6 Arrange the steps in the correct order to prove that a simple graph T is a tree if and only if it is connected, but the deletion of any of its edges produces a graph that is not connected. (First, prove that if a simple graph 7 is a tree, then it is connected and the deletion of any of its edges produces a graph that is not connected, and then prove the converse of it.) Rank the options below. Now, Twith {x,y) deleted has no path from x to y, since there was only one simple path from x to y in T, and the edge itself was it. Therefore, the graph with (x, y) deleted is not connected. Suppose that a simple connected graph 7 satisfies the condition that the removal of any edge will disconnect it. If T is not a tree, then it has a simple circuit, say, x1, x2. X. xq. If we delete the edge (x₁, x₁) from T, then the graph will remain connected, since wherever the deleted edge was used in forming paths between vertices we can instead use the rest of the circuit x1, x2. .x, or its reverse. Let (x, y) be an edge of Tsuch that x# y. This is a contradiction to the condition. Therefore, our assumption was wrong, and T is a tree. Suppose that T is a tree; then, by definition, it is connected. ▾
6 Arrange the steps in the correct order to prove that a simple graph T is a tree if and only if it is connected, but the deletion of any of its edges produces a graph that is not connected. (First, prove that if a simple graph 7 is a tree, then it is connected and the deletion of any of its edges produces a graph that is not connected, and then prove the converse of it.) Rank the options below. Now, Twith {x,y) deleted has no path from x to y, since there was only one simple path from x to y in T, and the edge itself was it. Therefore, the graph with (x, y) deleted is not connected. Suppose that a simple connected graph 7 satisfies the condition that the removal of any edge will disconnect it. If T is not a tree, then it has a simple circuit, say, x1, x2. X. xq. If we delete the edge (x₁, x₁) from T, then the graph will remain connected, since wherever the deleted edge was used in forming paths between vertices we can instead use the rest of the circuit x1, x2. .x, or its reverse. Let (x, y) be an edge of Tsuch that x# y. This is a contradiction to the condition. Therefore, our assumption was wrong, and T is a tree. Suppose that T is a tree; then, by definition, it is connected. ▾
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Arrange the steps in the correct order to prove that a simple graph T is a tree if and only if it is connected, but the deletion of any of its edges produces a graph that is not connected. (First, prove that if a simple graph T
is a tree, then it is connected and the deletion of any of its edges produces a graph that is not connected, and then prove the converse of it.)
Rank the options below.
Now, Twith {x,y) deleted has no path from x to y, since there was only one simple path from x to y in T, and the edge itself was it. Therefore, the graph with (x, y) deleted is not connected.
Suppose that a simple connected graph 7 satisfies the condition that the removal of any edge will disconnect it.
If T is not a tree, then it has a simple circuit, say, X₁, X2, ..., X₁, X1.
If we delete the edge [xr, x₁) from T, then the graph will remain connected, since wherever the deleted edge was used in forming paths between vertices we can instead use the rest of the
circuit x₁, x2, ..., Xr or its reverse.
Let {x, y) be an edge of Tsuch that x #y.
This is a contradiction to the condition. Therefore, our assumption was wrong, and T is a tree.
Suppose that T is a tree; then, by definition, it is connected.
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