T(m)=(1+)T(m-1)+2 for n>1 with T(1)=0 m 1n+1 xn=1+= →xnxn-1·…x1=n+1 n Tm=Tn-业+ 2 for n 1 n+1 n n+1 4口，1①，43，t夏，30Q0 Hengfeng Wei (hfweixinju.edu.cn) 2-5 Linear Recurrences farch26.20207/26
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T(n) = (1 + 1 n )T(n − 1) + 2 for n > 1 with T(1) = 0 xn = 1 + 1 n = n + 1 n =⇒ xnxn−1 · · · x1 = n + 1 T(n) n + 1 = T(n − 1) n + 2 n + 1 for n > 1 T(n) n + 1 = T(1) 2 + 2 X 3≤k≤n+1 1 k T(n) = 2(n + 1)(Hn+1 − 3 2 ) ≈ 2n ln n − 1.846n Hengfeng Wei (hfwei@nju.edu.cn) 2-5 Linear Recurrences March 26, 2020 7 / 26
T(m)=(1+)T(m-1)+2 for n>1 with T(1)=0 m xn=1+ 1n+1 →xnxn-1·…x1=n+1 T(n) T(n-1) .2 for n>1 n+1 m n+1 T(n)_T(1) n+1 2 2 ≤k≤n+1 4口·￥①，43，t夏，里Q0 Hengfeng Wei (hfweixinju.edu.cn) 2-5 Linear Recurrences farch26.20207/26
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T(n) = (1 + 1 n )T(n − 1) + 2 for n > 1 with T(1) = 0 xn = 1 + 1 n = n + 1 n =⇒ xnxn−1 · · · x1 = n + 1 T(n) n + 1 = T(n − 1) n + 2 n + 1 for n > 1 T(n) n + 1 = T(1) 2 + 2 X 3≤k≤n+1 1 k T(n) = 2(n + 1)(Hn+1 − 3 2 ) ≈ 2n ln n − 1.846n Hengfeng Wei (hfwei@nju.edu.cn) 2-5 Linear Recurrences March 26, 2020 7 / 26
T(m)=(1+)T(m-1)+2 for n>1 with T(1)=0 m 1n+1 xn=1+ →xnxn-1·…x1=n+1 T(n) T(n-1) 2 十 for n 1 n+1 n+1 T(n)T(1) n+1 +2 3≤k≤n+1 T(m)=2(n+1)(Hn+1- 3 ≈2mnn-1.846m 4口·￥①，43，t夏，里Q0 Hengfeng Wei (hfweixinju.edu.cn) 2-5 Linear Recurrences March26,20207/26
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T(n) = (1 + 1 n )T(n − 1) + 2 for n > 1 with T(1) = 0 xn = 1 + 1 n = n + 1 n =⇒ xnxn−1 · · · x1 = n + 1 T(n) n + 1 = T(n − 1) n + 2 n + 1 for n > 1 T(n) n + 1 = T(1) 2 + 2 X 3≤k≤n+1 1 k T(n) = 2(n + 1)(Hn+1 − 3 2 ) ≈ 2n ln n − 1.846n Hengfeng Wei (hfwei@nju.edu.cn) 2-5 Linear Recurrences March 26, 2020 7 / 26
T(n)=(1+)T(n-1)+2 for n>1 with T(1)=0 Tm)=2n+1(Hn+1- ≈2nnn-1.846m 4口，1①，43，t夏，里0Q0 Hengfeng Wei (hfweixinju.edu.cn) 2-5 Linear Recurrences farch26.20208/26
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T(n) = (1 + 1 n )T(n − 1) + 2 for n > 1 with T(1) = 0 T(n) = 2(n + 1)(Hn+1 − 3 2 ) ≈ 2n ln n − 1.846n T(n) = (n + 1) + 1 n X 1≤i≤n T(i − 1) + T(n − i) for n > 1 with T(0) = T(1) = 0 average number of comparisons of Quicksort Hengfeng Wei (hfwei@nju.edu.cn) 2-5 Linear Recurrences March 26, 2020 8 / 26
T(n)=(1+)T(n-1)+2forn>1 with T(1)=0 T(m)=2(n+1)(H+1- 3 ≈2nnn-1.846m Tm=m+1)+2∑(r6-1)+Tn-） 1≤i≤n for n >1 with T(0)=T(1)=0 average number of comparisons of QUICKSORT 4口，1①，43，t夏，里0Q0 Hengfeng Wei (hfweixinju.edu.cn) 2-5 Linear Recurrences March26,20208/26
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T(n) = (1 + 1 n )T(n − 1) + 2 for n > 1 with T(1) = 0 T(n) = 2(n + 1)(Hn+1 − 3 2 ) ≈ 2n ln n − 1.846n T(n) = (n + 1) + 1 n X 1≤i≤n T(i − 1) + T(n − i) for n > 1 with T(0) = T(1) = 0 average number of comparisons of Quicksort Hengfeng Wei (hfwei@nju.edu.cn) 2-5 Linear Recurrences March 26, 2020 8 / 26