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How computers used to do binary multiply and divide
12/7/2013 11:11:18 AM
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Hi, Tom:  As you said, multiply is a series of additions -- but not dependent on the magnitude of the multiplier, only the number of 1 bits after making both operands positive by complementing if negative.

It is a shift and add sequence starting with the low order bit of the multiplier if it is a 1 add the multiplicand to the double wide product high order and shift right 1 into low order. if multiplier bit is zero, just shift product right one.

Repeat until higher multiplier bits are zero or shifts equal to multipl;ier width. If the high bits are all 0s, then just shift for the remaining word width.

Division was trial subtraction by subtracting the divisore fron the dividend, if the result was positive shift 1 into the quotient high bit else shift 0.  Remainder is left in the reg that held the high order dividend and the quotient in the low order.

If the signs of the dividend and divisor were different, complement at the end.

This is best I remember, maybe a few details missing. The shifts amounted to *2 and /2 and the add/subtract would wind up in the appropriate power of two positions.

I think the constant used in the compiler is 1/10 so they are multiplying by the reciprical of 10 to effectively divide by 10.

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12/7/2013 10:01:00 AM
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Pretty cool math...I wonder whether they teach something like that in vlsi classes...Kris

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12/6/2013 7:56:58 PM
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There's some nice material about constant division on the companion website for "Hacker's Delight":

http://www.hackersdelight.org/divcMore.pdf

Also, VHDL has a nifty fixed point math package:

http://www.eda.org/fphdl/

-Brian

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Goldschmidt?
12/6/2013 7:40:07 PM
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Sounds a bit like Goldschmidt division: converting the factor 1/10 into X/(2^n). Multiply by X, shift N bits, done. Taking the upper 16 bits of a 32-bit word is equal to a 16-bit shift.

I suspect 1/10 doesn't translate perfectly to binary so, just like 1/3 becomes 0.33... you get a hex factor of 0x1999... in your division.

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