Homework Answers
Clearly it is true for n=2, since a tree having 2 vertices have both the vertices of degree one.
Let it be true for n =k, i.e. a tree having k vertices have atleast two vertices of degree one.
Now, for n =k+1, we have to add one more vertex to tree having k vertices.
Case 1: if we add (k+1)th vertex to a vertex having degree one then (k+1)th vertex will have degree one and we will end up having atleast two vertices of degree one.
Case 2: if we add (k+1)th vertex to any vertex having degree not equal two one, then we already have atleast 2 vertices of degree one( since statement is true for n=k).
Therefore, it is true for n=k+1, and hence by principle of mathematical induction, every tree having vertices greater than or equal to two vertices will have atleast two vertices of degree one.
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Suppose a graph has 7 vertices, each of degree 2 or 3. Is the graph
necessarily connected ?
b) Now the graph has 7 vertices, each degree 3 or 4. Is it
necessarily connected?
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square are graph which are not connected yet each vertex has degree
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Can you draw the tree diagram for this please 12. Let T be a tree with 8 edges that has exactl 5 vertices of degree 1Pro...
Can
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1. Use Pigeon hole principle to prove that any graph with at
least 2 vertices contains two vertices of the same degree. (Hint:
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necessarily connected ?
b) Now the graph has 7 vertices, each degree 3 or 4. Is it
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My professor gave an example in class. He said triangle and a
square are graph which are not connected yet each vertex has degree
2.
(Paul Zeitz, The Art and Craft of Problem...
Can
you draw the tree diagram for this please
12. Let T be a tree with 8 edges that has exactl 5 vertices of degree 1Prove that if v is a vertex of maximum degree in T, then 3 < deg(v) < 5
12. Let T be a tree with 8 edges that has exactl 5 vertices of degree 1Prove that if v is a vertex of maximum degree in T, then 3
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Prove by contradiction. (4 points)
2. Given (6 Points)
a. Prove the above equation using binomial theorem. (3
Points)
b. Give a combinatorial proof for the given equating. (3
Points)
4n = (0)2" + (1)2" +...+)2"-
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