Cosa significa

Cosa significa ?$\log^{O(1)}n$

Sono consapevole della notazione big-O, ma questa notazione non ha senso per me. Non riesco a trovare nulla al riguardo, perché non c'è modo che un motore di ricerca lo interpreti correttamente.

Per un po 'di contesto, la frase in cui l'ho trovata recita "[...] chiamiamo una funzione [efficiente] se usa lo spazio $O(\log n)$ e al massimo $\log^{O(1)}n$ per articolo."

Devi ignorare per un momento la forte sensazione che la " " sia nel posto sbagliato e proseguire con la definizione a prescindere. f ( n ) = log O ( 1 ) n significa che esistono costanti k e n 0 tali che, per tutti n ≥ n 0 , f ( n ) ≤ log k ⋅ 1 n = log $O$$f(n) = \log^{O(1)}n$$k$$n_0$$n\geq n_0$ .$f(n) \leq \log^{k\cdot 1}n = \log^k n$

Si noti che significa ( log n ) k . Le funzioni del log del modulo O ( 1 ) n sono spesso chiamate pollogaritmiche e si potrebbe sentire la gente dire " f è polilogo $\log^k n$$(\log n)^k$$\log^{O(1)}n$$f$ ".$n$

Noterai che è facile dimostrare che , poiché 2 n ≤ k n per tutti n ≥ 0 , dove k = 2 . Ci si potrebbe chiedere se 2 log n = log O ( 1 ) n . La risposta è sì poiché, per n abbastanza grande , registro n ≥ 2 , quindi 2 log n ≤ log 2 n per abbastanza grande $2n=O(n)$$2n\leq k n$$n\geq 0$$k=2$$2\log n = \log^{O(1)}n$$n$$\log n\geq 2$$2\log n \leq \log^2n$ .$n$

In una nota correlata, vedrai spesso polinomi scritti come $n^{O(1)}$: same idea.

Si tratta di un abuso della notazione che può essere interpretato dalla convenzione del segnaposto generalmente accettata : ogni volta che trovi un termine Landau , sostituiscilo (nella tua mente o sulla carta) con una funzione arbitraria g ∈ $O(f)$ .$g \in O(f)$

Quindi se lo trovi

$\qquad f(n) = \log^{O(1)} n$

devi leggere

per alcuni g ∈ O ( 1 ) $\qquad f(n) = \log^{g(n)} n$$g \in O(1). \hspace{5cm} (1)$

Note the difference from saying "$\log$ to the power of some constant": $g = n \mapsto 1/n$ is a distinct possibility.

Warning: The author may be employing even more abuse of notation and want you to read

$\qquad f(n) \in O(\log^{g(n)} n)$ for some $g \in O(1). \hspace{4.3cm} (2)$

Note the difference between (1) and (2); while it works out to define the same set of positive-valued functions here, this does not always work. Do not move $O$ around in expressions without care!

It means that the function grows at most as $\log$ to the power of some constant, i.e. $\log^2(n)$ or $\log^5(n)$ or $\log^{99999}(n)$...

"At most $\log^{O(1)} n$" means that there is a constant $c$ such that what is being measured is $O(\log^c n)$.

In a more general context, $f(n) \in \log^{O(1)} n$ is equivalent to the statement that there exists (possibly negative) constants $a$ and $b$ such that $f(n) \in O(\log^a n)$ and $f(n) \in \Omega(\log^b n)$.

It is easy to overlook the $\Omega(\log^b n)$ lower bound. In a setting where that would matter (which would be very uncommon if you're exclusively interested in studying asymptotic growth), you shouldn't have complete confidence that the author actually meant the lower bound, and would have to rely on the context to make sure.

The literal meaning of the notation $\log^{O(1)} n$ is doing arithmetic on the family of functions, resulting in the family of all functions $\log^{g(n)} n$, where $g(n) \in O(1)$. This works in pretty much the same as how multiplying $O(g(n))$ by $h(n)$ results in $O(g(n) h(n))$, except that you get a result that isn't expressed so simply.

Since the details of the lower bound are in probably unfamiliar territory, it's worth looking at some counterexamples. Recall that any $g(n) \in O(1)$ is bounded in magnitude; that there is a constant $c$ such that for all sufficiently large $n$, $|g(n)| < c$.

When looking at asymptotic growth, usually only the upper bound $g(n) < c$ matters, since, e.g., you already know the function is positive. However, in full generality you have to pay attention to the lower bound $g(n) > -c$.

This means, contrary to more typical uses of big-oh notation, functions that decrease too rapidly can fail to be in $\log^{O(1)} n$; for example,

A counterexample of a somewhat different sort is that $-1 \notin \log^{O(1)} n$.