As the title indicates, I want to test whether an $ n \times n$ matrix (numeric, symbolic,..) is diagonal and/or proportional to the $ n \times n$ identity matrix. I, of course, can test whether the $ n^2n$ individual offdiagonal entries are zero–but that’s, it would seem, is inefficient.
Tag: quick
Analysing worst case time complexity of quick sort for various cases
I am trying to understand worst case time complexity of quick sort for various pivots. Here is what I came across:

When array is already sorted in either ascending order or descending order and we select either leftmost or rightmost element as pivot, then it results in worst case $ O(n^2)$ time complexity.

When array is not already sorted and we select random element as pivot, then it gives worst case “expected” time complexity as $ O(n log n)$ . But worst case time complexity is still $ O(n^2)$ . ^{[1]}

When we select median of ^{[2]} first, last and middle element as pivot, then it results in worst case time complexity of $ O(n log n)$ ^{[1]}
I have following doubts
D1. Link 2 says, if all elements in array are same then both random pivot and median pivot will lead to $ O(n^2)$ time complexity. However link 1 says median pivot yields $ O(n log n)$ worst case time complexity. What is correct?
D2. How median of first, last and middle element can be median of all elements?
D3. What we do when random pivot is ith element? Do we always have to swap it with either leftmost or rightmost element before partitioning? Or is there any algorithm which does not require such swap?
Quick Clarification Question about Time Complexity in CLRS
I’m reading about the Hiring Problem in “Introduction to Algorithms” and read
Interviewing has a low cost, say $ c_i$ , whereas hiring is expensive, costing $ c_h$ . Letting $ m$ be the number of people hired, the total cost associated with this algorithm is $ O(c_in+c_hm)$ .
For clarification, this is the algorithm being referred to
HIREASSISTANT(n) best = 0 // candidate 0 is a leastqualified dummy candidate for i = 1 to n interview candidate i if candidate i is better than candidate best best = i hire candidate i
My question is, why does the book say the total cost is $ O(c_in+c_hm)$ and not $ \Theta(c_in+c_hm)$ ? For any input, since we must iterate over all $ n$ candidates and given that $ m$ are hired, couldn’t we also say that the asymptotic lower bound is $ \Omega(c_in+c_hm)$ therefore saying the cost for any input is $ \Theta(c_in+c_hm)$ ?
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Quick calculation for $x^y \bmod 2^d$
I need to calculate $ x^y \bmod 2^d$ in $ O(d)$ summations/bitwise operations and $ 1$ multiplication by $ y$ . $ x$ is restricted to be odd, $ d\geq 3$ .
$ a$ bit arithmetic (for any $ a$ ) is allowed, as this is just a theoretical question.
Recursion from a special case of quick sort
I have the following special case for merge sort:
“John wishes to perform a quicksort on an array, where he picks the partition as the $ \left \lfloor{\frac{2n}{3}}\right \rfloor th$ least element in the array (If the array has 10 elements, it will pick the element at the 6th position in the sorted array) such that this pivot selection procedure takes $ \Theta(n)$ time.
What is the worst case running time?”
I have tried forming a recurrence, but I end up having floors/ceilings and don’t know how to deal with them. I would like to reduce the case down to the recurrence:
$ T(n) = T(2n/3) + T(n/3) + \Theta(n) $
Where I’m then able to produce a recursion tree I can easily analyse.
However, when producing an upper bound I’m getting a ceiling on the $ n/3$ and a floor on the $ 2n/3$ .
For the purpose of producing a recursion tree, can I just assume these floors/ceilings are negligible for a very large n?
Thanks
Quick Edit is not enabled even after adding the clienttemplate.js as described below
We have added clienttemplate.js in Schema.xml in the required view. But still the Quick Edit is not enabled even after all the OOB features also. Can you please suggest any solution for this?
Thanks in Advance
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