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+24 votes
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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

  1. $\Theta(1)$
  2. $\Theta(\log n)$
  3. $\Theta(n)$
  4. $\Theta(n^2)$
asked in Digital Logic by Veteran (59.7k points) | 3.5k views

4 Answers

+27 votes
Best answer

Now to MULTIPLY these two Numbers:

 

  1. $4$ AND GATEs REQUIRED $B0$ MULTIPLY WITH $A3 \ A2 \ A1 \ A0$.
  2. $4$ AND GATEs REQUIRED $B1$ MULTIPLY WITH $A3 \ A2 \ A1 \ A0$.
  3. $4$ BIT ADDER
  4. $4$ AND GATEs REQUIRED $B2$ MULTIPLY WITH $A3 \ A2 \ A1 \ A0$.
  5. $4$ BIT ADDER
  6. $4$ AND GATEs REQUIRED $B3$ MULTIPLY WITH $A3 \ A2 \ A1 \ A0$.
  7. $4$ BIT ADDER


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

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

So, Total Delay $=\Theta(n).$

answered by Loyal (9.5k points)
edited by
+3

@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.

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 B0 with A3A2A1A0, 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.
0

for multiply NxN number we will need N-1 , N-bit adders.

one adder has delay of 2n-1 then n-1 will have delay of : (n-1)*(2n-1)=O(n2)

Please solve my doubt...

0

Dharmendra Lodhi

How r u saying that 1 Adder has a delay of 2n-1??

0

@akash.dinkar12

What is the delay per Adder

If it is Ripple carry Adder 

If it is lookahead Adder

+18 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)
answered by Veteran (61.2k points)
+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

0
how b1 multply with a1a2a3a4 take only 1 and .. it wil be 4 yes they can perform at same level
+8 votes
answered by Veteran (367k points)
0

 if Td = Θ(1) , then the total delay of multiplexer is Θ(n) 
worst case delay would be 
(2n+1)td .where td is the time delay of gates. , is it correct ???

+1
Can you give another link of array multiplier ? This link given is not working !
0
@arjun sir : another link for array multiplier please :(
+4 votes
Answer: C

The no. of gates used in a n bit array multiplier (n*n) is 2n-1.

So, if every single gate takes unit delay, then total delay O(2n-1) = O(n).
answered by Boss (34k points)
edited by
+1
How number of gates are 2n-1,I am getting n^2 Gates
Answer:

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