1. Consider a search graph in which every state has exactly 6 successors. Argue that there are exactly 1+6+ 6² + 63 + 6' paths of length at most 4 starting with the start state. In general, if d is a positive integer, then there are 1+6+6²+6³+…· +6ª paths of length at most d starting with the start state. You do not have to prove this, but it should be plausible now. Even more general, if b is the exact number of successors of each state, then there are 1+b+62...+Bd paths of length at most d starting from the start state. Again, you do not have to prove this, but it should be plausible now.
1. Consider a search graph in which every state has exactly 6 successors. Argue that there are exactly 1+6+ 6² + 63 + 6' paths of length at most 4 starting with the start state. In general, if d is a positive integer, then there are 1+6+6²+6³+…· +6ª paths of length at most d starting with the start state. You do not have to prove this, but it should be plausible now. Even more general, if b is the exact number of successors of each state, then there are 1+b+62...+Bd paths of length at most d starting from the start state. Again, you do not have to prove this, but it should be plausible now.
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Transcribed Image Text:1. Consider a search graph in which every state has exactly 6 successors.
Argue that there are exactly 1+6 + 62 + 63 + 6' paths of length at
most 4 starting with the start state.
In general, if d is a positive integer, then there are 1+6+6²+6³+. +6d
paths of length at most d starting with the start state. You do not
have to prove this, but it should be plausible now.
Even more general, if b is the exact number of successors of each state,
then there are 1+b+b?...+ paths of length at most d starting from
the start state. Again, you do not have to prove this, but it should be
plausible now.
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