+=== 1. ===
  * $ T_a(n) = 2T_a(n-1) + 1$

** nu se poate aplica master **
{{ :aa:lab:sol:10:3_1.jpg?600 |}}


  * $ T_b(n) = T_b(n-1) + 1$

** nu se poate aplica master **\\
Prin metoda arborelui obtinem n niveluri continand cate o constanta si un nod, deci $ T_b(n) = n * \Theta(1) = \Theta(n) $.


  * $ T_c(n) = 2T_c(n/2) + \log(n)$
** Metoda substitutiei** \\
Vrem sa demonstram ca $ T_c(n) = \Theta(n), \ presupunem \ T_c(1) = 1$ \\
$ <=> \exists c_1, c_2 > 0 $ a.i. $ c_1 n  \le T_c(n) \le c_2 n$\\
$ Ca \ sa \ demonstram \ ca \ T_c(n) = \Theta(n), \ vom \ demonstra \ ca \ T_c(n) = O(n) \ (1) \ si \ T_c(n) = \Omega(n) \ (2)$\\
(2) Caz de baza: n = 16 \\
$ c_1 * 16  \le T_c(16)$\\
$ c_1 * 16  \le 42 $ \\
$ => c_1 \le \frac{21}{8} $\\
$ (2) \ Pas \ inductiv $\\
$ Presupunem \ ca \ T_c(n/2) = \Omega(n/2) $\\
$ <=> \ \exists c_1 > 0 \ a.i. \ c_1 * \frac{n}{2} \le T_c(n/2) $\\
$ <=> c_1 * n \le 2 * T_c(n/2) $\\
$ <=> c_1 * n + \log n\le T_c(n) $\\
$ <=> c_1 * n \le c_1 * n + \log n\le T_c(n) $\\
$ => \ T_c(n) = \Omega(n) $\\
$ (1) Pentru \ usurinta, \ vom \ demonstra \ restrictia \ T_c(n) \le c_2 n - 2 \log n $\\
(1) Caz de baza: n = 16 \\
$ T_c(16) \le c_2 * 16 - 2 * \log 16 $\\
$ 42 \le c_2 * 16 - 8 $ \\
$ => c_2 \ge \frac{25}{8} $\\
$ (1) \ Pas \ inductiv $\\
$ Presupunem \ ca \ T_c(n/2) \le c_2 * \frac{n}{2} - 2 * \log \frac{n}{2} $\\
$ <=> 2 * T_c(n/2) \le c_2 * n - 4 * \log \frac{n}{2} $\\
$ <=> T_c(n) \le c_2 * n - 3 * \log n + 4$\\
$ <=> T_c(n) \le c_2 * n - 3 * \log n + 4 \le c_2 * n - 2 * \log n, \forall n \ge 16 $\\
$ => \ T_c(n) = O(n) $\\
$ T_c(n) \in \Theta(n) $\\
** Teorema Master  ** \\
$ a = 2, \ b = 2 => d = 1 $ \\
$ f(n) = \log n => \ f(n) = O(\sqrt n) => c = \frac{1}{2} \ (c < d) $\\
$ => T_c(n) \in \Theta(n) $\\
  * $ T_d(n) = T_d(n/9) + T_d(8n/9) + n$

** nu se poate aplica master **

{{ :aa:lab:sol:10:3_4.jpg?600 |}}

Desenand arborele, obtinem unele ramuri care se termina mai repede, altele mai incet, asa ca vom incadra timpul lui T intre 2 limite, $ S_1 $ si $ S_2 $

$ h_1 = \log_9n,  h_2 = \log_{\frac 9 8}n $\\

$ S_1 = \sum_0^{h_1-1} n + T(1) * 2 ^ {h_1} = n * h_1 +2 ^ {h_1} = n*\log_9n + 2^{\log_9n} \in \Theta(n\log n) $\\
$ S_2 = \sum_0^{h_2-1} n + T(1) * 1 = n * h_2 + 1 = n*\log_{\frac 9 8} n + 1 \in \Theta(n\log n) $\\

=> $ T(n) \in \Theta(n \log n)$


  * $ T_e(n) = T_e(2n/3) + 1$
Cazul 2 al Teoremei Master cu $ c = \log_{frac 3 2}1 = 0, k = 0 $ deci $ T_e(n) = \Theta(n^0 * \log^{0+1}n) \Theta(\log n) $
  * $ T_{Strassen}(n) = 7T_{Strassen}(n/2) + n^2 $

$ c = \log_27 \approx 2.81 $ => cazul 1 al teoremei master, deci $ T_{Strassen}(n) = \Theta(n ^ {\log_27}) \approx  \Theta(n ^ {2.81}) $
  * $ T_{Karatsuba}(n) = 3T_{Karatsuba}(n/2) + 1 $
$ c = \log_23 \approx 1.58 $ => cazul 1 al teoremei master, deci $ T_{Karatsuba}(n) = \Theta(n ^ {\log_23}) \approx  \Theta(n ^ {1.58}) $
  * $ T_{Quicksort}(n) = T_{Quicksort}(n-1) + O(n) $ 

** nu se poate aplica master **\\
Folosind metoda arborelui, fiecare nivel $ h $ va contine nodul $ T(n-h) $ si costul $ n-h $. Obtinem formula $ T_{Quicksort}(n) = \sum_{i=0}^{n-2} (n-i) + 1 * T(1) = \frac {n(n+1)} 2 \in \Theta(n^2) $ \\
  * $ T(n) = T(n/4) + T(3n/4) + n $
** nu se poate aplica master ** \\
Desenand arborele, obtinem unele ramuri care se termina mai repede, altele mai incet, asa ca vom incadra timpul lui T intre 2 limite, $ S_1 $ si $ S_2 $

$ h_1 = \log_4n,  h_2 = \log_{\frac 4 3}n $\\

$ S_1 = \sum_0^{h_1-1} n + T(1) * 2 ^ {h_1} = n * h_1 +2 ^ {h_1} = n*\log_4n + 2^{\log_4n} \in \Theta(n\log n) $\\
$ S_2 = \sum_0^{h_2-1} n + T(1) * 1 = n * h_2 + 1 = n*\log_{\frac 4 3} n + 1 \in \Theta(n\log n) $\\

=> $ T(n) \in \Theta(n \log n)$
=== 2. ===
    * $ T_a(n) = T_a(\sqrt{n}) + 1 $
      Notam \( n = 2^k \) \\
=> $ T_a(2^k) = T_a(\sqrt{2^k}) + 1$ \\
=> $ T_a(2^k) = T_a(2^ \frac{k}{2}) + 1$ \\
$ fie \ S(k) = T_a(2^k) $ \\
$ => S(k) = S(k/2) + 1 $ \\
**Teorema Master** \\
a = 1, b = 2 => d = 0 \\
c = d  = 0 (k = 0) =>  $ S(k) = \Theta(\log k) $ \\
=> $ T_a(2^k) = \Theta(\log k) $ \\
=> $ T_a(n) = \Theta(\log \log n) $ \\

    * $ T_b(n) = 2 * T_b(\sqrt{n}) + \log n $
      Notam \( n = 2^k \) \\
=> $ T_b(2^k) = 2 * T_b(\sqrt{2^k}) + k$ \\
=> $ T_b(2^k) = 2 * T_b(2^ \frac{k}{2}) + k $ \\
$ fie \ S(k) = T_b(2^k) $ \\
$ => S(k) = 2 * S(k/2) + k $ \\
**Teorema Master** \\
a = 2, b = 2 => d = 1 \\
c = d  = 1 (k = 0) =>  $ S(k) = \Theta(k \log k) $ \\
=> $ T_b(2^k) = \Theta(k \log k) $ \\
=> $ T_b(n) = \Theta(\log n * \log \log n) $ \\

    * $ T_c(n) = \sqrt{n} * T_c(\sqrt{n}) + n $
      Notam \( n = 2^k \) \\
=> $ T_c(2^k) = \sqrt{2^k} * T_c(\sqrt{2^k}) + 2^k$ \\
=> $ T_c (2^k) = \sqrt{2^k} * T_c(2^ \frac{k}{2}) + 2^k $ \\
$ fie \ S(k) = T_c(2^k) $ \\
$ => S(k) = \sqrt{2^k}  * S(k/2) + 2^k $ \\
$ fie \ S(k) = 2^k * P(k) $ \\
$ => \ 2^k * P(k) = \sqrt{2^k} * \sqrt{2^k} * P(k/2) + 2^k $ \\
$ => \ P(k) = P(k/2) + 1 $ \\
**Teorema Master** \\
a = 1, b = 2 => d = 0 \\
c = d = 0 (k = 0) =>  $ P(k) = \Theta(\log k) $ \\
=> $ S(k) = \Theta(2^k \log k) $ \\
=> $ T_c(2^k) = \Theta(2^k \log k) $ \\
=> $ T_c(n) = \Theta(n * \log \log n) $ \\