The discussion centers on understanding the proof of quick-sort's expected running time of O(n log n). The algorithm involves selecting a random pivot and recursively sorting the partitions. The expected time T(n) is derived using a recurrence relation that accounts for the time complexity of partitioning and the recursive calls. The participants explore how to establish that T(n) remains bounded by k n log n through mathematical induction. The conversation emphasizes the importance of the recurrence relation and the need for a structured approach to complete the proof.
Quicksort(S):
1. if S contains at most one element then return S
else
2. choose an element a randomly from S;
3. let S1, S2, and S3 be the sequences of elements in S less than, equal to, and greater than a, respectively;
4. return Quicksort(S1) followed by S2 followed by Quicksort(S3)
I have tried the following:
Let T(n) be the expected time required by the Quicksort algorithm in order to sort a sequence of n elements. We have that T(0)=T(1)=b for a constant b.
We suppose that the elements a is the i-th smallest elements of the n elements of the sequence S.
Then the 2 recursive calls of Quicksort have expected time T(i-1) and T(n-i), respectively.
Since the lines 2-3 require time O(n) and the line 1 requires time O(1), we have the following recurrence relation:
T(n) \leq cn+ \frac{1}{n} \sum_{i=1}^n \left[ T(i-1)+T(n-i)\right], n \geq 2
We will show that for n \geq 2, T(n) \leq k n \log n, for k=2(c+b) positive.
For n=2: From the relation (1) we have T(2) \leq c2+ 2b=2(c+b) \checkmark
We suppose that T(n) \leq k n \log n γfor some n \geq 2.
We will show that T(n+1) \leq k (n+1) \log (n+1) .
From the relation (1) we have T(n+1) \leq c(n+1)+ \frac{2}{n+1} \sum_{i=0}^{n} T(i) \leq c(n+1)+ \frac{2}{n+1} (n+1) T(n)= c(n+1)+2T(n) \leq c(n+1)+2 k n \log n \leq (n+1) (c+2k \log n)How could we continue? Or should we use strong induction?
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