```Lecture E
Introduction to Algorithms
Unit E1 – Basic Algorithms
Lesson E – Introduction to Algorithms
Slide 1 of 53.
Lesson or Unit Topic or Objective
Demonstrate the
notion of an
algorithm using two
classic ones
Lesson E – Introduction to Algorithms
Slide 2 of 53.
Computational problems
• A computational problem specifies an input-output
relationship


What does the input look like?
What should the output be for each input?
• Example:


Input: an integer number N
Output: Is the number prime?
• Example:


Input: A list of names of people
Output: The same list sorted alphabetically
• Example:


Input: A picture in digital format
Output: An English description of what the picture shows
Lesson E – Introduction to Algorithms
Slide 3 of 53.
Algorithms
• An algorithm is an exact specification of how to solve a
computational problem
• An algorithm must specify every step completely, so a
computer can implement it without any further
“understanding”
• An algorithm must work for all possible inputs of the
problem.
• Algorithms must be:


Correct: For each input produce an appropriate output
Efficient: run as quickly as possible, and use as little memory as
• There can be many different algorithms for each
computational problem.
Lesson E – Introduction to Algorithms
Slide 4 of 53.
Describing Algorithms
• Algorithms can be implemented in any programming
language
• Usually we use “pseudo-code” to describe algorithms
Testing whether input N is prime:
For j = 2 .. N-1
If j|N
Output “N is composite” and halt
Output “N is prime”
• In this course we will just describe algorithms in Java
Lesson E – Introduction to Algorithms
Slide 5 of 53.
Greatest Common Divisor
• The first algorithm “invented” in history was Euclid’s
algorithm for finding the greatest common divisor (GCD) of
two natural numbers
• Definition: The GCD of two natural numbers x, y is the
largest integer j that divides both (without remainder). I.e.
j|x, j|y and j is the largest integer with this property.
• The GCD Problem:


Input: natural numbers x, y
Output: GCD(x,y) – their GCD
Lesson E – Introduction to Algorithms
Slide 6 of 53.
Euclid’s GCD Algorithm
public static int gcd(int x, int y) {
while (y!=0) {
int temp = x%y;
x = y;
y = temp;
}
return x;
}
Lesson E – Introduction to Algorithms
Slide 7 of 53.
Euclid’s GCD Algorithm – sample run
while
int
x =
y =
}
(y!=0) {
temp = x%y;
y;
temp;
Example: Computing GCD(48,120)
After
After
After
After
After
0
1
2
3
4
rounds
round
rounds
rounds
rounds
temp
-72
48
24
0
x
72
120
72
48
24
y
120
72
48
24
0
Output: 24
Lesson E – Introduction to Algorithms
Slide 8 of 53.
Correctness of Euclid’s Algorithm
• Theorem: When Euclid’s GCD algorithm terminates, it
•
•
•
•
•
returns the mathematical GCD of x and y.
Notation: Let g be the GCD of the original values of x and
y.
Loop Invariant Lemma: For all k  0, The values of x, y
after k rounds of the loop satisfy GCD(x,y)=g.
Proof of lemma: next slide.
Proof of Theorem: The method returns when y=0. By the
loop invariant lemma, at this point GCD(x,y)=g. But
GCD(x,0)=x for every integer x (since x|0 and x|x). Thus
g=x, which is the value returned by the code.
Still Missing: The algorithm always terminates.
Lesson E – Introduction to Algorithms
Slide 9 of 53.
Proof of Lemma
• Loop Invariant Lemma: For all k  0, The values of x, y
after k rounds of the loop satisfy GCD(x,y)=g.
• Proof: By induction on k.


For k=0, x and y are the original values so clearly GCD(x,y)=g.
Induction step: Let x, y denote that values after k rounds and x’, y’
denote the values after k+1 rounds. We need to show that
GCD(x,y)=GCD(x’,y’). According to the code: x’=y and y’=x%y, so
the lemma follows from the following mathematical lemma.
• Lemma: For all integers x, y: GCD(x, y) = GCD(x%y, y)
• Proof: Let x=ay+b, where y>b 0. I.e. x%y=b.


(1) Since g|y, and g|x, we also have g|(x-ay), I.e. g|b. Thus
GCD(b,y)  g = GCD(x,y).
(2) Let g’=GCD(b,y), then g’|(x-ay) and g’|y, so we also have g’|x.
Thus GCD(x,y)  g’=GCD(b,y).
Lesson E – Introduction to Algorithms
Slide 10 of 53.
Termination of Euclid’s Algorithm
• Why does this algorithm terminate?



After any iteration we have that x > y since the new value of y is
the remainder of division by the new value of x.
In further iterations, we replace (x, y) with (y, x%y), and x%y < x,
thus the numbers decrease in each iteration.
Formally, the value of xy decreases each iteration (except, maybe,
the first one). When it reaches 0, the algorithm must terminate.
public static int gcd(int x, int y) {
while (y!=0) {
int temp = x%y;
x = y;
y = temp;
}
return x;
}
Lesson E – Introduction to Algorithms
Slide 11 of 53.
Square Roots
• The problem we want to address is to compute the square
root of a real number.
• When working with real numbers, we can not have
complete precision.


The inputs will be given in finite precision
The outputs should only be computed approximately
• The square root problem:


Input: a positive real number x, and a precision requirement 
Output: a real number r such that |r-x|
Lesson E – Introduction to Algorithms
Slide 12 of 53.
Square Root Algorithm
public static double sqrt(double x, double epsilon){
double low = 0;
double high = x>1 ? x : 1;
while (high-low > epsilon) {
double mid = (high+low)/2;
if (mid*mid > x)
high = mid;
else
low = mid;
}
return low;
}
Lesson E – Introduction to Algorithms
Slide 13 of 53.
Binary Search Algorithm – sample run
while (high-low > epsilon) {
double mid = (high+low)/2;
if (mid*mid > x)
high = mid;
else
low = mid;
}
Example: Computing sqrt(2) with precision 0.05:
mid
mid*mid
After 0 rounds
--After 1 round
1
1
After 2 rounds
1.5
2.25
After 3 rounds
1.25
1.56..
After 4 rounds
1.37..
1.89..
After 5 rounds
1.43..
2.06..
After 6 rounds
1.40..
1.97..
low
0
1
1
1.25
1.37..
1.37..
1.40..
high
2
2
1.5
1.5
1.5
1.43..
1.43..
Output: 1.40…
Lesson E – Introduction to Algorithms
Slide 14 of 53.
Correctness of Binary Search Algorithm
• Theorem: When the algorithm terminates it returns a value
r that satisfies |r-x|.
• Loop invariant lemma: For all k  0, The values of low,
high after k rounds of the loop satisfy: low  x  high.
• Proof of Lemma:


For k=0, clearly low=0  x  high=max(x,1).
Induction step: The code only sets low=mid if mid  x, and only
sets high=mid if mid>x.
• Proof of Theorem: The algorithm terminates when
high-low, and returns low. At this point, by the lemma:
low  x  high  low+. Thus |low-x|.
• Missing Part: Does the algorithm always terminate? How
Fast? We will deal with this later.
Lesson E – Introduction to Algorithms
Slide 15 of 53.
In General…
• This type of binary search can be used to find the roots of
any continuous function f.
• Mean Value Theorem: if f(low)<0 and f(high)>0 then for
some low<x<high, f(x)=0.
• In our case, to find 2, we solved f ( x )  x 2  2  0
Lesson E – Introduction to Algorithms
Slide 16 of 53.
Lecture E
Introduction to Algorithms
Unit E1 – Basic Algorithms
Lesson E – Introduction to Algorithms
Slide 17 of 53.
Lecture E
Introduction to Algorithms
Unit E2 – Running Time Analysis
Lesson E – Introduction to Algorithms
Slide 18 of 53.
Lesson or Unit Topic or Objective
Analysis of Running
Times of Algorithms
Lesson E – Introduction to Algorithms
Slide 19 of 53.
How fast will your program run?
• The running time of your program will depend upon:







The algorithm
The input
Your implementation of the algorithm in a programming language
The compiler you use
Maybe other things: temperature outside; other programs on your
computer; …
• Our Motivation: analyze the running time of an algorithm
as a function of only simple parameters of the input.
Lesson E – Introduction to Algorithms
Slide 20 of 53.
Basic idea: counting operations
• Each algorithm performs a sequence of basic operations:





Arithmetic:
Comparison:
Assignment:
Branching:
…
(low + high)/2
if ( x > 0 ) …
temp = x
while ( true ) { … }
• Idea: count the number of basic operations performed on
the input.
• Difficulties:



Which operations are basic?
Not all operations take the same amount of time.
Operations take different times with different hardware or
compilers
Lesson E – Introduction to Algorithms
Slide 21 of 53.
Testing operation times on your system
import java.util.*;
public class PerformanceEvaluation {
public static void main(String[] args) {
int i=0;
double d = 1.618;
SimpleObject o = new SimpleObject();
final int numLoops = 1000000;
long startTime = System.currentTimeMillis();;
for (i=0 ; i<numLoops ; i++){
// put here a command to be timed
}
long endTime = System.currentTimeMillis();
long duration = endTime - startTime;
double iterationTime = (double)duration / numLoops;
System.out.println("duration: "+duration);
System.out.println("sec/iter: "+iterationTime);
}}
class SimpleObject {
private int x=0;
public void m() { x++; }
}
Lesson E – Introduction to Algorithms
Slide 22 of 53.
Sample running times of basic Java operations
Operation
Loop Body
nSec/iteration
Sys1
;
Sys2
196
10
Double division
d = 1.0 / d;
400
77
Method call
o.m();
372
93
o=new
SimpleObject();
1080
110
Object Construction
Sys1: PII, 333MHz, jdk1.1.8, -nojit
Sys2: PIII, 500MHz, jdk1.3.1
Lesson E – Introduction to Algorithms
Slide 23 of 53.
Asymptotic running times
• Operation counts are only problematic in terms of constant
factors.
• The general form of the function describing the running
time is invariant over hardware, languages or compilers!
public static int myMethod(int N){
int sq = 0;
for(int j=0; j<N ; j++)
for(int k=0; k<N ; k++)
sq++;
return sq;
}
2
.
• We use “Big-O” notation, and say that the running time is
O(N 2 )
N
Lesson E – Introduction to Algorithms
Slide 24 of 53.
Asymptotic behavior of functions
Lesson E – Introduction to Algorithms
Slide 25 of 53.
Mathematical Formalization
• Definition: Let f and g be functions from the natural
numbers to the natural numbers. We write f=O(g) if there
exists a constant c such that for all n: f(n)  cg(n).
f=O(g)   c n: f(n)  cg(n)
• This is a mathematically formal way of ignoring constant
factors, and looking only at the “shape” of the function.
• f=O(g) should be considered as saying that “f is at most g,
up to constant factors”.
• We usually will have f be the running time of an algorithm
and g a nicely written function. E.g. The running time of
the previous algorithm was O(N^2).
Lesson E – Introduction to Algorithms
Slide 26 of 53.
Asymptotic analysis of algorithms
• We usually embark on an asymptotic worst case analysis
of the running time of the algorithm.
• Asymptotic:



Formal, exact, depends only on the algorithm
Ignores constants
Applicable mostly for large input sizes
• Worst Case:




Bounds on running time must hold for all inputs.
Thus the analysis considers the worst-case input.
Sometimes the “average” performance can be much better
Real-life inputs are rarely “average” in any formal sense
Lesson E – Introduction to Algorithms
Slide 27 of 53.
The running time of Euclid’s GCD Algorithm
• How fast does Euclid’s algorithm terminate?




After the first iteration we have that x > y. In each iteration, we
replace (x, y) with (y, x%y).
In an iteration where x>1.5y then x%y < y < 2x/3.
In an iteration where x  1.5y then x%y  y/2 < 2x/3.
Thus, the value of xy decreases by a factor of at least 2/3 each
iteration (except, maybe, the first one).
public static int gcd(int x, int y) {
while (y!=0) {
int temp = x%y;
x = y;
y = temp;
}
return x;
}
Lesson E – Introduction to Algorithms
Slide 28 of 53.
The running time of Euclid’s Algorithm
• Theorem: Euclid’s GCD algorithm runs it time O(N), where
N is the input length (N=log2x + log2y).
• Proof:






Every iteration of the loop (except maybe the first) the value of xy
decreases by a factor of at least 2/3. Thus after k+1 iterations the
k
value of xy is at most ( 2 / 3 ) the original value.
Thus the algorithm must terminate when k satisfies: xy ( 2 / 3 ) k  1
(for the original values of x, y).
Thus the algorithm runs for at most 1  log 3 / 2 xy iterations.
Each iteration has only a constant L number of operations, thus
the total number of operations is at most (1  log 3 / 2 xy ) L
Formally, (1  log 3 / 2 xy ) L  L (1  2 log 2 x  2 log 2 y )  3 LN
Thus the running time is O(N).
Lesson E – Introduction to Algorithms
Slide 29 of 53.
Running time of Square root algorithm
• The value of (high-low) decreases by a factor of exactly 2
each iteration. It starts at max(x,1), and the algorithm
terminates when it
public static double
sqrt(double x, double epsilon){
goes below .
double low = 0;
• Thus the number of
double high = x>1 ? x : 1;
while (high-low > epsilon) {
iterations is at most
double mid = (high+low)/2;
if (mid*mid > x)
high = mid;
else
low = mid;
log 2 (max( x ,1) /  )
• The running time is
O (log x  log 
1
)
}
return low;
}
Lesson E – Introduction to Algorithms
Slide 30 of 53.
Newton-Raphson Algorithm
public static double sqrt(double x, double epsilon){
double r = 1;
while ( Math.abs(r - x/r) > epsilon)
r = (r + x/r)/2;
return r;
}
Lesson E – Introduction to Algorithms
Slide 31 of 53.
Newton-Raphson – sample run
while ( Math.abs(r - x/r) > epsilon)
r = (r + x/r)/2;
Example: Computing sqrt(2) with precision 0.01:
After 0 rounds
After 1 round
After 2 rounds
r
1
1.5
1.41..
x/r
2
1.33..
1.41..
Output: 1.41…
Lesson E – Introduction to Algorithms
Slide 32 of 53.
Analysis of Running Time
• Correctness is clear since for every r the square root of x
is between and r and x/r.
• Here we will analyze the running time only for 1<x<2
• Denote: r '  ( r  x r ) / 2
r  2r x  x  4r x
4
r '  x  (r  x / r ) / 4  x 
2
2
2
2
4r
•
•
•
2
(r  x)
2

4r
2
2
, where  n  r  x after n loops
At the beginning  0  1 , and  1  1 / 4
n
2
In general we have that  n  2
2
At the end it suffices that  n   , since | r  x | | r  x |
Thus the algorithm terminates when n  log log   1
• Thus  n  
•
2
2
n 1
2
Lesson E – Introduction to Algorithms
Slide 33 of 53.
In General…
• The Newton-Raphson method can be used to find the
roots of any differentiable function f.
2
• In our case, to find 2, we solved f ( r )  r  2  0
2
f (r )
r 2
r2 r
• So,
r' r 
r
f ' (r )

2r
2
Lesson E – Introduction to Algorithms
Slide 34 of 53.
Lecture E
Introduction to Algorithms
Unit E2 – Running Time Analysis
Lesson E – Introduction to Algorithms
Slide 35 of 53.
Lecture E
Introduction to Algorithms
Unit E3 - Recursion
Lesson E – Introduction to Algorithms
Slide 36 of 53.
Lesson or Unit Topic or Objective
Using and
understanding
Recursion
Lesson E – Introduction to Algorithms
Slide 37 of 53.
Designing Algorithms
• There is no single recipe for inventing algorithms
• There are basic rules:


Understand your problem well – may require much mathematical
analysis!
Use existing algorithms (reduction) or algorithmic ideas
• There is a single basic algorithmic technique:
Divide and Conquer
• In its simplest (and most useful) form it is simple induction

In order to solve a problem, solve a similar problem of smaller size
• The key conceptual idea:

Think only about how to use the smaller solution to get the larger one
 Do not worry about how to solve to smaller problem (it will be solved using
an even smaller one)
Lesson E – Introduction to Algorithms
Slide 38 of 53.
Recursion
• A recursive method is a method that contains a call to
itself
• Technically:


All modern computing languages allow writing methods that call
themselves
We will discuss how this is implemented later
• Conceptually:


This allows programming in a style that reflects divide-n-conquer
algorithmic thinking
At the beginning recursive programs are confusing – after a while
they become clearer than non-recursive variants
Lesson E – Introduction to Algorithms
Slide 39 of 53.
Factorial
public static void Factorial {
public static void main() {
System.out.println(“5!=“ + factorial(5));
}
public static long factorial(int n){
if (n == 0)
return 1;
else
return n * factorial(n-1);
}
}
Lesson E – Introduction to Algorithms
Slide 40 of 53.
Elements of a recursive program
• Basis: a case that can be answered without using further
recursive calls

In our case:
if (n==0) return 1;
• Creating the smaller problem, and invoking a recursive
call on it

In our case: factorial(n-1)
• Finishing to solve the original problem

In our case: return n * /*solution of recursive call*/
Lesson E – Introduction to Algorithms
Slide 41 of 53.
Tracing the factorial method
System.out.println(“5!=“ + factorial(5))
5 * factorial(4)
4 * factorial(3)
3 * factorial(2)
2 * factorial(1)
1 * factorial(0)
return 1
return 1
return 2
return 6
return 24
return 120
Lesson E – Introduction to Algorithms
Slide 42 of 53.
Correctness of factorial method
• Theorem: For every positive integer n, factorial(n)
returns the value n!.
• Proof: By induction on n:
• Basis: for n=0, factorial(0) returns 1=0!.
• Induction step: When called on n>1, factorial calls
factorial(n-1), which by the induction hypothesis
returns (n-1)!. The returned value is thus n*(n-1)!=n!.
Lesson E – Introduction to Algorithms
Slide 43 of 53.
Raising to power – take 1
public static double power(double x, long n) {
if (n == 0) return 1.0;
return x * power(x, n-1);
}
Lesson E – Introduction to Algorithms
Slide 44 of 53.
Running time analysis
• Simplest way to calculate the running time of a recursive
program is to add up the running times of the separate
levels of recursion.
• In the case of the power method:

There are n+1 levels of recursion
• power(x,n), power(x,n-1), power(x, n-2), … power(x,0)


Each level takes O(1) steps
Total time = O(n)
Lesson E – Introduction to Algorithms
Slide 45 of 53.
Raising to power – take 2
public static double power(double x, long n) {
if (n == 0) return 1.0;
if (n%2 == 0) {
double t = power(x, n/2);
return t*t;
}
return x * power(x, n-1);
}
Lesson E – Introduction to Algorithms
Slide 46 of 53.
Analysis
• Theorem: For any x and positive integer n, the power
method returns x n .
• Proof: by complete induction on n.



Basis: For n=0, we return 1.
If n is even, we return power(x,n/2)*power(x,n/2). By the induction
n/2
n/2 2
n
hypothesis power(x,n/2) returns x
, so we return ( x )  x .
If n is odd, we return x*power(x,n-1). By the induction hypothesis
power(x,n-1) returns x n 1 , so we return x  x n 1  x n .
• The running time is now O(log n):



After 2 levels of recursion n has decreased by a factor of at least
two (since either n or n-1 is even, in which case the recursive call
is with n/2)
Thus we reach n==0 after at most 2log2n levels of recursion
Each level still takes O(1) time.
Lesson E – Introduction to Algorithms
Slide 47 of 53.
Reverse
public class Reverse {
static InputRequestor in = new InputRequestor():
public static void main(String[] args) {
printReverse();
}
public static void printReverse() {
int j = in.requestInt(“Enter another Number”+
” (0 for end of list):”);
if (j!=0){
printReverse();
System.out.println(j);
}
}
}
Lesson E – Introduction to Algorithms
Slide 48 of 53.
Recursive Definitions
• Many things are defined recursively.
• Fibonaci Numbers: 1, 1, 2, 3, 5, 8, 13, 21, …

fn = fn-1 + fn-2
• Arithmetic Expressions. E.g. 2+3*(5+(3-4))



A number is an expression
For any expression E: (E) is an expression
For any two expressions E1, E2: E1+E2, E1-E2, E1*E2, E1/E2 are
expressions
• Fractals
• In such cases recursive algorithms are very natural
Lesson E – Introduction to Algorithms
Slide 49 of 53.
Fibonaci Numbers
public class Fibonaci {
public static void main(String[] args)
for(int j = 1 ; j<20 ; j++)
System.out.println(fib(j));
}
{
public static int fib (int n) {
if (n <= 1) return 1;
return fib(n-1) + fib(n-2);
}
}
Lesson E – Introduction to Algorithms
Slide 50 of 53.
TurtleFractal
public class TurtleFractal {
static Turtle turtle = new Turtle();
public static void main(String[] args)
turtle.tailDown();
drawFractal(500,4);
}
public static void drawFractal(int
if (level==0)
turtle.moveForward(length);
else {
drawFractal(length/3, level-1)
turtle.turnLeft(60);
drawFractal(length/3, level-1)
turtle.turnRight(120);
drawFractal(length/3, level-1)
turtle.turnLeft(60);
drawFractal(length/3, level-1)
}
}}
{
length, int level){
;
;
;
;
Lesson E – Introduction to Algorithms
Slide 51 of 53.
FractalTurtle output
Lesson E – Introduction to Algorithms
Slide 52 of 53.
Lecture E
Introduction to Algorithms
Unit E3 - Recursion
Lesson E – Introduction to Algorithms
Slide 53 of 53.
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