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Q. Draw the Booth's algorithm and mutiply (-2 * -4) using Booth's algorithm

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Draw the Booth's algorithm and mutiply (-2 * -4) using Booth's algorithm

written 2.1 years ago by | modified 22 months ago by |

Q. Draw the Booth's algorithm and mutiply (-2 * -4) using Booth's algorithm

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written 22 months ago by |

states that “The value of series of 1’s of binary can be given as the weight of the bit preceding the series minus the weight of the last bit in the series.”*Booth’s Principle*- The booth’s multiplication algorithm multiplies the two signed binary integers.
- It is generally used to speed up the performance of the multiplication process.
- Booth’s Algorithm looks in the following manner in terms of flowchart representation:

**Terms Used in Booth's Algorithm**

**AC**stands forset as 0 initially.*Accumulator Counter***M**represents*Multiplicand Bits.***-M**represents*2’s Complement of M.***Q**represents*Multiplier Bits.***$Q_n$**represents the*Last Bit of Multiplier Q.***$Q_{n+1}$**represents the. But at the initial state, it is set as 0.*Incremented value of $Q_n$ by 1***SC**stands for. It represents a number of bits that is nothing but a total number of bits in the multiplier Q.*Sequential Counter*

- Set the Multiplicand and Multiplier values in binary bits formats as M and Q, respectively.
- At a starting, set the AC and $Q_{n + 1}$ registers values to 0.
- As an SC represent the number of bits in the multiplier (Q). According to that set the value of SC. The value of SC goes on decreasing with each repetition of the computation loop until its value reaches 0.
- The $Q_n$ represents the last bit of the Q, and the $Q_{n+1}$ shows the incremented bit of $Q_n$ by 1.
At each iteration Booth’s algorithm checks the condition of $Q_n Q_{n+1}$ bits.

**A] If the bits of $Q_n Q_{n+1}$ represent 01**- Then Multiplicand Bits (M) is added to the Accumulator Counter
**(AC = AC + M)**. - Then
**Arithmetic Right Shift**operation performs on AC $Q_n Q_{n+1}$ bits by 1. - Then decrease the value of Sequence Counter by 1.

**B] If the bits of $Q_n Q_{n+1}$ represent 10,**- Then Multiplicand Bits (-M) is subtracted from the Accumulator Counter
**(AC = AC - M).** - Then
**Arithmetic Right Shift**operation performs on AC $Q_n Q_{n+1}$bits by 1. - Then decrease the value of Sequence Counter by 1.

**C] If the bits of $Q_n Q_{n+1}$ represent 00 or 11**- Then directly
**Arithmetic Right Shift**operation performs on AC $Q_n Q_{n+1}$ bits by 1. - Then decrease the value of Sequence Counter by 1.

- Then Multiplicand Bits (M) is added to the Accumulator Counter
These checking iterations of bits value of $Q_n Q_{n+1}$continues until the value of Sequence Counter (SC) reaches 0.

- Finally when SC =0, the result of the multiplication is stored in AC and Q.

M = -2 = (1110) and –M = M’ +1 = 0010

Q = -4 = (1100)

Value of SC = 4, because the number of bits in Q is 4.

$Q_n = 1$ according to the last bit of Q and $Q_{n+1}$ set as 0 at initially.

As, (-2) * (-4) = 8

Value stored in AC & Q registers = **00001000**

(-2) * (-4) = $(00001000)_2$ = 8

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