Difference between Big-Theta and Big O notation in simple language

Solution 1:

Big O is giving only upper asymptotic bound, while big Theta is also giving a lower bound.

Everything that is Theta(f(n)) is also O(f(n)), but not the other way around.
T(n) is said to be Theta(f(n)), if it is both O(f(n)) and Omega(f(n))

For this reason big-Theta is more informative than big-O notation, so if we can say something is big-Theta, it's usually preferred. However, it is harder to prove something is big Theta, than to prove it is big-O.

For example, merge sort is both O(n*log(n)) and Theta(n*log(n)), but it is also O(n²), since n² is asymptotically "bigger" than it. However, it is NOT Theta(n²), Since the algorithm is NOT Omega(n²).

Omega(n) is asymptotic lower bound. If T(n) is Omega(f(n)), it means that from a certain n0, there is a constant C1 such that T(n) >= C1 * f(n). Whereas big-O says there is a constant C2 such that T(n) <= C2 * f(n)).

All three (Omega, O, Theta) give only asymptotic information ("for large input"):

• Big O gives upper bound
• Big Omega gives lower bound and
• Big Theta gives both lower and upper bounds

Note that this notation is not related to the best, worst and average cases analysis of algorithms. Each one of these can be applied to each analysis.

Solution 2:

I will just quote from Knuth's TAOCP Volume 1 - page 110 (I have the Indian edition). I recommend reading pages 107-110 (section 1.2.11 Asymptotic representations)

People often confuse O-notation by assuming that it gives an exact order of Growth; they use it as if it specifies a lower bound as well as an upper bound. For example, an algorithm might be called inefficient because its running time is O(n^2). But a running time of O(n^2) does not necessarily mean that running time is not also O(n)

On page 107,

1^2 + 2^2 + 3^2 + ... + n^2 = O(n^4) and

1^2 + 2^2 + 3^2 + ... + n^2 = O(n^3) and

1^2 + 2^2 + 3^2 + ... + n^2 = (1/3) n^3 + O(n^2)

Big-Oh is for approximations. It allows you to replace ~ with an equals = sign. In the example above, for very large n, we can be sure that the quantity will stay below n^4 and n^3 and (1/3)n^3 + n^2 [and not simply n^2]

Big Omega is for lower bounds - An algorithm with Omega(n^2) will not be as efficient as one with O(N logN) for large N. However, we do not know at what values of N (in that sense we know approximately)

Big Theta is for exact order of Growth, both lower and upper bound.

Solution 3:

If the running time is expressed in big-O notation, you know that the running time will not be slower than the given expression. It expresses the worst-case scenario.

But with Theta notation you also known that it will not be faster. That is, there is no best-case scenario where the algorithm will retun faster.

This gives are more exact bound on the expected running time. However for most purposes it is simpler to ignore the lower bound (the possibility of faster execution), while you are generally only concerned about the worst-case scenario.

Solution 4:

I am going to use an example to illustrate the difference.

Let the function f(n) be defined as

if n is odd f(n) = n^3
if n is even f(n) = n^2

From CLRS

A function f(n) belongs to the set Θ(g(n)) if there exist positive constants c1 and c2 such that it can be "sandwiched" between c1g(n) and c2g(n), for sufficiently large n.

AND

O(g(n)) = {f(n): there exist positive constants c and n0 such that 0 ≤ f(n) ≤ cg(n) for all n ≥ n0}.

AND

Ω(g(n)) = {f(n): there exist positive constants c and n0 such that 0 ≤ cg(n) ≤ f(n) for all n ≥ n0}.

The upper bound on f(n) is n^3. So our function f(n) is clearly O(n^3).

But is it Θ(n^3)?

For f(n) to be in Θ(n^3) it has to be sandwiched between two functions one forming the lower bound, and the other the upper bound, both of which grown at n^3. While the upper bound is obvious, the lower bound can not be n^3. The lower bound is in fact n^2; f(n) is Ω(n^2)

From CLRS

For any two functions f(n) and g(n), we have f(n) = Θ(g(n)) if and only if f(n) = O(g(n)) and f(n) = Ω(g(n)).

Hence f(n) is not in Θ(n^3) while it is in O(n^3) and Ω(n^2)

Solution 5:

Here's my attempt:

A function, f(n) is O(n), if and only if there exists a constant, c, such that f(n) <= c*g(n).

Using this definition, could we say that the function f(2^(n+1)) is O(2^n)?

1. In other words, does a constant 'c' exist such that 2^(n+1) <= c*(2^n)? Note the second function (2^n) is the function after the Big O in the above problem. This confused me at first.

2. So, then use your basic algebra skills to simplify that equation. 2^(n+1) breaks down to 2 * 2^n. Doing so, we're left with:

   2 * 2^n <= c(2^n)

3. Now its easy, the equation holds for any value of c where c >= 2. So, yes, we can say that f(2^(n+1)) is O(2^n).

Big Omega works the same way, except it evaluates f(n) >= c*g(n) for some constant 'c'.

So, simplifying the above functions the same way, we're left with (note the >= now):

2 * 2^n >= c(2^n)

So, the equation works for the range 0 <= c <= 2. So, we can say that f(2^(n+1)) is Big Omega of (2^n).

Now, since BOTH of those hold, we can say the function is Big Theta (2^n). If one of them wouldn't work for a constant of 'c', then its not Big Theta.

The above example was taken from the Algorithm Design Manual by Skiena, which is a fantastic book.

Hope that helps. This really is a hard concept to simplify. Don't get hung up so much on what 'c' is, just break it down into simpler terms and use your basic algebra skills.