/*! 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} Q24E A binary counter of unspecified ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A binary counter of unspecified length supports two operations: increment (which increases its value by one) and reset (which sets its value back to zero). Show that, starting from an initially zero counter, any sequence of n increment and reset operations takes time O(n); that is, the amortized time per operation is O(1) .

Short Answer

Expert verified

Binary counter is supports unclear length of two type of operations. On that basses increment and reset (1 & 0) is happen. It’s starts from zero counter and continue in sequence base on n increment & reset operations.

Step by step solution

01

Use of taxation technique

Be using the taxation technique of amortised analysis to establish approximatelybinary increase & resets operations on a binary counter.

This taxation approach applies taxes solely to specific operations, ensuring that the overall cost of those operations does not exceed the tax paid. The tax paid by those operations is the amortised cost.

02

Detail of taxation approach technique.

Description:

Always one bit gets coded as 1 when a counter is incremented.

• When change any bit of something like the binary counters to 1 , set amortised cost to$2.

• Switch it to 1 using $1and preserve some other$1for future reset operations.

• As a result, whenever a bit is set to 1 , a$1credit is provided.

• Assume the counter is reset after

1k =" " p =" "style = "box - sizing : border - box;">/k=" " style = "box - sizing : border - box;" >

• As a result, the reset procedure always has enough money to return the counter to zero.

• As a result, O(n) time will be required for n increment and reset operations.

As a result, the average cost per operation will be O(1) .

03

Final Conclusion.

1 bit is get coded as 1 when computer counting ingormation is increase. And binary counter is 1 when set amortised cost is $2 . So, as a result the reset procedure always has enough money to way back the counter till zero.

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 91Ó°ÊÓ!

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

Suppose you are given a weighted graph G=(V,E) with a distinguished vertex s and where all edge weights are positive and distinct. Is it possible for a tree of shortest paths from s and a minimum spanning tree in G to not share any edges? If so, give an example. If not, give a reason.

A long string consists of the four characters A,C,G,T ; they appear with frequency 31%,20%,9%and40% respectively. What is the Huffman encoding of these four characters?

Entropy: Consider a distribution overnpossible outcomes, with probabilities p1,p2,K,pn.

a. Just for this part of the problem, assume that each piis a power of 2 (that is, of the form 1/2k). Suppose a long sequence of msamples is drawn from the distribution and that for all 1≤i≤n, the ithoutcome occurs exactly times in the sequence. Show that if Huffman encoding is applied to this sequence, the resulting encoding will have length

∑i-1nmpilog1pi

b. Now consider arbitrary distributions-that is, the probabilities pi are noy restricted to powers of 2. The most commonly used measure of the amount of randomness in the distribution is the entropy.

∑i-1nmpilog1pi

For what distribution (over outcomes) is the entropy the largest possible? The smallest possible?

Give a linear-time algorithm that takes as input a tree and determines whether it has a perfect matching: a set of edges that touches each node exactly once.

A feedback edge set of an undirected graph G(V,E) is a subset of edgesE'⊂Ethat intersects every cycle of the graph. Thus, removing the edges will render the graph acyclic.

Give an efficient algorithm for the following problem:

Input: Undirected graph G(V,E) with positive edge weights we.

Output: A feedback edge set E'⊂Eminimum total weight ∑e∈E'we.

Let T be an MST of graph G. Given a connected subgraph H of G, show that T∩H is contained in some MST of H

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.