Instructor
Neelima Gupta
ngupta@cs.du.ac.in
Greedy Approach
A tool to design algorithms for optimization
problems
What is greedy approach?
 Choosing a current best solution without worrying
about future. In other words the choice does not
depend upon future sub-problems.
What is greedy approach?
 Such algorithms are locally optimal,
 For some problems, as we will see shortly, this local
optimal is global optimal also and we are happy.
General ‘Greedy’ Approach
 Step 1:
 Choose the current best solution.
 Step 2:
 Obtain greedy solution on the rest.
When to use?
 There must be a greedy choice to make.
 The problem must have an optimal substructure.
Activity Selection Problem
 Given a set of activities, S = {a1, a2, …, an} that need to
use some resource.
 Each activity ai has a possible start time si & finish time
fi, such that 0  si < fi < 
 We need to allocate the resource in a compatible
manner, such that the number of activities getting the
resource is maximized.
 The resource can be used by one and only one activity at
any given time.
.
Activity Selection Problem
 Two activities ai and aj are said to be compatible, if the
interval they span do not overlap. ..i.e. fi  sj or f j  si
 Example:
 Consider activities: a1, a2, a3, a4
s1--------f1
s2---------f2
s3------f3
s4------f4
 Here a1 is compatible with a3 & a4
 a2 is compatible with a3 & a4
 But a3 and a4 themselves are not compatible.
Activity Selection Problem
 Solution: Applying the general greedy algorithm
 Select the current best choice, a1 add it to the solution
set.
 Construct a subset S’ of all activities compatible with a1,
find the optimal solution of this subset.
 Join the two.
Lets think of some possible
greedy solutions
 Shortest Job First
 In the order of increasing start times
 In the order of increasing finish times
Shortest Job First
job1
job2
job3
0
1
2
3
4
5
6
7
8
9
Time
Thanks to: Navneet Kaur(22), MCA 2012
10
11
12
13
14
15
Shortest Job First
job1
job2
job3
0
1
2
3
4
5
6
7
8
9
Time
Thanks to: Navneet Kaur(22), MCA 2012
10
11
12
13
14
15
Shortest Job First
job1
job2
job3
0
1
2
3
4
5
6
7
SCHEDULE CHOSEN BY THIS
APPROACH
OPTIMAL SCHEDULE
8
9
Thanks to: Navneet Kaur(22), MCA 2012
10
11
12
13
14
Time
15
Increasing Start Times
job1
job2
job3
0
2
4
6
8
10
12
14
16
18
20
Time
Thanks to: Navneet Kaur(22),
MCA 2012
Increasing Start Times
job1
job2
job3
0
2
4
6
8
10
12
14
16
18
20
Time
Thanks to: Navneet Kaur(22),
MCA 2012
Increasing Start Times
job1
job2
job3
0
2
14
4
6
8
10
12
SCHEDULE CHOSEN BY THIS
APPROACH
OPTIMAL SCHEDULE
16
Thanks to: Navneet Kaur(22),
MCA 2012
18
20
Time
Increasing Finishing Times
i
2
1
3
4
5
Si
Fi
2
1
4
5
6
Thanks to Neha (16)
Pi
4
5
6
8
9
3
10
4
20
2
Increasing Finishing Times
P(1)=10
P(2)=3
P(3)=4
P(4)=20
P(5)=2
0
1
2
3
4
5
6
7
Thanks to Neha (16)
8
9
Time
Increasing Finishing Times
P(1)=10
P(2)=3
P(4)=20
P(5)=2
0
1
2
3
4
5
6
7
.
Thanks to Neha (16)
8
9
Time
ACTIVITY SELECTION PROBLEM
We include a₁ in the solution.
And then recurse on
S′ = {aₓ ԑ S-{a₁} : aₓ is compatible with a₁}
where S is input set of activities.
Thanks to: Navneet Kaur(22), MCA 2012
ACTIVITY SELECTION PROBLEM
CLAIM: If B′ is an optimal solution of S′, then
B=B′  {a1} is an optimal solution of S.
PROOF: Suppose  an imaginary solution B″,
which is optimal and includes a1 .
Suppose length of B″, i.e.,
|B″| = k″
Thanks to: Navneet Kaur(22), MCA 2012
ACTIVITY SELECTION PROBLEM
Now, we have to prove two things:
I. B is feasible.
II. |B| = k″
OR
we can prove that
|B′| = k″ - 1
Thanks to: Navneet Kaur(22), MCA 2012
ACTIVITY SELECTION PROBLEM
Proof of I. --B′ is a subset of S′.
And S′ is compatible with a₁ .
Hence, B is feasible.
Thanks to: Navneet Kaur(22), MCA 2012
ACTIVITY SELECTION PROBLEM
Proof of II. --Consider the set B″ - {a₁}
i.) Can |B′| ≥ k″ ?
If yes, then
|B′  {a₁}| ≥ k″ + 1
Thanks to: Navneet Kaur(22), MCA 2012
ACTIVITY SELECTION PROBLEM
But, this is contradiction to a problem that
B″ is optimal because |B″| = k″
And if the size of optimal solution is k″,
then we cannot have a solution of size
greater than k″ and this is giving a solution of
size k″+1, which is not possible.
Hence, statement (i) is wrong.
Thanks to: Navneet Kaur(22), MCA 2012
ACTIVITY SELECTION PROBLEM
(ii) Can |B′| < k″ - 1 ?
Consider B″- {a₁}.
This is a feasible solution of S′.
This implies that OPT(S′) ≥ k″ - 1
Hence,
Statement (ii) is wrong.
Thanks to: Navneet Kaur(22), MCA 2012
ACTIVITY SELECTION PROBLEM
From (i) and (ii), we get
|B′| = k″ - 1
This implies that |B| = k″.
Hence, B is optimal.
Thanks to: Navneet Kaur(22), MCA 2012
Activity Selection Problem
Statement:
 an optimal solution to a problem that contains a1
Proof:
Let A = {ak,…} be an optimal solution. Let ak be the first
activity in A i.e. the finishing time of ak is the least.
Construct another solution:
B = A – {ak}  {a1} = {a1,…}
Activity Selection Problem
Proof continued…
Clearly, f1  f k thus B is a set of compatible activities,
hence an optimal solution too.
Activity Selection Problem
Statement:
The solution is globally optimal.
Proof:
Suppose B = {a1…} has an optimal solution containing k+1
elements. (a1 being the first element)
Clearly, B – {a1} has an optimal solution with k elements.
Activity Selection Problem
Proof continued…
Now, suppose for B’ = B - {a1}  another optimal solution
containing more than k elements.
Then we can construct another optimal solution
B* = B’  {a1} with more than k+1 elements.
This is a contradiction to our assumption of an optimal
solution with k+1 elements.
FRACTIONAL KNAPSACK PROBLEM
Given a set S of n items, with value vi and weight wi and a
knapsack with capacity W.
Aim: Pick items with maximum total value but with weight
at most W. You may choose fractions of items.
GREEDY APPROACH
 Pick the items in the decreasing order of value per unit
weight i.e. highest first.
Example
knapsack capacity 50
Item 2
item 3
Item 1
30
20
10
● vi
= 60
vi = 100
vi/ wi
=6
vi/ wi
Thanks to:
Neha Katyal
vi
=5
vi/ wi
= 120
=4
Example
knapsack capacity 50
Item 2
●
20
item 3
30
10
●
vi = 100
vi/ wi = 5
Thanks to:
Neha
Katyal
vi = 120
vi/ wi = 4
60
Example
knapsack capacity 50
item 3
20
30
●
vi = 120
vi/ wi = 4
Thanks to:
Neha Katyal
+
10
●
100
60
Example
knapsack capacity 50
$80
20/
30
+
20
100
+
10
60
= 240
Thanks to:
Neha Katyal
Up Next
 Dynamic Programming
ACTIVITY SELECTION PROBLEM
Options that could be followed while scheduling the jobs:
 Shortest Job First
Eg. Three jobs to be scheduled:
Job1- start=5, end=10
Job2- start=1, end=7
Job3- start=8, end=15
Our shortest job first would schedule just job1
But the optimal algorithm would have scheduled 2 jobs job2 and job3.
So this approach is not working.
Thanks to: Navneet Kaur(22), MCA 2012
Next option that could be followed while scheduling the
jobs:
 Smallest start time first
Eg. Three jobs to be scheduled:
Job1- start=1, end=20
Job2- start=2, end=7
Job3- start=8, end=15
Our smallest start time first would schedule just job1
But the optimal algorithm would have scheduled 2 jobs job2 and job3.
So this approach is also not working.
Thanks to: Navneet Kaur(22), MCA 2012