Which of these functions is not O(n2)?
A. O(n2log2n)
B. O(log2(log2n)
C. O(nlog2n)
D. (log2n)2
I believe, that it is not A, because n2 takes dominance in that one, which would make it a O(n2) as well.
My question becomes how do you figure out which one is not a O(n2)? I have not worked much with logarithms, and I am still fuzzy about how to work with them.
Thanks.
Answer is A
Big-O is an upper bound and in programming represents the worst-case runtime. For example f(n) = n^2 means f(n) is O(n^2) and O(n^3) and O(n^100*logn). So O(n^2) is also O(n^3) but O(n^3) is not O(n^2)
In the case of your question, the only answer that is > n^2 is A. Graphing the functions can help you visualize it. As you can see A (red) is the only one increasing faster than n^2 (black)
This article may help you as well
https://en.wikipedia.org/wiki/Big_O_notation and go to Orders of common functions
Related
I have some confusion regarding the Asymptotic Analysis of Algorithms.
I have been trying to understand this upper bound case, seen a couple of youtube videos. In one of them, there was an example of this equation
where we have to find the upper bound of the equation 2n+3. So, by looking at this, one can say that it is going o be O(n).
My first question :
In algorithmic complexity, we have learned to drop the constants and find the dominant term, so is this Asymptotic Analysis to prove that theory? or does it have other significance? otherwise, what is the point of this analysis when it is always going to be the biggest n in the equation, example- if it were n+n^2+3, then the upper bound would always be n^2 for some c and n0.
My second question :
as per rule the upper bound formula in Asymptotic Analysis must satisfy this condition f(n) = O(g(n)) IFF f(n) < c.g(n) where n>n0,c>0, n0>=1
i) n is the no of inputs, right? or does n represent the number of steps we perform? and does f(n) represents the algorithm?
ii) In the following video to prove upper bound of the equation 2n+3 could be n^2 the presenter considered c =1, and that is why to satisfy the equation n had to be >= 3 whereas one could have chosen c= 5 and n=1 as well, right? So then why were, in most cases in the video, the presenter was changing the value of n and not c to satisfy the conditions? is there a rule, or is it random? Can I change either c or n(n0) to satisfy the condition?
My Third Question:
In the same video, the presenter mentioned n0 (n not) is the number of steps. Is that correct? I thought n0 is the limit after which the graph becomes the upper bound (after n0, it satisfies the condition for all values of n); hence n0 also represents the input.
Would you please help me understand because people come up with different ideas in different explanations, and I want to understand them correctly?
Edit
The accepted answer clarified all of the questions except the first one. I have gone through many articles on the web, and here I am documenting my conclusion if anyone else has the same question. This will help them.
My first question was
In algorithmic complexity, we have learned to drop the constants and
find the dominant term, so is this Asymptotic Analysis to prove that
theory?
No, Asymptotic Analysis describes the algorithmic complexity, which is all about understanding or visualizing the Asymptotic behavior or the tail behavior of a function or a group of functions by plotting mathematical expression.
In computer science, we use it to evaluate (note: evaluate is not measuring) the performance of an algorithm in terms of input size.
for example, these two functions belong to the same group
mySet = set()
def addToMySet(n):
for i in range(n):
mySet.add(i*i)
mySet2 = set()
def addToMySet2(n):
for i in range(n):
for j in range(500):
mySet2.add(i*j)
Even though the execution time of the addToMySet2(n) is always > the execution time of addToMySet(n), the tail behavior of both of these functions would be the same with respect to the largest n, if one plot them in a graph the tendency of that graph for both of the functions would be linear thus they belong to the same group. Using Asymptotic Analysis, we get to see the behavior and group them.
A mistake that I made assuming upper bound represents the worst case. In reality, The upper bound of any algorithm is associated with all of the best, average, and worst cases. so the correct way of putting that would be
upper/lower bound in the best/average/worst case of an
algorithm
.
We can't relate the upper bound of an algorithm with the worst-case time complexity and the lower bound with the best-case complexity. However, an upper bound can be higher than the worst-case because upper bounds are usually asymptotic formulae that have been proven to hold.
I have seen this kind of question like find the worst-case time complexity of such and such algorithm, and the answer is either O(n) or O(n^2) or O(log-n), etc.
For example, if we consider the function addToMySet2(n), one would say the algorithmic time complexity of that function is O(n), which is technically wrong because there are three factors bound, bound type, (inclusive upper bound and strict upper bound) and case are involved determining the algorithmic time complexity.
When one denote O(n) it is derived from this Asymptotic Analysis f(n) = O(g(n)) IFF for any c>0, there is a n0>0 from which f(n) < c.g(n) (for any n>n0) so we are considering upper bound of best/average/worst case. In the above statement the case is missing.
I think We can consider, when not indicated, the big O notation generally describes an asymptotic upper bound on the worst-case time complexity. Otherwise, one can also use it to express asymptotic upper bounds on the average or best case time complexities
The whole point of asymptotic analysis is to compare algorithms performance scaling. For example, if I write two version of the same algorithm, one with O(n^2) time complexity and the other with O(n*log(n)) time complexity, I know for sure that the O(n*log(n)) one will be faster when n is "big". How big? it depends. You actually can't know unless you benchmark it. What you know is at some point, the O(n*log(n)) will always be better.
Now with your questions:
the "lower" n in n+n^2+3 is "dropped" because it is negligible when n scales up compared to the "dominant" one. That means that n+n^2+3 and n^2 behave the same asymptotically. It is important to note that even though 2 algorithms have the same time complexity, it does not mean they are as fast. For example, one could be always 100 times faster than the other and yet have the exact same complexity.
(i) n can be anything. It may be the size of the input (eg. an algorithm that sorts a list) but it may also be the input itself (eg. an algorithm that give the n-th prime number) or a number of iteration, etc
(ii) he could have taken any c, he chose c=1 as an example as he could have chosen c=1.618. Actually the correct formulation would be:
f(n) = O(g(n)) IFF for any c>0, there is a n0>0 from which f(n) < c.g(n) (for any n>n0)
the n0 from the formula is a pure mathematical construct. For c>0, it is the n value from which the function f is bounded by g. Since n can represent anything (size of a list, input value, etc), it is the same for n0
I had seen in one of the videos (https://www.youtube.com/watch?v=A03oI0znAoc&t=470s) that, If suppose f(n)= 2n +3, then BigO is O(n).
Now my question is if I am a developer, and I was given O(n) as upperbound of f(n), then how I will understand, what exact value is the upper bound. Because in 2n +3, we remove 2 (as it is a constant) and 3 (because it is also a constant). So, if my function is f(n) where n = 1, I can't say g(n) is upperbound where n = 1.
1 cannot be upperbound for 1. I find hard understanding this.
I know it is a partial (and probably wrong answer)
From Wikipedia,
Big O notation characterizes functions according to their growth rates: different functions with the same growth rate may be represented using the same O notation.
In your example,
f(n) = 2n+3 has the same growth rate as f(n) = n
If you plot the functions, you will see that both functions have the same linear growth; and as n -> infinity, the difference between the 2 gets minimal.
In Big O notation, f(n) = 2n+3 when n=1 means nothing; you need to look at the trend, not discreet values.
As a developer, you will consider big-O as a first indication for deciding which algorithm to use. If you have an algorithm which is say, O(n^2), you will try to understand whether there is another one which is, say, O(n). If the problem is inherently O(n^2), then the big-O notation will not provide further help and you will need to use other criterion for your decision. However, if the problem is not inherently O(n^2), but O(n), you should discard any algorithm that happen to be O(n^2) and find an O(n) one.
So, the big-O notation will help you to better classify the problem and then try to solve it with an algorithm whose complexity has the same big-O. If you are lucky enough as to find 2 or more algorithms with this complexity, then you will need to ponder them using a different criterion.
I have a question reagarding to my lecture datastructures and algortihms.
I have to problem to understand how a algorithm grows. I dont unterstand the difference between the O Notations. And i dont understand the difference between them for example O(lgn)and O(nlgn).
I hope anyone can help me. thank you
To compare time complexities you should be able to make some mathematical proofs. In your example:
for every n>1 we have by multiplying with logn: nlogn>logn so nlogn is worse than logn. An easy way to understand this is by comparing the graphs of functions as suggested in comments or even try some big inputs to see the asymptotic behavior. For example for n=1000000 :
logn(1000000)=6 and 1000000log(1000000)=6000000 which is greater.
Also notice that you dount count constants in big O notation for example 4n is O(n) , n is O(n) and also cn+w is O(n) for c,w constants.
I'm trying to figure out the efficiency of my algorithms and I have little bit confusion.
Just need some expert idea to justify my answers or reference me to somewhere which is explaining about the being an element of not in asymptotic subject. (There is many resources but nothing about element of set is found by me)
When we say O(n^2) which is for two loops is it right to say:
n^2 is an element of O(n^3)
To my understanding big O is the worst case and omega is the best efficient case. If we put them on the graph all the cases of n^2 is part of O(n^3) so the first one is not right?
n^3 is an element of omega(n^2)
And also about the second one it is not right. Because some of the best cases of omega(n^2) is not in all the cases of n^3!
Finally is
2^(n+1) element of theta(2^n)
I have no idea how to measure that!
Big O, omega, theta in this context are all complexities. It's the functions with those complexities which form the sets you're thinking of.
Indeed, the set of functions with complexity O(n*n) is a subset of those with complexity O(n*n*n). Simply said, that's because O(n*n*n) means that the complexity is less than c*n*n*n as n goes to infinity, for some constant c. If a function has actual complexity 3*n*n + 7*n, then its complexity as n goes to infinity is obviously less than c*n*n*n, for any c.
Therefore, O(n*n*n) isn't just "three loops", it's "three loops or less".
Ω is the reverse. It's a lower bound for complexity, and c*n*n is a trivial lower bound for n*n*n as n goes to infinity.
The set of functions with complexity Θ(n*n) is the intersection of those with complexities O(n*n) and Ω(n*n). E.g. 3*n doesn't have complexity Θ(n*n) because it doesn't have complexity Ω(n*n), and 7*n*n*n doesn't have complexity Θ(n*n) because it doesn't have complexity O(n*n).
I will list the answers one by one.
1.) n^2 is an element of O(n^3)
True
To know more about Big-Oh read here.
2.) n^3 is an element of omega(n^2)
True
To understand omega notation read here.
3.) 2^(n+1) element of theta(2^n)
True
By now you would know why this is right.(Hint:Constant factor)
Please ask if you have any more questions.
I went through many lectures, videos and sources regarding Asymptotic notations. I understood what O, Omega and Theta were. But in algorithms, why do we use only Big Oh notation always, why not Theta and Omega (I know it sounds noobish, but please help me with this). What exactly is this upperbound and lowerbound in accordance with Algorithms?
My next question is, how do we find the complexity from an algorithm. Say I have an algorithm, how do I find the recurrence relation T(N) and then compute the complexity out of it? How do I form these equations? Like in the case of Linear Search using Recursive way, T(n)=T(N-1) + 1. How?
It would be great if someone can explain me considering me a noob so that I can understand even better. I found some answers but wasn't convincing enough in StackOverFlow.
Thank you.
Why we use big-O so much compared to Theta and Omega: This is partly cultural, rather than technical. It is extremely common for people to say big-O when Theta would really be more appropriate. Omega doesn't get used much in practice both because we frequently are more concerned about upper bounds than lower bounds, and also because non-trivial lower bounds are often much more difficult to prove. (Trivial lower bounds are usually the kind that say "You have to look at all of the input, so the running time is at least equal to the size of the input.")
Of course, these comments about lower bounds also partly explain Theta, since Theta involves both an upper bound and a lower bound.
Coming up with a recurrence relation: There's no simple recipe that addresses all cases. Here's a description for relatively simple recursive algorithmms.
Let N be the size of the initial input. Suppose there are R recursive calls in your recursive function. (Example: for mergesort, R would be 2.) Further suppose that all the recursive calls reduce the size of the initial input by the same amount, from N to M. (Example: for mergesort, M would be N/2.) And, finally, suppose that the recursive function does W work outside of the recursive calls. (Example: for mergesort, W would be N for the merge.)
Then the recurrence relation would be T(N) = R*T(M) + W. (Example: for mergesort, this would be T(N) = 2*T(N/2) + N.)
When we create an algorithm, it's always in order to be the fastest and we need to consider every case. This is why we use O, because we want to major the complexity and be sure that our algorithm will never overtake this.
To assess the complexity, you have to count the number of step. In the equation T(n) = T(n-1) + 1, there is gonna be N step before compute T(0), then the complixity is linear. (I'm talking about time complexity and not space complexity).