I would like to create a directed network graph using igraph c++ in which each node is randomly connected to exactly n other distinct nodes in the network (i.e. excluding connections to itself, and loops/multiple edges to the same loop). I was thinking of using the method igraph_erdos_renyi_game, but for some reason I do not get the desired degree distribution. In particular, if I set the arguments like this:
igraph_erdos_renyi_game(&g, IGRAPH_ERDOS_RENYI_GNM, n, m, true, false);
with n = 5, m = 1, I get this Adjacency Matrix (e.g.):
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
If I use igraph_k_regular_game(&g, n, m, true, false) instead, I get exactly the desired outcome, namely (e.g.):
0 0 0 0 1
1 0 0 0 0
0 0 0 0 1
0 1 0 0 0
0 0 0 1 0
So to summarize, I would like for each node to have 1 edge to n randomly selected agents. Am I misinterpreting the way the Erdos-Renyi method works, or am I passing the wrong arguments?
m is the total number of edges in erdos_renyi_game(). So just use k_regular_game(), it does exactly what you want.
Related
I have an array with sections of touching values in it. For example:
0 0 1 0 0 0 0 0 0 0
0 1 1 1 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 2 2 2 0 0
0 0 0 0 0 0 0 2 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 3 0 0 0 0 0 0 0
0 0 3 0 0 0 0 0 0 0
0 0 3 0 0 0 0 0 0 0
from this, I created a set of af::arrays: minX, maxX, minY, maxY. These define the box that encloses each group.
so for this example:
minX would be: [1,5,2] // 1 for label(1), 5 for label(2) and 2 for label(3)
maxX would be: [3,7,2] // 3 for label(1), 7 for label(2) and 2 for label(3)
minY would be: [0,3,7] // 0 for label(1), 3 for label(2) and 7 for label(3)
maxY would be: [1,4,9] // 1 for label(1), 4 for label(2) and 9 for label(3)
So if you take the i'th element from each of those arrays, you can get the upperleft/lowerright bounds of a box that encloses the corresponding label.
I would like use these values to pull out subarrays from this larger array. My goal is to put these values enclosed in the boxes into a flat list. In GPU memory, I also have calculated how many entries I would need for each box using the max/min X/Y values. So in this example - the result of the flat list should be:
result=[0 1 0 1 1 1 2 2 2 0 0 2 3 3 3]
where the first 6 entries are from the box
______
|0 1 0 |
|1 1 1 |
------
the second 6 entries are from the box
______
|2 2 2 |
|0 0 2 |
------
and the final three entries are from the box
___
| 3 |
| 3 |
| 3 |
---
I cannot figure out how to index into this af::array with min/max values in memory that resides on the GPU (and do not want to transfer them to the CPU). I was trying to see if gfor/seq would work for me, but it appears that af::seq cannot use array data, and everything I have tried with using af::index i could not get to work for me either.
I am able to change how I represent min/max (I could store indices for upper left/lower right) but my main goal is to do this efficiently on the GPU without moving data back and forth between the GPU and CPU.
How can this be achieved efficiently with ArrayFire?
Thank you for your help
How did you get there so far? which language are you using?
I guess you could be tiling the results to 3rd dimensions to handle each regions separately and end up with min/max vectors in GPU memory.
-2 -1 0
-1 1 1
0 1 2
This is 3x3 emboss kernel. How should I write this in 5x5?
As I understand, these filters take directional differences (see the wikipidea page).
We can decompose you filter into directions
0 -1 0 0 0 0 -2 0 0
0 0 0 -1 0 1 0 0 0
0 1 0 0 0 0 0 0 2
So, I think you can expand it over these 3 directions giving emphasis
0 0 -1 0 0 0 0 0 0 0 -2 0 0 0 0
0 0 -1 0 0 0 0 0 0 0 0 -2 0 0 0
0 0 0 0 0 -1 -1 0 1 1 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0 0 0 0 2 0
0 0 1 0 0 0 0 0 0 0 0 0 0 0 2
So, the final kernel would be
-2 0 -1 0 0
0 -2 -1 0 0
-1 -1 1 1 1
0 0 1 2 0
0 0 1 0 2
May be you can also try interpolating filter coefficients marked as x
-2 x -1 0 0
x -2 -1 0 0
-1 -1 1 1 1
0 0 1 2 x
0 0 1 x 2
The simple solution to fitting any lower-dimensional convolution kernel into a higher-dimensional matrix of the same rank is to surround it by zero weights. This is especially true when you're dealing with a concept like embossing, which is arguably more interested in immediate vector of change than the rate at which it is changing. That is, for this embossing matrix,
You could equivalently use this in 5 x 5:
Granted, this will get you a different visual effect than anything with any part of the matrix filled in; but sometimes, especially with edge-detection, immediate clarity is more important. We aren't always displaying it. If this were something like a Guassian blur kernel, having a greater range could improve the effect, but embossing isn't that different conceptually from Sobel-Feldman and it may be better to keep it tight.
How i can get the longest cycle in a undirected Graph (without BackTracking, it takes too long).
Example:
0 3 0 1 0
3 0 0 1 0
0 0 0 0 0
1 1 0 0 0
0 0 0 0 0
Solve: 3 + 3 + 1 => Out: 1 - 2 - 3 - 1.
If you can find the longest cycle, you can detect whether the graph has a Hamiltonian Cycle, which is an NP-complete problem, thus making your problem NP-hard.
That means no solution will be fundamentally better than backtracking unless P=NP.
I am trying to access the sparse mlf with the keys such as BEpos and BEneg where one key per line. Now the problem is that most commands are not meant to deal with too large input: bin2dec requires clean binary numbers without spaces but the regexp hack fails to too many rows -- and so on.
How to work with sparse keys to access sparse data?
Example
K>> mlf=sparse([],[],[],2^31,1);
BEpos=Cg(pos,:)
BEpos =
(1,1) 1
(2,3) 1
(2,4) 1
K>> mlf(bin2dec(num2str(BEpos)))=1
Error using bin2dec (line 36)
Binary string must be 52 bits or less.
K>> num2str(BEpos)
ans =
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 0
0 0 1 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
K>> bin2dec(num2str('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 0'))
Error using bin2dec (line 36)
Binary string must be 52 bits or less.
K>> regexprep(num2str(BEpos),'[^\w'']','')
Error using regexprep
The 'STRING' input must be a one-dimensional array
of char or cell arrays of strings.
Manually works
K>> mlf(bin2dec('1000000000000000000000000000000'))
ans =
All zero sparse: 1-by-1
Consider a different approach using manual binary to decimal conversions:
pows = pow2(size(BEpos,2)-1 : -1 : 0);
inds = uint32(BEpos*pows.')
I haven't benchmarked this, but it might work faster than bin2dec and cell arrays.
How it works
This is pretty simple: the powers of 2 are calculated and stored in pows (assuming the MSB is in the leftmost position). Then they are multiplied by the bits in the matching positions and summed to produce the corresponding decimal values.
Try to index with this:
inds = uint32( bin2dec(cellstr(num2str(BEpos,'%d'))) );
iterate through a 3d matrix
i need to check every possible solution to a certain predicament.
i have a matrix[x][y][z] that represents possible nodes to travel through. I already finished a method that should give me a set of solutions (it disables a single path every iteration and recalculates the entire solution, disable priority is based on travel capacity of the last solution)
but i need to see how effective my method is in terms of total time taken to calculate a set. For this i require a method calculate the solution on every permutation of these paths.
currently it only has a single layer in between 2 main layers (L1) where 0 is a free path and 1 is a non accessible path.
This here is the starting layout where i can toggle the values on layer L1 from 0 to 1 to disable a path and the basis of my shortest path search algorithm.
L0 0 0 0 0 0 L1 0 1 0 1 0 L2 0 0 0 0 0
0 1 0 1 0 1 1 1 1 1 0 1 0 1 0
0 0 0 0 0 0 1 0 1 0 0 0 0 0 0
0 1 0 1 0 1 1 1 1 1 0 1 0 1 0
0 0 0 0 0 0 1 0 1 0 0 0 0 0 0
how can i iterate through every possible combination of disabling the free paths when the dimensions of the matrix are non constant (meaning they are already user defined on compile time and can be changed whenever)? there are 2^n solutions where n is the number of free path on all mediary layers.
(a quick explanation in C or C++ would be best, even pseudo code is good) since there currently 9 free path to make combinations with there should be about 2^9 solutions which i need to test with. I havent done any brute force algorithms before so i have no idea how to make one.