Recently I have been reading Algorithm 4.It says knuth has prove that the average check of chars in msd algorithm is NlgN,so why the time-complexity of this algorithm is not log-linear but linear?
If we use check of chars as cost model,should the time-complexity change to log-linear?
Related
Good day! I saw a backtracking subset-generation algorithm in the link:
https://www.geeksforgeeks.org/backtracking-to-find-all-subsets/
which claims that the space complexity of the program is O(n). Yet, from what I've learned, the minimum complexity should be O(2^n) since it will be the size of our output. Is the given space complexity correct?
Find the unique mapping between elements of two same size arrays
This is quite known interview question and it is easy to find algorithm (using the idea of quicksort) that has O(NlogN) for average case and O(N^2) for worst case complexity. Also using the same techniques as for sorting problem we can show that any algorithm should do at least NlogN comparisons.
So the question I cant get answered, is there worst case O(NlogN) algorithm for this problem? Maybe it should be similar to merge sort.
Yes, as of 1995 there are worst-case O(n log n)-time algorithms known for this problem, but they appear to be quite complicated. Here are two citations from Jeff Erickson's algorithm notes:
János Komlós, Yuan Ma, and Endre Szemerédi, Sorting nuts and bolts in O(n log n) time, SIAM J. Discrete Math 11(3):347–372, 1998.
Phillip G. Bradford, Matching nuts and bolts optimally, Technical Report MPI-I-95-1-025, Max-Planck-Institut für Informatik, September 1995.
As Jeff remarks, "Both the algorithms and their analysis are incredibly technical and the constant hidden in the O(·) notation is quite large." He notes also that Bradford’s algorithm, which appeared second, is slightly simpler.
Hey just a quick question,
I've just started looking into algorithm analysis and I'm attempting to learn Big-Oh notation.
The algorithm I'm looking at contains a quicksort (of complexity O(nlog(n))) to sort a dataset, and then the algorithm that operates upon the set itself has a worst case run-time of n/10 and complexity O(n).
I believe that the overall complexity of the algorithm would just be O(n), because it's of the highest order, so it makes the complexity of the quicksort redundant. However, could someone confirm this or tell me if I'm doing something wrong?
Wrong.
Quicksort has worst case complexity O(n^2). But even if you have an O(nlogn) sort algorithm, this is still more than O(n).
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Plain English explanation of Big O
I am reading an the " Introduction to Algorithms" Book, but dont understand this.
O(100), O(log(n)), O(n*log(n)), O(n2), O(n3)
Ok Thanks, i dident even know what it was, so i am going to read that Big O post now.
But if anyone can explain this any further in layman's terms it would be much appreciated.
Thanks
That is the big O notation and an order of efficiency of algorithms:
O(1), not O(100) - constant time - whatever the input, the algorithm executes in constant time
O(log(n)) - logarithmic time - as input gets larger, so will the time, but by a decreasing amount
O(n*log(n)) - linear * logarithmic - increases larger than linear, but not as fast as the following
O(n^2), or generally O(n^k) where k is a constant - polynomial time, probably the worst of feasible algorithms
There are worse algorithms, that are considered unfeasible for non-small inputs:
O(k^n) - exponential
O(n!) - factorial
Algorithms that follow an Ackerman function...
This notation is orientative. For example, some algorithms in O(n^2) can perform, on average, faster than O(n*log(n)) - see quicksort.
This notation is also an upper bound, meaning it describes a worst case scenario.
It can be used for space complexity or time complexity, where n is the size of the input provided.
Big O (simplifying) indicates how long will a given algorithm to complete, n being the amount of entry.
For example:
O(100) -> will take 100 units to complete no matter how much entry.
O(log(n)) -> will take log(n) to complete
O(n2) -> will take n^2 (n * n) to complete
When I was reading about quantum algorithms I faced the Deutsch-Jozsa algorithm, I see that if we want to solve that problem in a non-quantum algorithm, our algorithm would have exponential time complexity. Now I want to know what is the time complexity of Deutsch-Jozsa algorithm as a quantum algorithm on quantum computers?
According to Wikipedia the complexity of the quantum algorithm is constant:
The Deutsch-Jozsa quantum algorithm produces an answer that is always correct with a single evaluation of f.
The algorithm itself are just some calculations on quantum states, without any iterations/... so complexity is O(1).