What is the difference between lower bound and tight bound?

Big O is the upper bound, while Omega is the lower bound. Theta requires both Big O and Omega, so that's why it's referred to as a tight bound (it must be both the upper and lower bound).

For example, an algorithm taking Omega(n log n) takes at least n log n time, but has no upper limit. An algorithm taking Theta(n log n) is far preferential since it takes at least n log n (Omega n log n) and no more than n log n (Big O n log n).

Θ-notation (theta notation) is called tight-bound because it's more precise than O-notation and Ω-notation (omega notation).

If I were lazy, I could say that binary search on a sorted array is O(n²), O(n³), and O(2ⁿ), and I would be technically correct in every case. That's because O-notation only specifies an upper bound, and binary search is bounded on the high side by all of those functions, just not very closely. These lazy estimates would be useless.

Θ-notation solves this problem by combining O-notation and Ω-notation. If I say that binary search is Θ(log n), that gives you more precise information. It tells you that the algorithm is bounded on both sides by the given function, so it will never be significantly faster or slower than stated.

If you have something that's O(f(n)) that means there's are k, g(n) such that f(n) ≤ k g(n).

If you have something that's Ω(f(n)) that means there's are k, g(n) such that f(n) ≥ k g(n).

And if you have a something with O(f(n)) and Ω(f(n)), then it's Θ(f(n).

The Wikipedia article is decent, if a little dense.

The phrases minimum time and maximum time are a bit misleading here. When we talk about big O notations, it's not the actual time we are interested in, it is how the time increases when our input size gets bigger. And it's usually the average or worst case time we are talking about, not best case, which usually is not meaningful in solving our problems.

Using the array search in the accepted answer to the other question as an example. The time it takes to find a particular number in list of size n is n/2 * some_constant in average. If you treat it as a function f(n) = n/2*some_constant, it increases no faster than g(n) = n, in the sense as given by Charlie. Also, it increases no slower than g(n) either. Hence, g(n) is actually both an upper bound and a lower bound of f(n) in Big-O notation, so the complexity of linear search is exactly n, meaning that it is Theta(n).

In this regard, the explanation in the accepted answer to the other question is not entirely correct, which claims that O(n) is upper bound because the algorithm can run in constant time for some inputs (this is the best case I mentioned above, which is not really what we want to know about the running time).

If I were lazy, I could say that binary search on a sorted array is O(n2), O(n3), and O(2n), and I would be technically correct in every case.

We can use o-notation ("little-oh") to denote an upper bound that is not asymptotically tight. Both big-oh and little-oh are similar. But, big-oh is likely used to define asymptotically tight upper bound.

Precisely the lower bound or $\omega $ bfon f(n) means the set of functions which are asymptotically less or equal to f(n) i.e U g(n)≤ cf(n) $\for all $`un≥ n' For some c, n' $\in $ $\Bbb{N}$

And the upper bound or $\mathit{O}$ on f(n) means the set of functions which are assymptotically greater or equal to f(n) which mathematically tells,

$ g(n)\ge cf(n) \for all n\ge n' $ , for some c,n' $\in $ $\Bbb{N}$.

Now the $\Theta $ is the intersection of the above written two

$\theta $

Like if a algorithm is like " exactly $\Omega\left( f(n)\ right$ " then it's better to say it's $\Theta\left(f(n)\right)$ .

Or , we can say also that it give us the actual speed where $ \omega $ gives us the lowest limit.