Introduction to Algorithms

Transcription

1 Itroductio to Algorithms 6.046J/8.40J/SMA5503 Lecture 9 Pro. Charles E. Leiserso

2 Biary-search-tree sort T Create a empty BST or i to do TREE-INSERT(T, A[i]) Perorm a iorder tree wal o T. Eample: A [ ] Tree-wal time O(), but how log does it tae to build the BST? Itroductio to Algorithms Day 7 L9.2

3 Aalysis o BST sort BST sort perorms the same comparisos as quicsort, but i a dieret order! The epected time to build the tree is asymptotically the same as the ruig time o quicsort. Itroductio to Algorithms Day 7 L9.3

4 Node depth The depth o a ode the umber o comparisos made durig TREE-INSERT. Assumig all iput permutatios are equally liely, we have Average ode depth E (# comparisos to isert ode i) i O( lg ) (quicsort aalysis) O(lg ). Itroductio to Algorithms Day 7 L9.4

5 Epected tree height But, average ode depth o a radomly built BST O(lg ) does ot ecessarily mea that its epected height is also O(lg ) (although it is). Eample. lg Ave. depth lg + O(lg ) 2 h Itroductio to Algorithms Day 7 L9.5

6 Height o a radomly built biary search tree Outlie o the aalysis: Prove Jese s iequality, which says that (E[X]) E[(X)] or ay cove uctio ad radom variable X. Aalyze the epoetial height o a radomly built BST o odes, which is the radom variable 2 X, where X is the radom variable deotig the height o the BST. Prove that 2 E[X ] E[2 X ] E[ ] O( 3 ), ad hece that E[X ] O(lg ). Itroductio to Algorithms Day 7 L9.6

7 Cove uctios A uctio : R R is cove i or all,β 0 such that + β, we have ( + βy) () + β(y) or all,y R. () + β(y) (y) () ( + βy) + βy y Itroductio to Algorithms Day 7 L9.7

8 Coveity lemma Lemma. Let : R R be a cove uctio, ad let {, 2,, } be a set o oegative costats such that. The, or ay set {, 2,, } o real umbers, we have ( Proo. By iductio o. For, we have, ad hece ( ) ( ) trivially. ). Itroductio to Algorithms Day 7 L9.8

9 Itroductio to Algorithms Day 7 L9.9 Proo (cotiued) + ) ( Iductive step: Algebra.

10 Itroductio to Algorithms Day 7 L9.0 Proo (cotiued) + + ) ( ) ( ) ( Iductive step: Coveity.

11 Itroductio to Algorithms Day 7 L9. Proo (cotiued) ) ( ) ( ) ( ) ( ) ( ) ( Iductive step: Iductio.

12 Itroductio to Algorithms Day 7 L9.2 Proo (cotiued) ) ( ) ( ) ( ) ( ) ( ) ( ) ( Iductive step: Algebra..

13 Jese s iequality Lemma. Let be a cove uctio, ad let X be a radom variable. The, (E[X]) E[ (X)]. Proo. ( E[ X ]) Pr{ X } Deiitio o epectatio. Itroductio to Algorithms Day 7 L9.3

14 Jese s iequality Lemma. Let be a cove uctio, ad let X be a radom variable. The, (E[X]) E[ (X)]. Proo. ( E[ X ]) Pr{ X } ( ) Pr{ X } Coveity lemma (geeralized). Itroductio to Algorithms Day 7 L9.4

15 Jese s iequality Lemma. Let be a cove uctio, ad let X be a radom variable. The, (E[X]) E[ (X)]. Proo. ( E[ X ]) Pr{ X } ( ) Pr{ X } E[ ( X )]. Tricy step, but true thi about it. Itroductio to Algorithms Day 7 L9.5

16 Aalysis o BST height Let X be the radom variable deotig the height o a radomly built biary search tree o odes, ad let 2 X be its epoetial height. I the root o the tree has ra, the X + ma{x,x }, sice each o the let ad right subtrees o the root are radomly built. Hece, we have 2 ma{, }. Itroductio to Algorithms Day 7 L9.6

17 Aalysis (cotiued) Deie the idicator radom variable Z as Z i the root has ra, 0 otherwise. Thus, Pr{Z } E[Z ] /, ad Z 2 ma{., } ( ) Itroductio to Algorithms Day 7 L9.7

18 Epoetial height recurrece [ ] E Z ( 2 ma{, ) E } Tae epectatio o both sides. Itroductio to Algorithms Day 7 L9.8

19 Epoetial height recurrece [ ] E Z ( 2 ma{, }) E E [ Z ( 2 ma{, })] Liearity o epectatio. Itroductio to Algorithms Day 7 L9.9

20 Itroductio to Algorithms Day 7 L9.20 Epoetial height recurrece [ ] ( ) ( ) [ ] E Z E Z E Z E E }], [ma{ ] [ 2 }, ma{ 2 }, ma{ 2 Idepedece o the ra o the root rom the ras o subtree roots.

21 Itroductio to Algorithms Day 7 L9.2 Epoetial height recurrece [ ] ( ) ( ) [ ] + E E Z E Z E Z E E ] [ 2 }], [ma{ ] [ 2 }, ma{ 2 }, ma{ 2 The ma o two oegative umbers is at most their sum, ad E[Z ] /.

22 Itroductio to Algorithms Day 7 L9.22 Epoetial height recurrece [ ] ( ) ( ) [ ] + 0 ] [ 4 ] [ 2 }], [ma{ ] [ 2 }, ma{ 2 }, ma{ 2 E E E Z E Z E Z E E Each term appears twice, ad reide.

23 Solvig the recurrece Use substitutio to show that E[ ] c 3 or some positive costat c, which we ca pic suicietly large to hadle the iitial coditios. 4 E[ 0 [ ] E ] Itroductio to Algorithms Day 7 L9.23

24 Solvig the recurrece Use substitutio to show that E[ ] c 3 or some positive costat c, which we ca pic suicietly large to hadle the iitial coditios. [ ] E E[ c 3 Substitutio. ] Itroductio to Algorithms Day 7 L9.24

25 Solvig the recurrece Use substitutio to show that E[ ] c 3 or some positive costat c, which we ca pic suicietly large to hadle the iitial coditios. [ ] E 4 4 4c E[ c 3 3 d ] Itegral method. Itroductio to Algorithms Day 7 L9.25

26 Solvig the recurrece Use substitutio to show that E[ ] c 3 or some positive costat c, which we ca pic suicietly large to hadle the iitial coditios. [ ] E 4 4 4c 4c E[ c d Solve the itegral. ] Itroductio to Algorithms Day 7 L9.26

27 Solvig the recurrece Use substitutio to show that E[ ] c 3 or some positive costat c, which we ca pic suicietly large to hadle the iitial coditios. [ ] E 4 4 4c 4c c E[ c d ] 4. Algebra. Itroductio to Algorithms Day 7 L9.27

28 The grad iale Puttig it all together, we have 2 E[X ] E[2 X ] Jese s iequality, sice () 2 is cove. Itroductio to Algorithms Day 7 L9.28

29 The grad iale Puttig it all together, we have 2 E[X] E[2 X ] E[ ] Deiitio. Itroductio to Algorithms Day 7 L9.29

30 The grad iale Puttig it all together, we have 2 E[X] E[2 X ] E[ ] c 3. What we just showed. Itroductio to Algorithms Day 7 L9.30

31 The grad iale Puttig it all together, we have 2 E[X] E[2 X ] E[ ] c 3. Taig the lg o both sides yields E[X ] 3lg +O(). Itroductio to Algorithms Day 7 L9.3

32 Post mortem Q. Does the aalysis have to be this hard? Q. Why bother with aalyzig epoetial height? Q. Why ot just develop the recurrece o X + ma{x,x } directly? Itroductio to Algorithms Day 7 L9.32

33 Post mortem (cotiued) A. The iequality ma{a, b} a + b. provides a poor upper boud, sice the RHS approaches the LHS slowly as a b icreases. The boud ma{2 a,2 b } 2 a + 2 b allows the RHS to approach the LHS ar more quicly as a b icreases. By usig the coveity o () 2 via Jese s iequality, we ca maipulate the sum o epoetials, resultig i a tight aalysis. Itroductio to Algorithms Day 7 L9.33

34 Thought eercises See what happes whe you try to do the aalysis o X directly. Try to uderstad better why the proo uses a epoetial. Will a quadratic do? See i you ca id a simpler argumet. (This argumet is a little simpler tha the oe i the boo I hope it s correct!) Itroductio to Algorithms Day 7 L9.34