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The longest alternating subsequence problem has also been studied in the setting of online algorithms, in which the elements of are presented in an online fashion, and a decision maker needs to decide whether to include or exclude each element at the time it is first presented, without any knowledge of the elements that will be presented in the future, and without the possibility of recalling ...
Barker codes of length N equal to 11 and 13 are used in direct-sequence spread spectrum and pulse compression radar systems because of their low autocorrelation properties (the sidelobe level of amplitude of the Barker codes is 1/N that of the peak signal). [15]
Compute a longest common subsequence of these two strings, and let , be the random variable whose value is the length of this subsequence. Then the expected value of λ n , k {\displaystyle \lambda _{n,k}} is (up to lower-order terms) proportional to n , and the k th Chvátal–Sankoff constant γ k {\displaystyle \gamma _{k}} is the constant ...
Download as PDF; Printable version; ... At each stage an alternating sequence of 1s and 0s is inserted between the terms of the previous sequence. ... Solution to ...
This table is used to store the LCS sequence for each step of the calculation. The second column and second row have been filled in with ε, because when an empty sequence is compared with a non-empty sequence, the longest common subsequence is always an empty sequence. LCS(R 1, C 1) is determined by comparing the first elements in each sequence.
Even in this best case, the low three bits of X alternate between two values and thus only contribute one bit to the state. X is always odd (the lowest-order bit never changes), and only one of the next two bits ever changes. If a ≡ +3, X alternates ±1↔±3, while if a ≡ −3, X alternates ±1↔∓3 (all modulo 8).
"there is an integer n such that if there is a sequence of rooted trees T 1, T 2, ..., T n such that T k has at most k+10 vertices, then some tree can be homeomorphically embedded in a later one" is provable in Peano arithmetic, but the shortest proof has length at least 1000 2, where 0 2 = 1 and n + 1 2 = 2 ( n 2) ( tetrational growth).
The longest increasing subsequence has also been studied in the setting of online algorithms, in which the elements of a sequence of independent random variables with continuous distribution – or alternatively the elements of a random permutation – are presented one at a time to an algorithm that must decide whether to include or exclude ...