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1 vote
0 answers
124 views

Been trying to prove that $n! = \Omega(n^{100})$. I was approaching it with the solution below. I am not really sure my assumptions

We know that: $$f(n) = n! = \Omega (g(n)$$ if $$g(n) = O(f(n))$$ then, $$f(n) \in O(g(n))$$ If $$ f(n) \leq c \cdot g(n) \quad \textbf{for all $n\geq 1$}$$ then, assuming $n=1$ and $c=100^{100}$: $$n^{...
Mike's user avatar
  • 21
1 vote
1 answer
86 views

Does one level of worst-case recursion in quicksort cost $\Theta(n^2)$ or $\Theta(n)$?

Page 180 (section 7.4.1) of CLRS 3rd edition says a worst-case split at every level of recursion in quicksort produces a $\Theta(n^2)$ running time It seems that $\Theta(n^2)$ should be $\Theta(n)$ ...
JJJohn's user avatar
  • 1,454
1 vote
2 answers
41 views

Searching/Sorting Algorithm

I just started studying algorithms and data structures and came across this problem: Given $x \in \mathbb{N}$ and two integer Arrays $A_1$ and $A_2$ each of the length $n$. Write an algorithm in ...
zartos's user avatar
  • 105
3 votes
4 answers
2k views

Merge sort seems to take the same number of comparisons for best and worst case.

I am trying to clear up my conceptions of merge sort. What I cannot understand how merge sort takes less number of comparisons during best case. Let me explain, looking at the merge procedure given ...
ng.newbie's user avatar
  • 1,025
0 votes
1 answer
33 views

Question regarding the $\mathcal{O}$ notation used on constant functions

I'm having difficulties to understand the following passage out of the CLRS book, when the Counting Sort algorithm is introduced: Counting sort assumes that each of the elements is an integer in ...
3nondatur's user avatar
  • 4,212
1 vote
2 answers
8k views

Comparison between $n\log n$ and $n^2$ sorting algorithms

Suppose we have two sorting algorithms which takes $O(n\log n)$ and $O(n^2)$ time. What can we say about it? Is it always better to choose $n\log n$ if the size $n$ is not given? Or can we say on an ...
ViX28's user avatar
  • 647
0 votes
1 answer
32 views

How big the maximal decrease in consecutive elements of a sequence?

Consider a sequence $(s_1, ..., s_k)$, and we have it sorted in decreasing order $(\tilde{s}_1, ..., \tilde{s}_k)= (\sigma(s_1), ..., \sigma(s_k))$. Define $k_{\max} = \max \left( \max_i \left( s_i ...
Daniel's user avatar
  • 2,670
1 vote
1 answer
103 views

correcting an invalid binary heap in $\Theta (n)$

We are given a binary max (every node is larger than its children) heap with $n$ elements. We now change $\frac{n}{4}$ of the elements at random. We don't know which ones and to which value. And so, ...
Oria Gruber's user avatar
  • 12.8k
-1 votes
2 answers
1k views

Need help analyzing a merge sort

Im working on trying to understand merge sorts better and had a professor give me this to try to help. It is finals week and I have been trying to walk through this problem but have been having ...
Sampson's user avatar
2 votes
1 answer
1k views

Lower bound for matrix sorting?

Consider the problem of sorting a $n$ by $n$ matrix i.e. the rows and columns are in ascending order. I want to find the lower and upper bound of this problem. I found that it is $O(n^2logn)$ by just ...
Jiyda Moussa's user avatar
2 votes
1 answer
89 views

Flawed twofold induction on an inequation (but where?)

The following induction is flawed because the result admits a counter-example, but I can't find where is the flaw. Please advise. [Edit: As pointed out in the answer, the error was in the base case. ...
Christian Rinderknecht's user avatar
2 votes
0 answers
88 views

On bounding the average cost of top-down merge sort

Let $A_n$ be the average number of comparisons to sort $n$ keys by merging them in a top-down fashion (see any algorithm textbook). It can he shown that $$ A_0 = A_1 = 0;\quad A_n = A_{\lfloor{n/2}\...
Christian Rinderknecht's user avatar
14 votes
3 answers
36k views

Quicksort Running Time

I am trying to refresh my knowledge (and hopefully learn more) about Algorithm Analysis. I took a course on this two years ago but I am trying to catch up on what I had learned back then. The way I ...
JakeClawson's user avatar