MCQs
`x``oplus` `y` = `x^2` + `y^2` = `y^2` + `x^2` = `y``oplus` `x`
`:.` commutative
Not associative, since, for example
( 1`oplus` 2 ) `oplus` 3 `ne` 1 `oplus` ( 2`oplus` 3 )
If matrix B is obtained from matrix A by replacing the `1^(th)` row by itself plus k times
the `m^(th)` row, for1 `ne` m then det(B)=det(A). With this property given matrix is
equal to the matrices
given in options (B),(C) and (D).
P(p < 3) = P(p = 0) +P(p = 1) + P(p = 2)
=`(e^(-lambda) lambda^0)/(0!)` + `(e^(-lambda) lambda^1)/(1!)` + `(e^(-lambda) lambda^2)/(2!)` (where `lambda` = 3)
= `e^(-3)` + `e^(-3)` x 3 + `(e^(-3) xx 9)/2`
= `e^(-3)` `(1 + 3 +9/2)` = `17/(2e^3)`
`-` `2^(8 - 1)` = -128 . Range is -`2^((n - 1))` to + `2^((n - 1))` - 1
4 to 2 priority encoder.
The maximum number of swaps that takes place in selection sort on n numbers is n
Concatenation of empty language with any language will give the empty language and `L_1`* = `Phi`* = `in` . Hence `L_1``L_2`*`U``L_1`* = {`in` }
To have maximum number of reduce moves, all the productions will be of the typeA`rightarrow` `alpha` `beta`
(where `alpha` and `beta` could be terminals or non-terminals). Consider the following illustration
then:
Input String :`a_1``a_2``a_3`........`a_(n - 2)` `a_(n - 1)` `a_n`
`uparrow`
`a_1``a_2``a_3`........`a_(n - 2)` A
`uparrow`
`a_1``a_2``a_3`................A n - 1 moves
`.`
`.`
`.`
`.`
`.`
`a_1` `a_2`
A
A scheduling algorithm assigns priority proportional to the waiting time of a process. Every
process starts with priority zero(the lowest priority). The scheduler re-evaluates the process
priorities every T time units and decides the next process to schedule. Which one of the
following is TRUE if the processes have no I/O operations and all arrive at time zero?
The given scheduling definition takes two parameters, one is dynamically assigned
processpriority and the other is 'T' time unit to re-evaluate the process priorities.
This dynamically assigned priority will be deciding processes order in ready queue
of round robin algorithm whose time quantum is same as ‘T’ time units. As all the
processes are arriving at the same time, they will be given same priority but soon
after first 'T` time burst
remaining processes will get higher priorities
For skewed binary search tree on n nodes, the tightest upper bound to insert a node is O(n)