# Mathematical formula implementation

Discussion in 'VHDL' started by vhdl_newbie, Mar 14, 2013.

1. ### vhdl_newbieGuest

Hello,
I am a beginner with learn VHDL and I need to implement a mathematical formula for a job interview.
I have one day to solve this problem so if there is any good samaritan that could help me with this problem, I would really appreciate it.

solve this formula for Y

Y=(P^2+K*P)/16

with

P = [0:63]
K = [0:4]
Y = [0:512]

Have a nice day !

vhdl_newbie, Mar 14, 2013

2. ### vhdl_newbieGuest

*I am a beginner with VHDL (sorry)

vhdl_newbie, Mar 14, 2013

3. ### FredxxGuest

That looks like 4 variables and a single equation linking them together.

Have a great interview.

Fredxx, Mar 14, 2013
4. ### vhdl_newbieGuest

Le jeudi 14 mars 2013 19:51:46 UTC-4, Fredxx a écrit :
Thank you for your interest, the problem may be unclear but that is the wayit was written.
I think they want me to design an architecture with 9 bits output (Y), 2 bits (K) input and 5 bits input (P)
Do you have any idea of the functions I could use for product, division andeponentiation ?
(I am new to VHDL)

vhdl_newbie, Mar 15, 2013
5. ### rickmanGuest

If this were a real requirement, I would say there was something wrong
with it. It could be coded, but it would be hard to actually build. If
P is a 64 bit integer, it has a range of 0 to 2^64-1. As an exponent of
K, the result would have 2^64+4 bits. I don't know how to represent
that in a way that VHDL would understand. Heck, my calculator threw up
when I tried to calculate it.

Are you sure this is the right question? I can't see any value in
asking a job candidate such a question.

WAIT! The notation is wonkey! You mean the variables have his range,
not that many bits, right? That's different. But it still doesn't fit.
You can't shove the result into the range for Y.

Maybe that's what they want you to tell them?

rickman, Mar 15, 2013
6. ### Paul UiterlindenGuest

Assuming [0:x] are ranges,

FUNCTION y
(
p: natural range 0 TO 63;
k: natural range 0 TO 4
) RETURN natural IS
VARIABLE result: natural;
BEGIN
result := (p**2 + k*p)/16;
IF result > 512 THEN
result := 512;
END IF;
RETURN result;
END FUNCTION y;

Is this synthesizable? No idea, probably not, but it was not a requirement.
And you!

Paul Uiterlinden, Mar 15, 2013
7. ### FredxxGuest

I would have treated P, K and Y as 64, 5 and 513 element arrays
respectively?

Fredxx, Mar 15, 2013
8. I think you're misreading. That's (P^P+K*P)/16, or P(P+K)/16

9. ### rickmanGuest

Yes, you are right. I misread the problem. Wouldn't be the first time
and won't be the last...

But the problem remains. If those are ranges, the general result won't
fit in the variable Y.

rickman, Mar 15, 2013
10. ### chrisabeleGuest

I think Rob's rewriting is really helpful (aside from his forgetting to
change the ^ to a *). And if we assume that the values in [] are value
ranges rather than bit index ranges then the maximum Y value would be
63*(63+4)/16 = 264 which certainly is within the range [0:512]. Of
course I've assumed that Y must be an integer and truncated (actually
rounded) the result, plus the intermediate result is certainly much
larger than 512. But given those assumptions the problem is not
unreasonable. So perhaps the code that Paul posted is the best answer.

Perhaps the point of the interview question was to see if the candidate
would raise exactly the issues that have been discussed this thread...

Chris

chrisabele, Mar 16, 2013
11. ### rickmanGuest

Yes, I forgot to divide by 16, I was only looking at the multiplications.

I still think it is a bit of an odd question for a job interview, but I
suppose it must relate to something the company does, or maybe the
interviewer just picked this as an example question without any real
reason.

had solved a very tough problem before. Immediately a problem came to
mind that I had done a couple of years before and had stumped me for a
few days before coming up with a solution. Unfortunately I couldn't
recall the solution! So much for thinking on your feet.

Needless to say, I didn't get the job...

rickman, Mar 16, 2013
12. ### rickmanGuest

into the interview trying to give the impression that you "know" VHDL.
You are clearly a beginner and that will show in the interview. What
you should consider is trying to learn a bit about VHDL and let them
know that you are coming up the learning curve.

One of the big difficulties beginners have with VHDL is the strong
typing. Learn how to convert between the different types, when you need
to do what type of conversion and why. If you can explain all this in
an interview I think it will impress anyone. If I were interviewing
someone as a VHDL designer and they could articulate a good approach to
type conversions I know I would be favorably impressed. I expect a lot
of people here can't do that.

rickman, Mar 16, 2013
13. ### vhdl_newbieGuest

Hello,
Thanks a lot for all of you answer, it was pretty helpful for me. So, for those asking, I already did the interview. It was only afterwards that they asked me to solve a VHDL problem. Also, I think the [] was meant to be the range for the integers. So I came up with this, from your what I had the time to learn in one day. Don't bother correcting it, I had to send it quickly so it is too late to change it now. I send it in case it could interest someone.
Thanks again for the help !

library IEEE;
use IEEE.Std_Logic_1164.all;
use IEEE.Numeric_Std.all;

entity formule is
port (
K : in Unsigned(2 downto 0); -- 3 bits for [0:4]
P : in Unsigned(5 downto 0); -- 6 bits for [0:64]
Y : out Unsigned(9 downto 0); -- 10 bits for [0:512]
);
end formule;

architecture arch of formule is

-- refactorizing : Y=P*(P+K)/16

Signal sum : Unsigned(6 downto 0); -- 7 bits for (P+K)
Signal numerator : Unsigned(12 downto 0); -- 13 bits for P*(P+K)

begin

sum <= P + K;
numerator <= P * sum;
Y <= numerator srl 4; -- I shift the numbers 4 bits to the right to divide by 16 (someone told me "/" doesnt work all the time)

end arch;

vhdl_newbie, Mar 18, 2013
14. ### rickmanGuest

I can't resist. This is pretty good, but you didn't allow enough range
for P. It is 0 to 64, not 0 to 63. So you need 7 bits. sum still only
needs 7 bits because the sum won't exceed this range given the input
constraints.

Otherwise this is good for a beginner.

rickman, Mar 19, 2013
15. ### AndyGuest

Me neither (can't resist).

Given the non-power-of two ranges involved, I would keep this in the integer domain:

K: in natural range 0 to 4;
P: in natural range 0 to 64;
Y: out natural range 0 to 64 * (64 + 4) / 16; -- 9 bits is enough
....

Y <= P * (P + K) / 16;

Synthesis tools recognize a multiply/divide by a constant power of two as a shift operation anyway.

Keep in mind, if you are trying to use built-in MAC blocks in an FPGA, this needs to be pipelined across multiple clock cycles. Otherwise a combinatorial implementation is going to be very slow and large.

Andy

Andy, Mar 19, 2013
16. ### vhdl_newbieGuest

Hello,
Thanks for the advices, Andy and Rick, i'll try the keep them in mind.
Again, thanks a lot for your help, you are a great community !

vhdl_newbie, Mar 22, 2013