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In combinatorial mathematics, a de Bruijn sequence of order n on a size-k alphabet A is a cyclic sequence in which every possible length-n string on A occurs exactly once as a substring (i.e., as a contiguous subsequence). Such a sequence is denoted by B(k, n) and has length k n, which is also the number of distinct strings of length n on A.
The exception to this result is a bit pattern the same as that of the generator polynomial. All uneven bit errors are detected by generator polynomials with even number of terms. 2-bit errors in a (multiple) distance of the longest bitfilter of even parity to a generator polynomial are not detected; all others are detected.
The longest common subsequence of sequences 1 and 2 is: LCS (SEQ 1,SEQ 2) = CGTTCGGCTATGCTTCTACTTATTCTA. This can be illustrated by highlighting the 27 elements of the longest common subsequence into the initial sequences: 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
A longest common subsequence (LCS) is the longest subsequence common to all sequences in a set of sequences (often just two sequences). It differs from the longest common substring : unlike substrings, subsequences are not required to occupy consecutive positions within the original sequences.
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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 ...
This subsequence has length six; the input sequence has no seven-member increasing subsequences. The longest increasing subsequence in this example is not the only solution: for instance, 0, 4, 6, 9, 11, 15 0, 2, 6, 9, 13, 15 0, 4, 6, 9, 13, 15. are other increasing subsequences of equal length in the same input sequence.
The LFSR is maximal-length if and only if the corresponding feedback polynomial is primitive over the Galois field GF(2). [3] [4] This means that the following conditions are necessary (but not sufficient): The number of taps is even. The set of taps is setwise co-prime; i.e., there must be no divisor other than 1 common to all taps.