Transcript Slides
CS 3343: Analysis of
Algorithms
Lecture 19: Introduction to Greedy
Algorithms
Outline
• Review of DP
• Greedy algorithms
– Similar to DP, not an actual algorithm, but a
meta algorithm
Two steps to dynamic programming
• Formulate the solution as a recurrence
relation of solutions to subproblems.
• Specify an order of evaluation for the
recurrence so you always have what you
need.
Restaurant location problem
• You work in the fast food business
• Your company plans to open up new restaurants in
Texas along I-35
• Many towns along the highway, call them t1, t2, …, tn
• Restaurants at ti has estimated annual profit pi
• No two restaurants can be located within 10 miles of
each other due to regulation
• Your boss wants to maximize the total profit
• You want a big bonus
10 mile
A DP algorithm
• Suppose you’ve already found the optimal solution
• It will either include tn or not include tn
• Case 1: tn not included in optimal solution
– Best solution same as best solution for t1 , …, tn-1
• Case 2: tn included in optimal solution
– Best solution is pn + best solution for t1 , …, tj , where j < n is the
largest index so that dist(tj, tn) ≥ 10
Recurrence formulation
• Let S(i) be the total profit of the optimal solution when the
first i towns are considered (not necessarily selected)
– S(n) is the optimal solution to the complete problem
S(n) = max
S(n-1)
j < n & dist (tj, tn) ≥ 10
S(j) + pn
Generalize
S(i) = max
S(i-1)
j < i & dist (tj, ti) ≥ 10
S(j) + pi
Number of sub-problems: n. Boundary condition: S(0) = 0.
Dependency:
S
j
i-1 i
Example
Distance (mi)
dummy 100
0
Profit (100k)
S(i)
5
2 2
6
7
6
3
6
3
6
7 9 8
3
6
7 9 9
10
10
7
4
3
12
2
4
12
5
12 12
14
26
26
Optimal: 26
S(i) = max
S(i-1)
S(j) + pi
j < i & dist (tj, ti) ≥ 10
Complexity
• Time: O(nk), where k is the maximum
number of towns that are within 10 miles
to the left of any town
– In the worst case, O(n2)
– Can be reduced to O(n) by pre-processing
• Memory: Θ(n)
Knapsack problem
• Each item has a value and a weight
• Objective: maximize value
• Constraint: knapsack has a weight limitation
Three versions:
0-1 knapsack problem: take
each item or leave it
Fractional knapsack problem:
items are divisible
Unbounded knapsack problem:
unlimited supplies of each item.
Which one is easiest to solve?
We studied the 0-1 problem.
Formal definition (0-1 problem)
• Knapsack has weight limit W
• Items labeled 1, 2, …, n (arbitrarily)
• Items have weights w1, w2, …, wn
– Assume all weights are integers
– For practical reason, only consider wi < W
• Items have values v1, v2, …, vn
• Objective: find a subset of items, S, such that
iS wi W and iS vi is maximal among all such
(feasible) subsets
A DP algorithm
• Suppose you’ve find the optimal solution S
• Case 1: item n is included
• Case 2: item n is not included
wn
Total weight limit:
W
Find an optimal solution using items
1, 2, …, n-1 with weight limit W - wn
wn
Total weight limit:
W
Find an optimal solution using items
1, 2, …, n-1 with weight limit W
Recursive formulation
• Let V[i, w] be the optimal total value when items 1, 2, …, i
are considered for a knapsack with weight limit w
=> V[n, W] is the optimal solution
V[n-1, W-wn] + vn
V[n, W] = max
V[n-1, W]
Generalize
V[i, w] =
V[i-1, w-wi] + vi
item i is taken
V[i-1, w]
item i not taken
max
V[i-1, w] if wi > w
item i not taken
Boundary condition: V[i, 0] = 0, V[0, w] = 0. Number of sub-problems = ?
Example
• n = 6 (# of items)
• W = 10 (weight limit)
• Items (weight, value):
2
4
3
5
2
6
2
3
3
6
4
9
w
0
1
2
3
4
5
6
7
8
9
10
0
0
0
0
0
0
0
0
0
0
i
wi
vi
0
1
2
2
0
2
4
3
0
3
3
3
0
4
5
6
0
5
2
4
0
6
6
9
0
wi
V[i-1, w-wi]
V[i-1, w]
V[i, w]
V[i-1, w-wi] + vi item i is taken
max
V[i, w] =
V[i-1, w]
V[i-1, w] if wi > w
item i not taken
item i not taken
w
0
1
2
3
4
5
6
7
8
9
10
i
wi
vi
0
0
0
0
0
0
0
0
0
0
0
1
2
2
0
0
2
2
2
2
2
2
2
2
2
2
4
3
0
0
2
2
3
3
5
5
5
5
5
3
3
3
0
0
2
3
3
5
5
6
6
8
8
4
5
6
0
0
2
3
3
6
6
8
9
9
11
5
2
4
0
0
4
4
6
7
7
10 10 12 13
6
6
9
0
0
4
4
6
7
9
10 13 13 15
V[i-1, w-wi] + vi item i is taken
max
V[i, w] =
V[i-1, w]
V[i-1, w] if wi > w
item i not taken
item i not taken
w
0
1
2
3
4
5
6
7
8
9
10
i
wi
vi
0
0
0
0
0
0
0
0
0
0
0
1
2
2
0
0
2
2
2
2
2
2
2
2
2
2
4
3
0
0
2
2
3
3
5
5
5
5
5
3
3
3
0
0
2
3
3
5
5
6
6
8
8
4
5
6
0
0
2
3
3
6
6
8
9
9
11
5
2
4
0
0
4
4
6
7
7
10 10 12 13
6
6
9
0
0
4
4
6
7
9
10 13 13 15
Optimal value: 15
Item: 6, 5, 1
Weight: 6 + 2 + 2 = 10
Value: 9 + 4 + 2 = 15
Time complexity
• Θ (nW)
• Polynomial?
– Pseudo-polynomial
– Works well if W is small
• Consider following items (weight, value):
(10, 5), (15, 6), (20, 5), (18, 6)
• Weight limit 35
– Optimal solution: item 2, 4 (value = 12). Iterate: 2^4 = 16 subsets
– Dynamic programming: fill up a 4 x 35 = 140 table entries
• What’s the problem?
– Many entries are unused: no such weight combination
– Top-down may be better
Events scheduling problem
e3
e1
e2
e6
e4
s8 e8 f8
e7
e5
s7
e9
f7
s9
f9
Time
• A list of events to schedule
– ei has start time si and finishing time fi
– Indexed such that fi < fj if i < j
• Each event has a value vi
• Schedule to make the largest value
– You can attend only one event at any time
Events scheduling problem
e3
e1
e2
e6
e4
s8 e8 f8
e7
e5
s7
e9
f7
s9
f9
Time
• V(i) is the optimal value that can be achieved
when the first i events are considered
V(n-1)
• V(n) = max {
V(j) + vn
en not selected
en selected
j < n and fj < sn
Restaurant location problem 2
• Now the objective is to maximize the
number of new restaurants (subject to the
distance constraint)
– In other words, we assume that each
restaurant makes the same profit, no matter
where it is opened
10 mile
A DP Algorithm
• Exactly as before, but pi = 1 for all i
S(i-1)
S(i) = max
S(j) + pi
j < i & dist (tj, ti) ≥ 10
S(i-1)
S(i) = max
S(j) + 1
j < i & dist (tj, ti) ≥ 10
Example
Distance (mi)
dummy 100
0
Profit (100k)
S(i)
5
2 2
6
6
3
6
10
7
1
1 1 1
1
1
1
1
1
1
1
1 1 1
2
2
2
3
4
4
Optimal: 4
S(i) = max
S(i-1)
S(j) + 1
j < i & dist (tj, ti) ≥ 10
• Natural greedy 1: 1 + 1 + 1 + 1 = 4
• Maybe greedy is ok here? Does it work for all cases?
Comparison
Dist(mi) 100
0
Profit (100k)
S(i)
5
2 2
6
6
3
6
10
7
1
1 1 1
1
1
1
1
1
1
1
1 1 1
2
2
2
3
4
4
Benefit of taking t1 rather than t2?
t1 gives you more choices for the future
Benefit of waiting to see t2?
None!
Dist(mi)
100
0
Profit (100k)
S(i)
5
2 2
6
6
6
7 9 8
3
6
7 9 9
10
3
3
6
10
7
2
4
12
5
12 12
14
26
26
Benefit of taking t1 rather than t2?
t1 gives you more choices for the future
Benefit of waiting to see t2?
t2 may have a bigger profit
Moral of the story
• If a better opportunity may come out next,
you may want to hold on your decision
• Otherwise, grasp the current opportunity
immediately because there is no reason to
wait …
Greedy algorithm
• For certain problems, DP is an overkill
– Greedy algorithm may guarantee to give you
the optimal solution
– Much more efficient
Formal argument
• Claim 1: if A = [m1, m2, …, mk] is the optimal solution to the
restaurant location problem for a set of towns [t1, …, tn]
– m1 < m2 < … < mk are indices of the selected towns
– Then B = [m2, m3, …, mk] is the optimal solution to the sub-problem
[tj, …, tn], where tj is the first town that are at least 10 miles to the
right of tm1
• Proof by contradiction: suppose B is not the optimal
solution to the sub-problem, which means there is a better
solution B’ to the sub-problem
– A’ = mi || B’ gives a better solution than A = mi || B => A is not
optimal => contradiction => B is optimal
A
A’
m1
m1
m2
B
B’ (imaginary)
mk
Implication of Claim 1
• If we know the first town that needs to be
chosen, we can reduce the problem to a
smaller sub-problem
– This is similar to dynamic programming
– Optimal substructure
Formal argument (cont’d)
• Claim 2: for the uniform-profit restaurant location
problem, there is an optimal solution that chooses t1
• Proof by contradiction: suppose that no optimal solution
can be obtained by choosing t1
– Say the first town chosen by the optimal solution S is ti, i > 1
– Replace ti with t1 will not violate the distance constraint, and the
total profit remains the same => S’ is an optimal solution
– Contradiction
– Therefore claim 2 is valid
S
S’
Implication of Claim 2
• We can simply choose the first town as
part of the optimal solution
– This is different from DP
– Decisions are made immediately
• By Claim 1, we then only need to repeat
this strategy to the remaining sub-problem
Greedy algorithm for restaurant
location problem
select t1
d = 0;
for (i = 2 to n)
d = d + dist(ti, ti-1);
if (d >= min_dist)
select ti
d = 0;
end
end
5
d
0
2 2
5 7 9
6
6
15
0
3
6
6
9
10
15
0
7
10
0
7
Complexity
• Time: Θ(n)
• Memory:
– Θ(n) to store the input
– Θ(1) for greedy selection
Events scheduling problem
e3
e1
e2
e6
e4
e5
e8
e7
e9
Time
•
•
•
•
Objective: to schedule the maximal number of events
Let vi = 1 for all i and solve by DP, but overkill
Greedy strategy: choose the first-finishing event that is compatible with
previous selection (1, 2, 4, 6, 8 for the above example)
Why is this a valid strategy?
–
–
–
–
–
–
•
Claim 1: optimal substructure
Claim 2: there is an optimal solution that chooses e1
Proof by contradiction: Suppose that no optimal solution contains e1
Say the first event chosen is ei => other chosen events start after ei finishes
Replace ei by e1 will result in another optimal solution (e1 finishes earlier than ei)
Contradiction
Simple idea: attend the event that will left you with the most amount of time
when finished
Knapsack problem
• Each item has a value and a weight
• Objective: maximize value
• Constraint: knapsack has a weight limitation
Three versions:
0-1 knapsack problem: take
each item or leave it
Fractional knapsack problem:
items are divisible
Unbounded knapsack problem:
unlimited supplies of each item.
Which one is easiest to solve?
We can solve the fractional knapsack
problem using greedy algorithm
Greedy algorithm for fractional
knapsack problem
• Compute value/weight ratio for each item
• Sort items by their value/weight ratio into
decreasing order
– Call the remaining item with the highest ratio the most
valuable item (MVI)
• Iteratively:
– If the weight limit can not be reached by adding MVI
• Select MVI
– Otherwise select MVI partially until weight limit
Example
item
Weight
(LB)
Value
($)
$ / LB
1
2
2
1
2
4
3
0.75
3
3
3
1
4
5
6
1.2
5
2
4
2
6
6
9
1.5
• Weight limit: 10
Example
item
Weight
(LB)
Value
($)
$ / LB
5
2
4
2
6
6
9
1.5
4
5
6
1.2
1
2
2
1
3
3
3
1
2
4
3
0.75
• Weight limit: 10
• Take item 5
– 2 LB, $4
• Take item 6
– 8 LB, $13
• Take 2 LB of item 4
– 10 LB, 15.4
Why is greedy algorithm for
fractional knapsack problem valid?
• Claim: the optimal solution must contain the MVI as
much as possible (either up to the weight limit or until
MVI is exhausted)
• Proof by contradiction: suppose that the optimal solution
does not use all available MVI (i.e., there is still w (w <
W) pounds of MVI left while we choose other items)
– We can replace w pounds of less valuable items by MVI
– The total weight is the same, but with value higher than the
“optimal”
– Contradiction
w
w
w
w
Elements of greedy algorithm
1. Optimal substructure
2. Locally optimal decision leads to globally
optimal solution
•For most optimization problems, greedy algorithm
will not guarantee an optimal solution
•But may give you a good starting point to use
other optimization techniques
•Starting from next week, we’ll study several
problems in graph theory that can actually be
solved by greedy algorithm