GATE CSE 2010 | Question: 8

Question

$P$ is a $16$-bit signed integer. The $2$'s complement representation of $P$ is $(F87B)_{16}$. The $2$'s complement representation of $8\times P$ is

• A. $(C3D8)_{16}$
• B. $(187B)_{16}$
• C. $(F878)_{16}$
• D. $(987B)_{16}$

Comments

Got it. @Manu Thakur sir. 

Thanks

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Is there any overflow

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just Left shift is needed 

Answer 1

Multiplication can be directly carried in $2$’s complement form.

$\textsf{F87B} = 1111 1000 0111 1011$ can be left shifted $3$ times to give $8P = 1100 0011 1101 1000 = \textsf{C3D8}.$

Alternate Method:

As MSB in $\textsf{(F87B)}$ is $1$, $P$ is a negative number.

So, $P = -1 \times 2$'s complement of $\textsf{(F87B)}$

$ P = -1 \times (0785) = -1 \times (0000\; 0111 \;1000 \;0101)$

$8 \ast P = -1 \times  (0011 \;1100 \;0010 \;1000)\;\;\; (P$ in binary left shifted $3$ times$)$

In $2$'s complement representation , this equals, $1100 0011 1101 1000 = \textsf{C3D8}$

Correct Answer: $A$

Comments

The first one was quite easier. :)

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why binary left shift why not right shift ?

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Left shift means adding a 0 at right end, which is equivalent to multiply by 2 in binary. Right shift will be doing division by 2.

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ya rgt thanks:)

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suppose iif the question asks 10*P then ??

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@Puja you can do it  (10*P = 8*P + 2*P) where (8*P) you shift 3 digit left and then for (2*P) you can shift it 1 digit left and then add it.

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sir I want to know why should be shift 3 times ?

if it is 9 then ?

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@hem chandra

Left shift  is equal to multiplication and right shift equals to division in binary numbers, for the factor of 2 like 2 , 2^2, 2^3 .............

Since here it is 8 , it is 2^3. So we need to shift 3 times

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if there is 9p then 8p+1p = 2^3p+2^0p (overall three left shifts)

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Thank You so much for this explanation :)

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@ Puja Mishra 

If you do 10*P then You will get

= 8*P + 2*P

= 1 1011 0100 1100 1110

This is the Case of Overflow as said in Question P is 16 Bit Signed Integer.

So i guess there will be one option of overflow in this scenario.

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@krishn.jh It won't be overflow because for overflow $C_{in} \neq C_{out}$ but here $C_{in} = C_{out} = 1$

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@Akhilesh so you mean to say

if you do 8*P + 2*P then no overflow will occur?

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It will be end carry, which is discarded in 2's complement.

And in 2's complement we can forget all 1s in MSBs, except one 1.
Eg. 111011 = 1011

Thus, we can forget the end carry in our result of 8P+2P because sign(negative here) as well as value is still preserved.

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@Akhilesh ok yes i know as concept says preceding 1's does not affect the values in 2's Complement as they make one partial block.

i got it as after discarding the carry 1, it is not going to affect the final result.

But Be careful  as if sum is 0 carry 1 in MSB place then surely there would be Overflow.

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1010=10

then after left shift by 1 bit it becomes 0100=4 ,why didnt i get 20?

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Great Answer!

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Can we do division directly on 2's complement form also?

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shifting mechanism can't be applied in case of 13*P, right ??

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@hem  Does it mean that the value of 8p and 9p will be same ? Please clear ,thank you !

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while performing Left shift Trailing bits will be filled with zeros, on right shift leading bits will filled with zeros.

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on right shift leading bits will filled with zeros – this happens only when number is unsigned

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"Multiplication can be directly carried in 2’s complement form."

Valid only for numbers multiplied by the powers of 2. i.e; 2,4,8,16, and so on.
Like, here the given 2's complement number is being multiplied by 2^3.

Correct me if i am wrong.

Answer 2

$P=(F87B)_{16}=(1111|1000|0111|1011)_{2}:$ 2's compliment representation

What is this number ?

$(0000|0111|1000|0100)_{2}+1=(0000|0111|1000|0101)_{2}=1925$

$So, P\ is =-1925$

$-1925\times 8=-15400$

Find 2's compliment representation of $-15400$

$+15400=(0011|1100|0010|1000)_{2}$

$(1100|0011|1101|0111)_{2}+1=(1100|0011|1101|1000)_{2}=(C3D8)_{16}$

 

Correct Answer is (A)

Answer 3

Other way to see.

Answer - A