Red-Black Trees 10/19/2009. Red-Black Trees. Example. Red-Black Properties. Black Height. Example

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1 lgorithms Red-lak Trees 13-2 Red-lak Trees Red-lak Trees ll binar searh tree operations take O(h) time, here h is the height of the tree Therefore, it is important to `balane the tree so that its height is as small as possible There are man as to ahieve this One of them: Red-lak trees ver node of suh a tree ontains one etra bit, its olor nother agreement: the pointer is treated as a leaf, an etra node The rest of the nodes are alled internal ata Strutures and lgorithms ndrei ulatov lgorithms Red-lak Trees 13-3 lgorithms Red-lak Trees 13-4 Red-lak Properties ample binar searh tree is a red-blak tree if it satisfies the folloing red-blak properties: ver node is either red or blak The root is blak ver leaf () is blak If a node is red, then both its hildren are blak For eah node, all paths from the node to desendant leaves ontain the same number of blak nodes lgorithms Red-lak Trees 13-5 lgorithms Red-lak Trees 13-6 ample lak Height The number of blak nodes on paths from node to its desendant leaves in a red-blak tree is alled its blak height, denoted T) nil[t] Lemma red-blak tree ith n internal nodes has height at most 2 log(n + 1) Proof ) We sho first that the subtree rooted at ontains at least 2 1 nodes Indution on ) ase ase: If ) = 0, then is a leaf, nil[t] ) 0 In this ase, 2 1 = 2 1 = 0 internal nodes 1

2 lgorithms Red-lak Trees 13-7 lgorithms Red-lak Trees 13-8 lak Height (ntd) Indutive hpothesis: the laim is true for an ith height less than that of Indutive ase: Let ) > 0 and has to hildren The blak height of the hildren is either ) or ) 1, depending on its olor Sine the hildren have smaller height, e an appl the indution hpothesis Thus the subtree rooted at ontains at least nodes ) 1 ) 1 ) (2 1) + (2 1) + 1 = 2 1 lak Height (ntd) Suppose h is the height of the tree the red-blak propert at least half of nodes on ever root-to-leaf path are blak (not inluding the root) Therefore the blak height of the root is at least h/2 Thus n 2 h / 2 1 log( n + 1) h / 2 Q lgorithms Red-lak Trees 13-9 lgorithms Red-lak Trees Rotations Sometimes e ill need to rearrange pointers inside an R-tree Left-Rotate(T,) Right-Rotate(T,) Rotations: Pseudoode Left-Rotate(T,) set :=right[] set right[]:=left[] set parent[left[]]:= set parent[]:=parent[] if parent[]=nil[t] then set root[t]:= else if =left[parent[]] then set left[parent[]]:= else right[parent[]]:= set left[]:= set parent[]:= lgorithms Red-lak Trees lgorithms Red-lak Trees Rotations: ample Insertion Insertion for R-trees is done in the same a as for ordinar binar searh trees. ept: e should be areful about links the ne node is olored red the resulting tree ma not be an R-tree, e need to fi it 2

3 lgorithms Red-lak Trees lgorithms Red-lak Trees Insertion: Pseudoode Insertion: FiUp R-Insert(T,) set :=[T], :=root[t] hile [T] do set := if ke[]<ke[] then set :=left[] else set :=right[] endhile set parent[]:= if =[T] then set root[t]:= else if ke[]<ke[] then set left[]:= else set right[]:= set left[]:=[t] right[]:=[t] olor[]:=r R-Insert-FiUp(T,) Step 1 Step 2 Step 3 lgorithms Red-lak Trees lgorithms Red-lak Trees FiUp: Pseudoode R-Insert-FiUp(T,) hile olor[parent[]]=r do if parent[]=left[parent[parent[]]] then do set :=right[parent[parent[]]] if olor[]=r then do set olor[parent[]]:=lk olor[]:=lk set olor[parent[parent[]]]:=r set :=parent[parent[]] else do if =right[parent[]] then do set :=parent[] Left-Rotate(T,) set olor[parent[]]:=lk set olor[parent[parent[]]]:=r else (same as then ith left and right sopped) olor[root[t]]:=lk Soundness Lemma Given an R-tree, algorithm R-Insert produes an R-tree Proof We sho that the main loop preserves the folloing loop invariant: (a) Node is red (b) If parent[] is the root then parent[] is blak () If there is a violation of the red-blak propert, there is at most one violation, and it is of either propert 2 or propert 4. If there is a violation of propert 2, it ours beause is the root and red If there is a violation of propert 4, it ours beause both and parent[] are red lgorithms Red-lak Trees lgorithms Red-lak Trees Soundness: Initialiation (a) When R-Insert-FiUp is alled, is a red node (b) If p[] is the root, it hasn t hanged et, and so is blak () Properties 1, 3, and 5 are true If propert 2 is violated, then the red root is just added, that is the root is. In this ase the tree does not have internal verties. Propert 4 is not violated in this ase If propert 4 is violated, then, sine the hildren of are blak sentinels and the tree had no prior violations, the violation must be beause both and parent[] are red. There are no other violations of red-blak properties. Soundness: Termination When the big loop of R-Insert-FiUp terminates parent[] is blak (if is the root then parent[] is the sentinel) Therefore there is no violation of propert 4. If propert 2 is violated, then the last line of R-Insert-FiUp restores it. 3

4 lgorithms Red-lak Trees lgorithms Red-lak Trees Soundness: Maintenane Soundness: Maintenane (ntd) There are several ases to onsider. We onl onsider those in hih parent[] is the left hild of parent[parent[]] This is the situation given in the pseudoode If parent[parent[]] does not eist then parent[] is the root, and b part (b) is blak. Therefore if the loop does not terminate here, parent[parent[]] eists oth and parent[] are red, and parent[parent[]] is blak We olor and parent[] blak, and parent[parent[]] red We then set parent[parent[]] to be the ne node ase 1. s unle is red OR Sho that the loop invariant is preserved (a) Sine e olor parent[parent[]] red, the ne (denoted ) is red lgorithms Red-lak Trees lgorithms Red-lak Trees Soundness: Maintenane (ntd) Soundness: Maintenane (ntd) (b) We have parent[ ] = parent[parent[parent[]]], and the olor of this node does not hange. If it is the root, it is blak () It is eas to see that ase 1 maintains propert 5, and does not violate properties 1 and 3 If is the root, then ase 1 orreted the onl violation of propert 4. Sine is red and it is the root, propert 2 beomes violated, and it is due to If is not the root, then ase 1 has not reated a violation of propert 2. ase 1 orreted the violation of propert 4 due to and parent[]. If parent[ ] is blak then e are done Otherise and parent[ ] reate the onl violation of propert 4 ase 2. s unle is blak and is a right hild ase 3. s unle is blak and is a left hild ase 3 ase 2 lgorithms Red-lak Trees lgorithms Red-lak Trees Soundness: Maintenane (ntd) ase 2. s unle is blak and is a right hild ase 3. s unle is blak and is a left hild In ase 2 the algorithm performs rotation and e get ase 3 Sine both nodes being rotated are red no R properties hange Moreover, parent[parent[]] eists and it does not hange hen ase 2 is transformed into ase 3 In ase 3 e perform same rotations and olor hanges that does not affet propert 5 Then sine there are no longer 2 red nodes in a ro, the hile loop is not eeuted again, for parent[] is no blak Soundness: Maintenane (ntd) (a) ase 2 makes points to parent[], hih is red Nothing hanges in ase 3 (b) ase 2 does not hange the olor of the root ase 3 makes parent[] blak, so if it is the root, it is blak () s in ase 1, properties 1, 3, and 5 are maintained in ases 2 and 3 Sine node is not the root, there is no violation of propert 2 ases 2 and 3 do not introdue an violation of propert 2, sine the onl node that beomes red also beomes a hild of a blak node b the rotation in ase 3 ases 2 and 3 orret the onl violation of propert 4 and do not introdue another violation 4

5 lgorithms Red-lak Trees lgorithms Red-lak Trees Running Time Sine the height of an R-tree ith n nodes is O(\log n), R-Insert takes O(log n) time, eept for R-Insert-FiUp In R-Insert-FiUp the hile loop repeats onl if ase 1 takes plae, and in this ase moves up 2 levels. The total number of iterations is O(log n) R-eletion R-elete also first runs the regular eletion algorithm, and then fies violations of the R properties R-elete(T,) if left[]=[t] or right[]=[t] then set := else set :=Tree-Suessor() if left[] [T] then set :=left[] else set :=right[] if [T] then set parent[]:=parent[] else set :=right[] if parent[]=[t] then set root[t]:= else if =left[parent[]]:= then left[parent[]:= else right[parent[]]:= if then do set ke[]:=ke[] op s data into if olor[]=lk then R-elete-FiUp(T,) return lgorithms Red-lak Trees lgorithms Red-lak Trees eletion: Pitures eletion: Pitures ase 1 ne ζ ase 2 ne ase 3 ase 4 ζ ne ζ ne = root[t] lgorithms Red-lak Trees ase 3 ase 4 ase 1 R-elete-FiUp R-elete-FiUp(T,) hile root[t] and olor[]=lk do if =left[parent[]] then do set :=right[parent[]] if olor[]=r then do set olor[]:=lk olor[parent[]]:=r Left-Rotate(T,parent[]) set :=right[parent[]] if olor[left[]]=lk and olor[right[]]=lk then set olor[]:=r :=parent[] else if olor[right[]]=lk then do set olor[left[]]:=lk olor[]:=r Right-Rotate(T,) set :=right[parent[]] set olor[]:=olor[parent[]] set olor[parent[]]:=lk set olor[right[]]:=lk Left-Rotate(T,parent[]) set :=root[t] else (same as then flipping left and right) set olor[]:=lk ase 2 5