CS 253: Algorithms. Chapter 4. Divide-and-Conquer Recurrences Master Theorem. Credit: Dr. George Bebis

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1 CS 5: Algorithms Chapter 4 Divide-ad-Coquer Recurreces Master Theorem Credit: Dr. George Bebis

2 Recurreces ad Ruig Time Recurreces arise whe a algorithm cotais recursive calls to itself Ruig time is represeted by a equatio or iequality that describes a fuctio i terms of its value o smaller iputs. T() = T(-) + What is the actual ruig time of the algorithm? i.e. T() =? Need to solve the recurrece Fid a explicit formula of the expressio Boud the recurrece by a expressio that ivolves

3 Example Recurreces T() = T(-) + Θ( ) Recursive algorithm that loops through the iput to elimiate oe item T() = T(/) + c Θ(lg) Recursive algorithm that halves the iput i oe step T() = T(/) + Θ() Recursive algorithm that halves the iput but must examie every item i the iput T() = T(/) + Θ() Recursive algorithm that splits the iput ito halves ad does a costat amout of other work

4 BINARY-SEARCH Fids if x is i the sorted array A[lo hi] Alg.: BINARY-SEARCH (A, lo, hi, x) if (lo > hi) retur FALSE mid (lo+hi)/ if x = A[mid] retur TRUE if ( x < A[mid] ) lo 5 BINARY-SEARCH (A, lo, mid-, x) if ( x > A[mid] ) BINARY-SEARCH (A, mid+, hi, x) mid 0 hi

5 Example A[8] = {,,, 4, 5, 7, 9, } lo = hi = 8 x = mid = 4, lo = 5, hi = mid = 6, A[mid] = x Foud!

6 Example I A[8] = {,,, 4, 5, 7, 9, } lo = hi = 8 x = low high lo =, hi = 8, mid = 4. A[4]= lo = 5, hi = 8, mid = 6, A[6]=7 low high lo = 5, hi = 5, mid = 5, A[5]= lo = 6, hi = 5 NOT FOUND! high low

7 Aalysis of BINARY-SEARCH Alg.: BINARY-SEARCH (A, lo, hi, x) if (lo > hi) retur FALSE mid (lo+hi)/ if x = A[mid] retur TRUE if ( x < A[mid] ) BINARY-SEARCH (A, lo, mid-, x) if ( x > A[mid] ) BINARY-SEARCH (A, mid+, hi, x) costat time: c costat time: c costat time: c same problem of size / same problem of size / T() = c + T(/)

8 Methods for Solvig Recurreces Iteratio method Recursio-tree method Master method 8

9 The Iteratio Method Covert the recurrece ito a summatio ad solve it usig a kow series Example: T() = c + T(/) T() = c + T(/) = c + c + T(/4) = c + c + c + T(/8) = c + c + c + c + T(/ 4 ) Assume = k the k = lg ad T() = c + c + c + c + c + + T(/ k ) (k times) T() = k * c + T() T() = clg

10 Iteratio Method Example T() = + T(/) Assume = k k = lg T() = + T(/) = + (/ + T(/4)) = + + 4T(/4) = + + 4(/4 + T(/8)) = T(/8) T() = + T(/ ) = k + k T(/ k ) = lg + T() T() = O(lg)

11 Methods for Solvig Recurreces Iteratio method Recursio-tree method Master method

12 The recursio-tree method Covert the recurrece ito a tree: Each ode represets the cost icurred at various levels of recursio Sum up the costs of all levels Used to guess a solutio for the recurrece

13 Example W() = W(/) + Subproblem size at level i = / i At level i: Cost of each ode = (/ i ) # of odes = i Total cost = ( / i ) h = Height of the tree / h = h = lg Total cost at all levels: W() = O( ) 0 lg 0 lg 0 ) ( W i i i i i i

14 Example T() = T(/4) + c Subproblem size at level i = /4 i At level i: Cost of each ode= c(/4 i ) # of odes= i Total cost =c (/6) i h = Height of the tree /4 h = h = log 4 Total cost at all levels: T( ) log 4 i0 6 T() = O( ) i c (last level has log 4 = log 4 odes) i log 4 log4 log4 c c O( ) i0 6 6

15 Example W() = W(/) + W(/) + The logest path from the root to a leaf: (/) (/) (/) i = i = log / Cost of the problem at level i = Total cost: lg W ( ) (log / ) O( lg ) lg( / ) via further aalysis W() =Θ(lg) 5

16 Methods for Solvig Recurreces Iteratio method Recursio-tree method Master method 6

17 Master Theorem Cookbook for solvig recurreces of the form: Idea: compare f() with log b a f() is asymptotically smaller or larger tha log b a by a polyomial factor OR T b ( ) at f ( ) where a, b,ad f() 0 f() is asymptotically equal with log b a

18 Master Theorem T b ( ) at f ( ) where a, b,ad f() 0 Case : if f() = O( log b a - ) for some > 0, the: T() = ( log b a ) Case : if f() = ( log b a ), the: T() = ( log b a lg) Case : if f() = ( log b a + ) for some > 0, ad if af(/b) cf() for some c < ad all sufficietly large, the: regularity coditio T() = (f()) 8

19 Example Case : if f() = O( log b a - ) for some > 0, the: T() = ( log b a ) Case : if f() = ( log b a ), the: T() = ( log b a lg) Case : if f() = ( log b a + ) for some > 0, ad if af(/b) cf() for some c < ad all sufficietly large, the: T() =(f()) T() = T(/) + a =, b =, log = Compare log b a = with f() = f() = ( log b a = ) Case T() = ( log b a lg) = (lg) 9

20 Example Case : if f() = O( log b a - ) for some > 0, the: T() = ( log b a ) Case : if f() = ( log b a ), the: T() = ( log b a lg) Case : if f() = ( log b a + ) for some > 0, ad if af(/b) cf() for some c < ad all sufficietly large, the: T() =(f()) a =, b =, log = T() = T(/) + Compare log = with f() = f() = ( log + ) Case (* eed to verify regularity cod.) a f(/b) c f() /4 c (½ c <) T() = (f()) = ( )

21 Example Case : if f() = O( log b a - ) for some > 0, the: T() = ( log b a ) Case : if f() = ( log b a ), the: T() = ( log b a lg) Case : if f() = ( log b a + ) for some > 0, ad if af(/b) cf() for some c < ad all sufficietly large, the: T() =(f()) T() = T(/) + a =, b =, log = Compare with f() = / f() = O( - ) Case T() = ( log b a ) = ()

22 Example 4 Case : if f() = O( log b a - ) for some > 0, the: T() = ( log b a ) Case : if f() = ( log b a ), the: T() = ( log b a lg) Case : if f() = ( log b a + ) for some > 0, ad if T() = T(/4) + lg a =, b = 4, log 4 = 0.79 Compare 0.79 with f() = lg f() = ( log 4 + ) Case af(/b) cf() for some c < ad all sufficietly large, the: T() =(f()) Check regularity coditio: (/4)lg(/4) (/4)lg = c f(), (/4 c < ) T() = (lg)

23 **here Example 5 Case : if f() = O(log b a - ) for some > 0, the: T() = ( log b a ) Case : if f() = ( log b a ), the: T() = ( log b a lg) Case : if f() = ( log b a + ) for some > 0, ad if af(/b) cf() for some c < ad all sufficietly large, the: T() =(f()) T() = T(/) + lg a =, b =, log = Compare with f() = lg Is this a cadidate for Case? Case : if f() = ( log b a + ) for some > 0 I other words, If lg. for some > 0 lg is asymptotically less tha for ay > 0!! Therefore, this i ot Case! (somewhere betwee Case & )

24 Exercise: T() = T(/) + lg Use Iteratio Method to show that T() = lg T() = lg + T(/) = lg + (/*lg(/) + T(/4)) = lg + lg(/) + T(/ ) =.

25 Chagig variables T() = T( ) + lg Try to trasform the recurrece to oe that you have see before Reame: m = lg = m T ( m ) = T( m/ ) + m Reame: k=m ad S(k) = T( k ) S(k) = S(k/) + k S(k) = (k*lgk) T() = T( m ) = S(m) = (mlgm)= (lg*lglg) **See Page 86 of the Textbook. 5

26 Exercise: T() = T( ) + lg Use Iteratio Method to show that T() = (lg*lglg) T() = lg + T( / ) = lg + (lg( / ) + T( /4 )) = lg + lg + T( /4 ) = lg + lg + + k T() k=lglg 6

27 Appedix A Summatios ) ( k k 6 ) )( ( k k 4 ) ( k k x x x x x x x x x k k k k the lim If 0 0 () l Harmoic Series O k H k

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