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Hot answers tagged upper-bound

8

So I'm not completely sure, but I think you're asking to count the number of strings of size $n$ (over the alphabet $\{a, b\}$) where the factor/substring $aa$ does not appear right? In this case, there are a few combinatorial approaches that you can take. Both Yuval and ADG have given simpler and more intuitive arguments, so I definitely suggest checking ...

7

Lee Gao's answer is excellent. Here is a different account. Consider the following automaton: This is an unambiguous finite automaton (UFA) without $\epsilon$ transitions: an NFA such that each word has exactly one accepting path. The number of words of length $n$ is thus the number of paths of length $n$ from the starting state to an accepting state (since ...

4

@Lee Gao's is too complex (I haven't even read the whole thing), here is a simplistic approach: Let f(n) be all desired strings out of which let a(n) be strings that end at a and b(n) be strings that end at b. Now for every string that ends at b we can directly add a to get ba in ending and a valid string: $$a(n)=b(n-1)\tag{1}$$ Note than we cannot add a ...

3

There are many functions that satisfy your condition. Here are a few. $a_i=1+c^{-i}$, for some constant $c\ge 2$. $a_i= 1+b_i/p_i$, where $p_i$ is the $i^{th}$ prime number and $b_i$ is any integer between 1 and $p_i-1$. $a_i =1+\alpha^i/\lceil\alpha^i\rceil$, for any positive transdental number $\alpha$. For example, you can take $\alpha = \pi$ or $\alpha=... 3 Finding a good analytic characterization of$n(N)$is tricky. Let's first consider the relaxation where$N = \frac{n}{\log n}$without the flooring restriction. Here's a somewhat nonintuitive approximation: let$m(z) = 1 + \frac{1}{z}$, let's see how$\frac{m(z)}{\log m(z)}$behaves as a function of$z$: $$\begin{array}{ccc} z = 1 & 10 & 100 & ... 3 You say that N=\lfloor\tfrac{n}{logn}\rfloor. But since you have a linear recurrence in N, not n, you really want n as a function of N. We have n ≈ N \log n. You substitute this for n and get n ≈ N \log (N \log n) = N \log N + N \log \log n. \log \log n is small compared to \log N, so we ignore it and get T(N)=T(N-1)+\mathcal{O}(N \log N) ... 3 Let us start by giving an alternative formula for \sum_{u,d} |L(u,d)|. For a node x, let \pi^{(d)}(x) be its d'th parent, and let h(x) be its height. Then$$ \sum_{u,d} |L(u,d)| = \sum_x |\{ L(\pi^{(d)}(x),d) : 0 \leq d \leq h(x)\}|. $$It suffices to show that |\{ L(\pi^{(d)}(x),d) : 0 \leq d \leq h(x)\}| = O(\sqrt{n}) for all nodes x. Let ... 2 The Risch algorithm is undecidable by Richardson Theorem if absolute value is allowed or semi-undecidable with \log_2, \pi, e^x, \sin x. Akitoshi Kawamura in his dissertation Computational Complexity in Analysis and Geometry has proved that integration is \#P{-}complete operation. There are multiple modifications of the Risch algorithm, but still nobody ... 2 Hint: Use a union bound. Note that for m \geq 2^k you cannot provide any non-trivial bound, since there are formulas with 2^k clauses of width k that are unsatisfiable. The same example shows that the bounded hinted above is tight (for all m \leq 2^k). 2 Let me write the definition in a little bit more detail. Suppose that f(n),g(n) are two functions from \mathbb{N} to \mathbb{R}_+, that is, they accept as input a natural number, and return as output a positive real number. We say that f(n) = O(g(n)) if there exist n_0 \in \mathbb{N} and C \in \mathbb{R}_+ such that for all n \geq n_0, ... 2 This kind of thing just doesn't work. For example, one of your intermediate terms is O(1)/O(\log n). However any function f can be written as g/h where g=O(1) and h=O(\log n). If f=O(1), then let g=f and h=1. If f=\Omega(1), then let g=1 and h=1/f. For a more formal treatment, try to rewrite each of your statements that involve ... 2 It seems that you have overlooked the fact that f(n) \in \Theta(n^4) already implies both an upper bound of f(n)\in O(n^4) and a lower bound of f(n)\in \Omega(n^4). Intuitively, \Theta- notation says that a function grows "as fast as" another function, which means both "at most as fast" and "at least as fast". 2 Let us notice the following:$$ |\sin(\pi \cdot n/2)| = \begin{cases} 1 & \text{if$n$is odd}, \\ 0 & \text{if$n$is even}. \end{cases} $$This implies that your function f(n) alternates between n^2 (for odd n) and 0 (for even n). This means, for example, that f(n) = O(n^2) (since f(n) \leq n^2 for all n) but f(n) \neq \Omega(n^2) ... 1 Let S be a maximal set of vertices in which any two vertices are at distance at least 3. (Note maximal just means that S cannot be enlarged by adding new vertices.) By design, any other vertex is at distance at most 2 from some vertex in S, and therefore S is a 2-dominating set. On the other hand, if we define the B_1(v) to consist of all vertices ... 1 The recurrence relation T(n) = T(n-1) + T(n-2), with initial conditions T(2) = T(1) = 1, has as solution the Fibonacci numbers, whose asymptotic growth is known to be \Theta(\phi^n), where \phi = \frac{1+\sqrt{5}}{2} < 2. The same bound holds for arbitrary positive initial conditions. 1 Here's what my search has come up with: There's a straightforward dynamic programming solution of the problem in this case, which requires \mathop{\Omega}\left( k \cdot n^3 \right) time; and with some thought one can avoid redundant computation of sums-of-squares by this algorithm, reducing the complexity to \mathop{\Omega}\left( k \cdot n^2 \right). ... 1 If you want to compute X (whatever X is), then you just need to compute X, and whatever intermediate results are required by your method for computing X. If your method doesn't need upper and lower bounds, there's no need to compute them. The point of upper and lower bounds is that they're usually easier to compute than the actual answer, and ... 1 There are lots of NP-complete problem where no algorithm will find an optimal solution - during your lifetime. Given the choice between a greedy algorithm that finds a solution quickly which is often good, and an algorithm that will find a guaranteed optimal solution, but not while you are waiting for it, what are you going to use? 1 Place T triangles next to each other such that the line l intersects them all. Observe how the number of intersections is T+1. Now substitute T with the upper bound on the number of triangles for any triangulation that you mentioned, and we obtain 2n-5+1=2n-4 as an upper bound on the number of intersections. The example provided in the question ... 1 Since n \geq 1 (since n is the input length), we have$$ 20n^2 \leq 20n^2 + 11n + 1 \leq (20+11+1)n^2 = 32n^2.$$Put differently, for all$n \geq 1$, the function$f(n) = 20n^2 + 11n + 1$is bounded from below by$20n^2$and from above by$32n^2$. This shows that$f(n) = \Theta(n^2)$. Note also that when$n$is large,$f(n)$gets very close to$20n^2$. ... 1 Remember that: We can view a nondeterministic computation as a directed acyclic graph of configurations indexed by time. I have a guess that in your eyes:$2^n$for both, one directs itself and one directs others and at most every has the$\Sigma$chance to go next state, maybe to itself, maybe the other state, so from$1$to$n$state, an NFA produces at ... 1 That guess is obviously wrong. If all items have weight 666 and value 1, then you can fit 3 items into a bin of size 2000, and only one into a bin of size 1000. If all weights are W <$w_i\$ ≤ 2W, then O = 0 and O' = highest value of a single item. There is very little that you can say except the obvious O ≤ O'. You might have "valuable" items with a ...

1

Tricky. As an approximation, assume that the number n is in the set with a probability p (n), and the sum of p (n) over all integers n is finite, so we may guess that the set is finite. You'd have to check enough values n to make some reasonable assumptions about p (n), and then based on those assumptions make a reasonable guess about the upper bound. ...

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