算法week4---algorithms-divide-conquer--part1(RSelect)

1. Randomized Selection
   1.1 R Select (array A, length n, order statistic i)
   (0): if n =1 return A[1]
   (1):choose pivot p from A uniformly at random.
   (2):partition A around p;
   let j = new index of p
   (3): if j=i, return p;
   (4): if j > i, return RSelect (1st part of A, j-1, i)
   (5):if j > i , return RSelect(1st part of A, j-1, i)
   (6):if j < i: return RSelect(2nd part of A, n-j, i-j)
2. Running Time of RSelect
   2.1 RSelect Theorem: for every input array of length n, the average running time of RSelect is $O(n)$.
   2.2 Tracking Progress via Phases.
   Note: RSelect uses $<= cn$ operations outside of the recursive call [for some constant c > 0]
   Notation: RSelect is in phase j if current array size between $(\frac{3}{4})^{j+1}n$ and $(\frac{3}{4})^j n$.
   $X_j =$ numbers of recursive calls during phase j.
   Note: running time of RSelect $<=\sum_{phase j} X_j \cdot c \cdot (\frac{3}{4})^j n$
   (1) $(\frac{3}{4})^jn$ : <= array size during phase $j$
   (2) $c\cdot (\frac{3}{4})^jn$: work per phase j subproblem.
   (3)$X_j$:of phase -j subproblems.
   2.3 Proof II: Reduction to Coin Flipping
   $X_j$ :# of recursive calls during phase $j$
   Note: if RSelect choose a pivot giving a $25\%-75\%$ spltit (or better): then current phase ends!(new subarray length at most 75% of old length).
   Recall: probability of $25\%-75\%$split or better is $50\%$.
   So: $E[x_j] <=$expected number of times you need to flip a fair coin to get one “head”.
   ($heads \approx good pivot, tails \approx bad pivot$)
   2.4 Proof III: Coin Flipping Analysis
   Let $N=$ number of coin flips until you get heads
   (a “geometric random varible”)
   Note:$E[N] = 1 + \frac{1}{\alpha} \cdot E[N]$
   $1$: 1st coin flip.
   $\frac{1}{\alpha}$: probablity of tails
   $E[N]:$ # of further coin flips needed in this case.
   Solution: $E[N] = \alpha$ (Recall $E[x_j] <= E[N]$)
   expected running time of RSelect:
   $$\begin{equation} \leq E[cn \sum_{phase_j}(\frac{3}{4})^jX_j] \\ \leq cn \sum_{phase j}(\frac{3}{4})^jE[X_j]\\ \leq 2 \cdot cn \sum_{phasej}(\frac{3}{4})^j\\ \leq 8 \cdot cn=O(n) \end{equation}$$
   其中： $E(of coin flips N)=2$
   $\sum_{phasej}(\frac{3}{4})^j \leq \frac{1}{1-\frac{3}{4}}=4$