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There are many variants of Rudrata’s problem, depending on whether the graph is undirected or directed, and whether a cycle or path is sought. Reduce the DIRECTED RUDRATA PATH problem to each of the following.(a)The (undirected) RUDRATA PATH problem.(b) The undirected RUDRATA PATH problem, which is just like RUDRATA PATH except that the endpoints of the path are specified in the input.

Short Answer

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2. Enumerating the source and destination solves the given problem.

Step by step solution

01

Explain the RUDTATA PATH problem

Consider a directed or undirected graph, In which there exists a path that visits each vertex exactly once. The path is known as RUDRATA PATH.

02

Reduce undirected RUDTATA PATH problem

Consider the directed graph G, in which each vertex v is categorized into new and old vertices i+o. In-degree vertices are represented as i and the out-degree vertices are represented as o. Consider that every directed edge is connected to the vertex that is transformed intoi+onew undirected edge of the vertex.

Assume that the vertex connected to the incoming edges is called the incoming vertex and the vertex connected to the outgoing edges is called the outgoing vertex. Connect each incoming vertex to undirected edges.

Thus, the graph G is transformed into G'. Consider the following example,

Therefore, the undirected graph is reduced to RUDATA PATH.

03

The undirected RUDRATA (s,t)− PATH problem.

Consider the directed graph G, in which each vertex is categorized into new and old vertices . In-degree vertices are represented as and the out-degree vertices are represented as . Consider that every directed edge is connected to the vertex that is transformed into new undirected edge of the vertex.

Based on the above graph information, enumerate the path as . Consider the vertex source as , and the vertex that ends the cycle is t.

Therefore, enumerating the source and destination solves the given problem.

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Most popular questions from this chapter

|ψ=12|00+12|11is one of the famous “Bell states,” a highly entangled state of its two qubits. In this question we examine some of its strange properties. (a) Suppose this Bell state could be decomposed as the (tensor) product of two qubits (recall the box on page ), the first in state α0|0+α1|1and the second in stateβ0|0+β1|1. Write four equations that the amplitudes α0,α1,β0andβ1must satisfy. Conclude that the Bell state cannot be so decomposed.

(b) What is the result of measuring the first qubit of |ψ?

(c) What is the result of measuring the second qubit after measuring the first qubit? (d) If the two qubits in state|ψ are very far from each other, can you see why the answer to (c) is surprising?

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

F0=0,F1=1,Fn=Fn1+Fn2.

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 Fn20.5nfor n6.

(b) Find a constant c<1such thatFn2cn for all n0. Show that your answer is correct.

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

Alice wants to throw a party and is deciding whom to call. She has n people to choose from, and she has made up a list of which pairs of these people know each other. She wants to pick as many people as possible, subject to two constraints: at the party, each person should have at least five other people whom they know and five other people whom they don’t know. Give an efficient algorithm that takes as input the list of n people and the list of pairs who know each other and outputs the best choice of party invitees. Give the running time in terms of n

Is there a faster way to compute the nth Fibonacci number than by fib2 (page 4)? One idea involves matrices.

We start by writing the equations F1=F1 and F2=F0+F1 in matrix notation:


role="math" localid="1659767046297" (F1F2)=(0111).(F0F1).

Similarly,

F2F3=(0111).(F1F2)=(0111)2.(F0F1)

And in general

(FnFn+1)=(0111)n.(F0F1)

So, in order to compute Fn, it suffices to raise this 2×2 matrix, call it X, to the nth power.

a. Show that two 2×2matrices can be multiplied using 4additions and 8multiplications.

But how many matrix multiplications does it take to compute Xn?

b. Show that O(logn) matrix multiplications suffice for computing Xn. (Hint: Think about computing X8.)

Thus, the number of arithmetic operations needed by our matrix-based algorithm, call it fib3, is just O(logn), as compared to O(n)for fib2. Have we broken another exponential barrier? The catch is that our new algorithm involves multiplication, not just addition; and multiplications of large numbers are slower than additions. We have already seen that, when the complexity of arithmetic operations is taken into account, the running time offib2becomes O(n).

c. Show that all intermediate results of fib3 are O(n) bits long.


d. Let M(n)be the running time of an algorithm for multiplying n-bit numbers, and assume that M(n)=O(n2) (the school method for multiplication, recalled in Chapter 1, achieves this). Prove that the running time of fib3 is O(M(n)logn).


e. Can you prove that the running time of fib3 is O(M(n))? Assume M(n)=Θ(na)for some 1a2. (Hint: The lengths of the numbers being multiplied get doubled with every squaring.)


In conclusion, whether fib3 is faster than fib2 depends on whether we can multiply n-bit integers faster thanO(n2) . Do you think this is possible? (The answer is in Chapter 2.) Finally, there is a formula for the Fibonacci numbers:

role="math" localid="1659768125292" Fn=15(1+52)n15(152)n.

So, it would appear that we only need to raise a couple of numbers to the nth power in order to computeFn . The problem is that these numbers are irrational, and computing them to sufficient accuracy is nontrivial. In fact, our matrix method fib3 can be seen as a roundabout way of raising these irrational numbers to the nth power. If you know your linear algebra, you should see why. (Hint: What are the eigenvalues of the matrix X?)

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?

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