Data Structures. Outline

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1 Data Structures Solvig Recurreces Tzachi (Isaac) Rose 1 Outlie Recurrece The Substitutio Method The Iteratio Method The Master Method Tzachi (Isaac) Rose 2 1

2 Recurrece A recurrece is a fuctio defied i terms of oe or more base cases, ad itself, with smaller argumets. Examples: Tzachi (Isaac) Rose 3 Outlie Recurrece The Substitutio Method The Iteratio Method The Master Method Tzachi (Isaac) Rose 4 2

3 The Substitutio Method Guess the form of the aswer, the use iductio to fid the costats ad show that solutio works. Tzachi (Isaac) Rose 5 Substitutio with Exact Fuctio Claim: T() = lg + Proof: Guess T () = lg + Basis: = 1 lg + = 1 = T () Iductive step: Iductive hypothesis is that T(k) = k lg k + k for all k <. Tzachi (Isaac) Rose 6 3

4 Substitutio with Order of Growth Iductio Paradigm: Name the costat i the additive term. Show the upper (O) ad lower (Ω) bouds separately. Example: Claim: Might eed to use differet costats for each. T () = Θ( lg ). Proof: Name the costat i the additive term : T () 2T(/2) + c, for some costat c. Guess ad Show by Iductio: T() d lg, for some positive costat d. Name the costat i the additive term : T () 2T(/2) + c, for some costat c. Guess ad Show by Iductio: T() d lg, for some positive costat d. Tzachi (Isaac) Rose 7 Substitutio with Order of Growth Upper boud: Name the costat i the additive term : T() 2T(/2) + c, for some costat tc. Guess ad Show by Iductio : T() d lg, for some positive costat d. Therefore, T() = O( lg ). Tzachi (Isaac) Rose 8 4

5 Substitutio with Order of Growth Lower boud: Name the costat i the additive term : T() 2T(/2) + c, for some costat c. Guess ad Show by Iductio : T() d lg, for some positive costat d. Therefore, T() = Ω( lg ). Tzachi (Isaac) Rose 9 Substitutio with Order of Growth Coclusio: T() = O( lg ) ad T() = Ω( lg ), Hece, T() = Θ( lg ). Tzachi (Isaac) Rose 10 5

6 Base Case What about base cases? T() is always costat for ay costat. We are ultimately iterested i a asymptotic solutio to a recurrece. Hece, it will always be possible to choose base cases that work. Whe we wat a exact solutio, the we have to deal with base cases. Tzachi (Isaac) Rose 11 Exact Form The proof should be exact form of our iductive hypothesis. Example: Give T() = 8T(/2) + Θ( 2 ) It ca be show that T() = O( 3 ). That is, T() 8T(/2) + c 2. But guessig T() d 3, will ot do. Sice, Tzachi (Isaac) Rose 12 6

7 Exact Form I the other had Guessig T() d 3 d 2 will do. Sice, Tzachi (Isaac) Rose 13 The Recursio Tree Method Use to geerate a guess. The verify by substitutio method Example: T() = T(/3) + T(2/3) + Θ(). There are log 3 full levels, ad after log 3/2 levels, the problem size is dow to 1. Each level cotributes c. Good guess: T() = Θ( lg ) Tzachi (Isaac) Rose 14 7

8 The Recursio Tree Method Upper boud: Guess: T () d lg. Therefore, T () = O( lg ). Tzachi (Isaac) Rose 15 The Recursio Tree Method Lower boud: Guess: T() d lg. Substitutio is the same as for the upper boud, but replacig by. Ed up with Therefore, T() = Ω( lg ). Coclusio T() = O( lg ) ad T() = Ω( lg ), Hece, T() = Θ( lg ). Tzachi (Isaac) Rose 16 8

9 Chagig Variables Cosider the recurrece T() = 2T( ) + lg Reame m = lg yields T(2 m ) = 2T(2 m/2 ) + m Reame S(m) = T(2 m ) yields S(m) = 2S(m/2)+m But, S(m) = 2S(m/2)+m = Θ(m lg m) Hece, T() = T(2 m ) = S(m) = Θ(m lg m) = Θ(lg lg lg ) Tzachi (Isaac) Rose 17 Outlie Recurrece The Substitutio Method The Iteratio Method The Master Method Tzachi (Isaac) Rose 18 9

10 The Iteratio Method Expad the recurrece Work some algebra to express as a summatio Evaluate the summatio Tzachi (Isaac) Rose 19 A Example Claim: 0 = 0 Let T ( ) = c + T ( 1) > 0 Proof: Tha T() = Θ() T() = c + T( 1) = c + c + T( 2) = = c + T(0) = c Tzachi (Isaac) Rose 20 10

11 A Example Claim: 0 = 0 Let T ( ) = + T ( 1) > 0 Proof: Tha T() = Θ( 2 ) T() = + T( 1) = T( 2) = = ( 1) + T(0) = = = = +1 2 Tzachi (Isaac) Rose 21 Outlie Recurrece The Substitutio Method The Iteratio Method The Master Method Tzachi (Isaac) Rose 22 11

12 The Master Method Cookbook" k" method dfor solvig divide ad icoquer recurreces of the d form T() = at(/b) + f(), with a 1, b > 1, ad f() asymptotic positive. Tzachi (Isaac) Rose 23 The Master Theorem Tzachi (Isaac) Rose 24 12

13 The Master Theorem Let T() = at(/b) + f(), such that a 1, b>1 ad f() asymptotic ti positive Θ T( ) = Θ logb a ( ) logb a ( log) f ( ) = O f ( ) = Θ logb a ε ( ) logb a ( ) log ( f ( ) ) b a ε f ( ) = Ω( ) Θ + af ( / b) < cf ( ) ad for large ε > 0 c < 1 Tzachi (Isaac) Rose 25 Case 1 T() = 9T(/3) + Θ(). We have a = 9, b = 3, f () = c, ad log a = log 9 = 2 b 3 All the coditios of the first case are satisfied: a = 9 1, b = 3 >1, f() = c asymptotic positive log b a ε 2 ε ad c = O ( ) = O ( ) for all 0< є 1 Hece, T() = Θ( 2 ). Tzachi (Isaac) Rose 26 13

14 Case 1 T() = 5T(/2) + Θ( 2 ) We have a = 5, b = 2, f () = 2, ad logb a = log2 5 All the coditios of the first case are satisfied: a = 5 1, b = 2 >1, f() = c 2 asymptotic positive 2 log 5 ε ad c = O ( 2 ) for some є > 0 log 5 Hece, T ( ) = 2 Tzachi (Isaac) Rose 27 T() = T(2/3) + 1 Case 2 We have a = 1, b = 3/2, f () = 1, logb a = log3 / 2 1 = 0 All the coditios of the secod case are satisfied: a = 1 1, b = 3/2 >1, f() = 1 log asymptotic positive ad b a 1 = Θ ( ) = Θ (1) log a Hece, T ( ) = Θ b log = Θ log ( ) ( ) Tzachi (Isaac) Rose 28 14

15 Geeralized Case 2 Whe, for some k 0, Tzachi (Isaac) Rose 29 Case 2 T() = 27 T(/3) + Θ( 3 lg ) We have a = 27, b = 3, f () = 3 lg, log a = log3 27 = 3 b All the coditios of the secod case are satisfied: a = 27 1, b = 3 >1, f() = 3 lg 3 3 asymptotic positive ad c lg = Θ ( lg ) Geeral case 2 with k = 1, T ( ) = Θ 3 log 2 ( ) Tzachi (Isaac) Rose 30 15

16 Case 3 T() = 3T(/4) + Θ( lg) We have a = 3, b = 4, f() = c lg, ad logb a = log4 3 All the coditios of the secod case are satisfied: a = 3 1, b = 4 >1, f() = c lg asymptotic positive ad log lg ( 4 3+ e c = Ω ) for some є > 0, ad for sufficietly large, af(/b) =c3(/4)lg(/4) c(3/4) lg =df() for d = 3/4c Hece, T() = Θ(lg ). Tzachi (Isaac) Rose 31 Case 3 The regularity coditio af(/b) cf() for some costat c < 1 ad all sufficietly large is always holds wheever for costat є > 0. Tzachi (Isaac) Rose 32 16

17 Case 3 T() = 5T(/2) + Θ( 3 ) We have a = 5, b = 2, f () = 3, logb a = log2 5 Coditios of the secod case are satisfied: a = 5 1, b = 2 >1, f() = c 3 asymptotic 3 log2 5 + ε positive ad for some є > 0 c = Ω ( Hece, T() = Θ( 3 ). ) Tzachi (Isaac) Rose 33 No Case Not implies to T() = 2T(/2) + lg, Sice a = 2, b = 2, ad log b a = But lg is ot polyomial larger tha. f ( ) ( lg) lg log a = = b Tzachi (Isaac) Rose 34 17

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