Q: If g is a simple graph with 15 edges?

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Proving this is simple. First, you prove that G has a spanning tree, it is connected, which is pretty obvious - a spanning tree itself is already a connected graph on the vertex set V(G), thus G which contains it as a spanning sub graph is obviously also connected. Second, you prove that if G is connected, it has a spanning tree. If G is a tree itself, then it must "contain" a spanning tree. If G is connected and not a tree, then it must have at least one cycle. I don't know if you know this or not, but there is a theorem stating that an edge is a cut-edge if and only if it is on no cycle (a cut-edge is an edge such that if you take it out, the graph becomes disconnected). Thus, you can just keep taking out edges from cycles in G until all that is left are cut-gees. Since you did not take out any cut-edges, the graph is still connected; since all that is left are cut-edges, there are no cycles. A connected graph with no cycles is a tree. Thus, G contains a spanning tree. Therefore, a graph G is connected if and only if it has a spanning tree!

Let S be a finite, non-empty set of positive integers. The divisor graph G(S) of S has S as its vertex set, and two distinct vertices i and j are adjacent if and only if either idivides j or j divides i. Let G be a simple graph. Then G is called a divisor graph if G is isomorphic to G(S) for some non-empty, finite set S of positive integers.REFERENCE :S. Ganesan, D. Uthayakumar, Corona of Bipartite Graphs with Divisor GRaphs Produce New Divosor Graphs,Bulletin of Kerala Mathematics AssociationVol.9, No.1, (2012, June) 219-226

cyclomatic number of a graph is e.n+1 where e is number of edge of graph and n is number of node in graoh g

1 kg = 1,000 g so 320 kg = 320*1000 = 320,000 g. Simple!1 kg = 1,000 g so 320 kg = 320*1000 = 320,000 g. Simple!1 kg = 1,000 g so 320 kg = 320*1000 = 320,000 g. Simple!1 kg = 1,000 g so 320 kg = 320*1000 = 320,000 g. Simple!

graph G(x)=[x]-1

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n-k-1

Sparse vs. Dense GraphsInformally, a graph with relatively few edges is sparse, and a graph with many edges is dense. The following definition defines precisely what we mean when we say that a graph ``has relatively few edges'': Definition (Sparse Graph) A sparse graph is a graph in which .For example, consider a graph with n nodes. Suppose that the out-degree of each vertex in G is some fixed constant k. Graph G is a sparse graph because .A graph that is not sparse is said to be dense:Definition (Dense Graph) A dense graph is a graph in which .For example, consider a graph with n nodes. Suppose that the out-degree of each vertex in G is some fraction fof n, . E.g., if n=16 and f=0.25, the out-degree of each node is 4. Graph G is a dense graph because .

Proving this is simple. First, you prove that G has a spanning tree, it is connected, which is pretty obvious - a spanning tree itself is already a connected graph on the vertex set V(G), thus G which contains it as a spanning sub graph is obviously also connected. Second, you prove that if G is connected, it has a spanning tree. If G is a tree itself, then it must "contain" a spanning tree. If G is connected and not a tree, then it must have at least one cycle. I don't know if you know this or not, but there is a theorem stating that an edge is a cut-edge if and only if it is on no cycle (a cut-edge is an edge such that if you take it out, the graph becomes disconnected). Thus, you can just keep taking out edges from cycles in G until all that is left are cut-gees. Since you did not take out any cut-edges, the graph is still connected; since all that is left are cut-edges, there are no cycles. A connected graph with no cycles is a tree. Thus, G contains a spanning tree. Therefore, a graph G is connected if and only if it has a spanning tree!

Proving this is simple. First, you prove that G has a spanning tree, it is connected, which is pretty obvious - a spanning tree itself is already a connected graph on the vertex set V(G), thus G which contains it as a spanning sub graph is obviously also connected. Second, you prove that if G is connected, it has a spanning tree. If G is a tree itself, then it must "contain" a spanning tree. If G is connected and not a tree, then it must have at least one cycle. I don't know if you know this or not, but there is a theorem stating that an edge is a cut-edge if and only if it is on no cycle (a cut-edge is an edge such that if you take it out, the graph becomes disconnected). Thus, you can just keep taking out edges from cycles in G until all that is left are cut-gees. Since you did not take out any cut-edges, the graph is still connected; since all that is left are cut-edges, there are no cycles. A connected graph with no cycles is a tree. Thus, G contains a spanning tree. Therefore, a graph G is connected if and only if it has a spanning tree!

A Compound Graph is an extension of a standard graph. Let G be a graph, G=(V,E) where V is a set of vertices and E is a set of edges, that is e = (v1, v2) in V2 A compound graph C is defined by a tree T=(V,F) where V is the same set as G and F are tree edges f=(v1,v2) in V2. C=(G,T) where G=(V,E) and T=(V,F) Furthermore, C has two additional constraints: e=(v1,v2) in E implies: 1) v1 is not on the path of v2 to the root of T AND 2) v2 is not on the path of v1 to the root of T. Intuitively, T defines a hierarchy. All the vertices sharing the same parent in T are in the same "group". The constraints state that you cannot have an edge connecting a vertex to one of its parent in the hierarchy.

The graph of g(x) is the graph of f(x) shifted 6 units in the direction of positive x.

ok here we go...Proof:If the some graph G has the same DFS and BFS then that means that G should not have any cycle(work out for any G with a cycle u will never get the same BFS and DFS .... and for a graph without any cycle u will get the same BFS/DFS).We will prove it by contradiction:So say if T is the tree obtained by BFS/DFS, and let us assume that G has atleast one edge more than T. So one more edge to T(T is a tree) would result in a cycle in G, but according to the above established principle no graph which has a cycle would result the same DFS and BFS, so out assumption is a contradiction.Hence G should have more edges than T, which implies that if the BFS and DFS for a graph G are the same then the G = T.Hope this helps u......................

g

Why

Graph that equation. If the graph pass the horizontal line test, it is an inverse equation (because the graph of an inverse function is just a symmetry graph with respect to the line y= x of a graph of a one-to-one function). If it is given f(x) and g(x) as the inverse of f(x), check if g(f(x)) = x and f(g(x)) = x. If you show that g(f(x)) = x and f(g(x)) = x, then g(x) is the inverse of f(x).

Let G be a complete graph with n vertices. Consider the case where n=2. With only 2 vertices it is clear that there will only be one edge. Now add one more vertex to get n = 3. We must now add edges between the two old vertices and the new one for a total of 3 vertices. We see that adding a vertex to a graph with n vertices gives us n more edges. We get the following sequence Edges on a graph with n vertices: 0+1+2+3+4+5+...+n-1. Adding this to itself and dividing by two yields the following formula for the number of edges on a complete graph with n vertices: n(n-1)/2.

Let S be a finite, non-empty set of positive integers. The divisor graph G(S) of S has S as its vertex set, and two distinct vertices i and j are adjacent if and only if either idivides j or j divides i. Let G be a simple graph. Then G is called a divisor graph if G is isomorphic to G(S) for some non-empty, finite set S of positive integers.REFERENCE :S. Ganesan, D. Uthayakumar, Corona of Bipartite Graphs with Divisor GRaphs Produce New Divosor Graphs,Bulletin of Kerala Mathematics AssociationVol.9, No.1, (2012, June) 219-226