A recurrence equation is just a recursive function definition. It defines a function at one input in terms of its value on smaller inputs.

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1 CS23 Algorithms Hadout #6 Prof Ly Turbak September 8, 200 Wellesley College RECURRENCES This hadout summarizes highlights of CLRS Chapter 4 ad Appedix A (CLR Chapters 3 & 4) Two-Step Strategy for Aalyzig Algorithms Characterize ruig-time (space, etc) of algorithm by a recurrece equatio 2 Solve the recurrece equatio, expressig aswer i asymptotic otatio Recurrece Equatios A recurrece equatio is just a recursive fuctio defiitio It defies a fuctio at oe iput i terms of its value o smaller iputs We use recurrece equatios to characterize the ruig time of algorithms T() typically stads for the ruig-time (usually worst-case) of a give algorithm o a iput of size Recurrece equatios for divide-ad-coquer algorithms typically have the form: Geeral Case ( ) T() = [# of subproblems] T ([size of each subproblem]) + [cost of divide&glue] Base Case ( < ) T() = 0 Because the base case is always the same, it is usually omitted whe defiig T() Derivig Recurrece Equatios: Examples Isert a elemet ito a sorted list of legth Isertio-sort a list of legth Biary search o a array of elemets Fid the maximum leaf of a balaced biary tree with umeric leaves Merge-sort a list of elemets Give a balaced biary tree with oegative leaves summig to S, fid a leaf such that all leaves to its left ad all leaves to its right both have sums S/2 Towers of Haoi T() = T( ) + T() = T( ) + T() = T( ) + T() = T( ) + T() = T( ) + T() = T( ) + T() = T( ) +

2 Solvig Recurrece Equatios: The Recursio-Tree Method There are several methods for solvig recurrece equatios (see CLRS for details) We will maily use the recursio-tree method (CLR also calls it the iteratio method) This is a twostep strategy for solvig recurrece equatios: Costruct a recursio tree by uwidig the recurrece equatio 2 Determie the cost of the etire tree by summig the costs of the odes Example : T() = T( - ) + Also derivable: T( - ) = T( - 2) + T( - 2) = T( - 3) + etc T() T(-) T(-2) 2 levels Total cost of odes = umber of odes = Example 2: T() = T( - ) + Also derivable: T( - ) = T( - 2) + ( - ) T( - 2) = T( - 3) + ( - 2) etc T() T(-) - T(-2) levels Total cost of odes = ( - 2) + ( - ) + = k = {Arithmetic series; see below} k= - -2

3 Arithmetic Series A series is arithmetic if a k = c + a (k-) Trick: Sum correspodig pairs of umbers: Examples: = a + a 2 + a 3 a -2 + a - + a a k = 2 (a + a ) k= Sum of first elemets of series Example 3: T() = T(/2) + Also derivable: T(/2) = T(/4) + T( /4) = T(/8) + etc T() T(/2) T(/4) lg() + levels Total cost of odes = umber of odes = lg() + = Θ(lg()) 3

4 Example 4: T() = 2T(/2) + Also derivable: T(/2) = 2T(/4) + /2 T(/4) = 2T(/8) + /4 T() level T(/2) T(/2) /2 /2 level 2 /2 /2 T(/4) T(/4) T(/4) T(/4) level lg() + level 3 /4 /4 /4 /4 Total cost = ()(umber of levels) = (lg() + ) = Θ( lg()) Example 5: T() = 2T(/2) + Also derivable: T(/2) = 2T(/4) + T(/4) = 2T(/8) + etc T() level 0 = 2 T(/2) T(/2) level 2 2 = 2 T(/4) T(/4) T(/4) T(/4) level lg() + Total cost = umber of odes = sum of odes at each level = lg() {Geometric series!} level 3 lg() = 2 k = k=0 2 4 = 2 = 2 lg() 4

5 Geometric Series A series is geometric if a k = c a (k-) Let S() stad for k =0 a 0 c k = a 0 + a 0 c + a 0 c 2 + a 0 c a 0 c (Note carefully: S() has + terms, ot terms!) Notice the followig: cs() = a 0 c + a 0 c 2 + a 0 c a 0 c + a 0 c + S() = a 0 + a 0 c + a 0 c 2 + a 0 c a 0 c (c ) S() = a 0 c + a 0 = a 0 (c + ) S() = a 0 (c+ ) (c - ) If 0 < c < ad, the above formula ca be rewritte as: a 0 lim S() = ( - c) if 0 < c < Examples: = lg() = /2 + /4 + /8 + = /3 + /9 + /27 + = {Paper tearig example}

6 Example 6: T() = T(/2) + Also derivable: T(/2) = T(/4) + /2 T(/4) = T(/8) + /4 T() T(/2) /2 T(/4) lg() + levels /2 /4 lg() Total cost = + /2 + /4 + + /2 lg() = 2 k < 2 k = k=0 k=0 Example 7: T() = 2T( - ) + Also derivable: T(-) = 2T(-2) + T(-2) = 2T(-3) + T() level 0 = 2 T(-) T(-) level 2 2 = 2 Total cost = = T(-2) T(-2) T(-2) T(-2) level level = 2 2-6

7 Solvig Recurrece Equatios: The Substitutio Method This is a alterative method for solvig recurrece equatios we shall see several times i this course I the substitutio method, we guess a solutio ivolvig various coefficiets, ad determie the coefficiets that lead to a solutio (if ay) Example : T() = T( - ) + Assume solutio has the form T() = a + b: What if we assumed the solutio had form T() = a 2 + b + c? Example 2: T() = T( - ) + Assume solutio has the form T() = a 2 + b + c: What if we assumed the solutio had form T() = a + b? Example 3: T() = 2T(( )/2) + Here a exact solutio is tricky Istead assume T() a(lg()), where What must be true of a? 7

8 Summary The followig table summarizes ad geeralizes the above examples Equivalece classes of fuctios are arraged i the table from "biggest" to "smallest" That is, if fuctio f appears above fuctio g i the table, the g is O(f) but f is ot O(g) (or, equivaletly, f is Ω(g) but g is ot Ω(f)) It is worth otig that there are may more equivalece classes tha are listed i the table, but the oes i the table are the oes most commoly ecoutered i this course Equvialece class of fuctios Θ(3 ) expoetial (base = 3) Name Typical Recurrece Equatios T() = 3 T( - a) + b, a > 0, b > 0 Θ(2 ) expoetial (base = 2) T() = 2 T( - a) + b, a > 0, b > 0 Θ( 2 ) quadratic T() = T( - a) + b, a > 0, b > 0 Θ( log()) log() T() = k T((-a)/k) + b, k >, a 0, b > Θ() liear T() = T( - a) + b, > 0, a > 0, b > 0 or T() = k T((-a)/k) + b, k >, a 0, b > 0 or T() = T((-a)/k) + b, k >, a 0, b > Θ(log()) logarithmic (base is T() = T(/k) + a, k >, a > 0 irrelevat) Θ() costat T() = a 8