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## Computer science

### Unit 1: Lesson 3

Asymptotic notation

# Big-θ (Big-Theta) notation

Let's look at a simple implementation of linear search:
var doLinearSearch = function(array, targetValue) {
for (var guess = 0; guess < array.length; guess++) {
if (array[guess] === targetValue) {
return guess;  // found it!
}
}
return -1;  // didn't find it
};
Let's denote the size of the array (array.length) by n. The maximum number of times that the for-loop can run is n, and this worst case occurs when the value being searched for is not present in the array.
Each time the for-loop iterates, it has to do several things:
• compare guess with array.length
• compare array[guess] with targetValue
• possibly return the value of guess
• increment guess.
Each of these little computations takes a constant amount of time each time it executes. If the for-loop iterates n times, then the time for all n iterations is c, start subscript, 1, end subscript, dot, n, where c, start subscript, 1, end subscript is the sum of the times for the computations in one loop iteration. Now, we cannot say here what the value of c, start subscript, 1, end subscript is, because it depends on the speed of the computer, the programming language used, the compiler or interpreter that translates the source program into runnable code, and other factors.
This code has a little bit of extra overhead, for setting up the for-loop (including initializing guess to 0) and possibly returning -1 at the end. Let's call the time for this overhead c, start subscript, 2, end subscript, which is also a constant. Therefore, the total time for linear search in the worst case is c, start subscript, 1, end subscript, dot, n, plus, c, start subscript, 2, end subscript.
As we've argued, the constant factor c, start subscript, 1, end subscript and the low-order term c, start subscript, 2, end subscript don't tell us about the rate of growth of the running time. What's significant is that the worst-case running time of linear search grows like the array size n. The notation we use for this running time is \Theta, left parenthesis, n, right parenthesis. That's the Greek letter "theta," and we say "big-Theta of n" or just "Theta of n."
When we say that a particular running time is \Theta, left parenthesis, n, right parenthesis, we're saying that once n gets large enough, the running time is at least k, start subscript, 1, end subscript, dot, n and at most k, start subscript, 2, end subscript, dot, n for some constants k, start subscript, 1, end subscript and k, start subscript, 2, end subscript. Here's how to think of \Theta, left parenthesis, n, right parenthesis:
For small values of n, we don't care how the running time compares with k, start subscript, 1, end subscript, dot, n or k, start subscript, 2, end subscript, dot, n. But once n gets large enough—on or to the right of the dashed line—the running time must be sandwiched between k, start subscript, 1, end subscript, dot, n and k, start subscript, 2, end subscript, dot, n. As long as these constants k, start subscript, 1, end subscript and k, start subscript, 2, end subscript exist, we say that the running time is \Theta, left parenthesis, n, right parenthesis.
We are not restricted to just n in big-Θ notation. We can use any function, such as n, squared, n, log, start base, 2, end base, n, or any other function of n. Here's how to think of a running time that is \Theta, left parenthesis, f, left parenthesis, n, right parenthesis, right parenthesis for some function f, left parenthesis, n, right parenthesis:
Once n gets large enough, the running time is between k, start subscript, 1, end subscript, dot, f, left parenthesis, n, right parenthesis and k, start subscript, 2, end subscript, dot, f, left parenthesis, n, right parenthesis.
In practice, we just drop constant factors and low-order terms. Another advantage of using big-Θ notation is that we don't have to worry about which time units we're using. For example, suppose that you calculate that a running time is 6, n, squared, plus, 100, n, plus, 300 microseconds. Or maybe it's milliseconds. When you use big-Θ notation, you don't say. You also drop the factor 6 and the low-order terms 100, n, plus, 300, and you just say that the running time is \Theta, left parenthesis, n, squared, right parenthesis.
When we use big-Θ notation, we're saying that we have an asymptotically tight bound on the running time. "Asymptotically" because it matters for only large values of n. "Tight bound" because we've nailed the running time to within a constant factor above and below.

This content is a collaboration of Dartmouth Computer Science professors Thomas Cormen and Devin Balkcom, plus the Khan Academy computing curriculum team. The content is licensed CC-BY-NC-SA.

## Want to join the conversation?

• I was confused about where k1 and k2 came from, as well as tight bound. I read this StackOverflow answer and it gave me a bit more to go on: http://stackoverflow.com/a/471292/17188 -- anyone else have a better explanation/clarification?
• k1 and k2 are simply real numbers that could be anything as long as f(n) is between k1*f(n) and k2*f(n).

Let's say that doLinearSearch(array, targetValue) runs at f(n)=2n+3 speed in microseconds on a certain computer (where n is the length of the array) and we're trying to prove that it has Θ(n) time complexity. We would need to find two real numbers k1, k2, and n0 such that k1*n<2n+3<k2*n for all n>n0. We need to solve for k1, k2, and n0. It happens that there are infinitely many solutions, but we only need to find one to complete our proof.

Now, it happens that the solution k1=1, k2=3, and n0=4 works. This means 1*n<2n+3<3n for all n>4. For example:
1*5<2*5+3<3*5 --> 5<13<15
1*6<2*6+3<3*6 --> 6<15<18
1*7<2*7+3<3*7 --> 7<17<21
1*4.1<2*4.1+3<3*4.1 --> 4.1<11.2<12.3
In mathematics, these examples alone wouldn't actually suffice for a proof, but there is a rigorous way to prove that this inequality is true. It's just rather complicated.

You might ask why I didn't pick n0=3 instead. After all, that seems to be the minimal possible answer, because 3 doesn't work (1*3<2*3+3<3*3 --> 3<9<9), but all n>3 works. However, while that's true, I don't need the minimal possible n0. I just need any n0 that works. If I find one solution, then I've proved what I've needed to.

I would imagine if you were taking a college level computer science course, you might need need to prove Big-Θ, Big-O, and Big-Ω statements. Also, the make-up of the actual function would probably be determined by something else because you don't usually work with actual microseconds or milliseconds in general time complexity. For example, maybe the algorithm has two loops that go from 0 to n (where n is the input of the algorithm), in which case the function might be f(n)=2n. In general, you won't need to rigorously prove any complexity theory statements because as I said before, it's rather complicated (although won't be if you have experience proving limits). Instead, you can just keep the rule of thumb for polynomials that you can drop all of the terms except the one with the highest exponent, and then drop the exponent.

I hope this helps!
• So it seems to me that Big O is the worst case scenario, Big Theta is the range of cases from best to worst and Big Omega I haven't got to yet. ...now why couldn't you guys just say that?
• They didn't say that, because it would be incorrect.
Big O, Big Theta and Big Omega are often all used to measure the worst case scenario (this is most often the case).
They could also all be used to measure an average case (this happens often).
They could also all be used to measure the best case (we rarely care about best cases)
The measurements of Big-O, Big-Theta, and Big-Omega would often be different depending on which case was picked.

Here's the simple version of what Big-O, Big-Theta, and Big-Omega are :
If you have a function f(N):
Big-O tells you which functions grow at a rate >= than f(N), for large N
Big-Theta tells you which functions grow at the same rate as f(N), for large N
Big-Omega tells you which functions grow at a rate <= than f(N), for large N
(Note: >= , "the same", and <= are not really accurate here, but the concepts we use in asymptotic notation are similar):

We often call Big-O an upper bound, Big-Omega a lower bound, and Big-Theta a tight bound. Often in computer science the function we are concerned with is the running time of an algorithm for inputs of size N.

Hope this makes sense
• What exactly does asymptotic mean? In maths at school I learned that an asymptote is a line that a curve never touches but the distance between them approaches zero. However, that definition isn't really helping me to understand this concept.
• When we are talking about "asymptotic" behavior here, we are talking about the behavior of functions when n is very large (approaching infinity).

We have 3 curves that we worry about.
A curve for the running time function f(n). (we often don't know exactly what this function looks like)
A curve that we know is "above" the running time function when n is large. ( Big-O )
A curve that we know is "below" the running time function when n is large. (Big-Omega)
(Note: the terms "above" and "below" aren't entirely accurate here, and Big-O and Big-Omega actually describe a set of curves rather than just one curve)

If we can squeeze the curves that are "above" and "below" the running time function close enough, then we can figure out Big-Theta. This is useful when we don't know exactly what the running time curve looks like, which is most often the case, because Big-Theta is just a "scaled" version of the running time curve (it can tell us the shape of the curve).
(Note: this isn't entirely accurate, it only scales the highest order behavior of f(n) (the parts that dominate as n increases) , and Big-Theta is also a set of curves)

Hope this makes sense
• I am just wondering what level is this course, I am 15 and it seems to be sorta confusing. I am trying hard to understand the Big-Theta notation, but it seems to get confusing rather quickly. I might just be confusing things in my head, not sure. I am going to continue working on this, but I would like to know what level this course is. Thanks
• It is difficult to give a grade level to this type of material since when and how it is taught varies so widely. Personally, I think it leans towards college level, especially the asymptotic notation calculations. It is important to know what types of things will slow your programs to the point of frustration. For the amateur coder, it is not so important to calculate and graph it! I don't think most software engineers approach issues that way, unless they need to pitch a project or they are really stuck on a difficult bottleneck.
People who are more recently in CS may disagree (I moved over to math), but I think the much more interesting aspects of algorithm study involve problem solving and translating solutions into code. And until you do a bunch of that, measuring how fast your subroutine goes won't be of much help.
Anyway, hope this is useful and you enjoy whichever areas you end up focusing on.
• Where I can find a good reading about how to compute the complexity of an algorithm? I want to understand how to figure out if an algorithm if n.log.n, n^2, etc.
• Here's the quick version of how to analyze running times:
(I use n below for the problem size )

-Replace any simple block of operations that are performed without calling a function with a constant
e.g.
x = 5* 4 + 2; y++;z = x * y;

will run in some constant amount of time
we usually label that time with some variable e.g. 'c1'

-If you have a for loop that loops n times, the running time is about n times the running time of the code inside the loop (this neglects the constant amount of time to initialize the for loop and the constant amount of time every loop to perform the comparison and increment)
e.g.
for(var i=0; i < n ; i++){	x = 5* 4 + 2; 	y++;	z = x * y;}

we determined above that the code inside of the for loop runs in 'c1'
so the running time of this code is about 'c1 * n'

e.g.
for(var j=0; j < n; j++){	for(var i=0; i < n ; i++){		x = 5* 4 + 2; 		y++;		z = x * y;	}}

we determined above that the code inside of our outermost for loop runs in about 'c1 * n'
so this code runs in about 'c1 * n * n' or 'c1 * n^2'

-If after each iteration of an algorithm the problem is '1/a' of the size before the iteration then the code will iterate about 'log_a(n)' times. Thus, the running time of the code will be about log_a(n) times the running time of the code run each iteration.
e.g.
while( n>1 ){	x = 5* 4 + 2; 	y++;	z = x * y;	n = n/2;}

the code inside of the while loop runs in some constant time 'c2'
the size of the problem size n, is reduce in 1/2 each iteration
so this code iterates about 'log_2(n)' times
so the running time of the code is about 'c2 * log_2(n)'

Beyond the above, I would look at the analysis in the articles here, and study how they find the running times to learn other techniques on how to figure out the running times of algorithms.

"Discrete Mathematics with Applications" by Susanna Epp, discusses how to analyze running times in the chapter on Efficiency of Algorithms.

Hope this makes sense
• I'm still not clear about what "tightly bound
" means in this context. I do, however understand everything else, including the definition of "Asymptotic".
• The upper bound tells us what grows asymptotically faster than or at the same rate as our function. The lower bound tells us what asymptotically grows slower than or at the same rate as our function. Our function must lie somewhere in between the upper and lower bound.

Suppose that we can squeeze the lower bound and our upper bound closer and closer together. Eventually they will both be at the same asymptotic growth rate as our function. That's what we would call tight bounds (where the upper bound and lower bound have the same growth rate as the function).
• I did not understand what you meant when you spoke about the constants k1 and k2; can you please elaborate some more?
• Suppose you have two functions, f(n) and g(n).
If,for large values of n, you are able to:
-squeeze f(n) between k1 * g(n) and k2 * g(n)
Then:
-you can say f(n) is Θ( g(n) )

You are allowed to pick any value you want for k1 and k2 as long as they are both greater than zero.

e.g.
Suppose: f(n) = 6n + 5 and g(n) = n + 2
We could pick k1 = 1 and k2 = 6 and n >= 1

We can see that: k1 * g(n) <= f(n) <= k2 * g(n)
because: 1 *(n+2) <= 6n+5 <= 6*(n+2)

We successfully squeezed f(n) between k1 * g(n) and k2 * g(n) so we can say:
f(n) is Θ( g(n))

Hope this makes sense
• In the graph why Θ(n) is not linear? the time taken for an increment in the for loop will be a constant always or will it be varying? and in the second graph why k1 and k2 are not constant? why it is curved?

is Θ(n) = c1​​⋅n+c​2
​​