Advanced Dynamic

Programming II

HKOI Training Team 2004

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In the previous lesson... What is DP? Some examples of DP Probably NOT enough for you to solve DP

problems in IOI/NOI Except those classic ones

To identify a DP problem, the keys are Recurrence Optimal substructure Experience (Chinglish(?) - “DP feel”)

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In this lesson... Dimension reduction DP on trees, graphs, etc. Game strategy - Minimax

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Dimension Reduction Reduce the memory complexity by one (or

more) dimension Usually a “rolling” array is employed

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Triangle revisited Only a 2x5 array is needed

3

1 4

2 5 8

9 5 6 1

5 2 3 6 6

A F

3

4 7

9 12 15

21 17 21 16

26 23 24 27 22

6

4 7

9 12 15

21 17 21 16

26 23 24 27 22

Rolling array

F

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4 7

9 12 15

21 17 21 16

26 23 24 27 22

F’

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LCS revisited Recall the recurrence F[i,j] = F[i-1,j-1]+1 if A[i]=B[j] F[i,j] = max{F[i-1,j],F[i,j-1]} if A[i]B[j] Note that F[i,?] only depends on F[i,?] and

F[i-1,?] Thus we can just keep 2 rows

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Non-rectangular structures DP can also be applied on graphs, trees,

etc. Usually structures with no cycles

Recurrence should not contain cycles! Rooted tree is a recursive structure Notation

C(v): the set of children of v (in a rooted tree)

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Path Counting A graph is a directed acyclic graph (DAG) if

it is directed and has no cycles

This is a DAG.

This is not a DAG.

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Path Counting Given a DAG G, and two vertices of G, s

and t, count the number of distinct paths from s to t What if I give you a graph with directed cycles?

How is the graph given to you? Adjacency matrix Adjacency list Other ways

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Path (example) s = A, t = E Paths:

ABDE, ACBDE, ACDE Answer = 3

A

B

C

FE

D

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Path (an attempt) Use DFS to find out all paths from s to t

Simple enough, but consider this graph How many paths from s to t?

24 = 16

s

t

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Path (solution) Obviously the three paths shown below

must be distinct Even if they meet at some intermediate

vertices!

C

A

t

...

... ...

s

B

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Path (solution) Topological order

s

t

1

23 4

5

6

7

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Path (solution) Number of paths from vertex to t

s

t 1

0

1

1

2

3

1

23 4

0

5

6

7

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Path (solution) Algorithm

Tsort the verticesSet F[v] = 0 for every vertex vSet F[t] = 1Following topological order, for each vertex v

For each outgoing edge (v, u)F[v] = F[v] + F[u]

Time complexity Tsort – O(V+E) DP – O(V+E) Total – O(V+E)

recurrencerelation

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Path (extensions) Longest path in DAG

Given a weighted DAG G, find the length of a longest path from s to t

Shortest path counting Given a weighted graph G, find the number of

shortest paths from s to t

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Longest Path in Tree I Given a weighted tree T, find the length of

the longest path from a given node s

s7

5

65

3

4

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Longest I (simple solution) Make s the root

s

7

5

65

34

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Longest I (simple solution) Calculate the nodes’ distances from s (in

pre-order/level-order)

s

7

5

65

34

0

5

12 11

10

14 13

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Longest I (another solution) A longest path must end at one of the

leaves

s

7

5

65

34

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Longest I (another solution) Let F[v] be the longest distance between v

to one of its descendant leaves For example, F[x] = 9

s

7

5

65

34

x

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Longest I (another solution) Compute F in post-order F[x] = 0 for every leaf x F[v] = max {F[u]+length(v,u)}

s

7

5

65

34

u C(v)

00

00

4

9

14 answer

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Longest I (another solution) Algorithm

Longest_One(vertex v) {if (v is a leaf)

F[v] 0else

F[v] 0for each child u of v do

Longest_One(u)if (F[u]+length(v,u)

F[v] F[u]+length(v,u)}

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Longest I (another solution) Time complexity – O(V) No overlapping subproblems F[] is redundant!

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Longest Path in Tree II Given a weighted tree T (all weights

positive), find the length of the longest path in T

7

5

65

3

4

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Longest II (solution) Take any node and make it root

756

53

4

7

56

5

34

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Longest II (solution) A longest path must be a leaf-to-leaf or a

root-to-leaf path Must it pass the root?

756

53

4

7

56

5

34

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Longest II (solution) Let z be the uppermost node in the longest

path Only two cases

z z

the only common node is z

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Longest II (solution) As in Longest I, let F[v] be the longest

distance between v to one of its descendant leaves

Define G as follows G[v] = F[v] if v has less than 2 children G[v] = max{F[u]+length(v,u)} +

second_max {F[w]+length(v,w)}

Note that max may equal second_max

u C(v)

w C(v)

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Longest II (demonstration) Computing G from F

75

6

5

3

4

0 0

0 0

7

12

0 000

0 0

(0+7)+(0+6) = 13

(7+5)+(0+4) = 16

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Longest II (demonstration) Computing G from F (again)

5

7

65

340 0

0 0

0 0

0 0

4

9

14

(0+4)+(0+3) = 7

(4+5)+(0+7) = 16

14

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Longest II (solution) Time complexity

Computing F – O(V) Computing G – O(V) Total – O(V)

F and G can be computed together Not quite a DP problem

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Simplified Gems Given a tree T with N nodes Each node is to be covered by a gemstone Costs of gemstones: 1, 2, 3, …, M There is an unlimited supply of gemstones Two adjacent nodes must contain

gemstones of different costs What is the minimum total cost?

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Gems (example) N = 8, M = 4

2

3

1

1

1

1

1

1

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Gems (attempt) Make the tree a rooted one first

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Gems (attempt) Let G[v] be the minimum cost to cover all

nodes in the subtree rooted at v How to set up the recurrence?

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Gems (solution) Let F[v,c] be the minimum cost to cover all

nodes in the subtree rooted at v and the cost of the gemstone covering v is c

Base cases F[x,c] = c for every leaf x and 1 c M

Progress F[v,c] = min {F[u,d]} + c

Post-order traversalu C(v) 1dM,dc

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Gems (demostration) M = 4

1

2

3

4

1

2

3

4

1

2

3

4

1

2

3

4

7

5

6

7

1

2

3

4

1

2

3

4

12

11

11

12

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Gems (solution) Algorithm (recursive, non-DP)

Gems(vertex v,integer c) {if (v is leaf) return cvalue cfor each child u of v do temp ∞ for d 1 to M do if (d c) temp min{temp, Gems(u,d)} value value + tempreturn value }

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Gems (solution) Algorithm (DP)

Gems_DP(vertex v) {if (v is a leaf) set base case and exitfor each child u of v do Gems_DP(u)for c 1 to M do F[v,c] c for each child u of v do temp ∞ for d 1 to M do if (d c) temp min{temp, F[u,d]} F[v,c] temp + c

}

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Gems (solution) Time complexity

Computing F[v,c] – O(M × #children of v) Computing F[v,c] for all vertices – O(MN) Computing all entries – (M2N)

The time complexity can be reduced to O(MN) with a “trick”

The original problem allows N to be as large as 10000 and M arbitrarily large Even O(N2) is too slow How to solve it??

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Game strategies Not closely related to DP Almost all game-type problems in

IOI/BOI/CEOI requires the concept of Minimax

DP is needed in most of these problems

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Game-type problems Usually interactive problems Write a program to play a simple two-

player game with a judging program e.g. play tic-tac-toe with the judging program

Often the judging program uses an optimal strategy

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Game tree A (finite or infinite) rooted tree showing

the movements of a game play

O

O O O

…

X

O

X

O

O X O X

… …X O X O

…

…

… … …

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Card Picking A stack of N cards with numbers on them Two players take turns to take cards from

the top of the stack 1, 2, or 3 cards can be taken in each turn Game ends when all cards have been

taken The player with a higher total score (sum

of numbers) wins

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Card (example)

2

1

9

7

1

4

3

4

A B

17 14

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Card (game tree) N = 4 Only 5 different states

1

2

3

4

2

3

4 4

3

4

3

4 4 NULL 4 NULL NULL

4 NULL NULL

NULL

A’s moveB’s move

NULL

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Minimax A recursive algorithm for choosing the

next move in a two-player game A value is associated with each state

e.g. in tic-tac-toe, all winning states may have value 1

We assume that the other player always chooses his best move

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Minimax Suppose A wants to maximize his final

score (value), which move should he make?

1 -1 4 3 2 1 7 -2

A’s moves

B’s moves

-1 1 -2

1

min min min

max

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Minimax Again!

1

3 9 8

-1 2 8 -4 -1 2 -2 7 97 5

A’s moveB’s move

1

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Minimax Answer: left move

2

2 1 1

2 3 8 1 -1 98 1

-1 2 3 8 9 8-1 -2 9

-1 2 8 -4 -1 2 -2 7 97 5

A’s moveB’s move

1

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Tic-tac-toe O wants to

maximize the value Is this a winning

state for O? O X

O O O

X X

O X

O O

O X X

O X

O O

X X

O O X

O O

X X

value: 1

X O X

O O

O X X

O X

X O O

O X X

O O X

X O O

X X

O O X

O O

X X X

value: -1

O O X

X O O

O X X

X O X

O O O

O X X

O O X

X O O

O X X

value: 0 value: 1 value: 0

O

X

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Card Picking revisited Let the F-value of a state be the maximum

difference (preserve +/- sign) between your score and your opponent’s score if the game starts from this state (assume that your opponent plays perfectly)

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Card Picking revisited A transition may alter the F-value

Two states that appear the same may have different F-values!

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NULLF-value: 0

F-value: 9

9

NULLF-value: 0

F-value: -9

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Card Picking revisited We can still apply the concept of Minimax

1

2

3

4

2

3

4 4

3

4

3

4 4 NULL 4 NULL

4 NULL NULL

NULL

NULL

0

0 0

0 0 0

-4

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NULL0

4

+7+3 +4 +4

-4

-2

-5 -9

-9

-3 -7 -4

-7-4

+1+3 +6

2

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Card Picking revisited A recurrence can be set up Many overlapping sub-problems, so DP! Find the optimal move by backtracking Most game-type problems in OI

competitions are similar to this game

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Conclusion Many DP problems discussed are now

classics More and more atypical DP problems in

competitions (esp. on trees) Still not enough for solving some difficult

IOI/NOI/BOI/CEOI DP problems We hope that those problems can be

covered in the summer vacation Practice, practice and practice

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The end Prepare for TFT (19 June)..

..as well as your exam Have a nice holiday!