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It's being said booth's algorithm produces the output exactly as normal binary multiplication while reducing the number of operations performed and can be used for both positive and negative numbers !

I tried multiplying two 4-bit numbers while I don't get the same result...Please guide what am doing wrong.

Multiplicand : 1101 , Multiplier : 1110,
Recorded Multiplier(Applying skipping over 1's) : 00-10

Normal Multiplication

Booth's way of Multiplication

The Result's are different Please Help !

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  • $\begingroup$ (Are you positive about the most significant zero(es) in the "Normal(?) Multiplication" result?) $\endgroup$
    – greybeard
    Commented Sep 4, 2020 at 12:36
  • $\begingroup$ Do you mean in the first row(r1 out of r1,r2,r3,r4) of the multiplication result ? I have done sign extension , since the MSB is Zero so the sign 0 will be extended further ! $\endgroup$
    – Dan
    Commented Sep 4, 2020 at 12:50
  • $\begingroup$ In Normal Multiplication we don't extend the sign so for Normal Multiplication the Result will be : 010110110(Correction) I took it by mistake , But the results are still not equal ! $\endgroup$
    – Dan
    Commented Sep 4, 2020 at 13:14
  • $\begingroup$ (I meant just summing the digits shown: there's a "double overflow" from bit 5, I think mechanically that should read 11010110.) For the overall approach, please visit en.wikipedia on signed binary multiplication and Booth encoding. $\endgroup$
    – greybeard
    Commented Sep 4, 2020 at 16:00

1 Answer 1

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When you use normal multiplication, multiplicand and multiplier are represented using (Sign + Magnitude) representation. So effectively 1101 is +(13) in Decimal and (1110) is +14 in decimal as they represent the magnitude. Sign bit would be separate. So the result is (+13)*(+14) = +182 which is 1011 0110 in binary.

When you use booth multiplication, operand are in 2's complement representation. So 1101 is -3 and 1110 is -2 in decimal. So the answer will be 0000 0110 that is +6 in decimal. The problem is with your representation of multiplicand and multiplier.

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  • $\begingroup$ My Understanding : If i want to use booth's algorithm for unsigned numbers then i can do that directly ! But If for Signed Numbers than they need to be represented in 2's complement representation . Now as Positive numbers are represented without any modifications (a sign bit(0) will be needed to represent positive numbers) and for negative i will first find out the 2's complement and then the reduced multiplier and fetch the result . $\endgroup$
    – Dan
    Commented Sep 5, 2020 at 6:12
  • $\begingroup$ Signed No's : To Multiply +13 and +14 using Booth's the procedure will be Multiplicand : 01101 and Multiplier : 01110, Reduced Multiplier : +100-10 and the Result will be : 010110110...Answer Match $\endgroup$
    – Dan
    Commented Sep 5, 2020 at 6:16
  • $\begingroup$ Focus on bits used to represent the number. In 4 bit, you can represent +7(0 111) to -7(1 111) using S+M representation. In the same 4 bits, you can represent +7(0111) to -8(1000). So when you are multiplying use appropriate bit size representation. You cannot represent 13 in 4 bits, using S+M representation. you need 1+4 bits. Sign bits are multiplied(XOR gate) separately in simple multiplication. Simple multiplication, multiplies only the magnitude part, which are straight binary representation of the operands. $\endgroup$
    – ajit
    Commented Sep 5, 2020 at 6:22
  • $\begingroup$ Thanks I learnt it $\endgroup$
    – Dan
    Commented Sep 6, 2020 at 12:00

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