Consider the following problem:
Given N points in plane and M line segments connecting them, find all polygons (convex or concave) that do not contain any other polygons inside.
For instance:
There are 5 polygons founded:
1 - 2 - 5 - 6
2 - 3 - 5
3 - 4 - 5
7 - 8 - 9
10 - 13 - 20 - 12 - 11
How can I identify these polygons and there corresponding vertices and edges? And what is the fastest solution for this?
Build graph with segment ends as vertices and segments as edges, then find cycles using DFS.
Note that the same edges might belong to multiple cycles (polygons) like your 2-5 and there might be many variants to select cycles. To exclude ambiguity, you can build fundamental set of cycles
Edit. As weston noticed in comments, resulted polygons might contains others. So sketch of more geometric approach:
Build lists of adjacency for graph.
Sort edges in ever list by polar angle.
Choose the most bottom vertex A.
If it's degree is 0, remove vertex, if 1, remove vertex and edge.
Otherwise get edge E with the smallest angle from this vertex.
Walk to pair vertex B.
Choose the most left edge F from B.
Move along to F edge into C.
If dead end, remove F and vertex C and return to B.
Repeat moving using left-hand rule until meet vertex A or dead end vertex.
In the walk process remove edges outgoing from vertices of degree 2 or mark them as used.
Related
Create a directed graph with 6 nodes (say) such that it has maximum number of strongly connected components.
For example, take complete graph with 4 nodes with all edges connected. This is graph has only 1 strongly connected component, i.e entire graph is a single component.
The objective is to maximise the number of components.
Your question does not make sense as posted.
Best guess: you want to create a new graph with a given number of nodes that will have the maximum number of strongly connected components. Edges can be added between any pairs of nodes.
The smallest strongly component is a pair of nodes with an edge going in each direction between them.
So the maximum number of components in a graph with N nodes will be N / 2
e.g. a 4 node graph with two components has edges between
1 2
2 1
3 4
4 3
1 3
I recently began working on r for social network analysis. Everything goes well and up until now, I found answers to my questions here or on google. But not this time!
I am trying to find a way to calculate "vertex reciprocity" (% of reciprocal edges of each actor of the network). On igraph, reciprocity(g) works fine to calculate the reciprocity of the whole network, but it doesn't help me with the score per actor. Does anybody know what I could do?
Thank you!
I am going to assume that you have a simple graph, that is no loops and no multiple links between nodes. In that case, it is fairly easy to compute this. What does it mean for a link to be reciprocated? When there is a link from a to b, there is a link back from b to a. That means that there is a path of length two from a to itself a->b->a. How many such paths are there? If A is the adjacency matrix, then the entries of AA gives the number of paths of length two. We only want the ones from a node to itself, so we want the diagonal of AA. This will only count a->b->a as one path, but you want to count it twice: once for the link a->b and once for b->a. So for each node you can get the number of reciprocated links from 2*diag(A*A). You want to divide by the total number of links to and from a which is just the degree.
Let me show the computation with an example. Since you do not provide any data, I will use the Enron email data that is available in the 'igraphdata' package. It has loops and multiple links which i will remove. It also has a few isolated vertices, which I will also remove. That will leave us with a connected, directed graph with no loops.
library(igraph)
library(igraphdata)
data(enron)
enron = simplify(enron)
## remove two isolated vertices
enron = delete_vertices(enron, c(72,118))
Now the reciprocity computation is easy.
EnronAM = as.matrix(as_adjacency_matrix(enron))
Path2 = diag(EnronAM %*% EnronAM)
degree(enron)
VertRecip = 2*Path2 / degree(enron)
Let's check it by walking through one node in detail. I will use node number 1.
degree(enron,1)
[1] 10
ENDS = ends(enron, E(enron))
E(enron)[which(ENDS[,1] == 1)]
+ 6/3010 edges from b72ec54:
[1] 1-> 10 1-> 21 1-> 49 1-> 91 1->104 1->151
E(enron)[which(ENDS[,2] == 1)]
+ 4/3010 edges from b72ec54:
[1] 10->1 21->1 105->1 151->1
Path2[1]
[1] 3
Node 1 has degree 10; 6 edges out and 4 edges in. Recip shows that there are three paths of length 2 from 1 back to itself.
1->10->1
1->21->1
1->151->1
That makes 6 reciprocated links and 4 unreciprocated links. The vertex reciprocity should be 6/10 = 0.6 which agrees with what we computed above.
VertRecip[1]
[1] 0.6
I was just barely introduced to .ply files and I don't understand how they work. The vertex list has only 3 values for each vertex: x,y,z. But each face has 5 values and I don't know what those 5 values mean. I just need a little explanation. Thanks!
This simply means that the face has 5 sides or 5 vertices and each vertex's position is specified in space by the 3 values x,y,z.
The above answer is not exactly right to my knowledge.
The first value is the number of vertices which define the face, in your case it should be 4, as there are only 4 numbers left. The four other numbers are indices of the vertices. Like often in computer science, the index of the first vertex is 0, not 1.
Some programs though support only triangular meshes, and faces with more than 3 edges are not supported.
Furthermore, if you use a visualization program which adds light and other rendering stuff, the face 3 0 1 2 is not the same as 3 0 2 1 as the triangle has another orientation.
I have a directed, unweighted, possibly cyclic graph that can contain loops and multiple duplicate edges (i.e. two edges from node 1 to node 2).
I would now like to find the length of the longest trail in this graph, i.e. the longest path that:
- uses no edge twice (but if there are multiple edges from node 1 to node 2, it can use every one of them)
- possibly visits nodes several time (i.e. it does not have to be a simple path)
In particular, is this problem NP-hard? I know that the longest simple path is NP-hard (reducing Hamiltonian Path to it) and the longest trail with edge reusal is in P (Bellman ford with weight -1 on every edge). However, with this problem, I am not quite sure and I could not find good information on it.
Although I am not completely sure, I think that this problem is NP-hard. As I understand, your question arises due to multiple edges between nodes. The graphs that has multiple edges between same nodes can be expanded to larger graphs with no multiple edges between them. Thus, a graph with multiple edges between same nodes has no difference than a graph without multiple edges.
Let me walkthrough a simple example to explain:
Let there be a graph with 3 nodes (A,B,C) and 5 edges between them (A to B, A to B, B to A, B to C, C to A)
This graph can be expanded and shown with 5 nodes and 7 edges.
Lets expand the node A to 3 different nodes (A1, A2, A3). When we adjust the edges according to previous edges, there exists 7 edges(A1 to B, A2 to B, B to A3, B to C, C to A1, C to A2, C to A3)
As a result, now we have a graph without multiple edges and can be evaluated with the help of Hamiltonian and Bellman Ford.
Hope I've at least cleared the problem a bit.
I have two graphs G, H labeled and I want to extract all common subgraph of two graphs, I got to a part that is:
1 - extract all the nodes that are in common, but I'm stuck on the part that includes:
2 - Step 1: Take the First vertex and store it in a set P = {first element} (which will be the set of all common subgraph), and go to 2nd if it is adjacent to the first of the two P graph G and H, we add it, and so on, but I do not know how to do it when i have more than 2
That is a NP-complete problem. See http://en.wikipedia.org/wiki/Subgraph_isomorphism_problem