/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q27E 聽Alice wants to throw a party a... [FREE SOLUTION] | 91影视

91影视

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鈥檛 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

Short Answer

Expert verified

Each vertex with each member inside the vertex set, such asuandv , indicates that the person v knows the person u . So, find the subset v where another vertex has a value greater than 5 , then repeat the process with both the modified degrees for every node until a network is formed in which neither vertex may be destroyed after the procedure is finished. As a result, the running time's total complexity has increased.

Step by step solution

01

Individual String ‘Fi’

Assuming that each individual attends the party with at least 5 different other things they trust and five others they don't. Throughout a few places in the question, every string 'fi' has been substituted with '?'. If you have any doubts about my answer, please leave a remark.

02

Vertex for Each Set

Assume the graph GV,E, which has a vertex for each person in the vertex set. If an edge e connects the vertices uandv, , it means that the person v is acquainted with the person u . So, in the induced network [where V ' is the new set of vertices], we may restrict the task to identifying a subset V of V whereby each vertex has a value more than 5 and less than|V'|-5 .

The continuous strategy can be used, in which all nodes are examined at first, and any node with a degree more than role="math" localid="1658920458569" |V'|-5orlessthan5 is eliminated. This graph is changed to be the induced graph, with the remaining vertices being assigned to the vertex set. This technique is repeated with the updated degrees for each node until a graph is formed in which no vertex may be deleted once the operation is completed. G' is the symbol for this graph. There are n iterations in the algorithm, and each iteration takes On time. As a result, the overall complexity of the running time is role="math" localid="1658920588246" O(n2).

03

Conclusion  

A vertex for each person in the vertex set like u and v means the person v is acquainted with the person u. So, identifying of sub set v where each vertex has value more than 5. same way this technique is repeated with the updated degrees for each node until a graph is form in which no vertex may deleted once the operation is complete. As a result overall complexity of the running time.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the following game. A 鈥渄ealer鈥 produces a sequence s1sn of 鈥渃ards,鈥 face up, where each card si has a value vi. Then two players take turns picking a card from the sequence, but can only pick the first or the last card of the (remaining) sequence. The goal is to collect cards of largest total value. (For example, you can think of the cards as bills of different denominations.) Assume n is even. (a) Show a sequence of cards such that it is not optimal for the first player to start by picking up the available card of larger value. That is, the natural greedy strategy is suboptimal. (b) Give an O(n2) algorithm to compute an optimal strategy for the first player. Given the initial sequence, your algorithm should precompute in O(n2) time some information, and then the first player should be able to make each move optimally in O(1) time by looking up the precomputed information.

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

For instance, the constraints.

x1=x2,x2=x3,x3=x4,x1x4

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

A vertex cover of a graph G=(V,E)is a subset of vertices SVthat includes at least one endpoint of every edge in E. Give a linear-time algorithm for the following task.

Input: An undirected tree T=(V,E).

Output: The size of the smallest vertex cover of T. For instance, in the following tree, possible vertex covers include{A,B,C,D,E,F,G}and{A,C,D,F}but not{C,E,F}.The smallest vertex cover has size 3: {B,E,G}.

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 22 matrix, call it X, to the nth power.

a. Show that two 22matrices 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)=(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?

See all solutions

Recommended explanations on Computer Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.