1. HKOI2009 Training (Advanced Group) (Reference: Powerpoint of Dynamic Programming II, HKOI Training 2005, by Liu Chi Man, cx)

2. Review  Recurrence relation  Dynamic programming State & Recurrence Formula Optimal substructure Overlapping subproblems 2

3. Outline  Dimension reduction (memory)  “Ugly” optimal value functions  DP on tree structures  Two-person games 3

4. Dimension reduction  Reduce the space complexity by one or more dimensions  “Rolling” array  Recall: Longest Common Subsequence (LCS)  Base conditions and recurrence relation: F i,0 = 0 for all i F 0,j = 0 for all j F i,j =F i-1,j-1 + 1(if A[i] = B[j]) max{ F i-1,j, F i,j-1 }(otherwise) 4

5. Dimension Reduction  A: stxc, B: sicxtc 5 0 0 0 0 0000000 0123456 1 2 3 4 0 111 111 11 12 111 112 22 22 1 2 2 3 0 0123456 0 000000 0 111111 1 2 111122 3 111222 4 112223

6. Dimension Reduction  We may discard old table entries if they are no longer needed  Instead of “rolling” the rows, we may “roll” the columns  Even less memory (5  2 entries)  Space complexity:  (min{N, M})  Drawback  Backtracking is difficult  That means we can get the number but not the sequence easily. 6

7. (Simplified) Cannoneer Base  How many non-overlapping cross pieces can be put onto a H  W grid?  W ≤ 10, H is arbitrary  A cross piece:  There may be patterns, but we just focus on a DP solution 7 Packing 8 cross pieces onto a 10  6 grid

8. (Simplified) Cannoneer Base  We place the pieces from top to bottom Phase k - putting all pieces centered on row k-1  In phase k, we only need to consider the occupied squares in rows k-2 and k-1 8 k -2 k -1 k ? Phase 3Phase 4Phase 5Phase 6Phase 7Phase 8Phase 9Phase 10

9. (Simplified) Cannoneer Base  The optimal value function C is defined by: C(k,S) = the max number of pieces after phase k, with rows k-1 and k giving the shape S  How to represent a shape? In a shape, each column can be Use 2, 1, 0 to represent these 3 cases A shape is a W-digit base-3 integer For example, the following shape is encoded as 010121 (3) = 97 (10) 9

10. (Simplified) Cannoneer Base  The recurrence relation is easy to construct  Max possible number of states = H  3 W That’s why W ≤ 10  Cannoneer Base appeared in NOI2001  Bugs Integrated, Inc. in CEOI2002 requires similar techniques 10

11. Dynamic Programming on Tree Structures  States may be (related to) nodes on a graph  Usually directed acyclic graphs  Topological order is the obvious order of recurrence evaluation  Trees are special graphs  A lot of graph DP problems are based on trees  Two major types:  Rooted tree DP  Unrooted tree DP 11

12. Rooted Tree Dynamic Programming  Base cases at the leaves  Recurrence at a node involves its child nodes only  Solution Evaluate the recurrence relation from leaves (bottom) to the root (top) Top-down implementations work well Time complexity is often  (N) where N is the number of nodes 12

13. Unrooted Tree Dynamic Programming  No explicit roots given  Two cases The problem can be transformed to a rooted one It can’t, so we try root every node  Case 2 increases the time complexity by a factor of N  Sometimes it is possible to root one node in O(N) time and each subsequent node in O(1) overall O(N) time 13

14. Node Heights  Given a rooted tree T  The height of a node v in T is the maximum distance between v and a descendant of v  For example, all leaves have height = 0  Find the heights of all nodes in T  Notations C(v) = the set of children of v p(v) = the parent of v 14

15. Node Heights  Optimal value function H(v) = height of node v  Base conditions H(u) = 0 for all leaves u  Recurrence H(v) = max { H(x) | x  C(v) } + 1  Order of evaluation All children must be evaluated before self Post-order 15

16. Node Heights  Example 16 A BC DEFG I H H(I) = 0 H(E) = 0H(F) = 0H(G) = 0H(H) = 0 H(D) = 1 H(B) = 2 H(C) = 1 H(A) = 3

17. Node Heights  Time complexity analysis  Naively  There are N nodes  A node may have up to N-1 children  Overall time complexity = O(N 2 )  A better bound  The H-value of a v is at most used by one other node – p(v)  The total number of H-values inside the “max {}”s = N-1  Overall time complexity =  (N) 17

18. Treasure Hunt  N treasures are hidden at the N nodes of a tree (unrooted)  The treasure at node u has value v(u)  You may not take away two treasures joined by an edge, otherwise missiles will fly to you  Find the maximum value you can take away 18

19. Treasure Hunt  Let’s see if the problem can be transformed to a rooted one  We arbitrarily root a node, say r  How to formulate? 19

20. Treasure Hunt  Optimal value function  Z(u,b) = max value for the subtree rooted at u and it is b that the treasure at u is taken away  b = true or false  Base conditions  Z(x,false) = 0 and Z(x,true) = v(x) for all leaves x  Recurrence  Z(u,true) =  Z(c, false) + v(u)  Z(u,false) =  max { Z(c,false), Z(c,true) }  Answer = max { Z(r,false), Z(r,true) } 20 c  C(u)

21. Treasure Hunt  Example (values shown in squares) 21 7 26 9351 1 2 false: 0 true: 1 false: 0 true: 3 false: 0 true: 5 false: 0 true: 1 false: 0 true: 2 false: 1 true: 9 false: 12 true: 3 false: 8 true: 6 false: 20 true: 27

22. Treasure Hunt  Our formulation does not exploit the properties of a tree root  Moreover the correctness of our formulation can be proven by optimal substructure  Thus the unrooted-to-rooted transformation is correct  Time complexity:  (N) 22

23. Unrooted Tree DP – Basic Idea  In rooted tree DP, a node asks (request) for information from its children; and then provides (response) information to its parent 23 Response Request

24. Unrooted Tree DP – Basic Idea  In unrooted tree DP, a node also makes a request to its parent and sends response to its children  Imagine B is the root  A sends information about the “subtree” {A,C} to B 24 A BC D Response Request E

25. Unrooted Tree DP – Basic Idea  Similarly we can root C, D, E and get different request-response flows  These flows are very similar  The idea of unrooted tree DP is to root all nodes without resending all requests and responses every time 25 A BC DE A BC DE

26. Unrooted Tree DP – Basic Idea  Root A and do a complete flow  A knows about subtrees {B,D,E} and {C}  Now B sends a request to A  A sends a response to B telling what it knows about {A,C}  B already knows about {D}, {E}  Rooting of B completes 26 A BC DE

27. Unrooted Tree DP – Basic Idea  Now let’s root D  D sends a request to B  B knows about {A,C}, {D}, and {E}; combining {A,C}, {E} and B itself, B knows about {B,A,C,E}, and sends a response to D  Rooting of D completes 27 A BC DE

28. Unrooted Tree DP – Basic Idea  Rooting a new node requires only one request and one response if its parent already knows about all its subtrees (including the “imaginary” parent subtree)  Further questions:  Fast computation of {B,A,C,E} from {A,C} and {E}? (rooting of D)  Fast computation of {B,A,C,E,D} from {A,C}, {E}, {D}? (rooting of B) 28 A BC DE

29. Shortest Rooted Tree  Given an unrooted tree T, denote its rooted tree with root r by T(r)  Find a node v such that T(v) has the minimum height among all T(u), u  T  The height of a tree = the height of its root  Solution  Just root every node and find the min height  We know how to find a height of a tree  Trivially this is  (N 2 )  Now let’s use what we learnt 29

30. Shortest Rooted Tree  Since parents and children are unclear now, we use slightly different notations  N(v) = the set of neighbors of v  H(v, u) = height of the subtree rooted at v if u is treated as the parent of v  H(v,  ) = height of the whole tree if v is root 30

31. Shortest Rooted Tree  Root A, complete flow  Height = 3 31 A BC DE F H(F,D) = 0 H(E,B) = 0 H(D,B) = 1 H(B,A) = 2H(C,A) = 0 H(A,  ) = 3

32. Shortest Rooted Tree  Root B Request: B asks A for H(A,B) How can A give the answer in constant time? 32 A BC DE F H(F,D) = 0 H(E,B) = 0 H(D,B) = 1 H(B,A) = 2H(C,A) = 0 H(A,  ) = 3

33. Shortest Rooted Tree  Suppose now B asks A for H(A,B), how can A give the answer in constant time?  Two cases B is the only largest subtree of A in T(A) B is not the only largest subtree, or B is not a largest subtree 33 A BCDEFGHIJK H(B,A)=7 H(C,A)=8 H(D,A)=1 H(E,A)=2 H(F,A)=9H(H,A)=4H(J,A)=3 H(G,A)=5H(I,A)=6H(K,A)=0 H(A,  )=10

34. Shortest Rooted Tree  (1) B is the only largest subtree of A in T(A) H(A,B) < H(A,  ) H(A,B) depends on the second largest subtree Trick: record the second largest subtree of A  (2) B is not the only largest subtree, or B is not a largest subtree H(A,B) = H(A,  ) 34

35. Shortest Rooted Tree  To distinguish case (1) from case(2), we need to record the two largest subtrees of A When? ○ When we evaluate H(A,  )  Back to our example 35

36. Shortest Rooted Tree  Root B Request: B asks A for H(A,B) Response: 1 36 A BC DE F H(F,D) = 0 H(E,B) = 0 H(D,B) = 1 H(B,A) = 2H(C,A) = 0 H(A,  ) = 3 1 st = B, 2 nd = C 1 st = D, 2 nd = E 1 st = , 2 nd =  1 st = F, 2 nd =  1 st = , 2 nd =  A H(A,B) = 1

37. Shortest Rooted Tree  Root B H(B,  ) = 2 can be calculated in constant time 37 A BC DE F H(F,D) = 0 H(E,B) = 0 H(D,B) = 1 H(B,A) = 2H(C,A) = 0 H(A,  ) = 3 1 st = B, 2 nd = C 1 st = D, 2 nd = E 1 st = , 2 nd =  1 st = F, 2 nd =  1 st = , 2 nd =  A H(A,B) = 1 H(B,  ) = 2

38. Shortest Rooted Tree  Root D Request: D asks B for H(B,D) Response: 2 H(D,  ) = 3 38 A BC DE F H(F,D) = 0 H(E,B) = 0 H(D,B) = 1 H(B,A) = 2H(C,A) = 0 H(A,  ) = 3 1 st = B, 2 nd = C 1 st = D, 2 nd = E 1 st = , 2 nd =  1 st = F, 2 nd =  1 st = , 2 nd =  A H(A,B) = 1 H(B,  ) = 2 BF H(D,  ) = 3 H(B,D) = 2

39. Shortest Rooted Tree  Root F, E, and C in the same fashion  In general, root the nodes in preorder  Time complexity analysis Root A –  (N) Root each subsequent nodes – O(1) Overall -  (N)  The O(1) is crucial for the linearity of our algorithm If rooting of a new node cannot be done fast, unrooted tree DP may not improve running time 39

40. Two-person Games  Often appear in competitions as interactive tasks Playing against the judge  Most of them can be solved by the Minimax method 40

41. Game Tree  A (finite or infinite) rooted tree showing the movements of a game play 41 … ……… … …… … o x o x o oooxoxooxox

42. Game Tree  This is a game  The boxes at the bottom show your gain (your opponent’s loss)  Your opponent is clever  How should you play to maximize your gain? 42 2945668176 End of game Your turn Her turn WHY?

43. Minimax  You assume that your opponent always try to minimize her loss (minimize your gain)  So your opponent always takes the move that minimize your gain  Of course, you always take the move that maximize your gain 43 2945668176 Your turn Her turn 9456 6 87 467 7

44. Minimax  Efficient? Only if the tree is small  In fact the game tree may in fact be an expanded version of a directed acyclic graph  Overlapping subproblems  memo(r)ization 44 A B D C 1 2 A BCD 12 D 122 D 12

45. Past Problems  IOI  2001 Ioiwari (game), Score (game), Twofive (ugly)  2005 Rivers(tree)  NOI  2001 炮兵陣地 (ugly), 2002 貪吃的九頭龍 (tree), 2003 逃學的小孩 (tree), 2005 聰聰與可可  IOI/NOITFT  2004 A Bomb Too Far (tree)  CEOI  2002 Bugs (ugly), 2003 Pearl (game)  Balkan OI  2003 Tribe (tree)  Baltic OI  2003 Gems (tree)  On HKOI Judge  1074 Christmas Tree 45