Can you please elucidate how it is computed? I am desperate .... i don't know what i am missing here.
Example:
Suppose the TA chooses p = 467 and q = 479, so n = 223693. Suppose also that b = 503 and Alice’s secret integer u = 101576. Then she will compute
Original_formula
How can I compute (u^(-1))^b to be larger than n (or at least larger than 0)? What am I missing?
Original source: http://flylib.com/books/en/3.230.1.96/1/
Related
Sorry for a possibly misleading title.
Given a floating point input "A", we need factor it as
Q = A + B/C
where A is integer part of Q, so basically fix(Q), and B & C must be coprime to one other.
Example:
Q = 14.7419 => Q = 14 + 23/31
So that
A = 14, B = 23, C = 31
Is there a way to get B and C with intrinsic matlab functions? If not, I would appreciate a guidance in the right direction :)
Thanks in advance.
The solution here would not be the rats function as suggested by beaker under the question, but the rat function:
Q = 14.7419;
A = fix( Q )
[B,C] = rat( Q-A, 0.001)
The second argument to rat is the tolerance.
For further details please refer to https://www.mathworks.com/help/matlab/ref/rat.html.
Can you please tell me what's wrong with the following code?
function [n]=calculate_n(p,delta)
e = 1.6*power(10,-19);
k = 1.38*power(10,-23);
T = 298;
co = 3.25*power(10,13)*e*power(10,4);
er=12.5;
eo=1.0;
Nv=3*power(10,13);
us = log((p*e)/sqrt(2*k*T*er*eo*Nv))*2*k*T;
tmp = delta+(e*e*p)/co+us;
n = 1/(exp((tmp))+1);
end
I am getting matrix dimension error while calculating n. Please help me.
Caller:
e = 1.6*power(10,-19);
x = logspace(13,18);
y=calculate_n(x,0.2*e);
semilogx(x,y,'-s');
grid on;
Just replace n = 1/(exp((tmp))+1); with n = 1./(exp(tmp)+1);. But beware, tmp is so small for these values that exp(tmp) will always be 1. Also, there is a surplus bracket around tmp, you might want to check if you put them correctly.
Edit:
The reason is that A/B tries to solve the system of linear equations A*x = B for x which is not what you wanted. It threw an error because it requires both variables to have the same number of columns. A./B performs element-wise matrix division which is what you wanted. However, if A and B are singular A/B = A./B. See the documentation for more info.
So on this problem it seems pretty straight forward we are given
mean of x = 10,281 and sigma of x = 4112.4
We are asked to determine P(X<15,000)
Now I thought the code for this in matlab should be super straightforward
mu = 10281
sigma = 4112.4
p = logncdf(15000,10281,4112.4)
However this gives
p = .0063
The given answer is .8790 and just looking at p you can tell it is wrong because we are at 15000 which is over the mean which means it should be above .5. What is the deal with this function?
I saw somewhere you might need to take the exp(15000) for x in the function that results in a probability of 1 which is too high.
Any pointers would be much appreciated
%If X is lognormally distributed with parameters:-
mu = 10281;
sigma = 4112.4;
%then log(X) is normally distributed with following parameters:
mew_actual = log((mu^2)/sqrt(sigma^2+mu^2));
sigma_actual = sqrt(log((sigma^2)/(mu^2) +1));
Now you can use either of the following to compute CDF:-
p = cdf('Normal',log(15000),mew_actual,sigma_actual)
or
p=logncdf(15000,mew_actual,sigma_actual)
which gives 0.8796
(which I believe is the correct answer)
The answer given to you is 0.8790 because if you solve the question by hand, you'll get something like: z = 1.172759 and when you look this value in the table, you can only find z = 1.17(without the rest of decimal places) and for which φ(z)=0.8790.
You can verify the exact answer using this calculator. The related screenshot is attached below:
I need to solve a min distance problem, to see some of the work which has being tried take a look at:
link: click here
I have four elements: two column vectors: alpha of dim (px1) and beta of dim (qx1). In this case p = q = 50 giving two column vectors of dim (50x1) each. They are defined as follows:
alpha = alpha = 0:0.05:2;
beta = beta = 0:0.05:2;
and I have two matrices: L1 and L2.
L1 is composed of three column-vectors of dimension (kx1) each.
L2 is composed of three column-vectors of dimension (mx1) each.
In this case, they have equal size, meaning that k = m = 1000 giving: L1 and L2 of dim (1000x3) each. The values of these matrices are predefined.
They have, nevertheless, the following structure:
L1(kx3) = [t1(kx1) t2(kx1) t3(kx1)];
L2(mx3) = [t1(mx1) t2(mx1) t3(mx1)];
The min. distance problem I need to solve is given (mathematically) as follows:
d = min( (x-(alpha_p*t1_k - beta_q*t1_m)).^2 + (y-(alpha_p*t2_k - beta_q*t2_m)).^2 +
(z-(alpha_p*t3_k - beta_q*t3_m)).^2 )
the values x,y,z are three fixed constants.
My problem
I need to develop an iteration which can give me back the index positions from the combination of: alpha, beta, L1 and L2 which fulfills the min-distance problem from above.
I hope the formulation for the problem is clear, I have been very careful with the index notations. But if it is still not so clear... the step size for:
alpha is p = 1,...50
beta is q = 1,...50
for L1; t1, t2, t3 is k = 1,...,1000
for L2; t1, t2, t3 is m = 1,...,1000
And I need to find the index of p, index of q, index of k and index of m which gives me the min. distance to the point x,y,z.
Thanks in advance for your help!
I don't know your values so i wasn't able to check my code. I am using loops because it is the most obvious solution. Pretty sure that someone from the bsxfun-brigarde ( ;-D ) will find a shorter/more effective solution.
alpha = 0:0.05:2;
beta = 0:0.05:2;
L1(kx3) = [t1(kx1) t2(kx1) t3(kx1)];
L2(mx3) = [t1(mx1) t2(mx1) t3(mx1)];
idx_smallest_d =[1,1,1,1];
smallest_d = min((x-(alpha(1)*t1(1) - beta(1)*t1(1))).^2 + (y-(alpha(1)*t2(1) - beta(1)*t2(1))).^2+...
(z-(alpha(1)*t3(1) - beta(1)*t3(1))).^2);
%The min. distance problem I need to solve is given (mathematically) as follows:
for p=1:1:50
for q=1:1:50
for k=1:1:1000
for m=1:1:1000
d = min((x-(alpha(p)*t1(k) - beta(q)*t1(m))).^2 + (y-(alpha(p)*t2(k) - beta(q)*t2(m))).^2+...
(z-(alpha(p)*t3(k) - beta(q)*t3(m))).^2);
if d < smallest_d
smallest_d=d;
idx_smallest_d= [p,q,k,m];
end
end
end
end
end
What I am doing is predefining the smallest distance as the distance of the first combination and then checking for each combination rather the distance is smaller than the previous shortest distance.
I have an array of Fa which contains values I found from a function. Is there a way to use interp1 function in Matlab to find the index at which a specific value occurs? I have found tutorials for interp1 which I can find a specific value in the array using interp1 by knowing the corresponding index value.
Example from http://www.mathworks.com/help/matlab/ref/interp1.html:
Here are two vectors representing the census years from 1900 to 1990 and the corresponding United States population in millions of people.
t = 1900:10:1990;
p = [75.995 91.972 105.711 123.203 131.669...
150.697 179.323 203.212 226.505 249.633];
The expression interp1(t,p,1975) interpolates within the census data to estimate the population in 1975. The result is
ans =
214.8585
- but I want to find the t value for 214.8585.
In some sense, you want to find roots of a function -
f(x)-val
First of all, there might be several answers. Second, since the function is piecewise linear, you can check each segment by solving the relevant linear equation.
For example, suppose that you have this data:
t = 1900:10:1990;
p = [75.995 91.972 105.711 123.203 131.669...
150.697 179.323 70.212 226.505 249.633];
And you want to find the value 140
val = 140;
figure;plot(t,p);hold on;
plot( [min(t),max(t)], [val val],'r');
You should first subtract the value of val from p,
p1 = p - val;
Now you want only the segments in which p1 sign changes, either from + -> -, or vice versa.
segments = abs(diff(sign(p1)==1));
In each of these segments, you can solve the relevant linear equation a*x+b==0, and find the root. That is the index of your value.
for i=1:numel(segments)
x(1) = t(segments(i));
x(2) = t(segments(i)+1);
y(1) = p1(segments(i));
y(2) = p1(segments(i)+1);
m = (y(2)-y(1))/(x(2)-x(1));
n = y(2) - m * x(2);
index = -n/m;
scatter(index, val ,'g');
end
And here is the result:
You can search for the value in Fa directly:
idx = Fa==value_to_find;
To find the index use find function:
find(Fa==value_to_find);
Of course, this works only if the value_to_find is present in Fa. But as I understand it, this is what you want. You do not need interp for that.
If on the other hand the value might not be present in Fa, but Fa is sorted, you can search for values larger than value_to_find and take the first such index:
find(Fa>=value_to_find,1);
If your problem is more complicated than that, look at Andreys answer.
Andrey's solution works in principle, but the code presented here does not. The problem is with the definition of the segments, which yields a vector of 0's and 1's, whereafter the call to "t(segments(i))" results in an error (I tried to copy & paste the code - I hope I did not fail in that simple task).
I made a small change to the definition of the segments. It might be done more elegantly. Here it is:
t = 1900:10:1990;
p = [75.995 91.972 105.711 123.203 131.669...
150.697 179.323 70.212 226.505 249.633];
val = 140;
figure;plot(t,p,'.-');hold on;
plot( [min(t),max(t)], [val val],'r');
p1 = p - val;
tn = 1:length(t);
segments = tn([abs(diff(sign(p1)==1)) 0].*tn>0);
for i=1:numel(segments)
x(1) = t(segments(i));
x(2) = t(segments(i)+1);
y(1) = p1(segments(i));
y(2) = p1(segments(i)+1);
m = (y(2)-y(1))/(x(2)-x(1));
n = y(2) - m * x(2);
index = -n/m;
scatter(index, val ,'g');
end
interpolate the entire function to a higher precision. Then search.
t = 1900:10:1990;
p = [75.995 91.972 105.711 123.203 131.669...
150.697 179.323 203.212 226.505 249.633];
precision = 0.5;
ti = 1900:precision:1990;
pi = interp1(t,p,ti);
now pi holds all pi values for every half a year. Assuming the values always increase you could find the year by max(ti(pi < x)) where x = 214.8585. Here pi < x creates a logical vector used to filter ti to only provide the years when p is less than x. max() is then used to take the most recent year, which will also be closest to x if the assumption that p is always increasing holds.
The answer to the most general case was given above by Andrey, and I agree with it.
For the example that you stated, a simple particular solution would be:
interp1(p,t,214.8585)
In this case you are solving for the year when a given population is known.
This approach will NOT work when there is more than one solution. If you try this with Andrey's values you will only get the first solution to the problem.