Introduction to Algorithm Analysis Concepts

15-211 Fundamental Data Structures and Algorithms

Aleks Nanevski and Margaret Reid-Miller

January 15, 2003

based on the lecture by Peter Lee

Plan

TodayIntroduction to some basic concepts in

the design of data structures lists, related algorithms and analysis

Reading:For today: Chapter 5 and 7.1-7.3For 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

Operations:create a new empty listreturn the length of the listadd an integer to the end of the list…

3 2 7 1 4

Implementing lists

How shall we implement this? What design process could we use?

One answer:Think mathematicallyThink inductively

Induction

Recall proofs by induction: If trying to prove that a property

P(n) holds for 0, 1, 2, …, thenProve 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 and must be defined inductively Example: factorial

fact(n) = 1 * 2 * ... * n Inductive definition:

fact(0) = 1fact(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 mathematicallyThink inductively

Inductive definitions

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

Base case

Inductive case

Integer lists in Java

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

use null

define a new IntList class

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

One possibility (not really the best):

Integer lists in Javapublic class IntList { int head; IntList tail;

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

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

Integer lists in Java Example:

L.head == 3 How to access 7?

L = 3 7 1 0 9head

tail

headtail

Integer lists in Java Example:

L.head == 3 How to access 7?

L.tail.head == 7

L.tail.tail.tail.head == ?

L = 3 7 1 0 9head

tail

headtail

Integer lists in Java Example:

L.head == 3 How to access 7?

L.tail.head == 7

L.tail.tail.tail.head == 0

L = 3 7 1 0 9head

tail

headtail

How about the length operation?

Another inductive definition

The length of a list L is0, if L is the empty list1 + length of the tail of L, otherwise

Implementing length()

public class ListOps {

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

The add operation Attach n onto the end of list L

Inductive definition: the singleton list containing n, if L is the

empty list

L = 3 7 1 0 9 n

The add operation Attach n onto the end of list L

Inductive definition: the singleton list containing n, if L is the

empty listotherwise, a list whose head is ? tail is ?

L = 3 7 1 0 9 n

The add operation Attach n onto the end of list L

Inductive definition: the singleton list containing n, if L is the empty listotherwise, a list whose

head is the head of L, and tail is M

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

L = 3 7 1 0 9 n

Implementing add()

public class ListOps { …

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

Running time

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

No, not “wall-clock” time.

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 timeThat is, its execution time should not

vary much when the size of the input varies

Constant-time operationspublic static int length(IntList l) { if (l == null) return 0; else return 1 + length(l.tail); }

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 operationspublic static int length(IntList 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 tailSo, the “steps” we want are the number of recursive calls

length()

How many steps for length()? Inductive reasoning:

0 steps (i.e. recursive calls), if l == null1 + #steps for l.tail, otherwise

For a list with N elements, length() requires N steps

We say that length() is linear time.

Why do we care about “steps”?n 100n sec 7n2 sec 2n sec

1 100 s 7 s 2 s5 .5 ms 175 s 32 s

10 1 ms .7 ms 1 ms45 4.5 ms 14 ms 1 year

100 100 ms 7 sec 1016 year1,000 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 otherNot 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 IntList add(int n, IntList l) { if (l == null) return new IntList(n, null); else return new IntList(l.head, add(n, l.tail));}

What about add()?

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

Answer: also linear time.

Let’s Try Something a Bit Harder…

Reverse

The reversal of a list L is:L, if L is emptyotherwise, the head of L added to the

end of M where M is the reversal of the tail of L

Implementing reverse()

public static IntList reverse(IntList l) { if (l == null) return null; else { IntList 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(n) = 1, if n=0t(n) = n + t(n-1)

time required to reverse the tail

time required to add head to the end

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

Reverset(0) = 1

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

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

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

Gauss 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 series 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 …

12

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 = 1Inductive 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

Gauss says that this is easy...Fold the sequence in half1 n sums up to: n+12 n-1 sums up to: n+13 n-2 sums up to: n+1

4 n-3 sums up to: n+1 5 n-4 ...6 n-5... ... total: (n+1)*n/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 at about the same rate as

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)

How about Sorting?

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

As always, think inductively:

sort(nil) = nilsort(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

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

But in the worst case this could take n steps.

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

Analysis of sort()sort(nil) = nilsort(L) = insert the head into the right

place in sort(tail(L))

t(0) = 1t(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…

for i=2, this sorts a[1:2]for i=3, this sorts a[1:3]...for i=n, this sorts a[1:n]

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 average time that we could ever hope to sort in? 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 summationLike 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

How does this series appear in programming?

Think of its recurrence equation:

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

t(0) = 1t(n) = 2 * t(n-1) + 1

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