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Lecture 8. Paradigm #6 Dynamic Programming
 Popularized by Richard Bellman ("Dynamic
Programming", Princeton University Press,
1957; call number QA 264.B36). Chapter 15 of
CLRS.
 Typically, dynamic programming reduces the
complexity of a problem from 2n to O(n3) or O(n2)
or even O(n).
 It does so by keeping track of already computed
results in a bottom-up fashion, hence avoiding
enumerating all possibilities.
 Typically applies to optimization problems.
Example 1. Efficient multiplication of matrices
(Section 15.2 of CLRS.)
 Suppose we are given the following 3 matrices:
M1 10 x 100

M2 100 x 5

M3 5 x 50
 There are two ways to compute M1*M2*M3: M1 (M2 M3) or (M1
M2) M3
 Since the cost of multiplying a p x q matrix by a q x r matrix is
pqr multiplications, the cost of M1 (M2 M3) is 100 x 5 x 50 + 10 x
100 x 50 = 75,000 multiplications, while the cost of (M1 M2) M3 is
10 x 100 x 5 + 10 x 5 x 50 = 7,500 multiplications: a difference of

a factor of 10.
Naïve approach
 We could enumerate all possibilities, and then take the
minimum. How many possibilities are there?
 The LAST multiplication performed is either M1*(M2 ... Mn), or
(M1 M2)*(M3 ... Mn), or ... (M1 M2 ...)(Mn). Therefore, W(n), the
number of ways to compute M1 M2 ... Mn, satisfies the following
recurrence:
W(n) = Σ1 ≤ k < n W(k)W(n-k)
--- Catalan number
 Now it can be proved by induction that W(n) = (2n-2 choose n1)/n. Using Stirling's approximation, which says that
n! = √(2πn) nn e-n (1 + o(1)),
we have (2n choose n) ~ 22n/√(π n),
 We conclude that W(n) ~ 4n n-3/2, which means our naive
approach will simply take too long (about 1010 steps when n =
20).
Dynamic Programming approach
 Let’s avoid all the re-computation of the recursive approach.
 Observe: Suppose the optimal method to compute M1 M2 ... Mn
were to first compute M1 M2 ... Mk (in some order), then compute
Mk+1 ... Mn (in some order), and then multiply these together.
Then the method used for M1 M2 ... Mk must be optimal, for
otherwise we could substitute a superior method and improve
the optimal method. Similarly, the method used to compute Mk+1
... Mn must also be optimal. The only thing left to do is to find the
best possible k, and there are only n choices for that.
 Letting m[i,j] represent the optimal cost for computing the
product Mi ... Mj, we see that
m[i,j] = min { m[i,k] + m[k+1,j] + p[i-1]p[k]p[j] }, i ≤ k < j
 k represents the optimal place to break the product Mi ... Mj into
two pieces. Here p is an array such that M1 is of dimension p[0]
× p[1], M2 is of dimension p[1] × p[2], ... etc.
Implementing it --- O(n3) time
 Like the Fibonacci number example, we cannot implement this
by recursion. It will be exponential time.

MATRIX-MULT-ORDER(p)
/* p[0..n] is an array holding the dimensions of the matrices; matrix i has
dimension p[i-1] x p[i] */
for i := 1 to n do
m[i,i] := 0
for d := 1 to n-1 do // d is the size of the sub-problem.
for i := 1 to n-d do
j := i+d
m[i,j] := infinity;
for k := i to j-1 do
q := m[i,k] + m[k+1,j] + p[i-1]*p[k]*p[j]
if q < m[i,j] then
m[i,j] := q
s[i,j] := k // optimal position for breaking m[i,j]
return(m,s)
Actually multiply the matrices
 We have stored the break points k’s in the array s. s[i,j]
represents the optimal place to break the product Mi ... Mj. We
can use s now to multiply the matrices:
 MATRIX-MULT(M, s, i, j)
/* Given the matrix s calculated by MATRIX-MULT-ORDER. The
list of matrices M = [M1, M2, ... , Mn]. Starting and finishing
indices i and j. This routine computes the product Mi ... Mj using
the optimal method */
if j > i then
X := MATRIX-MULT(M, s, i, s[i,j]);
Y := MATRIX-MULT(M, s, s[i,j]+1, j);
return(X*Y);
else return(Mi)
Longest Common Subsequence
(LCS)
Application: comparison of two DNA strings
Ex: X= {A B C B D A B }, Y= {B D C A B A}
Longest Common Subsequence:
X= AB
C
BDAB
Y=
BDCAB A
Brute force algorithm would compare each
subsequence of X with the symbols in Y
LCS Algorithm
 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)
 Notice that the LCS problem has optimal
substructure: solutions of subproblems are parts of
the final solution – often, this is when you can use
dynamic programming.
 Subproblems: “find LCS of pairs of prefixes of X
and Y”
LCS Algorithm
 First we’ll find the length of LCS. Later we’ll modify
the algorithm to find LCS itself.
 Let Xi, Yj be the prefixes of X and Y of length i and j
respectively
 Let c[i,j] be the length of LCS of Xi and Yj
 Then the length of LCS of X and Y will be c[m,n]
if x[i]  y[ j ],
c[i  1, j  1]  1
c[i, j ]  
 max(c[i, j  1], c[i  1, j ]) otherwise
LCS recursive solution
if x[i]  y[ j ],
c[i  1, j  1]  1
c[i, j ]  
 max(c[i, j  1], c[i  1, j ]) otherwise
 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
LCS recursive solution
if x[i]  y[ j ],
c[i  1, j  1]  1
c[i, j ]  
 max(c[i, j  1], c[i  1, j ]) otherwise
 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
LCS recursive solution
if x[i]  y[ j ],
c[i  1, j  1]  1
c[i, j ]  
 max(c[i, j  1], c[i  1, j ]) otherwise
 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, we take the maximum of LCS(Xi, Yj-1) and
LCS(Xi-1,Yj)
Think:
Why can’t we just take the length of LCS(Xi-1, Yj-1)12?
2/13/2015
LCS Length Algorithm
LCS-Length(X, Y)
1. m = length(X) // get the # of symbols in X
2. n = length(Y) // get the # of symbols in Y
3. for i = 1 to m c[i,0] = 0 // special case: Y0
4. for j = 1 to n
c[0,j] = 0 // special case: X0
5. for i = 1 to m
// for all Xi
6.
for j = 1 to n
// for all Yj
7.
if ( Xi == Yj )
8.
c[i,j] = c[i-1,j-1] + 1
9.
else c[i,j] = max( c[i-1,j], c[i,j-1] )
10. return c
LCS Example
We’ll see how LCS algorithm works on the following
example:
 X = ABCB
 Y = BDCAB
LCS(X, Y) = BCB
X=AB C B
Y= BD CAB
LCS Example (0)
j
i
0
1
0
1
Yj
B
2
3
4
5
D
C
A
B
Xi
A
2
B
3
C
4
B
X = ABCB; m = |X| = 4
Y = BDCAB; n = |Y| = 5
Allocate array c[5,4]
ABCB
BDCAB
LCS Example (1)
j
i
0
1
Xi
A
0
1
2
3
4
5
Yj
B
D
C
A
B
0
0
0
0
0
0
0
2
B
3
C
0
4
B
0
0
for i = 1 to m
for j = 1 to n
2/13/2015
ABCB
BDCAB
c[i,0] = 0
c[0,j] = 0
16
LCS Example (2)
j
i
0
1
Xi
A
0
1
2
3
4
5
Yj
B
D
C
A
B
0
0
0
0
0
0
0
0
2
B
3
C
0
4
B
0
ABCB
BDCAB
0
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
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17
LCS Example (3)
j
i
0
1
Xi
A
0
1
2
3
4
5
Yj
B
D
C
A
B
0
0
0
0
0
0
0
0
0
0
2
B
3
C
0
4
B
0
ABCB
BDCAB
0
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
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18
LCS Example (4)
j
i
0
1
Xi
A
0
1
2
3
4
5
Yj
B
D
C
A
B
0
0
0
0
0
0
0
0
0
0
1
2
B
3
C
0
4
B
0
ABCB
BDCAB
0
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
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19
LCS Example (5)
j
i
0
1
Xi
A
0
1
2
3
4
5
Yj
B
D
C
A
B
0
0
0
0
0
0
0
0
0
0
1
1
2
B
3
C
0
4
B
0
ABCB
BDCAB
0
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
2/13/2015
20
LCS Example (6)
j
i
0
1
Xi
A
0
1
2
3
4
5
Yj
B
D
C
A
B
0
0
0
0
0
0
0
0
0
0
1
1
0
1
2
B
3
C
0
4
B
0
ABCB
BDCAB
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
2/13/2015
21
LCS Example (7)
j
i
0
1
Xi
A
0
1
2
3
4
5
Yj
B
D
C
A
B
0
0
0
0
0
0
0
0
0
0
1
1
0
1
1
1
1
2
B
3
C
0
4
B
0
ABCB
BDCAB
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
2/13/2015
22
LCS Example (8)
j
i
0
1
Xi
A
0
1
2
3
4
5
Yj
B
D
C
A
B
0
0
0
0
0
0
0
0
0
0
1
1
0
1
1
1
1
2
2
B
3
C
0
4
B
0
ABCB
BDCAB
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
2/13/2015
23
LCS Example (10)
j
i
0
1
Xi
A
0
1
2
3
4
5
Yj
B
D
C
A
B
0
0
0
0
0
0
0
0
0
0
1
1
0
1
1
1
1
2
1
1
2
B
3
C
0
4
B
0
ABCB
BDCAB
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
2/13/2015
24
LCS Example (11)
j
i
0
1
Xi
A
0
1
2
3
4
5
Yj
B
D
C
A
B
0
0
0
0
0
0
0
0
0
0
1
1
0
1
1
1
1
2
1
1
2
2
B
3
C
0
4
B
0
ABCB
BDCAB
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
2/13/2015
25
LCS Example (12)
j
i
0
1
Xi
A
0
1
2
3
4
5
Yj
B
D
C
A
B
0
0
0
0
0
0
0
0
0
0
1
1
0
1
1
1
1
2
1
1
2
2
2
2
B
3
C
0
4
B
0
ABCB
BDCAB
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
2/13/2015
26
LCS Example (13)
j
i
0
1
Xi
A
0
1
2
3
4
5
Yj
B
D
C
A
B
0
0
0
0
0
0
0
0
0
0
1
1
0
1
1
1
1
2
1
2
2
2
2
B
3
C
0
1
4
B
0
1
ABCB
BDCAB
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
2/13/2015
27
LCS Example (14)
j
i
0
1
Xi
A
0
1
2
3
4
5
Yj
B
D
C
A
B
0
0
0
0
0
0
0
0
0
0
1
1
0
1
1
1
1
2
2
2
B
3
C
0
1
1
2
2
4
B
0
1
1
2
2
ABCB
BDCAB
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
2/13/2015
28
LCS Example (15)
j
i
0
1
Xi
A
0
1
2
3
4
5
Yj
B
D
C
A
B
0
0
0
0
0
0
0
0
0
0
1
1
0
1
1
1
1
2
2
B
3
C
0
1
1
2
2
2
4
B
0
1
1
2
2
3
ABCB
BDCAB
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
else c[i,j] = max( c[i-1,j], c[i,j-1] )
2/13/2015
29
LCS Algorithm Running Time
 LCS algorithm calculates the values of each entry of
the array c[m,n]
 So what is the running time?
O(m*n)
since each c[i,j] is calculated in
constant time, and there are m*n
elements in the array
2/13/2015
30
How to find actual LCS
 So far, we have just found the length of LCS, but
not LCS itself.
 We want to modify this algorithm to make it output
Longest Common Subsequence of X and Y
Each c[i,j] depends on c[i-1,j] and c[i,j-1]
or c[i-1, j-1]
For each c[i,j] we can say how it was acquired:
2
2
2
3
2/13/2015
For example, here
c[i,j] = c[i-1,j-1] +1 = 2+1=3
31
How to find actual LCS - continued
 Remember that
if x[i]  y[ j ],
c[i  1, j  1]  1
c[i, j ]  
 max(c[i, j  1], c[i  1, j ]) otherwise



So we can start from c[m,n] and go
backwards
Whenever c[i,j] = c[i-1, j-1]+1, remember
x[i] (because x[i] is a part of LCS)
When i=0 or j=0 (i.e. we reached the
beginning), output remembered letters in
reverse order
2/13/2015
32
Finding LCS
j
i
0
1
Xi
A
0
1
2
3
4
5
Yj
B
D
C
A
B
0
0
0
0
0
0
0
0
0
0
1
1
0
1
1
1
1
2
2
B
3
C
0
1
1
2
2
2
4
B
0
1
1
2
2
3
2/13/2015
33
Finding LCS (2)
j
i
0
1
Xi
A
0
1
2
3
4
5
Yj
B
D
C
A
B
0
0
0
0
0
0
0
0
0
0
1
1
0
1
1
1
1
2
2
B
3
C
0
1
1
2
2
2
4
B
0
1
1
2
2
3
LCS (reversed order): B C B
B C B
LCS (straight order):
(this string turned out to be a palindrome)
2/13/2015
34
If we have time, we will do some
exercises in class:
 Edit distance: Given two text strings A of length n
and B of length m, you want to transform A into B
with a minimum number of operations of the following
types: delete a character from A, insert a character
into A, or change some character in A into a new
character. The minimal number of such operations
required to transform A into B is called the edit
distance between A and B.
 Balanced Partition: Given a set of n integers each in
the range 0 ... K. Partition these integers into two
subsets such that you minimize |S1 - S2|, where S1
and S2 denote the sums of the elements in each of
the two subsets.

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