I have a treesort function which performs two distinct tasks, each with its own time complexity. I figured out the avg. case time complexity of the two tasks but how do I find the overall complexity of the algorithm.
For example the algorithm takes in a random list of "n" keys x:
Sort(x):
Insert(x):
#Time complexity of O(nLog(n))
Traverse(x):
#Time complexity of O(n)
Do I just add the two complexities together to give me O(n + nLog(n)) or do I take the dominant task (in this case Insert) and end up with an overall complexity of O(nLog(n))
In a simple case like this,
O((n) + (n log(n)) = O(n + n log(n))
= O(n (log(n) + 1))
= O(n log(n))
or do I take the dominant task (in this case Insert) and end up with an over complexity of O(nLog(n))
That's right. As n grows, first element in O(n + nLog(n)) sum will become less and less significant. Thus, for sufficiently large n, its contribution can be ignored.
You need to take the dominant one.
The whole idea of measuring complexity this way is based on the assumption that you want to know what happens with large ns.
So if you have a polynomial, you can discard all but the highest order element, if you have a logarithm, you can ignore the base and so on.
In everyday practice however, these differences may start to matter, so it's sometimes good to have a more precise picture of your algorithm's complexity, even down to the level where you assign different weights to different operations.
(Returning to your original questions, assuming you're using base 2 logarithms, at n=1048576, the difference between n+n*logn and n*logn is around 5%, which is probably not really worth worrying about.)
Related
In the Princeton tutorial on Coursera the lecturer explains the common order-of-growth functions that are encountered. He says that linear and linearithmic running times are "what we strive" for and his reasoning was that as the input size increases so too does the running time. I think this is where he made a mistake because I have previously heard him refer to a linear order-of-growth as unsatisfactory for an efficient algorithm.
While he was speaking he also showed a chart that plotted the different running times - constant and logarithmic running times looked to be more efficient. So was this a mistake or is this true?
It is a mistake when taken in the context that O(n) and O(n log n) functions have better complexity than O(1) and O(log n) functions. When looking typical cases of complexity in big O notation:
O(1) < O(log n) < O(n) < O(n log n) < O(n^2)
Notice that this doesn't necessarily mean that they will always be better performance-wise - we could have an O(1) function that takes a long time to execute even though its complexity is unaffected by element count. Such a function would look better in big O notation than an O(log n) function, but could actually perform worse in practice.
Generally speaking: a function with lower complexity (in big O notation) will outperform a function with greater complexity (in big O notation) when n is sufficiently high.
You're missing the broader context in which those statements must have been made. Different kinds of problems have different demands, and often even have theoretical lower bounds on how much work is absolutely necessary to solve them, no matter the means.
For operations like sorting or scanning every element of a simple collection, you can make a hard lower bound of the number of elements in the collection for those operations, because the output depends on every element of the input. [1] Thus, O(n) or O(n*log(n)) are the best one can do.
For other kinds of operations, like accessing a single element of a hash table or linked list, or searching in a sorted set, the algorithm needn't examine all of the input. In those settings, an O(n) operation would be dreadfully slow.
[1] Others will note that sorting by comparisons also has an n*log(n) lower bound, from information-theoretic arguments. There are non-comparison based sorting algorithms that can beat this, for some types of input.
Generally speaking, what we strive for is the best we can manage to do. But depending on what we're doing, that might be O(1), O(log log N), O(log N), O(N), O(N log N), O(N2), O(N3), or (or certain algorithms) perhaps O(N!) or even O(2N).
Just for example, when you're dealing with searching in a sorted collection, binary search borders on trivial and gives O(log N) complexity. If the distribution of items in the collection is reasonably predictable, we can typically do even better--around O(log log N). Knowing that, an algorithm that was O(N) or O(N2) (for a couple of obvious examples) would probably be pretty disappointing.
On the other hand, sorting is generally quite a bit higher complexity--the "good" algorithms manage O(N log N), and the poorer ones are typically around O(N2). Therefore, for sorting an O(N) algorithm is actually very good (in fact, only possible for rather constrained types of inputs), and we can pretty much count on the fact that something like O(log log N) simply isn't possible.
Going even further, we'd be happy to manage a matrix multiplication in only O(N2) instead of the usual O(N3). We'd be ecstatic to get optimum, reproducible answers to the traveling salesman problem or subset sum problem in only O(N3), given that optimal solutions to these normally require O(N!).
Algorithms with a sublinear behavior like O(1) or O(Log(N)) are special in that they do not require to look at all elements. In a way this is a fallacy because if there are really N elements, it will take O(N) just to read or compute them.
Sublinear algorithms are often possible after some preprocessing has been performed. Think of binary search in a sorted table, taking O(Log(N)). If the data is initially unsorted, it will cost O(N Log(N)) to sort it first. The cost of sorting can be balanced if you perform many searches, say K, on the same data set. Indeed, without the sort, the cost of the searches will be O(K N), and with pre-sorting O(N Log(N)+ K Log(N)). You win if K >> Log(N).
This said, when no preprocessing is allowed, O(N) behavior is ideal, and O(N Log(N)) is quite comfortable as well (for a million elements, Lg(N) is only 20). You start screaming with O(N²) and worse.
He said those algorithms are what we strive for, which is generally true. Many algorithms cannot possibly be improved better than logarithmic or linear time, and while constant time would be better in a perfect world, it's often unattainable.
constant time is always better because the time (or space) complexity doesn't depend on the problem size... isn't it a great feature? :-)
then we have O(N) and then Nlog(N)
did you know? problems with constant time complexity exist!
e.g.
let A[N] be an array of N integer values, with N > 3. Find and algorithm to tell if the sum of the first three elements is positive or negative.
What we strive for is efficiency, in the sense of designing algorithms with a time (or space) complexity that does not exceed their theoretical lower bound.
For instance, using comparison-based algorithms, you can't find a value in a sorted array faster than Omega(Log(N)), and you cannot sort an array faster than Omega(N Log(N)) - in the worst case.
Thus, binary search O(Log(N)) and Heapsort O(N Log(N)) are efficient algorithms, while linear search O(N) and Bubblesort O(N²) are not.
The lower bound depends on the problem to be solved, not on the algorithm.
Yes constant time i.e. O(1) is better than linear time O(n) because the former is not depending on the input-size of the problem. The order is O(1) > O (logn) > O (n) > O (nlogn).
Linear or linearthimic time we strive for because going for O(1) might not be realistic as in every sorting algorithm we atleast need a few comparisons which the professor tries to prove with his decison Tree- comparison analysis where he tries to sort three elements a b c and proves a lower bound of nlogn. Check his "Complexity of Sorting" in the Mergesort lecture.
If i have something with O(logN) and add it to something with O(1)
Is the overall complexity still logN?
thanks
Often big-O notation is an approximation. For example, you might say logN when the actual complexity is 4logN + 7. This is still considered to be logN time, because the major factor is the behaviour as N changes.
If you had some algorithm that is N^2 + logN, then the most significant term is N^2 and the logN quickly becomes unimportant as N increases... In that case, you might simply say it is O(N^2) because it describes the characteristic time complexity of the algorithm.
So it depends on your needs. If you simply need to describe the nature of the algorithm, then logN should suffice. If you need to completely categorize every part of it or compare with similar but optimized algorithms, then add in all the terms.
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.
I'm confused about how to do big-O analysis for the following problem -
find an element from an array of integers. ( an example problem)
my solution
sort the array using bubble sort ( n^2 )
binary search on the array for a given element (logn)
now the big-O for this is n^2 or n^2 + logn ? Should we only consider the higher term ?
Big-O for a problem is that of the best algorithm that exists for a problem. That for an algorithm made of two steps (like yours) is indeed the highest of the two, because e.g.
O(n^2) == O(n^2 + log n)
However, you can't say that O(n^2) is the correct O for your sample problem without proving that no better algorithm exists (which is of course not the case in the example;-).
Only the higher order term. The complexity is always the complexity of the highest term.
The way you did it, it would be O(n^2), since for large n, n^2 >>> logn
To put the analysis, well, more-practically (if you prefer, crudely) than Alex did, the added log n doesn't have an appreciable effect on the outcome. Consider analyzing this in a real-world system with one million inputs, each of which takes one millisecond to sort, and one millisecond to search (it's a highly-hypothetical example). Given O(n^2), the sort takes over thirty years. The search takes an additional 0.014 seconds. Which part do you care about improving? :)
Now, you'll see algorithms which clock in at O(n^2 x logn). The effect of multiplying n^2 by log n makes log n significant - in our example, it sees our thirty years and raises us four centuries.
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.