What is the difference between Θ(n) and O(n)?

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Sometimes I see Θ(n) with the strange Θ symbol with something in the middle of it, and sometimes just O(n). Is it just laziness of typing because nobody knows how to type this symbol, or does it mean something different?

Short explanation:

If an algorithm is of Θ(g(n)), it means that the running time of the algorithm as n (input size) gets larger is proportional to g(n).

If an algorithm is of O(g(n)), it means that the running time of the algorithm as n gets larger is at most proportional to g(n).

Normally, even when people talk about O(g(n)) they actually mean Θ(g(n)) but technically, there is a difference.


More technically:

O(n) represents upper bound. Θ(n) means tight bound. Ω(n) represents lower bound.

f(x) = Θ(g(x)) iff f(x) = O(g(x)) and f(x) = Ω(g(x))

Basically when we say an algorithm is of O(n), it's also O(n2), O(n1000000), O(2n), ... but a Θ(n) algorithm is not Θ(n2).

In fact, since f(n) = Θ(g(n)) means for sufficiently large values of n, f(n) can be bound within c1g(n) and c2g(n) for some values of c1 and c2, i.e. the growth rate of f is asymptotically equal to g: g can be a lower bound and and an upper bound of f. This directly implies f can be a lower bound and an upper bound of g as well. Consequently,

f(x) = Θ(g(x)) iff g(x) = Θ(f(x))

Similarly, to show f(n) = Θ(g(n)), it's enough to show g is an upper bound of f (i.e. f(n) = O(g(n))) and f is a lower bound of g (i.e. f(n) = Ω(g(n)) which is the exact same thing as g(n) = O(f(n))). Concisely,

f(x) = Θ(g(x)) iff f(x) = O(g(x)) and g(x) = O(f(x))


There are also little-oh and little-omega (ω) notations representing loose upper and loose lower bounds of a function.

To summarize:

f(x) = O(g(x)) (big-oh) means that the growth rate of f(x) is asymptotically less than or equal to to the growth rate of g(x).

f(x) = Ω(g(x)) (big-omega) means that the growth rate of f(x) is asymptotically greater than or equal to the growth rate of g(x)

f(x) = o(g(x)) (little-oh) means that the growth rate of f(x) is asymptotically less than the growth rate of g(x).

f(x) = ω(g(x)) (little-omega) means that the growth rate of f(x) is asymptotically greater than the growth rate of g(x)

f(x) = Θ(g(x)) (theta) means that the growth rate of f(x) is asymptotically equal to the growth rate of g(x)

For a more detailed discussion, you can read the definition on Wikipedia or consult a classic textbook like Introduction to Algorithms by Cormen et al.

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There's a simple way (a trick, I guess) to remember which notation means what.

All of the Big-O notations can be considered to have a bar.

When looking at a Ω, the bar is at the bottom, so it is an (asymptotic) lower bound.

When looking at a Θ, the bar is obviously in the middle. So it is an (asymptotic) tight bound.

When handwriting O, you usually finish at the top, and draw a squiggle. Therefore O(n) is the upper bound of the function. To be fair, this one doesn't work with most fonts, but it is the original justification of the names.

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one is Big "O"

one is Big Theta

http://en.wikipedia.org/wiki/Big_O_notation

Big O means your algorithm will execute in no more steps than in given expression(n^2)

Big Omega means your algorithm will execute in no fewer steps than in the given expression(n^2)

When both condition are true for the same expression, you can use the big theta notation....

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Rather than provide a theoretical definition, which are beautifully summarized here already, I'll give a simple example:

Assume the run time of f(i) is O(1). Below is a code fragment whose asymptotic runtime is Θ(n). It always calls the function f(...) n times. Both the lower and the upper bound is n.

for(int i=0; i<n; i++){
    f(i);
}

The second code fragment below has the asymptotic runtime of O(n). It calls the function f(...) at most n times. The upper bound is n, but the lower bound could be Ω(1) or Ω(log(n)), depending on what happens inside f2(i).

for(int i=0; i<n; i++){
    if( f2(i) ) break;
    f(i);
}

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Theta is a shorthand way of referring to a special situtation where the big O and Omega are the same.

Thus, if one claims The Theta is expression q, then they are also necessarily claiming that Big O is expression q and Omega is expression q.


Rough analogy:

If: Theta claims, "That animal has 5 legs." then it follows that: Big O is true ("That animal has less than or equal to 5 legs.") and Omega is true("That animal has more than or equal to 5 legs.")

It's only a rough analogy because the expressions aren't necessarily specific numbers, but instead functions of varying orders of magnitude such as log(n), n, n^2, (etc.).

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Comments
  • It's not obvious, but this question is a duplicate of this one stackoverflow.com/questions/464078/… from yesterday.
  • Possible duplicate of Difference between lower bound and tight bound?
  • If "If an algorithm is of O(g(n)), it means that the running time of the algorithm as n gets larger is at most proportional to g(n)." Then how do you say that "Basically when we say an algorithm is of O(n), it's also O(n2), O(n1000000), O(2n)," ??
  • @Andy897 It follows from the definition of "proportional". From Wikipedia: "In mathematics, two variables are proportional if a change in one is always accompanied by a change in the other, and if the changes are always related by use of a constant multiplier. The constant is called the coefficient of proportionality or proportionality constant."
  • What does >= \Omega(...) mean? I understand if we say it's a member of \Omega(...), but if it's greater than it? What sense does it make?
  • I usually never go below 3-4 answers on any questions. This was worth the ride. Thanks for sharing the trick. :D
  • But is wrong! The number of steps is bounded above by n^2 as n gets very large. However, an algorithm that runs in n^2 + c steps takes more than n^2 steps, but is still O(n^2). Big-O notation only describes asymptotic beahviour.
  • This is not a end all be all definition. It's just a launching point.... Since we are talking about asymptotic notations as n approaches infinity. The constant C becomes a non factor.
  • While I like the simplicity of this answer, it should be noted that an O(n^2) algorithm could very well take 1,000,000,000*n^2 steps to execute, which is certainly much larger than n^2. An algorithm being O(n^2) just means that it will take no more than k*n^2 steps to execute, where k is some positive real number.
  • What do you mean by "asymptotic runtime"?
  • Asymptotic in this context means "for large enough n". The runtime of code fragment whose asymptotic runtime is Θ(n) will grow linearly as n increases, e.g. runtime T can be expressed as T(n) = a*n+b. For small values of n (e.g. n=1 or 2) this may not be the best way of describing the behaviour - perhaps you have some initialization code that takes a lot longer than f(i).
  • You have messed up the words and graphs.
  • @kushalvm, thanks for your honesty. Could you kindly explain what do you mean specifically? For the sake of my learning and others that may get confused with this answer. :-)