# Squaring of a binary number

Discussion in 'VHDL' started by lokesh kumar, Jul 19, 2013.

1. ### lokesh kumarGuest

Hi,

Can anyone help me to design a code to square binary number?

Suppose "A" is a 5 bit number (a4a3a2a1a0)

If we do A x A then the output result will be "0 a4 0 a3 0 a2 0 a1 0 a0 " ( a 10 bit number)

For example : Suppose A= 11111
Then A x A = 11111 x 11111 = 0101010101 ( Xor operation is done for the adding to get the final result).

Could anyone please help me out how to make a generalized code for squaring of a 5-bit number to get an output like this?

Many Thanks!

lokesh kumar, Jul 19, 2013

2. ### 1999outbackGuest

1999outback, Jul 19, 2013

3. ### lokesh kumarGuest

Do not consider a simple binary addition. It is an XOR operation. You did the simple binary operation to get the result. But if you do XOR operation to add then you will get the same result as mine.

lokesh kumar, Jul 19, 2013
4. ### Nicolas MatringeGuest

Le 19/07/2013 21:08, lokesh kumar a écrit :
No matter how you do it, a multiplication is a multiplication and the
result doesn't change. You squared 31 (written in binary), the result IS
961 no matter how you got it.
Either your result is wrong, or what you want is not a square.

Nicolas

Nicolas Matringe, Jul 19, 2013
5. ### GaborSzakacsGuest

So in other words, you are doing _everything_ the same as a normal
other words as you add each column you throw away any carry bits?
So for your example you have:

11111
11111
11111
11111
11111
___________
101010101

What if I have a non-trivial case like 10110? Are you doing this
as a typical multiply like:

00000
10110
10110
00000
10110
__________
100010100

Where only my final "addition" becomes an XOR?

GaborSzakacs, Jul 19, 2013
6. ### lokesh kumarGuest

Yes, the final addition becomes XOR all the times. Because I am working on Galois field. So if I take a 5-bit number and do the square of it. Then I will get a 10 bit number. But I again I will have to use the irreducible polynomial to the 10 bit number to convert it to 5 bit. That is the concept ofGalois field. So basically I want to know how to make this 10 bit number by taking the square of a 5-bit number in VHDL.

lokesh kumar, Jul 19, 2013
7. ### FredxxGuest

All these numbers look bigger than 5 or 10 bits!
31 x 31 = 961 (11 1100 0001)

So clearly XORing is incorrect.
A square operation is precisely that, A * A. Most FPGAs have some
pretty good multipliers, best to use them.

Fredxx, Jul 19, 2013
8. ### GaborGuest

The syntax is not VHDL. He means for A to be a 5-bit vector
and the result ended up as:
'0' & A(4) \$ '0' & A(3) \$ '0' & A(2) \$ '0' & A(1) \$ '0' & A(0)
For the OP clearly "multiplication" or "squaring" is incorrect. He
apparently wants a different function that is similar to multiplication
but lacks any carries on the intermediate addition.

Gabor, Jul 20, 2013
9. ### rickmanGuest

If he is doing a calculation on a polynomial, I understand why he wants
a multiply with no carries. Each term of the polynomial has a
coefficient which is all the terms of the same value summed together mod
2 (XOR). But I don't understand his other statements. As you showed
earlier his general form above for the product is not accurate. Or he
is saying something we don't understand.

From what I understand (or think I understand) this should be code he
could use.

subtype binary_num_5 is std_logic_vector (4 downto 0);
signal A : binary_num_5;
signal B : binary_num_5;

function square (arg : std_logic_vector) return std_logic_vector is
constant arg_Hi : integer := arg'HIGH;
constant arg_Lo : integer := arg'LOW;
constant arg_Len : integer := arg'LENGTH;
variable prod : std_logic_vector ((arg_Hi + arg_Len) downto arg_L)
:= (others => '0');
begin
for i in arg'range loop
prod := prod XOR std_logic_vector (
SHIFT_LEFT (RESIZE (unsigned(arg), 2*arg_Len), i));
end loop;
return prod;
end square;

....

B <= square (A);

I think this will do the job but I haven't tested it, so many errors can
be present! If nothing else, it should give a good idea on how to
proceed. I will say the whole thing is a little bit simpler if it is
done with unsigned type signals rather than std_logic_vector. This
would eliminate the type casts in the loop assignment statement.

prod := prod XOR SHIFT_LEFT (RESIZE (arg, 2*arg_Len), i));

rickman, Jul 20, 2013
10. ### lokesh kumarGuest

library IEEE;
use IEEE.std_logic_1164.all;
use IEEE.std_logic_arith.all;
use IEEE.std_logic_unsigned.all;
--use work.my_package.all;
entity square_163_7_6_3 is
port (
a: in std_logic_vector(162 downto 0);
z: out std_logic_vector(162 downto 0)
);
end square_163_7_6_3;

architecture circuit of square_163_7_6_3 is

signal s, t, u, s_plus_t: std_logic_vector(162 downto 0);
signal xor1, xor2: std_logic;

begin

vector_s: for i in 0 to 80 generate s(2*i) <= a(i); s(2*i + 1) <= a(i+82); end generate;
s(162) <= a(81);

vector_t1: for j in 0 to 6 generate t(j) <= '0'; end generate;
t(7) <= a(82);
vector_t2: for i in 4 to 80 generate t(2*i) <= a(i+78); t(2*i + 1) <= a(i+79); end generate;
t(162) <= a(159);

xor1 <= a(160) xor a(161); xor2 <= a(161) xor a(162);
u(0) <= a(160); u(1) <= a(160) xor a(162); u(2) <= a(161); u(3) <= xor1;
u(4) <= a(82) xor a(160); u(5) <= xor2; u(6) <= a(83) xor xor1;
u(7) <= '0'; u(8) <= a(84) xor xor1; u(9) <= '0'; u(10) <= a(85) xor xor2;
u(11) <= '0'; u(12) <= a(86) xor a(162);
u(13) <= '0';
vector_u: for i in 7 to 80 generate u(2*i) <= a(i+80); u(2*i + 1) <= '0'; end generate;
u(162) <= a(161);

xor_gates1: for j in 0 to 162 generate s_plus_t(j) <= s(j) xor t(j); end generate;
xor_gates2: for j in 0 to 162 generate z(j) <= s_plus_t(j) xor u(j); end generate;

end circuit;

This the the exact code I found online, I think. But it is for 163-bit. So it is difficult to test and verify. Can you please help me to convert it for a 5-bit to make me understand it?

Many Thanks!

lokesh kumar, Jul 20, 2013
11. ### rickmanGuest

rickman, Jul 20, 2013
12. ### rickmanGuest

rickman, Jul 20, 2013
13. ### lokesh kumarGuest

It is a bit confusing to me.
Can I have your email id please? I can send you a relevant paper for the circuit with the algorithm. May be you will be able to understand it. I am unable to find more information related to it.

vector_s: for i in 0 to 80 generate s(2*i) <= a(i); s(2*i + 1) <= a(i+82); end generate;
s(162) <= a(81);

vector_t1: for j in 0 to 6 generate t(j) <= '0'; end generate;
t(7) <= a(82);
vector_t2: for i in 4 to 80 generate t(2*i) <= a(i+78); t(2*i + 1) <= a(i+79); end generate;
t(162) <= a(159);

Thanks!

lokesh kumar, Jul 20, 2013
14. ### rickmanGuest

Do you understand what this code is doing? It is generating the signal
s() from the signal a(). Is that clear?
This code generates t() from a().
I don't get it either. What are a(), s(), t() and z()? Is this
supposed to be squaring a() to get z()? If so, why is a() the same
length as z()? I can't find any of this code using Google.

I got your other email which was largely the same as your earlier post I
think. I prefer to discuss this here. There may be others who would
like to understand this or who can explain it. In fact, you might try
explaining what you are doing and ask how to do this in other groups
such as comp.dsp. Working from an undocumented code section is not a
great way to understand an algorithm.

I can't tell you anything about the code you posted. I have no idea why
they are doing all the calculations they are doing. I can read the
VHDL, but I can't read the mind of the person who wrote it.

I might be able to help you figure this out if you give more background.
What are you trying to do? What is the bigger picture? There is
often more than one way to skin a cat. If I understand what you are
trying to do with polynomials I think I have already explained what you
need to do to square a() and given you code to do it. If I don't
understand, perhaps you can explain better?

BTW, when you use Google Groups to post in newsgroups you need to fix
the quoted material and after a couple of quotes becomes unreadable.

rickman, Jul 20, 2013
15. ### FredxxGuest

This all looks like polynomials to me which are an art in themselves.
Without the writers intention it feels we're going to remain in the dark.

I confess that whenever I've need a polynomial to calculate or check a
CRC, I've used this rather invaluable site:
http://www.easics.be/webtools/crctool

Perhaps this might give Lokesh some insight?

Fredxx, Jul 20, 2013
16. ### lokesh kumarGuest

Lets consider a 5 bit number, A = 10100

If we do the squaring of it then we will get,

A = 10100
x 10100
-------------
00000
00000
10100
00000
10100
-------------------
c= 100010000 (XOR operation is performed to add)

Now Irreducible polynomial is used to reduce it to 5-bit. (The aim is: if the input is out 5-bit then we need to reduce the output to 5-bit)

So for a 5-bit number the irreducible polynomial is , F(x) = x^5 + x^2 + 1 (we can write it in binary form as 100101) (It is a standard value)

Now both c and F(x) are added to reduce it to 5-bit. (from the MSB)

100010000 (value of c that we got after the squaring)
xor 100101 ( value of F(x) that we calculated)
-------------------
000111000 ( it is not 5-bit)

So now again do the xor operation to the result with the irreducible polynomial.

111000 (Do not consider 3 zeros from the left)
xor 100101 (irreducible polynomial)
---------------
011101 ( now it is reduced to a 5-bit number,(do not consider the zero at left side))

If you take a close look on the result then, we have taken A = 10100 and we got c = 100010000 (before reduction)

so simply we insert one zero between all the bits of A, then we will also get the same result.

c = 0 1 0 0 0 1 0 0 0 0 ( zeros are indicated)
| | | | |

If you remove the indicated zeros then tht value is equal to "A"

So now my main aim is to design a generalised code for 5-bit. Suppose I am giving a 5-bit input, A = a4 a3 a2 a1 a0

Then after squaring, I am getting a result C = 0-a4-0-a3-0-a2-0-a1-0-a0 ( zeros are added between all the bits of A)

Now I have to use same irreducible polynomial (100101) to reduce it to 5-bit.

lokesh kumar, Jul 20, 2013
17. ### lokesh kumarGuest

Lets consider a 5 bit number, A = 10100

If we do the squaring of it then we will get,

A = 10100
x 10100
-------------
00000
00000
10100
00000
10100
-------------------
c= 100010000 (XOR operation is performed to add)

Now Irreducible polynomial is used to reduce it to 5-bit. (The aim is: if the input is out 5-bit then we need to reduce the output to 5-bit)

So for a 5-bit number the irreducible polynomial is , F(x) = x^5 + x^2 + 1 (we can write it in binary form as 100101) (It is a standard value)

Now both c and F(x) are added to reduce it to 5-bit. (from the MSB)

100010000 (value of c that we got after the squaring)
xor 100101 ( value of F(x) that we calculated)
-------------------
000111000 ( it is not 5-bit)

So now again do the xor operation to the result with the irreducible polynomial.

111000 (Do not consider 3 zeros from the left)
xor 100101 (irreducible polynomial)
---------------
011101 ( now it is reduced to a 5-bit number,(do not consider the zero at left side))

If you take a close look on the result then, we have taken A = 10100 and we got c = 100010000 (before reduction)

so simply we insert one zero between all the bits of A, then we will also get the same result.

c = 0 1 0 0 0 1 0 0 0 0 ( zeros are indicated)
| | | | |

If you remove the indicated zeros then tht value is equal to "A"

So now my main aim is to design a generalised code for 5-bit. Suppose I am giving a 5-bit input, A = a4 a3 a2 a1 a0

Then after squaring, I am getting a result C = 0-a4-0-a3-0-a2-0-a1-0-a0 ( zeros are added between all the bits of A)

Now I have to use same irreducible polynomial (100101) to reduce it to 5-bit.

lokesh kumar, Jul 20, 2013
18. ### rickmanGuest

Ok, we seem to be getting somewhere with this post. I am snipping
formatting problems.

This agrees with what we covered before.

Ah! That's the part that was missing.

Is this the same as dividing? I'd like to understand the theory behind
this... or maybe not... ;^) lol

But seriously, this rings a bell that division and multiplication are
similar if not the same in polynomial arithmetic.

Yes, this part actually doesn't require any hardware, just a bit of code
to assign different wires.

Yes, more or less. Applying the irreducible polynomial (ip) to the
result is not hard to do, but the iteration can be done a couple of
ways. Is there a way to *know* how many times it must be applied?
Obviously this is what the code you provided is doing. I'm guessing
there must be some way of knowing how many times the ip must be applied
and perhaps even it is always the same number? If so, I can easily
generate the code similar to the 163 bit wide solution. If not, the
code will have to loop on applying the ip I think.

We could have a value of c = 101010101

applying the ip
101010101
100101 - once
001111101
100101 - twice
0110111
100101 - three times
010010 result!

This took three iterations and the number is not fixed although I think
we can set a max limit at three. So we will need to loop and either
exit conditionally or loop for the max count and conditionally execute
the XOR. Better, just make the 1's in each ip the value of the msb in
the appropriate bit of each intermediate c value. I bet if you dig into
the 163 bit version that is what they are doing. I can't figure it out
myself. First thing they do is to fold the input vector so the msbs are
in the bit positions that would be filled with zeros. Then they
generate vectors s, t and u which are all then XOR'd together to produce
the output.

Do you have a preference about whether this is clocked in a register and
takes multiple clock cycles or goes through successive stages of logic
without registers or clocks?

Any reason to generalize this for N bit wide inputs? I think we have
all the pieces now so we could do that as long as we can define the ip
for each value of N. Just doing it for N=5 is simpler of course.

If you do this normalization on a few test values to make sure it gives
the right answer and tell me if you want it as straight logic,
registered iteratively or pipelined, I'll give you an implementation.

rickman, Jul 20, 2013
19. ### lokesh kumarGuest

Yes, I am also thinking there must be some ways to know about the number ofloops. We can calculate small numbers like 5-bit or 6-bit by using our hand. But it is very hard to calculate for 163-bit.

In your last message you told that, 3 loops are used to reduced. But it is not correct. After the first loop, you are getting two zeros in the beginning. So we do not need to apply the reduction polynomial value.So the 2nd loop is not needed. We can directly consider 1111101 and XOR with 100101 to get the 5-bit number.

http://www-brs.ub.ruhr-uni-bochum.de/netahtml/HSS/Diss/KumarSandeepS/diss.pdf

I am quite not sure if you can access this link.

If you are unable, then please google "elliptic curve cryptography for constrained devices" and check out the first link. On page number 132 (figure 8..2), the multiplier circuit for 163 bit is given. The reduction polynomial for 163 bit is, F(x) = x^163 + x^7 + x^6 + x^3 +1

So if you closely take a look on the circuit, then you can see that almost all the parts are similar. But there are 3 extra XOR gates are connected. The first extra XOR gate is connected after C2 register ( it is for x^3 which is a part of irreducible polynomial). The second extra XOR gate is connected after C5 register ( for x^6 which is a part of irreducible polynomial) and the third extra XOR gate is connected after C6 register ( for x^7 term in the polynomial). The output of C162 is directly connected to the XOR gate before C0. Because we do not need to add XOR gates for x^162 and 1.

Similarly for a 5-bit number, the reduction polynomial is F(x) = x^5 + x^2 + 1
So in this case we need to connect only one extra XOR gate after C1 (or before C0). And the output from c4 is directly connected to the XOR gate before C0.

In this case the reduction polynomial is attached. So the result if we takeboth A and B are of 5-bit numbers, then we will get the 5-bit number as anoutput.
This is the basic operation for the multiplier.

Now please look at the squaring circuit on page 133 (figure 8.3). I am unable to understand the circuit. I am not sure if the reduction polynomial is attached here. Please have a look on it and let me know if you understand.

-
Lokesh

lokesh kumar, Jul 21, 2013
20. ### lokesh kumarGuest

I think it is possible. We can use the shift registers and then using the irreducible polynomial.

lokesh kumar, Jul 21, 2013