Lecture-11-CS210-2012 (1).pptx

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Data Structures and Algorithms

(CS210/ESO207/ESO211)

Lecture 11 Data Structure queue:

Mathematical modeling, implementation using array

Two interesting problems (to be solved in the next class)

Proving correctness of 2-majority element.

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Queue: a new data structure

Data Structure Queue:

Mathematical Modeling of Queue

Implementation of Queue using arrays

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Stack

A special kind of list where all operations (insertion, deletion, query) take

place at one end only, called the top.

Behavior of Stack: Last in First out(LIFO)

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1

1

top

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Queue: a new data structure

A special kind of list based on First in First Out (FIFO) strategy.

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front rear

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Operations on a Queue

Query Operations

IsEmpty(Q): determine if Qis an empty queue.

Front(Q): returns the element at the front position of the queue.

Example:If Q is 1, 2, ,,thenFront(Q) returns ?? .

Update Operations

CreateEmptyQueue(Q): Create an empty queue.

Enqueue(x,Q): insert xat the endof the queue Q.

Example: If Q is 1, 2,, , then after Enqueue(x,Q), queue Qbecomes

??

Dequeue(Q): Delete element from the frontof the queue Q.

Example:If Q is 1, 2,, , then after Dequeue(Q), queue Qbecomes

??

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1,2,, ,

2,,

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How to access ith element from the front ?

To access ith element, we mustperform dequeue(hence

delete) the first i-1elements from the queue.

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An Important point you must remember for

every data structure

You can define any newoperation only in

terms of the primitive operations of the data

structures defined during its modeling.

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Implementation of Queue usingarray

Assumption:At any moment of time, the number of elements in queue is n.

Keep an array of Qsize n, and two variables front andrear.

front: the position of the firstelement of the queue in the array.

rear: the position of the lastelement of the queue in the array.

Enqueue(x,Q)

{ rearrear+1;

Q[rear]x

}Dequeue(Q)

{ frontfront+1}

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b d x y

front rear

u

rearfront

There is a serious

problem!

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8

t

rear

u h k

front

v

How to perform

Enqueue(t,Q) ?How to perform

Enqueue(x,Q) ?

x

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Enqueue(x,Q)

{ rearrear+1 ;

Q[rear]x

}

Dequeue(Q)

{ front front+1 }

IsEmpty(Q)

{ ?? }

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rear+1 mod n

(front+1) mod n

Do it as an exercise

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Two interesting problems

Applications of queue/stack

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8 queen problem

Place 8 queens on a chess board so that no two of them attack each other.

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Q

Q

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Shortest route in a grid

From a cell in the grid, we can move to any of its neighboring cell in one step.

From top left corner, find shortest route to each green cell avoiding obstacles.

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Proving correctness of algorithm

for 2-majority element

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Algorithm for 2-majority elementRevisiting from Lecture 4.

Key ideaon which the algorithmis based:

The majority element of an array remains preserved whenever we

eliminate/cancel a pair of distinct elements from the array.

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x

y

w

v

Majority

element

other

elements

>n/2< n/2

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Algorithm for 2-majority elementRevisiting from Lecture 4.

Algo-2-majority(A)

{ count0;

for(i=0to n-1)

{ if (count=0){ xA[i];

count1;

}

else if(x A[i]) countcount- 1;

else countcount+ 1;

}Count the occurrences of xin A, and ifit is more than n/2, then

print(xis 2-majority element) elseprint(there is no majority element in A)

}

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Proving the correctnessof

Algo-2-majority(A)

Question: What is to be proved ?

Answer:For every possible instance of A, the output of algorithm is correct.

Observation:If Adoes not have any majority element, the output of the

algorithm is correct. (It follows from the last statement of the algorithm).

Inference: To prove correctness of the algorithm, it suffices if we can prove

the following:

If Ahas a majority element, say , then at the end of the Forloop, x=

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To establish this, we certainly need to analyze the Forloop carefully.

So revisit the previous slide containing the algorithm for some time

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Algo-2-majority(A)

How to capture progress/status of the algorithm at the end of ith iteration ?

At the end of ith iteration, we have processed {A[0],A[1],,A[i-1]} .

Question:Can we state:

At the end of ith iteration, xis a majority element of {A[0],A[1],,A[i-1]} ?

Answer: No (convince yourself).

How to proceed then ?

Hint:Focus on how the algorithm is implementing the key idea (slide 14).

What does xand countrefer to at the end of ith iteration ? Ponder over this

question

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Algo-2-majority(A)

Insight into the For loop : (along with the key idea underlying the algo)

During ith iteration, we compare A[i-1] with x, and cancel both if they are

different, and increment countotherwise.

So if count=0, then all elements upto A[i-1] would have been eliminated

through distinct-elements pair formations. If count >0, then {x ,counttimes,x} elements would have still survived at the end of the ith iteration.

Notation:Let denote the multiset {x,count times,x, A[i],,A[n-1]}, where

countis the value of the variable countat the end of ith iteration of FORloop.

The insight into For Loop suggests that the following proposition has to holdat the end of ith iteration.

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P(i): is 2-majority element of nonemptymultiset .

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Algo-2-majority(A)

Question: What is P(n) ?

Answer:is 2-majority element of the nonempty multiset which is

{x,count times,x, A[n],,A[n-1]}.

What does it mean ?

is 2-majority element of {x,count times,x} and count>0. Hence =x.

But that it what we wanted to prove. Cool

So all we need to establish is that P(i) holds true for every i.

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Algo-2-majority(A)

Proof of P(i): (by mathematical induction on i)

Base case: (i=0).

Note the end of 0thiteration is the same as the beginning of 1st iteration.

Since count=0at this stage, so 0= {x,0times,x, A[0],,A[n-1]} is the same

as {A[0],,A[n-1]}, which is A. Since is majority element of A. So

P(0)holds true.

Induction hypothesis : P(i) holds true. That is,

is 2-majority element of nonempty multiset .

Induction step: We need to prove that P(i+1)holds at the end of i+1th

iteration. (Use your skills to prove it before studying the solution on the next slide.)

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Algo-2-majority(A)

In the beginning of i+1th iteration, using induction hypothesis,is 2-majority element of nonempty multiset = {x,count times,x, A[i],,A[n-1]}.

During the execution of i+1th iteration, there are the following two cases:

count=0or A[i]=x

In this situation, countgets incremented during the iteration. As a result,

observe that +1 is the same as . So using Induction hypothesis, is a

2-majority element of +1 . So P(i+1) holds.

A[i]x

In this case, we are cancelling an occurrence of xwith A[i] in . This givesset +1. Using the key idea (slide 14), the majority element is preserved in

this process. Secondly existence of 2-majority element in implies that this

cancellation does not eliminate all elements from the multiset. So +1is

nonempty and is its 2-majority element. Hence P(i+1) holds true in this

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Homework

1. Study and fully internalize the proof of correctness given in this lecture.

2. Do you realize that the proof is just formalizing the intuition you have

about the correctness of the algorithm using a better insight into the For

loop.

3. If you feel the proof is an overkill using unnecessary formalism, try todesign a better/simpler proof.

4. Design the proof of correctness for the algorithm for k-majority element

given in Assignment 2. In particular, what propositionholds at the end of

ith iteration. Then prove this proposition using principle of mathematical

induction.

5. Along similar lines, design a proof of correctness for other iterative

algorithms designed in the course. For example, local minima in array,

local minima in a grid, binary search.

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Lecture-11-CS210-2012 (1).pptx