# Recursion and Recurrences

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6 164 CHAPTER 5. RECURSION AND RECURRENCES Figure 5.4: A recursio tree diagram for Recurrece 5.5. log + 1 levels /2 /4 /2 /4 /8 /8 1 1 of work doe to solve the problem described by recurrece (5.4) is (log 2 + 1). The total work doe throughout the tree is the solutio to our recurrece, because the tree simply models the process of iteratig the recurrece. Thus the solutio to recurrece (5.3) is T () =(log + 1). More geerally, we ca cosider a recurrece that it idetical to (5.3), except that T (1) = a, for some costat a. Ithis case, T () =a + log, because a uits of work are doe at level 1 ad additioal uits of work are doe at each of the remaiig log levels. It is still true that, T () =Θ( log ), because the differet base case did ot chage the solutio to the recurrece by more tha a costat factor. Sice oe uit of time will vary from computer to computer, ad sice some kids of work might take loger tha other kids, it is the big-θ behavior of T () that is really relevat. Thus although recursio trees ca give us the exact solutios (such as T () =a+ log above) to recurreces, we will ofte just aalyze a recursio tree to determie the big-θ or eve, i complicated cases, just the big-o behavior of the actual solutio to the recurrece. I Problem?? we explore whether the value of T (1) actually iflueces the big-θ behavior of the solutio to a recurrece. Let s look at oe more recurrece. T () = T (/2) + if >1 1 if =1 (5.5) Agai, assume is a power of two. We ca iterpret this as follows: to solve a problem of size, wemust solve oe problem of size /2 ad do uits of additioal work. We draw the tree for this problem i Figure 5.4. We see i this figure that the problem sizes are the same as i the previous tree. The rest, however, is differet. The umber of subproblems does ot double, rather it remais at oe o each level. Cosequetly the amout of work halves at each level. Note that there are still log +1levels, as the umber of levels is determied by how the problem size is chagig, ot by how may subproblems there are. So o level i, wehave 1problem of size /2 i, for total work of /2 i uits.

7 5.1. GROWTH RATES OF SOLUTIONS TO RECURRENCES 165 We ow wish to compute how much work is doe i solvig a problem that gives this recurrece. Note that the additioal work doe is differet o each level, so we have that the total amout of work is ( + /2+/ = ( 1 log2 4 2) ) + +, which is times a geometric series. By Theorem 4.4, the value of a geometric series i which the largest term is oe is Θ(1). This implies that the work doe is described by T () =Θ(). We emphasize that there is exactly oe solutio to recurrece (5.5); it is the oe we get by usig the recurrece to compute T (2) from T (1), the to compute T (4) from T (2), ad so o. What we have doe here is show that T () =Θ(). We have ot actually foud a solutio. I fact, for the kids of recurreces we have bee examiig, oce we kow T (1) we ca compute T () for ay relevat by repeatedly usig the recurrece, so there is o questio that solutios do exist. What is ofte importat to us i applicatios is ot the exact form of the solutio, but a big-o upper boud, or, better, a Big-Θ boud o the solutio. Exercise Fid a big-θ boud for the solutio to the recurrece 3T (/3) + if 3 T () = 1 if <3 usig a recursio tree. Assume that is a power of 3. Exercise Solve the recurrece T () = 4T (/2) + if 2 1 if =1 usig a recursio tree. Assume that is a power of 2. Covert your solutio to a big-θ statemet about the behavior of the solutio. Exercise Ca you give a geeral big-θ boud for solutios to recurreces of the form T () =at (/2) + whe is a power of 2? You may have differet aswers for differet values of a. The recurrece i Exercise is similar to the mergesort recurrece. Oe differece is that at each step we divide ito 3 problems of size /3. Thus we get the picture i Figure 5.5. Aother differece is that the umber of levels, istead of beig log 2 +1isow log 3 +1,so the total work is still Θ( log ) uits. Note that log b = Θ(log 2 ) for ay b>1. Now let s look at the recursio tree for Exercise Now, we have 4 childre of size /2, ad we get the Figure 5.6 Let s look carefully at this tree. Just as i the mergesort tree there are log 2 +1levels. However, i this tree, each ode has 4 childre. Thus level 0 has 1 ode, level 1 has 4 odes, level 2 has 16 odes, ad i geeral level i has 4 i odes. O level i each ode correspods to a problem of size /2 i ad hece requires /2 i uits of additioal work. Thus the total work o level i is 4 i (/2 i )=2 i uits. Summig over the levels, we get log 2 i=0 log 2 2 i = i=0 2 i.

8 166 CHAPTER 5. RECURSION AND RECURRENCES Figure 5.5: The recursio tree diagram for the recurrece i Exercise /3 /3 + /3 + /3 = log + 1 levels /9 9(/9) = 1 (1) = There are may ways to simplify that expressio, for example from our formula for the sum of a geometric series we get. log 2 T () = i=0 2 i = 1 2(log 2 ) = = 2 2 = Θ( 2 ). Figure 5.6: The Recursio tree for Exercise /2 /2 + /2 + /2 + /2 = 2 log + 1 levels /4 16(/4) = 4 1 ^2(1) = ^2

10 168 CHAPTER 5. RECURSION AND RECURRENCES 2. Mergesort. I mergesort we sort a list of items that have some uderlyig order by dividig the list i half, sortig the first half (by recursively usig mergesort), sortig the secod half (by recursively usig mergesort), ad the mergig the two sorted list. For a list of legth oe mergesort returs the same list. 3. Recursio Tree. A recursio tree diagram for a recurrece of the form T () =at (/b)+g() has three parts. O the left, we keep track of the problem size, i the middle we draw the tree, ad o right we keep track of the work doe. We draw the diagram i levels, each level of the diagram represetig a level of recursio. The tree has a vertex represetig the iitial problem ad oe represetig each subproblem we have to solve. Each o-leaf vertex has a childre. The vertices are divided ito levels correspodig to (sub-)problems of the same size; to the left of a level of vertices we write the size of the problems the vertices correspod to; to the right of the vertices o a give level we write the total amout of work doe at that level by a algorithm whose work is described by the recurrece, ot icludig the work doe by ay recursive calls. 4. The Base Level of a Recursio Tree. The amout of work doe o the lowest level i a recursio tree is the umber of odes times the value give by the iitial coditio; it is ot determied by attemptig to make a computatio of additioal work doe at the lowest level. 5. Bases for Logarithms. We use log as a alterate otatio for log 2.Afudametal fact about logarithms is that log b = Θ(log 2 ) for ay real umber b>1. 6. Three behaviors of solutios. The solutio to a recurrece of the form T () =at (/2) + behaves i oe of the followig ways: (a) if a<2 the T () =Θ(). (b) if a =2the T () =Θ( log ) (c) if a>2 the T () =Θ( log 2 a ) Problems 1. Draw recursio trees ad fid big-θ bouds o the solutios to the followig recurreces. For all of these, assume that T (1) = 1 ad is a power of the appropriate iteger. (a) T () =8T (/2) + (b) T () =8T (/2) + 3 (c) T () =3T (/2) + (d) T () =T (/4) + 1 (e) T () =3T (/3) Draw recursio trees ad fid fid exact solutios to the followig recurreces. For all of these, assume that T (1) = 1 ad is a power of the appropriate iteger. (a) T () =8T (/2) + (b) T () =8T (/2) + 3 (c) T () =3T (/2) +

11 5.1. GROWTH RATES OF SOLUTIONS TO RECURRENCES 169 (d) T () =T (/4) + 1 (e) T () =3T (/3) Fid the exact solutio to recurrece Show that log b = Θ(log 2 ). 5. Recursio trees will still work, eve if the problems do ot break up geometrically, or eve if the work per level is ot c uits. Draw recursio trees ad ad fid the best big-o bouds you ca for solutios to the followig recurreces. For all of these, assume that T (1) =1. (a) T () =T ( 1) + (b) T () =2T ( 1) + (c) T () =T ( )+1(You may assume has the form =2 2i.) (d) T () =2T (/2) + log (You may assume is a power of 2.) 6. I each case i the previous problem, is the big-o boud you foud a big-θ boud? 7. If S() =as( 1) + g() ad g() <c with 0 c<a,how fast does S() grow (i big-θ terms)? S() =a i S( 1) + i 1 j=0 aj g( j) =a S(0) + 1 j=0 aj g( j) < a S(0) + 1 j=0 aj c j = a S(0) + c 1 ( a ) j j=0 c = a S(0)+Θ( ( ) a c )=Θ(a ) 8. If S() =as( 1) + g() ad g() >c with 0 <a c, how fast does S() grow i big-θ terms? 9. give a recurrece of the form T () =at (/b)+g() with T (1) = c>0 ad g() > 0 for all ad a recurrece of the form S() =as(/b) +g() with S(1) = 0 (ad the same g()), is there ay differece i the big-θ behavior of the solutios to the two recurreces? What does this say about the ifluece of the iitial coditio o the big-θ behavior of such recurreces?

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