Below is the exercise 5.25 in 《Introduction to algorithms, a creative approach》. After reading it several times, I still can't understand what it means. I can color a tree with 2 colors very easily and directly using the method it described, not 1+LogN colors.
《Begin》
This exercise is related to the wrong algorithm for determining whether a graph is bipartite, described in Section 5.11.In some sense, this exercise shows that not only is the algorithm wrong, but also the simple approach can not work. Consider the more general problem of graph coloring: Given an undirected graph G=(V,E), a valid coloring of G is an assignment of colors to the vertices such that no two adjacent vertices have the same color. The problem is to find a valid coloring, using as few colors as possible. (In general, this is a very difficult problem; it is discussed in Chapter 11.)
Thus, a graph is bipartite if it can be colored with two colors.
A. Prove by induction that trees are always bipartite.
B. We assume that the graph is a tree(which means that the graph is bipartite). We want to find a partition of the vertices into the two subsets such that there are no edges connecting vertices within one subset.
Consider again the wrong algorithm for determining whether a graph is bipartite, given in Section 5.11: We take an arbitrary vertex, remove it, color the rest(by induction), and then color the vertex in the best possible way. That is, we color the vertex with the oldest possible color, and add a new color only if the vertex is connected to vertices of all the old colors. Prove that, if we color one vertex at a time regardless of the global connections, we may need up to 1+logN colors.
You should design a construction that maximizes the number of colors for every order of choosing vertices. The construction can depend on the order in the following way.
The algorithm picks a vertex as a next vertex and starts checking the vertex’s edges. At that point, you are allowed to add edges incident to this vertex as you desire, provided that the graph remains a tree, such that, at the end, the maximal number of colors will be required. You can not remove an edge after it is put in(that would be cleanining the algorithm, which has already seen the edge). The best way to achieve this construction is by induction. Assume that you know a construction that requires<=k colors with few vertices, and build one that requires k+1 colors without adding too many new vertices.
《End》
Related
This problem may be related to this post.
This problem also asked here but with a different taste.
Consider an (undirected) square graph with a periodic boundary condition. Then find a complete cycle graph with length equal to 4. now I want to assign a unique representative to each cycle from its elements. Therefore in a square graph with n_v vertex i will find n_f=n_v 4-cycles and n_v representative for the cycles. For the square graph, everything is simple. just assign the bottom left vertex of each plaque(4-cycles).
(i just show first 4-cycle)
Now, I want to generalize it for other structures. consider (undirected) kagome graph with proper boundary condition,
(here I just show 3 distinct cycles)
In this case for assigning a vertex to cycles cover, you need three different length cycles. which show by similar color with the assigned vertex. However, now I want to generalize this to other complicated graphs. I want to know is this problem has a name and about its possibility or algorithm. For example, we cannot do it in a triangular graph:
This problem solved here.
I)Let show all faces and vertices by \alpha_i, where i contain
vertices and faces.
II) make a graph which relates \alpha_i (from i
in face group) to \alpha_j (to j in vertex group), if j(vortex)
belong to i(face).
III) find the independent edge set of this
graph gives for any vortex a face.
Please see here for additional information.
I want to solve a Sudoku puzzle in NetworkX by reducing it to a vertex coloring problem. The graph has a vertex for each cell of the Sudoku grid, and two vertices are adjacent if and only if the corresponding cells belong to the same row, column, or block. Clues are represented by additional edges in the graph, and a 9-coloring of the graph represents a solution to the puzzle.
However, it seems that all of the vertex coloring algorithms in NetworkX are heuristics, and they are not guaranteed to find a minimum vertex coloring. In my experiments, I get vertex colorings with 10 colors, even though I know that a 9-coloring exists.
How can I find a minimum vertex coloring with NetworkX?
Unfortunately, in networkx there are not exact algorithms for the chromatic number problem.
This means that you have to implement it by your own (or find some available implementations).
You might either use an exact algorithm (e.g., Lawler's Algorithm) or an ILP-based one.
Give an example of a graph that has the following properties. (Note that you need to give a single graph as the answer.)
The graph does not contain a triangle (that is, a clique of 3 vertices) as a subgraph.
Graph needs at least 4 colors for a proper vertex coloring
[If you think that such a graph is not possible, then prove that statement.]
1.Tomorrow we have a final exam, and this question may on the exam paper.
2.I think it is impossible to draw such a graph. But how to prove? Thank u very much.
A web search finds https://en.wikipedia.org/wiki/Gr%C3%B6tzsch_graph
The Grötzsch graph is a member of an infinite sequence of triangle-free graphs, each the Mycielskian of the previous graph in the sequence, starting from the null graph; this sequence of graphs was used by Mycielski (1955) to show that there exist triangle-free graphs with arbitrarily large chromatic number. Therefore, the Grötzsch graph is sometimes also called the Mycielski graph or the Mycielski–Grötzsch graph. Unlike later graphs in this sequence, the Grötzsch graph is the smallest triangle-free graph with its chromatic number (Chvátal 1974).
I've got a DAG of around 3.300 vertices which can be laid out quite successfully by dot as a more or less simple tree (things get complicated because vertices can have more than one predecessor from a whole different rank, so crossovers are frequent). Each vertex in the graph came into being at a specific time in the original process and I want one axis in the layout to represent time: An edge relation like a -> v, b -> v means that a and b came into being at some specific time before v.
Is there a layout algorithm for DAGs which would allow me to specify the positions (or at least the distances) on one axis and come up with an optimal layout regarding edge crossovers on the other?
You can make a topological sorting of the DAG to have the vertices sorted in a way that for every edge x->y, vertex x comes before than y.
Therefore, if you have a -> v, b -> v, you will get something like a, b, v or b, a, v.
Using this you can easily represents DAGs like this:
Yes, as #Arturo-Menchaca said a topological sorting may help to reduce overlapping count of edges. But it may be not optimal. There is no good algorithm for edge crossing minimization. Problem for crossing minimization is NP-complete. The heuristics are applied for solving this problem.
This StackOverflow link may help you: Drawing Directed Acyclic Graphs: Minimizing edge crossing?
I suppose your problem is related to an aesthetically pleasing way of the graph layout. Some heuristics are described in the articles Overview of algorithms for graph drawing, Force-Directed Drawing Algorithms. May be information about planar graph or almost planar graph can help you also.
Some review of the algorithms for checking and drawing planar graphs are described in the Wiki pages Planar graph, Crossing number (graph theory). The libraries and algorithms for planar graph drawing are described in the StackOverflow question How to check if a Graph is a Planar Graph or not? For example the author in the article GA for straight-line grid drawings of maximal planar graphs uses genetic algorithms for straight-line grid drawing.
Good descriptions for almost planar graphs are given in the articles Straight-Line Drawability of a Planar Graph Plus an Edge, On the Crossing Number of Almost Planar Graphs.
Try to modify the original algoritms using your condition with one axis alignment.
If I understood you correctly then you want to minimize the number of edge-crossings in your graph layout. If so, then the answer is "No", because this problem is proved to be NP-complete in the general case. See this, "Crossing Number is NP-Complete, Garey, Johnson".
If you need a not an optimal but just good enough solution, there are multiple articles on this topic because it is heavily related with circuit layouts. Probably googling "crossing number heuristics" and looking through the abstracts of some papers will solve your task better then me trying to guess blindly your requirements.
EDIT: Precisely, I am trying to find two disjoint independent sets of known size in a graph shaped like a triangular grid, which may have holes and has a variable perimeter shape.
I'm not very well versed in graph theory, so I'm not sure if there exists an efficient solution for this problem. Consider the following graphs:
The colors of any two nodes can be swapped. The goal is to ensure that no two red nodes are adjacent, and no two green nodes are adjacent. The edges marked with exclamation points are invalid. Basically, I need to write two algorithms:
Determine that the nodes in a given graph can be arranged so that red and green nodes are not adjacent to nodes of the same color.
Actually rearrange the nodes.
I'm a little lost on how to implement this. It's not too difficult to separate the nodes of one color, but repeating the process for the second color may mess up the first color. Without a way to determine whether the graph can actually be arranged properly, this process could loop forever.
Is there some kind of algorithm that I can use/write for this? I'm mainly interested in the first image's graph (a triangular grid), but a generic algorithm would work as well.
First, let's note that the problem is a variant of graph coloring.
Now, if you only dealing with 2 colors (red,green) - coloring a graph with 2 colors is fairly easy, and is basically done by finding out if the graph is bipartite, and coloring each "side" of the graph in one color. Finding if a graph is bipartite is fairly simple.
However, if you want more than two colors, the problem becomes NP-Complete, and is actually a variant of the Graph Coloring Problem.
Graph Coloring Problem:
Given a graph G=(V,E) and a number k determine if there is a
function c:V->{1,2.,,,.k} such that c(v) = v(u) -> (v,u) is not an
edge.
Informally, you can color the graph in k colors, and you need to determine if there is some coloring such that you never color 2 nodes that share an edge with the same color.
Note that while it seems your problem is slightly easier, since you already know what is the number of nodes in each color, it doesn't really make a difference.
Assume you have a polynomial time algorithm A that solves your problem.
Now, given an instance (G,k) of graph coloring - there are only O(n^3) possibilities to #color1,#color2,#color3 - so by examining each of these and invoking A on it, you can find a polynomial time solution to Graph-Coloring. This will mean P=NP, which is most likely (according to most CS researchers) not the case.
tl;dr:
For 2 colors: find out if the graph is bipartite - and give one color to each side of the graph.
For 3 or more colors: There is no known efficient solution, and the general belief is one does not exist.
I thought this problem would be easier for planar graph, but unfortunately it's not the case. Best match for this problem I was able to find is minimum sum coloring and largest bipartite subgraph.
For largest bipartite subgraph, assume that number of reds + number of greens exactly match the size of largest bipartite subgraph. Then this problem is equivalent to your. Paper claims that it's still NP-hard even for planar graphs.
For minimum sum coloring, assume that red color has weight 1, green color has color 2, and we have infinitely many blue* colors with some weight of >graph size. Then if answer is exactly minimal sum coloring, there is no polynomial algorithm to find it (although paper referes to such algorithm for chordal graphs).
Anyway, it seems that the closer your red+green count to the 'optimal' in some sense subgraph, the more difficult problem is.
If you can afford inexact solution, or relaxed solution then you only spearate, say, reds, you have an option. As I said in comment, approximate solution of maximum independent set problem for planar graph. Then color this set into red and blue colors, if it is large enough.
If you know that red+green is much less than total number of vertices, another approximation can work. Look at introduction chapter of this article. It claims that:
For graphs which are promised to have small chromatic number, a better
guarantee is known: given a k-colorable graph, an algorithm due to
Karger et al. [12] uses semidefinite programming (SDP) to find an
independent set of size about Ω(n/∆^(1−2/k)).
Your graph is for sure 4-colorable, so you can count on large enough independent set. The same article states that greedy solution already can find large enough independent set.