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- #DISCRETE MATHEMATICS WITH GRAPH THEORY 3RD EDITION SOLUTIONS SERIES#
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Graphs are key data structures used to represent networks, chemical structures, games etc. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.The study of discrete mathematics is one of the first courses on curricula in various educational disciplines such as Computer Science, Mathematics and Engineering. If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks. This article is contributed by Ankit Jain. Since the sum of degrees of vertices in the above problem is 9*3 = 27 i.e odd, such an arrangement is not possible. Sum of degrees of all vertices = 2* Number of Edges in the graph Now two vertices of this graph are connected if the corresponding line segments intersect. Here we need to consider a graph where each line segment is represented as a vertex.
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If we try to approach this problem by using line segments as edges of a graph,we seem to reach nowhere (This sounds confusing initially). We could think of solving it using graphs. Solution: This problem seems very difficult initially. Is it possible that each line segment intersects exactly 3 others? Problem 3: There are 9 line segments drawn in a plane. Hence, we can conclude that no matter what the final arrangement is not possible. There is no possible way for a knight to cross over (Two knights cannot exist on one vertex) the other in order to achieve the final state. However the order in which knights appear on the graph cannot be changed. We can only move the knights in a clockwise or counter-clockwise manner on the graph (If two vertices are connected on the graph: it means that a corresponding knight’s move exists on the grid). Note that in order to achieve the final state there needs to exist a path where two knights (a black knight and a white knight cross-over). Hence the initial state of the graph can be represented as : We use a hollow circle to depict a white knight in our graph and a filled circle to depict a black knight. Hence none of the edges connect to vertex 5. Clearly vertex 5 can’t be reached from any of the squares. Similarly we can draw the entire graph as shown below. We can say that vertex 1 is connected to vertices 6 and 8 in our graph. The reachable squares with valid knight’s moves are 6 and 8. There exists a edge between two vertices in our graph if a valid knight’s move is possible between the corresponding squares in the graph. Now we consider each square of the grid as a vertex in our graph. But what kind of a graph should we draw? Let each of the 9 vertices be represented by a number as shown below.
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However the above question can be solved using graphs.
#DISCRETE MATHEMATICS WITH GRAPH THEORY 3RD EDITION SOLUTIONS CRACK#
You might think you need to be a good chess player in order to crack the above question. Is it possible to reach the final state as shown below using valid knight’s moves ? Problem 2 – The figure below shows an arrangement of knights on a 3*3 grid. As a result we can conclude that our supposition is wrong and such an arrangement is not possible. Mathematics | Rings, Integral domains and Fields.Mathematics | Independent Sets, Covering and Matching.
#DISCRETE MATHEMATICS WITH GRAPH THEORY 3RD EDITION SOLUTIONS SERIES#