# Tag Info

108

Strictly speaking, $O(f(n))$ is a set of functions. So the value of $O(f(n))$ is simply the set of all functions that grow asymptotically not faster than $f(n)$. The notation $T(n) = O(f(n))$ is just a conventional way to write that $T(n) \in O(f(n))$. Note that this also clarifies some caveats of the $O$ notation. For example, we write that $(1/2) n^2 + ... 61 This algorithm can be re-written like this Scan A until you find an inversion. If you find one, swap and start over. If there is none, terminate. Now there can be at most$\binom{n}{2} \in \Theta(n^2)$inversions and you need a linear-time scan to find each -- so the worst-case running time is$\Theta(n^3)$. A beautiful teaching example as it trips up the ... 53 It's called linearithmic time, and is a special case of a more general class known as quasi linear. As the name may suggests, the algorithms that fall in this class almost run in linear time; in fact they have a lower complexity than algorithms which run in$O(n^k)$with$k > 1$. 50 Landau notation denotes asymptotic bounds on functions. See here for an explanation of the differences among$O$,$\Omega$and$\Theta$. Worst-, best-, average or you-name-it-case time describe distinct runtime functions: one for the sequence of highest runtime of any given$n$, one for that of lowest, and so on.. Per se, the two have nothing to do with ... 44 This is largely piggy-backing on the answers already posted, but may offer a different perspective. It's revealing that the question discusses "sufficiently small values of n". The whole point of Big-O is to describe how processing grows as a function of what's being processed. If the data being processed stays small, it's irrelevant to discuss the Big-O, ... 44$Ois a function \begin{align} O : (\mathbb{N}\to \mathbb{R}) &\to \mathbf{P}(\mathbb{N}\to \mathbb{R}) \\ f &\mapsto O(f) \end{align} i.e. it accepts a functionf$and yields a set of functions that share the asymptotic bound of (at most)$f$. And strictly speaking the correct notation is thus $$(n \mapsto T(n)) \in O(n\mapsto f(n))$$ ... 39 No, it's not possible: at least, not in an asymptotic sense, where you require the problem to keep getting strictly easier, forever, as$n \to \infty$. Let$T(n)$be the best possible running time for solving such a problem, where$n$is the size of the input. Note that the running time is a count of the number of instructions executed by the algorithm, so ... 28 Is what I wrote about big-O correct? Yes. How do big-Theta sets relate to each other, if they relate at all? They are a partition of the space of functions. If$\Theta(f)\cap \Theta(g)\not = \emptyset$, then$\Theta(f)=\Theta(g)$. Moreover,$\Theta(f)\subseteq O(f)$. Why do they relate to each other the way they do? The explanation is probably ... 26 Consider the following algorithm (or procedure, or piece of code, or whatever): Contrive(n) 1. if n = 0 then do something Theta(n^3) 2. else if n is even then 3. flip a coin 4. if heads, do something Theta(n) 5. else if tails, do something Theta(n^2) 6. else if n is odd then 7. flip a coin 8. if heads, do something Theta(n^4) 9. else if ... 26 Although it's not quite an answer to your question, the Boyer-Moore string search algorithm comes close. As Robert Moore says on his web page about the algorithm, Our algorithm has the peculiar property that, roughly speaking, the longer the pattern is, the faster the algorithm goes. In other words, generally speaking the algorithm searches for an ... 25 The issue comes down to ambiguous terminology.$(a^b)^c = a^{bc}$, but$a^{(b^c)} \neq a^{bc}$. In other words, exponents aren't associative. Conventionally, nested exponentials without parentheses are grouped in this second way, because it's more useful. So$2^{2^n} = 2^{(2^n)} \neq 2^{2n}$. If we wanted to talk about$(2^2)^n$, we could just write$2^{2n}...

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To rationalize how asymptotic notations ignore constant factors, I usually think of it like this: asymptotic complexity isn't for comparing performance of different algorithms, it's for understanding how performance of individual algorithms scales with respect to the input size. For instance, we say that a function that takes $3n$ steps is $O(n)$, because, ...

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Saying that a big-O bound is "asymptotically tight" basically means that the author should have written $\Theta(-)$. For example, $O(x^2)$ means that it's no more than some constant times $x^2$ for all large enough $x$; "asymptotically tight" means it really is some constant times $x^2$ for large enough $x$ and not, say, some constant ...

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All orders of magnitude involve a constant $C$, several of them actually. When the number of items are large enough that constant is irrelevant. The question is whether the number of items is small enough for that constant to dominate. Here's a visual way to think about it. All have a startup constant which determines their starting point on the Y axis. ...

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In mathematics, functions like this are called multilinear functions. But computer scientists probably won't generally know this terminology. This function should definitely not be called linear, either in mathematics or computer science, unless you can reasonably consider one of $m$ and $n$ a constant.

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Wikipedia says: An algorithm is said to be of polynomial time if its running time is upper bounded by a polynomial expression in the size of the input for the algorithm. $\mathcal{O}(\log n)$ is upper bounded by $\mathcal{O}(n)$, and $\mathcal{O}(n \log n )$ is upper bounded by $\mathcal{O}(n^2)$, therefore they are both in $P$.

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We have $$\frac{n!}{(n/2)!(n/4)!\cdots 2!} = \frac{n!}{(n/2)!(n/2)!} \frac{(n/2)!}{(n/4)!(n/4)!} \cdots \frac{4!}{2!2!} \frac{2!}{1!1!} = \\ \binom{n}{n/2} \binom{n/2}{n/4} \cdots \binom{4}{2} \binom{2}{1} \leq 2^n 2^{n/2} \cdots 2^4 2^2 = 2^{n+n/2+\cdots+2+1} <2^{2n} = 4^n.$$ Using Stirling's approximation we can get more refined asymptotics, but we ...

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Skiena provides a sorted list of the dominance relations between the most common functions in his book, The Algorithm Design Manual: $$n!\gg c^n \gg n^3 \gg n^2 \gg n^{1+\epsilon} \gg n \lg n \gg n \gg n^{1/2}$$ $$\gg \lg^2n \gg \lg n \gg \frac{\lg n}{\lg\lg n} \gg \lg\lg n \gg \alpha(n) \gg 1$$ Here $\alpha(n)$ denotes the inverse Ackermann function.

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Yes, of course. This is fine and perfectly acceptable. It is common and standard to see algorithms whose running time depends upon two parameters. For instance, you will often see the running time of depth-first search expressed as $O(n+m)$, where $n$ is the number of vertices and $m$ is the number of edges in the graph. This is perfectly valid. The ...

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linearithmic: adj. Of an algorithm, having running time that is O(N log N). Coined as a portmanteau of ‘linear’ and ‘logarithmic’ in Algorithms In C by Robert Sedgewick (Addison-Wesley 1990, ISBN 0-201-51425-7). http://catb.org/jargon/html/L/linearithmic.html

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In your comment you mentioned that you tried substitution but got stuck. Here's a derivation that works. The motivation is that we'd like to get rid of the $\sqrt{n}$ multiplier on the right hand side, leaving us with something that looks like $U(n) = U(\sqrt{n}) + something$. In this case, things work out very nicely: \begin{align} T(n) &= \sqrt{n}\ ... 16 There are many examples for such functions. Perhaps the easiest way to understand how to get such an example, is by manually constructing it. Let's start with function over the natural numbers, as they can be continuously completed to the reals. A good way to ensure that f\neq O(g) and g\neq O(f) is to alternate between their orders of magnitude. For ... 16 It's not actually an equation f = O(g) is a lazy shorthand that should be written f \in O(g). So if you look back at the definition of O, you should be able to see what \log n \in O(n^{c}) for any c > 0 means: For every c > 0, there exists n_{0} \geq 0 and k \geq 0 such that \log n \leq k\cdot n^{c} for all n \geq n_{0}. Always ... 16 You need to ignore for a moment the strong feeling that the "O" is in the wrong place and plough on with the definition regardless. f(n) = \log^{O(1)}n means that there exist constants k and n_0 such that, for all n\geq n_0, f(n) \leq \log^{k\cdot 1}n = \log^k n. Note that \log^k n means (\log n)^k. Functions of the form \log^{O(1)}n are ... 16 First off, both algorithms "work" for all inputs. The question is about performance. The answers to that question are kind of crappy. One way to say one algorithm is asymptotically more efficient than another is if there is some (problem-specific) input size such that for any larger input size the more efficient algorithm will take fewer "computational ... 16 a^{(b^c)} is not the same as (a^b)^c. When people write 2^{2^k}, they usually mean 2^{(2^k)}, not (2^2)^k. 16 In order to compare 2 complexities just calculate a limit of their ratios as below: \displaystyle\begin{align*} \lim_{n\to\infty}\frac{n^2log(n)}{n^2\sqrt{n}} &= \lim_{n\to\infty}\frac{log(n)}{\sqrt{n}} = \lim_{n\to\infty}\frac{log(\sqrt{n})^2}{\sqrt{n}} = \lim_{n\to\infty}\frac{2log(\sqrt{n})}{\sqrt{n}} \\ &\underset{\left| k = \... 15 Here is another (admittedly rather constructed) example, but still one I find remarkable. It is intended to show that plots can be very misleading for judging asymptotic growth. The following plots show two functions f and g — disclosed below ;) — in different ranges. Remarkably, both functions are monotonically (strictly) increasing, (... 15 The Akra-Bazzi method The Akra-Bazzi method gives asymptotics for recurrences of the form: T(x) = \sum_{1 \le i \le k} a_i T(b_i x + h_i(x)) + g(x) \quad \text{for $x \ge x_0$}  This covers the usual divide-and-conquer recurrences, but also cases in which the division is unequal. The "fudge terms" $h_i(x)$ can cater for divisions that don't come out ...

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We first need to clarify what we mean by "does this hold if we have an infinite chain?". We interpret it as an infinite sequence of functions $\{f_i:\mathbb{N}\to\mathbb{N} \mid 1\leq i\}$ such that for all $i$ we have $f_i(n) = O(f_{i+1}(n))$. Such a sequence might not have a last function. We can look at the limit of the functions in the sequence, i.e. \$...

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