**Unformatted text preview: **A- { x } = { 1 , 3 , 4 } , y = 4, Pr ( y ≥ x ) = 1 Case 3: x = 3, A- { x } = { 1 , 2 , 4 } , y = 4, Pr ( y ≥ x ) = 1 Case 4: x = 4, A- { x } = { 1 , 2 , 3 } , y = 3, Pr ( y ≥ x ) = 0 So, we know that Pr ( y ≥ x ) = 0 if x is maximal among all n numbers, otherwise Pr ( y ≥ x ) = 1. Also, Pr ( y < x ) = 1-Pr ( y ≥ x ), Thus, we have ∑ ∀ x [ Pr ( y ≥ x )(1) + Pr ( y < x )( n )] = 1 + 1 + 1 .... + 1 + n = ( n-1)(1) + n = 2 n-1 T ( n ) = T ( n-1) + 1 n ∑ ∀ x [ Pr ( y ≥ x )(1) + Pr ( y < x )( n )] T ( n ) = T ( n-1) + 1 n (2 n-1) T ( n ) < T ( n-1) + 2 T ( n ) = O ( n ) c ± 2006 Chung Kai Lun Peter. Comments are welcomed. Email: [email protected] 1...

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- Spring '10
- may
- English, Analysis of algorithms, Computational complexity theory, Best, worst and average case, Chung Kai Lun Peter, RandM ax