1.1 data structure and algorithm lecture 7 dynamic programming topics reference: introduction to...

1.1Data Structure and Algorithm

Lecture 7

Dynamic Programming

Topics

Reference: Introduction to Algorithm by Cormen

Chapter 15: Dynamic Programming

1.2Data Structure and Algorithm

Longest Common Subsequence (LCS)

Biologists need to measure similarity between DNA and thus determine how closely related an organism is to another.

They do this by considering DNA as strings of letters A,C,G,T and then comparing similarities in the strings.

Formally, they look at common subsequences in the strings.

Example X = ABCBDAB, Y=BDCABA Subsequences may be: ABA, BCA, BCB, BBA BCBA BDAB etc.

But the Longest Common Subsequences (LCS) are

BCBA and BDAB. How to find LCS efficiently?

1.3Data Structure and Algorithm

Longest Common Subsequence( Brute Force Approach )

if |X| = m, |Y| = n, then there are 2m subsequences of X; we must compare each with Y (n comparisons)

So the running time of the brute-force algorithm is O(n 2m)

Making it impractical for long sequences.

LCS problem has optimal substructure:

solutions of subproblems are parts of the final solution.

Subproblems: “find LCS of pairs of prefixes of X and Y”

1.4Data Structure and Algorithm

LCS: Optimal Substructure

111 and of LCSan is and then , If nmknmknm YXZyxzyx

1 and of LCSan is that implies

then , If

n

nknm

YXZ

yzyx

. and

of LCSany be ,,,let and sequences

be ,,, and ,,,Let

21

2121

YX

zzzZ

yyyYxxxX

k

nm

YXZ

xzyx

m

mknm

and of LCSan is that implies

then , If

1

1.5Data Structure and Algorithm

LCS: Setup for Dynamic Programming

First we’ll find the length of LCS, along the way we will leave “clues” on finding subsequence.

Define Xi, Yj to be the prefixes of X and Y of length i and j respectively.

Define c[i,j] to be the length of LCS of Xi and Yj

Then the length of LCS of X and Y will be c[m,n]

1.6Data Structure and Algorithm

LCS recurrence

Notice the issues…This recurrence is exponential. We don’t know max ahead of time. The subproblems overlap, to find LCS we need to find

LCS of c[i, j-1] and of c[i-1, j]

ji

ji

yxjijicjic

yxjijic

ji

jic

and 0, if ]),1[],1,[max(

, and 0, if 1]1,1[

0or 0 if 0

],[

1.7Data Structure and Algorithm

LCS Algorithm

First we’ll find the length of LCS. Later we’ll modify the algorithm to find LCS itself.

Recall we want to let Xi, Yj to be the prefixes of X and Y of length i and j respectively

And that Define c[i,j] to be the length of LCS of Xi and Yj

Then the length of LCS of X and Y will be c[m,n]

otherwise]),1[],1,[max(

],[][ if1]1,1[],[

jicjic

jyixjicjic

1.8Data Structure and Algorithm

LCS Recursive Solution

We start with i = j = 0 (empty substrings of x

and y)

Since X0 and Y0 are empty strings, their LCS is

always empty (i.e. c[0,0] = 0)

LCS of empty string and any other string is

empty, so for every i and j: c[0, j] = c[i,0] = 0

1.9Data Structure and Algorithm

LCS Recursive Solution

When we calculate c[i,j], we consider two

cases:

First case: x[i]=y[j]: one more symbol in

strings X and Y matches, so the length of LCS

Xi and Yj equals to the length of LCS of

smaller strings Xi-1 and Yi-1 , plus 1

1.10Data Structure and Algorithm

LCS Recursive Solution

Second case: x[i] != y[j]

As symbols don’t match, our solution is not

improved, and the length of LCS(Xi , Yj) is the

same as before (i.e. maximum of LCS(Xi, Yj-1)

and LCS(Xi-1,Yj)

1.11Data Structure and Algorithm

LCS Example

We’ll see how LCS algorithm works on the following

example:

X = ABCB

Y = BDCAB

What is the Longest Common Subsequence

of X and Y?

LCS(X, Y) = BCB

X = A B C B

Y = B D C A B

1.12Data Structure and Algorithm

LCS Example (0)j 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

B

Yj BB ACD

X = ABCB; m = |X| = 4Y = BDCAB; n = |Y| = 5Allocate array c[6,5]

1.13Data Structure and Algorithm

LCS Example (1)j 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

B

Yj BB ACD

0

0

0

0

for i = 1 to m c[i,0] = 0

1.14Data Structure and Algorithm

LCS Example (2)j 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

B

Yj BB ACD

0

0

00000

0

0

0

for j = 0 to n c[0,j] = 0

1.15Data Structure and Algorithm

LCS Example (3)j 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

B

Yj BB ACD

0

0

00000

0

0

0

0

case i=1 and j=1 A != B but, c[0,1]>=c[1,0] so c[1,1] = c[0,1], and b[1,1] =

1.16Data Structure and Algorithm

LCS Example (4)j 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

B

Yj BB ACD

0

0

00000

0

0

0

0

case i=1 and j=2 A != D but, c[0,2]>=c[1,1] so c[1,2] = c[0,2], and b[1,2] =

0

1.17Data Structure and Algorithm

LCS Example (5)j 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

B

Yj BB ACD

0

0

00000

0

0

0

0

case i=1 and j=3 A != C but, c[0,3]>=c[1,2] so c[1,3] = c[0,3], and b[1,3] =

0 0

1.18Data Structure and Algorithm

LCS Example (6)j 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

B

Yj BB ACD

0

0

00000

0

0

0

0 0 0 1

case i=1 and j=4 A = A so c[1,4] = c[0,3]+1, and b[1,4] =

1.19Data Structure and Algorithm

LCS Example (7)j 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

B

Yj BB ACD

0

0

00000

0

0

0

000 1 1

case i=1 and j=5 A != B this time c[0,5]<c[1,4] so c[1,5] = c[1, 4], and b[1,5] =

1.20Data Structure and Algorithm

LCS Example (8)j 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

B

Yj BB ACD

0

0

00000

0

0

0

0 0 10 1

1

case i=2 and j=1 B = B so c[2, 1] = c[1, 0]+1, and b[2, 1] =

1.21Data Structure and Algorithm

LCS Example (9)j 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

B

Yj BB ACD

0

0

00000

0

0

0

0 0 10 1

1

case i=2 and j=2 B != D and c[1, 2] < c[2, 1] so c[2, 2] = c[2, 1] and b[2, 2] =

1

1.22Data Structure and Algorithm

LCS Example (10)j 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

B

Yj BB ACD

0

0

00000

0

0

0

0 0 10 1

1

case i=2 and j=3 B != D and c[1, 3] < c[2, 2] so c[2, 3] = c[2, 2] and b[2, 3] =

1 1

1.23Data Structure and Algorithm

LCS Example (11)j 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

B

Yj BB ACD

0

0

00000

0

0

0

0 0 10 1

1

case i=2 and j=4 B != A and c[1, 4] = c[2, 3] so c[2, 4] = c[1, 4] and b[2, 2] =

1 1 1

1.24Data Structure and Algorithm

LCS Example (12)j 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

B

Yj BB ACD

0

0

00000

0

0

0

0 0 10 1

1

case i=2 and j=5 B = B so c[2, 5] = c[1, 4]+1 and b[2, 5] =

1 1 1 2

1.25Data Structure and Algorithm

LCS Example (13)j 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

B

Yj BB ACD

0

0

00000

0

0

0

0 0 10 1

1

case i=3 and j=1 C != B and c[2, 1] > c[3,0] so c[3, 1] = c[2, 1] and b[3, 1] =

1 1 1 2

1

1.26Data Structure and Algorithm

LCS Example (14)j 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

B

Yj BB ACD

0

0

00000

0

0

0

0 0 10 1

1

case i=3 and j= 2 C != D and c[2, 2] = c[3, 1] so c[3, 2] = c[2, 2] and b[3, 2] =

1 1 1 2

1 1

1.27Data Structure and Algorithm

LCS Example (15)j 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

B

Yj BB ACD

0

0

00000

0

0

0

0 0 10 1

1

case i=3 and j= 3 C = C so c[3, 3] = c[2, 2]+1 and b[3, 3] =

1 1 1 2

1 1 2

1.28Data Structure and Algorithm

LCS Example (16)j 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

B

Yj BB ACD

0

0

00000

0

0

0

0 0 10 1

1

case i=3 and j= 4 C != A c[2, 4] < c[3, 3] so c[3, 4] = c[3, 3] and b[3, 3] =

1 1 1 2

1 1 2 2

1.29Data Structure and Algorithm

LCS Example (17)j 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

B

Yj BB ACD

0

0

00000

0

0

0

0 0 10 1

1

case i=3 and j= 5 C != B c[2, 5] = c[3, 4] so c[3, 5] = c[2, 5] and b[3, 5] =

1 1 1 2

1 1 2 22

1.30Data Structure and Algorithm

LCS Example (18)j 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

B

Yj BB ACD

0

0

00000

0

0

0

0 0 10 1

1

case i=4 and j=1 B = B so c[4, 1] = c[3, 0]+1 and b[4, 1] =

1 1 1 2

1 1 2 2

1

2

1.31Data Structure and Algorithm

LCS Example (19)j 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

B

Yj BB ACD

0

0

00000

0

0

0

0 0 10 1

1

case i=4 and j=2 B != D c[3, 2] = c[4, 1] so c[4, 2] = c[3, 2] and b[4, 2] =

1 1 1 2

1 1 2 2

11

2

1.32Data Structure and Algorithm

LCS Example (20)j 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

B

Yj BB ACD

0

0

00000

0

0

0

0 0 10 1

1

case i=4 and j= 3 B != C c[3, 3] > c[4, 2] so c[4, 3] = c[3, 3] and b[4, 3] =

1 1 1 2

1 1 2 2

11 2

2

1.33Data Structure and Algorithm

LCS Example (21)j 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

B

Yj BB ACD

0

0

00000

0

0

0

0 0 10 1

1

case i=4 and j=4 B != A c[3, 4] = c[4, 3] so c[4, 4] = c[3, 4] and b[3, 5] =

1 1 1 2

1 1 2 2

11 22

2

1.34Data Structure and Algorithm

LCS Example (22)j 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

B

Yj BB ACD

0

0

00000

0

0

0

0 0 10 1

1

case i=4 and j=5 B= B so c[4, 5] = c[3, 4]+1 and b[4, 5] =

1 1 1 2

1 1 2 2

11 22 3

2

1.35Data Structure and Algorithm

LCS Algorithm

LCS-Length(X, Y) m = length(X), n = length(Y)

for i = 1 to m do c[i, 0] = 0 for j = 0 to n do c[0, j] = 0 for i = 1 to m do for j = 1 to n do

if ( xi = = yj ) then c[i, j] = c[i - 1, j - 1] + 1 else if c[i - 1, j]>=c[i, j - 1] then c[i, j] = c[i - 1, j]

else c[i, j] = c[i, j - 1] return c and b

"" , jib

"" , jib

"" , jib

1.36Data Structure and Algorithm

LCS Algorithm Running Time

LCS algorithm calculates the values of each entry of the array c[m,n] So the running time is clearly O(mn) as each entry is done in 3 steps. Now how to get at the solution? We use the arrows we created to guide us. We simply follow arrows back to base case 0

1.37Data Structure and Algorithm

Finding LCSj 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

Yj BB ACD

0

0

00000

0

0

0

1000 1

1 21 1

1 1 2

1

22

1 1 2 2 3B

1.38Data Structure and Algorithm

Finding LCS (2)j 0 1 2 3 4 5

0

1

2

3

4

i

Xi

A

B

C

Yj BB ACD

0

0

00000

0

0

0

1000 1

1 21 1

1 1 2

1

22

1 1 2 2 3B

B C BLCS:

1.39Data Structure and Algorithm

Finding LCS (3)

Print_LCS (X, i, j) if i = 0 or j = 0 then return if b[i, j] = “ “ then

Print_LCS (X, i-1, j-1) Print X[i]

elseif b[i, j] = “ “ then Print_LCS (X, i-1, j)

else Print_LCS (X, i, j-1)

Cost: O(m+n)

1.40Data Structure and Algorithm

Element of Dynamic Programming

Optimal Substructure

Overlapping Subproblems

1.41Data Structure and Algorithm

Optimal Substructure

A problem exhibits optimal substructure if an optimal solution contains optimal solutions to its subproblems.

Build an optimal solution from optimal solutions to subproblems

solutions of subproblems are parts of the final solution.

Example :Longest Common Subsequence - An LCS contains within it optimal solutions to the prefixes

of the two input sequences.

Common with Greedy Solution

1.42Data Structure and Algorithm

Overlapping Subproblems

Divide-and-Conquer is suitable when generating brand-new problems at each step of the recursion.

Dynamic-programming algorithms take advantage of overlapping subproblems by solving each subproblem once and then storing the solution in a table where it can be looked up when needed, using constant time per lookup

1.43Data Structure and Algorithm

Dynamic VS Greedy

Dynamic programming uses optimal substructure in a bottom-up fashion First find optimal solutions to subproblems

and, having solved the subproblems, we find an optimal solution to the problem

Greedy algorithms use optimal substructure in a top-down fashion First make a choice – the choice that looks

best at the time – and then solving the resulting subproblem