Breaking News
Blog

# Doing Math in FPGAs, Part 5 (Binary Division)

NO RATINGS
3 saves
Page 1 / 2 Next >
User Rank
Blogger
Re: Doing division
2/21/2014 3:52:51 PM
NO RATINGS
@betajet:

I was thinking specifically of Newton-Raphson (N-R).  Wikipedia covers the gory details of the theory behind it so I won't repeat it here.

Yeah, I have actually read through the whole thing.  I really like that it finishes in log(N) steps (if your special cases are handled, and you have a good initial guess).  The problem with it is the generalization to "any" implementation - I don't see a good way to package up an initial guess - I suppose I could just make it half of 2^N, but then it might not be saving me anything...  Of course this is because I want to generalize it for fixed-point.  I see that it may be a bit easier for floating point.

User Rank
CEO
Re: Doing division
2/21/2014 3:43:36 PM
NO RATINGS
I was thinking specifically of Newton-Raphson (N-R).  Wikipedia covers the gory details of the theory behind it so I won't repeat it here.

Basically, to do floating-point division with N-R, you want to find the reciprocal of the denominator D.  To use N-R, you first estimate the reciprocal X[0] and then refine the estimate using the N-R iteration X[i+1] = X[i] * (2 - D * X[i]) which has only multiples and adds.  Each iteration reduces the error quadratically, so if you start with an estimate that's accurate to 13 bits, one iteration gives you 26 bits (single precision) and two gives you 52 bits (double precision).  It you have high-speed multiplication hardware, this is a great way to go.

You find the initial estimate using a look-up ROM.  The denominator D is a binary floating-point number of the form M * 2^E, where mantissa M is between 0.5 and 1.0.  To calculate your first estimate X[0], look up the 12-16 MSbs of M in a ROM to get a number between 2.0 and 1.0, divided by 2 to renormalize it to between 0.5 and 1.0.  The exponent is easy: just negate it and add 1 to compensate for the mantissa look-up.  In other words, X[0] = XM * 2^(1-E) where XM is the value read from the ROM.

You need to handle some special cases:  If M is zero or an unnormalized fraction, 1/D is +infinity.  If D is +infinity, 1/D is +0.

There is also a good N-R iteration for calculating square roots.  In this case, you actually calculate 1/sqrt(x) and then multiply by x to get sqrt(x).

Integer square root using the "long division" method is also loads of fun.  It's a simplified form of the decimal "long division" method that was taught in USA public schools before 1950 and dropped once TV started to rot the minds of USA youth :-)

User Rank
Blogger
Re: Doing division
2/20/2014 7:26:34 AM
NO RATINGS
Actually, the next post doesn't talk about such things, and I am interested in your thoughts.  I know of the SVT algorithm, but haven't yet quite wrapped my mind around it.  I also took a look at Newton-Rhapson, but the need for a scaling factor throws me in that I haven't figured out how to make it generic - that is, if I instantiate a QN = 8,12. but want to then instantiate a QN=3,15, the scaling factors will need to be different (I think).

I'm interested in thoughts...

User Rank
CEO
Re: Doing division
2/19/2014 2:00:40 PM
NO RATINGS
If you have floating-point numbers, it's actually easier to do division.  The mantissas are already normalized, so you don't need the pre-shifting business, and floating-point uses sign/magnitude representation so you only need an XOR to take care of signs.

If you have fast multiplication hardware -- which you usually do nowadays -- you can use an extremely clever technique to calculate the reciprocal.  I think the author is working up to this, so I won't say more.

User Rank
Manager
Computer binary division
2/19/2014 1:36:43 PM
NO RATINGS
I am trying to remmember IBM 360/75 divide.

Dividend occupied even odd register pair to form 64 bit width  Divisor in third 32 bit register.

make both positive by complement if negative, remember if signs were different and make quotient negative at end

do a trial subtraction of divisor from high half of dividend, if positive write result into dividend and shift one into dividend low bit.  If negative result shift dividend left and zero into low bit.  (Aligns left bits and generates zero high quotient bits)

repeat 32 times(word width) so the 32 bit quotient is shifted into the low register of the dividend pair.  The shift and subtract reduces the dividend to a 32 bit remainder in the register that held the high order dividend 32 bits.

Note that shifting zero into the emptied bit of the dividend until the divisor can be successfully subtracted accomplishes the alignment of the operands while reducing the magnitude of the quotient. (As I already wrote...)

Multiples of the divisor (2x,4x, 5x?) were generated and decoding of dividend and divisor bits to generate multiple quotient bits to reduce cycles.  That is too complicated to remember.

The ALU and working registers were 64 bits for floating point so no big cost for divide and multiply 64 bit product.

User Rank
Blogger
Doing division
2/19/2014 12:07:34 PM
NO RATINGS
Nice and determistic, but not necessarily the fastest way?  Anybody out there got anything better? I'd like to hear it.