Đánh giá độ phúc tạp : Dynamic programming 01

Đánh giá độ phúc tạp

Dynamic programming

From the D&C Theorem, we can see that a recursive algorithm is likely to be polynomial

if the sum of the sizes of the subproblems is bounded by kn (Using the variables of that Theorem, k = a/c) If, however, the obvious division of a paroblem of size n results in n problems of size n-1 then the recursive algorithm is likely to have exponential growth

Dynamic programming can be thought of as being the reverse of recursion Recursion is

a top-down mechanism we take a problem, split it up, and solve the smaller problems that are created Dynamic programming is a bottom-up mechanism we solve all possible small problems and then combine them to obtain solutions for bigger problems

The reason that this may be better is that, using recursion, it is possible that we may solve the same small subproblem many times Using dynamic programming, we solve it once

Evaluation of the product of n matrices

We wish to determine the value of the product ∏i = 1 to n Mi, where Mi has r i-1 rows and r i

columns

The order of which matrices are multiplied together first can significantly affect the time required

To multiply M×N, where matrix M is p×q and matrix N is q×r, takes pqr operations if

we use the "normal" matrix multiplication algorithm Note that the matrices have a

common dimension value of q This makes the matrices have the property of

compatibility, without which it would not be possible for them to be multiplied Also

note that, while matrix multiplication is associative, matrix multiplication is not

commutative That is, N×M might not equal M×N and, in fact, N×M might not even be defined because of a lack of compatibility

EXAMPLE:

Calculate M = M1 × M2 × M3 × M4,

where the dimensions of the matrices are

M1: 10,20 M2: 20,50 M3: 50,1 M4: 1,100

Calculating M = M1 × ( M2 × ( M3 × M4 ) ) requires 125000 operations

Calculating M = ( M1 × ( M2 × M3 ) ) × M4 requires 2200 operations

We could figure out how many operations each possible order will take and then use the one having the minimum number of operations, but there are there are an exponential

number of orderings Any of the n-1 multiplications could be first and then any of the remaining n-2 multiplications could be next and so on, leading to a total of (n-1)!

orderings

We can find the best order in time O(n3) by using dynamic programming

If m i ,j is the minimum cost of evaluating the product Mi × × Mj then:

m i, j = 0, if i = j, and

m i, j = MINi ≤ k < j { m i,k + m k+1,j + r i-1 r k r j }, if i < j

The algorithm:

for i := 1 to n do

m i,i := 0

for length := 1 to n-1 do

for i := 1 to n-length do

j := i + length

m i,j = MIN i≤k<j {m i,k + m k+1,j + r i-1 r k r j }

In the above listing, length refers to the number of matrix multiplications in a

subproblem An alternative approach would be to use size (equal to length+1) as the number of matrices in the subproblem

For the example given above, we would calculate:

m1,1 = 0, m2,2 = 0, m3,3 = 0, m4,4 = 0

m1,2 = 10000, m2,3 = 1000, m3,4 = 5000

m1,3 = 1200, m2,4 = 3000

m1,4 = 2200

Longest common substring problem

Given two sequences of letters, such as A = HELLO and B = ALOHA,

find the longest contiguous sequence appearing in both

One solution: (assume strings have lengths m and n)

For each of the m starting points of A, check for the longest common string starting at each of the n starting points of B

The checks could average Θ(m) time → a total of Θ(m2n) time

Dynamic programming solution:

Let Li, j = maximum length of common strings that end at A[i] & B[j] Then,

A[i] = B[j] → L i, j = 1 + Li-1, j-1

A[i] ≠ B[j] → L i, j = 0

LONGEST COMMON SUBSTRING(A,m,B,n)

for i := 0 to m do Li,0 := 0

for j := 0 to n do L0,j := 0

len := 0

answer := <0,0>

for i := 1 to m do

for j := 1 to n do

if Ai ≠ B j then

L i,j := 0

else

L i,j := 1 + L i-1,j-1

if Li,j > len then

len := L i,j

answer = <i,j>

Example:

A L O H A

H 0 0 0 1 0

E 0 0 0 0 0

L 0 1 0 0 0

L 0 1 0 0 0

O 0 0 2 0 0

Longest common subsequence

String C is a subsequence of string A if string C can be obtained by deleting 0 or more symbols from string A

Example:

houseboat

h ous e bo at

ousbo is a subsequence of houseboat

String C is a common subsequence of strings A and B if C is a subsequence of A and also

C is a subsequence of B

Example:

houseboat

computer

h ou seboa t

c mp ut er

out is a common subsequence of houseboat and computer

String C is a longest common subsequence (LCS) of strings A and B if C is a common subsequence of A & B and there is no other common subsequence of A & B that has

greater length

Let Li, j be the length of an LCS of A[1 i] & B[1 j], i.e., the prefixes of strings A & B of lengths i and j

Li, j = Li-1, j-1 + 1, if A i = B i

Li, j = Max{ L i-1, j, Li, j-1 }, if A i ≠ B i

LONGEST COMMON SUBSEQUENCE(A,m,B,n)

for i := 0 to m do Li,0 := 0

for j := 0 to n do L0,j := 0

for i := 1 to m do

for j := 1 to n do

if Ai = B j then

L i,j := 1 + L i-1,j-1

else

L i,j := Max{ L i-1,j , L i,j-1 }

length := L m,n

Dan Hirschberg

Computer Science

University of California, Irvine, CA 92697-3425

dan at ics.uci.edu

Last modified: Oct 28, 2003