I am learning about complexity theory and have a question asking to show truth/falsehood of a number of Big-O statements.
I've done the first few e.g. showing 2^(n+1) is in O(2^n) by finding a constant and N value. But now they are asking more abstract things, for example:
If f(n) is O(g(n)), is log f(n) in O(log g(n))?
Is 2^(f(n)) in O(2^(g(n)))
These both seem like they would be true but I don't know how to express them formally with a constant and a N value. If you can give an example of how I could show these I can go do the rest of the problems.
The comments are both accurate. Here are some notes along the lines you are probably looking for.
Assume f(n) is O(g(n)). Then there exist n0 and c such that f(n) <= cg(n) for n >= n0. Take the logarithm of both sides. log(f(n)) <= log(cg(n)). We can use the laws of logarithms to rewrite this as log(f(n)) <= log(c) + log(g(n)). If g(n) is greater than 1, then log(c) + log(g(n)) <= (1+log(c))*log(g(n)), so we can choose c' = 1 + log(c) and get the desired result. Otherwise, note that for g(n) = 1 we're still good since any choice for c' works.
The second one is not true. Choose f(n) = 2n and g(n) = n. We see f(n) is O(g(n)) by choosing c = 3. However, 2^(2n) = 4^n is not O(2^n). To see that, assume we had n0 and c. Then 4^n <= c*2^n. Dividing by 2^n gives 2^n <= c. But this can't be correct since n can increase indefinitely whereas c is fixed.
Related
My professor recently brushed over the formal definition of Big O:
To be completely honest even after him explaining it to a few different students we all seem to still not understand it at its core. The problems in comprehension mostly occurred with the following examples we went through:
So far my reasoning is as follows:
When you multiply a function's highest term by a constant, you get a new function that eventually surpasses the initial function at a given n. He called this n a "witness" to the function O(g(n))
How is this c term created/found? He mentioned bounds a couple of times but didn't really specify what bounds signify or how to find them/use them.
I think I just need a more solid foundation of the formal definition and how these examples back up the definition.
I think that the way this definition is typically presented in terms of c values and n0's is needlessly confusing. What f(n) being O(g(n)) really means is that when you disregard constant and lower order terms, g(n) is an asymptotic upper bound for f(n) (for a function to g to asymptotically upper bound f just means that past a certain point g is always greater than or equal to f). Put another way, f(n) grows no faster than g(n) as n goes to infinity.
Big O itself is a little confusing, because f(n) = O(g(n)) doesn't mean that g(n) grows strictly faster than f(n). It means when you disregard constant and lower order terms, g(n) grows faster than f(n), or it grows at the same rate (strictly faster would be "little o"). A simple, formal way to put this concept is to say:
That is, for this limit to hold true, the highest order term of f(n) can be at most a constant multiple of the highest order term of g(n). f(n) is O(g(n)) iff it grows no faster than g(n).
For example, f(n) = n is in O(g(n) = n^2), because past a certain point n^2 is always bigger than n. The limit of n^2 over n is positive, so n is in O(n^2)
As another example, f(n) = 5n^2 + 2n is in O(g(n) = n^2), because in the limit, f(n) can only be about 5 times larger than g(n). It's not infinitely bigger: they grow at the same rate. To be precise, the limit of n^2 over 5n^2 + 3n is 1/5, which is more than zero, so 5n^2 + 3n is in O(n^2). Hopefully this limit based definition provides some intuition, as it is completely equivalent mathematically to the provided definition.
Finding a particular constant value c and x value n0 for which the provided inequality holds true is just a particular way of showing that in the limit as n goes to infinity, g(n) grows at least as fast as f(n): that f(n) is in O(g(n)). That is, if you've found a value past which c*g(n) is always greater than f(n), you've shown that f(n) grows no more than a constant multiple (c times) faster than g(n) (if f grew faster than g by more than a constant multiple, finding such a c and n0 would be impossible).
There's no real art to finding a particular c and n0 value to demonstrate f(n) = O(g(n)). They can be literally whatever positive values you need them to be to make the inequality true. In fact, if it is true that f(n) = O(g(n)) then you can pick any value you want for c and there will be some sufficiently large n0 value that makes the inequality true, or, similarly you could pick any n0 value you want, and if you make c big enough the inequality will become true (obeying the restrictions that c and n0 are both positive). That's why I don't really like this formalization of big O: it's needlessly particular and proofs involving it are somewhat arbitrary, distracting away from the main concept which is the behavior of f and g as n goes to infinity.
So, as for how to handle this in practice, using one of the example questions: why is n^2 + 3n in O(n^2)?
Answer: because the limit as n goes to infinity of n^2 / n^2 + 3n is 1, which is greater than 0.
Or, if you're wanting/needing to do it the other way, pick any positive value you want for n0, and evaluate f at that value. f(1) will always be easy enough:
f(1) = 1^2 + 3*1 = 4
Then find the constant you could multiply g(1) by to get the same value as f(1) (or, if not using n0 = 1 use whatever n0 for g that you used for f).
c*g(1) = 4
c*1^2 = 4
c = 4
Then, you just combine the statements into an assertion to show that there exists a positive n0 and a constant c such that cg(n) <= f(n) for all n >= n0.
n^2 + 3n <= (4)n^2 for all n >= 1, implying n^2 + 3n is in O(n^2)
If you're using this method of proof, the above statement you use to demonstrate the inequality should ideally be immediately obvious. If it's not, maybe you want to change your n0 so that the final statement is more clearly true. I think that showing the limit of the ratio g(n)/f(n) is positive is much clearer and more direct if that route is available to you, but it is up to you.
Moving to a negative example, it's quite easy with the limit method to show that f(n) is not in O(g(n)). To do so, you just show that the limit of g(n) / f(n) = 0. Using the third example question: is nlog(n) + 2n in O(n)?
To demonstrate it the other way, you actually have to show that there exists no positive pair of numbers n0, c such that for all n >= n0 f(n) <= cg(n).
Unfortunately showing that f(n) = nlogn + 2n is in O(nlogn) by using c=2, n0=8 demonstrates nothing about whether f(n) is in O(n) (showing a function is in a higher complexity class implies nothing about it not being a lower complexity class).
To see why this is the case, we could also show a(n) = n is in g(n) = nlogn using those same c and n0 values (n <= 2(nlog(n) for all n >= 8, implying n is in O(nlogn))`), and yet a(n)=n clearly is in O(n). That is to say, to show f(n)=nlogn + 2n is not in O(n) with this method, you can't just show that it is in O(nlogn). You would have to show that no matter what n0 you pick, you can never find a c value large enough such that f(n) >= c(n) for all n >= n0. Showing that such a pair of numbers does not exist is not impossible, but relatively speaking it's a tricky thing to do (and would probably itself involve limit equations, or a proof by contradiction).
To sum things up, f(n) is in O(g(n)) if the limit of g(n) over f(n) is positive, which means f(n) doesn't grow any faster than g(n). Similarly, finding a constant c and x value n0 beyond which cg(n) >= f(n) shows that f(n) cannot grow asymptotically faster than g(n), implying that when discarding constants and lower order terms, g(n) is a valid upper bound for f(n).
Let's say you were given two logarithmic functions like
and you were asked to find if f(n) is O(g(n)) Ω(g(n)) or Θ(g(n)), how would you go about it? I found questions like these easier when you were comparing two exponential equations, because for example with x(n) = n^2 and p(n) = n^2 you could find a c > 0 (ex 3) where x(n) <= cp(n) for all n greater than some n>0 and that would prove that x(n) = O(p(n)). However, comparing two logarithmic functions seems much more difficult for some reason. Any help is appreciated, thanks!
f(n) is O(g(n)) iff there is a constant c and n_0 such that f(n) <= c * g(n) for each n >= n_0.
f(n) is Ω(g(n)) iff there is a constant c and n_0 such that f(n) >= c * g(n) for each n >= n_0.
Now, f(n) is Θ(g(n)) iff f(n) is O(g(n)) and f(n) is Ω(g(n)).
So, in your cases, we have:
f(n) = log (n^2) = 2logn
which means, g(n) is logn and c = 2, which means f(n) <= 2 * logn and f(n) >= 2 * logn, which makes it Ω(logn).
Btw. its also f(n) <= n and f(n) >= 1, so f(n) can be O(n), but we don't use it, since we can find a better O(g(n)). In this case we don't have the same function in both notations, to for those values we don't have Ω. However, we just need one option for g(n) to declare Ω. In cases we can't find it, we say its not Ω. Note the word "we say".
In second case, we care only for "highest growing value", logn part. Now, c = 1, and g = log(n), so in this case, its also Ω(logn).
As the title says, can someone name an f(n) and a g(n) where they aren't upper bounds of each other. I had absolutely no idea and put two random constants:
f(n) = 8
g(n) = 3
Still no idea
In your example, they are both O(1). I'd say they're both "equivalent" and both upper/lower bounds of each other.
I'm pretty sure
f(n) = sin(n)
g(n) = cos(n)
will work. If you take the limit as n approaches infinity, f(n)/g(n) does not converge, and neither will g(n)/f(n) converge. Therefore, neither is an upper bound of the other.
Please post in a comment if you're not sure about why limits are being used here, and I can explain in greater depth.
Take f(n) be any positive value depending on n and
g(n) = n*f(n) if n is even else f(n)/n
Then, there is no constant A such that for n large enough g(n) <= A f(n), and
there is no constant B such that for n large enough f(n) <= B g(n). Thus g is not O(f) and f is not O(g).
I was doing study on Asymptotic Notations Topic, i recon that its formula is so simple yet it tells nothing and there are couple of things i don't understand.
When we say
f(n) <= c.g(n) where n >= n₀
And we don't know the value of c =? and n=? at first but by doing division of f(n) or g(n) we get the value of c. (here is where confusion lies)
First Question: How do we decide which side's 'n' has to get divided in equation f(n) or g(n)?
Suppose we have to prove:
2n(square) = O(n(cube))
here f(n) = 2(n(square)) and g(n)=n(cube)
which will form as:
2(n(square)) = c . n(cube)
Now in the notes i have read they are dividing 2(n(square)) to get the value of c by doing that we get c = 1;
But if we do it dividing n(cube) [which i don't know whether we can do it or not] we get c = 2;
How do we know what value we have to divide ?
Second Problem: Where does n₀ come from what's its task ?
Well by formula we know n >= n(0) which means what ever we take n we should take the value of n(0) or should be greater what n is.
But i am confuse that where do we use n₀ ? Why it is needed ?
By just finding C and N can't we get to conclusion if
n(square) = O(n(cube)) or not.
Would any one like to address this? Many thanks in advance.
Please don't snub me if i ask anything stupid or give -1. Address it please any useful link which covers all this would be enough as well:3
I have gone through the following links before posting this question this is what i understand and here are those links:
http://openclassroom.stanford.edu/MainFolder/VideoPage.php?course=IntroToAlgorithms&video=CS161L2P8&speed=
http://faculty.cse.tamu.edu/djimenez/ut/utsa/cs3343/lecture3.html
https://sites.google.com/sites/algorithmss15
From the second url in your question:
Let's define big-Oh more formally:
O(g(n)) = { the set of all f such that there exist positive constants c and n0 satisfying 0 <= f(n) <= cg(n) for all n >= n0 }.
This means, that for f(n) = 4*n*n + 135*n*log(n) + 1e8*n the big-O is O(n*n).
Because for large enough c and n0 this is true:
4*n*n + 135*n*log(n) + 1e8*n = f(n) <= O(n*n) = c*n*n
In this particular case the [c,n0] can be for example [6, 1e8], because (this is of course not valid mathematical proof, but I hope it's "obvious" from it):
f(1e8) = 4*1e16 + 135*8*1e8 + 1e16 = 5*1e16 + 1080*1e8 <= 6*1e16 = 6*1e8*1e8 =~= O(n*n). There are of course many more possible [c,n0] for which the f(n) <= c*n*n holds true, but you need to find only one such pair to prove the f(n) has O(f(n)) of O(n*n).
As you can see, for n=1 you need quite a huge c (like 1e9), so at first look the f(n) may look much bigger than n*n, but in the asymptotic notion you don't care about the first few initial values, as long as the behaviour since some boundary is as desired. That boundary is some [c,n0]. If you can find such boundary ([6, 1e8]), then QED: "f(n) has big-O of n*n".
The n >= n₀ means that whatever you say in the lemma can be false for some first k (countable) parameters n' : n' < n₀, but since some n₀ the lemma is true for all the rest of (bigger) integers.
It says that you don't care about first few integers ("first few" can be as "little" as 1e400, or 1e400000, ...etc... from the theory point of view), and you only care about the bigger (big enough, bigger than n₀) n values.
Ultimately it means, that in the big-O notation you usually write the simplest and lowest function having the same asymptotic notion as the examined f(n).
For example for any f(n) of polynomial type like f(n) = ∑aini, i=0..k the O(f(n)) = O(nk).
So I did throw away all the lower 0..(k-1) powers of n, as they stand no chance against nk in the long run (for large n). And the ak does lose to some bigger c.
In case you are lost in that i,k,...:
f(n) = 34n4 + 23920392n2 has O(n4).
As for large enough n that n4 will "eclipse" any value created from n2. And 34n4 is only 34 times bigger than n4 => 34 is constant (relates to c) and can be omitted from big-O notation too.
Just started learning algorithms. So the exercise is to find if statement is always/sometimes true or false. Em, where does my logic fails here?
f(n) != O(g(n)) and g(n) != O(f(n))
O-notation is 0 <= f(n) <= cg(n) where c is some constant. So not equal here means:
f(n) > cg(n) and g(n) > cf(n)
If f(n) = g(n) = 1, and let's say c = 1/2:
1 > (1/2)*1 and 1 > (1/2)*1
So it is true in this case. But the book says it's false in this particular case. What part do I misunderstand?
Big-O is not 0 <= f(n) <= c g(n) for some constant, per se. It's that there exists a number c such that the relation holds for "large enough" values of n. (This is the "asymptotic" that we refer to when we call Big-O an asymptotic notation, the other common ones being Big-Theta and Big-Omega.)
For example, let's say there's an algorithm that operates on some data structure with n elements, and takes 3n^2 + 7n + 18 steps. Call this f(n). We say that the Big-O of this expression is O(n^2) because there exists a constant (in this case anything larger than 3) such that for all "large enough" values of n, f(n) <= c n^2.