COMP 482: Design and
Analysis of Algorithms
Spring 2013
Lecture 7
Prof. Swarat Chaudhuri
Recap: Interval Scheduling
Interval scheduling.
Job j starts at sj and finishes at fj.
Two jobs compatible if they don't overlap.
Goal: find maximum subset of mutually compatible jobs.
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Time
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Recap: Interval Scheduling
Algorithm. Greedy algorithm; objective is Earliest Finish Time.
Argument for optimality: Greedy stays ahead
job ir+1 finishes before jr+1
Greedy:
i1
i1
ir
OPT:
j1
j2
jr
ir+1
jr+1
...
why not replace job jr+1
with job ir+1?
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Also applicable to Truck Driver’s problem (selection of breakpoints)
Variant: 24-7 Interval Scheduling
You have a processor that can operate 24-7. People submit requests to
run daily jobs on the processor. Each such job comes with a start
time and an end time; if the job is accepted it must run continuously
for the period between the start and end times, EVERY DAY. (Note
that some jobs can start before midnight and end after midnight.)
Given a list of n such jobs, your goal is to accept as many jobs as
possible (regardless of length), subject to the constraint that the
processor can run at most one job at any given point of time. Give
an algorithm to do this.
For example, here you have four jobs (6pm, 6am), (9pm, 4am), (3am,
2pm), (1pm, 7pm).
The optimal solution is to pick the second and fourth jobs.
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Solution
Let I1,…,In be the n intervals. We call an Ij-restricted solution one that
contains the interval Ij.
Here’s an algorithm, for fixed j, to compute an Ij-restricted solution of
maximum size. Let x be a point in Ij. First delete Ij and all intervals
that overlap it. The remaining intervals do not contain the point x,
so we can cut the timeline at x and produce an instance of the
Interval Scheduling Problem from class. This takes O(n) time
assuming intervals are sorted by ending time.
Now, the algorithm for the full problem is to compute an Ij-restricted
solution of maximum size for each j = 1,…,n. This takes a total of
O(n2) time. We now pick the largest of these solutions and claim
that it is the optimal.
Why? Consider the optimal solution to the full problem. Suppose this
produces a set of intervals S. There must be SOME Ij in S, so the
solution is an optimal Ij-restricted solution. But then our algorithm
would find it.
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Recap: Interval Partitioning
Interval partitioning.
Lecture j starts at sj and finishes at fj.
Goal: find minimum number of classrooms to schedule all lectures so
that no two occur at the same time in the same room.
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Ex: This schedule uses 4 classrooms to schedule 10 lectures.
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Recap: Interval Partitioning
Interval partitioning.
Lecture j starts at sj and finishes at fj.
Goal: find minimum number of classrooms to schedule all lectures so
that no two occur at the same time in the same room.
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Greedy algorithm. Consider lectures in increasing order of start time.
Assign lecture to any compatible existing classroom. If not possible,
open new classroom.
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Interval Partitioning: Lower Bound on Optimal Solution
Def. The depth of a set of open intervals is the maximum number that
contain any given time.
Key observation. Number of classrooms needed by any algorithm 
depth.
Argument for optimality. Show that depth  number of classrooms
opened by greedy algorithm.
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Greedy Analysis Strategies
Greedy algorithm stays ahead. Show that after each step of the
greedy algorithm, its solution is at least as good as any other
algorithm's.
Structural. Discover a simple "structural" bound asserting that every
possible solution must have a certain value. Then show that your
algorithm always achieves this bound.
Exchange argument. Gradually transform any solution to the one found
by the greedy algorithm without hurting its quality.
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Q1: Coin Changing
Goal. Given currency denominations: 1, 5, 10, 25, 100, devise a method
to pay amount to customer using fewest number of coins.
Ex: 34¢.
Cashier's algorithm. At each iteration, add coin of the largest value
that does not take us past the amount to be paid.
Ex: $2.89.
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Q1: Coin-Changing: Greedy Algorithm
Cashier's algorithm. At each iteration, add coin of the largest value
that does not take us past the amount to be paid.
Sort coins denominations by value: c1 < c2 < … < cn.
coins selected
S  
while (x  0) {
let k be largest integer such that ck  x
if (k = 0)
return "no solution found"
x  x - ck
S  S  {k}
}
return S
Q1. Is cashier's algorithm optimal?
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Coin-Changing: Analysis of Greedy Algorithm
Theorem. Greed is optimal for U.S. coinage: 1, 5, 10, 25, 100.
Pf. (by induction on x)
Consider optimal way to change ck  x < ck+1 : greedy takes coin k.
We claim that any optimal solution must also take coin k.
– if not, it needs enough coins of type c1, …, ck-1 to add up to x
– table below indicates no optimal solution can do this
Problem reduces to coin-changing x - ck cents, which, by induction, is
optimally solved by greedy algorithm. ▪
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k
ck
All optimal solutions
must satisfy
Max value of coins
1, 2, …, k-1 in any OPT
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1
P4
-
2
5
N1
4
3
10
N+D2
4+5=9
4
25
Q3
20 + 4 = 24
5
100
no limit
75 + 24 = 99
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Coin-Changing: Analysis of Greedy Algorithm
Observation. Greedy algorithm is sub-optimal for US postal
denominations: 1, 10, 21, 34, 70, 100, 350, 1225, 1500.
Counterexample. 140¢.
Greedy: 100, 34, 1, 1, 1, 1, 1, 1.
Optimal: 70, 70.
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Greedy Analysis Strategies
Greedy algorithm stays ahead. Show that after each step of the
greedy algorithm, its solution is at least as good as any other
algorithm's.
Structural. Discover a simple "structural" bound asserting that every
possible solution must have a certain value. Then show that your
algorithm always achieves this bound.
Exchange argument. Gradually transform any solution to the one found
by the greedy algorithm without hurting its quality.
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4.2 Scheduling to Minimize Lateness
Scheduling to Minimizing Lateness
Minimizing lateness problem.
Single resource processes one job at a time.
Job j requires tj units of processing time and is due at time dj.
If j starts at time sj, it finishes at time fj = sj + tj.
Lateness: j = max { 0, fj - dj }.
Goal: schedule all jobs to minimize maximum lateness L = max j.
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Ex:
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dj
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14
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lateness = 2
d3 = 9
0
1
d2 = 8
2
d6 = 15
3
4
d1 = 6
5
6
7
lateness = 0
max lateness = 6
d5 = 14
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d4 = 9
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Minimizing Lateness: Greedy Algorithms
Greedy template. Consider jobs in some order.
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[Shortest processing time first] Consider jobs in ascending order
of processing time tj.
[Earliest deadline first] Consider jobs in ascending order of
deadline dj.
[Smallest slack] Consider jobs in ascending order of slack dj - tj.
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Minimizing Lateness: Greedy Algorithms
Greedy template. Consider jobs in some order.
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[Shortest processing time first] Consider jobs in ascending order
of processing time tj.
1
2
tj
1
10
dj
100
10
counterexample
[Smallest slack] Consider jobs in ascending order of slack dj - tj.
1
2
tj
1
10
dj
2
10
counterexample
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Minimizing Lateness: Greedy Algorithm
Greedy algorithm. Earliest deadline first.
Sort n jobs by deadline so that d1  d2  …  dn
t  0
for j = 1 to n
Assign job j to interval [t, t + tj]
sj  t, fj  t + tj
t  t + tj
output intervals [sj, fj]
max lateness = 1
d1 = 6
0
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d2 = 8
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d3 = 9
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d4 = 9
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d5 = 14
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d6 = 15
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Minimizing Lateness: No Idle Time
Observation. There exists an optimal schedule with no idle time.
d=4
0
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d = 12
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d = 12
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Observation. The greedy schedule has no idle time.
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Minimizing Lateness: Inversions
Def. An inversion in schedule S is a pair of jobs i and j such that:
di < dj but j scheduled before i.
inversion
before swap
j
i
Observation. Greedy schedule has no inversions.
Observation. All schedules with no idle time and no inversions have
same maximum lateness.
Observation. If a schedule (with no idle time) has an inversion, it has
one with a pair of inverted jobs scheduled consecutively.
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Minimizing Lateness: Inversions
Def. An inversion in schedule S is a pair of jobs i and j such that:
di < dj but j scheduled before i.
inversion
fi
j
before swap
i
i
after swap
j
f'j
Claim. Swapping two adjacent, inverted jobs reduces the number of
inversions by one and does not increase the max lateness.
Pf. Let  be the lateness before the swap, and let  ' be it afterwards.
'k = k for all k  i, j
'i  i
¢j = fj¢ - d j
(definitio n)
If job j is late:
= f -d
( j finishes at time f )
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j
i
£ fi - di
£
i
(i < j )
i
(definitio n)
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Minimizing Lateness: Analysis of Greedy Algorithm
Theorem. Greedy schedule S is optimal.
Pf. Define S* to be an optimal schedule that has the fewest number of
inversions, and let's see what happens.
Can assume S* has no idle time.
If S* has no inversions, then S = S*.
If S* has an inversion, let i-j be an adjacent inversion.
– swapping i and j does not increase the maximum lateness and
strictly decreases the number of inversions
– this contradicts definition of S* ▪
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Q2: Subsequences
Suppose you have a collection of possible events (e.g., possible
transactions) and a sequence S of n events. A given event may occur
multiple times—e.g., you could have an event “buy Google stock”
multiple times in a log of transactions.
A sequence S’ is a subsequence of a sequence S if there’s a way to
delete certain events from S such that the remaining sequence
equals S’. For example, the reason to do this could be patternmatching.
Give an algorithm that takes two sequences of events—S’ of length m
and S of length n—and decides in time O(m+n) whether S’ is a
subsequence of S.
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Solution
Greedy algorithm: Let the i-th event of S be S(i). Find the first event
in S that matches S’(1), then the second event in S that matches
S’(2), and so on. The running time is O(m+n).
It is easy to show that if the algorithm finds a match, then S’ is in fact
a subsequence of S.
More difficult direction: if the algorithm does not find a match, then
no match exists.
The proof of this is by contradiction. Suppose S’ matches the
subsequence S(l1).S(l2)…S(lm). Suppose GREEDY produces the
sequence S(k1).S(k2)…. Show that greedy can produce a match all
the way up to S(km) and also ki ≤ li for all i. This is done in a way
similar to the proof in interval scheduling.
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