I've received a task to find an algorithm which divides a graph G(V,E) into pairs of neighboring edges (colors the graph, such that every pair of neighboring edges has the same color).
I've tried to tackle this problem by drawing out some random graphs and came to a few conclusions:
If a vertex is connected to 2(4,6,8...) vertices of degree 1, these make a pair of edges.
If a vertex of degree 1 is directly connected to a cycle, it doesn't matter which edge of the cycle is paired with the lone edge.
However, I couldn't come up with any other conclusions, so I tried a different approach. I thought about using DFS, finding articulation points and dividing graph into subgraphs with an even number of edges, because those should be dividable by this rule as well, and so on until I end up with only subgraphs of |E(G')| = 2.
Another thing I've come up with is to create a graph G', where E(G) = V(G') and V(G) = E(G'). That way I could get a graph, where I could remove pairs of vertices (former edges) either via DFS or always starting with leaf vertices along with their adjacent vertices.
The last technique is most appealing to me, but it seems to be the slowest one. Any feedback or tips on which of these methods would be the best is much appreciated.
EDIT: In other words, imagine the graph as a layout of a town. Vertices being crossroads, edges being the roads. We want to decorate (sweep, color) each road exactly once, but we can only decorate two connected roads at the same time. I hope this helps for clarification.
For example, having graph G with E={ab,bd,cd,ac,ae,be,bf,fd}, one of possible pair combinations is P={{ab,bf},{ac,cd},{ae,eb},{bd,df}}.
One approach is to construct a new graph G where:
A vertex in G corresponds to an edge in the original graph
An edge in G connects vertices a and b in G where a and b represent edges in the original graph that meet at a vertex in the original graph
Then, if I have understood the original problem correctly, the objective for G is to find the maximum matching, which can be done, for example, with the Blossom algorithm.
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I am looking back at some of my algorithms homework(exam soon haha), and I am having troubles understanding the solution to one of the questions.
The Question
Tutor Solution
I am having difficulties visualizing the solution, I understand that if you have an odd number of cycles than your graph cannot be bipartite. But as I stated, I don't understand the shortest path from s to u and v to s.
Let's say G is bipartite. Then the vertex set can be divided into V1 and V2 s.t. every edge of G includes a vertex from each set. Then the vertices of every path in G alternate between V1 and V2, so the parity of the length of every path between every pair of vertices is the same. I.e., if G is bipartite and v,w are two vertices of G, then the length of every path connecting v & w is either even or odd.
If there is an edge (u,v) connecting two vertices in the same layer then this is violated. The BFS path to v and u have the same length, so same parity, since v & u are in the same layer. The edge between v & u gives us a path longer by 1, so of a different parity. Contradiction.
The tutor's solution isn't as clear as it could be, since it talks about cycles, and the two paths don't necessarily form a cycle since they can share vertices. And the definition of bipartite graph is slightly wrong (or uses non-standard definitions of edges, cross-product etc. where the things aren't pairs but 2-element sets).
But instead of the definition as given, I'd just say that a graph is bipartite if the vertices can be coloured black or white, such that each edge goes between different-coloured vertices. (Equivalently you can simply say "the graph is 2-colourable").
From this definition, it follows that if there's an even length path between two vertices they must be the same colour, otherwise different colours. In the BFS on a bipartite graph, a vertex u is layer i has a path of length i from s to u, so all vertices in the same layer have the same colour. Thus there can be no edge between two vertices in the same layer.
Is there a way to compute (accurate or hevristics) this problem on medium sized (up to 1000 nodes) weighted graph?
Place n (for example 5) sensors in nodes of the graph in such way that the sum of distances from every other node to the closest sensor will be minimal.
I'll show that this problem is NP-hard by reduction from Vertex Cover. This applies even if the graph is unweighted (you don't say whether it's weighted or not).
Given an unweighted graph G = (V, E) and an integer k, the question asked by Vertex Cover is "Does there exist a set of at most k vertices such that every edge has at least one endpoint in this set?" We will build a new graph G' = (V', E), which is the same as G except that all isolated vertices have been discarded, solve your problem on G', and then use it to answer the original question about Vertex Cover.
Suppose there does exist such a set S of k vertices. If we consider this set S to be the locations to put sensors in your problem, then every vertex in S has a distance of 0, and every other vertex is at a distance of exactly 1 away from a vertex that is in S (because if there was some vertex u for which this wasn't true, it would mean that none of u's neighbours are in S, so for each such neighbour u, the edge uv is not covered by the vertex cover, which would be a contradiction.)
This type of problem is called graph clustering. One of the popular methods to solve it is the Markov Cluster (MCL) Algorithm. A web search should provide some implementation examples. However it does not generally provide the optimal solution.
A subgraph Sn of a graph G is a sunflower graph, if consists of a Cycle Cn = {v1,v2,..,vn} of n vertices together with other n independent vertices {u1,u2,...,un} such that for each i, ui is adjacent to vi and vj, where j = i-1(mod n).
You could think of a sunflower - in the sense of the question - as a cycle of triangles. In time O(N^3) you can check each triple of points to see if it is a triangle and create a new graph whose vertices denote triangles in the original graph and where two vertices are linked if the two triangles share one or more vertices.
Then a depth first search looking for back edges should find cycles in this graph. Not all cycles are good. I think it may be enough to check that no two successive edges in the supposed cycle in the derived graph are produced by the same vertex in the original graph, and that you can check this as part of the depth first search. It may take some detailed analysis of cases to establish this, unless you can find a neat proof.
Given a undirected graph. How to find the size of maximum subset of vertices of the graph in which each vertex has at degree atleast p, where the degree in subset is find among the vertices in subset only.
Vertices of degree less than p can never be part of the solution. Remove them entirely, including their edges. Look at the new graph and repeat, etc.
When this process stops, all vertices have degree at least p.
Then, look at the connected components of that graph and pick the largest one! (As Evgeny Kluev correctly points out, this is unnecessary of course. In my head, the remaining subgraph should have been connected, but of course the original problem makes no such demands.)
I have a graph G which consists only of star graphs. A star graph consists of one central node having edges to every other node in it. Let H1, H2,…,Hn be different star graphs of different sizes which are present in G. We call the set of all nodes which are centres in any star graph R.
Now suppose these star graphs are building edges to other star graphs such that no edge is incident between any nodes in R. Then, how many edges exist at maximum between the nodes in R and the nodes which are not in R, if the graph should remain planar?
I want the upper bound on the number of such edges. One upper bound that I have in mind is: consider them as bipartite planar graph where R is one set of vertices and rest of the vertices form another set A. We are interested in edges between these sets (R and A). Since it is planar bipartite, the number of such edges is bounded by twice the number of nodes in G.
What I feel is that is there a better bound, maybe twice the nodes in A plus the number of nodes in R.
In case you can disprove my intuition, then that would also be good. Hopefully some of you can come up with a good bound along with some relevant arguments.
That's the best you can do. Take any planar graph G and construct its face-vertex incidence graph H, whose faces all have 4 edges. Let R be the set of faces of G and construct stars any which way using edges in H. This achieves the bound for bipartite planar graphs.