I recently read a wikipedia article on Manacher's algorithm and after seeing the sample implementation and dozens of other implementations... I'll be honest I have no clue how this algorithm is linear. The way I see it, it's rather in the best case O(n+n/2) but that's not linear, is it?
http://en.wikipedia.org/wiki/Longest_palindromic_substring
For each character within the original string we're trying to expand the P array in both directions until we either reach a string boundary or the symmetrical property is not satisfied. If it would only be like so this'd mean O(n^2) but with the extra observations will be less than that. Still at most I could get my head down to O(n+n/2) but not to O(n) as that would esentially mean the internal nested loop o(1). Anything higher than that and it's higher breaks the linearity for the whole algorithm.
so in a nutshell, how is this algorithm linear?
O(n + n/2) is linear, O(n+n/2) ~ O(n)
The time is still proportional to n.
Or, being more precice, the limit of (n+n/2) / n as n goes to infinity (and also when it doesn't) is a finite constant. So O(n+n/2) and O(n) are equivalent.
Related
Suppose I have an algorithm that runs in O(n) for every input of size n, but only after a pre-computation step of O(n^2) for that given size n. Is the algorithm considered O(n) still, with O(n^2) amortized? Or does big O only consider one "run" of the algorithm at size n, and so the pre-computation step is included in the notation, making the true notation O(n+n^2) or O(n^2)?
It's not uncommon to see this accounted for by explicitly separating out the costs into two different pieces. For example, in the range minimum query problem, it's common to see people talk about things like an ⟨O(n2), O(1)⟩-time solution to the problem, where the O(n2) denotes the precomputation cost and the O(1) denotes the lookup cost. You also see this with string algorithms sometimes: a suffix tree provides an O(m)-preprocessing-time, O(n+z)-query-time solution to string searching, while Aho-Corasick string matching offers an O(n)-preprocessing-time, O(m+z)-query-time solution.
The reason for doing so is that the tradeoffs involved here really depend on the use case. It lets you quantitatively measure how many queries you're going to have to make before the preprocessing time starts to be worth it.
People usually care about the total time to get things done when they are talking about complexity etc.
Thus, if getting to the result R requires you to perform steps A and B, then complexity(R) = complexity(A) + complexity(B). This works out to be O(n^2) in your particular example.
You have already noted that for O analysis, the fastest growing term dominates the overall complexity (or in other words, in a pipeline, the slowest module defines the throughput).
However, complexity analysis of A and B will be typically performed in isolation if they are disjoint.
In summary, it's the amount of time taken to get the results that counts, but you can (and usually do) reason about the individual steps independent of one another.
There are cases when you cannot only specify the slowest part of the pipeline. A simple example is BFS, with the complexity O(V + E). Since E = O(V^2), it may be tempting to write the complexity of BFS as O(E) (since E > V). However, that would be incorrect, since there can be a graph with no edges! In those cases, you will still need to iterate over all the vertices.
The point of O(...) notation is not to measure how fast the algorithm is working, because in many specific cases O(n) can be significantly slower than, say O(n^3). (Imagine the algorithm which runs in 10^100 n steps vs. the one which runs in n^3 / 2 steps.) If I tell you that my algorithm runs in O(n^2) time, it tells you nothing about how long it will take for n = 1000.
The point of O(...) is to specify how the algorithm behaves when the input size grows. If I tell you that my algorithm runs in O(n^2) time, and it takes 1 second to run for n = 500, then you'll expect rather 4 seconds to for n = 1000, not 1.5 and not 40.
So, to answer your question -- no, the algorithm will not be O(n), it will be O(n^2), because if I double the input size the time will be multiplied by 4, not by 2.
I am studying algorithms, but the calculations to find Time Complexity are not that much easy for me, it is hard to remember when to use log n, n log n, n^2, n^3, 2n, etc, my doubt is all about how to consider these input functions while computing the complexity, is their any specific way to calculate the complexity ,like using for loop take's this much complexity always and so on....?
Log(n): when you are using recursion and a tree is generated use log(n).
I mean in divide and conquer when you are diving problem into 2-halfs actually you are generating a recursive tree.
its complexity is Log(n), why ? because its a binary tree in nature and for binary tree we use Log(Base2)(n).
try yourself: suppose n=4(Elements) so log(base2)(4)=2, you divide it into equal half.
nLog(n): remember Log(n) its was division till single element. after that you start merging sorted elements that take liner time
in other words Merging of elements has complexity "n" so total complexity will be n(Merging) + Log(n)(Dividing) which is finally become nLog(n).
n^2:
when you see a problem is solved in two nested loop then Complexity is n^2.
i.e Matrix/2-D arrays they computed in 2 Loops. one loop inside the outer Loop.
n^3: oh 3-D arrays, this is for 3 nested loops. loop inside loop inside loop.
2n: thanks you did not forgot to write "2" with this "n" otherwise I forgot to explain this.
so "2" in here with "n" is constant just ignore it. why ?. because if you travel to other city by AIR. you will count only hours taken by flight not the hours consumed in reaching AIR port. I mean this is minor we remove constant.
and for "n" just remember this word "Linear" i.e Big-O(n) is linear complexity. Sadly I discovered there is no Algorithm that sort elements in Linear time. i.e just in one loop.(Single array traversal).
Things To Remember:
Nominal Time: Linear Time, Complexity Big-O(n)
Polynomial Time: Not Linear Time, Complexity Big-O[ log(n), nlog(n), n^2, n^3, n^4, n^5).
Exponential Time: 2^n, n^n i.e this problem will solve in exponential time i.e N^power(n) (These are bad bad bad, not called algorithm)
There are many links on how to roughly calculate Big O and its sibling's complexity, but there is no true formula.
However, there are guidelines to help you calculate complexity such as these presented below. I suggest reviewing as many different programs and data structures to help familiarize yourself with the pattern and just study, study, study until you get it! There is a pattern and you will see it the more you study it.
Source: http://www.dreamincode.net/forums/topic/125427-determining-big-o-notation/
Nested loops are multiplied together.
Sequential loops are added.
Only the largest term is kept, all others are dropped.
Constants are dropped.
Conditional checks are constant (i.e. 1).
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).
So most everyone should know that max_element(unsorted_array) can be solved in O(n) time. I realized that while that it is easy to compute that, it seems it would be much harder to solve it in a less than optimal solution, such as n*log(log(n)) time. Now obviously an algorithm could simply be O(n + n*log(log(n)) ), where the more time consuming part of the algorithm has no real purpose. At the same time, you could just use the regular O(n) algorithm log(log(n)) times. Neither of these are very interesting.
So my question is, is there an algorithm that can find the max element in a set of numbers (stored in the container of your choice) in a way that there is no redundant loops or operations, but is Θ(n*log(log(n))) ?
Van Emde Boas Trees?
There is a basic misconception here:
O(n + n*log(log(n)) ) is exactly identical to O(n log(log(n)))
Please read the wiki page carefully: http://en.wikipedia.org/wiki/Big_O_notation
The Big-O notation is asymptotic. This means that O(f(n) + g(n)) = O(max(f(n), g(n))) for all functions f, g. This is not a trick, they are really equal.
Symbols like O(n^2), O(n), etc., are not functions, they are sets; specifically O(f(n)) means "the set of all functions which are asymptotically less than or equal to a constant times f(n)". If f(n) >= g(n), then O(f(n)) contains O(g(n)) and so adding g(n) into that equation changes nothing.
How about proof that it cannot be done.
Theory: It is possible to determine the maximum element of an unsorted array without examining every element.
Assume you have examined all but one element of the unsorted array of n (n>1) items.
There are two possibilities for largest element of the array.
The largest element you have yet seen (out of n-1).
The one element you have not seen
The unexamined element could be bigger (unless an examined element is the absolute maximum value representable); the array is unsorted.
Result: Contradiction. You must examine the nth element in order to determine the maximum (in a mathematics context; you can take a shortcut in computer science under one probably rare circumstance)
Since it doesn't matter what value n has for this, it should apply for all n except the degenerate case (n=1)
If this isn't a valid response, I may be unclear on the requirements... ?
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.