Division by some constants are very easy in binary math. Good bets are:
Robert A. LaBudde, PhD, PAS, Dpl. ACAFS
If you only need to divide by five, and you don't want to write a standard division routine, you can do the following: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 ... 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.)
Nikolai Golovchenko says
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 161 bits. 16th bit can be used to round multiplication result.
3.578 = 4  1/2 + 1/16 + 1/64Step4.Now shift v and add and sub...... ;)
on Signed divide:
rlf reg, w ;sign extension rrf reg, f How about fixing it by adding 1 to the dividend before shifts if the dividend is negative: 1/2 = 0 (1+1) >> 1 = 0 2/2 = 1 (2+1) >>1 = 1 3/2 = 1 (3+1) >> 1 = 1
James Newton has written a small QBASIC program to generate the sequence of operations required for a given Divisor and # of bits of precision. Updated with help from Nicholai:
INPUT "enter number to divide by: ", in INPUT "bits precision: ", bits accum = 1 / in i = 1 j = 1 WHILE j < bits ni = i / 2 IF accum < 0 THEN 'neg IF ABS(accum + ni) > ABS(accum + i) THEN PRINT "Add dividend to accumulator (dividend /"; 1 / i ;")" accum = accum + i END IF ELSE IF ABS(accum  ni) > ABS(accum  i) THEN PRINT "Subtract dividend from accumulator (dividend /"; 1 / i ;")" accum = accum  i END IF END IF PRINT "Shift dividend right. Shift#"; j j = j + 1 i = ni WEND IF accum <> 0 THEN PRINT "Final error:"; accum; "would require 1/"; 1 / accum; "th of dividend to be added to accumulator" ELSE PRINT "no error" ENDIF
Nikolai and James are working on a code generator for the
PIC Microcontroller using this basic
method with (really nice) optimizations by Nikolai: See:
http://www.piclist.com/codegen
http://www.piclist.com/codegen/constdivmul
David Parker says:
The closest trick that I have seen for constants that are not powers of two is to multiply by 2^32/(integer constant), making sure that 2^32/(integer constant) is rounded towards infinity. For example [on the x86 32bit], to divide Edx by 10, you could use the following:; Edx = unsigned integer. Mov Eax,(1 Shl 31)/5+1 ; 2^32/10 = 2^31/5, plus 1 for rounding. Mul Edx ; Multiply Edx by 1/10. ; Edx = Edx/10, rounded toward 0, Eax = destroyed.On many machines the Mul instruction isn't all that speedy, so this trick is of limited usefulness. On the other hand, if you are using floating point numbers then multiplying by 1/constant is usually much more efficient than dividing by the constant.
See also:
Comments:
As another approach to this problem, you can do a "full divide" but take advantage of the fact that you know certain things about the number with which you are dividing. Just as an example, here is a MACRO I use for dividing an eight bit number by a constant {on the PIC microcontroller}:;DIVIDEL divides WREG by a literal value. It gives the quotient in reg and ;the remainder in WREG. ;reg: the register to place the quotient ;lit: the literal with which WREG is divided ; DIVIDEL macro reg, lit clrf reg ;Set at zero to begin local divi, cnt divi = lit cnt = 0 while (divi & 0x80) == 0 ;Shift literal until MSB is in bit 7 divi <<= 1 cnt++ ;Keep up with number of shifts endw while cnt >= 0 addlw 0  divi ;Subtract shifted literal btfsc STATUS, C ;If positive then bsf reg, cnt ; set appropriate bit (i.e. divides once) btfss STATUS, C ;else addlw divi ; add back to restore number divi >>= 1 ;shift literal and continue until done cnt endw endmIt doesn't approximate the divide with a shifted multiply and it also has the side effect of placing the remainder in WREG. It does, however, generate more instructions and take a little more time. Divide by 3 takes 35 cycles (and instruction words) instead of 17 cycles.
I use this one for dividing by 10.
For Y = X/10: Y = (X + X/2 + X/8  X/64) / 16
This will work with 8 bit registers only, for numbers up to 159 without underflow or overflow or any of those bad things.
The X/64 term can be dropped, giving a operating range of up to 88, which is handy for converting seconds and minutes (which only go up to 59).
Ashley Preston
ashley@ieee.org +
Questions:
I have a Microchip PIC16F627, and need to add 3, 8 bit analogue numbers together, then divide by 3. The divide is giving me a headache!. I've tryed all the MATH divide links and they all fail. I'm using the assembler instruction set in MPLAB. Can anyone email me a short, simple piece of code that works ?
James Newton replies: http://www.piclist.com/cgibin/constdivmul.exe?Acc=ACC&Bits=8&endian=little&Const=.333333&ConstErr=0.5&Temp=TEMP&cpu=pic16 Seems to work for me...+
Code:
unsigned divu10(unsigned n) { unsigned q, r; q = (n >> 1) + (n >> 2); // q=n/2+n/4 = 3n/4 q = q + (q >> 4); // q=3n/4+(3n/4)/16 = 3n/4+3n/64 = 51n/644 q = q + (q >> 8); // q=51n/64+(51n/64)/256 = 51n/64 + 51n/16384 = 13107n/16384 q = q + (q >> 16); // q= 13107n/16384+(13107n/16384)/65536=13107n/16348+13107n/1073741824=858993458n/1073741824 // note: q is now roughly 0.8n q = q >> 3; // q=n/8 = (about 0.1n or n/10) r = n  (((q << 2) + q) << 1); // rounding: r= n2*(n/10*4+n/10)=n2*5n/10=n10n/10 return q + (r > 9); // adjust answer by error term }+
file: /Techref/method/math/divconst.htm, 10KB, , updated: 2020/6/28 21:41, local time: 2020/9/25 23:42,

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