dynamic programming

Pangfeng Liu, Advanced Computer Programming 2004, National Taiwan University

Advanced ComputerProgrammingDynamic Programming


Dynamic Programming Elements

Table Element DefinitionRecursive FormulaComputing Sequence


Fundamental Elements in CS

RecursionStructural Complexity

Mathematical InductionLogic

Divide-and-ConquerProgramming Language


Fibonacci Numbers

DefinitionF[i] = F[i-1] + F[i-2]F[0] = 1F[1] = 1


DP Elements

Table Element DefinitionF[i] is the i-th element in the sequence.

Recursive FormulaF[i] = F[i-1] + F[i-2]

Computing SequenceCompute F[2], F[3], … etc.


Comparison

A direct loop implementation.A recursive implementation.One major advance of DP is to reuse the

elements that we know already.


Partition

Given a set of numbers ei, determine whetherwe can partition it into two subsets, so that thesum of these two subset of numbers are equal.Given 1, 3, 3, 4, 5, the answer is “yes”.One set has 1, 3, and 4.The other set has 3, and 5.


DP Elements

Table Element DefinitionLet P(i, j) be 1 if it is possible to select from the

first i elements so that the sum is j.

Recursive FormulaP(i, j) is 1 if P(i –1, j) is 1, or P(i -1, j –ei) is 1.

Computing SequenceCompute one row of P at a time.


Table Computation

The numbers are 1, 3, 3, 4, 5

111110

11111101511111101401101101300001101300000001187654321


Approximate String Matching

Given two string S1 and S2, find a shortest sequenceof operations that transform s1 to S2.

The operations include the following.Substitution

cat to bat

Insertioncat to cats

Deletioncat to at


An example

bronzebrownzebrownzbrowncrown


DP Elements

Table Element DefinitionLet E(i, j) be the minimum number of operations

to transform the first i character to the first jcharacters of S2.Let the i-th char of S1 be c and the j-th char of S2

be d.


DP Elements

Recursive FormulaIf c is d, E(i, j) = E(i-1, j-1)If c is not d, E(i, j) is the minimum of the following.

E(i –1, j –1) + 1, (substitute c by d).E(i –1, j) + 1, (delete c).E(i, j –1) + 1, (insert d).

Computing SequenceCompute one row of E at a time.


Table Computation

432345n432234w432123o543212r654321ceznorb


Longest Common Subsequence

Given two string S1 and S2, find the longestsubstring of both S1 and S2.A is a substring of B if we can find all of A’s

characters in B in the same order they appearin A.abc is a substring of gafbfc.


An example

The longest common substring of “bronze”and “crown”is “ron”.The LCS is not necessarily unique.


DP Elements

Let L(i, j) be the length of LCS of the first icharacter to the first j characters of S2.Let the i-th char of S1 be c and the j-th char

of S2 be d.


DP Elements

If c is d, L(i, j) = L(i, j) + 1If c is not d, E(i, j) is the maximum of the

following.E(i –1, j)E(i, j –1)

Computing SequenceCompute one row of L at a time.


Table Computation

333210n222210w222210o111110r000000ceznorb


Longest Increasing Sequence

Given a sequence of numbers, find thelongest subsequence in which each element isat least as large as the previous one.If the sequence is 9, 5, 2, 8, 7, 3, 1, 6, and 4,

the answer is 3.2, 3, 62, 3, 4


DP Elements

Let L(i) be the length of the longestsubsequence for the first i numbers that endsat the i-th number.L(i) is the maximum of L(j) + 1 for all j that

the i-th number is greater than the j-th number.


Table Computation

9, 5, 2, 8, 7, 3, 1, 6, and 4.The final answer is the maximum of all L(i).

2, 3, 12, 3, 1N/A22, 52, 5N/AN/AN/A

331222111

461378259


Matrix Multiplication

Given a sequence of matrixes M1, … Mn, find theorder to multiply them so that the number ofoperations is minimized.

The number of operations of multiplying a a bmatrix to an b c matrix is ac(2b-1).Multiplication:

abc

Additiona(b-1)c


An Example

(((A B) C) D)3 2 4 = 243 5 2 = 303 4 5 = 60

Total is 114.

1 1 1 11 1 1 11 1 1 1

1 1 1 1 11 1 1 1 1

1 11 11 11 1

1 1 1 11 1 1 11 1 1 11 1 1 11 1 1 1

A B C D


DP Elements

Let M(i, j) be the minimum number ofmultiplications to compute the product form the i-thto the j-th matrix.

M(i, j) = mink (M(i, k) + M(k + 1, j) + rickcj)k is where we “break”the product into two halves.ri is the number of rows in the i-th matrix.ck is the number of columns in the k-th matrix.cj is the number of columns in the j-th matrix.


Table Computation

432i = 1

0j = 1

03 4 2 = 242

04 2 5 = 40Min(24 + 3 2 5,40 + 3 4 5) = 54

3

02 5 4 = 40Min(40 + 4 5 4,40 +4 2 5) = 80

Min(54 + 3 5 4,24 + 40 + 3 2 4,80 + 3 4 4) = 88

4

1 1 1 11 1 1 11 1 1 1

1 1 1 1 11 1 1 1 1

1 11 11 11 1

1 1 1 11 1 1 11 1 1 11 1 1 11 1 1 1


Grammar and Words

Given a binary operator on an alphabet A,and a table of results of , and a string ofcharacters taken from A, find all the possibleresuls.


An Example

A = {a, b, c} ((b b) b) a

= (a b) a= c a = c

(b (b b)) a= (b a) a= a a = a

There is no way toconstruct “b”. cccc

baab

ccaa

cba


DP Elements

Let C(i, j) be the set of characters we canconstruct from the substring of S from the i-thto the j-th character.C(i, j)=m{c1c2|c1C(i, m), c2C(m+1, j)}We “break”the product into two halves at m.


Table Computation

2

baa aac, a4

43i = 1

bj = 1

a2

bc, a3

cccc

baab

ccaa

cba

S = b b b a


DP Elements

We could also use the following definition.Let C(i, j, k) be 1 if we can construct the character k

with the substring of S from the i-th to the j-thcharacter.

C(i, j, k) = 1There exist k1, k2 in A, and m between i and j so that both

C(i, m, k1) is 1 and C(m+1, j, k2) is 1, and k1k2 = k.We “break”the product into two halves at m.


Minimum Length Triangulation

Given a convex polygon, find thetriangulation that has the minimum totallength.A Triangulation is a set of non-intersecting

edges that partition a polygon into triangles.


A Triangulation Example


A Triangle

Every polygon edge is associated with one triangle,which can be identified by a vertex.

e e e

e e


Recursion

Let T(i, j) be the minimum triangulation from vi tovj.

T(i, j) = min(T(i, k) + T(k, j) + dik + dkj), for all kbetween i and j.

i

j

k


One-dimensional Shortest Path

Given a directed graph with all of its edgegoing from left to right, find a shortest pathfrom the leftmost node to the rightmost node.


An Example

4 8

6

4 3 24


DP Elements

Let L(i) be the length of the shortest path thatends at the i-th node from the left.L(i) = minj L(j) + distance from j to i


Table Computation

4 8

6

4 3 2

Min(4+8, 6+2)Min(0+6, 8+3)Min(0+4, 4+4)40

86840

54321

4


Stereo Sets

We want to ship stereo sets. Each stereo sethas a receiver and a pair of speakers.A receiver weights R, and a speaker weights S.We ship the sets by C containers. The i-th

container can ship up to Wi in weight.Find the maximum number of set that we can

ship.


DP Elements

Let N(i, r) be the maximum number of speakers wecan ship with the first i container when the numberof receivers is r.

N(i, r)=maxk (S(i –1, r –k) + Wi –k R)/S)k is the number of receiver we placed in the i-th container.

The answer is the maximum r such that N(C, r) 2r.


Table Computation

R = 3, S = 2, W1 = 8, W2 = 8, W3 = 5.

5

0

0

Don’t care

4

0

2

max(4+(5-3)/2,2+5/2) = 5

3

0

max(2+(8-6)/2,1+ (8–3)/2,0+ 8/2) = 4

max(5+(5-3)/2,4+5/2) = 6

21r = 0

124i = 1

max(4+(8-6)/2,2+(8-3)/2,1+8/2) = 5

max(4+(8-3)/2,2+(8/2)) = 6

4+4 = 82

max(6+(5-3)/2,5+5/2) = 7

max(8+(5-3)/2,6+(5/2)) = 10

8+2= 103


DP Elements

Let N(i, r) be the maximum number of speakers wecan ship with the first i container when the numberof receivers is r.

N(i, r)=maxk (S(i –1, r –k) + Wi –k R)/S)k is the number of receiver we placed in the i-th container.

The answer is the maximum r such that N(C, r) 2r.


Fast Food Chain

There are n fast food restaurants along ahighway. Their positions are given as p1,p2, …pn.Select k restaurants as suppliers, so that the

sum of distance between restaurants and theircorresponding suppliers is minimized.


An Example

n = 7, k = 3p1=0, p2 = 1, p3 = 3, p4=6, p5 = 7, p6 = 9, p7 = 10.By selecting p2 , p4 , and p6 ,the sum of distance is

1 + 0 + 2 + 0 + 1 + 0 + 1 = 5


DP Elements

Let C(i, j) be the minimum cost of selecting jsuppliers in the first i restaurants.C(i, j)=mink (C(i –k, j - 1)+Mk+1,i)Mk+1,i is the minimum cost of selecting a supplier

from the pk+1 to pi.


Table Computation

765432i = 1

2218128310j = 1

Not usedmin(0+12,1+7,3+3,8+2,12+0)=6

min(0+9,1+4,3+1,8+0)= 4

min(0+5,1+3,3+0) = 3

min(0+2,1+0)=1

0N/A2

min(0+9,1+6,3+3,4+1,6+0)=5

Not usedNot usedNot used0N/AN/A3


Just Talking

There are m male students, f female students, and ateacher in a classroom.

The teacher wants to compute the total amount ofmoney in the classroom by a “reduction”procedure.

A student tells his/her amount of money to anotherstudent (or to the teacher). The receiver computesthe sum, and relay the information to the nextreceiver.


Reduction

Everyone can only speak to one person,everyone can only listen to one person, andno one can speak and listen at the same time.It takes 3 seconds for a female student to tell

a number, and 5 seconds for a male student.How to do this in the least amount of time?


Examples

We have the teacher, Tom, John and Mary in theclassroom.

Tom tells John, Mary tells the teacher, then Johntells the teacher.10 seconds.

Tom tells teacher, John tells Mary, then Mary tellsteacher.8 seconds.


Minimum Critical Path

A path is a route from a leaf to the root.A path is critical if it has the longest length, where

the length is defined as the sum of node costs.The minimum critical path problem is to arrange a

set of nodes into a binary tree so that the length ofthe critical path is minimized.


Observation

It is always possible to find an optimalarrangement where the cost of every parent isno greater than both of its children.This implies we can always put the smallest

node at the root.


An Example

3

5 5

8 8

5

5 3

10 8


Two Classes of Gossips

If there are only two kinds of nodes, theminimum critical path problem can be solvedby dynamic programming.


DP Elements

Let C(f, s) be the minimum critical path for a treewith f fast talkers and s slow talkers.

If f > 0, C(f, s) = T(f)+min(C(fl, sl) + C(fr, sr))fl, and sl are the number of fast/slow talkers in the left

subtree.fr, and sr are the number of fast/slow talkers in the right

subtree.

If f is 0, C(f, s) = log(s+1)


Cutting Pieces

IOI 2004Given a plate of size w by h, we want to cut it

into pieces so that minimum area is wasted..There are only k sizes we can sell these pieces.Every cut must go through the entire width or

the entire length.


An Example

The size of the plate is5 by 7.

The sellable sizes are 2by 3, 5 by 2 and, 7 by10.


DP Elements

Let M(w, h) be the minimum waste area.M(w, h) = min (M(wl, h) + M(wr, h), M(w, hl)

+ M(w, hu))wl and wr are the width after a vertical cut.hl and hu are the height after a horizontal cut.


Possible Saving

Do not cut a piece that is too narrow or tooshort.Cut till the middle is fine due to symmetry.Is it possible to prove that there exist an

optimal solution that we alwaysCut horizontally with the one of the height?Cut vertically with one of the width?


A Similar Problem

What if the problem is reduced to onedimension?We have a problem similar to partition, butThe elements can be selected multiple times.We are minimize the “leftover”.

In this case you can prove the “pick one ofthe sizes”is a correct strategy.


The Largest Square

We have a rectangular construction site, inwhich we want to build a square house.Some part of the land is not suitable for

construction.What is the largest house you can build?


An Example


DP Elements

Let S(i, j) be the maximum size one can build for ahouse whose lower left corner is at (i, j).

S(i+1,j) == S(i,j+1)S(i,j) = S(i,j+1)+1 if (i+S(i,j+1), j+S(i,j+1)) is subitable

for construction.S(i,j) = S(i,j+1) otherwise.

S(i+1,j) != S(i,j+1)S(i,j) = min(S(i+1,j), S(i,j+1)) + 1


An Illustration

(i,j) (i,j)


The Result

1


An Example

122233201

111222112

101112012

122012211

111211201

101201222

111111111


Maximum Triangle

The problem is similar to the maximumsquare.Find the maximum sized triangle in a

triangular area.The only complication is that you might have

a “up-side-down”triangle.


An Example

111

21

11

2


Postage Stamps Problem

We have infinite amount of stamps ofdifferent values.What is the smallest postage amount that

could not be paid by these stamps?

Stamps : 1 , 3 , 5

Need : 14 = 5 + 5 + 3 + 1


DP Elements

Let A(i) be 1 if we could pay for the postageof i units, 0, otherwise.Let the postage stamps be of s1, s2, … sk units.A(i) = 1 if and only if there is an j such that

A(i –sj) is 1, where j is between 1 and k.The smallest i such that A(i) is 0 is the answer.


The Messenger

There is a messenger and N recipients on a unit gridmesh.

The messenger can talk to a recipient as long as theyhave the same x or y coordinate, so he only needs tomove to the same row or column of the recipient.

Given the coordinates of all recipients, how to movethe messenger so that he could send the message toall recipients with the shortest moving distance.


An Illustration

R

R

R

R

M

R

dynamic programming

Documents

jth number

j characters of s2

jth char of s2

ith element

ith number

ith char of s1

i elements

set of numbers ei