# Computer Organization | Booth’s Algorithm

Booth algorithm gives a procedure for **multiplying binary integers** in signed 2â€™s complement representation **in efficient way**, i.e., less number of additions/subtractions required. It operates on the fact that strings of 0â€™s in the multiplier require no addition but just shifting and a string of 1â€™s in the multiplier from bit weight 2^k to weight 2^m can be treated as 2^(k+1 ) to 2^m. As in all multiplication schemes, booth algorithm requires examination **of the multiplier bits** and shifting of the partial product. Prior to the shifting, the multiplicand may be added to the partial product, subtracted from the partial product, or left unchanged according to following rules:

- The multiplicand is subtracted from the partial product upon encountering the first least significant 1 in a string of 1â€™s in the multiplier
- The multiplicand is added to the partial product upon encountering the first 0 (provided that there was a previous ‘1’) in a string of 0â€™s in the multiplier.
- The partial product does not change when the multiplier bit is identical to the previous multiplier bit.

**Hardware Implementation of Booths Algorithm –** The hardware implementation of the booth algorithm requires the register configuration shown in the figure below.

**Booth’s Algorithm Flowchart –**

We name the register as A, B and Q, AC, BR and QR respectively. Qn designates the least significant bit of multiplier in the register QR. An extra flip-flop Qn+1is appended to QR to facilitate a double inspection of the multiplier.The flowchart for the booth algorithm is shown below.

AC and the appended bit Qn+1 are initially cleared to 0 and the sequence SC is set to a number n equal to the number of bits in the multiplier. The two bits of the multiplier in Qn and Qn+1are inspected. If the two bits are equal to 10, it means that the first 1 in a string has been encountered. This requires subtraction of the multiplicand from the partial product in AC. If the 2 bits are equal to 01, it means that the first 0 in a string of 0â€™s has been encountered. This requires the addition of the multiplicand to the partial product in AC. When the two bits are equal, the partial product does not change. An overflow cannot occur because the addition and subtraction of the multiplicand follow each other. As a consequence, the 2 numbers that are added always have a opposite signs, a condition that excludes an overflow. The next step is to shift right the partial product and the multiplier (including Qn+1). This is an arithmetic shift right (ashr) operation which AC and QR to the right and leaves the sign bit in AC unchanged. The sequence counter is decremented and the computational loop is repeated n times. Product of negative numbers is important, while multiplying negative numbers we need to find 2’s complement of the number to change its sign, because it’s easier to add instead of performing binary subtraction. product of two negative number is demonstrated below along with 2’s complement.

**Example –** A numerical example of booth’s algorithm is shown below for n = 4. It shows the step by step multiplication of -5 and -7.

BR = -5 = 1011, BR' = 0100, <-- 1's Complement (change the values 0 to 1 and 1 to 0) BR'+1 = 0101 <-- 2's Complement (add 1 to the Binary value obtained after 1's complement) QR = -7 = 1001 <-- 2's Complement of 0111 (7 = 0111 in Binary) The explanation of first step is as follows: Qn+1 AC = 0000, QR = 1001, Qn+1 = 0, SC = 4 Qn Qn+1 = 10 So, we do AC + (BR)'+1, which gives AC = 0101 On right shifting AC and QR, we get AC = 0010, QR = 1100 and Qn+1 = 1

OPERATION | AC | QR | Qn+1 | SC |
---|---|---|---|---|

0000 | 1001 | 0 | 4 | |

AC + BR’ + 1 | 0101 | 1001 | 0 | |

ASHR | 0010 | 1100 | 1 | 3 |

AC + BR | 1101 | 1100 | 1 | |

ASHR | 1110 | 1110 | 0 | 2 |

ASHR | 1111 | 0111 | 0 | 1 |

AC + BR’ + 1 | 0100 | 0111 | 0 | |

ASHR | 0010 | 0011 | 1 | 0 |

Product is calculated as follows:

Product = AC QR Product = 0010 0011 = 35

**Best Case and Worst Case Occurrence: **Best case is when there is a large block of consecutive 1’s and 0’s in the multipliers, so that there is minimum number of logical operations taking place, as in addition and subtraction. Worst case is when there are pairs of alternate 0’s and 1’s, either 01 or 10 in the multipliers, so that maximum number of additions and subtractions are required. **GATE Practice Questions –**

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