Simple Graph. General Graph

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1 Graph Theory A graph is a collection of points (also called vertices) and lines (also called edges), with each edge ending at a vertex In general, it is allowed for more than one edge to have the same two endpoints (these edges are called multiple edges) or for an edge to use the same endpoint twice (such an edge is a loop) Sometimes loops and multiple edges are not allowed; graphs with no loops or multiple edges are called simple graphs Examples: General Graph Simple Graph Graphs need not be connected The two graphs pictured above could be considered as a single, but non-connected, graph As a natural example, the highway system in the United States is a non-connected graph, because of Hawaii and Alaska A path in a graph is an alternating sequence of vertices and edges, beginning and ending with a vertex For example, using the labeling below, one path is v 1, e 1, v 2, e 2, v 3, e 5, v 2, e 1, v 1, e 4, v 4, e 6, v 4 v 2 e 1 v 1 the graph, then at every point other than the starting point (which is also the ending point) there must be an even number of edges converging on the point, since whenever you arrive at a point you must be able to leave on an unused edge Also, the common starting and ending point must have an even number of edges converging on it, since you use one to leave the first time and one to arrive the last time Since an odd number of edges meet at the vertices v 1 and v 2, this graph cannot have an Euler cycle The number of edges that are incident at a vertex is called the degree of the vertex So we know that if there is an Euler cycle, all the degrees must be even, and likewise it is easy to see that if there is an Euler path there can be at most two odd degrees (Notice that a loop contributes 2 to the degree of the vertex it touches) So exactly which graphs have Euler paths or cycles? Clearly the graph must be connected if it is to have either Is it true that any connected graph in which all degrees are even has an Euler cycle? Remarkably, the answer is yes We saw in class that this is true Briefly: if you trace any path at random in a graph with all even degrees, eventually you will get stuck at your starting point At that point you have a cycle, but you might have missed some edges However, it is always possible by the same method to find an additional cycle and paste it in to the first one For example, consider this graph: v 2 v 5 e 8 e 9 e 1 v 1 e 5 e 2 e 7 e 4 e 5 e 2 e 7 e 4 v 3 v 4 e 3 e 6 v 3 e 3 v 4 e 6 In general it is permissible to use vertices and edges more than once, but sometimes this is forbidden A path that uses vertices and edges at most once each is a simple path A cycle is a path with the same starting and ending vertex; it is simple if vertices and edges are used at most once each (the common starting and ending vertex counts as one use) Sometimes it is unnecessary to list all the vertices and edges of a path or cycle; for example, if there are no multiple edges, just listing the vertices is enough Euler Paths and Cycles An Euler path is a path that uses every edge in a graph exactly once; of course, this path will usually have to use vertices many times each, so it will not be a simple path The labeled graph above does have an Euler path: v 1, e 1, v 2, e 2, v 3, e 7, v 1, e 4, v 4, e 6, v 4, e 3, v 3, e 5, v 2 An Euler cycle is an Euler path that starts and ends at the same vertex It is not hard to see that the labeled graph above has no Euler cycle Imagine that the edges in the graph represent actual footpaths If you could follow an Euler cycle through 1 You might first find the cycle v 2, e 2, v 3, e 5, v 2, e 8, v 5, e 9, v 1, e 1, v 2 Then you can start over at v 1, and you might discover v 1, e 4, v 4, e 6, v 4, e 3, v 3, e 7, v 1 Now you can paste the second cycle into the first where v 1 occurs in the first cycle: v 2, e 2, v 3, e 5, v 2, e 8, v 5, e 9, v 1, e 4, v 4, e 6, v 4, e 3, v 3, e 7, v 1, e 1, v 2 This cycle uses all edges exactly once; if it didn t, we could keep finding new cycles and pasting them in until all the edges were used up So now we know that any connected graph with all even degrees has an Euler cycle What if a connected graph has exactly two odd degrees is there an Euler path? Yes, and it s easy to see: add one new edge between the two odd degree vertices, so now the graph has all even degrees Then there is an Euler cycle in the new graph, and when the extra edge is removed it is turned into an Euler path that starts at one endpoint of the extra edge and ends at the other endpoint Suppose for example that the two odd degree vertices are u and v We add a new edge connecting them, so now u and v have even degree, so there is an Euler cycle, something like this: x u v x or x v u x 2

2 If we remove the edge between u and v we get a path like v x u or u x v Using the same sort of reasoning, we can see that if there are exactly 2k vertices with odd degree, then we can find k paths that together use all the edges exactly once For example, if there are 4 vertices with odd degree then we can find 2 paths that together use all the edges exactly once Here s an example, with one path shown as solid edges and the other as dashed edges More generally, suppose there are 6 vertices with odd degree, say u, v, w, x, y, and z We add three new edges, between u and v, w and x, y and z Now all vertices have even degree, so there is an Euler cycle like t u v w x y z t Now remove the added edges to get t u v w x y z t This at first looks like 4 separate paths, but since t is at both ends, it is really three paths: z t u v w x y It works the same way for any even number of vertices of odd degree What if a graph has an odd number of vertices with odd degree? This can t happen! Consider the sum of all the degrees: d 1 + d d n This counts every edge twice, once at each end, so d 1 +d 2 + +d n = 2e If an odd number of degrees were odd, the sum d 1 + d d n would be odd, but in fact the sum is even, namely 2e, so the number of odd degrees must be even Note that this also gives an easy way to compute the number of edges in any graph: add up the degrees and divide by 2 This means that all graphs with the same degrees have the same number of edges It is tempting to guess that two graphs with the same degrees are really the same graph For example, these two graphs have the same degrees and look different: But these graphs are really the same: in each graph every point is adjacent (connected directly by an edge) to every other This is the only simple graph with degree sequence 3, 3, 3, 3, but it s not the only graph; for example: It is also possible for two really different simple graphs to have the same degree sequence: Why are these two graphs really different? In the first, the vertex with degree 4 is adjacent to just one of the vertices with degree 3; in the second the degree 4 vertex is adjacent to both vertices of degree 3 In general, it can be quite difficult to decide if two apparently different graphs are really different or just drawn differently Hamilton Paths and Cycles A Hamilton path is a path that uses every vertex in a graph exactly once; of course, such a path will be able to use edges at most once, and typically many edges not at all Since loops and multiple edges are no help, we generally assume that we are talking about simple graphs when we are concerned with Hamilton paths A Hamilton cycle is a Hamilton path that starts and ends at the same vertex All the graphs pictured above have Hamilton paths, and all but two have Hamilton cycles Consider again these two graphs: 3 4

3 sum of the numbers on the edges in the Hamilton cycle to be as small as possible This is the Traveling Salesman Problem, or the TSP This also turns out to be a very difficult problem Unlike the Hamilton cycle problem, we know in advance that there is a solution, but also we have the option of settling for a less than optimal solution: we can try to find a Hamilton cycle that doesn t cost too much more than the cheapest one possible One easy way to attempt such a task is to use a greedy algorithm This is a name used for any approach that proceeds by doing what is good in the short run and hoping it will be good in the long run There are two reasonable ways to do this for the TSP The one on the left has a Hamilton path but no Hamilton cycle; the one on the right has a Hamilton cycle How can we see that the one on the left has no Hamilton cycle? At vertices 1, 2, and 4 we have no choice about what edges to use in a Hamilton cycle, which forces us to use 3 edges incident at vertex 3, but no Hamilton cycle can use more than two edges incident at a single vertex Unfortunately, there is no known easy test to see whether or not a connected graph has a Hamilton path or cycle Sometimes it is possible to spot a simple reason that a Hamilton path or cycle cannot exist, but no single method works for all graphs It is generally true that graphs with lots of edges have Hamilton cycles, but some graphs without lots of edges also have Hamilton cycles For example, a graph consisting of just one long cycle has only a few edges (relative to the number of vertices) but obviously has a Hamilton cycle On the other hand, a graph like the one below can have many edges but no Hamilton path v Here you should imagine that the ellipses are filled with lots of vertices and edges, but it s easy to see that there s no way to get a path that uses all vertices without using the central vertex v more than once The Traveling Salesman Problem A complete graph is a graph in which every two vertices are adjacent A complete graph has many Hamilton cycles, since from each vertex you can go to any other vertex Suppose that in a complete graph we have associated numbers with the edges For example, if the vertices represent cities, the numbers might be the driving distances between cities, or the flight times, or the cost of sending a package Out of the many Hamilton cycles in the graph, we would like to find the one with smallest total cost or distance we want the 5 In the first method, pick any vertex; call it v 1 Next pick the vertex connected to v 1 by the lowest cost edge; call it v 2 Next pick the vertex connected to v 2 by the lowest cost edge, but of course do not pick v 1 ; call it v 3 In general, if you ve picked v 1, v k, pick the vertex that is not already on the list and is connected to v k by the lowest cost edge Eventually all vertices will be on the list, and then of course you have just one choice for the last edge back to v 1 This is the nearest neighbor algorithm: at each step we go to the nearest unused neighboring vertex The second method is similar but a little surprising Start by picking the lowest cost edge Then pick the next lowest cost edge, even if it is not connected to the first one Continue in this way, picking the next lowest cost edge no matter where it is, except never pick an edge that forms a cycle (until the very last edge, of course) and never pick an edge incident at a vertex that already has two chosen edges This is the cheapest link algorithm There are more sophisticated ways to look for low cost Hamilton cycles, but these two usually work remarkably well, and are very easy to do In fact they are so easy, that you might as well do them all: Use the nearest neighbor algorithm with every possible starting point, then do the cheapest link algorithm (just once there is no choice about how to start) If the graph has n vertices this will give you n + 1 Hamilton cycles, and you can choose the one with lowest cost Spanning Trees Here is a problem superficially similar to the Hamilton cycle problem: in a graph G pick just barely enough edges so that every vertex is connected by a path to every other vertex In other words, throw away as many edges as possible without disconnecting the graph This is a very easy problem: we can throw away any edge in any cycle without disconnecting the graph So we can repeatedly discard edges in cycles until there are no cycles left A graph with no cycles is called a tree, and a tree that connects all the vertices of some graph G, using edges from the original G, is called a spanning tree for G It is easy to see that a tree with n vertices has exactly n 1 edges Notice first that every tree must have a vertex with degree 1: just start at any vertex and follow edges at random, without repeating any edge Since there are no cycles, you can never revisit a vertex, so eventually you must get stuck, and the only way to get stuck is to end up at a vertex of degree 1 Now starting with a tree on n vertices, you can remove a vertex of 6

4 degree 1, and its edge After doing this n 1 times you will be down to 1 vertex and no edges, and you will have removed n 1 edges, so the original tree had n 1 edges Of more importance in practice than this simple problem is one that is similar to the TSP: if the edges in a graph have a cost attached, find a spanning tree with the smallest total cost Remarkably the greedy algorithms that fail to find a cheapest Hamilton cycle do find cheapest spanning trees The nearest neighbor algorithm for spanning trees is called Prim s Algorithm Start with any vertex, say v 1, and pick the cheapest edge to a neighbor, say v 2 Now among all the edges from either v 1 or v 2 pick the cheapest, connecting either v 1 or v 2 to v 3 At each stage, pick the cheapest edge from one of the vertices already picked to a vertex not yet picked After n 1 edges have been picked, you are done, and have a guaranteed cheapest spanning tree Here is an example, making edges solid as they are added 55 S S The cheapest link algorithm for spanning trees is called Kruskal s algorithm As before, pick the cheapest edge in the entire graph, then the next cheapest, and so on, but never pick an edge that will form a cycle After n 1 edges have been picked, the spanning tree is complete, and is guaranteed to have minimum possible total cost 7 Graph Coloring A coloring of a graph is a coloring of the vertices so that no two adjacent vertices are the same color Colors need not literally be used, of course; labels will do, either letters or numbers A loop in a graph makes it impossible to color, since the vertex with the loop is adjacent to itself; multiple edges have no effect at all; hence, we assume when talking about coloring that all graphs are simple Usually we want to color a graph with the smallest possible number of colors, called the chromatic number of the graph and written χ(g) Here s an example of a graph colored with 3 colors, R, B, G: R G B R B It is easy to see that this graph requires 3 colors because there is a triangle But it is possible for 3 colors to be required even without a triangle; in fact any odd length cycle requires 3 colors: 8

5 Likewise, it is possible for a graph to require 4 colors even if the graph does not contain 4 vertices all connected to each other, and so on Here is a graph with chromatic number 4 but no 4 vertices all connected to each other: As with Hamilton cycles, greedy coloring algorithms do not always give a coloring with the smallest number of colors, but usually do reasonably well Here s one greedy coloring algorithm: list the vertices in some order, say v 1, v 2,,v n Also list the colors that will be used: R, B, G, Y, Color the first vertex with the first color For each vertex after that, use the first color on the list that works, ie, that does not cause two adjacent vertices to have the same color Here is an example: v 1 :R v 2 :B v 5 :Y v 4 :G v 3 :R The greedy algorithm uses 4 colors, even though it is possible to color the graph with 3 How badly can the greedy algorithm perform? The largest degree in a graph is called ; in the graph above is 4 The greedy algorithm will never use more than + 1 colors, and this is quite easy to see Start the greedy algorithm with a list of + 1 colors Whenever it comes time to color a vertex, that vertex has at most neighbors which can use at most colors Thus there is at least one color on the list that can be used to color the next vertex An interesting special case occurs with graphs that can be colored with just 2 colors: a slightly different greedy algorithm always works, because in fact there is essentially only one way to color such a graph with two colors As soon as one vertex is assigned a color, say R, all of its neighbors must get the other color, B, then all neighbors of those vertices must be R, and so on So if it is possible to color the graph with 2 colors, there is no choice about how to do it, and a greedy approach will succeed We do have to make sure that the vertices are ordered so that each is connected to at least one of the previous vertices Try coloring the following graph with 2 colors, for example, by coloring one vertex, then all of its adjacent vertices, and so on 9 The following graph is also 2-colorable, but if the greedy algorithm colors the vertices in the order shown it will fail, because vertex 2 is not adjacent to any vertex before it (namely, vertex 1) The most famous fact about coloring graphs is the Four Color Theorem, first proposed in the late 1800s but not proved until 1977 A planar graph is one that can be drawn on a flat plane so that edges do not touch each other except at their endpoints Here is a planar graph drawn so that its edges cross, and again so they do not Both are the same since both consist of 4 points all connected to each other On the other hand, the complete graph on 5 vertices cannot be drawn without crossing edges, as you can see by trying it for a while The Four Color Theorem says that every planar graph can be colored with 4 colors This is much too difficult to prove (it has been done only with the aid of computers), but it is not too hard to see that all planar graphs can be colored with 5 colors Here is a sketch of how to do it: First, let n be the number of vertices in a planar graph, e the number of edges, and r the number of regions into which the plane is divided by the vertices and edges In the 10

6 graph above drawn without intersecting edges there are 4 regions, three triangles and the infinite region around the graph In the 11 vertex graph before that, there are 5 regions It is always true that n e + r = 2; this is called Euler s formula Each region has some number of edges on its boundary For each of the r regions we count the number of edges on the boundary, and call these numbers f 1, f 2,, f r Each edge has a region on both sides, so each edge is counted twice in this list, so f 1 +f 2 +f f r = 2e Since there are no loops or multiple edges, each f i is at least 3, so This means that 2e/3 r 2e = f 1 + f 2 + f f r = 3r Now write Euler s formula as r = e n + 2 Then 2e/3 r = e n + 2, so 2e/3 e n + 2 We can simplify this by multiplying by 3 and getting the two e s together: e 3n 6; this also means 2e 6n 12 Recall that the sum of the degrees is also twice the number of edges: d 1 +d 2 + +d n = 2e Is it possible that every one of the degrees is 6 or more? If it were possible, then there would be a planar graph in which this is true: 6n d 1 + d d n = 2e 6n 12 But this says that 6n is smaller than 6n 12, which is not true This means that the degrees cannot all be 6 or more Put another way, every planar graph has a vertex of degree less than or equal to 5 Now we re ready to describe how to color any planar graph with 5 colors In the planar graph, find a vertex of degree 1, 2, 3, 4, or 5, and remove it This creates a new planar graph, and it too must have a vertex of degree at most 5; remove that one Continue to remove vertices of small degree until there are only five vertices left, and color those five vertices with 5 colors Now put the removed vertices back one at a time If you replace a vertex of degree 1, 2, 3, or 4, it is connected to at most 4 vertices that use at most 4 colors, so the replaced vertex can be colored with the fifth color If the replaced vertex has degree 5, but its 5 neighbors are colored with 4 or fewer colors, then it is possible to color the replaced vertex with the fifth color The only hard case is if you put back a degree 5 vertex and it is connected to vertices that use all 5 colors it appears that you will need a sixth color Suppose the picture looks like this in the vicinity of the replaced vertex: Ṛ Start at the vertex colored R and look for a path of alternating R and Y vertices that ends at the Y vertex in the diagram If there isn t one, then starting at the R vertex, switch the R to Y, then all neighboring Y s to R, and so on, until the graph is correctly colored again The original Y will still be color Y, so the replaced vertex at the center can be colored R On the other hand, if there is an alternating R-Y path from the R vertex to the Y vertex, then there cannot be an alternating B-P path from the B vertex to the P vertex This means the B can be changed to P and the replaced vertex can be colored B B G Y P 11 12

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