04greedy 2x2

1
Chapter 4
Greedy
Algorithms
Slides by Kevin Wayne.
Copyright © 2005 Pearson-Addison Wesley.
All rights reserved.
4.1 Interval Scheduling
3
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.
Time
0 1 2 3 4 5 6 7 8 9 10 11
f
g
h
e
a
b
c
d
4
Interval Scheduling: Greedy Algorithms
Greedy template. Consider jobs in some natural order.
Take each job provided it's compatible with the ones already taken.
 [Earliest start time] Consider jobs in ascending order of sj.
 [Earliest finish time] Consider jobs in ascending order of fj.
 [Shortest interval] Consider jobs in ascending order of fj - sj.
 [Fewest conflicts] For each job j, count the number of
conflicting jobs cj. Schedule in ascending order of cj.

5
Interval Scheduling: Greedy Algorithms
Greedy template. Consider jobs in some natural order.
counterexample for earliest start time
counterexample for shortest interval
counterexample for fewest conflicts
6
Greedy algorithm. Consider jobs in increasing order of finish time.
Implementation. O(n log n).
 Remember job j* that was added last to A.
 Job j is compatible with A if sj ≥ fj*.
Sort jobs by finish times so that f1 ≤ f2 ≤ ... ≤ fn.
A ← φ
for j = 1 to n {
if (job j compatible with A)
A ← A ∪ {j}
}
return A
set of jobs selected
Interval Scheduling: Greedy Algorithm
7
Interval Scheduling: Analysis
Theorem. Greedy algorithm is optimal.
Pf. (by contradiction)
 Assume greedy is not optimal, and let's see what happens.
 Let i1, i2, ... ik denote set of jobs selected by greedy.
 Let j1, j2, ... jm denote set of jobs in the optimal solution with
i1 = j1, i2 = j2, ..., ir = jr for the largest possible value of r.
j1 j2 jr
i1 i2 ir ir+1
. . .
Greedy:
OPT: jr+1
why not replace job jr+1
with job ir+1?
job ir+1 finishes before jr+1
8
j1 j2 jr
i1 i2 ir ir+1
Interval Scheduling: Analysis
 Let i1, i2, ... ik denote set of jobs selected by greedy.
 Let j1, j2, ... jm denote set of jobs in the optimal solution with
i1 = j1, i2 = j2, ..., ir = jr for the largest possible value of r.
. . .
Greedy:
OPT:
solution still feasible and optimal,
but contradicts maximality of r.
ir+1
job ir+1 finishes before jr+1

4.1 Interval Partitioning
10
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.
Ex: This schedule uses 4 classrooms to schedule 10 lectures.
Time
9 9:30 10 10:30 11 11:30 12 12:30 1 1:30 2 2:30
h
c
b
a
e
d g
f i
j
3 3:30 4 4:30
1
2
3
4
11
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.
Ex: This schedule uses only 3.
Time
9 9:30 10 10:30 11 11:30 12 12:30 1 1:30 2 2:30
h
c
a e
f
g i
j
3 3:30 4 4:30
d
b
1
2
3
12
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 ≥ depth.
Ex: Depth of schedule below = 3 ⇒ schedule below is optimal.
Q. Does there always exist a schedule equal to depth of intervals?
Time
9 9:30 10 10:30 11 11:30 12 12:30 1 1:30 2 2:30
h
c
a e
f
g i
j
3 3:30 4 4:30
d
b
a, b, c all contain 9:30
1
2
3

13
Interval Partitioning: Greedy Algorithm
Greedy algorithm. Consider lectures in increasing order of start time:
assign lecture to any compatible classroom.
Implementation. O(n log n).
 For each classroom k, maintain the finish time of the last job added.
 Keep the classrooms in a priority queue.
Sort intervals by starting time so that s1 ≤ s2 ≤ ... ≤ sn.
d ← 0
for j = 1 to n {
if (lecture j is compatible with some classroom k)
schedule lecture j in classroom k
else
allocate a new classroom d + 1
schedule lecture j in classroom d + 1
d ← d + 1
}
number of allocated classrooms
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Interval Partitioning: Greedy Analysis
Observation. Greedy algorithm never schedules two incompatible
lectures in the same classroom.
Pf.
 Let d = number of classrooms that the greedy algorithm allocates.
 Classroom d is opened because we needed to schedule a job, say j,
that is incompatible with all d-1 other classrooms.
 These d jobs each end after sj.
 Since we sorted by start time, all these incompatibilities are caused
by lectures that start no later than sj.
 Thus, we have d lectures overlapping at time sj + ε.
 Key observation ⇒ all schedules use ≥ d classrooms. ▪
4.2 Scheduling to Minimize Lateness
16
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: lj = max { 0, fj - dj }.
 Goal: schedule all jobs to minimize maximum lateness L = max lj.
Ex:
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
d5 = 14d2 = 8 d6 = 15 d1 = 6 d4 = 9d3 = 9
lateness = 0lateness = 2
dj 6
tj 3
1
8
2
2
9
1
3
9
4
4
14
3
5
15
2
6
max lateness = 6

17
Minimizing Lateness: Greedy Algorithms
Greedy template. Consider jobs in some order.
 [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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Greedy template. Consider jobs in some order.
 [Shortest processing time first] Consider jobs in ascending order
of processing time tj.
 [Smallest slack] Consider jobs in ascending order of slack dj - tj.
counterexample
counterexample
dj
tj
100
1
1
10
10
2
dj
tj
2
1
1
10
10
2
Minimizing Lateness: Greedy Algorithms
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0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
d5 = 14d2 = 8 d6 = 15d1 = 6 d4 = 9d3 = 9
max lateness = 1
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]
Minimizing Lateness: Greedy Algorithm
Greedy algorithm. Earliest deadline first.
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Minimizing Lateness: No Idle Time
Observation. There exists an optimal schedule with no idle time.
Observation. The greedy schedule has no idle time.
0 1 2 3 4 5 6
d = 4 d = 6
7 8 9 10 11
d = 12
0 1 2 3 4 5 6
d = 4 d = 6
7 8 9 10 11
d = 12

21
Minimizing Lateness: Inversions
Def. Given a schedule S, an inversion is a pair of jobs i and j such that:
i < j but j scheduled before i.
Observation. Greedy schedule has no inversions.
Observation. If a schedule (with no idle time) has an inversion, it has
one with a pair of inverted jobs scheduled consecutively.
ijbefore swap
fi
inversion
[ as before, we assume jobs are numbered so that d1 ≤ d2 ≤ … ≤ dn ]
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Minimizing Lateness: Inversions
Def. Given a schedule S, an inversion is a pair of jobs i and j such that:
i < j but j scheduled before i.
Claim. Swapping two consecutive, inverted jobs reduces the number of
inversions by one and does not increase the max lateness.
Pf. Let l be the lateness before the swap, and let l ' be it afterwards.
 l 'k = lk for all k ≠ i, j
 l 'i ≤ li
 If job j is late:
ij
i j
before swap
after swap
!
"l j = "fj # dj (definition)
= fi # dj ( j finishes at time fi )
$ fi # di (i < j)
$ li (definition)
f'j
fi
inversion
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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* ▪
24
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.
Other greedy algorithms. Kruskal, Prim, Dijkstra, Huffman, …

4.3 Optimal Caching
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Optimal Offline Caching
Caching.
 Cache with capacity to store k items.
 Sequence of m item requests d1, d2, …, dm.
 Cache hit: item already in cache when requested.
 Cache miss: item not already in cache when requested: must bring
requested item into cache, and evict some existing item, if full.
Goal. Eviction schedule that minimizes number of cache misses.
Ex: k = 2, initial cache = ab,
requests: a, b, c, b, c, a, a, b.
Optimal eviction schedule: 2 cache misses.
a b
a b
c b
c b
c b
a b
a
b
c
b
c
a
a ba
a bb
cacherequests
red = cache miss
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Optimal Offline Caching: Farthest-In-Future
Farthest-in-future. Evict item in the cache that is not requested until
farthest in the future.
Theorem. [Bellady, 1960s] FF is optimal eviction schedule.
Pf. Algorithm and theorem are intuitive; proof is subtle.
a b
g a b c e d a b b a c d e a f a d e f g h ...
current cache: c d e f
future queries:
cache miss eject this one
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Reduced Eviction Schedules
Def. A reduced schedule is a schedule that only inserts an item into
the cache in a step in which that item is requested.
Intuition. Can transform an unreduced schedule into a reduced one
with no more cache misses.
a x
an unreduced schedule
c
a d c
a d b
a c b
a x b
a c b
a b c
a b c
a
c
d
a
b
c
a
a
a b
a reduced schedule
c
a b c
a d c
a d c
a d b
a c b
a c b
a c b
a
c
d
a
b
c
a
a
a b ca a b ca

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Reduced Eviction Schedules
Claim. Given any unreduced schedule S, can transform it into a reduced
schedule S' with no more cache misses.
Pf. (by induction on number of unreduced items)
 Suppose S brings d into the cache at time t, without a request.
 Let c be the item S evicts when it brings d into the cache.
 Case 1: d evicted at time t', before next request for d.
 Case 2: d requested at time t' before d is evicted. ▪
t
t'
d
c
t
t'
c
S'
d
S
d requested at time t'
t
t'
d
c
t
t'
c
S'
e
S
d evicted at time t',
before next request
e
doesn't enter cache at requested
time
Case 1 Case 2
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Farthest-In-Future: Analysis
Theorem. FF is optimal eviction algorithm.
Pf. (by induction on number or requests j)
Let S be reduced schedule that satisfies invariant through j requests.
We produce S' that satisfies invariant after j+1 requests.
 Consider (j+1)st request d = dj+1.
 Since S and SFF have agreed up until now, they have the same cache
contents before request j+1.
 Case 1: (d is already in the cache). S' = S satisfies invariant.
 Case 2: (d is not in the cache and S and SFF evict the same element).
S' = S satisfies invariant.
Invariant: There exists an optimal reduced schedule S that makes
the same eviction schedule as SFF through the first j+1 requests.
31
j
Pf. (continued)
 Case 3: (d is not in the cache; SFF evicts e; S evicts f ≠ e).
– begin construction of S' from S by evicting e instead of f
– now S' agrees with SFF on first j+1 requests; we show that having
element f in cache is no worse than having element e
same f same fee
S S'
j same d same fde
S S'
j+1
32
Let j' be the first time after j+1 that S and S' take a different action,
and let g be item requested at time j'.
 Case 3a: g = e. Can't happen with Farthest-In-Future since there
must be a request for f before e.
 Case 3b: g = f. Element f can't be in cache of S, so let e' be the
element that S evicts.
– if e' = e, S' accesses f from cache; now S and S' have same cache
– if e' ≠ e, S' evicts e' and brings e into the cache; now S and S'
have the same cache
same e same f
S S'
j'
Note: S' is no longer reduced, but can be transformed into
a reduced schedule that agrees with SFF through step j+1
must involve e or f (or both)

33
Let j' be the first time after j+1 that S and S' take a different action,
and let g be item requested at time j'.
 Case 3c: g ≠ e, f. S must evict e.
Make S' evict f; now S and S' have the same cache. ▪
same g same g
S S'
j'
otherwise S' would take the same action
same e same f
S S'
j'
must involve e or f (or both)
34
Caching Perspective
Online vs. offline algorithms.
 Offline: full sequence of requests is known a priori.
 Online (reality): requests are not known in advance.
 Caching is among most fundamental online problems in CS.
LIFO. Evict page brought in most recently.
LRU. Evict page whose most recent access was earliest.
Theorem. FF is optimal offline eviction algorithm.
 Provides basis for understanding and analyzing online algorithms.
 LRU is k-competitive. [Section 13.8]
 LIFO is arbitrarily bad.
FF with direction of time reversed!
4.4 Shortest Paths in a Graph
shortest path from Princeton CS department to Einstein's house
36
Shortest Path Problem
Shortest path network.
 Directed graph G = (V, E).
 Source s, destination t.
 Length le = length of edge e.
Shortest path problem: find shortest directed path from s to t.
Cost of path s-2-3-5-t
= 9 + 23 + 2 + 16
= 50.
cost of path = sum of edge costs in path
s
3
t
2
6
7
4
5
23
18
2
9
14
15
5
30
20
44
16
11
6
19
6

37
Dijkstra's Algorithm
Dijkstra's algorithm.
 Maintain a set of explored nodes S for which we have determined
the shortest path distance d(u) from s to u.
 Initialize S = { s }, d(s) = 0.
 Repeatedly choose unexplored node v which minimizes
add v to S, and set d(v) = π(v).
,)(min)(
:),(
e
Suvue
udv l+=
!=
"
s
v
u
d(u)
S
le
shortest path to some u in explored
part, followed by a single edge (u, v)
38
Dijkstra's Algorithm
Dijkstra's algorithm.
 Maintain a set of explored nodes S for which we have determined
the shortest path distance d(u) from s to u.
 Initialize S = { s }, d(s) = 0.
 Repeatedly choose unexplored node v which minimizes
add v to S, and set d(v) = π(v).
,)(min)(
:),(
e
Suvue
udv l+=
!=
"
s
v
u
d(u)
shortest path to some u in explored
part, followed by a single edge (u, v)
S
le
39
Dijkstra's Algorithm: Proof of Correctness
Invariant. For each node u ∈ S, d(u) is the length of the shortest s-u path.
Pf. (by induction on |S|)
Base case: |S| = 1 is trivial.
Inductive hypothesis: Assume true for |S| = k ≥ 1.
 Let v be next node added to S, and let u-v be the chosen edge.
 The shortest s-u path plus (u, v) is an s-v path of length π(v).
 Consider any s-v path P. We'll see that it's no shorter than π(v).
 Let x-y be the first edge in P that leaves S,
and let P' be the subpath to x.
 P is already too long as soon as it leaves S.
l(P) ≥ l (P') + l (x,y) ≥ d(x) + l (x, y) ≥ π(y) ≥ π(v)
nonnegative
weights
inductive
hypothesis
defn of π(y) Dijkstra chose v
instead of y
S
s
y
v
x
P
u
P'
40
Dijkstra's Algorithm: Implementation
For each unexplored node, explicitly maintain
 Next node to explore = node with minimum π(v).
 When exploring v, for each incident edge e = (v, w), update
Efficient implementation. Maintain a priority queue of unexplored
nodes, prioritized by π(v).
† Individual ops are amortized bounds
PQ Operation
Insert
ExtractMin
ChangeKey
Binary heap
log n
log n
log n
Fib heap †
1
log n
1
Array
n
n
1
IsEmpty 1 11
Priority Queue
Total m log n m + n log nn2
Dijkstra
n
n
m
n
d-way Heap
d log d n
d log d n
log d n
1
m log m/n n
!
"(v) = min
e = (u,v) : u# S
d(u) + le .
!
"(w) = min { "(w), "(v)+le }.

41
Edsger W. Dijkstra
The question of whether computers can think is like the
question of whether submarines can swim.
Do only what only you can do.
In their capacity as a tool, computers will be but a ripple
on the surface of our culture. In their capacity as
intellectual challenge, they are without precedent in the
cultural history of mankind.
The use of COBOL cripples the mind; its teaching should,
therefore, be regarded as a criminal offence.
APL is a mistake, carried through to perfection. It is the
language of the future for the programming techniques
of the past: it creates a new generation of coding bums.
Extra Slides
Coin Changing
Greed is good. Greed is right. Greed works.
Greed clarifies, cuts through, and captures the
essence of the evolutionary spirit.
- Gordon Gecko (Michael Douglas)
44
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.

45
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.
Q. Is cashier's algorithm optimal?
Sort coins denominations by value: c1 < c2 < … < cn.
S ← φ
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
coins selected
46
Coin-Changing: Analysis of Greedy Algorithm
Theorem. Greedy algorithm 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. ▪
1
ck
10
25
100
P ≤ 4
All optimal solutions
must satisfy
N + D ≤ 2
Q ≤ 3
5 N ≤ 1
no limit
k
1
3
4
5
2
-
Max value of coins
1, 2, …, k-1 in any OPT
4 + 5 = 9
20 + 4 = 24
4
75 + 24 = 99
47
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.
Selecting Breakpoints

49
Selecting Breakpoints
Selecting breakpoints.
 Road trip from Princeton to Palo Alto along fixed route.
 Refueling stations at certain points along the way.
 Fuel capacity = C.
 Goal: makes as few refueling stops as possible.
Greedy algorithm. Go as far as you can before refueling.
Princeton Palo Alto
1
C
C
2
C
3
C
4
C
5
C
6
C
7
50
Truck driver's algorithm.
Implementation. O(n log n)
 Use binary search to select each breakpoint p.
Selecting Breakpoints: Greedy Algorithm
Sort breakpoints so that: 0 = b0 < b1 < b2 < ... < bn = L
S ← {0}
x ← 0
while (x ≠ bn)
let p be largest integer such that bp ≤ x + C
if (bp = x)
return "no solution"
x ← bp
S ← S ∪ {p}
return S
breakpoints selected
current location
51
Selecting Breakpoints: Correctness
 Let 0 = g0 < g1 < . . . < gp = L denote set of breakpoints chosen by greedy.
 Let 0 = f0 < f1 < . . . < fq = L denote set of breakpoints in an optimal
solution with f0 = g0, f1= g1 , . . . , fr = gr for largest possible value of r.
 Note: gr+1 > fr+1 by greedy choice of algorithm.
. . .
Greedy:
OPT:
g0 g1 g2
f0 f1 f2 fq
gr
fr
why doesn't optimal solution
drive a little further?
gr+1
fr+1
52
Selecting Breakpoints: Correctness
 Let 0 = g0 < g1 < . . . < gp = L denote set of breakpoints chosen by greedy.
 Let 0 = f0 < f1 < . . . < fq = L denote set of breakpoints in an optimal
solution with f0 = g0, f1= g1 , . . . , fr = gr for largest possible value of r.
 Note: gr+1 > fr+1 by greedy choice of algorithm.
another optimal solution has
one more breakpoint in common
⇒ contradiction
. . .
Greedy:
OPT:
g0 g1 g2
f0 f1 f2 fq
gr
fr
gr+1

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