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CE 221Data Structures & Algorithms

Chapter 1 Introduction

Text: Read Weiss, §1.1 – 1.3

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Course Policy

• Syllabus• Grading• Labs always in C++ (and/or C)

programming language• Each assignment starts and ends in the same

Lab session. Late assignments will not be accepted.

• Study hard!

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Introduction

• See that how a program performs for reasonably large input is just as important as its performance on moderate amounts of input

• Summarize basic mathematical background needed

• Review recursion

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Motivating Examples: Selection• Selection problem: you have a group of N numbers and

would like to determine the kth largest.• I: read them into an array. Sort them in decreasing order.

Return the kth element.• II: read the first k elements into the array. Sort them in

decreasing order. Next read the remaining elements one by one. If the new element read is smaller than the last, ignore it otherwise place in the correct spot in the array bumping one element out of the array.

• A simulation with a random file of 10 million elements and k = 5,000,000 shows that each requires several days of computer processing.

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Motivating Examples: Word Puzzles• Solving a popular word puzzle: Input consists of a two

dimensional array of letters and a list of words. The objective is to find the words lying horizontally, vertically or diagonally in either direction.

• I: for each word in the word list, check (row, column, orientation)

• II: for each ordered quadruple (row, column, orientation, number of characters), test whether the word is in the word list.

• {this, two, fat, that}

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Math Review - Exponents

• XAXB = XA+B

• XA/XB = XA-B

• (XA)B = XAB

• XN+XN = 2XN !=X2N

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Math Review – Logarithms I• In computer science, all logarithms are to the base 2 unless

specified otherwise.• Definition 1.1. XA = B iff logXB=A• Theorem 1.1. logAB = logCB/logCA where A, B, C > 0, A != 1

• Proof: Let X=logCB, Y=logCA, Z=logABCX=B, CY=A, AZ=B by Definition 1.1.B=CX=(CY)Z.Therefore, X=YZ• Theorem 1.2. log AB = logA + logB where A, B > 0

• Proof: X=logA, Y=logB, and Z=logAB,2X=A, 2Y=B, and 2Z=AB, 2X2Y=AB=2Z.Therefore, X+Y=Z

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Math Review – Logarithms II

• log A/B = logA – logB

• log(AB)=BlogA

• logX < X for all X > 0

• log1 = 0, log2 = 1, log1024 = 10

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Math Review – Series I• Geometric Series

• If 0 < A < 1, then and as N tends to , the sum approaches 1/(1-A)

• S=1+A+A2+A3+A4...• AS=A+A2+A3+A4+A5... • S-AS= 1 which implies S=1/(1-A)

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Math Review – Series II

• Arithmetic Series

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Math Review – Series III

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Math Review – Modular Arithmetic• A B (mod N) meansA is congruent to B modulo N,If N divides A-B(remainders are the same)

• Example:81 61 1 (mod 10)

• if A B (mod N), then• A + C B + C (mod N) and• AD BD (mod N)

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The P Word – Proof by Induction• There are various ways of proving statements in

data structures analysis• Proof by Induction: It has two standart parts: The

first step is proving a base case. Establishing that a theorem is true for some small (usually degenerate) value(s). This step is almost always trivial.

• Next, an inductive hypothesis is assumed. Generally this means that the theorem is assumed to be true for all cases up to some limit k.Using the assumption, the theorem is then shown to be true for the next value, typically k+1.

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The P Word – Induction Example I• Example: Prove that Fibonacci Numbers F0=1, F1=1, and

Fi=Fi-1+Fi-2 for i > 1, satisfy Fi < (5/3)i for i ≥ 1.

• Proof: Verify that the theorem is true for the trivial cases (base cases): F1=1 < (5/3)1 and, F2 = 2 < (5/3)2. These prove the basis. We now assume that the theorem is true for i = 1, 2, ..., k; this is the inductive hypothesis. To prove the theorem, we need to prove Fk+1<(5/3)k+1.

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• Example: If N ≥ 1 then

• Proof: For the base case, the theorem is true when N = 1. For the inductive hypothesis, assume the theorem is true for 1 ≤ k ≤ N. Let’s try to prove that it is true for N + 1

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The P Word – Induction Example II

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• Proof by Counterexample: Best way for proving that a statement is false.

• Example: The statement Fk ≤ k2 is false. The easiest way to prove this is to compute F11 = 144 > 112 = 121

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Proof by Counterexample

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Proof by Contradiction• It proceeds by assuming that the theorem is false and

showing that this assumption implies that some known property is false, and hence the original assumption is erroneous.

• Example: Prove that there is an infinite number of primes.

• Proof: Assume the theorem is false, so that there is some largest prime Pk. Let P1, P2, ..., Pk be all the primes in order and consider N = P1P2...Pk + 1. Clearly, N > Pk, so by assumption N can not be prime.

However, none of P1, P2, ..., Pk divides N exactly, because remainders are all 1. This is a contradiction: numbers are either prime or a product of primes. Hence the original assumption is false implying that the theorem is true.

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A Brief Introduction to Recursion• A function that is defined in terms of itself is called

recursive. Not all mathematically recursive functions are correctly or efficiently implemented by recursion.

• Example: for all integers x ≥ 0 with f(0)=0

#include <stdio.h>

int F( int X ) { if( X == 0 ) /* base case */ return 0; else return 2 * F( X - 1 ) + X * X;}

main( ){ printf( "F(4) = %d\n", F( 4 ) ); return 0;}

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• If F is called with a value of 4, then 2*F(3)+4*4 will be required to be computed. Thus a call is made to find F(3).

• F(4)=2*F(3)+4*4

• F(3)=2*F(2)+3*3

• F(2)=2*F(1)+2*2

• F(1)=2*F(0)+1*1

• F(0)=0 base case. Recursive calls until a base case. F(-1)=?

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Recursion - Bad

#include <stdio.h>

int Bad( unsigned int N ) { if( N == 0 ) return 0; else return Bad( N / 3 + 1 ) + N - 1;}

main( ){ printf( "Bad is infinite recursion\n" ); return 0;}

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• Bad(0)=0, Bad(N)=Bad(N/3 + 1) + N – 1• To compute Bad(1), the computer will repeatedly make

calls to Bad(1). Eventually, it will run out of space

Fundamental Rules of Recursion - I

• 1) Base cases: You must always have some base cases, which can be solved without recursion.

• 2) Making progress: For the cases to be solved recursively, the recursive call must always be to a case that makes progress toward a base case.

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Recursion - Induction

#include <stdio.h>#define PrintDigit( Ch ) ( putchar( ( Ch ) + '0' ) )

/* Print nonnegative N */void PrintOut( unsigned int N ) { if( N >= 10 ) PrintOut( N / 10 ); PrintDigit( N % 10 );}

main( ){ PrintOut( 1369 ); putchar( '\n' ); return 0;}

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• PrintOut(N) prints out positive integers. PrintDigit(Ch) take a single digit number as character and output it to the terminal

Theorem: Recursive number-printing algorithm is correct for N ≥ 0.Proof (by induction): If N has one digit, then it is correct. Assume it works for all numbers of k or fewer digits. A number of k+1 digits is expressed by its first k digits followed by its least significant digit

By the inductive hypothesis, first part is printed correctly and then the last digit is appended.

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• 3) Design rule: Assume that all the recursive calls work. This rule is important. It relieves you of the burden of thinking about the details of bookkeeping.

• 4) Compound interest rule: Never duplicate work by solving the same instance of a problem in separate calls.

• Hidden bookkeeping costs are mostly justifiable. However; It should never be used as a substitute for a simple loop.

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Fundamental Rules of Recursion - II

Homework Assignments

• 1.5, 1.7, 1.8.a, 1.8.b, 1.8.c, 1.9, and 1.12.

• You are requested to study and solve the exercises. Note that these are for you to practice only. You are not to deliver the results to me.

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