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The final remainder will be in A and quotient will be in Q. Step 4: Decrease counter if counter > 0, repeat process from step 2 else stop the process. If sign of A is 1, set Q 0 to zero and add M back to A (restore A). Step 3: Subtract M from A placing answer back in A. Step 2: Shift A, Q left one binary position. Step 1: Initialize A, Q and M registers to zero, dividend and divisor respectively and counter to n where n is the number of bits in the dividend. Restoring Division (Unsigned Binary Division) The process continues until all the bits of the dividend are exhausted.
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The divisor is subtracted from this number to produce a new partial remainder. At each cycle, additional bits from the dividend are appended to the partial remainder until the result is greater than or equal to the divisor. The result is referred to as a partial remainder. When the event occurs, a 1 is placed in the quotient and the divisor is subtracted from the partial dividend. Until this event occurs, 0s are placed in the quotient from left to right. The operation involves repetitive shifting and addition or subtraction.įirst, the bits of the dividend are examined from left to right, until the set of bits examined represents a number greater than or equal to the divisor this is referred to as the divisor being able to divide the number. ĭivision is somewhat more than multiplication but is based on the same general principles. When count reaches to zero, result resides into AQ in the form of signed integer. Following the addition or subtraction the arithmetic right shift occurs. If the two bits differ then the multiplicand is added to or subtracted from the A register depending on weather the two bits are 01 or 10. A and Q -1 are initialized to zero if two bits (Q 0 and Q -1) are the same (11 or 00) then all the bits of A, Q and Q -1 registers are shifted to the right 1 bit. The result of multiplication will appear in A and Q resister. There is also one bit register placed logically to the right of the least significant bit Q 0 of the Q register and designated as Q -1.
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Multiplier and multiplicand are placed in Q and M register respectively. Signed Multiplication (Booth Algorithm) – 2’s Complement Multiplication Step 4: Check for completion if not completed, go to step 2. Step 3: Logical Shift the content of X left one position and content of Y right one position. Step 2: Test Y 0 if it is 1, add content of X to the accumulator A. Place the multiplicand in X and multiplier in Y. If the multiplier bit is 0, the partial product is zero if the multiplier bit is 1, the multiplicand is partial product.
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The process is repeated for each bit of the original multiplier. If Q 0 is 0, no addition is performed just do the shift. Then all the bits of CAQ are shifted to the right 1 bit so that C bit goes to A n-1, A0 goes to Q n-1 and Q 0 is lost. If Q 0 is 1, the multiplicand is added to the register A and is stored back in register A with C bit used for carry. Now, the control logic reads the bits of the multiplier one at a time. C is the 1-bit register which holds the carry bit resulting from addition. A third register A is initially set to zero. The multiplier and multiplicand bits are loaded into two registers Q and M. Multiplication Algorithm & Division Algorithm