I am working on Graphite. Its is in C++ built-in libraries.
I am trying to make a surface which would be a sphere. The sphere is in form of a mesh.
How can I make adjacency matrix and degree matrix for a surface so that I can compute the Laplacian matrix?
I am working on a surface which I need to deform it later for that I need an adjacency matrix and degree matrix for the surface.
I am using C++.
Thanks in advance.
How to create adjacency matrix
Here is how you can create an adjacency matrix:
#include <vector>
vector < vector<int> > adjacencyMatrix(vertices);
How to derive degree matrix
Consider the following adjacency matrix that represents an unweighted directed graph:
1 0 1
1 0 0
1 0 0
The index of the row represents the vertex. So, row 0 represents vertex 0, 1 represents vertex 1...
The vertex 0 is connected to itself and has an incoming edge from vertex 2. Similarly, 1 has an incoming edge from 0, and 2 has one with 1.
We have to find out how many edges end at each vertex. Using that information, we create following degree matrix:
2 0 0
0 1 0
0 0 1
The above shows that vertex 0 has 2 edges ending on it, vertex 1 has 1, and vertex 2 has 1.
Since each row in the adjacency matrix represents the incoming connections for that vertex, all you have to do it sum up each row and store them in another matrix (i.e. degree matrix). Since row 0 had a sum of 2, that means you store at the (0, 0) position of the degree matrix the value 2. Similarly, since row 1 had a sum of 1, you store that value at the (1, 1) position...
Let me know if you need me to actually code this. I'm assuming you understand and can take it from here.
Note: the above works for an adjacency matrix for unweighted directed graph. You will have to modify it slightly for other types of graphs.
If you have a undirected graph with N nodes, you need an NxN matrix (initialized to 0).
For each edge (a,b) in your graph mark the entry at (a,b) and (b,a) with 1 in the adjacency matrix, while the degree matrix is incremented with 1 at (a,a) and at (b,b).
For a simple graph (no multiple edges and self-loops) you have L = D - A.
Related
If there is a graph with N nodes and I'm only given a N*N matrix of the distances from each node to every other (the diagonal is of course 0), what would be the most efficient way of generating a graph with as few edges as possible?
for n = 4 and the matrix
0 1 2 3
1 0 3 4
2 3 0 5
3 4 5 0
only having 3 edges would be enough, all connected to the 1st node.
the edge from 1 and 2 would have length 1
the edge from 1 and 3 would have length 2
the edge from 1 and 4 would have length 3
" N*N matrix of the distances from each node to every other (the diagonal is of course 0)"
This is called the adjacency matrix. It completely specifies the graph. Each non-sero value defines an edge between two vertices. You cannot reduce the number of edges below the number of non zero value without changing the graph.
Simply create a full graph using the matrix as the incidence matrix, then remove unneeded edges.
Edge (a,b) is unneeded if and only if there is another path from a to b of the same length as (a,b). That is, if there is vertex c different from both a and b such that
distance(a,b) = distance(a,c) + distance(c,b)
This will not work if some distances not on the diagonal are zero or negative.
What is the best way in C++ to calculate the coordinate of a rectangle (hyper-rectangle) in n-dimension?
I have dimensions definition of the rectangle in a 1d vector like:
[min1, max1, min2, max2, ...., minN, maxN]
For example, in 2d the dimensions, the vector is
[min1, max1, min2, max2]
And the corner coordinates I am looking for are
[min1, min2], [min1, max2], [max1, min2], [max1, max2]
How do we do this for a hyper-rectangle in n-dimension?
That hyper-rectangle has 2^N vertices.
To compute the coordinates of i-th vertex where i in [ 0 .. 2^N-1 ] interval, loop over bits in the i from 0 (lowest one) to N-1. If the bit is set, use maximum coordinate for that dimension, if that bit is 0, use minimum coordinate for that dimension.
For example, for cube N=3, it has 8 vertices.
The first one has index 0, 0b000, you’ll get minimum values for all 3 coordinates.
The last one has index 7, 0b111, you’ll get maximum values for all 3 coordinates.
The rest of the vertices are in between, you’ll get some combination of min and max coordinates.
I have computed the trifocal tensor and corresponding projection matrices P_0, P_1 and P_2 from line correspondences over 3 views, according to 'Multiple View Geometry by Hartley & Zisserman, 2nd edition', Chapter 16. The computed matrices are:
P_0 =
[1 0 0 0
0 1 0 0
0 0 1 0]
P_1 =
[-0.284955 -0.129918 -0.0276358 0.922516
0.122053 0.560496 0.061383 0.385913
0.00455229 -0.0114709 -0.607497 0.00589735]
P_2 =
[0.21558 -0.10182 0.00499782 0.998876
0.0079606 0.11325 0.0226247 0.047112
0.006613 -0.00260303 -0.130705 0.00512245]
Now I want to compute the 3D (plücker) lines from these projection matrices. I know the intrinsic camera matrix K. What I don't understand is, how to include the intrinsic matrix K with the normalized projection matrices from the trifocal tensor P_1, P_2 and P_3 in order to get correct 3D information. More specifically, I want to follow the triangulation procedure described by Bartoli and Sturm (Section 4, Triangulation).
I appreciate your help.
What do you mean with correct 3D information? The whole coordinate system is only computable up to a scale.
Which algorithm exactly did you use for the computation? Algorithm 16.2 in that chapter?
Why don't you use the triangulation algorithm here:
http://www.robots.ox.ac.uk/~vgg/hzbook/code/vgg_multiview/vgg_line3d_from_lP_lin.m
http://www.robots.ox.ac.uk/~vgg/hzbook/code/vgg_multiview/vgg_line3d_from_lP_nonlin.m
I have some degenerate tree (it looks like as array or doubly linked list). For example, it is this tree:
Each edge has some weight. I want to find all equal paths, which starts in each vertex.
In other words, I want to get all tuples (v1, v, v2) where v1 and v2 are an arbitrary ancestor and descendant such that c(v1, v) = c(v, v2).
Let edges have the following weights (it is just example):
a-b = 3
b-c = 1
c-d = 1
d-e = 1
Then:
The vertex A does not have any equal path (there is no vertex from left side).
The vertex B has one equal pair. The path B-A equals to the path B-E (3 == 3).
The vertex C has one equal pair. The path B-C equals to the path C-D (1 == 1).
The vertex D has one equal pair. The path C-D equals to the path D-E (1 == 1).
The vertex E does not have any equal path (there is no vertex from right side).
I implement simple algorithm, which works in O(n^2). But it is too slow for me.
You write, in comments, that your current approach is
It seems, I looking for a way to decrease constant in O(n^2). I choose
some vertex. Then I create two set. Then I fill these sets with
partial sums, while iterating from this vertex to start of tree and to
finish of tree. Then I find set intersection and get number of paths
from this vertex. Then I repeat algorithm for all other vertices.
There is a simpler and, I think, faster O(n^2) approach, based on the so called two pointers method.
For each vertix v go at the same time into two possible directions. Have one "pointer" to a vertex (vl) moving in one direction and another (vr) into another direction, and try to keep the distance from v to vl as close to the distance from v to vr as possible. Each time these distances become equal, you have equal paths.
for v in vertices
vl = prev(v)
vr = next(v)
while (vl is still inside the tree)
and (vr is still inside the tree)
if dist(v,vl) < dist(v,vr)
vl = prev(vl)
else if dist(v,vr) < dist(v,vl)
vr = next(vr)
else // dist(v,vr) == dist(v,vl)
ans = ans + 1
vl = prev(vl)
vr = next(vr)
(By precalculating the prefix sums, you can find dist in O(1).)
It's easy to see that no equal pair will be missed provided that you do not have zero-length edges.
Regarding a faster solution, if you want to list all pairs, then you can't do it faster, because the number of pairs will be O(n^2) in the worst case. But if you need only the amount of these pairs, there might exist faster algorithms.
UPD: I came up with another algorithm for calculating the amount, which might be faster in case your edges are rather short. If you denote the total length of your chain (sum of all edges weight) as L, then the algorithm runs in O(L log L). However, it is much more advanced conceptually and more advanced in coding too.
Firstly some theoretical reasoning. Consider some vertex v. Let us have two arrays, a and b, not the C-style zero-indexed arrays, but arrays with indexation from -L to L.
Let us define
for i>0, a[i]=1 iff to the right of v on the distance exactly i there
is a vertex, otherwise a[i]=0
for i=0, a[i]≡a[0]=1
for i<0, a[i]=1 iff to the left of v on the distance exactly -i there is a vertex, otherwise a[i]=0
A simple understanding of this array is as follows. Stretch your graph and lay it along the coordinate axis so that each edge has the length equal to its weight, and that vertex v lies in the origin. Then a[i]=1 iff there is a vertex at coordinate i.
For your example and for vertex "b" chosen as v:
a--------b--c--d--e
--|--|--|--|--|--|--|--|--|-->
-4 -3 -2 -1 0 1 2 3 4
a: ... 0 1 0 0 1 1 1 1 0 ...
For another array, array b, we define the values in a symmetrical way with respect to origin, as if we have inverted the direction of the axis:
for i>0, b[i]=1 iff to the left of v on the distance exactly i there
is a vertex, otherwise b[i]=0
for i=0, b[i]≡b[0]=1
for i<0, b[i]=1 iff to the right of v on the distance exactly -i there is a vertex, otherwise b[i]=0
Now consider a third array c such that c[i]=a[i]*b[i], asterisk here stays for ordinary multiplication. Obviously c[i]=1 iff the path of length abs(i) to the left ends in a vertex, and the path of length abs(i) to the right ends in a vertex. So for i>0 each position in c that has c[i]=1 corresponds to the path you need. There are also negative positions (c[i]=1 with i<0), which just reflect the positive positions, and one more position where c[i]=1, namely position i=0.
Calculate the sum of all elements in c. This sum will be sum(c)=2P+1, where P is the total number of paths which you need with v being its center. So if you know sum(c), you can easily determine P.
Let us now consider more closely arrays a and b and how do they change when we change the vertex v. Let us denote v0 the leftmost vertex (the root of your tree) and a0 and b0 the corresponding a and b arrays for that vertex.
For arbitrary vertex v denote d=dist(v0,v). Then it is easy to see that for vertex v the arrays a and b are just arrays a0 and b0 shifted by d:
a[i]=a0[i+d]
b[i]=b0[i-d]
It is obvious if you remember the picture with the tree stretched along a coordinate axis.
Now let us consider one more array, S (one array for all vertices), and for each vertex v let us put the value of sum(c) into the S[d] element (d and c depend on v).
More precisely, let us define array S so that for each d
S[d] = sum_over_i(a0[i+d]*b0[i-d])
Once we know the S array, we can iterate over vertices and for each vertex v obtain its sum(c) simply as S[d] with d=dist(v,v0), because for each vertex v we have sum(c)=sum(a0[i+d]*b0[i-d]).
But the formula for S is very simple: S is just the convolution of the a0 and b0 sequences. (The formula does not exactly follow the definition, but is easy to modify to the exact definition form.)
So what we now need is given a0 and b0 (which we can calculate in O(L) time and space), calculate the S array. After this, we can iterate over S array and simply extract the numbers of paths from S[d]=2P+1.
Direct application of the formula above is O(L^2). However, the convolution of two sequences can be calculated in O(L log L) by applying the Fast Fourier transform algorithm. Moreover, you can apply a similar Number theoretic transform (don't know whether there is a better link) to work with integers only and avoid precision problems.
So the general outline of the algorithm becomes
calculate a0 and b0 // O(L)
calculate S = corrected_convolution(a0, b0) // O(L log L)
v0 = leftmost vertex (root)
for v in vertices:
d = dist(v0, v)
ans = ans + (S[d]-1)/2
(I call it corrected_convolution because S is not exactly a convolution, but a very similar object for which a similar algorithm can be applied. Moreover, you can even define S'[2*d]=S[d]=sum(a0[i+d]*b0[i-d])=sum(a0[i]*b0[i-2*d]), and then S' is the convolution proper.)
I have 2D transformations stored in a plain ol' 3x3 Matrix. How can I re-format that one to a Matrix I can shove over to OpenGL in order to transform orthogonal shapes, polygons and suchlike.
How do I have to put the values so the transformations are preserved?
(On an unrelated note, is there a fast way to invert a 3x3 Matrix?)
Some explanation about transformation matrices: All the columns, except the last one, describe the orientation of a new coordinate system in the base of the current coordinate system. So the first column is the X vector of the new coordinate system, as seen from the current, the second is the new Y vector and the 3rd is the new Z. So far this only covers the rotation. The last column is used for the relative offset. The last row and the bottom most right value are used for the homogenous transformations. It's best to leave the last row 0, ..., 0, 1
In your case you're missing the Z values, so we just insert a identity transform there, so that incoming values are left as they are.
Say this is your original matrix:
xx xy tx
yx yy ty
0 0 1
This matrix is missing the Z transformation. Inserting identity means: Leave Z as is, and don't mix with the rest. So ·z = z· = 0, except zz = 1. This gives you the following matrix
↓
xx xy 0 tx
yx yy 0 ty
0 0 1 0 ←
0 0 0 1
You can apply that onto the current OpenGL matrix stack with glMultMatrix if OpenGL version is below 3 core profile. Be aware that OpenGL numbers the matrix in column major order i.e, the indices in the array go like this (hexadecimal digits)
0 4 8 c
1 5 9 d
2 6 a e
3 7 b f
This contrary to the usual C notation which is
0 1 2 3
4 5 6 7
8 9 a b
c d e f
With OpenGL-3 core and later you've to do matrix management and manipulation yourself, anyway.
EDIT for second part of question
If by inverting one means finding the matrix M^-1 for a given matrix M, so that M^1 * M = M * M^1 = 1. For 3×3 matrices the determinant inversion method requires less operations than Gauss-Jordan elemination and is thus the most efficient way to do it. Already for 4×4 matrices determinant inversion is slower than every other method. http://www.sosmath.com/matrix/inverse/inverse.html
If you know that your matrix is orthonormal, then you may just transpose the upper left part except bottom row and rightmost column, and negate the sign of the rightmost column, except the very bottom right element. This exploits the fact that for orthonormal matrices M^-1 = M^T.
Just add the fourth row and column. For example given
2 3 3
3 2 4
0 0 1
Create the following
2 3 3 0
3 2 4 0
0 0 1 0
0 0 0 1
The transformation still occurs on the x-y plane even though it is now in three-space.