GATE CSE 1996 | GATE CSE Year Wise Previous Years Questions

GATE CSE 1996

MCQ (Single Correct Answer)

-0.6

Which one of the following is false? Read $$ \wedge $$ as AND, $$ \vee $$ as OR, $$ \sim $$ as NOT, $$ \to $$ as one way implication and $$ \leftrightarrow $$ two way implication.

$$\left( {\left( {x \to y} \right) \wedge x} \right) \to y$$

$$\left( {\left( { \sim x \to y} \right) \wedge \left( { \sim x \to \sim y} \right)} \right) \to x$$

$$\left( {x \to \left( {x \vee y} \right)} \right)$$

$$\left( {\left( {x \vee y} \right) \leftrightarrow \left( { \sim x \to \sim y} \right)} \right)$$

GATE CSE 1996

MCQ (Single Correct Answer)

-0.3

A ROM is sued to store the table for multiplication of two $$8$$-bit unsigned integers. The size of ROM required is

$$256 \times 16$$

$$64\,K \times 8$$

$$4\,K \times 16$$

$$64\,K \times 16$$

GATE CSE 1996

MCQ (Single Correct Answer)

-0.6

A solution to the Dining Philosophers Problem which avoids deadlock is

ensure that all philosophers pick up the left fork before the right fork.

ensure that all philosophers pick up the right fork before the left fork

ensure that one particular philosopher picks up the left fork before the right fork, and that all other philosophers pick up the right fork before the left fork

None of the above.

GATE CSE 1996

Subjective

-0

The concurrent programming constructs fork and join are as below: Fork (label) which creates a new process executing from the specified label join (variable) which decrements the specified synchronization variable (by 1) and terminates the process if the new value is not 0. Show the precedence graph for S1, S2, S3, S4 and S5 of the concurrent program below.
N = 2
M = 2
fork L3
fork L4
S1
L1 : join N
S3
L2: join M
S5
L3:S2
goto L1
L4:S4
goto L2
next:

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