So, welcome to this class onHardware Security.
So, today we shall be talking aboutdifferent
topic,we will be talking about afinite field
architecture like last last class we discussed
about multipliers. In today's class, we will
discuss about another important class of arithmetic
circuits in finite fields, we are called as
finite field inverse. So, how to compute finite
field inversions?
So,in more specifically, we shall be talking
about multiplicative inverse in GF 2 to the
power of m, which is a characteristic to field.
We shall be talking about Itoh-Tsujiiinversion
algorithm,we shall be talking about a generalization
of the Itoh-Tsujiiinversion algorithm. And
time permitting we shall be discussing about
hardware design and also performance of the
corresponding hardware circuit.
So, to start withhow do we compute a multiplicative
inverse or what is the definition of a multiplicative
inverse.Multiplicative inverse means that
given an element a, we want to find out it
is inverse, which is denoted as a inverse
or a to the power of minus 1 in the fieldsuch
that if I multiply a with a inverse, I get
the unit or I get the unit element in the
field or I get 1.
So,so to start so you know like there are
different techniques for finite field inversion
computations. The most important ones I have
kind of written here like the first one is
the extended Euclidean algorithm, which is
a more generalization of the Euclidean algorithm,
which is used to compute the greatest common
divisor of two numbers. There is another algorithm
or there is another result, which is very
famously known as the Fermat's little theorem,
which also can be used to compute multiplicative
inversions.
So, sothe here is that the Fermat's little
theorem is more efficient on hardware than
the Euclidean technique. Because, the Euclidean
technique as you probably know is that it
will be it will it will be resulting in a
inin a computation, which requires moremore
intensive operations, and also essentially
willwill require more number of clock cycles.
So, essentially right we we resort to Fermat's
little theorem, and let us state how a Fermat's
little theorem works. So, when we want to
compute a to the power of minus 1, the idea
is that and I want to do a modulo of the field,
then our modulo of the size of the field.
Then what we do is we try to compute a to
the power of minus 1 as a to the power of
n minus 2 modulo n. Note here that the definition
of n is such that n can be any number as such,
but a has to be co-prime with n ok that means,
the greatest common divisor of a and n has
to be 1 or in another another sense, we can
also say this as a belongs to the multiplicative
group, which is created by n, which is denoted
of n as z n star ok.
So, you can also say for simplicity that let
us choose n as a prime number in which case
all the elements, which are non-zero like
starting from 1 to n minus 1 belong to its
multiplicative group. So, just let us take
a small example. So, suppose I take n equal
to 5, which is a prime number, so then you
can easily see that 3 to the power of 5 minus
2 modulo 5 should be it is inverse, you can
check it that this turns out to be 2. And
therefore, if I multiply 3 with 2, then I
get 6 which modulo 5 is 1 ok. So, therefore
2 is indeed the multiplicative inverse of
3 modulo 5.
So, just to generalize this,wethe generalization
is based upon a particular parameter a particular
function, which is called as the Euler's Totient
function often denoted as phi of m ok. So,
suppose that a and m are integers such that
the greatest common divisor of a and m is
1 that means, a and m are mutually co-prime
or relatively prime. And then the number of
integers in Z m that are relatively prime
to m, and does not exceed m is often denoted
as phi m ok. And this is called also as the
Euler's Totient function or the phi function
ok.
So, just to define write the phi of 1 is taken
as 1, this is an initialize the phi function
ok. And subsequently write if I for example,
1 to compute 26, m equal to 26, and I want
to calculate phi of 26, then it turns out
to be 12 that means, there are 12 numbers
are co-prime with 26. If p is prime, then
it is natural to understand that phi of p
will be phi will be p minus 1 ok, just as
we saw in the example.
And if n is equal to let us say here, I have
just tried to kind of tabulate the values
of phi n. So, you will see that phi n is a
pretty much irregular function ok, it is neither
monotonically increasing nor monotonically
decreasing, it is kind of an erratic function.
But, luckily likeif you know the prime factors
of n then, you can compute phi n quite efficiently
ok.
However, I am not going into that definition
here, but rather we will take a look into
Fermat's little theorem, which is stated here
using the phi function ok. So, for example
if I take the greatest common divisor of a
and m as 1 ok, and then I can write a to the
power of phi m is congruent to 1 modulo m,
this equation is also called as the Euler's
function or Euler's theorem ok. Fermat's little
theorem is actually a special case of Euler's
theorem, where m is a p which is a prime number
ok. And if it is a prime number, then I can
write phi of p as p minus 1, so what we get
is a to the power of p minus 1 is congruent
to 1 modulo p.
If you have the field, which is GF to the
power of m, therefore right where there are
2 to the power of m elements in the field.
So, this is the extension field that we saw
in the last class ok, where you extend from
GF 2 to the GF 2 to the power of m using an
irreducible polynomial whose dimension is
or degree is m ok. So, then we know that a
to the power 2 to the power of m minus 1 should
be congruent to 1, which means that if I want
to compute a to the power of minus 1, then
I need to compute a to the power of 2 to the
power of m minus 2 ok, because essentially
that should be giving me the multiplicative
inverse ok. So, therefore right how will I
compute this a to the power of 2 to the power
of m minus 2 is what we need to discuss. If
I want to compute a to the power of minus
1 or the multiplet or the multiplicative inverse
of a.
So, here there is a technique,which is called
which is basically based on what is called
as an addition chain ok. So, if I want to
knifely compute a to the power 2 to the power
of m minus 2, then you can see that I will
require m minus 1 squarings and m minus 2
field multiplications ok.
Now, multiplication is costly, so we do not
apply this knife method. Instead, we essentially
can apply an algorithm, which is called as
the Itoh-Tsujii inversion algorithm, which
uses the addition chain ok. So, let us define
an addition chain how are what is the definition
of addition chain, so addition chain for a
positive integer n is a sequence of nsequence
of natural numbers ok starting from u 0, u
1, and so on to say u l that means, I will
say that l is the length of the addition chainsuch
that I will initialize u 0 to 1 and u l to
n that means, since is the addition chain
for n the starting point is 1, and the end
point is n ok. The in between values, we need
to determine ok.
But, there is a restriction the restriction
is that any sequence in this addition chain
should be obtained by the sum of previous
two numbers ok that means, for example if
that is u l or say u i, then u i should be
essentially expressible as a sum of u j plus
u k that means, should be expressed as a sum
of the previous two numbers ok, and u such
that u j and u k both belongs to this addition
chain ok. So, therefore I should be able toexpress
you know like u u i as the sum of u j plus
u k ok.
So, for example right if I want to compute
say you know like for example, you can see
in this addition chain that this is an addition
chain for 162, so I have started with 1 my
endpoint is 162, you can observe that in this
addition chain 2 can be obtained by adding
1 with itself ok. Likewise 4 can be obtained
by adding 2 with itself, likewise 5 can you
obtain by adding 4 with 1 ok, 10 can be obtained
with 5 with 5, 20 with 10 with 10, 40 with
20 with 20, 80 with 40 with 40, 81 with 80
and 1, and 162 where we add 81 with 81 ok.
So, these kind of addition chain where one
of the numbers is the previous number is also
often called as a brewer chain ok. So, this
is a specific name given better which is called
as a brewer chain, which is essentiallywhere
your previous number is just also also a part
of the addition chain ok. And what we want
right therefore, if you if you want to compute
this a to the power 2 to the power of n minus
2, then what we want is that this addition
chain should be optimal in length ok. Although,
it is a hard problem or this difficult problem
to find out like what is the optimal chain,
but there are somepropositions in literature,
which essentially gives us efficient chains
ok, which we can use to compute this multiplicative
inverse.
So, taking thisbackground or taking this addition
chain into account, what we can do is this
so if I want to now compute e to the power
2 to the power of m minus 2 which was my inverse,then
what I can do is I candefine avariable, which
I denote as beta k ok. So, beta k is nothing
but a to the power 2 to the power of k minus
1 ok, so beta k is defined as a to the power
of 2 to the power of k minus 1.
So, if I want to calculate a to the power
of minus 1, then therefore I as I said from
Fermat's little theorem, I need to calculatea
to the power 2 to the power of m minus 2,
which is nothing but the square of a to the
power 2 to the power of m minus 1 minus 1
whole square ok. So, what is this inside the
parentheses is nothing but beta m minus 1
a ok.
So, therefore I need an efficient way to calculate
beta m minus 1, and then finally I will square
to get the inverse ok. So, this beta m minus
1 interestingly can be computed using a nice
recursive formulation. So, what I can do is
I can write beta k plus j, which is like some
position in the addition chainand express
them in terms of beta k and beta j ok.So,
therefore what I do is I try toyou know like
derive beta k plus j, which is a higher beta
value from the smaller beta values ok.
And therefore, I can do it efficiently.So,
why does it work, so it is very easy to see
like beta k plus j is nothing but e to the
power 2 to the power of k plus j minus 1.
So, therefore, what I can write is I can for
for convenience, I can introduce this minus
2 to the power of k, and then ad plus 2 to
the power of k, so basically it is the same
minus 1.
And then from the first two factors, I take
2 to the power of k common ok. So, therefore
I get 2 to the power j minus 1 into 2 to the
power of k, and the other term is a to the
power 2 to the power of k minus 1. So, you
can note easily that this that is a to the
power 2 to the power of j minus 1 is nothing
but beta j, and I have done a beta j whole
to the power 2 to the power of k, whereas
the other term is beta k itself.
So, therefore I have been able to to express
beta k plus j in terms of smaller beta j and
beta k values ok. Interestingly, you can also
observe that I can decompose this in either
you know like, I can also write this as in
the other form like, I can also write this
as beta k to the power 2 to the power of j
into beta j right, soso which one will I choose
that may depend upon the fact that I probably
would like to reduce the number of squarings,
which is required here ok, because here I
am doing beta j to the power 2 to the power
of k. So, depending upon whether j is k small
or k small, I can write thewrite write the
form accordingly ok. So, so if k is small,
then I will write in this form. If j is small,
then I will write this as beta k whole to
the power 2 to the power of j multiplied by
beta j ok, which is also correct.
So,so therefore for example let us take an
example, suppose I want to calculate for m
equal to 233 last day we saw about a Karatsuba
representation for 233. So, we will take that
same multiplier applied here, so what we want
to do is therefore, I want to compute a to
the power of minus 1, which is equal to a
to the power 2 to the power of m minus 2,
where m equal to 233.
So, what do I need therefore, I need an addition
chain right I need an addition chain for 232
ok, which is m minus 1. So, therefore here
it is an addition chain for 232. So, you can
you can observe that it satisfies all the
properties of being an addition chain. And
therefore, right what I can do is I can apply
my recursive formulation, and from there I
can essentially calculate the value of beta
232 ok. If I know beta 232, then I will just
do a squarings to get the corresponding multiplicative
inverse of a ok. So, let us try to see step
by step how we can calculate these beta values
to arrive at beta 232, and then finally get
the inverse ok.
So,so what we do here ishere is an corresponding
tabular representation of how we can do that,
so we start with beta 1 a which is a. So,
we start with a and then we calculate beta
2 a ok. So, you can see that this like the
if you observe thesuffix here, so 1, 2, 3,
6, 7, 14 28, 29, 58, 116, 232, this essentially
stands for your addition chain ok. So, these
are the values, which you had in the addition
chain.
So, now what you can do is that you can calculate
beta 2 in terms of beta 1, because you have
you can express 2 as 1 plus 1 oklike what
so ever therefore, if you can if you want
to calculate this, therefore you can apply
the recursive formulation, so therefore beta
1 plus 1 is nothing but beta 1to the power
of 2 to the power of 1 into beta 1 and therefore
you can calculate this corresponding value.Likewise,
you can take any beta for example, let us
take the example of say beta29 for example
if I want to calculate beta 29, then I can
express 29 being in the addition chain as
term 20 plus 1 ok.
And therefore, what I will do is I will write
beta 28 whole to the power 2 to the power
1 into beta 1. Note I could have also done
that as beta 1 to the power 2 the other way
round, but that would have incurred more number
of squarings ok. So, therefore these are very
natural thing that I will do to reduce the
number of squarings. So, all in all here we
need 232 squaring, so maybe you can count
it down and you need 10 multiplications.
So, you can count for example, 1, 2, 3, 4,
5, 6, 7, 8, 9, and 10, so you need 10 multiplications,
and if you add up the number of squarings,
which you are doing here.So, like for example,
here you are doing is squaring, here you are
doing squaring, here you are doing 3 squarings
and so on, you will get 232 squarings which
you are essentially doing. ah
Now, again remember remember that squarings
in GF 2 to the power of m arithmetic is easy
ok. So, therefore this is a nice trade-off
in that sense, where you are basically optimizing
the number ofmultiplications at the expense
of squarings, which is actually quite conveniently
implement in GF 2 arithmetic.
So,so how will I realize this by an hardware
circuit, so here is a proposal. So, of course
if I want to do a knife implementation right,
then I would have just have some squarer's,
and then I would have repeatedly applied those
squarings functions right. But, then that
may not be efficient, because you can observe
that in the addition chain as you go up higher,
there are some places, where you are doing
significantly in a large number of squarings
like here you are doing say 116 squarings
ok.
So, in the general case it may happen that
you are doing, so it since we are doing a
large number of squarings that would although
the squarings circuit isefficiently implemented.
But, still if you are cascading, so many squarings
operations, then at the end of the day your
performance will get affected ok.So, therefore
what you can do is you can try to trade off
between hardware and the clock cycles, and
rather than implementing 1 squarer, you can
actually have a cascade of squarer's. So,
you can have say U s a number of squarer's
ok.
So, U s essentially is a value in the addition
chain, which you have chosen by some kind
of analysis I will try to elaborate later
on, but imagine that you have got U s number
of squarer's, which you have cascaded. And
then you have followed it by a multiplexer
circuit ok. So, now suppose I wantyou like
to calculate a squarings I want to calculates
sudden number of squarings, and if the number
of squarings is less than U s, then what I
can do is that I can just correspondingly
multiplex out the output right.
For example, if I have got say 2 squarings
required, then I will do I will pass the input
here, and what I will do is I will give an
appropriate control here. So, that this signal,
which is essentially after 2 squarings operations
guess gets tapped out ok. On the other hand
right if the number of squarings is more than
U s, then you have to essentially feed this
back result back ok.
And you have to do say U i by U s seal number
of cycles ok. So, you have to basically keep
on applying this 1 after the other, and you
have to get the corresponding result ok. So,
depending upon you know like how many number
of squarings you have, you essentially have
to do this configurability of or you have
to configure your hard work ok, you have to
either give the output after 1 clock cycle
or you have to spend more clock cycles for
getting the result ok.
So, so now what we yeah if we can estimate
the number of clock cycles required here,
so this expression shows the number of clock
cycles. So, you can see that there are three
components here. The l minus 1 term is due
to l minus 1 multiplications. As I said that
there will be l minus 1 multiplications, which
you are doing. And if you employ a combinational
circuit like the Karatsuba multiply that we
saw in the last class for every multiplication
you are basically expending one clock cycle.
So, you are doing l minus 1 clock cycles.
You need one extra clock cycle, because if
the extra clock cycle which you need for the
final squarings operation, because you have
to do a final squarings at the end . And this
summation gives you the number of squarings
for doing all the independent all the iterations
which were there in the previous table ok
. So, you can observe that the number of squaring,
which you were doing right isis u i minus
u i plus 1. So, u i is 1 addition chain u
i plus 1 is the next addition chain.So, you
are doing u i minus u i minus 1 number of
squarings ok.
So, if you go back to this table for example,
you can see that the number of squarings which
you are essentially doing essentially is nothing
but the summation of 1 is a subtraction or
the difference from one u i value with the
previous one. Like suppose it is 28, it is
twenty 28 and 14, so you are doing 14 squarings
operations here ok . So, likewise if these
3 and 2 you are doing 1 squarings operations
here; if it is 232 and116, here you are doing
116 squarings operation. So, the difference
ok.
So, therefore, right here you are, so therefore,
the number of squarings that you need to do
is u i minus u i minus 1 . And since often
you know like it may be more than u s. So,
as I said that you have to divide it by u
s, and then take the you know like the the
seal of that, and therefore, that gives you
the number of clock cycles per iterations
. And you have got l number of iterations,
so youyou know likevary i 2 to l, so that
gives you a rough total number of clock cycles
which you expect ok .
So, therefore, right once you have this right
I mean its fine, but at the same time you
would like to optimize it ok. So, here is
a possible optimization that you can try to
sort of implement . So,so as you see that
in the normal Itoh-Tsujii the inversion algorithm
there are a lot of squarings operation ok.
So, we are . So, you can actually try to do
an interesting comparison that is little you
can what you can try is you can try to compare
a squarer circuit with a quad circuit ok.
So, what is the quad circuit a quad circuit
is nothing but which something which computes
a to the power of 4 ok, and squarer is something
which computes a to the power of 2 ah.
So, here is an example which we try to work
out for GF 2 to the power of 9 ok. So, you
can take an irreducible polynomial in x to
the power of 9 plus and so on . And then you
can calculate or create this field GF 2 to
the power of 9, and compute the complexities
or you know like the number of operations
which you need to do, squarings and quads
in these 2for this field .
So, here isyou know like the corresponding
bits like 0 to 8, because there are 9 bits
which are there in GF to the power of 8. And
what we observe here is the circuits or the
circuit complexity for calculating the squares
ok. This one is for the quad operations . So,
you can observe easily that quite intuitively
the number of operations in the quad is more
than the number of operations in square, because
in square you are just doing power of 2, whereas
in quad you are doing power of 4.
But interestingly if you look now in terms
of LUTs, you will see that and if you remember
that in the last class some of the classes
we discussed that in a lookup table, you essentially
can fitting. Like suppose if I take a look
up table with 4 inputs you can actually fit
in 4 a Boolean function with 4 input variables.
If we have got a Boolean function with 1 input
variable or 2 input variable, then also that
LUT gets consumed ok . So, therefore, right
if I want to calculate the number of lookup
tables, you see that here there is not no
function done, so there is no lookup table
requirement.But if you have got b 1 plus b
5, then you have got 1 lookup table requirement
ok.Likewise b 6 there is no lookup table requirement.
If for 4 b 2 plus b 6 we have got 1 lookup
table requirement; like this we have got 4
look up tables which are being used ok .
On the other hand for the quad circuit, you
will see that you still have got one lookup
tables, but even for this value like b 1 plus
b 3 plus b 5 plus b 7, where you are doing
more computation you are still using one lookup
table ok. So, here at the end you are using
6 lookup tables ah. So, now, if we compare
a quad circuit with a cascade of 2 squarers,
where you are doing squarings and then squaring,
then you can see that there is an improvement.
Because here you are you are acquiring 6 look
up tables, but here you would have required
4 into 2 that is 8 look up tables. But in
terms of delay, if you compare these two circuits
both of them have got a delay of one lookup
table ok, because they are consuming one look
up tables ok. So, therefore, there seems to
be an opportunity for trade off where you
can try to optimize and have a better performing
inversion circuit ok. ah
So, what we observed is we try to kind of
see it for other fields like larger fields.
For example, if you see for GF 233 you will
see that a squarer circuit will consume some
153 look up tables, whereas a quad will consume
233 look up tables . But if you compare with
again with 2 times 153, then you see that
there is 25 percentage advantage; delay wise
again both of them are same ok . So, therefore,
naturally it tells us that you know like going
from our squarer circuit to a quad circuit
may give me an advantage, because I can utilize
the look-up tables better . But the question
is you also need to generalize like the Itoh-Tsujii
algorithm to work for quad circuits ok .
So, therefore, what we do is we essentially
try to look into trying or trying to generalize
this this theorem and it turns out that the
first one is very straight forward ok. So,
now here instead of doingor defining beta,
we try to define alphas ok. So, alpha k 1
is nothing but alpha to the power 2 to the
power of n k 1 minus 1 ok. So, in the previous
case, what did we have we had a to the power
of 2 to the power of k 1 minus 1 but now we
have generalized this and put in n equal n
ok. So, we have generalize it to any integer
n . So, likewise if you want to make it work
as a quad circuit right, then I will make
n equal to 2 ok.
And that will work as a to the power of 4
to the power of k right . And essentially
we do power of 4 then power of 2 . So, likewise
if you take alpha k 2 which is a to the power
2 to the power of n k 2 minus 1, then we can
see that you can actually calculate alpha
k 1 plus k 2 in terms of alpha k 1 and alpha
k 2 ok.Now, this proof is very straight forward
similar to what we have already seen in the
context of beta. So, I am not going to that.
What about inverse? So, once you have done
this right and you have computed thisseries,
then you can actually calculate e to the power
of minus 1 also in a very straightforward
manner. So, what you can do is depending upon
you know like whether n divides n minus 1.
So, n what is n, n is the parameter through
which you are generalizing this quad or you
know like higher powers of 2 ok. So, this
n is essentially here. ah
So, therefore, when n divides m minus 1, because
n may divided m minus 1, it may not be divided
m minus 1. If m divides n minus 1, then alpha
m minus 1 by n is defined, because m minus
1 by n is there an integer ok . So, then if
you can calculate alpha m minus 1 by n, then
if you just squared it you will get a inverse
ok. But if it is not defined, then you can
write m minus 1 as n q plus r where n q and
r are all natural numbers or integers, then
you can write this a inverse as alpha q to
the power 2 to the power of r into beta r
a and then square the entire operation ok.
ah
So, you can try to observe thisvery see a
very, very clearly here that is suppose you
have got alpha m minus 1 by n . So, then alpha
m minus 1 by n is nothing but you can write
this as a to the power 2 to the power of m
minus 1 by n multiplied by n. So, therefore,
this n and n cancel. So, you have got alpha
2 to the power of 2 to the power of n minus
1 minus 1, which essentially is nothing but
the inverse from for Fermat's little theorem
ok. When n does not divide m minus 1, then
you cannot write this in this form, because
m minus 1 by n is not an integer.
So, there you see that alpha q that I mean
is not there the right hand side of the previous
equation which is alpha q a whole to the power
2 to the power of r into beta r is nothing
but I can elaborate this alpha q as a to the
power 2 to the power of n q minus 1and then
raise it to the power 2 to the power of r
multiplied it with a to the power 2 to the
power of r minus 1. Note that it is beta actually
not not alpha so they and then square it.
So, therefore, what is this this is nothing
but alpha or rather a to the power 2 to the
power of n q plus r ok minus 1 and this is
squared ok.
So, therefore, here this 2 to the power of
r cancels with this minus 2 to the power of
r. And therefore, you have got 2 to the power
ofn q plus r and n q and n q plus r was m
minus 1 ok . So, therefore, I can write this
as a to the power 2 to the power of m minus
1 minus 1 and which I which a very square
I again get a inverse ok. So, therefore, in
both casesa inverse is correctly computed.
So, what we try to do is that we just try
to see if I apply quad Itoh-Tsujii now ok.
And then we will see that how or what is the
amount ofadvantage or improvement that you
get over the normal Itoh-Tsujii inversion
algorithm . So, if you implement this, you
we will see that now, we will start alpha
1, we will initialize alpha 1 to say a cube
that is a to the power of 3, and then I will
try to do the addition chain, but right now
I do not need to go to 232, but rather I can
stop at 116 ok. So, what I will basically
now do is, I will try to calculate alpha 2
as alpha 1 plus 1 alpha 2 plus 1 exactly like
what we have done previously,but now I have
changed this to this power of force ok.
So, therefore, the computation is being done
fast actually it is being done at a faster
rate, but again as you observe that if you
compare a quad with us 2 squarings, then the
delay is same ok . So, therefore, right in
the initial pre-computation here, there is
an amount of cost that you have to pay, because
in the previous case you were just initializing
with a, but now you are initializing with
a cube ok. And remember that you do not have
a squarings circuit. So, therefore, if you
want to implement a power of 3, you need 3
multiplication operations ok. So, you need
to actually you need 2 clock cycles, because
you have to do a into a into a. So, you need
2 clock cycles .
And you also have to do a final squarings
operation which you do with a multiplication,
because you do not have a squarings ok. So,
you have got 3 extra operations which you
are doing ok. And likewise so and in total
right you will see that there are 12 multiplications,
because there are 9 multiplications here and
if you also accommodate these extra 3 multiplications
then you need to do 12 multiplications ok
. Number of quads operations, you have to
do 115 quad operation instead of 232 ok, which
you did in the corner spawn when we have when
we were operating with beta . And you also
say save 7 clock cycles ok in the process.
So, therefore, right if you just write this
entire stuff, then you will see that the number
of clock cycles is as follows l plus 1 plus
sigma now i goes from 2 to l minus 1, in the
previous case I was running from 2 to l . So,
therefore, the number of savings in terms
of clock cycles is quite interesting it is
u l minus u l minus 1 divided by us minus
1 ok that is this extra step. And there is
one extra clock cycle here, so I subtract
out minus 1 to accommodate that ok.So, therefore,
right you see that this essentially can be
as high as m minus 1 by 2 for GF 2 to the
power m. So, therefore, if you have got larger
dimensions, then actually this advantage could
be even more ok .
So,with this I will you know like briefly
stop here. And I willstart discussing about
the corresponding hardware architecture,when
I am trying to realize this in the form of
an architecture in my nextnext class.
Thank you.
