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Greedy Strategy

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Most of these problems have n inputs and require us to obtain a

subset that satisfies some constraints

Greedy Strategy :

Any Subset that satisfies the given constraints is called a feasible

solution

We are required to find a feasible solution that either maximizes or

minimizes a given objective function

A feasible solution that does this is called an

OPTIMAL SOLUTION

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For Greedy Strategy :

We can suggest a two stage algorithm :

1. Arrange the input in the best possible sequence

i.ethe most appropriate elements would be available

starting from the beginning of the list with the most deserving / the

most meritoriouselement appearing as the first element

2. Test element by element whether to include the current element

into the SOLUTION set or not.

Preprocessing

Solving

Greedy Sequencing

Creating the solution

set

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For Greedy Strategy :

Control abstract of greedy strategy based algorithm is given below :

Procedure greedy (A, n)

{

Solution for i 1 to n

{ x Select(A)

If feasible(solution , x) then solution U(Solution,x)

}

}

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O/1 Knapsack: Maximizepixiwhere xi= 0 or 1Subject to wixiM

O/multiple 1 Knapsack: Maximizepixiniif xi= 1, then ni= 1, 2, 3, 4, . . . . N pieces of objects

Subject to wixiniM

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Illustration :

Consider M = 20 , n = 3 (I, II, III)

(p1,p2,p3) = (25, 24, 15)

(w1,w2,w3) = (18, 15, 10)

Some four feasible solutions

S.No (x1, x2, x3) wixi pixi1 ( 1/2, 1/3,1/4)

18 *(1/2) + 15 * (1/3) + 10 *(1/4)

= 9 + 5 + 2.5 = 16.5

25 *(1/2) + 24 * (1/3) + 15 *(1/4)

= 12.5 + 8 + 3.75 = 24.25

2 (1, 2/15, 0)18 * 1 + 15 * (2/15) + 10 * 0

= 18 + 2 = 20

25 * 1 + 24 * (2/15) + 15 * 0

= 25 + 3.2 = 28.2

3 (0, 2/3, 1)

18 * 0 + 15 * (2/3) + 10 * 1

= 0 + 10 + 10 = 20

25 * 0 + 24 * (2/3) + 15 * 1

= 0 + 16 + 15 = 31

4 (0, 1, 1/2 )18 * 0 + 15 * 1 + 10 * (1/2)

= 0 + 15 + 5 = 20

25 * 0 + 24 * 1 + 15 * (1/2)

= 0 + 24 + 7.5 = 31.5

Observation 1 :Surely Solution 1 is sub optimal because wixi < MObservation 2 :Solution 4 appears to be an OPTIMALsolution

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Let us try different possible greedy sequencing strategy

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Strategy 1 : ( Profit based greedy sequencing)

We are interested in the largest profit.

Hence we shall fill the largest profit payable object as much

as possible ( 1)Sequence (p1, p2, p3) = (25, 24, 15)

is (I, II, III) (w1, w2, w3) = (18, 15, 10)

S.No Consider How Much ? Occupancy

wixi

Cumulative

Occupancy wixiM - wixi Profitpixi Cumulative

Profitpixi1 I 1 18 * 1 = 18 18 2018 = 2 25 * 1 = 25 25

2 II 2/15 (2/15) * 15= 2

20 0 24 * (2/15)=3.2

28.2

3 III Cannot be included as M = 0 [Profit = 28.2]

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S.No Consider How Much ? Occupancy

wixi

Cumulative


Profitpixi1 I 1 18 * 1 = 18 18 2018 = 2 25 * 1 = 25 25

2 II 15 (2/15) * 15

= 2

20 0 24 * (2/15)

=3.2

28.2

3 III Cannot be included as M = 0 [Profit = 28.2]

Solution is ( I 1, II 2/15, III 0 ) = (1, 2/15, 0)

Profit = 28.2

From the earlier table it is clear that this profit is not the best profit

WHY ?

Even though the objective function value took on large increases at

each step, the number of steps was few, as the knapsack capacity was

used up at a rapid rate

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Strategy 2 : Filling more objects by conserving more capacity

Greedy Sequencing : Objects in order of non decreasing weights

(III, II, I) (10, 15, 18) (w3, w2, w1)

(15, 24, 25) (p3, p2, p1)

S.No Consider wi Occupancy

wixi

Cumulative


Profitpixi1 III 1 10 10 2010

= 10

15 15

2 II 10/15 = 2/3 15 * (2/3)

= 10

20 0 (2/3) * 24

=16

31

3 I Cannot be included as M = 0 [Profit = 31]

Solution is ( I 0, II 2/3, III 1) (0 , 2/3, 1) PROFIT = 31

Even this solution is not the best solution

WHY ?

This time even though the capacity was used slowly, profits were not

coming rapidly enough

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Strategy 2 : A balanced strategy

Balance between the rate at which profit increases and the

rate at which capacity is used.

Sequence (II, III, I) (15, 10, 18) [w]

(24, 15, 25) [p]

Index is profit per unit capacity

Greedy Sequencing :Starting from that object which gives

maximum profit per unit capacity

[pi/wi] = [p1/w1, p2/w2, p3/w3] = [25/18, 24/15, 15/10]

= (1.389, 1.6, 1.5)

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S.No Consider wi Occupancy

wixi

Cumulative


Profitpixi1 II 1 15 15 5 24 24

2 III 5/10 = 1/2 (1/2) * 10

= 5

20 0 (1/2) * 15

=7.5

31.5

3 I Cannot be included as M = 0 [Profit = 31.5]

Solution is (0, 1, 1/2) PROFIT = 31.5

This is the optimal solution

Sequence (II, III, I) (15, 10, 18) [w]

(24, 15, 25) [p]

[pi/wi] = [p1/w1, p2/w2, p3/w3] = [25/10, 24/15, 15/10]

= (1.389, 1.6, 1.5)

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Algorithm :

1. Greedy sequence the i/p data based on pi/wiratio

2. Fill up the knapsack considering object by object filling up

as much as possible (1) till wixi = M

Filling up the knapsack O(n) - (1)

Most important is Greedy Sequencing the input

Arriving at the greedy policy

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3. Job Sequencing with deadlines

One Object / Unit time

The job cannot exceed beyond the Maximum dead line

Let n be the number of jobs & d1, d2, d3. . . ., dn be the maximum

deadlines for completing the jobs J1, J2, J3. . ., Jn

Therefore, Max no. of jobs that can be taken up =Max no. of time units available = Min(n, max(di))

If ( p1, p2, . . . , pn) are the profits brought by the

jobs (J1, J2, J3. . ., Jn) then what should be the sequence of jobs to be

executed to maximize the profit ?

Jobs should be sequenced in the order of profits to maximize the

profit

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N = 5

(p1, p2, p3, p4, p5) = (20, 15, 10, 5, 1)

deadlines (d1, d2, d3, d4,d5) = ( 2, 2, 1, 3, 3 )

Max.no of time units available = Min(n, max(di)) = Min(5,3) = 3

This decides the available slots

[01] [ 12] [23]

S.No.

Jobs

considered

in the

sequence of

profit

Deadline

Allocation Made

Remark pi pi[0-1] [1,2] [2,3]

1 1 2 1 Allocate in the

latest slot becauseearlier slot can be

useful for an object

with higher profit

and lower deadline

20 20

2 2 2 2 1 15 35

3 3 1 x Not possible - 35

4 4 3 2 1 4 5 40

J b

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S.No.

Jobs

considered

in the

sequence of

profit

Deadline

Allocation Made

Remark pi pi[0-1] [1,2] [2,3]

1 1 2 1

Allocate in the

latest slot because

earlier slot can be

useful for an object

with higher profit

and lower deadline

20 20

2 2 2 2 1 15 35

3 3 1 x Not possible - 35

4 4 3 2 1 4 5 40

All slots filled up

Hence allocation process is completed

Solution set (1, 1, 0, 1, 0) with the system slots (2, 1, X, 3, X)

(1, 2, 4) ( II, I, III)THINK:1. If execution of a job is a prerequisite to execute some other job

2. If duration of job execution is also a specification

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Transportation Problem

Demand

P4

P3

P2

Cost of transport from pito ci

Goods should be moved such that transportation

cost is minimized

P1

CapacityC5C4C3C2C1

Consumer

P

roducer

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J1 J2 J3 .. ..

E1 Efficiency of employee Eiin executing job Ji

One job to be allocated to one employee such

that overall efficiency is maximized

E2

E3

:

:

Assignment Problem

G d St t

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Greedy Strategy :

5. Minimum Spanning Tree - MST

Consider a weighted graph G(n, e, ) where wkis the

weight of edge ek ek (vi, vj)

10

11

10

10

12

12

10

9

9

88

7

1

2

3

4

5

6

7

ek

wk

vi

vj

W May correspond to traversal time if it is a road network

May correspond to volume of flow if it is a flow network

P bl

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Problem :

If one of the nodes can be chosen as the pump then

choose the edges such that total distance traversed by the

fluid to various nodes is minimum

If such a minimal distance network can be obtained, then

which node should be the pump ?

Such a minimum distance network should span through all

vertices.

In such a network there should not be a circuit, because it

provides alternate path to reach any vertex and ambiguity

exists

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Such a network should consist of minimum number of

edges.

Since the pump has to reach all vertices such a networkshould still remain connected

Hence such a network should be a TREE

Recall :

A tree is a circuit less connected graph

A tree has one and only one path between every pair ofvertices

In a tree there are (n-1) edges if n represents the total

number of vertices

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A erte in the tree ith minim m eccentricit is called the

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A vertex in the tree with minimum eccentricity is called the

treecentre.

What is eccentricity e(vi) ?

Eccentricity of a vertex vi, e(vi) is defined as the distance of

the vertex to the farthest vertex from it.

11

10

109

8

7

1

2

3

4

5

6

7

e(1) = 5

e(3) = 4

e(2) = 3

e(4) = 3

e(5) = 4

e(7) = 5

e(6) = 4

Algorithm :

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Algorithm :

Compute the path length from the vertex to all pendant

vertices in the tree

Choose the max path length This gives the eccentricity of the vertex

Path Length Should it be theoretically 1 for every edge ?

Or Should it be cost of the edge ? (Practically)

In this example e(4) = 3 is the minimum

IIy e(2) = 3 is also minimum

Therefore , Vertex 4 can be the tree centre or even vertex 2

i.e. PUMP can be located in vertex 4 or in 2

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Can there be a simpler algorithm to locate the tree centre ?

YES!!

11

10

109

8

7

1

2

3

4

5

6

7

Remove all pendant vertices

10

9 7

2

3

4

6

Continue the process

of removing pendant

vertices

9

2

4Algorithm terminates

giving the tree centres

ICASE 2:

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D

A

E

X C

B

H

I

F

G

A

X C

B

XAssignment :

1. Why should this method of peeling pendant vertices work?

Justify!!

2. Can a tree be unicentred or bicentred and if bicentred then

should the centers be adjacent

CASE 2:

Problem :

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Problem :

Given a graph G(n,e),

we have to choose (n-1) edges out of e edges [(n-1) < e]

such that

the resultant graph is connected and circuit less

and the sum of the weights of the participating edges should

be minimum.

Solution Set : is a subset of dataset

Objective Function :Minimize wi i = 1,2, . . .(n-1)Such that resulting graph is connected and circuit less

Can we sequence the edges in the most appropriate way ?

Greedy Sequencing KRUSKAL ALGORITHM

KRUSKALs Algorithm :

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KRUSKAL s Algorithm :

STEP 1 : Arrange the edges in the non-decreasing sequence

of edge weightsgreedy sequencing

STEP 2 : Insert the next edge iff it doesnt make a circuit, with

the edges already present.

Step 2 is (n-1)O(e) algorithm

What about overall algorithm?

Got to consider STEP 1

What can be max e ?

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What can be max e ?

Note:

(i) Loop edges & Parallel edges should be eliminated.

(ii) If parallel edges have to be replaced by an edge

which should be replaced ? Min Weighted edge

(iii) Then max_e n(n-1)/2

Therefore, (n-1) - O(n(n-1)/2) for step 2

But e < n(n-1)/2

Step 1 O(n2) - Sorting

Step 2 O(n2

) - Subset Creation

- HOW?

why?

Illustration :

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Illustration :

Consider

A

B

E

D

FC

5

5

4

4

4

4

6

2

1

3

3

6

D

B

A E

FC

5

5

4

4

4

4

6

23

3

A General graph is

reduced to a simp le

graph (w ithout paral lel

edges and self loop )

Greedy Sequencing :

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Greedy Sequencing :

Sequence Edge Weight

9 (A,B) 5

8 (A,E) 5

7 (A,C) 4

6 (B,D) 4

5 (C,D) 4

4 (E,F) 4

3 (B,C) 3

2 (C,E) 3

1 (D,F) 2

B

A E

FC

5

5

4

4

4

4

6

23

3

D

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n = 6 n-1 = 5 |mst| = 3

B

A E

FC

23

3

D

9 (A,B) 5

8 (A,E) 5

7 (A,C) 4

6 (B,D) 4

5 (C,D) 4

4 (E,F) 4

3 (B,C) 3

2 (C,E) 3

1 (D,F) 2


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n = 6 n-1 = 5 |mst| = 4

B

A E

FC

4

23

3

D

9 (A,B) 5

8 (A,E) 5

7 (A,C) 4

6 (B,D) 4

5 (C,D) 4

4 (E,F) 4

3 (B,C) 3

2 (C,E) 3

1 (D,F) 2


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n = 6 n-1 = 5 |mst| = 4

4

B

A E

FC

4

23

3

D4 Cannot be insertedbecause a circui t is

formed

9 (A,B) 5

8 (A,E) 5

7 (A,C) 4

6 (B,D) 4

5 (C,D) 4

4 (E,F) 4

3 (B,C) 3

2 (C,E) 3

1 (D,F) 2


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n = 6 n-1 = 5 |mst| = 4

B

A E

FC

4

23

3

D

9 (A,B) 5

8 (A,E) 5

7 (A,C) 4

6 (B,D) 4

5 (C,D) 4

4 (E,F) 4

3 (B,C) 3

2 (C,E) 3

1 (D,F) 2


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n = 6 n-1 = 5 |mst| = 5

4

B

A E

FC

4

23

3

D

9 (A,B) 5

8 (A,E) 5

7 (A,C) 4

6 (B,D) 4

5 (C,D) 4

4 (E,F) 4

3 (B,C) 3

2 (C,E) 3

1 (D,F) 2

MST is formed

Wi= 2 + 3 + 3 + 4 + 4

= 16

PRIMs Algorithm : STEP 1: Transfo rm a

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PRIM s Algorithm : STEP 1: Transfo rm a

general graph to a

simple graph.

A

B

E

D

FC

5

5

4

4

4

4

6

2

1

3

3

6

DB

A E

FC

5

5

4

4

4

4

6

23

3

PRIMs Algorithm : STEP 2: Create

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PRIM s Algorithm : STEP 2: Create

distance matr ix

(Adjacency Matrix )

B

A E

FC

5

5

4

4

4

4

6

23

3

A B C D E F

A - 5 4 5 B 5 - 3 4 C 4 3 - 4 3 D 4 4 - 6 2E 5 3 - 4F 2 4 -

PRIMs Algorithm : STEP 3: Start w ith tw o

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g

sets

MST set

|nm| = 0

B

A E

FC

D

RESIDUAL set

|nr| = 6

PRIMs Algorithm : Bring A into mst-set

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g g

MST set RESIDUAL set

|nm| = 1 |nr| = 5

B

A E

FC

D

PRIMs Algorithm : STEP 3: Bring A in to

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g

B

E

FC

D

g

mst-set


|nm| = 1 |nr| = 5

A

There are many edges that connect these two sets :

Like (A,B) = 5, (A,C) = 4, (A,E) = 5

PRIMs Algorithm : Choose the m inimum of al l

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g

B

E

FC

D

these connected edges

(A,B) = 5, (A,C) = 4, (A,E) = 5


|nm| = 1 |nr| = 5

A

Therefore, C moves into the MST set along with the edge AC

PRIMs Algorithm : Choose the m inimum of al l

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g

B

E

F

D

these connected edges

(A,B) = 5, (A,C) = 4, (A,E) = 5


|nm| = 2 |nr| = 4

A

Therefore, C moves into the MST set along with the edge AC

C

4

|nm| = 2 |nr| = 4

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B

E

F

D


| m| | r|

A

C

4

A B C D E F

A 5 5 B - 4 C 3 4 3 D 4 - 6 2

E 3 - 4F 2 4 -

This amounts to

deriving the following

residual distance

matrix

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Repeat the previou s operat ions

PRIMs Algorithm :Th d th t

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B

E

F

D

RESIDUAL set

|nr| = 4

MST set

|nm| = 2

A

C

4

(A,B) = 5

(A,E) = 5

(C,B) = 3

(C,D) = 4

(C,E) = 3

There are many edges that

connect these two sets

Choose the minimum i.e (C,B) or (C,E)

Say (C,B)

PRIMs Algorithm :

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E

F

D

RESIDUAL set

|nr| = 3

MST set

|nm| = 3

A

C

4

PRIM s Algorithm :

(A,B) = 5

(A,E) = 5

(C,B) = 3

(C,D) = 4

(C,E) = 3

There are many edges that


Choose the minimum i.e (C,B) or (C,E)

Say (C,B)

B

3

|nr| = 4|nm| = 2

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E

F

D

RESIDUAL setMST set

| m|

A

C

4

B3

A B C D E F

A 5 B 4 C 4 3 D - 6 2

E 3 - 4

F 2 4 -

Reduced distance

matr ix

PRIMs Algorithm :

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PRIM s Algorithm :There are many edges that


Choose the minimum i.e. (C,E)

E

F

D

RESIDUAL set

|nr| = 3

MST set

|nm| = 3

A

C

4

(A,E) = 5

(B,D) = 4

(C,D) = 4

(C,E) = 3B

3

PRIMs Algorithm :

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Choose the minimum i.e. (C,E)

F

D

RESIDUAL set

|nr| = 2

MST set

|nm| = 4

A

C

4

(A,E) = 5

(B,D) = 4

(C,D) = 4

(C,E) = 3B

3

E

3

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PRIMs Algorithm :

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Choose the minimum i.e. Say (B,D)

F

D

RESIDUAL set

|nr| = 2

MST set

|nm| = 4

A

C

4

(B,D) = 4

(C,D) = 4

(E,F) = 4B

3

E

3

PRIMs Algorithm :

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Choose the minimum i.e. Say (B,D)

F

RESIDUAL set

|nr| = 1

MST set

|nm| = 5

A

C

4

(B,D) = 4

(C,D) = 4

(E,F) = 4B

3

E

D4

3

|nr| = 1|nm| = 5

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F

RESIDUAL setMST set

m

A

C

4

B3

E

D

A B C D E F

A B C D 2

E 4

F -

Reduced distance

matr ix

4

3

PRIMs Algorithm :

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Choose the minimum i.e. (D,F)

F

RESIDUAL set

|nr| = 1

MST set

|nm| = 5

A

C

4

(C,F) = (D,F) = 2

(E,F) = 4B

3

E

D4

(B,F) = (A,F) =

3

PRIMs Algorithm :

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s go tThere are many edges that


Choose the minimum i.e. (D,F)

RESIDUAL set

|nr| = 0

MST set

|nm| = 6

A

C

4

(C,F) = (D,F) = 2

(E,F) = 4B

3

E

D4

(B,F) = (A,F) =

F2

Since nr= the algorithm terminates

Wi = 4 + 3 + 3 + 4 + 2 = 16

3

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6. Compare and contrast Prims algorithm & Kruskalss algorithm.

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7. Show that a mst shall always contain the most minimum weighted

edge and the pendant edge irrespective of its weight (if present).

8. If the signal level falls while reaching different vertices from PUMP

vertex (tree centre) then BOOSTERS have to be inserted atappropriate nodes (vertices). This requires Tree-Vertex-Splitting.

Tree Vertex Splitting is also a greedy algorithm. Study this algorithm

(HSR)

9. Implement tree-centre locating algorithms

10.How do you argue that Prims algorithm is also a Greedy Strategy

based ?

11.There is one more algorithm to generate mst called randomized mst

algorithm. Study this.

12.Propose a solution

In a locality a closed circuit cable TV connection has to be provided.

Propose the layout. Which should be the transmitter nodes? Where

should the amplifier (boosters) be placed ?

Additional Reference : N.DEO, Graph Theory

Extension to Clustering

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Zahns Algorithm

STEP 1 :Given m samples in n-d space. Obtain the adjacencyor distance matrix

1 2 . . m

1

2

:

m

STEP 2 Obt i MST ( ll P i l ith i d)

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STEP 2 :Obtain MST (generally Prims algorithm is used)

STEP 3 :Define the Undesiredbranches in the MST

whose weights are more than some threshold

(Perhaps average of all branch weights)

STEP 4 :Remove these undesiredbranches. A MST breaks

into a forest of connected components. Each

connected component is a cluster.

6 Single Source Shortest Paths:

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6. Single Source Shortest Paths:

Given a network find the shortest path from a source point to

a destination point. (Dijkstras Algorithm)Illustration :

V0

V2

V1

V3

V4

V5

50 10

1020

15

20

15 35

30

3

45

Source: V0 ( V1, V2, V3, V4, V5) Shortest Path ?

STEP 1 : 45

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DISTANCE MATRIX

V0 V1 V2 V3 V4 V5

V0 - 50 10 45 V1 - 15 10 V2 20 - 15 V3 20 - 35 V4 30 - V

5 3 -

V0

V2

V1

V3

V4

V5

50 10

10

20

15

2015 35

30

3

45

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From V0to

V1 V2 V3 V4 V5

50 10 45

Min is V0 V2 = 10

V0

V2

V1

V3

V4

V5

50 10

10

20

15

2015 35

30

3

STEP 2 :

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From V0to

to V1 V3 V4 V5

V0 Vw 50 45 V0 Vw

Via V2

V0 V2 Vw

10 + =

10 + 15

= 25

10 + =

10 + =

Choose the

minimum

50 25 45

Therefore , The minimum is 25 V0 V3

Which is V0 V2 V3

STEP 3 : V5V4V1to

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From V0to

Choose the minimum 45

Shortest Path V0 V3 V1 i.e 45

25 + =

25 + 35

= 60

25 + 20

= 45V0 V3 Vw

V0 V2 Vw

4550V0 Vw

4545Choose theminimum

V0

V2

V1

V3

V4

V5

50 10

1020

15

2015 35

30

3

45

STEP 4 : to V4 V5

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From V0to

Min_Path V0 V4

Min_Cost 45

V0 Vw 45 V0 V2 Vw V0 V3 Vw 60 V0 V1 Vw

45 + 10

= 55

45 + =

Choose the

minimum45

V0

V2

V1

V3

V4

V5

50 10

1020

15

2015 35

30

3

45

STEP 5 : to V5

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From V0to

Min V0 V5

V0 Vw V0 V2 Vw V0 V3 Vw V0 V1 Vw V0 V4 Vw

Hence the shortest paths from vo :

V1 V2 V3 V4 V5

45 10 25 45

V0

V2

V1

V3

V4

V5

50 10

1020

15

2015 35

30

3

45

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1. Formulate the algorithm. Implement it. Estimate the timeand space required.

2. Obtain different paths covering all vertices from V0

3. Obtain all paths from every vertex to all other vertices

4. Obtain different paths from every vertex to one destination

vertex

5. Check what would happen if the weight of one edge isve?

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Greedy Method 16-11-2013