Lecture 8:Dynamic Programming

Shang-Hua Teng

First Example: n choose k

• Many combinatorial problems require the calculation of the binomial coefficient.

• This is equivalent to given n objects how many different ways can you choose k of them.

• Mathematically we have

• or,

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!

knk

n

k

n

nkk

nk

k

n

k

n

k

n

0

0

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int choose(n,k) if (n = = k) return 1;if (n < k) or (k < 0) return 0;return choose(n-1, k-1) + choose(n - k, k);

Consider Divide and Conquer

How long does this take?

T(n,k) = T(n-1, k-1) + T(n-k, k) )( kn

Remember Pascal’s Trianglefunction choose (n,k)

1 1 1 1 2 1 1 3 3 1 1 4 6 4 11 5 10 10 5 1

By adding two numbers from previous row we can calculate the triangle quickly.

Consider triangle shape

11 11 2 11 3 3 11 4 6 4 11 5 10 10 5 1

012345

0 1 2 3 4 5

• Looks like a two dimensional array.

• Create the triangle in this array and output A(i,j) for

j

i

Do it algorithmically

Create an 2-d array A of size (n+1) by (k +1 ).

Set A[i, 0] 1 for i = 0,1,2,.., nSet A[i, i ] 1 for i = 0,1,2,...., n

for i = 1 to nfor j = 1 to k

A[ i, j ] = A[ i-1, j ] + A[ i –1, j-1 ];

Output A[n,k]

Runtime is O(nk)

Matrix Chain Multiplication - Review

• Recall how to multiply matrices.

• Given two matrices

• The product is

• Time is

AB

CX =

)( edA )( feB

d

e

e

f

1

0

],[],[],[e

k

jkkiji BAC

d

f

)(defO

Matrix Chaining contd.• What about multiplying multiple matrices?

• We know matrix multiplication is associative. Can we use this to help minimize calculations?

• Sure, consider

• If we multiply we do 4000 operations

• But instead if we try we do 1575 operations

54321 AAAAAA

)35( A )1003( B )5100( C

CBA

)( CBA

How do we parenthesize

• Brute force - try all ways to parenthesize

• Find out which has smallest number of operations by doing multiplication, then pick the best one.

• But how many ways can we parenthesize these?

• Equivalent to the number of ways to make a binary tree with n nodes. This is see Catalan numbers.

• See homework Problem 12-4, ex15.2-3 in the book.

nAAAAA 321

)2( n

Be greedy

• We just learned that a greedy algorithm can sometimes work, let’s try.

• Why not try starting with the product with the most operations.

• Consider

• If we do our greedy method we would compute • This is 2000 operations.• Try instead, which takes 1000

operations.

)105()510()105()510( DCBA

DCBA

DCBA

Another Greedy way

• Select the product with the fewest operations.

• Consider

• Here the new greedy method would compute

• This is 109989+9900+108900=228789 operations

• Try

• Here we only need 9999+89991+89100=189090

)99100()1009()911()11101( DCBA

DCBA

DCBA

Optimality of Subproblems

• Consider general question again, we want to parenthesize

• Now suppose somehow we new that the last multiplication we need to do was at the ith position.

• Then the problem is the same as knowing the best way to parenthesize and

nAAAA 321

)()( 121 nii AAAAA

)( 21 iAAA )( 1 ni AA

Overlapping Subproblems

• We know how to break problem into smaller pieces, why doesn’t divide and conquer work?

• We need to optimize each piece why doesn’t greedy work.

• Again suppose we know where the final multiply is and we want to generalize the cost with a recurrence, consider

• But notice these subproblems overlap.

11,1,, min jkijkki

jkiji dddNNN

Dynamic Programming

• Since subproblems overlap we can not use divide and conquer method.

• Instead work from the “bottom up.”

• We know how to parenthesize one matrix, two matrices, we can build from there…

Matrix Chain Order

sm

kjis

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jik

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lij

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and return

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What have we noticed

• Optimal solution depends on optimal subproblems

• Subproblems overlap.

• So a “top down” divide and conquer doesn’t seem to work.

• Somehow we need to build up from bottom by remembering solution from subproblem.

Another Example

• Biologists need to measure how similar strands of DNA are to 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 = AGTCAACGTT, Y=GTTCGACTGTG

• Both S = AGTG and S’=GTCACGT are subsequences

• How to do find these efficiently?

Brute Force

• 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.

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

111 and of LCSan is and then , If nmknmknm YXZyxzyx

1 and of LCSan is that implies

then , If

n

nknm

YXZ

yzyx

Some observations

•

•

•

•

. 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

Setup

• 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]

The recurrence

• 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

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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

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