Post by thread:16-BIT DIVISION

Hallo all! I am looking for a routine in ASM that make a division of a 16-bit binary number for 5 . For example. 015Eh / 05 = 046h So my 16-bit binary number will never be larger than 03DEH, with this I will need only 1 byte for the result. Any idea? Any help would be appreciated Thanks in advanced Luis F.

<x-flowed>At 07:19 PM 10/27/99 -0200, Luis wrote: >Hallo all! >I am looking for a routine in ASM that make a division of a 16-bit binary >number for 5 . >For example. >015Eh / 05 = 046h > >So my 16-bit binary number will never be larger than 03DEH, with this I will >need only 1 byte for the result. If you only need to divide by five, and you don't want to write a standard division routine, you can do the following: 1. Form y = x/4. (You shift your dividend by 2 bits to the right.) 2. Add y to sum. 2. Form z = y/4. Subtract from sum. 3. Form w = z/4. Add to sum. 4. Form v = w/4. Subtract from sum. (This is sufficient for 3DE or less.) 5. Form u = v/4. Add to sum. 6. Form t = u/4. Add to sum. 7. Form s = u/4. Add to sum. (This is sufficient for FFFF dividend.) For your 3DE or less, you need to add four terms, each of which is shifted 2 bits from the previous value. You can create a macro for this, or a subroutine. ================================================================ Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: spam_OUTralTakeThisOuTlcfltd.com Least Cost Formulations, Ltd. URL: http://lcfltd.com/ 824 Timberlake Drive Tel: 757-467-0954 Virginia Beach, VA 23464-3239 Fax: 757-467-2947 "Vere scire est per causae scire" ================================================================ </x-flowed>

Hey, I guessed! If you have to divide or multiply a number by a constant there is a possibility to optimize this routine for the given constant. Multiplication and division are treated the same, because the constant can be fractional and regarded as multiplier in both cases. Example: Assume constant multiplier c=3.578 and variable v is 16 bit long. Step1. Convert c to binary fractional form: 3.578(dec) = 11.1001 0011 1111 0111 1100 ...(bin) Step2. Replace series of ones with difference All series of two and more one's can be replaced by differences. For example, 1111 = 10000 - 1. The difference requires only one substraction instead of four additions. If there are no such series than optimization not possible. 3.578(dec) = 100.0001 0100 0000 0000 0000..(bin) - 0.1000 0000 0000 0000 0100..(bin) = = 4 - 1/2 + 1/16 + 1/64 - ...(dec). Step3. Limit fractional part of positive and negative constant multiplier to 16-1 bits. 16th bit can be used to round multiplication result. 3.578 = 4 - 1/2 + 1/16 + 1/64 Step4.Now shift v and add and sub...... ;) Am I right? Bye. Nikolai

<x-flowed>At 10:24 AM 10/28/99 +0300, you wrote: >Hi Robert! > >I've got interested by this way of division. How do you decompose 1/5 into >(1/4-1/16+1/64-1/256...)? It may be useful in digital filter design with >fixed weights :) > >Regards, >Nikolai X / 5 = X/(4+1) = (X/4) /(1+1/4)= (X/4) * (1 - 1/4 + 1/16 - 1/64 + 1/256 ...) = X/4 - X/16 + X/64 - X/256 + X/1024 - X/4096 ... ================================================================ Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: .....ralKILLspam@spam@lcfltd.com Least Cost Formulations, Ltd. URL: http://lcfltd.com/ 824 Timberlake Drive Tel: 757-467-0954 Virginia Beach, VA 23464-3239 Fax: 757-467-2947 "Vere scire est per causae scire" ================================================================ </x-flowed>

"Robert A. LaBudde" wrote: {Quote hidden}> > At 10:24 AM 10/28/99 +0300, you wrote: > >Hi Robert! > > > >I've got interested by this way of division. How do you decompose 1/5 into > >(1/4-1/16+1/64-1/256...)? It may be useful in digital filter design with > >fixed weights :) > > > >Regards, > >Nikolai > > X / 5 = X/(4+1) = (X/4) /(1+1/4)= (X/4) * (1 - 1/4 + 1/16 - 1/64 + 1/256 ...) > = X/4 - X/16 + X/64 - X/256 + X/1024 - X/4096 ... > > ================================================================ > Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: ralKILLspamlcfltd.com > Least Cost Formulations, Ltd. URL: http://lcfltd.com/ > 824 Timberlake Drive Tel: 757-467-0954 > Virginia Beach, VA 23464-3239 Fax: 757-467-2947 > > "Vere scire est per causae scire" > ================================================================ Hello, An alternative method would be: result = 0; while(X>=5) { X = X - 5; result = result + 1; } -Andre

In my text companding project, I'll be dividing by 96 allot so I wrote this little thing: INPUT "enter number to divide by: ", in INPUT "bit precision: ", bits accum = -1 / in i = .5 j = 1 WHILE j < bits ni = i / 2 PRINT "shift dividend right. Shift#"; j IF accum < 0 THEN 'neg IF ABS(accum + ni) > ABS(accum + i) THEN PRINT "add dividend to accumulator after shift#"; j; "(dividend /"; 1 / i accum = accum + i END IF ELSE IF ABS(accum - ni) > ABS(accum - i) THEN PRINT "subtract dividend from accumulator after shift#"; j; "(dividend /"; 1 / i accum = accum - i END IF END IF j = j + 1 i = ni WEND PRINT "Final error:"; accum; "would require 1/"; 1 / accum; "th of dividend to be added to accumulator" I thought it might be of use to someone. James Newton .....jamesnewtonKILLspam.....geocities.com phone:1-619-652-0593 http://techref.homepage.com NOW OPEN (R/O) TO NON-MEMBERS! Members can add private/public comments/pages ($0 TANSTAAFL web hosting) PIC/PICList FAQ: http://www.piclist.com {Original Message removed}