Transcript Slides

CS 3343: Analysis of
Algorithms
Lecture 18: More Examples on Dynamic
Programming
Review of Dynamic Programming
• We’ve learned how to use DP to solve
– a special shortest path problem
– the longest subsequence problem
– a general sequence alignment
• When should I use dynamic
programming?
• Theory is a little hard to apply
– More examples would help
Two steps to dynamic programming
• Formulate the solution as a recurrence
relation of solutions to subproblems.
• Specify an order to solve the subproblems
so you always have what you need.
A special shortest path problem
S
m
n
G
Each edge has a length (cost). We need to get to G from S. Can only
move right or down. Aim: find a path with the minimum total length
Recursive thinking
n
• Suppose we’ve found the shortest path
• It must use one of the two edges:
– (m, n-1) to (m, n)
– (m-1, n) to (m, n)
Case 1
Case 2
• If case 1
m
– find shortest path from (0, 0) to (m, n-1)
– SP(0, 0, m, n-1) + dist(m, n-1, m, n) is the overall shortest path
• If case 2
– find shortest path from (0, 0) to (m-1, n)
– SP(0, 0, m, n-1) + dist(m, n-1, m, n) is the overall shortest path
• We don’t know which case is true
– But if we’ve find the two paths, we can compare
– Real shortest path is the one with shorter overall length
Recursive formulation
Let F(i, j) = SP(0, 0, i, j). => F(m, n) is length of SP from (0, 0) to (m, n)
n
F(m-1, n) + dist(m-1, n, m, n)
F(m, n) = min
F(m, n-1) + dist(m, n-1, m, n)
Generalize
m
F(i-1, j) + dist(i-1, j, i, j)
F(i, j) = min
F(i, j-1) + dist(i, j-1, i, j)
Boundary condition: i = 0 or j = 0.
Easy to figure out manually.
i = 1 .. m, j = 1 .. n
Number of subproblems = m * n
determines structure of DP table
(i, j)
Data dependency
determines order to compute
Longest Common Subsequence
• Given two sequences x[1 . . m] and y[1 . . n], find a
longest subsequence common to them both.
“a” not “the”
x: A
B
C
B
D
A
y: B
D
C
A
B
A
B
BCBA =
LCS(x, y)
functional notation,
but not a function
Recursive thinking
m
x
n
y
• Case 1: x[m]=y[n]. There is an optimal LCS that matches
x[m] with y[n].
Find out LCS (x[1..m-1], y[1..n-1])
• Case 2: x[m] y[n]. At most one of them is in LCS
– Case 2.1: x[m] not in LCS
– Case 2.2: y[n] not in LCS
Find out LCS (x[1..m-1], y[1..n])
Find out LCS (x[1..m], y[1..n-1])
Recursive thinking
m
x
n
y
• Case 1: x[m]=y[n]
Reduce both sequences by 1 char
– LCS(x, y) = LCS(x[1..m-1], y[1..n-1]) || x[m]
• Case 2: x[m]  y[n]
concatenate
– LCS(x, y) = LCS(x[1..m-1], y[1..n]) or
LCS(x[1..m], y[1..n-1]), whichever is longer
Reduce either sequence by 1 char
Recursive formulation
Let c[i, j] be the length of LCS(x[1..i], y[1..j])
=> c[m, n] is the length of LCS(x, y)
c[m, n] =
c[m–1, n–1] + 1
max{c[m–1, n], c[m, n–1]}
if x[m] = y[n],
otherwise.
Generalize
c[i, j] =
c[i–1, j–1] + 1
max{c[i–1, j], c[i, j–1]}
if x[i] = y[j],
otherwise.
i = 1 .. m
j = 1 .. n
Boundary condition: i = 0 or j = 0.
Easy to figure out manually.
Number of subproblems = m * n
(i, j)
Order to compute?
Another DP example
• You work in the fast food business
• Your company plans to open up new restaurants in
Texas along I-35
• Towns along the highway called 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 some regulation
• Your boss wants to maximize the total profit
• You want a big bonus
10 mile
Brute-force
•
•
•
•
•
Each town is either selected or not selected
Test each of the 2n subsets
Eliminate subsets that violate constraints
Compute total profit for each remaining subset
Choose the one with the highest profit
• Θ(n 2n)
Natural greedy 1
• Take first town. Then the next town >= 10 miles
• Can you give an example that this algorithm
doesn’t return the correct solution?
100k
100k
500k
Natural greedy 2
• Almost take a town with the highest profit and are not
<10 miles of another selected town
• Can you give an example that this algorithm doesn’t
return the correct solution?
300k
300k
500k
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
• Natural greedy 1: 6 + 3 + 4 + 12 = 25
• Natural greedy 2: 12 + 9 + 3 = 24
Complexity
• Time: (nk), where k is the maximum
number of towns that are within 10 miles
to the left of any town
– In the worst case, (n2)
– Can be improved to (n) with some
preprocessing tricks
• 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 study the 0-1 problem today.
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
iS wi  W and iS vi is maximal among all such
(feasible) subsets
Naïve algorithms
• Enumerate all subsets.
– Optimal. But exponential time
• Greedy 1: take the item with the largest value
– Not optimal
– Give an example
• Greedy 2: take the item with the largest
value/weight ratio
– Not optimal
– Give an example
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
A few more examples
Longest increasing subsequence
• Given a sequence of numbers
125329493568
• Find a longest subsequence that is nondecreasing
– E.g. 1 2 5 9
– It has to be a subsequence of the original list
– It has to in sorted order
=> It is a subsequence of the sorted list
Original list: 1 2 5 3 2 9 4 9 3 5 6 8
LCS:
Sorted:
122334556899
1234568
Events scheduling problem
e3
e1
e2
e6
e4
e5
e8
e7
e9
Time
•
•
•
A list of events to schedule (or shows to see)
– 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
•
Very similar to the new restaurant location problem
– Sort events according to their finish time
– Consider: if the last event is included or not
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
Coin change problem
• Given some denomination of coins (e.g., 2, 5, 7,
10), decide if it is possible to make change for a
value (e.g, 13), or minimize the number of coins
• Version 1: Unlimited number of coins for each
denomination
– Unbounded knapsack problem
• Version 2: Use each denomination at most once
– 0-1 Knapsack problem
Use DP algorithm to solve new
problems
• Directly map a new problem to a known problem
• Modify an algorithm for a similar task
• Design your own
– Think about the problem recursively
– Optimal solution to a larger problem can be computed
from the optimal solution of one or more subproblems
– These sub-problems can be solved in certain
manageable order
– Works nicely for naturally ordered data such as
strings, trees, some special graphs
– Trickier for general graphs
• The text book has some very good exercises.