Introduction to Algorithm Analysis Concepts

15-211 Fundamental Data Structures and Algorithms

Peter Lee

January 15, 2004

Plan

Today

Introduction to some basic concepts in the design of data structures

Reading:

For today: Chapter 5 and 7.1-7.3

For next time: Chapter 18 and 19

Homework 1 is available!

See the Blackboard

Due Monday, Jan.19, 11:59pm

A First Data Structure

Lists of integers

Let’s start with a very simple data structure

Lists of integers, with operations such as:

create a new empty list

return the length of the list

add an integer to the end of the list

…

Implementing lists

How shall we implement this?

What design process could we use?

One answer:

Think mathematically

Think inductively

Induction

Recall proofs by induction:

If trying to prove that a property P(n) holds for all natural numbers 0, 1, 2, …, then

Prove the base case of P(0)

For n>0, assume P(n-1), show that P(n) holds

Inductive definitions

A great deal of computer science can be defined inductively

For example, we can define the factorial function as follows:

fact(0) = 1

fact(n) = n * fact(n-1), for n>0

Base case

Inductive case

Implementing lists

How shall we implement this?

What design process could we use?

One answer:

Think mathematically

Think inductively

Inductive definitions

An integer list is either

an empty list, or

an integer paired with an integer list

Base case

Inductive case

Integer lists in Java

An integer list is either

an empty list, or

an integer paired with an integer list

use null

define a new ListCell class

The inductive definition gives us guidance on ways to implement integer lists in Java

One possibility (not really the best):

Integer lists in Java

public class List { int head; List tail;

public List(int n, List l) { head = n; tail = l; }}

An integer list is eitheran empty list, oran integer paired with an integer list

How about the length operation?

Another inductive definition

The length of a list L is

0, if L is the empty list

1 + length of the tail of L, otherwise

Implementing length()

public class ListOps {

public static int length (List l) { if (l==null) return 0; else return 1 + length(l.tail); }}

The add operation

The add of n onto the end of list L is

the singleton list containing n, if L is the empty list

otherwise, a list whose head is the head of L and the tail is M

where M is the result of adding n onto the end of the tail of L

Implementing add()

public class ListOps {

…

public static List add (int n, List l) { if (l==null) return new List(n, null); else return new List(l.head, add(n, l.tail)); }}

Running time

How much time does it take to compute length()?

and also add()?

The “step”

In order to abstract from a particular piece of hardware, operating system, and language, we will focus on counting the number of steps of an algorithm

A “step” should execute in constant time

That is, it’s execution time should not vary much when the size of the input varies

public class ListOps {

public static int length (List l) { if (l==null) return 0; else return 1 + length(l.tail); }}

Constant-time operations

This is the only operation in length() that does not run in a constant amount of time.Hence, we want to know how many times this operation is invoked.

Constant-time operations

public static int length(List l) { if (l==null) return 0; else return 1 + length(l.tail);}

Each call to length() requires at most a constant amount of time plus the time for a recursive call on the tail

So, the “steps” we want are the number of recursive calls

length()

How many steps for length()?

for a list with N elements, length() requires N-1 steps

Since length() requires ~N steps for an “input” of size N, we say that length() runs in linear time

Why do we care about “steps”?

n 100n sec 7n2 sec 2n sec

1 100 s 7 s 2 s

5 .5 ms 175 s 32 s

10 1 ms .7 ms 1 ms

45 4.5 ms 14 ms 1 year

100 100 ms 7 sec 1016 year

1,000 1 sec 12 min --

10,000 10 sec 20 hr --

1,000,000 1.6 min .22 year --

Our goal

Our goal is to compare algorithms against each other

Not compute the “wall-clock” time

We will also want to know if an algorithm is “fast”, “slow”, or maybe so slow as to be impractical

What about add()?

public static List add(int n, List l) { if (l==null) return new List(n, null); else return new List(l.head, add(n, l.tail));}

Let’s Try Something a Bit Harder…

Reverse

The reversal of a list L is:

L, if L is empty

otherwise, the head of L added to the end of M

where M is the reversal of the tail of L

Implementing reverse()

public static List reverse(List l) { if (l==null) return null; else { List r = reverse(l.tail); return add(l.head, r); }}

How many steps?

How many “steps” does reverse take?

Think back to the inductive definition:The reversal of a list L is:

L, if L is empty

otherwise, the head of L added to M• where M is the reversal of the tail of L

Running time for reverse

The running time is given by the following recurrence equation:

t(0) = 0

t(n) = n + t(n-1)

Solving for t would tell us how many steps it takes to reverse a list

time required to reverse the tail

time required to add head to the end

public static List reverse(List l) { if (l==null) return null; else { List r = reverse(l.tail); return add(l.head, r); }}

Reverse

t(0) = 0

t(n) = n + t(n-1)

Solving recurrence equations

A common first step is to use repeated substitution:

t(n) = n + t(n-1)

= n + (n-1) + t(n-2)

= n + (n-1) + (n-2) + t(n-3)

and so on…

= n + (n-1) + (n-2) + (n-3) + … + 1

Klaus says that this is easy…

t(n) = n + (n-1) + (n-2) + … 1 = n(n+1)/2

But how on earth did he come up with this beautiful little closed-form solution?

Incrementing series

By the way, this is an arithmetic seires that comes up over and over again in computer science, because it characterizes many nested loops:

for (i=1; i<n; i++) { for (j=1; j<i; j++) { f(); }}

Mathematical handbooks

For really common series like this one, standard textbooks and mathematical handbooks will usually provide closed-form solutions.

So, one way is simply to look up the answer.

Another way is to try to think visually…

Visualizing it

n

n

0 1 2 3 …

1

2

3

…Area: n2/2

Area of the leftovers: n/2

So: n2/2 + n/2 = (n2+n)/2 = n(n+1)/2

Proving it

Yet another approach is to start with an answer or a guess, and then verify it by induction.t(1) = 1(1+1)/2 = 1

Inductive case: for n>1, assume t(n-1) = (n-1)(n-1+1)/2

= (n2 – n) /2

then t(n) = n + (n2 – n) /2

= (n2 + n)/2

= n(n+1)/2

Summations

Arithmetic and geometric series come up everywhere in analysis of algorithms.

Some series come up so frequently that every computer scientist should know them by heart.

Quadratic time

Very roughly speaking,

f(n) = n(n+1)/2

grows no faster than

g(n) = n2

In such cases, we say that reverse() runs in quadratic time

(we’ll be more precise about this later in the course)

n2 is an upper bound

0

20

40

60

80

100

120

0 1 2 3 4 5 6 7 8 9 10

f(n)g(n)

How about Sorting?

Everybody knows how to sort an array, but we have singly linked lists.

As always, think inductively:

sort(nil) = nil

sort(L) = insert the head into the right place in sort(tail(L))

Ordered Insert

Need to insert element in order, in an already sorted lists.

2 5 10 20 50

12 2 5 12 10 20 50

Code for ordered insert

public List order_insert(int x, List l) { if (x <= l.head) return new List(x, l); List t = order_insert(x, l.tail); return new List(l.head, t);}

The running time depends on the position of x in the new list.

But in the worst case this could take n steps.

Analysis of sort()

sort(nil) = nil

sort(L) = insert the head into the right place in sort(tail(L))

t(0) = 0

t(n) = n + t(n-1)

which we already know to be “very roughly” n2, or quadratic time.

Insertion sort

for i = 2 to n do

insert a[i] in the proper place

in a[1:i-1]

This is yet another example of a doubly-nested loop…

How fast is insertion sort?

We’ve essentially counted the number of computation steps in the worst case.

But what happens if the elements are nearly sorted to begin with?

A preview of some questions

Question: Insertion sort takes n2 steps in the worst case, and n steps in the best case. What do we expect in the average case? What is meant by “average”?

Question: What is the fastest that we could ever hope to sort? How could we prove our answer?

Worst-case analysis

We’ll have much more to say, later in the course, about “worst-case” vs “average-case” vs “expected case” performance.

Better sorting

The sorting algorithm we have just shown is called insertion sort.

It is OK for very small data sets, but otherwise is slow.

Later we will look at several sorting algorithms that run in many fewer steps.

Quiz Break

Doubling summation

Like the incrementing summation, sums of powers of 2 are also encountered frequently in computer science.

What is the closed-form solution for this sum? Prove your answer by induction.

Hint 1: Visualizing it

Imagine filling a glass by halves…

2n-1

2n-2

2n-32n-42n-5

Hint 2: Visualizing it

A somewhat geekier hint:

term in binary20 121 1022 10023 100024 10000…

What is the sum of this column of binary numbers?

Proving it

Base case:When n=1, then 20 = 21-1

Induction step, when n>1.Assume true for n’<n, consider n

By the IH, then

Can you think of an algorithm whose running time is characterized by this series?

Summary

Counting constant-time “steps” of computation

Linear time and quadratic time

Recurrence equations

Sums of geometric series

Simple list algorithms

Next time: Programming tips