We are going over the master theorem in my algorithms class, and for one problem, I'm trying to compare nlogn vs 1 to figure out which case of the MT it falls under. But I'm having a hard timing figuring out which is bigger.
Edit: This is for solving a recurrence problem. The equation is T(n) = 2T(n/4) + N*LogN. Just threw this in incase it helps.
Think about it this way:
O(N*LogN) will increase with N in such a way that for any X, no matter how large, you can find a value of N such that N*LogN is greater than X.
O(1) will stay the same, no matter what N is.
This means that O(1) is asymptotically better, i.e. for some (perhaps very high) value of N the O(N*LogN) will become slower.
If an algorithm is O(NlogN) that means that there exists a number A and a quantity of execution time B, such that for any input size N greater than A, the execution time will be less than B times NlogN.
If an algorithm is O(1), that would mean that there exists some fixed amount of time C in which the algorithm would be guaranteed to complete regardless of the input size.
In comparing two algorithms, one of which is O(NlgN) and one of which is O(1), one will generally discover that the O(1) algorithm is faster for values of N that are sufficiently large, but in many cases the O(NlgN) algorithm may be faster for small values of N.
Indeed, while something like an O(N^3) or O(N^4) algorithm would generally seem pretty bad, it's possible that even an O(N^4) algorithm may outperform an O(1) algorithm if N is usually a small number (e.g. 1-5 or so) and never gets very big (even an occasional value of 50 could seriously dog performance).
Related
When talking about complexity in general, things like O(3n) tend to be simplified to O(n) and so on. This is merely theoretical, so how does complexity work in reality? Can O(3n) also be simplified to O(n)?
For example, if a task implies that solution must be in O(n) complexity and in our code we have 2 times linear search of an array, which is O(n) + O(n). So, in reality, would that solution be considered as linear complexity or not fast enough?
Note that this question is asking about real implementations, not theoretical. I'm already aware that O(n) + O(n) is simplified to O(n)?
Bear in mind that O(f(n)) does not give you the amount of real-world time that something takes: only the rate of growth as n grows. O(n) only indicates that if n doubles, the runtime doubles as well, which lumps functions together that take one second per iteration or one millennium per iteration.
For this reason, O(n) + O(n) and O(2n) are both equivalent to O(n), which is the set of functions of linear complexity, and which should be sufficient for your purposes.
Though an algorithm that takes arbitrary-sized inputs will often want the most optimal function as represented by O(f(n)), an algorithm that grows faster (e.g. O(n²)) may still be faster in practice, especially when the data set size n is limited or fixed in practice. However, learning to reason about O(f(n)) representations can help you compose algorithms to have a predictable—optimal for your use-case—upper bound.
Yes, as long as k is a constant, you can write O(kn) = O(n).
The intuition behind is that the constant k doesn't increase with the size of the input space and at some point will be incomparably small to n, so it doesn't have much influence on the overall complexity.
Yes - as long as the number k of array searches is not affected by the input size, even for inputs that are too big to be possible in practice, O(kn) = O(n). The main idea of the O notation is to emphasize how the computation time increases with the size of the input, and so constant factors that stay the same no matter how big the input is aren't of interest.
An example of an incorrect way to apply this is to say that you can perform selection sort in linear time because you can only fit about one billion numbers in memory, and so selection sort is merely one billion array searches. However, with an ideal computer with infinite memory, your algorithm would not be able to handle more than one billion numbers, and so it is not a correct sorting algorithm (algorithms must be able to handle arbitrarily large inputs unless you specify a limit as a part of the problem statement); it is merely a correct algorithm for sorting up to one billion numbers.
(As a matter of fact, once you put a limit on the input size, most algorithms will become constant-time because for all inputs within your limit, the algorithm will solve it using at most the amount of time that is required for the biggest / most difficult input.)
I've recently finished two tests for a data a structures class and I've got a question related to O(n) vs O(n^2) wrong twice. I was wondering if I could get help understanding the problem. The problem is:
Suppose that Algorithm A has runtime O(n^2) and Algorithm B has runtime O(n). What can we say about the runtime of these two algorithms when n=17?
a) We cannot say anything about the specific runtimes when n=17
b) Algorithm A will run much FASTER than Algorithm B
c) Algorithm A will run much SLOWER than Algorithm B
For both tests I answered C based on: https://en.wikipedia.org/wiki/Big_O_notation#Orders_of_common_functions. I knew B made no sense based on the link provided. Now I am starting to think that its A. I'm guessing its A because n is small. If that is the cases I am wondering when is n sufficiently larger enough that C would true.
There are actually two issues here.
The first is the one you mentioned. Orders of growth are asymptotic. They just say that there exists some n0 for which, for any n > n0, the function is bounded in some way. They say nothing about specific values of n, only "large enough" ones.
The second problem (which you did not mention), is that O is just an upper bound (as opposed to Θ), and so even for large enough n you can't compare the two. So if A = √n and B = n, then obviously B grows faster than A. However, A and B still fit the question, as √ n = O(n2) and n = O(n).
The answer is A.
Big Oh order of a function f(x) is g(x) if f(x)<=K*g(x) forall x>some real number
Big Oh of 3*n+2 and n is O(n) since 4*n is greater than both functions for all x>2 . since both the Big oh notation of the functions are same we cannot say that they run in the same time for some value.For example at n=0 the value of first function is 2 and the second one is 0
So we cannot exactly relate the running times of two functions for some value.
The answer is a): You can't really say anything for any specific number just given the big O notation.
Counter-example for c: B has a runtime of 1000*n (= O(n)), A has a runtime of n^2.
When doing algorithm analysis, specifically Big Oh, you should really only think about input sizes tending towards infinity. With such a small size (tens vs. thousands vs. millions), there is not a significant difference between the two. However, in general O(n) should run faster than O(n^2), even if it the difference is less than few milliseconds. I suspect the key word in that question is much.
My answer is based on my experience in competitive programming, which require a basic understanding of the O or called Big O.
When you talk about which one is faster and which one is slower, of course, basic calculation is done that. O(n) is faster than O(n^2), big oh is used based on worst case scenario.
Now when exactly that happen? Well, in competitive programming, we used 10^8 thumb rule. It's mean if an algorithm complexity is O(n) and then there is around n = 10^8 with time limit around 1 second, the algorithm can solve the problem.
But what if the algorithm complexity is O(n^2)? No, then, it will need around (10^8)^2 which is more than 1 second. (1-second computer can process around 10^8 operation).
So, for 1 second time, the max bound for O(n^2) is around 10^4 meanwhile for O(n) can do up to 10^8. This is where we can clearly see the different between the two complexity in 1 second time pass on a computer.
Question
Hi I am trying to understand what order of complexity in terms of Big O notation is. I have read many articles and am yet to find anything explaining exactly 'order of complexity', even on the useful descriptions of Big O on here.
What I already understand about big O
The part which I already understand. about Big O notation is that we are measuring the time and space complexity of an algorithm in terms of the growth of input size n. I also understand that certain sorting methods have best, worst and average scenarios for Big O such as O(n) ,O(n^2) etc and the n is input size (number of elements to be sorted).
Any simple definitions or examples would be greatly appreciated thanks.
Big-O analysis is a form of runtime analysis that measures the efficiency of an algorithm in terms of the time it takes for the algorithm to run as a function of the input size. It’s not a formal bench- mark, just a simple way to classify algorithms by relative efficiency when dealing with very large input sizes.
Update:
The fastest-possible running time for any runtime analysis is O(1), commonly referred to as constant running time.An algorithm with constant running time always takes the same amount of time
to execute, regardless of the input size.This is the ideal run time for an algorithm, but it’s rarely achievable.
The performance of most algorithms depends on n, the size of the input.The algorithms can be classified as follows from best-to-worse performance:
O(log n) — An algorithm is said to be logarithmic if its running time increases logarithmically in proportion to the input size.
O(n) — A linear algorithm’s running time increases in direct proportion to the input size.
O(n log n) — A superlinear algorithm is midway between a linear algorithm and a polynomial algorithm.
O(n^c) — A polynomial algorithm grows quickly based on the size of the input.
O(c^n) — An exponential algorithm grows even faster than a polynomial algorithm.
O(n!) — A factorial algorithm grows the fastest and becomes quickly unusable for even small values of n.
The run times of different orders of algorithms separate rapidly as n gets larger.Consider the run time for each of these algorithm classes with
n = 10:
log 10 = 1
10 = 10
10 log 10 = 10
10^2 = 100
2^10= 1,024
10! = 3,628,800
Now double it to n = 20:
log 20 = 1.30
20 = 20
20 log 20= 26.02
20^2 = 400
2^20 = 1,048,576
20! = 2.43×1018
Finding an algorithm that works in superlinear time or better can make a huge difference in how well an application performs.
Say, f(n) in O(g(n)) if and only if there exists a C and n0 such that f(n) < C*g(n) for all n greater than n0.
Now that's a rather mathematical approach. So I'll give some examples. The simplest case is O(1). This means "constant". So no matter how large the input (n) of a program, it will take the same time to finish. An example of a constant program is one that takes a list of integers, and returns the first one. No matter how long the list is, you can just take the first and return it right away.
The next is linear, O(n). This means that if the input size of your program doubles, so will your execution time. An example of a linear program is the sum of a list of integers. You'll have to look at each integer once. So if the input is an list of size n, you'll have to look at n integers.
An intuitive definition could define the order of your program as the relation between the input size and the execution time.
Others have explained big O notation well here. I would like to point out that sometimes too much emphasis is given to big O notation.
Consider matrix multplication the naïve algorithm has O(n^3). Using the Strassen algoirthm it can be done as O(n^2.807). Now there are even algorithms that get O(n^2.3727).
One might be tempted to choose the algorithm with the lowest big O but it turns for all pratical purposes that the naïvely O(n^3) method wins out. This is because the constant for the dominating term is much larger for the other methods.
Therefore just looking at the dominating term in the complexity can be misleading. Sometimes one has to consider all terms.
Big O is about finding an upper limit for the growth of some function. See the formal definition on Wikipedia http://en.wikipedia.org/wiki/Big_O_notation
So if you've got an algorithm that sorts an array of size n and it requires only a constant amount of extra space and it takes (for example) 2 n² + n steps to complete, then you would say it's space complexity is O(n) or O(1) (depending on wether you count the size of the input array or not) and it's time complexity is O(n²).
Knowing only those O numbers, you could roughly determine how much more space and time is needed to go from n to n + 100 or 2 n or whatever you are interested in. That is how well an algorithm "scales".
Update
Big O and complexity are really just two terms for the same thing. You can say "linear complexity" instead of O(n), quadratic complexity instead of O(n²), etc...
I see that you are commenting on several answers wanting to know the specific term of order as it relates to Big-O.
Suppose f(n) = O(n^2), we say that the order is n^2.
Be careful here, there are some subtleties. You stated "we are measuring the time and space complexity of an algorithm in terms of the growth of input size n," and that's how people often treat it, but it's not actually correct. Rather, with O(g(n)) we are determining that g(n), scaled suitably, is an upper bound for the time and space complexity of an algorithm for all input of size n bigger than some particular n'. Similarly, with Omega(h(n)) we are determining that h(n), scaled suitably, is a lower bound for the time and space complexity of an algorithm for all input of size n bigger than some particular n'. Finally, if both the lower and upper bound are the same complexity g(n), the complexity is Theta(g(n)). In other words, Theta represents the degree of complexity of the algorithm while big-O and big-Omega bound it above and below.
Constant Growth: O(1)
Linear Growth: O(n)
Quadratic Growth: O(n^2)
Cubic Growth: O(n^3)
Logarithmic Growth: (log(n)) or O(n*log(n))
Big O use Mathematical Definition of complexity .
Order Of use in industrial Definition of complexity .
Hi I would really appreciate some help with Big-O notation. I have an exam in it tomorrow and while I can define what f(x) is O(g(x)) is, I can't say I thoroughly understand it.
The following question ALWAYS comes up on the exam and I really need to try and figure it out, the first part seems easy (I think) Do you just pick a value for n, compute them all on a claculator and put them in order? This seems to easy though so I'm not sure. I'm finding it very hard to find examples online.
From lowest to highest, what is the
correct order of the complexities
O(n2), O(log2 n), O(1), O(2n), O(n!),
O(n log2 n)?
What is the
worst-case computational-complexity of
the Binary Search algorithm on an
ordered list of length n = 2k?
That guy should help you.
From lowest to highest, what is the
correct order of the complexities
O(n2), O(log2 n), O(1), O(2n), O(n!),
O(n log2 n)?
The order is same as if you compare their limit at infinity. like lim(a/b), if it is 1, then they are same, inf. or 0 means one of them is faster.
What is the worst-case
computational-complexity of the Binary
Search algorithm on an ordered list of
length n = 2k?
Find binary search best/worst Big-O.
Find linked list access by index best/worst Big-O.
Make conclusions.
Hey there. Big-O notation is tough to figure out if you don't really understand what the "n" means. You've already seen people talking about how O(n) == O(2n), so I'll try to explain exactly why that is.
When we describe an algorithm as having "order-n space complexity", we mean that the size of the storage space used by the algorithm gets larger with a linear relationship to the size of the problem that it's working on (referred to as n.) If we have an algorithm that, say, sorted an array, and in order to do that sort operation the largest thing we did in memory was to create an exact copy of that array, we'd say that had "order-n space complexity" because as the size of the array (call it n elements) got larger, the algorithm would take up more space in order to match the input of the array. Hence, the algorithm uses "O(n)" space in memory.
Why does O(2n) = O(n)? Because when we talk in terms of O(n), we're only concerned with the behavior of the algorithm as n gets as large as it could possibly be. If n was to become infinite, the O(2n) algorithm would take up two times infinity spaces of memory, and the O(n) algorithm would take up one times infinity spaces of memory. Since two times infinity is just infinity, both algorithms are considered to take up a similar-enough amount of room to be both called O(n) algorithms.
You're probably thinking to yourself "An algorithm that takes up twice as much space as another algorithm is still relatively inefficient. Why are they referred to using the same notation when one is much more efficient?" Because the gain in efficiency for arbitrarily large n when going from O(2n) to O(n) is absolutely dwarfed by the gain in efficiency for arbitrarily large n when going from O(n^2) to O(500n). When n is 10, n^2 is 10 times 10 or 100, and 500n is 500 times 10, or 5000. But we're interested in n as n becomes as large as possible. They cross over and become equal for an n of 500, but once more, we're not even interested in an n as small as 500. When n is 1000, n^2 is one MILLION while 500n is a "mere" half million. When n is one million, n^2 is one thousand billion - 1,000,000,000,000 - while 500n looks on in awe with the simplicity of it's five-hundred-million - 500,000,000 - points of complexity. And once more, we can keep making n larger, because when using O(n) logic, we're only concerned with the largest possible n.
(You may argue that when n reaches infinity, n^2 is infinity times infinity, while 500n is five hundred times infinity, and didn't you just say that anything times infinity is infinity? That doesn't actually work for infinity times infinity. I think. It just doesn't. Can a mathematician back me up on this?)
This gives us the weirdly counterintuitive result where O(Seventy-five hundred billion spillion kajillion n) is considered an improvement on O(n * log n). Due to the fact that we're working with arbitrarily large "n", all that matters is how many times and where n appears in the O(). The rules of thumb mentioned in Julia Hayward's post will help you out, but here's some additional information to give you a hand.
One, because n gets as big as possible, O(n^2+61n+1682) = O(n^2), because the n^2 contributes so much more than the 61n as n gets arbitrarily large that the 61n is simply ignored, and the 61n term already dominates the 1682 term. If you see addition inside a O(), only concern yourself with the n with the highest degree.
Two, O(log10n) = O(log(any number)n), because for any base b, log10(x) = log_b(*x*)/log_b(10). Hence, O(log10n) = O(log_b(x) * 1/(log_b(10)). That 1/log_b(10) figure is a constant, which we've already shown drop out of O(n) notation.
Very loosely, you could imagine picking extremely large values of n, and calculating them. Might exceed your calculator's range for large factorials, though.
If the definition isn't clear, a more intuitive description is that "higher order" means "grows faster than, as n grows". Some rules of thumb:
O(n^a) is a higher order than O(n^b) if a > b.
log(n) grows more slowly than any positive power of n
exp(n) grows more quickly than any power of n
n! grows more quickly than exp(kn)
Oh, and as far as complexity goes, ignore the constant multipliers.
That's enough to deduce that the correct order is O(1), O(log n), O(2n) = O(n), O(n log n), O(n^2), O(n!)
For big-O complexities, the rule is that if two things vary only by constant factors, then they are the same. If one grows faster than another ignoring constant factors, then it is bigger.
So O(2n) and O(n) are the same -- they only vary by a constant factor (2). One way to think about it is to just drop the constants, since they don't impact the complexity.
The other problem with picking n and using a calculator is that it will give you the wrong answer for certain n. Big O is a measure of how fast something grows as n increases, but at any given n the complexities might not be in the right order. For instance, at n=2, n^2 is 4 and n! is 2, but n! grows quite a bit faster than n^2.
It's important to get that right, because for running times with multiple terms, you can drop the lesser terms -- ie, if O(f(n)) is 3n^2+2n+5, you can drop the 5 (constant), drop the 2n (3n^2 grows faster), then drop the 3 (constant factor) to get O(n^2)... but if you don't know that n^2 is bigger, you won't get the right answer.
In practice, you can just know that n is linear, log(n) grows more slowly than linear, n^a > n^b if a>b, 2^n is faster than any n^a, and n! is even faster than that. (Hint: try to avoid algorithms that have n in the exponent, and especially avoid ones that are n!.)
For the second part of your question, what happens with a binary search in the worst case? At each step, you cut the space in half until eventually you find your item (or run out of places to look). That is log2(2k). A search where you just walk through the list to find your item would take n steps. And we know from the first part that O(log(n)) < O(n), which is why binary search is faster than just a linear search.
Good luck with the exam!
In easy to understand terms the Big-O notation defines how quickly a particular function grows. Although it has its roots in pure mathematics its most popular application is the analysis of algorithms which can be analyzed on the basis of input size to determine the approximate number of operations that must be performed.
The benefit of using the notation is that you can categorize function growth rates by their complexity. Many different functions (an infinite number really) could all be expressed with the same complexity using this notation. For example, n+5, 2*n, and 4*n + 1/n all have O(n) complexity because the function g(n)=n most simply represents how these functions grow.
I put an emphasis on most simply because the focus of the notation is on the dominating term of the function. For example, O(2*n + 5) = O(2*n) = O(n) because n is the dominating term in the growth. This is because the notation assumes that n goes to infinity which causes the remaining terms to play less of a role in the growth rate. And, by convention, any constants or multiplicatives are omitted.
Read Big O notation and Time complexity for more a more in depth overview.
See this and look up for solutions here is first one.
What is the difference between O(n^2) and O(n.log(n))?
n^2 grows in complexity more quickly.
Big O calculates an upper limit of running time relative to the size of a data set (n).
An O(n*log(n)) is not always faster than a O(n^2) algorithm, but when considering the worst case it probably is. A O(n^2)-algorithm takes ~4 times longer when you duplicate the working set (worst case), for O(n*log(n))-algorithm it's less. The bigger your data set is the more it usually gets faster using an O(n*log(n))-algorithm.
EDIT: Thanks to 'harms', I'll correct a wrong statement in my first answer: I told that when considering the worst case O(n^2) would always be slower than O(n*log(n)), that's wrong since both are except for a constant factor!
Sample: Say we have the worst case and our data set has size 100.
O(n^2) --> 100*100 = 10000
O(n*log(n)) --> 100*2 = 200 (using log_10)
The problem is that both can be multiplied by a constant factor, say we multiply c to the latter one. The result will be:
O(n^2) --> 100*100 = 10000
O(n*log(n)) --> 100*2*c = 200*c (using log_10)
So for c > 50 we get O(n*log(n)) > O(n^2), for n=100.
I have to update my statement: For every problem, when considering the worst case, a O(n*log(n)) algorithm will be quicker than a O(n^2) algorithm for arbitrarily big data sets.
The reason is: The choice of c is arbitrary but constant. If you increase the data set large enough it will dominate the effect of every constant choice of c and when discussing two algorithms the cs for both are constant!
You'll need to be a bit more specific about what you are asking, but in this case O(n log(n)) is faster
Algorithms that run in O(nlog(n)) time are generally faster than those that run in O(n^2).
Big-O defines the upper-bound on performance. As the size of the data set grows (n) the length of time it takes to perform the task. You might be interested in the iTunes U algorithms course from MIT.
n log(n) grows significantly slower
"Big Oh" notation gives an estimated upper bound on the growth in the running time of an algorithm. If an algorithm is supposed to be O(n^2), in a naive way, it says that for n=1, it takes a max. time 1 units, for n=2 it takes max. time 4 units and so on. Similarly for O(n log(n)), it says the grown will be such that it obeys the upper bound of O(n log(n)).
(If I am more than naive here, please correct me in a comment).
I hope that helps.