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Results tagged with Search options answers only user 22714
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Questions related to the (computational) complexity of solving problems

Each element in the sequence is part of the combination or not (a binary choice). Since there are $n$ elements, this gives $2^n$ possible combinations. Using a gray code to iterate over the elements …
answered Jul 21 '17 by orlp
Trivial arbitrarily hard example based on the existence of an one-way function $f : \mathbb{N}^k \to \{0, 1\}$. The problem: Given a set $S$ of size $n$, can we choose a tuple $t$ formed from $k$ …
answered Oct 1 '18 by orlp
Your question is ill-defined, and it's precisely that that I believe to be your misunderstanding. There isn't an inherent cost in Landau notation, or any complexity analysis measure for that matter. …
answered Feb 7 '18 by orlp
Suppose that you have a CNF formula $x$ with input size $n$, and that you can solve $\text{SAT}$ in $O(n^c)$ time for some constant $c$. We construct a new formula $y$. For each clause $l_i$ in $x$, …
answered Dec 2 '18 by orlp
A regular expression can be transformed into an NFA as you say. And an NFA can be transformed into a DFA. This latter transformation is exponential in the worst case (in terms of the size of the origi …
You can solve this problem in polynomial time. For simplicity let's assume the DFA has a single accepting state. If it has multiple you can just run this algorithm for each accepting state until you r …
Counterexample: $f(n) = 2^n$. Then $f(2n) = f(n)^2$.
This problem is exactly solved by the k-d tree, which organizes points in $k$ dimensions with logarithmic lookup time.
In the Sedgewick-Bentley talk "QUICKSORT IS OPTIMAL" they give an (informal) proof that Quicksorts average case is optimal. So the average case lower bound for sorting is $\Omega(n\lg{n})$.
As per a comment of Albert Hendriks, we need to find $i, j, k$ such that $-A[i] = 4A[j] = 5A[k]$. We can actually do this in a single linear pass. We sweep $i$ from the left, such that $-A[i]$ can on …
Then it yields that $SAT \in P$ which itself then follows that $SAT \in TIME(n^k)$. Sure. As stands, we are able to do reduce every language in $NP$ to $SAT$. Therefore, $NP \subseteq TIME(n^ … answered Apr 25 by orlp Note that there's always a prime between$p$and$2p$for any$p$thus we can eliminate$n$as a complexity parameter, as for any non-trivial instance we have$n < p$. The complement (there exists a … answered Apr 7 by orlp The crucial data structure here is an interval tree, that can be pre-computed at compile time, which has a complexity of$O(\log n + k)$to find all$k\$ ranges that contain a certain point, assuming t …