Just like when you\'re plotting y =f(x) you need to do two things- define f and

Question

Just like when you're plotting y =f(x) you need to do two things- define f and define x (x is usually defined using linspace) Defining f: You can do this as an anonymous function. Stan with something simple: f(s, t) = cos(s) * sin(t) Defining s and t: We're going to do this in two steps. Find, we're going to define the values s and t go over: s = linspace(0, 4*pi, 200); t = linspace(0, 2*pi, 100); The naive thing to try is just evaluating f(s, t) and plotting it z = f(s, t); surf(s, t, z); What happened? What error message did you get? What are the matrix dimensions of s, t, and z? Now try plot f(s, t, z) What did that plot? Why? The problem is that we need a grid of values to pass to f, not just two arrays. This is where the meshgrid function comes in: [S, T] = meshgrid (s, t) Z = [S, T] What are the matrix dimensions of S and T?Z? Now try plotting surf(S, T, Z) Here are some other fun things to try: surf(S, T, Z, 'FaceColour', [1 0 1], 'EdgeColor', 'none'): This makes the surface purple (red, green, blue) and nums off the edges (black grid lines). You'll get a uniform gray surface camlight right: lighting Phong: This adds surface shading and a light Self check:

Explanation / Answer

odular multiplication, want conversions from RNS to binary or RNS to AN alternative RNS. This transformation is
not trivial, and most of the algorithms rely of it. it's the rationale why we have a tendency to target this operation, specially
from RNS to RNS that is employed in cryptography,11 analyzing the influence of the selection of the coprime n-uple
defining the RNS basis.
We initial gift the classical conversion algorithms from RNS to RNS or alternative representations. Then, we
analyze these ways through the purpose of read of the RNS bases. we have a tendency to propose some criteria of bases choice
to obtain economical conversions. And we end, within the conclusion, with a discussion around RNS.
2. SOME GENERALITIES ON THE CONVERSION
When we propose to use RNS to represent numbers, the primary question is: what's the price of the conversion to
and from these systems? it's clear that the interpretation from a classical illustration to RNS is just like a
modular reduction. one amongst the foremost trivial means is to store all the powers of the number modulo every part of
the RNS basis.12–14 Thus, if we have a tendency to think about the RNS basis (m1, m2, . . . , mn) with M =
Qn
i=1 mi < T wherever is
the number of the classical illustration, then we've got to store T n values of size T /n digits, in alternative words T
2
digits, for exploitation the subsequent formula:
xi = |X|mi
=

X
T
j=1
j ×
j

mi
=

X
T
j=1
j ×

j

mi

mi
(1)
We assume that the standard multiplier-adder is that the basic operator of RNS computing. Hence, the conversion
from illustration to RNS are often worn out T iterations. currently if we have a tendency to think about a full parallelization, with n
modular multiplier-adders, this conversion are often index (ie O(log(T))). however during this paper we have a tendency to focus our
attention to the conversion from RNS to RNS or to illustration.
To modify our future discussions, we have a tendency to think about that = two, that each one the mi have constant variety t of bits,
and for the RNS to RNS conversion the 2 bases have constant variety of moduli.
We note B = (m1, m2, . . . , mn) and Be = (mf1, mf2, . . . , mfn) the 2 RNS bases that we are going to use.
2.1. From the Chinese Remainder Theorem
In the proof of the Chinese Remainder Theorem we will show, that an answer of the subsequent system:



X x1 (mod m1)
... ... ...
X xn (mod mn)
(2)
can be created from the formula:
X =

x1 |M1|
1
m1 M1 + x2 |M2|
1
m2 M2 + . . . + xn |Mn|
1
mn Mn

M
(3)
Where, Mi =
M
mi
for one i n and |Mi
|
1
mi
represents the inverse of Mi modulo mi
.
It is obvious that the analysis of a such expression is serious. To modify a bit bit the equation (3) we have a tendency to
consider i = xi
|Mi
|
1
mi mod mi
, thus we have a tendency to get a replacement expression for X:
X =
Xn
i=1
iMi
!
mod M (4)
Despite this transformation, every term stays an excellent price of constant vary than M. thus we have a tendency to deduce that the
cost of the conversion from RNS to binary illustration is akin to n standard product modulo AN mi
, n
products of an outsized variety by alittle one, and n one additions of enormous numbers and a discount modulo M.
Here again, we will put to get a index time quality, however the operators are going to be Brobdingnagian and
not realistic. Now, if the amount n of moduli is little, like for a few DSP, the equation (4) offers a decent
solution.15, 16
The conversion RNS to RNS is analogous, we are going to build constant quite operations on every modulo of the new
basis Be, the equation (3) becomes for j :
X =

x1 |M1|
1
m1
|M1|mfj
+ x2 |M2|
1
m2
|M2|mfj
+ . . . + xn |Mn|
1
mn
|Mn|mfj

mfj
(5)
Thus, this analysis desires n standard product modulo AN mi
, and for every modulo of the new RNS basis
n standard product (of tiny numbers), and n one standard additions of residues and a discount modulo M.
So if we have a tendency to don't take under consideration the ultimate reduction, we want n(n + 1) standard product and n(n 1) standard
additions on residues. we've got to note that we will have an excellent degree of parallelization.
Now, the reduction modulo M is that the main downside of this approach. in reality most of the strategy projected
to find such that:
X + M =
Xn
i=1
iMi
!
(6)
Hence, once is decided, M are often subtracted from the second member of equation (6). This remark is
also obtainable for the RNS to RNS conversion.
A. P. Shenoy and R. Kumaresan have projected to use an additional modulo to seek out this issue.17 during this technique
all the computing is finished with integers. AN alternative purpose of read is to seek out AN approximation of with floating
point cell.9, 18, 19
2.1.1. Shenoy Kumaresan
This technique relies on the information of the residue modulo an additional co-prime American state. thus X is outline by its RNS
representation (x1, x2, . . . , xn), and that we apprehend Xe = X mod American state. Hence, from expression (6) we will deduce the
residue of X + M modulo me:
= (X + M) mod American state =

Xn
i=1
i
|Mi
|me

me
(7)
Thus, we have
= ( xe) × |M|
1
me mod American state (8)
This analysis desires n standard product and n1 standard additions on residues. Then the ultimate reduction
is done with a product by of M (|M|mfj
for RNS to RNS) and a subtraction.
But this condition on the additional modulo can't be continuously glad. it's not obvious of maintaining the
knowledge of the residue modulo American state throughout the various calculuses. So, we have a tendency to should think about AN approach obtainable
just with the RNS illustration

Related Questions

Navigate

• Browse (All)
• Browse J
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.