1 / 8

1 2 3 4 5 6 7 8 9 0 1 b a a b c b b a b c $ a a b c b b a b c $ a b c b b a b c $ b c b b a b c $

1 2 3 4 5 6 7 8 9 0 1 b a a b c b b a b c $ a a b c b b a b c $ a b c b b a b c $ b c b b a b c $ c b b a b c $ b b a b c $ b a b c $ a b c $ b c $ c $ $. Provided by 89902098 吳亭範. [11,-]. [5,5]. [1,1]. [2,2]. [5,-]. 1. 1. 1. [2,-]. [11,-]. [3,-]. [5,-]. 1. [2,2]. [5,5].

kirti
Download Presentation

1 2 3 4 5 6 7 8 9 0 1 b a a b c b b a b c $ a a b c b b a b c $ a b c b b a b c $ b c b b a b c $

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 1 2 3 4 5 6 7 8 9 0 1 b a a b c b b a b c $ a a b c b b a b c $ a b c b b a b c $ b c b b a b c $ c b b a b c $ b b a b c $ b a b c $ a b c $ b c $ c $ $ Provided by 89902098吳亭範 [11,-] [5,5] [1,1] [2,2] [5,-] 1 1 1 [2,-] [11,-] [3,-] [5,-] 1 [2,2] [5,5] [1,-] [4,5] [11,-] [6,-] [2,-] [11,-] 1 2 1 [7,-] [4,-] [6,-] Case 2: 節外生枝 [6,-] [3,-] [9,-] Case 3: 若無其事

  2. Create Minimal Tree in Linear TimeSymbol definition • S = given sequence of numbers. • “Right branch” = The tree branch can be travered from root to right-most node by following only right-child • I construct the tree by directly reading the sequence one by one for every new-coming number • Please view in play mode for animation

  3. New A[i+1] A[n] General Case • Assume right branch is referred as A[0](root), A[1], A[2]…A[n](right most node) • For each new coming node, linear search the right branch for proper place where S[A[i+1]]>S[new] and S[A[i]]<S[new] • Insert new between A[i] ,A[i+1] set that new = Right-Child[A[i]] and A[i+1] = Left-Child[new] A[0] A[1] A[i] < <

  4. New A[i+1] A[n] Boundary CaseValue(New) > Value(A[n]) • Attach New at right-hand side of A[n] • Let new = Right-Child[A[n]] A[0] A[1] A[i] <

  5. New A[i+1] A[n] Boundary CaseValue(New)<Value(A[0]) • Attach A[0] at left-hand side of New • Let A[n] = Left-Child[New] • The new “Right Branch” created. < A[0] A[1] A[i]

  6. Proof of Correctness • In order traversal sequence • At each step, the newly added number is always at right-most side, thus being an appending to original sequence of “in-order traversal”. • The property of sequence S[1…n-1] will not be changed while new number appended at the end. • Heap like Optimal substructure • The minimal tree is like a Heap. Each S[parent] is smaller-equal than S[child] • Each Step Keep the optimality • As we insert New number between A[i],A[i+1]. • S(A[i])<=S(New)<=S(A[i+1]) • <=S(Subtree rooted by A[i+1]) • And S(0) <= S(1) <= S(2) <= …. <= S(A[i]) <= S(New) • Due to the correct traversal sequence and the number relation, this algorithm is guaranteed to generate a minima tree for the current string S[1…n].

  7. Proof of Linearity • For each stepk • Costk = O(Search from Right) + O(Attach) • = Ф(n-i) + O(1) • = Фk • Total Cost • Фi= {Numbers of Nodes that < Value(New), which was removed from “Right brancn” and becoming child of New} At Step I • Each node is compared to New number only one time before removed. So |Comparisons of Search|= O(|S|) • Costall = Ф1+ Ф2+ …+ Ф|s| • = O(|S|)

  8. Acknowledgements • Thanks to 王姵瑾 and 李沛倫 for helpful discussion.

More Related