Instructor
Neelima Gupta
[email protected]
Table of Contents
Greedy Algorithms
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.
 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
Proving the Optimality
Scheduling Jobs with Processing
times and Deadlines
 Jobs are given with processing times pi and deadlines
di .
 There is no specified start time. A job can be
scheduled at any time. Algorithm decides the start
time ( and hence the finish time) of a job.
 Let fi be the time at which a job finishes as per some
schedule. Then its lateness is defined as
li = fi – di
if fi>di
0
, otherwise
 Aim : minimize max lateness i.e. minimize max{li}
Figures from Anjali, Hemant
Possible Greedy Approaches
 Shortest Job First: completely ignores half of the input
data viz. the deadlines
 Doesn’t work :
 t1 = 1, d1=100, t2=10, d2=10
 Minimum Slackness First
 Doesn’t work :
 t1 = 1, d1=2, t2=10, d2=10
 Earliest Deadline First: completely ignores the other
half of the input data viz. the processing time….but it
works….gives the optimal
SJF fails: figure from Anjali,
Hemant
back
Minimum Slackness First
 Let si be the time by which the job must be assigned
to meet the deadline. i.e. si = di – pi
 Let the last job scheduled finishes at time t. Then slacktime for job i is defined as sti = si – t.
 Thus slack time represents, how much we can wait/defer
to schedule the ith job.
 We should schedule the next job for which this time is
minimum. i.e. sti is minimum. Since t is same for all the
jobs, the job with minimum si is scheduled next.
MSF fails: figure from Anjali,
Hemant
back
Earliest Deadline First
d1<=d2<=d3…..<=dn
s2=f1
j1
j2
f1=p1
s3=f2
j3
f2=f1+p2
Exchange Argument
• Let O be an optimal solution and S be the
solution obtained by our greedy.
• Gradually we transform O into S without
hurting its (O’s) quality. Thereby implying
that |O| = |S|.
• Hence proving that greedy is optimal.
Inversion
We say that a schedule A has an inversion if
a job i with deadline di is scheduled before
another job j with earlier deadline( dj<di).
 Idle Time
-The time that passes during a gap
-There is work to be done, yet for some
reason the machine is sitting idle.
Earliest Deadline First
 Claim: Our schedule has no inversions and no idle
time….trivial
 Clearly Optimal has no idle time.
 1. All schedules with no inversion and no idle time has
same maximum lateness
 Proof:
 2. There is an optimal schedule with no inversion and
no idle time.
 A. If O (an optimal) has an inversion then there is a pair
of jobs I an j such that j is scheduled immediately after I
and d_j < d_i….i.e. pair of inverted jobs that are
consecutive.
 Proof:
 B. The new swapped schedule has maximum lateness no
larger than that of O.
 Proof:
Proof of 1. :Anjali, Hemant
Network Design Problems
 Given a set of routers placed at n locations
V = {v1 … vn}
 We want to connect them in a cheapest way.
 Claim: The minimum cost solution to the above
problem is a tree.
Minimum Spanning Tree
 Assume that all edges have distinct edge costs. We’ll
remove this restriction later.
 Greedy Approaches:
 Kruskal
 Prim’s
 Reverse Delete
 Cut Property: Let S be a non-trivial subset of V, let e =
(v, w) be the minimum weight edge with one end in S
and the other end in V – S. Then every MST contains e.