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Chapter 0: 12 E (page 11)

The k−SPANNING TREE problem is the following.Input: An undirected graph G=(V,E) Output: A spanning tree of G in which each node has degree ≤k, if such a tree exists.Show that for any k≥2:

  1. k− SPANNING TREE is a search problem.
  2. k− SPANNING TREE is NP-complete. (Hint: Start with k=2 and consider the relation between this problem and RUDRATA PATH.)

Short Answer

Expert verified

1.k− SPANNING TREE is a search problem for any k≥2.

2.k−SPANNING TREE is NP-complete.

Step by step solution

01

Explain Spanning tree

Consider that the spanning tree is a subset of a Graph G that covers all of the vertices with the fewest number of edges feasible. It can be deduced from this definition that every linked and undirected Graph Gcontains at least one spanning tree

02

To prove that k−  SPANNING TREE is a search problem

Consider the given input and output with k≥2.

Here, it is important to demonstrate that given a solution S to the spanning tree problem that can be checked in polynomial time whether it is in fact a k-spanning tree. This comments to verifying that every node in the original graph is used in S such that S have no cycle because it is a tree.

Every node in the tree has a maximum degree k . All of these can be checked efficiently and therefore the k− spanning tree is a search problem.

Therefore, it can be concluded that for k≥2, the k−spanning tree is a search problem.

03

To prove that k -SPANNING TREE is NP-Complete problem

Any of a class of computer problems for which no efficient solution algorithm has been developed is known as an NP-complete issue.From part (a) it is known that the k−spanning tree is in NP.

In the Rudrata path algorithm, assume G is an unweighted undirected graph. Add weights equal to 1 on every edge of G while executing the Rudrata path algorithm with k=2 .It is observed that a tree that has each vertex with a degree at most 2 is a path. Hence, there is no path without loops that reaches all the vertices so there will be no Rudrata path.

Therefore, it can be concluded that the Rudrata path is reduced to a k−spanning tree along with the fact that the k−spanning tree is in NP.

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Most popular questions from this chapter

Mean and median. One of the most basic tasks in statistics is to summarize a set of observations x1,x2,…,xn⊆R by a single number. Two popular choices for this summary statistic are:

• The median, which we’ll callμ1

• The mean, which we’ll callμ2

(a) Show that the median is the value of μthat minimizes the function

∑i|xi-μ|

You can assume for simplicity that is odd. (Hint: Show that for any , the function decreases if you move either slightly to the left or slightly to the right.)

(b) Show that the mean is the value of μ that minimizes the function

∑i(xi-μ)2

One way to do this is by calculus. Another method is to prove that for any μ∈R,

∑i(xi-μ)2=∑i(xi-μ2)2+n(μ-μ2)2

Notice how the function for μ2 penalizes points that are far from much more heavily than the function for μ1 . Thus μ2 tries much harder to be close to all the observations. This might sound like a good thing at some level, but it is statistically undesirable because just a few outliers can severely throw off the estimate of μ2 . It is therefore sometimes said that μ1 is a more robust estimator than μ2 . Worse than either of them, however, is μ∞ , the value of μthat minimizes the function

maxi|xi-μ|

(c) Show that μ∞ can be computed in O(n) time (assuming the numbers are xismall enough that basic arithmetic operations on them take unit time).

Here’s a problem that occurs in automatic program analysis. For a set of variablesx1,......,xn, you are given some equality constraints, of the form “ xi=xj” and some disequality constraints, of the form “ xi≠xj.” Is it possible to satisfy all of them?

For instance, the constraints.

x1=x2,x2=x3,x3=x4,x1≠x4

cannot be satisfied. Give an efficient algorithm that takes as input m constraints over n variables and decides whether the constraints can be satisfied.

Question: 0.1. In each of the following situations, indicate whether f=O(g),orf=Ω(g),or both (in which case f=⊙(g))

Question: An Eulerian tourin an undirected graph is a cycle that is allowed to pass through each vertex multiple times, but must use each edge exactly once.

This simple concept was used by Euler in to solve the famous Konigsberg bridge problem, which launched the field of graph theory. The city of Konigsberg (now called Kaliningrad, in western Russia) is the meeting point of two rivers with a small island in the middle. There are seven bridges across the rivers, and a popular recreational question of the time was to determine whether it is possible to perform a tour in which each bridge is crossed exactly once. Euler formulated the relevant information as a graph with four nodes (denoting land masses) and seven edges (denoting bridges), as shown here.

Notice an unusual feature of this problem: multiple edges between certain pairs of nodes.

(a) Show that an undirected graph has an Eulerian tour if and only if all its vertices have even degree. Conclude that there is no Eulerian tour of the Konigsberg bridges.

(b) An Eulerian pathis a path which uses each edge exactly once. Can you give a similar if-and-only-if characterization of which undirected graphs have Eulerian paths?

(c) Can you give an analog of part (a) for directedgraphs?

The Fibonacci numbers F0,F1,F2,... are defined by the rule

F0=0,F1=1,Fn=Fn−1+Fn−2.

In this problem we will confirm that this sequence grows exponentially fast and obtain some bounds on its growth.

(a) Use induction to prove that Fn≥20.5nfor n≥6.

(b) Find a constant c<1such thatFn≥2cn for all n≥0. Show that your answer is correct.

(c) What is the largestc you can find for which Fn=Ω(2cn)?
See all solutions

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