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I have a question about linear codes.
Let's say we have two (n,k) linear codes C1 and C2 with parity check matrix H1 and H2. Is the intersection of C1 and C2 still a linear code? If so, what is its parity check matrix H3 given H1 and H2? C3 is the intersection of C1 and C2 means H1c3=0 and H2c3=0 for all c3\in C3.
Yes. It is also a linear code.
A linear code of length n and rank k is a linear subspace C with dimension k of the vector space V.
Given subspaces U and W of a vector space V, then their intersection U ∩ W := {v ∈ V : v is an element of both U and W} is also a subspace of V.
To obtain H dimension this statement may be used:
Let (G,+G,∘)K be a K-vector space.
Let M and N be finite-dimensional subspaces of G.
Then M+N and M∩N are finite-dimensional, and:
dim(M+N) + dim(M∩N) = dim(M) + dim(N)
so:
dim(M+N) + dim(M∩N) = k1 + k2
where dim(M∩N) is new k of the intersection.
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solution of the bit error probability problem
prob=function(E,m) #--- prob is the estimated error probabity for given values of signal to
#---noise ratio E and sample size m
{
stopifnot(E>=0 & m>0) #--- this says that the function won't accept negative values of E and
#---m shoulde be at least 1
n=rnorm(m) #--- this says that n is a random sample of size m from N(0,1)
#---distribution
m=mean(n< -sqrt(E)) #--- this says that m is the proportion of values in n which are less
#---than the negative root of E
return(m) #--- this gives us the value of m, which is the estimated error
#---probability
}
E=seq(0,2,by=0.001)
sam=1000
y=sapply(E,prob,m=sam)
p=10*log10(E)
plot(p, log(y),
main="Graph For The Error Probabilities",
xlab=expression(10*log[10](E)),
ylab="log(Error Probability)",
type="l")
You can try the MATLAB code like below
clc;
clear;
close all;
function y = prob(E,m)
assert(E>=0 & m>0);
n = randn(1,m);
y = mean(n+sqrt(E)<0);
end
E = 0:0.001:2;
sam = 1000;
y = arrayfun(#(x) prob(x,sam),E);
p = 10*log10(E);
plot(p,log(y));
title("Graph For The Error Probabilities");
xlabel('10\log_{10}(E)');
ylabel("log(Error Probability)");
OUTPUT (MATLAB)
OUTPUT (R)
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Given the equation (depth = log(c * p + 1) / log(c * 1000 + 1) * p)
How can I find p?
I wanted to get an equation with p = sqrt(exp(... or something like this
The isn't an analytical solution to problems like this because they contain the variable both inside and outside a transcendental function like log().
With some quick algebra, you can simplify the expression to
p*log(1 + c*p) = b
where b=depth*log(1+1000*c) is a constant.
I propose using a single point iteration, to get a numeric result.
Start with some guess of p=1 or something and then do a loop until you converge to a value of (c# shown below).
b = depth*log(1+1000*c);
p = 1;
do
{
p_old = p;
p = b/log(1+c*p);
}
while( abs(p-p_old)> 1e-6);
and hope it converges soon.
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I wonder if there is some method to get correct remainder of N after dividing with Q (or modulo Q) from the given remainder of N after dividing with P(or modulo P).
In other words you have N%P, P and Q. You have to find out N%Q, given that gcd(P,Q) = 1.
Example. Suppose P = 19 and Q = 15. Now, for N =100, we can compute N%P = (100 % 19) = 5
and N % Q = (100 % 15) = 10. But, suppose if you are not given 100, then how can you get N % Q = 10 from the given P(=19) and remainder (N % P) = 5.
You can not get a definite answer because there are many possible N which will give the specified remainder. Let's consider your example. We know that N % P = N % 19 = 5. The possible values of N are: 24 (19+5), 43 (19+19+5), 62, (19+19+19+5), 81, 100, 119... Taking this into account the possible remainders of N % Q = N % 15 are 9,13,2,6,10,14... and 10 is only one of possible solutions (more precisely there are Q possible remainders (solutions) of dividing by Q).
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How to demonstrate that all multiplicative orders divide the order (size) of the multiplicative group F of F13.
.
You show that the cyclic group <x> generated by an element x is a subgroup of IF* and that "u~v iff u^(-1)*v in <x>" is an equivalence relation that divides the multiplicative group into equivalence classes of equal size.
So that you get
[size of IF*]
= [size of <x>] * [number of equivalence classes]
which means that the order of x = [size of <x>] is a divisor of the number of invertible elements, i.e., the size of the multiplicative group of IF
See also the little theorem of Fermat.
Since the group is abelian, the simplest thing is to use that multiplication by any element is a bijection. Let F = {g1, g2, g3, ..., gn} and let h be an arbitrary element. Then also F = {h*g1, h*g2, ..., h*gn}. Hence multiplying all elements together we get g1 * g2 * g3 * ... * gn = h*g1 * h*g2 * ... * h*gn. But the latter equals h^n * g1 * g2 * ... * gn. Now use the cancellation law to conclude that h^n = 1 from which the result follows.
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I am trying to solve a Mathematical equation in one of my geometric modelling problem.
Let's say if I have 2 vectors, A and B, and I have the following equation:
A x B = c (c is a scalar value).
If I know the coordinate of my vector B (7/2, 15/2); and I know the value of c, which is -4.
How can I calculate my vector A, to satisfy that equation (A X B = c) ?
The problem is underdetermined; there isn't a unique such A. I assume that by "multiplication" you mean the cross product.
A = (x,y)
B = (7/2, 15/2)
A×B = x(15/2) - y(7/2)
-4 = (15x-7y)/2
15x - 7y = -8
This gives a line along which points A=(x,y) can lie. Specifically, for any real number t,
x = -1 + 7t
y = -1 + 15t
gives a solution.