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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 ...
A Reed–Solomon code (like any MDS code) is able to correct twice as many erasures as errors, and any combination of errors and erasures can be corrected as long as the relation 2E + S ≤ n − k is satisfied, where is the number of errors and is the number of erasures in the block.
The longest increasing subsequence problem is closely related to the longest common subsequence problem, which has a quadratic time dynamic programming solution: the longest increasing subsequence of a sequence is the longest common subsequence of and , where is the result of sorting.
Comparison of two revisions of an example file, based on their longest common subsequence (black) A longest common subsequence (LCS) is the longest subsequence common to all sequences in a set of sequences (often just two sequences).
The degree of all nodes is 1, so each node is selected with probability 1/2, and with probability 1/4 both nodes in a component are not chosen. Hence, the number of nodes drops by a factor of 4 each step, and the expected number of steps is log 4 ( n ) {\displaystyle \log _{4}(n)} .
The first phase of patience sort, the card game simulation, can be implemented to take O(n log n) comparisons in the worst case for an n-element input array: there will be at most n piles, and by construction, the top cards of the piles form an increasing sequence from left to right, so the desired pile can be found by binary search. [1]
A decision problem is EXPSPACE-complete if it is in EXPSPACE, and every problem in EXPSPACE has a polynomial-time many-one reduction to it. In other words, there is a polynomial-time algorithm that transforms instances of one to instances of the other with the same answer. EXPSPACE-complete problems might be thought of as the hardest problems ...
SEQ 1 = A CG G T G TCG T GCTATGCT GA T G CT G ACTTAT A T G CTA SEQ 2 = CGTTCGGCTAT C G TA C G TTCTA TT CT A T G ATT T CTA A. Another way to show this is to align the two sequences, that is, to position elements of the longest common subsequence in a same column (indicated by the vertical bar) and to introduce a special character (here, a dash ...