I. Strict 2PL guaranteed conflict serializable because of 2PL condition and also strict recoverable.
II. Thomas Write timestamp ordering ensures serializable. Thomas write rule timestamp ordering allowed to execute schedule which is view equal serial schedule based on timestamp ordering.
PREVIOUS YEAR SOLVED PAPERS MCQ - GATE 2019
In a complete graph we can traverse the n vertices in any order and return to the starting vertex and form a Hamiltonian cycle. The number of such cycles will be n!
However, since circular rotations will have to ignored. Since for example K4 with vertices {1, 2, 3, 4}, the cycle 1-2-3-4 is same as 2-3-4-1 is same as 3-4-1-2 etc. we now get only (n – 1)! distinct Hamiltonian cycles. Further, the cycle 1-2-3-4 and 1-4-3-2 are also same (clockwise and anticlockwise).
So ignoring this orientation also we finally get 
distinct Hamiltonian cycles.
distinct Hamiltonian cycles.
23
Compute
form.
So apply L’H rule
B+ tree non-leaf node has a pointer to data records is a false statement.
B+ tree non-leaf node consists of only keys and tree pointers (node pointers).
Below is the structure of B+ tree non leaf node
L = {a2 + 3k or b10 + 12k} for k ≥ 0
= a2 (a3)* or b10 (b12)*
= {a2, a5, a8, ..., b10, b22, b234 .....}
The pumping length is p, than for any string w ∈ L with ⏐w⏐≥ p must have a repetition
i.e. such a string must be breakable into w = xyz such that ⏐y⏐≥ 0 and y can be pumped
indefinitely, which is same as saying xyz ∈ L ⇒ xy*z ∈ L.
The minimum pumping length in this language is clearly 11, since b10 is a string which
has no repetition number, so upto 10 no number can serve as a pumping length.
Minimum pumping length is 11. Any number at or above minimum pumping length can
serve as a pumping length. The only number at or above 11, in the choice given is 24.
So correct answer is option (B)
