Recurrence Examples.

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Presentation transcript:

Recurrence Examples

Recursion Tree for T(n)=3T(n/4)+(n2) cn2 cn2 c(n/4)2 T(n/16) (c) T(n/4) T(n/4) T(n/4) (a) (b) cn2 cn2 (3/16)cn2 c(n/4)2 c(n/4)2 c(n/4)2 log 4n (3/16)2cn2 c(n/16)2 c(n/16)2 c(n/16)2 c(n/16)2 c(n/16)2 c(n/16)2 c(n/16)2 c(n/16)2 c(n/16)2 (nlog 43) T(1) T(1) T(1) T(1) T(1) T(1) 3log4n= nlog 43 Total O(n2) (d)

Solution to T(n)=3T(n/4)+(n2) The height is log 4n, #leaf nodes = 3log 4n= nlog 43. Leaf node cost: T(1). Total cost T(n)=cn2+(3/16) cn2+(3/16)2 cn2+  +(3/16)log 4n-1 cn2+ (nlog 43) =(1+3/16+(3/16)2+  +(3/16)log 4n-1) cn2 + (nlog 43) <(1+3/16+(3/16)2+  +(3/16)m+ ) cn2 + (nlog 43) =(1/(1-3/16)) cn2 + (nlog 43) =16/13cn2 + (nlog 43) =O(n2).

Recursion Tree of T(n)=T(n/3)+ T(2n/3)+O(n)

Recursion tree for T(n)=aT(n/b)+f(n)

General Divide-and-Conquer Recurrence T(n) = aT(n/b) + f (n) where f(n)  (nd), d  0 Master Theorem: If a < bd, T(n)  (nd) If a = bd, T(n)  (nd log n) If a > bd, T(n)  (nlog b a ) Examples: T(n) = T(n/2) + n  T(n)  (n ) Here a = 1, b = 2, d = 1, a < bd T(n) = 2T(n/2) + 1  T(n)  (n log 2 2 ) = (n ) Here a = 2, b = 2, d = 0, a > bd

Examples T(n) = T(n/2) + 1  T(n)  (log(n) ) Here a = 1, b = 2, d = 0, a = bd T(n) = 4T(n/2) + n  T(n)  (n log 2 4 ) = (n 2 ) Here a = 4, b = 2, d = 1, a > bd T(n) = 4T(n/2) + n2  T(n)  (n2 log n) Here a = 4, b = 2, d = 2, a = bd T(n) = 4T(n/2) + n3  T(n)  (n3) Here a = 4, b = 2, d = 3, a < bd