Greedy Algorithms: Activity Selection, Fractional Knapsack, and Coin Change Making, Slides of Design and Analysis of Algorithms

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Lecture No. 25

Greedy Algorithms

•^

Activity Selection Problem– Example– Recursive algorithm– Iterative Algorithm

-^

Fractional Knapsack Problem– Problem Analysis– Greedy Approach for Fractional Knapsack

-^

Coin Change Making Problem– Analysis– Greedy Algorithm

Today Covered

Proof (Part B)First we prove second part because it is bit simpler•^

Suppose that S

im

is nonempty

•^

It means there is some activity a

such that:k^

f^ i^

≤^

s^ k

< f

≤k^

s

m^

< f

.m

f^ k

< f

.m

•^

Then a

is also in Sk^

and it has an earlier finishij^

time than a

, which contradicts our choice of am

.m

Hence S

im

is empty, proved

Theorem

S

im

S

mj

s^ m

f^ m S a^ m

ij

a^ k

Part A•^

To prove first part, suppose that A

is a maximum-ij^

size subset of mutually compatible activities of S

,ij^

•^

Order A

monotonic increasing order of finish timeij

•^

Let a

be the first activity in Ak^

.ij^

Case 1•^

If a

= ak^

, then we are done, since we have shownm

that a

m^

is used in some maximal subset of

mutually compatible activities of S

.ij^

Theorem

Why is this Theorem Useful?

-^

Making the greedy choice i.e., the activity with the earliestfinish time in S

ij

• Reduce the number of subproblems and choices– Solved each subproblem in a top-down fashion -^

Only one subproblem left to solve.

Dynamicprogramming

Using the theorem

Number of subproblemsin the optimal solutionNumber of choices toconsider

2 subproblems:S^ ik

, S

kj

j – i – 1 choices

1 choice: the activitywith the earliestfinish time in S

ij

1 subproblem: S

mj

S^ im

= 

Recursive-Activity-Selector

( s, f, i, j

)

1

m

←

i^

• 1

2

while

m

<

j^

and

s

m^

<^

fi^



Find the first activity in

S

. ij

3

do

m

←

m

• 1

4

if^

m

<

j

5

then

return

{ a

} m



Recursive-Activity-Selector (

s, f, m,

j ) 6

else return

Ø

A Recursive Greedy Algorithm

i^ = 0, j^ =

n

• 1 = 12 m

←

i^

• 1

←

0 + 1 = 1

m

<

j^

(1 < 12) and

s

< 1

f^0

(But 1>0)

if^

m

<

j^

(1 < 12)

return {

a^1

}^ 

Recursive-Activity-Selector (

s, f,

^1

, 12)

A Recursive Greedy Algorithm

1

2

3

4

6

7

8

5

9

10

11

12

13

14

0

time

a^0

a^1

1

i^ = 1, m

i^

m

j^

(2 < 12) and

s

2

<^

f^1

m

m

A Recursive Greedy Algorithm 2

3

4

6

7

8

5

9

10

11

12

13

14

0

time

a^2

a^1

1 m

j^

(4 < 12) and

s

4

<^

f^1

(But 5 > 4)

if^

m

j^

return {

a^4

}^ 

Recursive-Activity-Selector(

s, f,

A Recursive Greedy Algorithm 2

3

4

6

7

8

5

9

10

11

12

13

14

0

time

a^4

a^1

1

A Recursive Greedy Algorithm 2

3

4

6

7

8

5

9

10

11

12

13

14

0

time

a^4

a^1 i^ = 4, m

i^

m

j^

(5 < 12) and

s

5

<^

f^4

m

m

a^5

1

A Recursive Greedy Algorithm 2

3

4

6

7

8

5

9

10

11

12

13

14

0

time

a^4

a^1 m

j^

(7 < 12) and

s

7

<^

f^4

m

m

a^7

1

A Recursive Greedy Algorithm 2

3

4

6

7

8

5

9

10

11

12

13

14

0

time

a^4

a^1

a^8

m

j^

(8 < 12) and

s

f^1

(But 8 > 7)

if^

m

j^

return {

a^8

}^

Recursive-Activity-Selector (

s, f, 8,

1

A Recursive Greedy Algorithm 2

3

4

6

7

8

5

9

10

11

12

13

14

0

time

a^4

a^1

a^8

a^10

m

j^

(10 < 12) and

s

10

f^8

m

m

1

A Recursive Greedy Algorithm 2

3

4

6

7

8

5

9

10

11

12

13

14

0

time

a^4

a^1

a^8

a^11

m

j^

(11 < 12) and

s 11

f^8

(But 12 > 11)

if^

m

j^

return {

a^11

}^

Recursive-Activity-Selector (

s, f, 11,