@VS ji. Thanks for valuable effort. Just small addition if look ahead carry adder is used then 4-Unit of delay in each adder. But it will not change answer because 4 is a constant with respect to n.

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+27 votes

Consider an array multiplier for multiplying two $n$ bit numbers. If each gate in the circuit has a unit delay, the total delay of the multiplier is

- $\Theta(1)$
- $\Theta(\log n)$
- $\Theta(n)$
- $\Theta(n^2)$

+30 votes

Best answer

Now to MULTIPLY these two Numbers:

- $4$ AND GATEs REQUIRED $B0$ MULTIPLY WITH $A3 \ A2 \ A1 \ A0$.
- $4$ AND GATEs REQUIRED $B1$ MULTIPLY WITH $A3 \ A2 \ A1 \ A0$.
- $4$ BIT ADDER
- $4$ AND GATEs REQUIRED $B2$ MULTIPLY WITH $A3 \ A2 \ A1 \ A0$.
- $4$ BIT ADDER
- $4$ AND GATEs REQUIRED $B3$ MULTIPLY WITH $A3 \ A2 \ A1 \ A0$.
- $4$ BIT ADDER

For $4$ bits, Total Delay$= 3+4=7$

For $n$ bits, Total Delay$=n-1+n = 2n-1$

So, Total Delay $=\Theta(n).$

Correct Answer: $C$

+3

0

16 AND gates are used, but in the calculation 4 are considered...is it because every time 4 AND gates will work simultaneously? (Like for multiplication of B_{0} with A_{3}A_{2}A_{1}A_{0}, the AND gates used work simultaneously and so the total delay is 1?)

0

As per the figure, all AND gates can perform their task independently, making all ANDing operation O(1). Only Only adders will face delay of O(n) as last adder will wait for all the above adder to add.

+19 votes

Take A = A1 A2 A3 A4

B= B1 B2 B3 B4

NOW TO MULTIPLY THESE TWO NUMBER .

1 AND GATE REQUIRE B1 MULTIPLY WITH A1 A2 A3 A4.

1 AND GATE REQUIRE B2 MULTIPLY WITH A1 A2 A3 A4.

1 AND GATE REQUIRE B3 MULTIPLY WITH A1 A2 A3 A4.

1 AND GATE REQUIRE B4 MULTIPLY WITH A1 A2 A3 A4.

NOW 3 OR GATE REQUIRE.

TOTAL 7 GATE REQUIRE FOR 4 BIT TAKE N BIT U FIND 2N-1.

SO TIME COMPLEXITY WILL BE = ϴ(n)

B= B1 B2 B3 B4

NOW TO MULTIPLY THESE TWO NUMBER .

1 AND GATE REQUIRE B1 MULTIPLY WITH A1 A2 A3 A4.

1 AND GATE REQUIRE B2 MULTIPLY WITH A1 A2 A3 A4.

1 AND GATE REQUIRE B3 MULTIPLY WITH A1 A2 A3 A4.

1 AND GATE REQUIRE B4 MULTIPLY WITH A1 A2 A3 A4.

NOW 3 OR GATE REQUIRE.

TOTAL 7 GATE REQUIRE FOR 4 BIT TAKE N BIT U FIND 2N-1.

SO TIME COMPLEXITY WILL BE = ϴ(n)

+2

Diagramatic answer seems to helpful. But still dont know how to drive the equation "So the delay is approxiamtely sqrt(2)*(2n-1)."

http://stackoverflow.com/questions/27095216/time-complexity-in-n-bit-array-multiplication

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