WYSE Academic Challenge Sectional Computer Science 2005 SOLUTION SET

Similar documents
Basic allocator mechanisms The course that gives CMU its Zip! Memory Management II: Dynamic Storage Allocation Mar 6, 2000.

Security of Bluetooth: An overview of Bluetooth Security

Message Integrity and Hash Functions. TELE3119: Week4

Announcements. Reading. Project #4 is on the web. Homework #1. Midterm #2. Chapter 4 ( ) Note policy about project #3 missing components

Copyright 2016 Ramez Elmasri and Shamkant B. Navathe

2005 Academic Challenge

CMSC Computer Architecture Lecture 12: Virtual Memory. Prof. Yanjing Li University of Chicago

CSC 220: Computer Organization Unit 11 Basic Computer Organization and Design

. Written in factored form it is easy to see that the roots are 2, 2, i,

Computer Science Foundation Exam. August 12, Computer Science. Section 1A. No Calculators! KEY. Solutions and Grading Criteria.

Elementary Educational Computer

COSC 1P03. Ch 7 Recursion. Introduction to Data Structures 8.1

Python Programming: An Introduction to Computer Science

Python Programming: An Introduction to Computer Science

Τεχνολογία Λογισμικού

Operating System Concepts. Operating System Concepts

Morgan Kaufmann Publishers 26 February, COMPUTER ORGANIZATION AND DESIGN The Hardware/Software Interface. Chapter 5.

Chapter 1. Introduction to Computers and C++ Programming. Copyright 2015 Pearson Education, Ltd.. All rights reserved.

Lecture 28: Data Link Layer

Math Section 2.2 Polynomial Functions

Deadlocks Detection and Avoidance. Prof. Sirer CS 4410 Cornell University

The Magma Database file formats

Copyright 2016 Ramez Elmasri and Shamkant B. Navathe

Hash Tables. Presentation for use with the textbook Algorithm Design and Applications, by M. T. Goodrich and R. Tamassia, Wiley, 2015.

Homework 1 Solutions MA 522 Fall 2017

Appendix D. Controller Implementation

Lower Bounds for Sorting

EE 459/500 HDL Based Digital Design with Programmable Logic. Lecture 13 Control and Sequencing: Hardwired and Microprogrammed Control

How do we evaluate algorithms?

9.1. Sequences and Series. Sequences. What you should learn. Why you should learn it. Definition of Sequence

Polynomial Functions and Models. Learning Objectives. Polynomials. P (x) = a n x n + a n 1 x n a 1 x + a 0, a n 0

Lecture 2. RTL Design Methodology. Transition from Pseudocode & Interface to a Corresponding Block Diagram

Java Inheritance. Class ADT (Abstract Data Type) Interface. Classes implement the concept of ADT: Interfaces define interaction contracts: Rui Moreira

Chapter 9. Pointers and Dynamic Arrays. Copyright 2015 Pearson Education, Ltd.. All rights reserved.

Recursive Procedures. How can you model the relationship between consecutive terms of a sequence?

Architectural styles for software systems The client-server style

CIS 121 Data Structures and Algorithms with Java Spring Stacks, Queues, and Heaps Monday, February 18 / Tuesday, February 19

Probability of collisions in Soft Input Decryption

Improvement of the Orthogonal Code Convolution Capabilities Using FPGA Implementation

CS200: Hash Tables. Prichard Ch CS200 - Hash Tables 1

Copyright 2016 Ramez Elmasri and Shamkant B. Navathe

Chapter 11. Friends, Overloaded Operators, and Arrays in Classes. Copyright 2014 Pearson Addison-Wesley. All rights reserved.

Reliable Transmission. Spring 2018 CS 438 Staff - University of Illinois 1

Chapter 4. Procedural Abstraction and Functions That Return a Value. Copyright 2015 Pearson Education, Ltd.. All rights reserved.

SERIAL COMMUNICATION INTERFACE FOR ESA ESTRO

Τεχνολογία Λογισμικού

Chapter 10. Defining Classes. Copyright 2015 Pearson Education, Ltd.. All rights reserved.

Chapter 3 Classification of FFT Processor Algorithms

Sorting in Linear Time. Data Structures and Algorithms Andrei Bulatov

Lecture 1: Introduction and Strassen s Algorithm

End Semester Examination CSE, III Yr. (I Sem), 30002: Computer Organization

Extending The Sleuth Kit and its Underlying Model for Pooled Storage File System Forensic Analysis

6.854J / J Advanced Algorithms Fall 2008

An Improvement of the Basic El-Gamal Public Key Cryptosystem

Ones Assignment Method for Solving Traveling Salesman Problem

Evaluation scheme for Tracking in AMI

Chapter 4 The Datapath

Solution printed. Do not start the test until instructed to do so! CS 2604 Data Structures Midterm Spring, Instructions:

Arithmetic Sequences

Copyright 2016 Ramez Elmasri and Shamkant B. Navathe

Civil Engineering Computation

AN EFFICIENT GROUP KEY MANAGEMENT USING CODE FOR KEY CALCULATION FOR SIMULTANEOUS JOIN/LEAVE: CKCS

CIS 121 Data Structures and Algorithms with Java Spring Stacks and Queues Monday, February 12 / Tuesday, February 13

Chapter 3. Floating Point Arithmetic

Media Access Protocols. Spring 2018 CS 438 Staff, University of Illinois 1

IMP: Superposer Integrated Morphometrics Package Superposition Tool

n We have discussed classes in previous lectures n Here, we discuss design of classes n Library design considerations

Throughput-Delay Scaling in Wireless Networks with Constant-Size Packets

Running Time ( 3.1) Analysis of Algorithms. Experimental Studies. Limitations of Experiments

Analysis of Algorithms

Multiprocessors. HPC Prof. Robert van Engelen

The isoperimetric problem on the hypercube

1.2 Binomial Coefficients and Subsets

Weston Anniversary Fund

3D Model Retrieval Method Based on Sample Prediction

3.1 Overview of MySQL Programs. These programs are discussed further in Chapter 4, Database Administration. Client programs that access the server:

Chapter 4 Threads. Operating Systems: Internals and Design Principles. Ninth Edition By William Stallings

GE FUNDAMENTALS OF COMPUTING AND PROGRAMMING UNIT III

Running Time. Analysis of Algorithms. Experimental Studies. Limitations of Experiments

MOTIF XF Extension Owner s Manual

What are we going to learn? CSC Data Structures Analysis of Algorithms. Overview. Algorithm, and Inputs

Cubic Polynomial Curves with a Shape Parameter

Message Authentication Codes. Reading: Chapter 4 of Katz & Lindell

Data Structures and Algorithms. Analysis of Algorithms

Lecturers: Sanjam Garg and Prasad Raghavendra Feb 21, Midterm 1 Solutions

Examples and Applications of Binary Search

IS-IS in Detail. ISP Workshops

DEFINITION OF CELL BEHAVIOUR. Actions and Behaviour. CELL = a CELL CELL = b CELL

MapReduce and Hadoop. Debapriyo Majumdar Data Mining Fall 2014 Indian Statistical Institute Kolkata. November 10, 2014

Data Structures Week #5. Trees (Ağaçlar)

Heaps. Presentation for use with the textbook Algorithm Design and Applications, by M. T. Goodrich and R. Tamassia, Wiley, 2015

Analysis Metrics. Intro to Algorithm Analysis. Slides. 12. Alg Analysis. 12. Alg Analysis

Morgan Kaufmann Publishers 26 February, COMPUTER ORGANIZATION AND DESIGN The Hardware/Software Interface. Chapter 5

Panel for Adobe Premiere Pro CC Partner Solution

The CCITT Communication Protocol for Videophone Teleconferencing Equipment

Term Project Report. This component works to detect gesture from the patient as a sign of emergency message and send it to the emergency manager.

Big-O Analysis. Asymptotics

Data diverse software fault tolerance techniques

CMSC Computer Architecture Lecture 2: ISA. Prof. Yanjing Li Department of Computer Science University of Chicago

Abstract. Chapter 4 Computation. Overview 8/13/18. Bjarne Stroustrup Note:

Transcription:

WYSE Academic Challege Sectioal Computer Sciece 2005 SOLUTION SET 1. Correct aswer: a. Hz = cycle / secod. CPI = 2, therefore, CPI*I = 2 * 28 X 10 8 istructios = 56 X 10 8 cycles. The clock rate is 56 X 10 8 cycles / 7 secods = 8 X 10 8 Hz (= cycles / secod) = 800 X 10 6 MHz 2. Correct aswer: e. The CRC is oe of the most used techiques for error detectio i data commuicatios, it is used i modems, it is a type of hash fuctios based o polyomials. 3. Correct aswer: a. NOT(NOT(X OR Y) OR NOT(X OR NOT Y)) = (X OR Y) AND (X OR NOT Y). Clearly this is equal to the value of X. 4. Correct aswer: a. Iheritace - Orgaizes ad facilitates polymorphism ad ecapsulatio by permittig objects to be defied ad created that are specialized types of already-existig objects - these ca share (ad exted) their behavior without havig to reimplemet that behavior. This is typically doe by groupig objects ito classes, ad defiig classes as extesios of existig classes, thus ad groupig classes ito trees or lattices reflectig behavioral commoality Polymorphism - Refereces to ad collectios of objects may refer to objects of differet types, ad ivokig a operatio o a referece will produce behavior depedig o the actual type of the referet. Ecapsulatio - Also called iformatio hidig: Esures that objects caot chage the iteral state of other objects i uexpected ways; oly the object's ow iteral methods are allowed to access its state. Each type of object exposes a iterface to other objects that specifies how other objects may iteract with it. This prevets users from breakig the ivariats of the program. Abstractio - The ability for a program to igore some aspects of the iformatio it's maipulatig, i.e. the ability to focus o the essetial. Each object i the system serves as a model of a abstract "actor" that ca perform work, report o ad chage its state, ad "commuicate" with other objects i the system, without revealig how these features are implemeted. Processes, fuctios or methods may also be so abstracted, ad whe they are, a variety of techiques are required to exted a abstractio: 5. Correct aswer: e. All the statemets are true for XML.

Computer Sciece 6. Correct aswer: a. i f() = ( k + 1) = (k+1)((k+1) -1)/k = (1+1/k)(k+1) (1+1/k) i= 1 Here k is costat ad k is a iteger greater tha 1. Igorig the costat terms i (1+1/k), yields f() = ( k + 1) = Θ ((1 + 1/ k )( k + 1) ) = Θ (( k + 1) ) i= 1 Formally, we take the costats c 1 = 1, c 2 = 1+1/k, 0 = 1, 0<= c 1 *(k+1) (1+1/k)(k+1) (1+1/k) c 2 *(k+1) for all 0 Therefore, f() = Θ (( k + 1) ) By def. f() = T(g()) iff f() = O(g()) ad g() = O(f())). 7. Correct aswer: b. After 1 st roud, X=4, Y=2; After 2 d roud, X=20, Y=3; After 4 th roud, X=2 20 + 8 > 1000, Y=4 8. Correct aswer: c. Postorder traversal: leftchild, rightchild, root 9. Correct aswer: d. For hold ad wait: To prevet deadlock avoid hold ad wait. Need to be sure a process does ot hold oe resource while requestig aother. For mutual exclusio: oly oe process may use a resource at a time. To prevet deadlock do ot require mutual exclusio. For pagig: Pagig has othig to do with deadlock. It does ot fall uder the coditios for deadlock. For preemptio: To prevet deadlock allow preemptio. If a process holdig certai resources is deied a further request, that process must release its origial resources. For circular wait: To prevet deadlock do t go ito a circular wait situatio! 10. Correct aswer: a. Loss-less joi is oly performed o primary key ad foreig key attributes. Joi o ay other commo attributes will geerate spurious tuples. 11. Correct aswer: b. (NOT ((X OR Y) AND ((NOT (Y AND Z)) AND X))) =(NOT ((T OR F) AND ((NOT (F AND F)) AND T))) =(NOT (T AND ((NOT F) AND T))) =(NOT (T AND (T AND T))) =(NOT (T AND T)) =(NOT T) =FALSE

Computer Sciece 12. Correct aswer: d. Each task is assumed to sped 1 uit of time. For four processors system S, Graph 4, T1 ad T2, T3 ad T4, T5 ad T6, T7 ad T8 ca be executed o the four processes simultaeously. Total time is 2 uits. Graph 1: time is 8 uits. Graph 2: time is 5 uits. Graph 3: time is 4 uits. Graph 5: time is 4 uits. 13. Correct aswer: c. CPU register is the fastest memory. The order is CPU register, Cache, Ram, Magetic disk, Magetic Tape. 14. Correct aswer: b. 101001111 = 14F, as takig from the right 4 bits each, 1111=F, 0100=4, 1=1 15. Correct aswer: d. The seek time is, time eeded to mechaically positio the read/write head o the correct track. 16. Correct aswer: c. Graph is used for layig electric cable or to put cable i house. Graph helps to fid differet poits, for example to fid earest poits, or to cover all the poits. 17. Correct aswer: b. QUEUE is First i first out (FIFO). Put 6 i queue S, remove 6 from queue S, put 2 i queue T, put 9 i queue 1 2 3 4 5 6 2 9 2 5 9 2 5 9 6 S S T T T T 7 8 9 3 1 3 4 5 9 S S T 18. Correct aswer: d. FAT is file allocatio table, which idicates the physical locatio of hard disk (secodary storage). 19. Correct aswer: a. Job cotrol laguage (JCL) is used for iteractio amog the hardware. For example: Assembly laguage. It is ot java, c or fortra.

20. Correct aswer: c. Computer Sciece 21. Correct aswer: e. Add(4) = 4 + Add(3) + Add(2) Add(3) = 3 + Add(2) + Add(1) Add(2) = 2 + Add(1) + Add(0) Add(0) = 0 Add(1) = 0 Add(2) = 2 Add(3) = 5 Add(4) = 11 22. Correct aswer: e. I sequece e, K would be iserted prior to P, hece it would ot be a child of P, but rather P would be a child of K. 23. Correct aswer: e.

First build the biary tree. 15 / \ 12 27 / \ / \ 3 14 18 33 \ 10 Computer Sciece Postorder traversal: leftchild, rightchild, root. 24. Correct aswer: c. Assume A=11, B=10, C=10, (A>B) && (C<=B) = TRUE I,!(A<B) &&!(C>=B) = FALSE II, (A>B) && (B>C) = FALSE III,!((A<=B) (B<C)) = TRUE 25. Correct aswer: c. Whe >1, Strage()=Strage(-2)+1=Strage(-2*2)+1*2=Strage(- 2*3)+1*3= =Strage(-2*/2)+1*/2=Strage(0)+/2=1+/2 26. Correct aswer: c. TDMA: Time Divisio Multiple Access, a techology for deliverig digital wireless service usig time-divisio multiplexig (TDM). TDMA works by dividig a radio frequecy ito time slots ad the allocatig slots to multiple calls. I this way, a sigle frequecy ca support multiple, simultaeous data chaels. TDMA is used by the GSM digital cellular system. FDMA: Frequecy Divisio Multiple Access, is the oldest ad most importat of the three mai ways for multiple radio spectrum. CDMA: Short for Code-Divisio Multiple Access, a digital cellular techology that uses spreadspectrum techiques. GSM: Short for Global System for Mobile Commuicatios, oe of the leadig digital cellular systems. GSM was first itroduced i 1991. As of the ed of 1997, GSM service was available i more tha 200 coutries ad has become the de facto stadard i Europe ad Asia. Referece: http://www.gsmworld.com/techology/gsm.shtmlhttp://www.gsmworld.com/techology/g sm.shtml 27. Correct aswer: d. The cocept of public-key cryptography was itroduced i 1976 by Whitfield Diffie ad Marti Hellma i order to solve the key maagemet problem. I their cocept, each

Computer Sciece perso gets a pair of keys, oe called the public key ad the other called the private key. Each perso s public key is published while the private key is kept secret. The eed for the seder ad receiver to share secret iformatio is elimiated; all commuicatios ivolve oly public keys, ad o private key is ever trasmitted or shared. No loger is it ecessary to trust some commuicatios chael to be secure agaist eavesdroppig or betrayal. The oly requiremet is that public keys are associated with their users i a trusted (autheticated) maer (for istace, i a trusted directory). Ayoe ca sed a cofidetial message by just usig public iformatio, but the message ca oly be decrypted with a private key, which is i the sole possessio of the iteded recipiet. Furthermore, public-key cryptography ca be used ot oly for privacy (ecryptio), but also for autheticatio (digital sigatures). 28. Correct aswer: d. I is true sice the graph is directed graph ad oe edge is oly represeted with oe ode s adjacecy list. So the sum of legths of all the adjacecy lists should be E II is wrog sice the sum of legths of all the adjacecy lists is E from above reaso, ot 2 E III is true sice the graph G is a fiite directed acyclic graph. If G has ot ay vertex with o icomig edge, that is all vertices must have icomig edge, G must be a cyclic graph. 29. Correct aswer: c. The output of left-top: (NOT A) AND (NOT B) The output of left-bottom: NOT (A AND C AND (NOT D)) 30. Correct aswer: d. 10*16+3=163 7*8+7=63 1*2 7 +1*2 5 +1*2 3 +1*2 1 =128 +32+8+2=170 10*16+15=175 99