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How many colors are required to color the provinces of Costa Rica? A common visual aid for maps is to color the regions of the map differently, so that no two regions which share a border also share a color. For example, to the right is a map of the provinces of Costa Rica where the author is presently spending his vacation.

It is colored with eight different colors, one for each province. This raises the obvious question, what is the minimum number of colors required to color Costa Rica in a way that no two provinces which share a common border share a common color? With a bit of fiddling, the reader will see that the answer is obviously three. Hence even these three alone cannot be colored with just two. This problem becomes much trickier, and much more interesting to we mathematicians, as the number of provinces increases.

So let us consider the map of the arrondissements of Paris. The twenty arrondissements of Paris, which are arranged in a strangely nautilusy spiral. With so many common borders, we require a more mathematically-oriented description of the problem.

Enter graph theory. Let us construct an undirected graph for Parisin which the vertices are the arrondissements, and two vertices are connected by an edge if and only if they share a common border.

In general, when we color graphs we do not consider two regions to share a border if their borders intersect in a single point, as in the four corners of the United States. Performing this construction on Paris, we get the following graph.

Recognizing that the positions of the corresponding vertices are irrelevant, we shift them around to have a nicely spread-out picture. Our first map of Costa Rica was 3-colorable, and we saw indeed that.

We now turn our attention to coloring Paris. For the sake of the pseudocode, we represent colors by natural numbers. Obviously, the quality of the coloring depends on the order in which we investigate the vertices. Further, there always exists an ordering of the vertices for which this greedy algorithm produces a coloring which achieves the minimum required number of colors.

Applying this to Pairs, we get the following 4-coloring:. But this is not enough. Can we do it with just three colors? The call of the wild beckons! We challenge the reader to find a 3-coloring of Paris, or prove its impossibility. In the mean time, we have other matters to attend to. Colloquially this means a fast solution is believed not to exist, and to find one or prove this belief is true would immediately make one the most famous mathematician in the world.

The only general solutions take at worst an exponential or factorial amount of time with regards to the number of vertices in the graph.

greedy coloring algorithm pseudocode

In other words, trying to run a general solution on a graph of 50 vertices is likely to take 50! Now we turn to a different aspect of graph theory, which somewhat simplifies the coloring process. The maps we have been investigating thus far are special. Specifically, their representations as graphs admit drawings in which no two edges cross. This is obvious by our construction, but on the other hand we recognize that not all graphs have such representations.

For instance, try to draw the following graph so that no two edges cross:. For now, we simply recognize that not all graphs have such drawings. This calls for a definition! And after a lifetime of cartography, we might notice that every map we attempt to color admits a 4-coloring. We now have a grand conjecture, that every planar graph is 4-colorable.

The audacity of such a claim!By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service. The dark mode beta is finally here.

Graph Coloring | Set 2 (Greedy Algorithm)

Change your preferences any time. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. The solution provided to the Interval Scheduling Problem was this:. An extension of the problem, also known as the Interval Coloring Problem, can be explained as follows.

We are given a set of intervals, and we want to colour all intervals so that intervals given the same colour do not intersect and the goal is to try to minimize the number of colours used.

I thought of a better one, but I'm not sure if it is correct. Basically, we modify the algorithm for Interval Scheduling. I might have a silly bug in the pseudocode but I hope you understand what my algorithm is trying to do. Basically I'm peeling off all possible colourings by repeatedly running the standard interval scheduling algorithm. So, shouldn't my algorithm print all the various colourings of the intervals?

Is this more efficient than the previous algorithm, if it is a correct algorithm? If it is not, could you please provide a counter-example? Your algorithm can produce a suboptimal solution. Here's an example of what your algorithm will produce:. Clearly the following 2-colour solution which will be produced by the proven-optimal algorithm is possible for the same input:.

Here's an O n log n algorithm: Instead of looping through all n intervals, loop through all 2n interval endpoints in increasing order. Maintain a heap priority queue of available colours ordered by colour, which initially contains n colours; every time we see an interval start point, extract the smallest colour from the heap and assign it to this interval; every time we see an interval end point, reinsert its assigned colour into the heap.

Each endpoint start or end processed is O log n time, and there are 2n of them. This matches the time complexity for sorting the intervals, so the overall time complexity is also O n log n.

Learn more. Asked 3 years, 5 months ago. Active 3 years, 5 months ago. Viewed 1k times. Active Oldest Votes. But aren't the intervals already sorted in the algorithm from the book, based on their starting time?

Thanks though! You can formulate the problem either way; if you ask for a list of intervals that are already sorted by starting time then yes, you can of course omit the sorting step. That might cause the time complexity to drop to e. O ndepending on how efficient the main loop is.By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service. The dark mode beta is finally here. Change your preferences any time. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information.

In interval scheduling, the algorithm is to pick the earliest finish time. But in interval colouring the former does not work.

Kruskal's algorithm

Is there an example or explanation on why picking earliest finish time won't work for interval colouring? This can be thought of as the interval partitioning problem if it makes more sense. The interval scheduling problem that i'm referring to is: If you go to a theme park and there are many shows, the start and finish time of each show is an interval, and you are the resource. You want to attend as many shows as possible.

If you only need a counter example of greedy algorithm on coloring, btilly provides one already. First, for scheduling problem, you can indeed prove greedy algorithm works.

The idea is like this:. They are just different category of problem. Define depth be the maximum of conflicts at all time. Logically we know depth is the lower bound, but we have to proof it is the upper bound as well by contradiction.

Assume the depth of the interval set is dand the answer is greater than d. Note that if you sort by start time descending, or finish time ascending, then you cannot use the same reasoning. Now we know depth is the answer, we have to find it.

Concept-wise, it DOES NOT matter if you find by using start time or finish time, ascending or descending, all options can give you the depth of the interval set. But that's another story, it's for implementation, concept-wise, it does not matter, you only want to find the depth of the interval set. For interval scheduling problem, the greedy method indeed itself is already the optimal strategy; while for interval coloring problem, greedy method only help to proof depth is the answer, and can be used in the implementation to find the depth but not in the way as shown in btilly's counter example.

This is just a case of playing around with pictures until you find an example. The first picture I drew that showed the problem had the following partitioning:.We introduced graph coloring and applications in previous post. As discussed in the previous post, graph coloring is widely used. Unfortunately, there is no efficient algorithm available for coloring a graph with minimum number of colors as the problem is a known NP Complete problem.

There are approximate algorithms to solve the problem though. Following is the basic Greedy Algorithm to assign colors. Color first vertex with first color. Do following for remaining V-1 vertices. If all previously used colors appear on vertices adjacent to v, assign a new color to it.

Also, the number of colors used sometime depend on the order in which vertices are processed. For example, consider the following two graphs.

Note that in graph on right side, vertices 3 and 4 are swapped. If we consider the vertices 0, 1, 2, 3, 4 in left graph, we can color the graph using 3 colors.

But if we consider the vertices 0, 1, 2, 3, 4 in right graph, we need 4 colors. So the order in which the vertices are picked is important.

Many people have suggested different ways to find an ordering that work better than the basic algorithm on average. The most common is Welsh—Powell Algorithm which considers vertices in descending order of degrees. Here d is the maximum degree in the given graph. Since d is maximum degree, a vertex cannot be attached to more than d vertices.

When we color a vertex, at most d colors could have already been used by its adjacent. To color this vertex, we need to pick the smallest numbered color that is not used by the adjacent vertices. If colors are numbered like 1, 2, …. This can also be proved using induction. See this video lecture for proof. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. Writing code in comment? Please use ide.

Minimum cost to traverse from one index to another in the String D'Esopo-Pape Algorithm : Single Source Shortest Path Minimum number of edges that need to be added to form a triangle Minimum number of colors required to color a graph Check if given path between two nodes of a graph represents a shortest paths Maximum number of edges that N-vertex graph can have such that graph is Triangle free Mantel's Theorem Find the minimum spanning tree with alternating colored edges Shortest path with exactly k edges in a directed and weighted graph Set 2.

m Coloring Problem | Backtracking-5

Basic Greedy Coloring Algorithm: 1. Graph g1 5. Graph g2 5. Graph int v. Improved By : ChamanJhinga. Load Comments.Kruskal's algorithm is a minimum-spanning-tree algorithm which finds an edge of the least possible weight that connects any two trees in the forest.

If the graph is not connected, then it finds a minimum spanning forest a minimum spanning tree for each connected component. This algorithm first appeared in Proceedings of the American Mathematical Societypp. At the termination of the algorithm, the forest forms a minimum spanning forest of the graph.

If the graph is connected, the forest has a single component and forms a minimum spanning tree. The following code is implemented with disjoint-set data structure :. Kruskal's algorithm can be shown to run in O E log E time, or equivalently, O E log V time, where E is the number of edges in the graph and V is the number of vertices, all with simple data structures.

These running times are equivalent because:. We can achieve this bound as follows: first sort the edges by weight using a comparison sort in O E log E time; this allows the step "remove an edge with minimum weight from S " to operate in constant time.

Next, we use a disjoint-set data structure to keep track of which vertices are in which components. We need to perform O V operations, as in each iteration we connect a vertex to the spanning tree, two 'find' operations and possibly one union for each edge. Even a simple disjoint-set data structure such as disjoint-set forests with union by rank can perform O V operations in O V log V time. The proof consists of two parts.

First, it is proved that the algorithm produces a spanning tree. Second, it is proved that the constructed spanning tree is of minimal weight.

We show that the following proposition P is true by induction : If F is the set of edges chosen at any stage of the algorithm, then there is some minimum spanning tree that contains F.

Kruskal's algorithm is inherently sequential and hard to parallelize. It is, however, possible to perform the initial sorting of the edges in parallel or, alternatively, to use a parallel implementation of a binary heap to extract the minimum-weight edge in every iteration. A variant of Kruskal's algorithm, named Filter-Kruskal, has been described by Osipov et al.

The basic idea behind Filter-Kruskal is to partition the edges in a similar way to quicksort and filter out edges that connect vertices of the same tree to reduce the cost of sorting. The following pseudocode demonstrates this. Filter-Kruskal lends itself better for parallelization as sorting, filtering, and partitioning can easily be performed in parallel by distributing the edges between the processors.

Finally, other variants of a parallel implementation of Kruskal's algorithm have been explored. Examples include a scheme that uses helper threads to remove edges that are definitely not part of the MST in the background, [6] and a variant which runs the sequential algorithm on p subgraphs, then merges those subgraphs until only one, the final MST, remains.

From Wikipedia, the free encyclopedia. This article needs additional citations for verification.Given an undirected graph and a number m, determine if the graph can be colored with at most m colors such that no two adjacent vertices of the graph are colored with the same color. Here coloring of a graph means the assignment of colors to all vertices. Input: 1 A 2D array graph[V][V] where V is the number of vertices in graph and graph[V][V] is adjacency matrix representation of the graph.

A value graph[i][j] is 1 if there is a direct edge from i to j, otherwise graph[i][j] is 0. Output: An array color[V] that should have numbers from 1 to m.

greedy coloring algorithm pseudocode

The code should also return false if the graph cannot be colored with m colors. Following is an example of graph that can be colored with 3 different colors. Naive Algorithm Generate all possible configurations of colors and print a configuration that satisfies the given constraints. Backtracking Algorithm The idea is to assign colors one by one to different vertices, starting from the vertex 0. Before assigning a color, we check for safety by considering already assigned colors to the adjacent vertices.

If we find a color assignment which is safe, we mark the color assignment as part of solution. If we do not a find color due to clashes then we backtrack and return false. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. Writing code in comment? Please use ide. If exist then checks whether the color to.

greedy coloring algorithm pseudocode

It mainly uses graphColoringUtil to solve the problem. It returns. Please note that there. This initialization is needed. It mainly uses graphColoringUtil. It returns false if the m. Python program for solution of M Coloring. A utility function to check if the current color assignment.

Graph Theory: 64. Vertex Colouring

A recursive utility function to solve m. Print the solution. This code is contributed by Divyanshu Mehta. Please note that there may be more than one. WriteLine "Solution does not exist". WriteLine. Load Comments.The Greedy algorithm could be understood very well with a well-known problem referred to as Knapsack problem. Although the same problem could be solved by employing other algorithmic approaches, Greedy approach solves Fractional Knapsack problem reasonably in a good time.

Let us discuss the Knapsack problem in detail. Given a set of items, each with a weight and a value, determine a subset of items to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible.

The knapsack problem is in combinatorial optimization problem. It appears as a subproblem in many, more complex mathematical models of real-world problems. One general approach to difficult problems is to identify the most restrictive constraint, ignore the others, solve a knapsack problem, and somehow adjust the solution to satisfy the ignored constraints.

In many cases of resource allocation along with some constraint, the problem can be derived in a similar way of Knapsack problem.

Following is a set of example. A thief is robbing a store and can carry a maximal weight of W into his knapsack. There are n items available in the store and weight of i th item is w i and its profit is p i. What items should the thief take? In this context, the items should be selected in such a way that the thief will carry those items for which he will gain maximum profit. Hence, the objective of the thief is to maximize the profit. In this case, items can be broken into smaller pieces, hence the thief can select fractions of items.

In this version of Knapsack problem, items can be broken into smaller pieces. So, the thief may take only a fraction x i of i th item. It is clear that an optimal solution must fill the knapsack exactly, otherwise we could add a fraction of one of the remaining items and increase the overall profit. Here, x is an array to store the fraction of items. After sorting, the items are as shown in the following table.

First all of B is chosen as weight of B is less than the capacity of the knapsack. Next, item A is chosen, as the available capacity of the knapsack is greater than the weight of A. Now, C is chosen as the next item. However, the whole item cannot be chosen as the remaining capacity of the knapsack is less than the weight of C. Now, the capacity of the Knapsack is equal to the selected items.

Hence, no more item can be selected. This is the optimal solution.

DAA - Fractional Knapsack

We cannot gain more profit selecting any different combination of items. Previous Page. Next Page. Previous Page Print Page.