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Advanced Dynamic Programming II. HKOI Training Team 2004. 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 - PowerPoint PPT Presentation
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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
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
33
4 7
9 12 15
21 17 21 16
26 23 24 27 22
F’
7
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
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 a DAG.
This is not 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
A
B
C
FE
D
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
s
t
13
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
15
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
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
s7
5
65
3
4
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Longest I (simple solution) Make s the root
s
7
5
65
34
20
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
21
Longest I (another solution) A longest path must end at one of the
leaves
s
7
5
65
34
22
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
23
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
24
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)}
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
7
5
65
3
4
27
Longest II (solution) Take any node and make it root
756
53
4
7
56
5
34
28
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
29
Longest II (solution) Let z be the uppermost node in the longest
path Only two cases
z z
the only common node is z
30
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)
31
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
32
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
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
2
3
1
1
1
1
1
1
36
Gems (attempt) Make the tree a rooted one first
37
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?
38
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
39
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
40
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 }
41
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
}
42
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??
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)
2
1
9
7
1
4
3
4
A B
17 14
48
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
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 -1 4 3 2 1 7 -2
A’s moves
B’s moves
-1 1 -2
1
min min min
max
51
Minimax Again!
1
3 9 8
-1 2 8 -4 -1 2 -2 7 97 5
A’s moveB’s move
1
52
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
53
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
54
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)
55
Card Picking revisited A transition may alter the F-value
Two states that appear the same may have different F-values!
9
NULLF-value: 0
F-value: 9
9
NULLF-value: 0
F-value: -9
56
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
47
NULL0
4
+7+3 +4 +4
-4
-2
-5 -9
-9
-3 -7 -4
-7-4
+1+3 +6
2
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!