Advanced Dynamic Programming II

1
Advanced Dynamic Programming II

• HKOI Training Team 2004

2
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)

3
In this lesson...

• Dimension reduction
• DP on trees, graphs, etc.
• Game strategy - Minimax

4
Dimension Reduction

• Reduce the memory complexity by one (or more)
  dimension
• Usually a rolling array is employed

5
Triangle revisited

• Only a 2x5 array is needed

3
1 4
2 5 8
9 5 6 1
5 2 3 6 6
3
4 7

9 12 15


21 17 21 16

26 23 24 27 22
A
F
6
Rolling array
3

3
9 12 15
26 23 24 27 22
4 7

4 7
21 17 21 16
9 12 15
F
21 17 21 16
26 23 24 27 22
F
7
LCS revisited

• Recall the recurrence
• Fi,j Fi-1,j-11 if AiBj
• Fi,j maxFi-1,j,Fi,j-1 if Ai?Bj
• Note that Fi,? only depends on Fi,? and
  Fi-1,?
• Thus we can just keep 2 rows

8
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)

9
Path Counting

• A graph is a directed acyclic graph (DAG) if it
  is directed and has no cycles

This is not a DAG.
This is a DAG.
10
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

11
Path (example)

• s A, t E
• Paths
• ABDE, ACBDE, ACDE
• Answer 3

12
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

13
Path (solution)

• Obviously the three paths shown below must be
  distinct
• Even if they meet at some intermediate vertices!

s
C
B
A
...
...
...
t
14
Path (solution)

• Topological order

7
s
6
5
t
2
3
4
1
15
Path (solution)

• Number of paths from vertex to t

3
7
s
6
2
1
5
t
1
2
3
4
1
0
1
0
16
Path (solution)

• Algorithm
• Tsort the vertices
• Set Fv 0 for every vertex v
• Set Ft 1
• Following topological order, for each vertex v
• For each outgoing edge (v, u)
• Fv Fv Fu
• Time complexity
• Tsort O(VE)
• DP O(VE)
• Total O(VE)

17
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

18
Longest Path in Tree I

• Given a weighted tree T, find the length of the
  longest path from a given node s

s
5
7
4
5
6
3
19
Longest I (simple solution)

• Make s the root

s
5
7
5
6
3
4
20
Longest I (simple solution)

• Calculate the nodes distances from s (in
  pre-order/level-order)

0
s
5
5
7
5
6
10
3
4
12
11
14
13
21
Longest I (another solution)

• A longest path must end at one of the leaves

s
5
7
5
6
3
4
22
Longest I (another solution)

• Let Fv be the longest distance between v to one
  of its descendant leaves
• For example, Fx 9

s
x
5
7
5
6
3
4
23
Longest I (another solution)

• Compute F in post-order
• Fx 0 for every leaf x
• Fv max Fulength(v,u)

u ? C(v)
s
14
answer
5
9
7
5
6
4
3
4
0
0
0
0
24
Longest I (another solution)

• Algorithm
• Longest_One(vertex v)
• if (v is a leaf)
• Fv ? 0
• else
• Fv ? 0
• for each child u of v do
• Longest_One(u)
• if (Fulength(v,u)
• Fv ? Fulength(v,u)

25
Longest I (another solution)

• Time complexity O(V)
• No overlapping subproblems
• F is redundant!

26
Longest Path in Tree II

• Given a weighted tree T (all weights positive),
  find the length of the longest path in T

5
7
4
5
6
3
27
Longest II (solution)

• Take any node and make it root

28
Longest II (solution)

• A longest path must be a leaf-to-leaf or a
  root-to-leaf path
• Must it pass the root?

29
Longest II (solution)

• Let z be the uppermost node in the longest path
• Only two cases

the only common node is z
30
Longest II (solution)

• As in Longest I, let Fv be the longest distance
  between v to one of its descendant leaves
• Define G as follows
• Gv Fv if v has less than 2 children
• Gv maxFulength(v,u) second_max
  Fwlength(v,w)
• Note that max may equal second_max

u ? C(v)
w ? C(v)
31
Longest II (demonstration)

• Computing G from F

12
(75)(04) 16
7
(07)(06) 13
0
0
0
0
0
0
0
0
0
0
32
Longest II (demonstration)

• Computing G from F (again)

14
14
9
(45)(07) 16
4
(04)(03) 7
0
0
0
0
0
0
0
0
33
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

34
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?

35
Gems (example)

• N 8, M 4

36
Gems (attempt)

• Make the tree a rooted one first

37
Gems (attempt)

• Let Gv be the minimum cost to cover all nodes
  in the subtree rooted at v
• How to set up the recurrence?

38
Gems (solution)

• Let Fv,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
• Fx,c c for every leaf x and 1 ? c ? M
• Progress
• Fv,c ? min Fu,d c
• Post-order traversal

u ? C(v)
1?d?M,d?c
39
Gems (demostration)

• M 4

12
11
11
12
7
5
6
7
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
40
Gems (solution)

• Algorithm (recursive, non-DP)
• Gems(vertex v,integer c)
• if (v is leaf) return c
• value ? c
• for each child u of v do
• temp ? 8
• for d ? 1 to M do
• if (d ? c)
• temp ? mintemp, Gems(u,d)
• value ? value temp
• return value

41
Gems (solution)

• Algorithm (DP)
• Gems_DP(vertex v)
• if (v is a leaf)
• set base case and exit
• for each child u of v do
• Gems_DP(u)
• for c ? 1 to M do
• Fv,c ? c
• for each child u of v do
• temp ? 8
• for d ? 1 to M do
• if (d ? c)
• temp ? mintemp, Fu,d
• Fv,c ? temp c

42
Gems (solution)

• Time complexity
• Computing Fv,c O(M children of v)
• Computing Fv,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??

43
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

44
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

45
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








46
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

47
Card (example)
A
B
2
1
9
7
1
4
3
17
14
4
48
Card (game tree)

• N 4
• Only 5 different states

As move
1
2
3
4
Bs move
2
3
4
3
4
4
3
4
4
NULL
4
NULL
NULL
4
NULL
NULL
NULL
NULL
49
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

50
Minimax

• Suppose A wants to maximize his final score
  (value), which move should he make?

1
max
-1
1
-2
min
min
min
51
Minimax

• Again!

As move
Bs move
1
3
9
8
1
-1
2
8
-4
-1
2
-2
7
9
7
5
52
Minimax

• Answer left move

As move
Bs move
2
2
1
1
2
3
8
1
-1
9
8
1
-1
2
3
8
9
8
-1
-2
9
1
-1
2
8
-4
-1
2
-2
7
9
7
5
53
Tic-tac-toe

• O wants to maximize the value
• Is this a winning state for O?

O X
O O
X X
O X
O O O
X X
O X
O 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
O

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 0
value 1
54
Card Picking revisited

• Let the F-value of a state be the maximum
  difference (preserve /- sign) between your score
  and your opponents score if the game starts from
  this state (assume that your opponent plays
  perfectly)

55
Card Picking revisited

• A transition may alter the F-value
• Two states that appear the same may have
  different F-values!

56
Card Picking revisited

• We can still apply the concept of Minimax

1
2
3
4
2
1
6
3
2
3
4
3
4
-9
-7
4
-4
-2
-9
-3
-7
-4
-5
3
4
7
4
NULL
4
4
4
NULL
NULL
0
0
0
7
3
4
4
4
NULL
NULL
-4
NULL
0
0
0
-4
NULL
0
57
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

58
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

59
The end

• Prepare for TFT (19 June)..
• ..as well as your exam
• Have a nice holiday!