Computer Science Test - 2

The correct answer is option 3.

Concept:

Schedules in which transactions commit only after all transactions whose changes they read commit are called recoverable schedules. In other words, if some transaction T₂ is reading value updated or written by some other transaction T₁, then the commit of T₁ must occur after the commit of T₂.

Example:

R1(x), W1(x), R2(x), R1(y), R2(y), W2(x), W1(y), C1, C2;

Hence the correct answer is recoverable.

Additional Information

Strict Schedule:

A schedule is strict if for any two transactions T₁, T₂, if a write operation of T₁ precedes a conflicting operation of T₂ (either read or write), then the commit or abort event of T₁ also precedes that conflicting operation of T₂.

In other words, T₂ can read or write updated or written values of T₁ only after T₁ commits/aborts.

Example:

Cascadeless Schedule:

Schedules in which transactions read values only after all transactions whose changes they are going to read commit are called cascadeless schedules. Avoids that a single transaction abort leads to a series of transaction rollbacks.

Example:

Serializable schedule:

The Serializable schedule is a type of schedule where one transaction is executed completely before starting another transaction.Example: R₁(X), W₁(X) Commit₁, W₂(X) ,R₂(X), Commit₂.

Database Schedules Relationship:

Option 1:

Since 1111 1111 1110 1011 is in 2’ complement number system it can be written as 101011

Decimal of this binary -32 x 1 + 16 x 0 + 8 x 1 + 4 x 0 + 2 x 1 + 1 x 1 = -32 + 8 + 2 + 1 = -21

Option 2:

Since 1111 1111 1110 1001 is in 2’ complement number system it can be written as 10 1001

Decimal value of above binary: -32 x 1+ 16 x 0 + 8 x 1 + 4 x 0 + 2 x 0 + 1 x 1 = -32 + 8 + 1 = -23

Option 3:

Since 1111 1111 1110 0111 is in 2’ complement number system it can be written as 10 0111

Decimal value of above binary: -32 x 1 + 16 x 0 + 8 x 0 + 4 x 1 + 2 x 1 + 1 x 1 = -32 + 4 + 2 + 1 = -25

Option 4:

Since 1111 1111 1110 1000 is in 2’ complement number system it can be written as 101000

Decimal value of above binary: -32 x 1 + 16 x 0 + 8 x 1 + 0 + 0 + 0 = -32 + 8 = -24

Therefore option 4 is the correct answer.

Let two relations –

A be a set of all values of column M = {1, 2, 3, 4}

B be a set of all non-NULL values of column N = {1, 3}

Or in another example column N can have the same values as column M.

So A = B

By taking both the cases –

B is a subset of A

B ⊆ A

Here X denotes it may present or not and pn denotes If (x,y) present then (y,x) is present or not or both not present.

(x,x)= present  or not = 2 choices.

(x,y) (y,x) = possible choices are both are present , (x,y) is present and (y,x) is not present, or (y,x) is present and (x,y) is not present = 3 choices.

Concept –

rer(513, 2) = 1 + rer(256, 2)

rer(256, 2) = 0 + rer(128, 2)

rer(128, 2) = 0 + rer(64, 2)

rer(64, 2) = 0 + rer(32, 2)

rer(32, 2) = 0 + rer(16, 2)

rer(16, 2) = 0 + rer(8, 2)

rer(8, 2) = 0 + rer(4, 2)

rer(4, 2) = 0 + rer(2, 2)

rer(2, 2) = 0 + rer(1, 2)

rer(1, 2) = 1 + rer(0, 2)

rer(0, 2) = 0

Now, adding the values in bottom up function after returning from rer(0, 2), we get,

rer(1, 2) = 1 + 0 = 1

rer(2, 2) = 0 + 1 = 1

rer(4, 2) = 0 + 1 = 1

rer(8, 2) = 0 + 1 = 1

rer(16, 2) = 0 + 1 = 1

rer(32, 2) = 0 + 1 = 1

rer(64, 2) = 0 + 1 = 1

rer(128, 2) = 0 + 1 = 1

rer(256, 2) = 0 + 1 = 1

rer(513, 2) = 1 + 1 = 2

Hence, the function returns 2 i.e., sum of bits when 513 represented in binary.

Concept:

In cycle stealing mode of DMA, the DMA controller steals the CPU cycles to transfer data between the IO device and the main memory, so the CPU is free to perform other tasks during the data transfer.

Explanation:

Data Preparation Time (X):

S₁: ∀x P(x) ∨ ∀xQ(x) and ∀x(P(x) ∨ Q(x)) are not logically equivalent.

This statement is correct. Consider an example: x – is a number. P(x) denotes x is even and Q(x) denotes x is odd. 

∀x P(x) ∨ ∀xQ(x)- it means for all numbers x is even or for all numbers x is odd.

∀x(P(x) ∨ Q(x)) – it means for all x, x is either odd or even.

From this, it is clear that both these are not logically equivalent.

S₂: ∃x P(x) ∧ ∃x Q(x) and ∃x (P(x) ∧ Q(x)) are not logically equivalent

This statement is correct.  Consider the same example as of first case.

∃x P(x) ∧ ∃x Q(x) – It means there exists a number which is odd and there exists a number which is even also which is true.

∃x (P(x) ∧ Q(x))- It means, there exists a number which is both even and odd which is not possible. It is false.

So, these are not logically equivalent.

The correct answer is 3.

Key Points:

• Semaphores: Semaphores are integer variables that are utilized to address the critical section problem by utilizing two atomic actions for process synchronization, wait and signal.
• Wait: If the parameter S is positive, the wait action decrements its value. If S is negative or zero, no operation is carried out.
• Signal: The signal operation increments the value of its argument S.

Explanation:

The given data:

A: down(s); cs; up (s);
B: down(s); cs; up (s);
C: down (s); cs; up(s);
D: up(s); cs; down(s);
E: up(s); cs; down(s);
F: up(s); cs; down(s);

Processes D, E, and F can do signal (Up) operation without any intermediate down operation and thus S can go up to value 3.

Hence the correct answer is 3.