Today s Outline. One More Operation. Priority Queues. New Operation: Merge. Leftist Heaps. Priority Queues. Admin: Priority Queues

Transcription

1 Tody s Outline Priority Queues CSE Dt Structures & Algorithms Ruth Anderson Spring 4// Admin: HW # due this Thursdy / t :9pm Printouts due Fridy in lecture. Priority Queues Leftist Heps Skew Heps 4// One More Opertion Priority Queues Merge two heps. Ides (Leftist Heps) 4// 4// 4 New Opertion: Merge Given two heps, them into one hep first ttempt: insert ech element of the smller hep into the lrger. second ttempt: conctente inry heps rrys nd run uildhep. Leftist Heps Ide: Focus ll hep mintennce work in one smll prt of the hep Leftist heps:. Most nodes re on the left. All the merging work is done on the right 4// 4// 6

2 Definition: Null Pth Length null pth length (npl) of node x = the numer of nodes etween x nd null in its sutree OR npl(x) = min distnce to descendnt with or children npl(null) = - npl(lef, k zero children) = npl(node with one child) = Equivlent definitions:. npl(x) is the height of lrgest perfect sutree rooted t x. npl(x) = + min{npl(left(x)), npl(right(x))} 4// Leftist Hep Properties Hep-order property prent s priority vlue is to childrens priority vlues result: minimum element is t the root Leftist property For every node x, npl(left(x)) npl(right(x)) result: tree is t lest s hevy on the left s the right Are leftist trees complete lnced 4// Are These Leftist Right Pth in Leftist Tree is Short (#) Clim: The right pth is s short s ny in the tree. Proof: (By contrdiction) Every sutree of leftist tree is leftist! 4// 9 Pick shorter pth: D < D Sy it diverges from right pth t x npl(l) D - ecuse of the pth of length D - to null npl(r) D - ecuse every node on right pth is leftist 4// D Leftist property t x violted! L x R D Right Pth in Leftist Tree is Short (#) Clim: If the right pth hs r nodes, then the tree hs t lest r - nodes. Proof: (By induction) Bse cse : r=. Tree hs t lest - = node Inductive step : ssume true for r < r. Prove for tree with right pth t lest r.. Right sutree: right pth of r- nodes r- - right sutree nodes (y induction). Left sutree: lso right pth of length t lest r- (y previous slide) r- - left sutree nodes (y induction) Totl tree size: ( r- -) + ( r- -) + = r - 4// Why do we hve the leftist property Becuse it gurntees tht: the right pth is relly short compred to the numer of nodes in the tree A leftist tree of N nodes, hs right pth of t most log (N+) nodes Ide perform ll work on the right pth 4//

3 Merge two heps (sic ide) Put the smller root s the new root, Hng its left sutree on the left. Recursively its right sutree nd the other tree. Merging Two Leftist Heps (T,T ) returns one leftist hep contining ll elements of the two (distinct) leftist heps T nd T T L R < L R T 4// 4// 4 Merge Continued If npl(r ) > npl(l ) L R R = Merge(R, T ) R L 4 Merge Exmple 4 4// (specil cse) 4// 6 Sewing Up the Exmple Finlly Done 4 4 4// 4//

4 Merge Two Leftist Heps Other Hep Opertions 6 insert 9 deletemin 4// 9 Student Activity 4// Opertions on Leftist Heps with two trees of totl size n: O(log n) insert with hep size n: O(log n) pretend node is size leftist hep insert y merging originl hep with one node hep deletemin with hep size n: O(log n) remove nd return root left nd right sutrees 4// Good Bd Leftist Heps: Summry 4// Amortized Time m or tized time: Running time limit resulting from writing off expensive runs of n lgorithm over multiple chep runs of the lgorithm, usully resulting in lower overll running time thn indicted y the worst possile cse. If M opertions tke totl O(M log N) time, mortized time per opertion is O(log N) Difference from verge time: Skew Heps Prolems with leftist heps extr storge for npl extr complexity/logic to mintin nd check npl right side is often hevy nd requires switch Solution: skew heps lindly djusting version of leftist heps lwys switches children when fixing right pth mortized time for:, insert, deletemin = O(log n) however, worst cse time for ll three = O(n) 4// 4// 4 4

5 T Merging Two Skew Heps L R T < 4// R Only one step per itertion, with children lwys switched L 4 Exmple 4// Skew Hep Code void (hep, hep) { cse { hep == NULL: return hep; hep == NULL: return hep; hep.findmin() < hep.findmin(): temp = hep.right; hep.right = hep.left; hep.left = (hep, temp); return hep; otherwise: return (hep, hep); } 4// } Runtime Anlysis: Worst-cse nd Amortized No worst cse gurntee on right pth length! All opertions rely on worst cse complexity of ll ops = Amortized Anlysis (Chpter ) Result: M s tke time M log n mortized complexity of ll ops = 4// Binry Heps Compring Priority Queues Leftist Heps d-heps Skew Heps 4// 9 Student Activity