Download Dynamic Programming

Designing a Dynamic
Programming Algorithm for an
Optimization Problem
• Step 1. Characterize the structure of optimal
solution
• Step 2. Recursively define the value of an
optimal solution
• Step 3. Compute the value of an optimal
solution in a bottombottom-up fashion
• Step 4. Construct an optimal solution from
computed information
Dynamic Programming: Matrix Chain Multiplication
Matrix Chain Multiplication Problem
Given a chain A1 , A2 , … , An of n matrices,
where (for i = 1, 2,… , n) A is a pi × pi
i
matrix
Fully parenthesize the product
A1 A2
Time Complexity of Matrix Multiplication
pxq
qxr
Matrix-Multiply(A, B)
if columns[A]
≠ rows[B]
then error “incompatible dimensions”
pqr
else for iÅ1 to rows[A] = p do
{{
rq
for j Å 1 to columns[B] = r do
C[I, j] Å 0
q for k Å 1 to columns[A] = q do
C[I, j] Å C[I, j] + A[I, k]*B[k, j]
An
so that the number of scalar multiplications
is minimum
Example
Two ways to
parenthesize
( A1
10
100
A2
)
5
A1
10
A3
50
A2
100
A1 ( A2
10
100
A3
5
50
A3 )
5
50
= (10)(100)(5) + (10)(5)(50)
= (100)(5)(50) + (10)(100)(50)
= 5000 + 2500 = 7500
= 25000 + 50000 = 75000
{
return C
1
A1 A2 A3 A4
Let P ( n ) = number of ways to parenthesize a
product of n matrices, then
Can be parenthesized in 5 ways
( A ( A ( A A )))
1
2
3
if n = 1
1
 n
P ( n) = 
 ∑ P ( k ) P ( n − k ) if n ≥ 2
 k =1
It can be shown that P ( n ) = C ( n − 1) ,
where C ( n − 1) is the (n-1)-th Catalan
4
( A (( A A ) A ))
1
2
3
4
( ( A1 A2 ) ( A3 A4 ) )
(( A ( A A )) A )
1
2
3
number
4
( ( ( A1 A2 ) A3 ) A4 )
Notation
m[i, j] =
Ai… j = Ai Ai + 1
C ( n) =
Objective
Aj
Min # of mult. to evaluate
1  2n 
 4n 
= Ω  3/2 


n + 1 n 
n 
Ai… j
if
0
m[ i , j ] = 
min m + mk +1, j + pi −1 pk p j ) if
 i ≤ k ≤ j ( i ,k
i= j
i< j
Find a way to fully parenthesize the product
A1…n = A1 A2
An
in such a way that the cost is minimum
s[i, j] =
that k such that
m [ i , j ] = m [ i , k ] + m [ k + 1, j ] + pi −1 pk p j
2
• Optimal
Substruc. Property
( --- ) ( --- )
•Overlapping Subproblems Property
Recursive formula for m[I, j] leads to an
exponential problem
Observation: Relatively few problems
Set of problems
{ i…j }
So # of subprobs. is the # of ways of
choosing I, j such that 1 ≤ i ≤ j ≤ n ,
which is
 n
 2 + n
 
⇒
Overlap
((
)
)( (
)
)
3
Analysis of RecursiveRecursive-MatrixMatrix-Chain
T ( n) =
time to compute optimal parenth. of
chain of n matrices
T (1) ≥ 1

n −1

T
(
n
)
1
≥
+
( T ( k ) + T ( n − k ) + 1)
∑


k =1
T (n) ≥ 1 +
n −1
∑ (T ( k ) + T ( n − k ) + 1 )
k =1
n −1
= 2∑ T (k ) + n
k =1
We use math induction to prove that
T ( n) = Ω ( 2 n )
The basis step is easy:
T (1) ≥ 1 = 2
0
Assuming the Induction hypothesis
for ≤ n − 1, we have:
n −1
n− 2
k =1
k =0
for
n>1
T( ) ≥ 2
−1
T ( n ) ≥ 2∑ 2 k − 1 + n = 2∑ 2 k + n
= 2( 2
n −1
− 1) + n = 2n − 2 + n
≥ 2n −1
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Memoization
Memoization
• Create a table for solns to subprobs so far
computed
• Initialize each table entry to a symbol
indicating subprob not yet computed
• Each time a subproblem is first encountered,
its soln is computed and placed in the table
• Each subsequent time it is encountered, its
previously computed soln is simply retrieved
from the table
How to Design a Dynamic Prog.
Prog. Algorithm
1. Identify the subproblems –
• Subprobs must be of same type as original probs
How to Design a Dynamic Prog.
Prog. Algorithm
3. Apply “generic dynamic prog. algorithm –
• Subprobs must be smaller
• Use bottom-up version
• How to specify each subprob?
• use top-down version (Memoization)
•What is the base case? What is the recursive case?
2. Characterize the op. soln as a combination
of op solns to subprobs –
• Focus on value of soln rather than description of soln
4. Modify your algorithm to compute
description of soln to value of soln, if needed
• Add additional table
• Introduce notation to rep. value of op soln to each subprob
• Consider trying all possible solns to prob
5
Longest Common Subsequence
Longest Common Subsequence
(LCS)
If
{
Def. A seq
X = A, B , C , B , D, A, B
increasing seq
such that xi =
then common subsequences are, for
example,
and
is a subseq of
X = x1 , x2, … , xn if there is a strictly
Y = B , D, C , A, B , A
BCA
Z = z1 , z2, … , zk
(LCS)
j
i1 , i2 ,… , ik
zj
of the indices of
X
LCS Problem: Given two seqs, find an LCS
BCBA
Both are common subsequences (CS). But
only the last is a LCS.
Notation: Given X = x1 , x2, … , xn , the i-th
prefix X i is defined as the subseq
x1 , x2, … , xi
Thm 16 (p351). (Optimal substructure of LCS)
Let  X = x1 , x2 ,… , xm 
be seqs,


Y
=
y
,
y
,
…
,
y

1
2
n 
and let
1.
Z = z1, z2,…, zk = LCS ( X,Y )
xm = yn ⇒zk = xm = yn
xm ≠ yn 

2. &
 ⇒
z k ≠ x m 
3.
xm ≠ y
&
zk ≠ yn
n


 ⇒


and
So …
, then
?
xm = yn
Zk−1 = LCS( Xm−1,Yn−1)
Z = LC S
(X
m − 1
Z = LC S
(X
,Y
,Y
n − 1
)
)
Yes
.
.
zk = xm = yn
&
Zk−1 = LCS( Xm−1,Yn−1 )
NO
?
zk ≠ x m
&
Z = LCS ( X m −1 , Y )
z k ≠ yn
&
Z = LCS ( X ,Yn−1 )
6
Notation
c[i, j] =
Length of
LCS ( X i ,Y j )
0
if i =0or j = 0

c[ i, j] = c[ i −1, j −1] +1
if i, j >0&xi = yj

Max( c[ i, j −1] ,c[ i −1, j]) if i, j >0&xi ≠ yj
[
[
]
b i , j points to the entry of c 0… i ,0…
corresponding to the optimal subprob soln
chosen when computing c i , j
[
j]
]
7