Booth's algorithm Question : Binary Number Arithmetic (Multiplication) 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


The Result's are different Please Help !
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.