56. Consider the minterm list form of a Boolean function F given below.

Here, m denotes a minterm and d denotes a don’t care term. The number of essential prime implicants of the function F is ______ .Note: Numerical Type question.

- Option : C
- Explanation :

F(P, Q, R, S) = Σm(0, 2, 5, 7, 9, 11) + d(3, 8, 10, 12, 14)

Number of EPI = 3

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- Option : A
- Explanation :

Given, total number of instructions (n) = 100

Number of stages (k) = 5

Since, if n instructions take c cycle, so (c-1) stalls will occur for these instructions.

Therefore, the number of clock cycles required = Total number of cycles required in general case + Extra cycles required (here, in PO stage)

= (n + k – 1) + Extra cycles

= (100 + 5 -1) + 40*(3-1)+35*(2-1)+20*(1-1)

= (100 + 4) + 40*2+35*1+20*0

= 104 + 115

= 219 cycles

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59. Given a language L, define L^{i} as follows:

The order of a language L is defined as the smallest k such that L

The order of is ___________.

Note – Numerical Type question

- Option : A
- Explanation :

We need to find smallest value of k which satisfies

L_{1}^{k}= L_{1}^{k+1}

L_{1}= ε + (00)

Try k = 0; L_{1}^{0}= L_{1}^{1}

ε = L_{1}which is false.

So order is not 0.

Try k = 1: L_{1}^{1}= L_{1}^{2}

L_{1}^{2}= L_{1}Now, L_{1}^{2}=(ε + 0(00)*)(ε + 0(00)*)

= ε + 0(00)* + 00(00)* = 0*

Clearly L_{1}^{2}≠ L_{1}

So order is not 1.

Try k = 2: L_{1}^{2}= L_{1}^{3}

Now, L_{1}^{3}= L_{1}^{2}.L_{1}

= 0*(ε + 0(00)*) = 0*

Clearly L_{1}^{3}= L_{1}^{2}=0*

(so, order of L_{1}is 2)

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- Option : A
- Explanation :

T(N) =(N-1)Ck * T(k) * T(N-k-1), where k = number of nodes on left subtree

T(1) = 1

T(2) = 1

T(3) = 2

T(4) = 3C2 * T(2) * T(1) = 3

T(5) = 4C3 * T(3) * T(1) = 8

T(6) = 5C3 * T(3) * T(2) = 20

T(7) = 5C3 * T(3) * T(3) = 80

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