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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 ...
Signed binary angle measurement. Black is traditional degrees representation, green is a BAM as a decimal number and red is hexadecimal 32-bit BAM. In this figure the 32-bit binary integers are interpreted as signed binary fixed-point values with scaling factor 2 −31, representing fractions between −1.0 (inclusive) and +1.0 (exclusive).
A bijection with the sums to n is to replace 1 with 0 and 2 with 11. The number of binary strings of length n without an even number of consecutive 0 s or 1 s is 2F n. For example, out of the 16 binary strings of length 4, there are 2F 4 = 6 without an even number of consecutive 0 s or 1 s—they are 0001, 0111, 0101, 1000, 1010, 1110. There is ...
Let p be an interior point of the disk, and let n be a multiple of 4 that is greater than or equal to 8. Form n sectors of the disk with equal angles by choosing an arbitrary line through p, rotating the line n / 2 − 1 times by an angle of 2 π / n radians, and slicing the disk on each of the resulting n / 2 lines.
A regular hexadecagon is a hexadecagon in which all angles are equal and all sides are ... by eight mirrors can alternate long and ... {16/5} {16/6} or 2{8/3}
A mathematical constant is a key number whose value is fixed by an unambiguous definition, often referred to by a symbol (e.g., an alphabet letter), or by mathematicians' names to facilitate using it across multiple mathematical problems. [1]
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.
For n = 5, the Schur cover of the alternating group is given by SL(2, 5) → PSL(2, 5) ≅ A 5, which can also be thought of as the binary icosahedral group covering the icosahedral group. Though PGL(2, 5) ≅ S 5 , GL(2, 5) → PGL(2, 5) is not a Schur cover as the kernel is not contained in the derived subgroup of GL(2 ,5).