Introduction to Algorithms Chapter 4 Recurrences. Solving Recurrences

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1 Itroductio to Algorithms Chpter 4 Recurreces 4 -- Solvig Recurreces A recurrece is equtio or iequlity tht descries fuctio i terms of itself y usig smller iputs The expressio: c T ( ) 2T + c 2 is recurrece. >

2 Solvig Recurreces Solvig Recurreces Exmples: Exmples: T() 2T(/2) + Θ() T() Θ( lg ) T() 2T(/2) + T() Θ( lg ) T() 2T(/2) T() Θ( lg ) Three methods for solvig recurreces Three methods for solvig recurreces Sustitutio method Itertio method Mster method Recurrece Exmples Recurrece Exmples > ) ( 0 ) ( T c T > ) ( c T c T > + ) ( c T c T > + 0 ) ( 0 0 ) ( T T

3 Sustitutio Method The sustitutio method mkig good guess method Guess the form of the swer, the use iductio to fid the costts d show tht solutio works Our gol: show tht T() 2T(/2) + O( lg ) Sustitutio Method T() ) 2T(/2) 2 + O( lg ) Thus, we eed to show tht T() c lg with pproprite choice of c Iductive hypothesis: ssume T(/2) c (/2) lg (/2) Sustitute ck ito recurrece to show tht T() c lg follows, whe c T() 2 T(/2) + 2 (c (/2) lg (/2)) + c lg(/2) + c lg c lg 2 + c lg c + c lg for c O( lg ) for c

4 Sustitutio Method Cosider T ( ) 2T ( ) + lg Simplify it y lettig m lg 2 m m T ( 2 ) m / 2 2T ( 2 ) + m Reme S(m) T(2 m ) S(m) 2S(m/2) + m O(m lg m) Chgig ck from S(m) to T(), we oti T() T(2 m ) S(m) O(m lg m) O( lg lg lg ) Itertio Method Itertio method: Expd the recurrece k times Work some lger to express s summtio Evlute the summtio

5 0 T ( ) c + T ( ) 0 > 0 T() c + T(-) c + c + T(-2) 2c + T(-2) 2c + c + T(-3) 3c + T(-3) 3) kc + T(-k) ck + T(-k) So fr for k we hve T() ck + T(-k) To stop the recursio, we should hve - k 0 k T() c + T(0) c Thus i geerl T() ) O() T ( ) + T ( ) 0 > 0 T() + T(-) T(-2) T(-3) T(-4) (-k+)( + T(-k) i T ( k) for k i k + + To stop the recursio, we should hve - k 0 k i + T (0) i i i + 2 T ( ) O( )

6 c T ( ) 2T + c 2 > T() 2 T(/2) + c 2(2 T(/2/2) + c) ) + c T(/2 ) + 2c 2 + c 2 2 (2 T(/2 /2) + c) ) + (2 2 -) )c T(/2 ) + 4c 4 + 3c3 2 3 T(/2 ) + (2 3 -) )c 2 3 (2 T(/2 /2) + c) ) + 7c T(/2 ) + (2 4 -) )c 2 k T(/2 ) + (2 k - )c k 4 -- c T ( ) 2T + c 2 > So fr for k we hve T() 2 k T(/2 k ) + (2 k -)c To stop the recursio, we should hve /2 k k lg T() 2 lg T(/2 lg ) + (2 lg -)c T(/) + ( -)c T() + (-)c c + (-)c c + c c 2c c c c/2 c Ο() for ll ½

7 c T ( ) T + c > T() ) T(/) ) + c (T(//) ) + c/) ) + c 2 2 T(/ 2 ) + c/ + c 2 T( / 2 ) + c(/ + ) 2 (T(/ /)) + c/ 2 ) + c(/ + ) 3 3 T( / 3 ) + c( 2 / 2 ) + c(/ + ) 3 T(/ 3 ) + c( 2 / 2 + / + ) k T(/ k ) + c( k- / k- + k- 2 / k / 2 + / + ) k c T ( ) T + c > So we hve T() k T(/ k ) + c( k- / k / 2 +/+) To stop the recursio, we should hve / k k k log T() k T() + c( k- / k / 2 + /+) k c + c( k- / k / 2 + / + ) c k + c( k- / k / 2 + / + ) c k / k + c( k- / k / 2 +/+) c( k / k / 2 + / + )

8 c T ( ) T + c > So with k log T() c( k / k / 2 + / + ) Wht if? T() c(+ +++) //k+ times c(k + ) c(log + ) Θ( log ) c T ( ) T + c > So with k log T() c( k / k / 2 + / + ) Wht if <? k k + Recll tht Σ(x k + x k- + + x + ) (x k+ -)/(x-) So: k k + L+ + k + ( ) ( ) T() c Θ() Θ() k + ( ) < ( )

9 So with k log c T ( ) T + c T() c( k / k / 2 + / + ) > Wht if >? k k k + ( ) + + L + + Θ k k ( ) T() c Θ( k / k ) c Θ( log / log ) c Θ( log / ) recll logrithm fct: log log c Θ( log / ) Θ(c log / ) Θ( log ) k ( ) ) So c T ( ) T + c > T ( ) Θ Θ( ) ( log ) Θ log ( ) < >

10 The Mster Theorem Give: divide d coquer lgorithm A lgorithm tht divides the prolem of size ito suprolems, ech of size / Let the cost of ech stge (i.e., the work to divide the prolem + comie solved suprolems) e descried y the fuctio f() The, the Mster Theorem gives us cookook for the lgorithm s ruig time: The Mster Theorem if T() T(/) + f() where & > the log log ε Θ( ) f ( ) O( ) log log ε > 0 T( ) Θ( lg) f ( ) Θ( ) c < log + Θ( f ( ) ) ε f ( ) Ω( )& f ( / ) < cf ( ) for lrge

11 Uderstdig Mster Theorem I ech of the three cses, we re comprig f() with log, the solutio to the recurrece is determied y the lrger of the two fuctios. I cse, if the fuctio log is the lrger, the the solutio T() ) Θ( log ). log ). I cse 3, if the fuctio f() ) is the lrger, the the solutio is T() ) Θ(f()). I cse 2, if the two fuctios re the sme size, the the solutio is T() ) Θ( log lg ) ) Θ(f() lg ) Uderstdig Mster Theorem log, I cse, ot oly must f() ) e smller th log it must e polyomilly smller. Tht is f() ) must e symptoticlly smller th log y fctor of ε for some costt ε > 0. I cse 3, ot oly must f() ) e lrger th log it must e polyomilly lrger d i dditio stisfy the regulrity coditio tht: f(/) c f(). log,

12 Uderstdig Mster Theorem It is importt to relize tht the three cses do ot cover ll the possiilities for f(). There is gp etwee cses d 2 whe f() ) is smller th log ut ot polyomilly smller. There is gp etwee cses 2 d 3 whe f() ) is lrger th log ut ot polyomilly lrger. If f() ) flls ito oe of these gps, or if the regulrity coditio i cse 3 fils to hold, the mster method cot e used to solve the recurrece Usig The Mster Method Cse T() 9T(/3) + 9, 3, f() log log 3 9 Θ( 2 ) Sice f() O( log ε ) O( ) O(.5 ) where ε0.5 cse pplies: T( ) Θ log log ε ( ) whe f ( ) O( ) Thus the solutio is T() Θ( 2 )

13 Usig The Mster Method Cse 2 T() T(2/3) +, 3/2, f() log log 3/2 0 Sice f() Θ( log ) Θ() cse 2 pplies: log log ( lg ) whe f ( ) ( ) T ( ) Θ Θ Thus the solutio is T() Θ(lg ) Usig The Mster Method Cse 3 T() 3T(/4) + lg 3, 4, f() lg log log Sice f() Ω( log 43+ε ) Ω( ) Ω() where ε 0.2, d for sufficietly lrge,. f(/) 3(/4) lg(/4) < (3/4) lg for c 3/4 cse 3 pplies: T ( ) Θ Thus the solutio is T() Θ( lg ) log +ε ( f ( ) ) whe f ( ) Ω( )

14 Whe the Mster Method does ot pply to recurrece T() ) 2T(/2) + lg 2, 2, f() lg, d log Note tht f() lg is symptoticlly lrger th log. The prolem is tht it is ot polyomilly lrger. The rtio f() / log ( lg ) / lg is symptoticlly less th ε for y positive costt ε. The recurrece flls ito the gp etwee cse 2 d cse