Restoring Division Algorithm For Unsigned Integer
Algorithm
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Restoring Division Algorithm For Unsigned Integer Steps
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Step-1:
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Initialize registers
Q = Dividend, M = Divisor, A = 0, n = number of bits in dividend
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Step-2:
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Content of register A and Q is shifted left as by considering them as a single unit AQ
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Step-3:
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Content of register M is subtracted from A and result is stored in A (i.e. A = A - M)
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Step-4:
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The MSB (most significant bit) of the A is checked if it is 0 then LSB (least significant bit) of Q is set to 1 otherwise if it is 1 then LSB of Q is set to 0 and value of register A is restored.
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Step-5:
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The value of counter n is decremented by 1
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Step-6:
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If value of n becomes 0 then goto next step otherwise repeat from step-2
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Step-7:
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Now register Q contains Quotient value and A contains Remainder value
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Example-1
1. Find 11 divided by 3 using restoring division Algorithm method
Solution:
Dividend = 11
Divisor = 3
First the registers are initialized with corresponding values (Q = Dividend, M = Divisor, A = 0, n = number of bits in dividend)
| n | M | A | Q | Operation |
| 4 | 00011 | 00000 | 1011 | initialize |
| 4 | 00011 | 00001 | 011_ | shift left AQ |
| 00011 | 11110 | 011_ | A=A-M |
| 00011 | 00001 | 0110 | Q[0]=0 And restore A |
| 3 | 00011 | 00010 | 110_ | shift left AQ |
| 00011 | 11111 | 110_ | A=A-M |
| 00011 | 00010 | 1100 | Q[0]=0 And restore A |
| 2 | 00011 | 00101 | 100_ | shift left AQ |
| 00011 | 00010 | 100_ | A=A-M |
| 00011 | 00010 | 1001 | Q[0]=1 |
| 1 | 00011 | 00101 | 001_ | shift left AQ |
| 00011 | 00010 | 001_ | A=A-M |
| 00011 | 00010 | 0011 | Q[0]=1 |
register Q contain the quotient 3 and register A contain remainder 2
This material is intended as a summary. Use your textbook for detail explanation.
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