Average-Case Analysis of QuickSort

Transcription

1 Average-Case Aalysis of QuickSort Comp 363 Fall Semester 003 October 3, 003 The purpose of this documet is to itroduce the idea of usig recurrece relatios to do average-case aalysis. The average-case ruig time of QuickSort is obtaied as a applicatios of this idea.. Average-Case Aalysis Usig Recurrece Relatios I class we showed that whe the search key X is i the list L of size, the c A () + where A is Liear Search. Ofte the aalysis ca be doe usig recurrece relatios rather tha evaluatig a sum directly. For coveiece, let f() c A (). We ca write a recurrece relatio for f() uder the assumptio that X is i the list, amely, ½,if f(). f( ),if + How did we arrive at this recurrece? Thik about applyig Liear Search to a usorted list. We must compare the Þrst elemet of the list to the search key X. I other words, the probability of ispectig the Þrst elemet of the list is. How may comparisos do we make whe we ispect the Þrst elemet of the list? We make exactly oe compariso. Now havig compared the Þrst elemet of the list with X, how may comparisos do we make o average to search the rest of the list? Well, it depeds o whether the search key X was foud i positio oe of the list. The probability that we eve have to search the rest of the list is sice the probability of the search key beig i the Þrst positio is /. The expressio f( ) is the average-case ruig time of A o a list of size. Therefore, the quatity f( ) represets the average-case ruig

2 time of A o the remaiig positios i L discouted by the probability that we will eve have to search the last positios. Puttig it all together, we obtai that the average-case ruig time of Liear Search o a list of size is + f( ). Let us ow solve this recurrece by urollig it! f() + + f( ) + f( ) ( + + +( )) + ( ) + + Notice that we obtai the same result that we did whe we evaluated the sum directly from the deþitio.. Average-Case Aalysis of QuickSort We are ow ready to tackle the mai task at had, a aalysis of the averagecase ruig time of QuickSort. We will adopt the approach of the last Sectio, amely, to develop a recurrece relatio for the average-case ruig time ad the solve it. Let A be QuickSort, ad for coveiece, let f() c A (). We ca write a recurrece relatio for the average-case ruig time of QuickSort as 0,if f() P ( ) + (f(i ) + f( i)),if i To uderstad this recurrece, we must cosider it term by term. The term is eeded sice method Split does exactly comparisos. Recall that method Split returs the positio (speciþcally, pivotloc) where the list will be

3 partitioed. Certaily, o radom data, this positio could be the Þrst oe, the secod oe, etc. Sice there are possible positios ad each oe is equally likely, the probability that pivotloc will be ay oe of the possible positios is. What about the term f(i ) + f( i)? (Notice that we are multiplyig this term by.) To uderstad what this sum meas, let s cosider a speciþc example. Suppose 0, ad suppose Split returs the value pivotloc 5. (Let s further suppose that the array is idexed startig at, ot0. Therefore, the idices rage from to 0, ot 0 to 9.) QuickSort is ow called recursively, Þrst o the left subarray of legth four ad the o the right subarray of legth Þve. This situatio is accouted for by takig i 5i the sum of the recurrece formula. That is, we will be addig i the term 0 (f(5 ) + f(0 5)) (f(4) + f(5)). 0 to the total. The quatities f(4) ad f(5) represet the average-case ruig time of QuickSort o arrays of size four ad Þve respectively. However, we discout this term by multiplyig it by the probability that the array will be partitioed i this particular maer! As i rus through its values from to, the list is beig partitioed i all the possible ways. The recurrece relatio is ow see to correctly represet the average-case ruig time of QuickSort o lists of size. Now for the fu part! We must solve the recurrece. The Þrst step is to simplify as much as possible the sum i the recurrece. X (f(i ) + f( i)) i (f(0) + f( )) + (f() + f( )) + + (f( ) + f()) + (f( ) + f(0)) (f() + f()) + (f(3) + f(3)) + + (f( ) + f( )) X f(i) Therefore, we have simpliþed the sum cosiderably as i X (f(i ) + f( i)) X f(i) 3

4 So ow the recurrece has the form 0,if f() ( ) + P f(i),if Let s Þrst multiply through by to clear fractios. X f() ( ) + f(i) To have ay hope of solvig the recurrece, we must get rid of the sum. A trick that is ofte used is to subtract successive values of the recurrece. I other words, let us compute the quatity X f() ( )f( ) ( ) + f(i) X ( )( ) f(i) ( ) + f( ) Notice that all but oe of the terms of the sum cacelled out! have gotte rid of a sum completely! So ow we have I this way, we f() ( )f( ) ( ) + f( ) Therefore, So we ow have f() ( )f( ) + f( ) + ( ) ( +)f( ) + ( ) f() ( +)f( ) + ( ) Dividig both sides through by ( +),weget f() f( ) ( ) + + ( +) 4

5 What we have actually ow doe is formulate a ew recurrece relatio! g() f(). The, clearly, + ( 0,if g() g( ) + ( ),if (+) Let Let s ow solve this recurrece relatio by urollig it. ( ) g() g( ) + ( +) ( ) ( ) g( ) + + ( ) ( +) g() + ( ) ( +) ( ) + + (sice g() 0) 3 ( +) X (i ) i(i +) X i i(i +) If we ca evaluate this last sum, we would be almost doe. Let us ow try to do this. Computig, we obtai X i i(i +) X i + i + X X i + i (we use the method of partial fractios here)

6 Let H The umber H is called the th harmoic umber. (We have ecoutered this umber before i class.) Cotiuig, H H + Therefore, we have show that X Gettig back to g(), g() X i i(i +) H + i i(i +) H H We are ow almost doe! Recall that Therefore, we have show that f() ( +)g() ( +) H 4 + ( +)H 4 H +H 4 f() H +H 4 Now it was proved o Test # (see your test) that H Θ(l ). Sice the logarithms all have the same growth rate, H Θ(log ). Therefore, the expressio H +H 4 Θ( log ). Therefore, we have proved that c A () f() Θ( log ) So, the average-case ruig time of QuickSort is Θ( log ). This ruig time is dramatically differet from its worst-case ruig time which, as we kow, is Θ( ). 6

7 The istructor has doe some computig ad has produced the followig table with the help of Maple c. log c A () ( +)H 4 c A () log It is iterestig to cosider c A () log H +H 4 log H log + H log 4 log H O Test #, it was show that l. Therefore, sice log log e l, So, H log + Computig, we obtai H log log e H log c A () log log e 4 log log e +0+0 log e Note that our aswer makes perfectly good sese give our computatioal results! Therefore, c A () log e ( log ) 7