Practical Session 2 Constants Don t Matter!!!

Transcription

1 Practical Sessio 2 Costats Do t Matter!!! Algorithm Aalysis f() = O(g()) Big O f() = Ω(g()) Big Omega There exist c > 0 ad 0 > 0 such that: 0 f() cg() for each 0 There exist c > 0 ad 0 > 0 such that: 0 cg() f() for each 0 lim f() g() < lim f() g() > 0 f() = Θ(g()) Big Theta There exist c1, c2 > 0 ad 0 > 0 such that: 0 c1g() f() c2g() for each 0 lim f() g() = c 0 < c < שימו לב: פונקציות המתארות זמן ריצה הינן אי-שליליות אסימפטוטית. במהלך הקורס בד"כ נתייחס לפונקציות מסוג זה. אינטואיטיבית: O(g()) f() = אומר ש-( g( הוא מעין חסם עליון ל-( f(, כלומר f() לא יכולה "גרועה" )במונחי זמן ריצה( מ-( g( )כפול קבוע(. Ω(g()) f() = אומר ש-( g( הוא מעין חסם תחתון ל-( f(, כלומר f() לא יכולה "טובה" )במונחי זמן ריצה( מ-( g( )כפול קבוע(. תכונות של Ω, Θ, O עבור שתי פונקציות,f g מתקיים טרנזיטיביות: f() = O(g()), g() = O(h()) f() = O(h()) f() = Ω(g()), g() = Ω(h()) f() = Ω(h()) f() = Θ(g()), g() = Θ(h()) f() = Θ(h()) סימטריות: f() = Θ(g()) g() = Θ(f()) "סימטריות זוגית" symmetry( :)Traspose f() = O(g()) g() = Ω(f()) נסו להוכיח תכונות אלו בבית, ע"י שימוש בהגדרות הבסיסיות

2 Some Basics Prove: 100 * + 12 = Θ(). c c 1 = 1, true for every c 2 c c 2 = 104, true for every 3 Prove: = O( 3 ) c 3 c = 2, true for every 2 Questio 1 Prove : (log ) log = Ω( 3/2 ). c 3/2 = c(2 log ) 3/2 = c(2 3/2 ) log (log ) log c 3/2 (log ) log c = 1 0 = 8 Questio 2 Prove that log(!) = Θ( log ). log(!) = log(1 2 3 ) = log 1 + log 2 + log log log i log log(!) = O( log) i=1 log i log i 2 log ( 2 ) 2 (log log 2) = 2 log 2 c log i=1 i= 2 log(!) = Ω( log )

3 Explaatio of the iequality: * 1 1 : log c log => log clog log 2 1 clog 2 c = = 8 log(!) = Θ( log ) Questio 3 Prove : (log )! = Ω(log(!)) By the previous questio, it is sufficiet to prove that (log )! = Ω( log ). Why? The proof uses trasitivity of Ω ad traspose symmetry. If we prove that (log )! = Ω( log ), the usig trasitivity ad traspose symmetry we get (log )! = Ω(log(!)) נסו להוכיח בבית, ע"י שימוש בטרנזיטיביות ובסימטריה חילופית, את הטענה המרכזית. Observe that for a iteger k 4, we have k! > 2 k. Hece, (log)! = (log)(log -1)! > (log) 2 log-1 = (log) 2 log 2-1 = (log) (/2) c log. (log)! = Ω( log). Questio 4 Give a example for two fuctios g() ad f() such that either g() = O(f()) or g() = Ω(f()). Solutio A: f() =, g() = 1-cos. Solutio B: f() = 2 1 if = 1 g() = { g( 1) if g( 1) 3 otherwise

4 Questio 5 Give the mootoe icreasig fuctios f(): N N, g(): N N such that f() = Θ(g()), prove log(f()) = Θ(log(g())) פתרון: מהנתון, קיימים > 0 0 c 1, c 2, כך שלכל 0 מתקיים כי: g() c 1 g() f() c 2 0 נתון כי הפונקציות מונוטוניות עולות, ולכן קיים N כך ש > 0 g() f(), עבור כל. אזי, לכל מתקיים כי: log(c 1 g()) log(f()) log(c 2 g()) אי שיוויון שמאלי: עבור מספיק גדול: log(f()) log(g()) log(c 1 ) + log(g()) = log(c 1 g()) 0.5 אי שיוויון ימני: עבור מספיק גדול: log(g()) log(f()) log(c 2 g()) = log(c 2 ) + log(g()) 2 Questio 6 Prove or disprove: 1. For ay fuctios f() ad g(), f() = O(g()) g() = Ω(f()). 2. For ay fuctio f(), f() = (f ( 2 )). 1. f() = O(g()) f() cg() g() 1 f() g() = Ω(f()). c 2. False, e.g., f() = 2. The f() = 2 (2 2) = f ( 2 ). We will prove that 2 O (2 2 ): Suppose there exists a costat c such that: 2 c 2 2 for each 0. The 2 2 c for each 0 (divide both sides by 2 2). So c is ot a costat, because it is bigger tha a divergig expressio. Cotradictio! Questio 7 Give a sorted array A of distict itegers. a) Describe a O(1) time algorithm that determies whether or ot there exists a iteger elemet x, such that A[1] < x < A[] ad x is ot i A. b) Describe a O(log ) time algorithm that fids such a x. c) Describe a efficiet algorithm that determies whether or ot there exists a elemet i i A, such that A[i] = i (1 i ). Aalyze the ruig time of the algorithm.

5 I aswerig this questio we assume that A's idexes ru from 1 to a) We describe the algorithm SimpleCheck(A,r,q) where r<q are atural umbers. SimpleCheck(A,r,q) retur (A[q] A[r] < q-r) To solve a we ru SimpleCheck(A,1,). b) We describe the algorithm SolveB(A,r,q) where r<q are atural umbers. SolveB(A,r,q) if (r + 1 = q) retur A[r] + 1 m (r + q)/2 if (SimpleCheck(A,r,m) = true) retur SolveB(A,r,m) else retur SolveB(A,m,q) To solve b we use the followig algorithm: if (SimpleCheck(A,1,) = false) Prit "There is o such elemet i A" else retur SolveB(A,1,) c) We describe a recursive algorithm SolveC(A,r,q) where r<q are atural umbers. SolveC(A,r,q) if (q < r) retur false m (r + q)/2 if (A[m] = m) retur true if (A[m] > m) //A's itegers are distict, so A[m+1] > m+1, A[m+2] > m+2, ad so o. retur SolveC(A,r,m-1) else //if (A[m] < m). A's itegers are distict, so A[m-1] < m-1, A[m-2] < m-2, ad so o. retur SolveC(A,m+1,q) To solve C we use the call SolveC(A,1,).

6 Questio 8 You have two lamps made from a special secret material. You are i a tall buildig with N floors, ad you must determie what is the top floor M from which you ca drop a lamp to the groud so that it does't break o impact. You ca throw a lamp out a widow as may times as you wat as log as it does't break. Describe a strategy for fidig the highest safe rug that requires you to drop a lamp at most f() times, for some fuctio f() that grows slower tha liearly. (I other words, it should be f() the case that lim = 0). Suppose you have Oe lamp what will be your strategy? What is f()? Suppose you have lamps what will be your strategy? What is f()? Oe is tempted to do a biary search with the first lamp, but this is ot that efficiet, sice if you drop your first lamp from the N/2'th floor ad it breaks, you eed to do a liear search usig the secod lamp from the 1st floor up - which could mea N/2-1 total throws i the worst case. The solutio is to throw the first lamp from floors i N for icreasig values of i. Oce it breaks you have a possible rage of N values of M, so i total you ever do more tha f() = 2 1 throws. Improvemet: I the previous solutio, if for example the buildig was 100 stories tall, we would eed 19 throws i the worst case. The solutio below allows us to do it i 14 throws: Throw from floor #14 If it breaks, we have 13 throws (floor #1 floor #13). Icludig the first throw from floor #14 we used 14 throws. If it does ot break throw from floor #27 (14+13). If it breaks, we have 12 throws (floor #15 floor #26). Icludig the first throw from floor #14 ad the secod throw from floor #27 we used 14 throw. If it does ot break throw from floor #39 ( ). If it breaks, we have 11 throws (floor #28 floor #38). Icludig the first throw from floor #14, the secod throw from floor #27 ad the third throw from #39 we used 14 throw. Cotiue i the same maer util we reach #100. I summary, we start from floor #14 ad each throw reduce 1 from the distace to the ext jump util it breaks, the throw all the optio i the remaiig rage. Where did the 14 come from? 14 is the smallest umber that gives: If the umber of floors is, we eed x x(x+1) x=