Recursion-tree method

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Presentation transcript:

Recursion-tree method A recursion tree models the costs (time) of a recursive execution of an algorithm. The recursion-tree method can be unreliable. The recursion-tree method promotes intuition, however.

Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n2:

Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n2: T(n)

Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n2: n2 T(n/4) T(n/2)

Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n2: n2 (n/4)2 (n/2)2 T(n/16) T(n/8) T(n/8) T(n/4)

Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n2: n2 (n/4)2 (n/2)2 (n/16)2 (n/8)2 (n/8)2 (n/4)2 … Q(1)

Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n2: n2 (n/4)2 (n/2)2 (n/16)2 (n/8)2 (n/8)2 (n/4)2 … Q(1)

Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n2: n2 (n/4)2 (n/2)2 (n/16)2 (n/8)2 (n/8)2 (n/4)2 … Q(1)

Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n2: n2 (n/4)2 (n/2)2 (n/16)2 (n/8)2 (n/8)2 (n/4)2 … … Q(1)

Example of recursion tree Solve T(n) = T(n/4) + T(n/2) + n2: n2 (n/4)2 (n/2)2 (n/16)2 (n/8)2 (n/8)2 (n/4)2 … … Q(1) Total = = Q(n2) geometric series

Appendix: geometric series for x ¹ 1 for |x| < 1

The master method The master method applies to recurrences of the form T(n) = a T(n/b) + f (n) , where a ³ 1, b > 1, and f(n)  (nd), d  0

Idea of master theorem Recursion tree: f (n) f (n) a f (n/b) … h = logbn a … f (n/b) f (n/b) f (n/b) a … f (n/b2) f (n/b2) f (n/b2) … #leaves = ah = alogbn = nlogba nlogbaT (1) T (1)

Three common cases Compare a with bd: If a > bd, logba > d nlogba > nd We also know that f(n)  (nd) f (n) grows polynomially slower than nlogba. Solution: T(n) = Q(nlogba) .

Idea of master theorem Recursion tree: f (n) f (n) a … f (n/b) f (n/b) h = logbn … a2 f (n/b2) f (n/b2) f (n/b2) f (n/b2) … … CASE 1: The weight increases geometrically from the root to the leaves. The leaves hold a constant fraction of the total weight. nlogbaT (1) T (1) Q(nlogba)

Three common cases Compare a with bd: If a = bd, logba = d Nlogba = nd We also know that f(n)  (nd) f (n) and nlogba grow at similar rates. Solution: T(n) = Q( f(n) log n) .

Idea of master theorem Recursion tree: f (n) f (n) a … f (n/b) f (n/b) h = logbn … a2 f (n/b2) f (n/b2) f (n/b2) f (n/b2) … … CASE 2: The weight is approximately the same on each of the logbn levels. nlogbaT (1) T (1) Q(nlogbalg n)

Three common cases (cont.) Compare a with bd: If a < bd, logba < d nlogba < nd We also know that f(n)  (nd) f (n) grows polynomially faster than nlogba. Solution: T(n) = Q( f (n) ) = Q(nd) .

Idea of master theorem Recursion tree: f (n) f (n) a … f (n/b) f (n/b) h = logbn … a2 f (n/b2) f (n/b2) f (n/b2) f (n/b2) … … CASE 3: The weight decreases geometrically from the root to the leaves. The root holds a constant fraction of the total weight. nlogbaT (1) T (1) Q( f (n))

Examples Ex. T(n) = 4T(n/2) + n a = 4, b = 2, d = 1  nlogba = n2; f (n) = n. (a=4) >( bd =21) CASE 1:  T(n) = Q(n2). Ex. T(n) = 4T(n/2) + n2 a = 4, b = 2, d=2  nlogba = n2; f (n) = n2. (a =4) = (bd=22) CASE 2:  T(n) = Q(n2log n).

Examples Ex. T(n) = 4T(n/2) + n3 a = 4, b = 2, d=3  nlogba = n2; f (n) = n3. (a=4) < (bd = 23) CASE 3:  T(n) = Q(nd) = Q(n3).