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DESIGN & ANALYSIS OF ALGORITHM
DYNAMIC PROGRAMMING
Informatics Department
Parahyangan Catholic University
INTRODUCTION
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We have seen some algorithm design principles,
such as divide-and-conquer, brute force, and
greedy
Brute force is widely applicable, but inefficient
Divide-and-conquer and greedy are fast, but only
applicable on very specific problems.
Dynamic Programming is somewhere in between
them, while still providing polynomial time
complexity, it is widely applicable.
DYNAMIC PROGRAMMING (DP)
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Similar to divide-and-conquer, DP solves problem
by combining the solutions of its sub-problems.
The term “programming” here refers to a tabular
method
DP solves a sub-problem, then saves its solution
in a table
FINDING N-TH FIBONACCI NUMBER
F(n) =
0
1
F(n-1) + F(n-2)
n=0
n=1
n>1
Recursive solution :
FIBONACCI (n)
if (n==0)
return 0
else if (n==1)
return 1
else
return FIBONACCI(n-1) + FIBONACCI(n-2)
FINDING N-TH FIBONACCI NUMBER
Recursive solution :
F(5)
F(4)
F(3)
F(3)
F(2)
F(1)
F(2)
F(1)
F(0)
F(1)
F(2)
F(0)
F(1)
F(1)
F(0)
FINDING N-TH FIBONACCI NUMBER
Memoization :
Maintain a table to store sub-problem’s solution
solution with memoization:
// Initially : Arr[0] = 0
//
Arr[1] = 1
//
Arr[2..n] =
-1
FIBONACCI (n)
if (Arr[n] != -1)
return Arr[n]
else
Arr[n]=FIBONACCI(n-1) + FIBONACCI(n-2)
return Arr[n]
FINDING N-TH FIBONACCI NUMBER
solution with memoization:
F(5)
F(4)
F(3) =2
F(2) =1
F(3)
F(2)
F(1)
F(1)
F(0)
n 0
1
2
3
4
5
F(n) 0
1
-1
1
-1
2
-1
3
-1
5
FINDING N-TH FIBONACCI NUMBER
Bottom-up solution :
Use the natural ordering of sub-problems, solve them
one-by-one starting from the “smallest” one
Bottom-up solution:
FIBONACCI (n)
Arr[0] = 0
Arr[1] = 1
if(n>1)
for i=2 to n do
Arr[i] = Arr[i-1] + Arr[i-2]
return Arr[n]
TIME COMPLEXITY
Recursive solution :
FIBONACCI (n)
if (n==0)
return 0
else if (n==1)
return 1
else
return FIBONACCI(n-1) + FIBONACCI(n-2)
Height = n
F(n)
F(n-1)
F(n-2)
F(n-3)
Every instance has ≤ 2
recursive calls
F(n-2)
F(n-3)
F(n-4)
Therefore, time
complexity is
O(2n)
TIME COMPLEXITY
Bottom-up solution:
FIBONACCI (n)
Arr[0] = 0
Arr[1] = 1
if(n>1)
for i=2 to n do
Arr[i] = Arr[i-1] + Arr[i-2]
return Arr[n]
There is a loop that iterates ≤ n times, each doing
a constant amount of work.
So the time complexity is O(n)
ROD CUTTING PROBLEM
Serling Enterprises buys long steel rods and cuts them into
shorter rods, which it then sells. Each cut is free, however
different rod length sells for different price. The management
of Sterling Enterprises wants to know the best way to cut up
the rods.
We assume that we know :
length
1
2
3
4
5
6
7
8
9
10
price
1
5
8
9
10
17
17
20
24
30
Example : n=4
(n-1) possible cut locations  2n-1 ways of cutting
ROD CUTTING PROBLEM
length
1
2
3
4
5
6
7
8
9
10
price
1
5
8
9
10
17
17
20
24
30
Example : n=4
1
9
1
5
=7
=9
1
8
1
5
1
=9
BEST 5
=7
5
5
=10
8
1
1
=7
1
=9
1
1
1
1
=4
ROD CUTTING PROBLEM
1
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i
n
Consider a rod of length n, and we cut a rod of
length i
Then we left with a rod of length n-i
Naturally, we want to optimize the selling price
of the remaining rod
ROD CUTTING PROBLEM
Recursive solution :
ROD-CUTTING (n)
if (n==0)
return 0
else
best = -∞
for i=1 to n do
best = MAX(best, price[i]+ROD-CUTTING(n-i))
return best
ROD CUTTING PROBLEM
Recursive solution :
RC(4)
RC(2)
RC(3)
RC(2)
RC(1)
RC(0)
RC(0)
RC(1)
RC(0)
RC(0)
RC(1)
RC(0)
RC(1)
RC(0)
RC(0)
Same problem as recursive Fibonacci
RC(0)
ROD CUTTING PROBLEM
Solution with memoization:
// Initially : Arr[0] = 0
//
Arr[1..n] = -∞
ROD-CUTTING (n)
if Arr[n] ≥ 0
return Arr[n]
else
best = -∞
for i=1 to n do
best = MAX(best, price[i]+ROD-CUTTING(n-i))
Arr[n] = best
return best
Exercise
Write a bottom-up solution for Rod Cutting problem !
TIME COMPLEXITY
Recursive solution :
ROD-CUTTING (n)
if (n==0)
return 0
else
best = -∞
for i=1 to n do
best = MAX(best, price[i]+ROD-CUTTING(n-i))
return best
Every instance has ≤ n recursive calls, and the
depth of the recursive tree is O(n).
So the time complexity is O(nn)
What is the time complexity for
the bottom-up solution ?
SHORTEST PATH IN DAG
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DAG = Directed Acyclic Graph
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Example :
1
6
A
B
1
4
S
2
C
2
3
D
E
1
SHORTEST PATH IN DAG
6
A
1
B
1
4
S
2
2
3
C
E
1
D
Sort using topological sort
3
S
2
C
1
4
A
6
B
1
D
2
1
E
SHORTEST PATH IN DAG
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Starting from S, suppose we want to reach node D
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The only way to reach D is either through B or C
dist(D) = MIN(dist(B)+1 , dist(C)+3)
3
S
2
C
1
4
A
6
B
1
D
2
1
E
SHORTEST PATH IN DAG
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A similar relation can be written for every node.
As we have seen before, it is best to use bottom-up
way of computing dist, that is from “left most”
node to “right most” node
DAG-SHORTEST-PATH ()
initialize all dist[.] to ∞
dist[S] = 0
for each vertex v except S in left-to-right order do
for each edge (u,v) do
dist[v] = MIN(dist[v] , dist[u] + weight(u,v))
DYNAMIC PROGRAMMING
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DAG’s shortest path is a very general technique. We model
many other problems into DAG problem.
Example #1: Fibonacci
0
1
0
1
2
3
4
3
4
Example #2 : Rod Cutting
0
0
1
2
5
PROPERTIES OF DYNAMIC PROGRAMMING
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Problem can be divided into smaller
sub-problems
Optimal substructure:
an optimal solution to the problem contains within
its optimal solutions to sub-problems
Overlapping sub-problems:
the space of sub-problems is “small”, in the sense
that a recursive algorithm for the problem solves
the same sub-problems over and over, rather than
always generating new sub-problems
DYNAMIC PROGRAMMING
V.S. DIVIDE-AND-CONQUER
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Divide-and-Conquer
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Dynamic Programming
And so on…
LONGEST INCREASING SUBSEQUENCE
Given a sequence of numbers a1, a2, a3, … , an. A
subsequence is any subset of these numbers taken in
order of the form ai1, ai2, ai3…, aik, where 1 ≤ i1 < i2 <
… < ik ≤ n, and an increasing subsequence is one
which the numbers are getting strictly larger. The
task is to find the increasing subsequence of greatest
length.
Example: 5, 2, 8, 6, 3, 6, 9, 7
LONGEST INCREASING SUBSEQUENCE
How do we model this problem into a DAG ?
any number ai can precede aj iff ai < aj
5
2
8
6
3
Consider this number,
LIS that ends here must be either:
• subsequence consist of “6” alone
• LIS that ends at “5” + “6”
• LIS that ends at “2” + “6”
• LIS that ends at “”3” + “6”
6
9
7
LONGEST INCREASING SUBSEQUENCE
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How do we find the LIS of sequence
5, 2, 8, 6, 3, 6, 9, 7 ?
Does the optimal solution always ends at “7” ?
// Initially L[1..n] = 1
LONGEST-INCREASING-SUBSEQUENCE ()
for j=2 to n do
for each i<j such that ai < aj do
if(L[j] < 1 + L[i]) then
L[j] = 1 + L[i]
return maximum of L[1..n]
This algorithm only gives the sequence’s length.
How do we find the actual sequence ?
RECONSTRUCTING A SOLUTION
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We can extend the dynamic programming
approach to record not only the optimal value for
each sub-problem, but also the choice that led to
that value
// Initially L[1..n] = 1
// Initially prev[1..n] = 0
LONGEST-INCREASING-SUBSEQUENCE ()
for j=2 to n do
for each i<j such that ai < aj do
if(L[j] < 1 + L[i]) then
L[j] = 1 + L[i]
prev[j] = i
return maximum of L[1..n] and array prev
EXERCISE : YUCKDONALD’S
Yuckdonald’s is considering opening a series of restaurants
along Quaint Valley Highway (QVH). The n possible locations
are along a straight line, and the distance of these locations
from the start of QVH are, in miles and in increasing order, m1,
m2, …, mn. The constraints are as follows:
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At each location, Yuckdonald’s may open at most one restaurant. The
expected profit from opening a restaurant at location i is pi > 0
Any two restaurants should be at least k miles apart, where k is a
positive integer