rkgit.weebly.comrkgit.weebly.com/uploads/9/0/9/0/9090685/unit_i_daa.d… · web viewunit- i 1.1...

DAA 1 (ECS-502)

Unit- I

1.1 Algorithms

1.2 Growth of Functions

1.3 Asymptotic Notations

EXERCISE

1.4 Solving Recurrences

1.4.1 Master Method

1.4.2 Substitution Method

1.4.3 Change of Variable Method

1.4.4 Iteration Method

1.4.5. Recursion Tree Method

EXERCISE

Contd………

Prepared By: NIKUNJ KUMAR RKGIT

DAA 2 (ECS-502)

1.1 Algorithms:

An algorithm is any well-defined computational procedure that takes some value, or set of

values, as input and produces some value, or set of values, as output. An algorithm is thus a

sequence of computational steps that transform the input into the output.

We can also view an algorithm as a tool for solving a well-specified computational

problem.

An Algorithm must satisfy the following criteria:

i. It requires one or more than one inputs.

ii. It produces at least one output.

iii. Each instruction should be clear and unambiguous.

iv. It must terminate after a finite number of times/ steps.

How to design an algo.: There are five fundamental techniques, which are used to design an

algo:

i. Divide and Conquer Approach

Binary Search

Multiplication of two n bit numbers

Computing xn

V. Stressen’s matrix multiplication

Quick Sort

Merge Sort

Heap Sort

ii. Greedy method

Knapsack Problem

Minimum cost Spanning Tree (Kruskal’s and Prim’s Algo)

Single Source Shortest Path

iii. Dynamic Programming

All Pair Shortest Path

Chain Matrix Multiplication

Lohgest Common Subsequesnce (LCS)


DAA 3 (ECS-502)

Optimal Binary Search Tree (OBST)

Travelling Salesman Problem (TSP)

iv. Backtracking

N-Queen’s Problem

Sum of Subsets

v. Branch and Bound

Assignment Problem

1.2 Growth of Functions:

Given an array with n elements, we want to rearrange them in ascending order. Sorting

algorithms such as the Bubble, Insertion and Selection Sort all have a quadratic time complexity O

(n2), that limits their use when an array is very big. We now need an algorithm which is more

efficient as compared to insertion sort and has linear time complexity. Merge Sort, a divide-and

conquer algorithm to sort an n element array had time complexity O(nlogn). That is, we are

concerned with how the running time of algorithm increases with the size of input.

Graph shown below shows Merge Sort algorithm is significantly faster than Insertion Sort

algorithm if array is of greater size. Merge sort is 24 to 241 times faster than Insertion Sort. And the

running time gap goes on increasing very fast when input size n goes on increasing


S.No

Number of elements

to be sorted

Total testing time in seconds

Merge Sort

Insertion Sort

1. 3000 0.031 1.046 2. 4000 0.046 1.906 3. 5000 0.062 2.984 4. 6000 0.078 4.281 5. 7000 0.094 5.75 6. 10000 0.125 12.062 7. 15000 0.203 28.093 8. 20000 0.281 49.312 9. 25000 0.343 76.781

10. 35000 0.781 146.4

DAA 4 (ECS-502)

The above tests were run on PC running Windows XP and the following specifications: Intel Core 2

Duo CPU E8400 at 3.00 GHz with 2 GB of RAM. Algorithms were run in Java.

As can be noticed from the table that time gap between the running of

insertion sort and merge sort becoming larger and larger as we increase the size of array n. To find

the complexity or running time of algorithm we are not interested in knowing the exact time of

algorithm, because if we know the exact time its of no use since this algorithm with the same input

can be run on slower machine or can be run on faster machine and this time can differ.

Since there are many possible algorithms or programs that computes the

same results. But we would like to use the one that is fastest. How do we decide how fast an

algorithm is? Since knowing how fast an algorithm run's for a certain input does not reveal

anything about how fast it run's on the other input. Here we need a formula that relates input size n

to the running time of the algorithm i.e we only specify how the running time scales as the size of

input grows. For example the running time for algorithm with input size n is 4n2, we say that it's

running time scales as n2 or to make it simple if size is n = 10, its running time is 100, if input size is

n = 100, its running time is 10000 and so on. We use asymptotic notations to show the running

time of algorithm.

1.3 Asymptotic Notations (Analysis of efficiency):

1. It is a way to describe the behavior of function.

2. It describes the growth of function.

3. How the running time scales with the size of input.

Here we use five asymptotic notations and also define them one by one.


DAA 5 (ECS-502)

O –Notation:

Big O notation provides the upper bound on the growth of

function. Upper bound means, it is the maximum time (Worst Case) that can be

taken by the algorithm. we define the big-O notation as

O(g(n)) = f(n): there exist positive constants c and no such that 0 ≤ f(n) ≤ cg(n) for all

n ≥ no.

Ω –Notation:

Big -notation provides a lower bound on growth ofΩ

function. Lower bound means it is the minimum time (Best Case) that can be taken

by the algorithm. We define the big - Ω notation as

Ω(g(n)) = f(n): there exist positive constants c and no such that 0 ≤ cg(n)≤ f(n) for all n

≥ no.

–Notation:Θ

This notation provides both lower and upper bound on the

function f(n) i.e. average case. Here exist two constants c1 and c2 such that c1g(n) is

a lower bound on f(n) and c2g(n) is an upper bound on f(n). We define big- Θ

notation as

(g(n)) = f(n): there exist three positive constants cΘ 1, c2 and no such that 0 ≤ c1g(n)≤

f(n) ≤ c2g(n) for all n ≥ no.


DAA 6 (ECS-502)

–Notation:ο This is used to provide loose upper bound.

(g(n)) = f(n): there exist positive constants c>0 and nο o>0 such that 0 ≤ f(n) < cg(n) for all n ≥ no.

–Notation:ω This is used to provide loose lower bound.

(g(n)) = f(n): there exist positive constants c>0 and no>0 such that 0 ≤ cg(n) < f(n) for all n ≥ no.ω

Some Properties of Asymptotic Notations:

Transitivity:

f(n) = (Θ g(n)) and g(n) = (Θ h(n)) imply f(n) = (Θ h(n)),f(n) = O(g(n)) and g(n) = O(h(n)) imply f(n) = O(h(n)),f(n) = (Ω g(n)) and g(n) = (Ω h(n)) imply f(n) = (Ω h(n)),f(n) = o(g(n)) and g(n) = o(h(n)) imply f(n) = o(h(n)),f(n) = ω(g(n)) and g(n) = ω(h(n)) imply f(n) = ω(h(n)).Reflexivity:

f(n) = (Θ f(n)),f(n) = O(f(n)),f(n) = (Ω f(n)).

Symmetry:f(n) = (Θ g(n)) if and only if g(n) = (Θ f(n)).

Transpose symmetry:f(n) = O(g(n)) if and only if g(n) = (Ω f(n)),f(n) = o(g(n)) if and only if g(n) = ω(f(n)).

Limit Rule for calculating Asymptotic Notation:

–Notationοwhere c ≥ 0 (nonnegative but not infinite)

O–Notation

–Notationω(either a strictly positive constant or infinity)

Ω –Notation

–NotationΘ


DAA 7 (ECS-502)

C > 0 (strictly positive but not infinity)

Tutorial Sheet (Asymptotic Notations)

1. Find O notation for the following

a) f(n) = 7n + 8

b) f(n) = 27n2 + 16n

c)f(n) = 2n3 + n2 + 2n

2. If f(n) = 3n2 + 5 and g(n) = 2n2 . Proof that f(n) = g(n)Θ

3. For f(x) = 3x2 + 2x +1 State true or false in the following:

a) f(x) = O(x3)

b) x3 = O(f(x))

c) x4 = O(f(x))

d) f(x) = O(f(x))

e) f(x) = O(x4)

f) f(x) = Ω (x3)

g) f(x) = Ω (x2)

h) x3 = Ω(f(x))

i) f(x) = Ω(x4)

j) x2 = Ω(f(x))

4. Show that maxf(n),g(n) = (f(n)+g(n)).Θ

5. State True or False for the following:


DAA 8 (ECS-502)

a) log n.log n = O ((n log n)/100)

b) √log n = O (log(log n))

c) n2 = O(n3)

d) n2 = Ω(n3)

e) f(n) = O(g(n)) implies g(n) = O(f(n))

f) f(n)+g(n) = (min(f(n),g(n)))Θg) f(n) + O f(n) = f(n)Θ

h) f(n) = Og(n) implies log (f(n)) = O(log(g(n)))

where log(g(n)) ≥ 1

i) f(n) = Og(n) implies 2f(n) = O(2g(n))

j) f(n) = O f(n2)

k) f(n) = Og(n) implies g(n)= Ω(f(n))

l) f(n) = O(f(n/2))

Solutions

1.a) f(n) = 7n + 8

7n + 8 O(n), we have to find c and no such that

7n + 8 ≤ cn

7n + 8 ≤ 7n + n

7n + 8 ≤ 8n , treating it as an equality we got n = 8 and c = 8

Therefore, 7n + 8 ≤ cn for n = 8 and c = 8.

So, we can say that 7n + 8 = O(n) or f(n) = O(n) for n = 8 and c = 8

b) f(n) = 27n2 + 16n

c) f(n) = 2n3 + n2 + 2n


DAA 9 (ECS-502)

2. If f(n) = 3n2 + 5 and g(n) = 2n2 . Proof that f(n) = g(n)Θ

f(n) = g(n)Θc1 g(n) ≤ f(n) ≤ c2 g(n)

L.H.S. c1 g(n) ≤ f(n) c1 2n2 ≤ 3n2+5 is satisfy for c1 = 1 and n ≥ 1

R.H.S f(n) ≤ c2 g(n) 3n2 + 5 ≤ c2 2n2 is satisfy for c2 = 1 and n ≥ 3

Now solution is c1 = 1, c2 = 1 and n ≥ 3

3. a) T b) F c) F

d) Te) T

f) Fg) Th) Ti) Fj) T

4. Show that maxf(n),g(n) = (f(n)+g(n)).Θ

maxf(n),g(n) = (f(n)+g(n))Θf(n) = g(n) Θ

for notation Θ c1 g(n) ≤ f(n) ≤ c2 g(n)then c1 (f(n)+g(n)) ≤ maxf(n),g(n) ≤ c2 (f(n)+g(n))

L.H.S. c1 (f(n)+g(n)) ≤ maxf(n),g(n)c1 (f(n)+g(n)) ≤ h(n)

where h(n) = f(n) if f(n) > g(n)and h(n) = g(n) if g(n) > f(n)

so we can say it satisfy for c1 = ½ and n ≥ 1

R.H.S. maxf(n),g(n) ≤ c2 (f(n)+g(n))h(n) ≤ c2 (f(n)+g(n))

where h(n) = f(n) if f(n) > g(n)and h(n) = g(n) if g(n) > f(n)

so we can say it satisfy for c2 = 1 and n ≥ 1

so, the solution is c1 = ½, c2 = 1 and n ≥ 1

5. a) T b) F

g) Tg) F


DAA 10 (ECS-502)

c) T d) F e) F f) F

i) Tj) Tk) Tl) T

1.4 Solving Recurrences:

1.4.1 Master Method

The master method is applicable only if the recurence is given in the form

T(n) = aT(n/b) + f(n)

a: How many parts we are dividing the problem

n/b: size of each problem

f(n) = cost of dividing the problem D(n)+cost of combining the problem C(n)

Where a > 1 and b > 1 be constants. Now as a first step we have to calculate the value of nlogb

a. The

solution of given recurrence can be given by comparing these two values polynimialy nlogb

a and f(n).

Case 1: If value of nlogb

a is polynomially greater then f(n) then case 1 of master theorem holds which

can be given as:

If f(n) = O(nlogb

a - Ԑ ) for some constant > 0Ԑ , then the solution of the recurrence will be T(n) = Θ

( nlogb

a )


DAA 11 (ECS-502)


a is polynomially and asymptotically equal to f(n) then case 2 of master

theorem holds which can be given as:

If f(n) = (nΘ logb

a ) ,then the solution of the recurrence will be T(n)= ( nΘ logb

a log n)


a is polynomially smaller then f(n) then case 3 of master theorem holds which

can be given as:

If f(n) = Ω(nlogb

a + Ԑ) for some constant Ԑ > 0 and af(n/b) ≤ c.f(n) for some constant c<1, then the

solution of the recurrence will be T(n) = ( f(n))Θ

1.4.2 Substitution Method

1.4.3 Change of Variable Method

1.4.4 Iteration Method

1.4.5. Recursion Tree Method

Tutorial Sheet (Solving Recurrences)

1. Solve the following Recurrences using Master Method:

a) T (n) = 4T (n/2) + n

b) T (n) = 9T (n/3) + n

c) T (n) = 8T (n/2) + n

d) T (n) = 2T (n/2) + n

e) T (n) = T (n/2) + 1

f) T (n) = T (2n/3) + 1

g) T (n) = 2T (n/2) + n2

h) T (n) = 3T (n/4) + ( n2)

i) T (n) = 3T (n/4) + ( nlog n)

j) T (n) = 4T(n/2) + n

k) T (n) = 4T(n/2) + n2

l) T (n) = 4T(n/2) + n3

m) The running time of an algorithm A is described by the recurrence T(n) =

7T(n/2) + n2. A competing algorithm A' has a running time of T`(n) = aT'(n/4) + n 2. What is the

largest integer value for a such that A' is asymptotically faster than A?

n) Use the master method to show that the solution to the recurrence T(n) = T(n/2) + (1) of binary

search is T(n) = (1g n)

o) Can the master method be applied to the recurrence T (n) = 4T (n/2) + n 2logn? Why or why not?

Give an asymptotic upper bound for this recurrence.

p) Can the master method be applied to the recurrence T (n) = 2T (n/2) + n log n? Why or why not?

Give an asymptotic upper bound for this recurrence.

q) T(n) = 5T(n/2) + (nθ 2)


DAA 12 (ECS-502)

r) T(n) =27T(n/3) + ( n3 log n)

s) T(n) = 5T(n/2) + (n3)

2. Solve the following Recurrences using Substitution Method:

a) Determine the upper bound of T(n) = 2T(n/2) + n is O(nlogn).

b) T (n) = 2T ( n/2 + 17) + n is O(nlogn).

c) T(n) = T(n – 1) + n is is O(n2).

3. Solve the following Recurrences using Change of Variable Method:

a) T (n) = 2T (√n) + log n

b) T (n) = 3T (√n) + log n, Your solution should be asymptotically tight. Do not worry

about whether values are integral.

c) T (n) = 2T (√n) + 1

4. Solve the following Recurrences using Iteration Method:

a) T(n) = T(n – 1) + 1 and T(1) = (1)Ө

b) T(n) = T(n – 1) + n and T(1) = (1)Ө

c) T(n) = 2T(n – 1) + 1

5. Solve the following Recurrences using Recursion Tree Method:

a) T(n) = 2T(n/2) + cn

b) T(n) = 4T(n/2) + cn

c) T (n) = 3T (⌊n/2⌋) + n

d) T (n) = T (n/3) + T (2n/3) + cn

e) T (n) = 3T (⌊ n/4⌋) + (n2)θf) T(n) = T(n – 1) + n

g) T(n) = 2T(n – 1) + 1

h) T(n) = T(n – 1) + 1

i) T(n) = T(n/2) + T(n/4) + T(n/8) + n


DAA 13 (ECS-502)

Solutions

1. a) Here a = 4, b = 2, f(n) = n

And nlogb

a = nlog2

4 = n2

Therefore Ԑ > 1 exist such that f(n) = O(nlogb

a - Ԑ )

Or for Ԑ = 1 f(n) = O(nlog2

4 - Ԑ ) = O(nlog2

4 - 1) = O(n2 - 1) = O(n ) = f(n)

Therefore Case 1 of Master theorem holds and we conclude the solution of given recurrence as

T(n) = ( nΘ logb

a) = ( nΘ log2

4) = ( nΘ 2 )

T(n) = ( nΘ 2 )

b) Here a = 9, b = 3, f(n) = n

And nlogb

a = nlog3

9 = n2


a - Ԑ )


9 - Ԑ ) = O(nlog3

9 - 1) = O(n2 - 1) = O(n ) = f(n)


T(n) = ( nΘ logb

a) = ( nΘ log3

9) = ( nΘ 2 )

T(n) = ( nΘ 2 )

c) Here a = 8, b = 2, f(n) = n

And nlogb

a = nlog2

8 = n3


a - Ԑ )


8 - Ԑ) = O(nlog2

8 - 2) = O(n3 - 2) = O(n ) = f(n)


T(n) = ( nΘ logb

a) = ( nΘ log2

8) = ( nΘ 3 )


DAA 14 (ECS-502)

T(n) = ( nΘ 3 )

d) Here a = 2, b = 2, f(n) = n

And nlogb

a = nlog2

2 = n

Here f(n) = (nΘ log2

2 ) = (n ) = f(n)ΘTherefore Case 2 of Master theorem holds and we conclude the solution of given recurrence as

T(n) = ( nΘ logb

a log n ) = ( nΘ log2

2 log n) = ( n log n )Θ T(n) = ( n log n)Θe) Here a = 1, b = 2, f(n) = 1

And nlogb

a = nlog2

1 = n0 = 1

Here f(n) = (nΘ log2

1 ) = (nΘ 0 ) = f(n) = (1)


T(n) = ( nΘ logb

a log n ) = ( nΘ log2

1 log n) = ( nΘ 0 log n) = ( log n )Θ T(n) = ( log n)Θf) Here a = 1, b = 3/2, f(n) = 1

And nlogba = nlog3/2

1 = n0 = 1

Here f(n) = (nΘ log3/2

1 ) = (nΘ 0 ) = f(n) = (1)


T(n) = ( nΘ logb

a log n ) = ( nΘ log3/2

1 log n) = ( nΘ 0 log n ) = ( log n )Θ T(n) = ( log n )Θg) Here a = 2, b = 2, f(n) = n2

And nlogb

a = nlog2

2 = n

Therefore Ԑ > 1 exist such that f(n) = Ω (nlogb

a + Ԑ )

Or for Ԑ = 1 f(n) = Ω(nlog2

2 + Ԑ) = Ω(nlog2

2 + 1) = Ω(n1+ 1) = Ω(n2 ) = f(n)


T(n) = ( f(n) ) = ( nΘ 2)

T(n) = ( nΘ 2 )

h) Here a = 3, b = 4, f(n) = n2

And nlogb

a = nlog4

3 = n0. 756

Therefore Ԑ > 1 exist such that f(n) = (nlogb

a + Ԑ )

Or for Ԑ = 1.25 f(n) = Ω(nlog4

3 +Ԑ ) = Ω(nlog4

3 + 1.25) = Ω(n0.75+ 1.25) = Ω(n2 ) = f(n)


T(n) = ( f(n) ) = ( nΘ 2) so solution will be T(n) = ( n2 )Θi) Here a = 3, b = 4, f(n) = n log n

And nlogb

a = nlog4

3 = n0. 756


DAA 15 (ECS-502)


a + Ԑ)

Or for Ԑ = 0.25 f(n) = Ω(nlog4

3 + Ԑ) = Ω(nlog4

3 + 0.25) = Ω(n0.75+ 0.25) = Ω(n ) = f(n)


T(n) = ( f(n) ) = ( n log n) Θ T(n) = ( n log n)Θj) T(n) = 4T(n/2) + n

Here a = 4, b = 2, f(n) = n.

Therefore nlogb

a = nlog2

4 = n2

Since Case 1: If f(n) = O(nlogb

a - Ԑ) for some constant Ԑ > 0, Here Ԑ = 1

Therefore Case I of Master Theorem Holds and T(n) = (nΘ logb

a )

T(n) = (nθ 2)

k) T (n) = 4T(n/2) + n2

Here a = 4, b = 2, f(n) = n2

Therefore nlogb

a = nlog2

4 = n2

Since f(n) = nlog24 = n2 therefore Case 2: If f(n) = (nlogba ) Of Master Theorem Holds and theΘ

solution is

T(n) = ( nΘ logb

a log n)

T(n) = (nθ 2 log n)

l) T (n) = 4T(n/2) + n3

Here a = 4, b = 2, f(n) = n3

Therefore nlogb

a = nlog2

4 = n2

Since, Case 3 If f(n) = (nlogb

a + Ԑ) for some constant Ԑ > 0, Here = 1,Therefore Case III of Masterε

Theorem Holds

and the solution is T(n) = ( f(n) ).Θ T(n) = (nθ 3)

m) Algorithm A is given as T(n) = 7T(n/2) + n2

Here a = 7, b = 2, f (n) = n2

Therefore nlogb

a = nlog2

7 , Since 2 < log27 < 3 , therefore there exists constant such thatεCase 1: If f(n) = O(nlog

ba - Ԑ ) for some constant Ԑ > 0, Here will have some decimal value (0.2453).ε

Therefore Case I of Master Theorem Holds and the solution is T(n) = ( nlogb

a )

T(n) = (nθ log2

7) --------(1)


DAA 16 (ECS-502)

For algorithm A` , T'`(n) = aT`'(n/4) + n2

If largest value of a = 16 then

nlogb

a = nlog4

16 = n2 . For this Case 2: If f(n) = (n logb

a ) , of Master Theorem Holds and the solution of

recurrence will be

T(n) = ( nlogb

a log n)

Therefore , T(n) = (nθ 2 log n) ---------(2)

But for a = 16 A` will be slower than A as can be seen from solution (1) and (2)

We will now try for smaller values of a

If largest value of a = 15 then

nlogba = nlog4

15 , Since here 1 < nlog4

15 < 2, Therefore Case 3 If f(n)=(nlogb

a + Ԑ ) for some constant Ԑ >

0, of Master Theorem Holds and the solution is T(n) = ( f(n) ). Therefore solution of recurrence will

be

T (n) = (nθ 2) ----------(3)

n) Here a = 1, b = 2, f(n) = 1

Therefore nlogb

a = nlog2

1 = n0 = 1

Here, Case 2: If f(n) = (nΘ logb

a ) of Master Theorem holds , and the solution of the recurrence can

be given as

T(n) = ( nΘ logb

a log n)

Therefore T(n) = (nθ logb

a log n) = (1. log n) = (log n)θ θT(n) = (log n)θ

o) For T (n) = 4T (n/2) + n2 log n

a = 4, b = 2, f(n) = n2 log n

nlogb

a = nlog2

4 = n2 . Here f(n) is asymptotically larger then n logb

a , but not polynomially larger. This

case falls into the gap between Case 2 and Case 3. The solution of this recurrence can be given by

the theorem as stated under.

If f(n) = (nθ logb

a logkn), where k ≥ 0, then the master recurrence has solution T(n) = (nθ logb

a logk+1n).

Therefore the solution of the given recurrence will be

T(n) = (nθ 2 log2n).

p) For T (n) = 2T (n/2) + n log n

a = 2, b = 2, f(n) = nlog n


DAA 17 (ECS-502)

nlogb

a = nlog2

2 = n . Here f(n) is asymptotically larger then nlogb




If f(n) = (nθ logb


a logk+1n).


T(n) = (n logθ 2n)

q) Here a = 5, b = 2, f(n) = n2

And nlogb

a = log25 , Since here 2 < log25 < 3, Therefore Case 1 of Master theorem holds since Ԑ >

1 exist such that f(n) = O(nlogb

a - Ԑ )

f(n) = O(nlog2

5 - Ԑ) = O(nlog2

5 – 0.245) = O(n2 ) = f(n)

Therefore T(n) = ( nΘ logb

a) = ( nΘ log2

5)

T(n) = ( nΘ log2

5 )

r) Here a = 27, b = 3, f(n) = n3 log n

nlogb

a = nlog3

27 = n3 . Here f(n) is asymptotically larger then nlogb




If f(n) = (nθ logb


a logk+1n).


T(n) = (nθ 3 log2n).

s) Here a = 5, b = 2, f(n) = n3

And nlogb

a = nlog2

5 = n , Since here 2 < log25 < 3,


a +Ԑ )


T(n) = ( f(n) ) = ( nΘ Θ 3)

T(n) = ( nΘ 3)

2. a) Here we guess that the solution is T(n) = O(n log n). Our method is to prove that T(n) ≤ cn log

n for an appropriate choice of constant c>0.

Given recurrence is

T(n) = 2T(n/2)+ n -------- (1)

Solution we guess T(n) ≤ cn log n --------- (2)


DAA 18 (ECS-502)

Since we need T(n/2) to substitute in equation (1), so we calculate T(n/2) from equation (2) by

substituting n as n/2 in equation (2) we get

T(n/2)≤ cn/2 log n/2 --------- (3)

Now substituting the value of T(n/2) from equation (3) in recurrence relation (1) we get

T(n) ≤ 2(cn/2 log (n/2) ) + n

≤ 2cn/2 log (n/2) + n

≤ cn ( log n – log 2 ) + n

≤ cn (log n – 1 ) + n

= cn log n – cn + n This will hold as long as c ≥ 1.

≤ cn log n

Therefore the upper bound for T (n) = 2T (n/2)+ n is T(n) = O(n log n) for c ≥ 1

b) Here we had the solution T(n) = O(n log n). Our method is to prove that T(n) ≤ cn log n for an

appropriate choice of constant c>0.

Given recurrence is

T(n) = 2T(n/2 + 17) + n -------- (1)

Solution T(n) ≤ cn log n -------- (2)

Since we need T(n/2 + 17) to substitute in equation (1), so we calculate T(n/2 + 17) from equation

(2) by substituting n as n/2 + 17 in equation (2) we get

T(n/2 + 17)≤ c(n/2 + 17) log( n/2 + 17) --------- (3)

Now substituting the value of T(n/2 + 17) from equation (3) in recurrence relation (1) we get

T(n) ≤ 2(cn/2 + 17) log (n/2 + 17) + n

n/2 + 17, can be written as n/2 if n is very large, and the same is for cn + 32, it can be written as cn

≤ (cn + 34) log (n/2) + n

≤ cn ( log n – log 2 ) + n

≤ cn (log n – 1 ) + n

= cn log n – cn + n This will hold as long as c ≥ 1.

≤ cn log n

Therefore the upper bound for T (n) = 2T (n/2 + 17) + n is T(n) = O(n log n) for c ≥ 1

c) Here we are given solution as T(n) = O(n2) which can also be written as

T(n) = c n2

T(n – 1) = c(n – 1)2

= c(n2 + 1 - 2n)


DAA 19 (ECS-502)

Placing the value of T(n – 1) in the given recurrence we get

T(n) = c(n2 + 1 - 2n) + cn

= cn2 + c – 2cn + cn

Ignoring lower order terms we get

T(n) = cn2

Or T(n) = O(n2).

3. a) The recurrence T (n) = 2T (√n) + log n on the first instance looks to be the difficult one. But we

can simplify this recurrence by changing (renaming) variables. Here we assume that √n will be a

integer.

Introduce one new variable m and assign it value log n. Now renaming log n as m we got

i.e. m = log n, which implies :- 2m = n which yields

T(2m) = 2T(2m/2) + m

We now rename S(m) = T(2m) to produce new recurrence

S(m) = 2S(m/2) + m --------(1)

The recurrence which we now got is very much similar to the recurrence

T(n) = 2T(n/2) + n which had a solution O(n log n)

Therefore recurrence (1) has a same solution

S(m) = O(m log m)

Now changing S(m) to T(n), we obtain

T(n) = T(2m) = S(m) = O(m log m) = O( log n log log n).

b) O(log2n)log23

c) The recurrence T (n) = 2T (√n) + 1 on the first instance looks to be the difficult one. But we can

simplify this recurrence by changing (renaming) variables. Here we assume that √n will be a

integer.

Introduce one new variable m and assign it value log n. Now renaming log n as m we got

i.e. m = log n, which implies :- 2m = n which yields

T(2m) = 2T(2m/2) + 1

We now rename S(m) = T(2m) to produce new recurrence

S(m) = 2S(m/2) + 1 --------(1)

The recurrence which we now got is very much similar to the recurrence

T(n) = 2T(n/2) + 1 which had a solution O(n)

Therefore recurrence (1) has a same solution


DAA 20 (ECS-502)

S(m) = O(m)

Now changing S(m) to T(n), we obtain

T(n) = T(2m) = S(m) = O(m) = O(log n).

4. a) T(n) = T(n – 1) + 1

T(n – 1) = T(n – 2) + 1

T(n ) = T(n – 2) + 1 + 1

T(n ) = T(n – 2) + 2

T(n – 2) = T(n – 3) + 1

T(n ) = T(n – 3) + 1 + 2

T(n ) = T(n – 3) + 3

T(n – 3) = T(n – 4) + 1

T(n ) = T(n – 4) + 1 + 3

T(n ) = T(n – 4) + 4

………………………..

………………………..

T(n ) = T(n – k) + k

When k = n – 1

T(n – k) = T(n – (n - 1))

T(n ) = T(n – (n - 1)) + n – 1

T(n ) = T(1) + n – 1

T(n ) = Ө(1) + n – 1

T(n ) = Ө(n)

b) T(n) = T(n – 1) + n

T(n – 1) = T(n – 2) + n – 1

T(n) = T(n – 2) + n + n – 1

T(n – 2) = T(n – 3) + n – 2

T(n) = T(n – 3) + n + n – 1 + n – 2

T(n – 3) = T(n – 4) + n – 3

T(n) = T(n – 4) + n + n – 1 + n – 2 + n – 3

……………………………………………..

……………………………………………..

T(n) = T(n – k) + n + n – 1 + n – 2 + n – 3 + ……..+ (n – (k – 1))


DAA 21 (ECS-502)

For k = n – 1

T(n) = T(n – (n – 1)) + n + n – 1 + n – 2 + n – 3 + ……..+ (n – ((n – 1) – 1))

T(n) = T(1) + n + n – 1 + n – 2 + n – 3 + ……..+ 2

T(n) = T(1) + n + n – 1 + n – 2 + n – 3 + ……..+ 2

T(1) = Ө(1) and,n + n – 1 + n – 2 + n – 3 + ……..+ 2 =

T(n) = Ө(1) +

T(n) = Ө(n2)

c) T(n) = 2T(n – 1) + 1

T(n – 1) = 2T(n – 2) + 1

T(n) = 2[2T(n – 2) + 1] + 1

T(n) = 2.[2T(n – 2)] + 2 + 1

T(n) = 2.[2T(n – 2)] + 1 + 2

T(n – 2) = 2T(n – 3) + 1

T(n) = 2.2[2T(n – 3) + 1] + 1 + 2

T(n) = 2.2[2T(n – 3)] + 2.2 + 1 + 2

T(n) = 22 [2T(n – 3)] + 1 + 2 + 22

T(n – 3) = 2T(n – 4) + 1

T(n) = 23 [2T(n – 4) + 1] + 1 + 2 + 22

T(n) = 23 [2T(n – 4)] + 1 + 2 + 22 + 23

…………………………………….

…………………………………….

T(n) = 2k[2T(n – (k – 1))] + 1 + 2 + 22 + 23 +…….+ 2k

For k = n - 1

T(n) = 2n - 1[2T(n – ((n – 1) – 1))] + 1 + 2 + 22 + 23 +…….+ 2n - 1

T(n) = 2n - 1[2T(2)] + 1 + 2 + 22 + 23 +…….+ 2n – 1

T(n) = Ө(2n)

5.a)

Step 1: Draw the recurrence tree It is drawn

below


DAA 22 (ECS-502)

Step 2. To calculate the height of the tree ( or the lowest level of the tree )

n/2i = 1

Which implies n = 2i

Therefore i = log2n

Thus the height of the tree = log2n

We know that:

Total levels of the tree = Height of the tree + 1

Therefore, Total levels of the tree = log2n + 1

Step 3. Cost at each level of the tree

From the recursion tree of T(n) = 2T(n/2) + cn we can easily conclude that cost of each level is cn.

Step 4. Total cost of the recursion tree

Total cost = Total no. of levels x cost of each level

= (log2n + 1) x cn

= cn log2n + cn

By ignoring lower order terms we get

Total cost = cn log2n

T(n) = Ө(n log2n)

b) Step 1: The recurrence tree for the recursive relation T(n) = 4T(n/2) + cn can be easilty drawn

(try it yourself)

Step 2: To calculate the height of the tree ( or the lowest level of the tree )

n/2i = 1


Therefore i = log2n (lowest level i is the height of the tree)



DAA 23 (ECS-502)

We know the Total levels of the tree = Height of the tree + 1

Therefore the Total levels of the tree = log2n + 1

Step 3 & 4:

Total cost of last level = Total number of leaves x Cost of each Leave

= 4log2n x T(1)

= nlog24 x T(1)

= n2 .c

= cn2

= O(n2)

Cost of each level = Cost of each node at that level x Total number of nodes at that level.

= n + 2n + 4n + 8n + 16n + ………….. + log2n – 1

= n(1 + 2 + 4 + 8 + 16 + …………….) + log2n – 1

= n(20 + 21 + 22 + 23 + 24 + ……….) + log2n – 1

This is a Geometric series

= n(2 log2

n – 1 + 1 – 1) / ( 2 – 1)

= n(n log2

2 – 1)

= n(n1 – 1)

= n(n – 1)

Total cost of tree = Cost of 0, 1, 2, 3, …….., log2n – 1 levels + Cost of last level

= n(n – 1) + O(n2)

= n2 -1 + cn2

= O(n2)

Tight asymptotic bound for the given recurrence T(n) = 4T(n/2) + cn can be given as

T (n) = O (n2)


DAA 24 (ECS-502)

c) Step 1: Draw the recurrence tre It is drawn as shown

Step 2. To calculate the height of the tree ( or the lowest

level of the tree )

n/2i = 1


Therefore i = log2n (lowest level i is the height of the

tree)


We know the Total levels of the tree = Height of the tree

+ 1

Therefore the Total levels of the tree = log2n + 1

Step 3:


DAA 25 (ECS-502)


= 3log2

n x T(1)

= nlog2

3 x T(1) = O(nlog2

3)


Total cost of recursion tree = cost of 0,1,2,...,(log2n – 1) levels + cost of last level

= x cn + O(nlog2

3)

Therefore, the solution of recurrence T (n) = 3T (⌊n/2⌋) + n is T(n) = O(nlog2

3).

d) Step 1: Draw the recurrence tree It is drawn

below

Step 2. To calculate the height of the tree ( or

the lowest level of the tree )

here there are two order of reductions i.e. 1/3

and 2/3. Now among these two we have to take

the order of reduction that will go to the deeper

level, since that will decide the height of tree.

Therefore among these two order of reductions

1/3 and 2/3, we select 2/3, since it reaches

more deeper then 1/3.

Therefore, n/(2/3)i = 1

Which implies n = (3/2)i

Therefore i = log3/2n


Therefore, Total levels of the tree = log3/2n + 1


DAA 26 (ECS-502)


From the recursion tree of T (n) = T (n/3) + T (2n/3) + cn we can easily conclude that cost of each

level is cn.


Total cost = Total no. of levels x cost of each level

= (log3/2n + 1) x cn

= cn log3/2n + cn

By ignoring lower order terms we get

Total cost = cn log3/2n, by rounding the base value of we get T(n) = (n log n).Ωe) Step 1: Draw the recurrence tree It is drawn as shown


level of the tree )

n/4i = 1


Therefore i = log4n

(lowest level i is equal to the height of the tree)




Step 3:


= 3log4

n x T(1)

= nlog4

3 x T(1) = O(nlog4

3)


Total cost of recursion tree = cost of 0, 1, 2, …… (log4n – 1) levels + cost of last level


DAA 27 (ECS-502)

= x cn2 + O(nlog4

3) Here we got an equation, now we to find the

highest degree of polynomial, it is nlog2

3 i.e. T(n) = O(n2).

f) Step 1: Draw the recurrence tree It is drawn as shown


level of the tree )

From the recurrence tree it can be easily noted that the

height of this tree in n. Since at each step its size is

reduced by only one.


From the recursion tree of T(n) = T(n – 1) + n we can

easily see the cost of each level if we sum up the cost of

each level we got,

1 + (n – 1) + (n – 2) + (n – 3) +……… + n


Total cost of recursion tree is

1 + (n – 1) + (n – 2) + (n – 3) +……… + n

We know that the sum of this series is

Therefore, we finally got the solution as T(n) = Ө(n2)

h) Hint: cost of last level = 1

Cost of remaining levels = n – 1

Total cost = n – 1 + 1

T(n) = Ө(n)


DAA 28 (ECS-502)

g) Step 1: Draw the recurrence tree It is drawn as shown

Step 2. To calculate the height of the tree ( or the lowest level of

the tree )

From the recurrence tree it can be easily noted that the height

of this tree in n. Since at each step its size is reduced by only

one.


From the recursion tree

Total number of nodes at last level = 2n


= 2n x T(1)

= O(2n)



Total cost of recursion tree = cost of 0, 1, 2, …… (n – 1) levels + cost of last level

= 2n- 1 + O(2n)

Therefore, the solution of recurrence T(n) = 2T(n – 1) + 1 is

T(n) = Ө(2n).

i) Step 1: Draw the recurrence tree It is drawn


level of the tree )

there are three order of reductions i.e. ½, ¼, and 1/8.

Now among these three we have to take the order of


DAA 29 (ECS-502)

reduction that will go to the deeper level, since that will decide the height of tree. Therefore among

these three order of reductions ½, ¼, and 1/8, we select ½, since it reaches more deeper then ¼,

and 1/8.

n/2i = 1 Therefore i = log2n Thus the height of the tree = log2n

We know that:




First we find the cost of last level.

Total number of nodes at last level = 3log2

n


= 3log2

n x T(1) = nlog2

3 x T(1) = O(nlog2

3)

Therefore, the solution of recurrence T(n) = T(n/2) + T(n/4) + T(n/8) + n is

T(n) = O(nlog2

3).


rkgit.weebly.comrkgit.weebly.com/uploads/9/0/9/0/9090685/unit_i_daa.d… · web viewunit- i 1.1...

Documents