(30, 6) = {0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,}
gen_printf(30, 7) = {0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,1,}
gen_printf(30, 8) = {0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,}
gen_printf(30, 9) = {0,0,1,0,0,1,0,0,0,1,0,0,1,0,0,1,0,0,0,1,0,0,1,0,0,1,0,0,0,1,}
gen_printf(30, 10) = {0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,}
gen_printf(30, 11) = {0,1,0,0,1,0,0,1,0,1,0,0,1,0,0,1,0,0,1,0,1,0,0,1,0,0,1,0,0,1,}
gen_printf(30, 12) = {0,1,0,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,0,1,0,1,0,0,1,}
gen_printf(30, 13) = {0,1,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,1,0,0,1,}
gen_printf(30, 14) = {0,1,0,1,0,1,0,1,0,1,0,1,0,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,0,1,}
gen_printf(30, 15) = {0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,}
gen_printf(30, 16) = {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,}
gen_printf(30, 17) = {1,0,1,0,1,0,1,1,0,1,0,1,0,1,1,0,1,0,1,0,1,1,0,1,0,1,0,1,0,1,}
gen_printf(30, 18) = {1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,1,0,1,0,1,}
gen_printf(30, 19) = {1,0,1,1,0,1,1,0,1,0,1,1,0,1,1,0,1,1,0,1,0,1,1,0,1,1,0,1,0,1,}
gen_printf(30, 20) = {1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,}
gen_printf(30, 21) = {1,1,0,1,1,0,1,1,0,1,1,1,0,1,1,0,1,1,0,1,1,1,0,1,1,0,1,1,0,1,}
gen_printf(30, 22) = {1,1,0,1,1,1,0,1,1,1,0,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,0,1,}
gen_printf(30, 23) = {1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,}
gen_printf(30, 24) = {1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,}
gen_printf(30, 25) = {1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,0,1,1,1,1,1,0,1,}
gen_printf(30, 26) = {1,1,1,1,1,1,0,1,1,1,1,1,1,0,1,1,1,1,1,1,1,0,1,1,1,1,1,1,0,1,}
gen_printf(30, 27) = {1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,0,1,}
gen_printf(30, 28) = {1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,}
gen_printf(30, 29) = {1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,}
gen_printf(30, 30) = {1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,}
#edit - better evenly spaced pattern.
Explanation:
So let's take an array of 8 ints and we want to have 5 ones. The ideal ratio of (ones / zeros) in a sequence with 8 elements and 5 ones, well would be (5 / 3). We will never approach such ratio, but we can try.
The idea is to loop through the array and remember the number of ones and zeros we have written in the array. If the ratio of (written ones / written zeros) is lower then the destination ratio (ones / zeros) we want to achieve, we need to put a one to the sequence. Otherwise we put zero in the sequence. The ratio changes and we make the decision next time. The idea is to pursue the ideal ratio of ones per zeros in each slice of the array.
A simple way to do this would be to round the ideal fractional positions.
truths = randBetween(1, 8)
values = [0,0,0,0,0,0,0,0]
offset = randBetween(0, 8 * truths - 1)
for(i = 0; i < truths; i++) {
pos = (offset + (i * 8)) / truths
values[pos % 8] = 1
}
This is an application of Bresenham's line-drawing algorithm. I use it not because it's fast on old hardware, but it places true values exactly.
#include <iostream>
#include <stdexcept>
#include <string>
#include <random>
int main(int argc, char **argv) {
try {
// Read the argument.
if(argc != 2) throw std::invalid_argument("one argument");
int dy = std::stoi(argv[1]);
if(dy < 0 || dy > 8) throw std::out_of_range("[0..8]");
int values[8] = {0};
// https://en.wikipedia.org/wiki/Bresenham%27s_line_algorithm
int dx = 8;
int delta = 2 * dy - dx; // Balance the line. Permute it up later.
for(int x = 0; x < dx; x++) {
if(delta > 0) {
values[x] = 1;
delta -= 2 * dx;
}
delta += 2 * dy;
}
for(int x = 0; x < dx; x++)
std::cout << (x ? ", " : "") << values[x];
std::cout << std::endl;
// Rotate the number by a random amount.
// I'm sure there is an easier way to do this.
// https://stackoverflow.com/questions/7560114/random-number-c-in-some-range
std::random_device rd; // obtain a random number from hardware
std::mt19937 eng(rd()); // seed the generator
std::uniform_int_distribution<> distr(0, dx - 1);
int rotate = distr(eng);
bool first = true;
int x = rotate;
do {
std::cout << (first ? "" : ", ") << values[x];
first = false;
x = (x + 1) % dx;
} while(x != rotate);
std::cout << std::endl;
} catch(const std::exception &e) {
std::cerr << "Something went wrong: " << e.what() << std::endl;
return 1;
}
return 0;
}
Once you have an exact solution, rotate it by a random amount.
0, 1, 0, 0, 1, 0, 1, 0
1, 0, 0, 1, 0, 0, 1, 0
You need to calculate distance dynamically. One element is clear, that can reside at arbitrary location
2 elements is clear, too, distance needs to be 4.
4 elements need a distance of 2
8 elements a distance of 1
More difficult are numbers that don't divide the array:
3 requires a distance of 2.66.
5 requires a distance of 1.6
7 requires a distance of 0.875
Errm... In general, if you have a distance of X.Y, you will have to place some of the elements at distances of X and some at distances of X + 1. X is simple, it will be the result of an integer division: 8 / numberOfElements. The remainder will determine how often you will have to switch to X + 1: 8 % numberOfElements. For 3, this will result in 2, too, so you will have 1x distance of 2 and 2x distance of 3:
[ 1 0 1 0 0 1 0 0 ]
2 3 3 (distance to very first 1)
For 5, you'll get: 8/5 = 1, 8%5 = 3, so: 2x distance of 1, 3x distance of 2
[ 1 1 1 0 1 0 1 0 ]
1 1 2 2 2
For 7 you'll get: 8/7 = 1, 8%7 = 1, so: 7x distance of 1, 1x distance of 2
[ 1 1 1 1 1 1 1 0 ]
1 1 1 1 1 1 2
That will work for arbitrary array length L:
L/n = minimum distance
L%n = number of times to apply minimum distance
L-L%n = number of times to apply minimum distance + 1
Mathematical metrics won't reveal any difference between first applying all smaller distances then all larger ones, human sense for aesthetics, though, might prefer if you alternate between larger and smaller as often as possible – or you apply the algorithm recursively (for larger array length), to get something like 2x2, 3x3, 2x2, 3x3 instead of 4x2 and 6x3.

Sorted Float Vectors for Scatter Plotting (C++ / QCustomPlot)

Questions:
1 - Sort multiple float vectors in the same order (keeping correspondance)
2 - QCustomPlot (QCP) plots ONLY outer boundary of a scatter plot.
(answering either of these 2 questions would solve my problem)
Situation:
I have 3 vectors for plotting:
std::vector<float> x, y;
std::vector<int> hits;
The resulting plot is a hit or miss scatter plot. The resulting plot is used by QCustomPlot's curve, which ends up as a circular "scribble." It just needs to look similar to a circle with no "scribbling" inside. I need this plot to overlay over another plot.
I don't have much control over the initial order of x, y, or hits.
x and y are sorted in a traditional grid indexing:
x = -8, -8, -8, -8, -8, -4, -4, -4, -4, -4, ... 8
y = -8, -4, 0, 4, 8, -8, -4, 0, 4, 8, ... 8
The hits are based on wether (let's just say archer's arrow) a hit was successful based on a range and speed of a fast target (let's just say bird).
The resulting plot is the outer boudary of hits based on the bird as a center reference.
The data vectors can be very large.
Method 1: I can calculate range and angle. Then doing a sort on float vectors: Sort the angles in order, so that when QCustomPlot plots the outer boundary without 'scribbles' inside. However, I need to know how to keep the corresponding x & y values together based on sorting the angle.
// Make range and angle vectors for sorting
std::vector<float> range, angle;
for(int i = 0; i < x.size(); ++i {
float r = sqrt(x[i]*x[i] + y[i]*y[i]);
range.push_back(r);
float a = 0;
if(y < 0)
a = -acos(x[i]/r);
else
a = acos(x[i]/r);
angle.push_back(a);
}
// Sort all vectors by ascending angle vector.
/* Do stuff here! */
// Set up boundary plot data
QVector<float> plot_x, plot_y;
for(int i = 0; i < x.size(); ++i {
if(hits[i]) {
plot_x.push_back(x[i]);
plot_y.push_back(y[i]);
}
}
// curve is a QCPCurve object already existing.
curve->addData(plot_x, plot_y); // Already sorted QVectors
Method 2: Get QCustomPlot curve->addData(x, y) member to only plot an "perimeter line" of the scatter plot's hits. I have tried using QCPScatterStyle, .setCustomPath, but have not been successful.
Thank you in advance!
-John
If you want to order several vectors using some criteria and all the indices correspond create a new vector that are the indices, and order it, then use those indices to create the new vectors:
#include <cmath>
#include <QDebug>
static float calc_angle(float x, float y){
float r = sqrt(x*x + y*y);
float angle = acos(x/r);
return y<0 ? -angle : angle;
}
int main(int argc, char *argv[])
{
std::vector<int> hits{0, 1, 2, 1, 0, 1, 2, 1, 0, 1};
std::vector<float> x{-8, -8, -8, -8, -8, -4, -4, -4, -4, -4};
std::vector<float> y{-8, -4, 0, 4, 8, -8, -4, 0, 4, 8};
Q_ASSERT(x.size() == y.size() && y.size() == hits.size());
std::vector<int> indexes(x.size());
std::iota(indexes.begin(), indexes.end(), 0);
std::sort(indexes.begin(), indexes.end(), [&](const int & i, const int & j) -> bool{
return calc_angle(x[i], y[i]) < calc_angle(x[j], y[i]);
});
QVector<float> plot_x, plot_y;
QVector<int> new_hits;
for(const int & index : indexes){
plot_x<<x[index];
plot_y<<y[index];
new_hits<<hits[index];
}
qDebug()<<indexes;
qDebug()<< plot_x;
qDebug()<<plot_y;
qDebug()<<new_hits;
return 0;//a.exec();
}
Output:
std::vector(8, 0, 1, 2, 3, 4, 5, 6, 7, 9)
QVector(-4, -8, -8, -8, -8, -8, -4, -4, -4, -4)
QVector(4, -8, -4, 0, 4, 8, -8, -4, 0, 8)
QVector(0, 0, 1, 2, 1, 0, 1, 2, 1, 1)

Eigen matrix rowwise addition

I have an Nx3 Eigen matrix representing a bunch of locations of vertices in 3d space.
I'm trying to add a 1x3 matrix to each row in the Nx3 to move every point a given direction and distance.
#include <Eigen/Dense>
int N = 20; //for example
MatrixXf N3(N, 3);
N3.fill(2);
MatrixXf origin(1, 3);
origin << 1, 2, 3;
Now I want to add origin to each row in N3 so N3 becomes 3, 4, 5 in each row. (The values in N3 are all different 3d vertex locations in the real code.)
3, 4, 5
3, 4, 5
3, 4, 5 etc...
you may just write
N3 += origin.replicate(N,1);
note that no temporary matrix is created, replicate() returns an expression.
Try this (untested)
for (int i = 0 ; i < 3 ; i++)
N3.block(i, 0, 1, 3) = N3.block(i, 0, 1, 3) + origin
I do not remember if += is supported
MatrixXf result = N3 + MatrixXf::Constant(1, N, 1) * origin;
Should be simple as that.

CSR Matrix - Matrix multiplication

I have two square matrices A and B
I must convert B to CSR Format and determine the product C
A * B_csr = C
I have found a lot of information online regarding CSR Matrix - Vector multiplication. The algorithm is:
for (k = 0; k < N; k = k + 1)
result[i] = 0;
for (i = 0; i < N; i = i + 1)
{
for (k = RowPtr[i]; k < RowPtr[i+1]; k = k + 1)
{
result[i] = result[i] + Val[k]*d[Col[k]];
}
}
However, I require Matrix - Matrix multiplication.
Further, it seems that most algorithms apply A_csr - vector multiplication where I require A * B_csr. My solution is to transpose the two matrices before converting then transpose the final product.
Can someone explain how to compute a Matrix - CSR Matrix product and/or a CSR Matrix - Matrix product?
Here is a simple solution in Python for the Dense Matrix X CSR Matrix. It should be self-explanatory.
def main():
# 4 x 4 csr matrix
# [1, 0, 0, 0],
# [2, 0, 3, 0],
# [0, 0, 0, 0],
# [0, 4, 0, 0],
csr_values = [1, 2, 3, 4]
col_idx = [0, 0, 2, 1]
row_ptr = [0, 1, 3, 3, 4]
csr_matrix = [
csr_values,
col_idx,
row_ptr
]
dense_matrix = [
[1, 3, 3, 4],
[1, 2, 3, 4],
[1, 4, 3, 4],
[1, 2, 3, 5],
]
res = [
[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0],
]
# matrix order, assumes both matrices are square
n = len(dense_matrix)
# res = dense X csr
csr_row = 0 # Current row in CSR matrix
for i in range(n):
start, end = row_ptr[i], row_ptr[i + 1]
for j in range(start, end):
col, csr_value = col_idx[j], csr_values[j]
for k in range(n):
dense_value = dense_matrix[k][csr_row]
res[k][col] += csr_value * dense_value
csr_row += 1
print res
if __name__ == '__main__':
main()
CSR Matrix X Dense Matrix is really just a sequence of CSR Matrix X Vector product for each row of the dense matrix right? So it should be really easy to extend the code you show above to do this.
Moving forward, I suggest you don't code these routines yourself. If you are using C++ (based on the tag), then you could have a look at Boost ublas for example, or Eigen. The APIs may seem a bit cryptic at first but it's really worth it in the long term. First, you gain access to a lot more functionality, which you will probably require in the future. Second these implementations will be better optimised.