This is related to the discussion of average case of quick sort. Given that we have a sorted sub-array $Z_{ij} = i, i+1, \dots, j$ where $i < j$.

Claim: $i$ and $j$ are compared if and only if, among all elements in $Z_{ij}$, the first element to be picked as a pivot is either $i $ or $ j$.

Forward direction Proof: The “only if”: suppose the first one picked as pivot is some $k $ that is between $i$ and $j$, then $i $ and $ j$ will be separated into different partitions and will never meet each other.

Problem 1: For backward direction above, I understand that it should be that if the first element to be picked as a pivot is either $i $ or $ j$, then $i$ and $j$ are compared. So I am not sure why the "only if" part was stated as above please?

Backward direction Proof: The “if”: if $i$ is chosen as pivot (the first one among $Z_{ij}$), then $ j$ will be compared to pivot $i$ for sure, because nobody could have possibly separated them yet!

Then we can proceed and define $X$ to be the total number of comparisons and so on until we have $O(n\log{n})$ complexity, which is clear once the above claim is understood, which I have some issues with please.

If $i$ that is chosen as a pivot guarantees based on the proof that we will get $O(n\log n)$ complexity based on the condition that $i$ is the first element to be a pivot and $j$ to be after $i$, then we can see that $E[X_{ij}] = \Pr(i \text{ or } j)$, where $\Pr$ back means the probability that either $i$ or $j$ to be the first pivot and both at the beginning and at the end of the array.

$$E[X] = \Sigma_{i}\Sigma_{j=i+1}E[X_{ij}] = \frac{2}{j-1+1}$$

Claim 2: Now in the randomized case, we can assume a less clever assumption (which I don't understand) that: Given that $n$ is the number of elements to be sorted and $T(n)$ is expected time to sort $n$ elements. First pivot chooses $i$th smallest element, all equally likely. Then:

$$T(n) = (n-1) + \frac 1n \sum_{i=0}^{n-1}(T(i)+T(n-i-1))$$ $$= (n-1) + \frac 2n \sum_{i=1}^{n-1}(T(i))$$

Problem 2: how the previous claim is less clever than the first one please? And how $T(n)$ above is formulated?

Problem 3: how $T(n)$ becomes $T(n)= (n-1) + \frac 2n \sum_{i=1}^{n-1}(T(i))$

  • $\begingroup$ Problem 2 is unanswerable – which claim is more clever is completely subjective. $\endgroup$ Oct 2, 2021 at 10:57
  • 1
    $\begingroup$ Problem 3 you can solve on your own. Try harder. $\endgroup$ Oct 2, 2021 at 10:57

1 Answer 1


The claim is proved as follows (in reverse order):

  • If the first element among $Z_{ij}$ to be chosen as pivot is $i$ or $j$, then $i$ and $j$ are compared.
  • If the first element among $Z_{ij}$ to be chosen as pivot is not $i$ or $j$, then $i$ and $j$ are not compared.

The forward direction is the contrapositive of the "only if" implication.

Being clever or not is subjective rather than technical. I would just ignore this designation.

Perhaps unrolling the sums would help you prove the identity:

$$ T(0) + T(n-1) + T(1) + T(n-2) + \cdots + T(n-1) + T(0) = 2(T(0) + T(1) + \cdots + T(n-1)). $$

  • $\begingroup$ Thank you very much. I see that $T(0) + T(n-1), ...T(n-1)+T(0)$, so have 2 of each and thus we got 2 multiplied by sum please? If this is true, so this "less clever" approach is basically assumin pivot can be any element and thus $T(i) + T(n-i-1)$ please? $\endgroup$
    – Avv
    Oct 2, 2021 at 16:36

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