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I have N-time steps and want to create N real numbers corresponding to each time step as follows: the numbers should be in the range [a, b] with a switching probability of p.
Ie, if kth number (corresponding to kth time step) is n, then the probability that k+1th number (corresponding to k+1th time step) is any other number different than n is p. All created numbers should be in the range [a,b].
How can this be done in matlab?
Thanks.
I am not really sure that I got all of your requirements but this might be the script you are searching for:
N = 10; % count of numbers
p = 0.2; % switching probability
a = 5;
b = 20;
% init empty numbers and get the first random number
numbers = zeros(N,1);
numbers(1) = rand(1) * (b-a) + a;
for iNumber = 2:N
% test if number must change
if rand(1) < (1-p)
% the number must be the same than the previous
% copy the value and go to next number
numbers(iNumber) = numbers(iNumber-1);
continue;
else
% a new number must be found
while 1
% get a new random number
numbers(iNumber) = rand(1) * (b-a) + a;
% check if the new random number is different from the previous
if numbers(iNumber) ~= numbers(iNumber-1)
% in case they are different, the while 1 can be stopped
break;
end
end % while 1
end % if rand(1) < (1-p)
end % for iNumber = 2:N
Guaranteed super random:
N = randi(100);
p = rand;
l1 = rand*randi(100);
l2 = rand*randi(100);
if l2 < l1
[l2 l1] = deal(l1,l2);
end
out = []
while isempty(out)
if rand>rand
n = 2:N
a = rand([N,1])*(l2-l1)+l1;
x = rand([N-1,1])<p;
n(x==0)=n(x==0)-1;
n = [1 n];
out = a(n);
end
end
Related
I want to compute the above summation, for a given 'x'. The summation is to be carried out over a block of lengths specified by an array , for example block_length = [5 4 3]. The summation is carried as follows: from -5 to 5 across one dimension, -4 to 4 in the second dimension and -3 to 3 in the last dimension.
The pseudo code will be something like this:
sum = 0;
for i = -5:5
for j = -4:4
for k = -3:3
vec = [i j k];
tv = vec * vec';
sum = sum + 1/(1+tv)*cos(2*pi*x*vec'));
end
end
end
The problem is that I want to find the sum when the number of dimensions are not known ahead of time, using some kind of variable nested loops hopefully. Matlab uses combvec, but it returns all possible combinations of vectors, which is not required as we only compute the sum. When there are many dimensions, combvec returning all combinations is not feasible memory wise.
Appreciate any ideas towards solutions.
PS: I want to do this at high number of dimensions, for example 650, as in machine learning.
Based on https://www.mathworks.com/matlabcentral/answers/345551-function-with-varying-number-of-for-loops I came up with the following code (I haven't tested it for very large number of indices!):
function sum = fun(x, block_length)
sum = 0;
n = numel(block_length); % Number of loops
vec = -ones(1, n) .* block_length; % Index vector
ready = false;
while ~ready
tv = vec * vec';
sum = sum + 1/(1+tv)*cos(2*pi*x*vec');
% Update the index vector:
ready = true; % Assume that the WHILE loop is ready
for k = 1:n
vec(k) = vec(k) + 1;
if vec(k) <= block_length(k)
ready = false;
break; % v(k) increased successfully, leave "for k" loop
end
vec(k) = -1 * block_length(k); % v(k) reached the limit, reset it
end
end
end
where x and block_length should be both 1-x-n vectors.
The idea is that, instead of using explicitly nested loops, we use a vector of indices.
How good/efficient is this when tackling the suggested use case where block_length can have 650 elements? Not much! Here's a "quick" test using merely 16 dimensions and a [-1, 1] range for the indices:
N = 16; tic; in = 0.1 * ones(1, N); sum = fun(in, ones(size(in))), toc;
which yields an elapsed time of 12.7 seconds on my laptop.
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For instance: a, b, c and n are the three constants, which are required to be calculated by using data fitting method in a particular equation.
How can I calculate the statistics (mean, standard deviation, variance, skewness value and student t-test value) of the parameters as of a custom equation, for example the quadratic plateau equation?
Example:
x=[0,40,80,100,120,150,170,200],
y=[1865,2855,3608,4057,4343,4389,4415,4478]
y=a*(x+n)^2+b*(x+n)+c, x < xc(Ymax) ....(1) y=yp, x >= xc(Ymax) ....(2)
I have fitted this equation by given code:
yf = #(b,x) b(1).*(x+n).^2+b(2)*(x+n)+b(3); B0 = [0.006; 21; 1878];
[Bm,normresm] = fminsearch(#(b) norm(y - yf(b,x)), B0); a=Bm(1);
b=Bm(2); c=Bm(3); xc=(-b/(2*a))-n; p=p=a*(xc+n)^2+b*(xc+n)+c;
if (x < xc)
yfit = a.*(x+n).^2+ b*(x+n)+c;
else
yfit = p;
end
plot(x,yfit,'*')
hold on; plot(x,y); hold off
Note: I have already used the polyfit command, it was helpful and provided me the results. However, I really don’t find it suitable, as there is no option to customize the equation. Can I find these statistics by any code?
Questions 1, 2 and 4)
Good practice is to set initial values close to the final result if you have previous knowledge about the equation system:
What you have is an overdetermined system of linear equations.
y(1) = a*x(1)^2 + b*x(1) + c
y(2) = a*x(2)^2 + b*x(2) + c
y(3) = a*x(3)^2 + b*x(3) + c
…
y(n) = a*x(n)^2 + b*x(n) + c
or in general:
y = A*X, where
A = [a; b; c]
X = [x(1)^2 x(1) 1;
x(2)^2 x(2) 1;
x(3)^2 x(2) 1;
...
x(n)^2 x(n) 1]
One of the common practices to fit the overdetermined system (since it has no solution) is "least square fit" (mldivide,\ (link))
x=[0; 40; 80; 100; 120; 150; 170; 200];
y=[1865; 2855; 3608; 4057; 4343; 4389; 4415; 4478];
X = [x.^2 x ones(numel(x),1)];
A = y\X;
a0=A(1); %- initial value for a
b0=A(2); %- initial value for b
c0=A(3); %- initial value for c
You can customize equation, when you customize your X and A
but you also can set initial values to ones, it should have neglectable small impact on the result. More related to Question 4
a0=1;
b0=1;
c0=1;
or to random values
rng(10);
A = rand(3,1);
a0=A(1);
b0=A(2);
c0=A(3);
Question 3 - Statistics
If you need more control on monitoring of optimization process, use more general form of writing anonymous function (in code below> myfun) to save all intermediate values of parameters (a_iter, b_iter, c_iter)
function Fiting_ex()
global a_iter b_iter c_iter
a_iter = 0;
b_iter = 0;
c_iter = 0;
x=[0; 40; 80; 100; 120; 150; 170; 200];
y=[1865; 2855; 3608; 4057; 4343; 4389; 4415; 4478];
X = [x.^2 x ones(numel(x),1)];
A = y\X;
a0=A(1);
b0=A(2);
c0=A(3);
B0 = [a0; b0; c0];
[Bm,normresm] = fminsearch(#(b) myfun(b,x,y),B0);
a=Bm(1);
b=Bm(2);
c=Bm(3);
xc=-b/(2*a);
p=c-(b^2/(4*a));
yfit = zeros(numel(x),1);
for i=1:numel(x)
if (x(i) < xc)
yfit(i) = a.*x(i).^2+ b*x(i)+c;
else
yfit(i) = p;
end
end
plot(x,yfit,'*')
hold on;
plot(x,y);
hold off
% Statistic on optimization process
a_mean = mean(a_iter(2:end)); % mean value
a_var = var(a_iter(2:end)); % variance
a_std = std(a_iter(2:end)); % standard deviation
function f = myfun(Bm, x, y)
global a_iter b_iter c_iter
a_iter = [a_iter Bm(1)];
b_iter = [b_iter Bm(2)];
c_iter = [c_iter Bm(3)];
yf = Bm(1)*(x).^2+Bm(2)*(x)+Bm(3);
a=Bm(1);
b=Bm(2);
c=Bm(3);
xc=-b/(2*a);
p=c-(b^2/(4*a));
yfit = zeros(numel(x),1);
for i=1:numel(x)
if (x(i) < xc)
yfit(i) = a.*x(i).^2+ b*x(i)+c;
else
yfit(i) = p;
end
end
f = norm(y - yfit);
I am trying to implement a simplex algorithm following the rules I was given at my optimization course. The problem is
min c'*x s.t.
Ax = b
x >= 0
All vectors are assumes to be columns, ' denotes the transpose. The algorithm should also return the solution to dual LP. The rules to follow are:
Here, A_J denotes columns from A with indices in J and x_J, x_K denotes elements of vector x with indices in J or K respectively. Vector a_s is column s of matrix A.
Now I do not understand how this algorithm takes care of condition x >= 0, but I decided to give it a try and follow it step by step. I used Matlab for this and got the following code.
X = zeros(n, 1);
Y = zeros(m, 1);
% i. Choose starting basis J and K = {1,2,...,n} \ J
J = [4 5 6] % for our problem
K = setdiff(1:n, J)
% this while is for goto
while 1
% ii. Solve system A_J*\bar{x}_J = b.
xbar = A(:,J) \ b
% iii. Calculate value of criterion function with respect to current x_J.
fval = c(J)' * xbar
% iv. Calculate dual solution y from A_J^T*y = c_J.
y = A(:,J)' \ c(J)
% v. Calculate \bar{c}^T = c_K^T - u^T A_K. If \bar{c}^T >= 0, we have
% found the optimal solution. If not, select the smallest s \in K, such
% that c_s < 0. Variable x_s enters basis.
cbar = c(K)' - c(J)' * inv(A(:,J)) * A(:,K)
cbar = cbar'
tmp = findnegative(cbar)
if tmp == -1 % we have found the optimal solution since cbar >= 0
X(J) = xbar;
Y = y;
FVAL = fval;
return
end
s = findnegative(c, K) %x_s enters basis
% vi. Solve system A_J*\bar{a} = a_s. If \bar{a} <= 0, then the problem is
% unbounded.
abar = A(:,J) \ A(:,s)
if findpositive(abar) == -1 % we failed to find positive number
disp('The problem is unbounded.')
return;
end
% vii. Calculate v = \bar{x}_J / \bar{a} and find the smallest rho \in J,
% such that v_rho > 0. Variable x_rho exits basis.
v = xbar ./ abar
rho = J(findpositive(v))
% viii. Update J and K and goto ii.
J = setdiff(J, rho)
J = union(J, s)
K = setdiff(K, s)
K = union(K, rho)
end
Functions findpositive(x) and findnegative(x, S) return the first index of positive or negative value in x. S is the set of indices, over which we look at. If S is omitted, whole vector is checked. Semicolons are omitted for debugging purposes.
The problem I tested this code on is
c = [-3 -1 -3 zeros(1,3)];
A = [2 1 1; 1 2 3; 2 2 1];
A = [A eye(3)];
b = [2; 5; 6];
The reason for zeros(1,3) and eye(3) is that the problem is inequalities and we need slack variables. I have set starting basis to [4 5 6] because the notes say that starting basis should be set to slack variables.
Now, what happens during execution is that on first run of while, variable with index 1 enters basis (in Matlab, indices go from 1 on) and 4 exits it and that is reasonable. On the second run, 2 enters the basis (since it is the smallest index such that c(idx) < 0 and 1 leaves it. But now on the next iteration, 1 enters basis again and I understand why it enters, because it is the smallest index, such that c(idx) < 0. But here the looping starts. I assume that should not have happened, but following the rules I cannot see how to prevent this.
I guess that there has to be something wrong with my interpretation of the notes but I just cannot see where I am wrong. I also remember that when we solved LP on the paper, we were updating our subjective function on each go, since when a variable entered basis, we removed it from the subjective function and expressed that variable in subj. function with the expression from one of the equalities, but I assume that is different algorithm.
Any remarks or help will be highly appreciated.
The problem has been solved. Turned out that the point 7 in the notes was wrong. Instead, point 7 should be
Figure 1. Hypothesis plot. y axis: Mean entropy. x axis: Bits.
This Question is in continuation to a previous one asked Matlab : Plot of entropy vs digitized code length
I want to calculate the entropy of a random variable that is discretized version (0/1) of a continuous random variable x. The random variable denotes the state of a nonlinear dynamical system called as the Tent Map. Iterations of the Tent Map yields a time series of length N.
The code should exit as soon as the entropy of the discretized time series becomes equal to the entropy of the dynamical system. It is known theoretically that the entropy of the system, H is log_e(2) or ln(2) = 0.69 approx. The objective of the code is to find number of iterations, j needed to produce the same entropy as the entropy of the system, H.
Problem 1: My problem in when I calculate the entropy of the binary time series which is the information message, then should I be doing it in the same base as H? OR Should I convert the value of H to bits because the information message is in 0/1 ? Both give different results i.e., different values of j.
Problem 2: It can happen that the probality of 0's or 1's can become zero so entropy correspondng to it can become infinity. To prevent this, I thought of putting a check using if-else. But, the loop
if entropy(:,j)==NaN
entropy(:,j)=0;
end
does not seem to be working. Shall be greateful for ideas and help to solve this problem. Thank you
UPDATE : I implemented the suggestions and answers to correct the code. However, my logic of solving was not proper earlier. In the revised code, I want to calculate the entropy for length of time series having bits 2,8,16,32. For each code length, entropy is calculated. Entropy calculation for each code length is repeated N times starting for each different initial condition of the dynamical system. This appraoch is adopted to check at which code length the entropy becomes 1. The nature of the plot of entropy vs bits should be increasing from zero and gradually reaching close to 1 after which it saturates - remains constant for all the remaining bits. I am unable to get this curve (Figure 1). Shall appreciate help in correcting where I am going wrong.
clear all
H = 1 %in bits
Bits = [2,8,16,32,64];
threshold = 0.5;
N=100; %Number of runs of the experiment
for r = 1:length(Bits)
t = Bits(r)
for Runs = 1:N
x(1) = rand;
for j = 2:t
% Iterating over the Tent Map
if x(j - 1) < 0.5
x(j) = 2 * x(j - 1);
else
x(j) = 2 * (1 - x(j - 1));
end % if
end
%Binarizing the output of the Tent Map
s = (x >=threshold);
p1 = sum(s == 1 ) / length(s); %calculating probaility of number of 1's
p0 = 1 - p1; % calculating probability of number of 0'1
entropy(t) = -p1 * log2(p1) - (1 - p1) * log2(1 - p1); %calculating entropy in bits
if isnan(entropy(t))
entropy(t) = 0;
end
%disp(abs(lambda-H))
end
Entropy_Run(Runs) = entropy(t)
end
Entropy_Bits(r) = mean(Entropy_Run)
plot(Bits,Entropy_Bits)
For problem 1, H and entropy can be in either nats or bits units, so long as they are both computed using the same units. In other words, you should use either log for both or log2 for both. With the code sample you provided, H and entropy are correctly calculated using consistant nats units. If you prefer to work in units of bits, the conversion of H should give you H = log(2)/log(2) = 1 (or using the conversion factor 1/log(2) ~ 1.443, H ~ 0.69 * 1.443 ~ 1).
For problem 2, as #noumenal already pointed out you can check for NaN using isnan. Alternatively you could check if p1 is within (0,1) (excluding 0 and 1) with:
if (p1 > 0 && p1 < 1)
entropy(:,j) = -p1 * log(p1) - (1 - p1) * log(1 - p1); %calculating entropy in natural base e
else
entropy(:, j) = 0;
end
First you just
function [mean_entropy, bits] = compute_entropy(bits, blocks, threshold, replicate)
if replicate
disp('Replication is ON');
else
disp('Replication is OFF');
end
%%
% Populate random vector
if replicate
seed = 849;
rng(seed);
else
rng('default');
end
rs = rand(blocks);
%%
% Get random
trial_entropy = zeros(length(bits));
for r = 1:length(rs)
bit_entropy = zeros(length(bits), 1); % H
% Traverse bit trials
for b = 1:(length(bits)) % N
tent_map = zeros(b, 1); %Preallocate for memory management
%Initialize
tent_map(1) = rs(r);
for j = 2:b % j is the iterator, b is the current bit
if tent_map(j - 1) < threshold
tent_map(j) = 2 * tent_map(j - 1);
else
tent_map(j) = 2 * (1 - tent_map(j - 1));
end % if
end
%Binarize the output of the Tent Map
s = find(tent_map >= threshold);
p1 = sum(s == 1) / length(s); %calculate probaility of number of 1's
%p0 = 1 - p1; % calculate probability of number of 0'1
bit_entropy(b) = -p1 * log2(p1) - (1 - p1) * log2(1 - p1); %calculate entropy in bits
if isnan(bit_entropy(b))
bit_entropy(b) = 0;
end
%disp(abs(lambda-h))
end
trial_entropy(:, r) = bit_entropy;
disp('Trial Statistics')
data = get_summary(bit_entropy);
disp('Mean')
disp(data.mean);
disp('SD')
disp(data.sd);
end
% TO DO Compute the mean for each BIT index in trial_entropy
mean_entropy = 0;
disp('Overall Statistics')
data = get_summary(trial_entropy);
disp('Mean')
disp(data.mean);
disp('SD')
disp(data.sd);
%This is the wrong mean...
mean_entropy = data.mean;
function summary = get_summary(entropy)
summary = struct('mean', mean(entropy), 'sd', std(entropy));
end
end
and then you just have to
% Entropy Script
clear all
%% Settings
replicate = false; % = false % Use true for debugging only.
%H = 1; %in bits
Bits = 2.^(1:6);
Threshold = 0.5;
%Tolerance = 0.001;
Blocks = 100; %Number of runs of the experiment
%% Run
[mean_entropy, bits] = compute_entropy(Bits, Blocks, Threshold, replicate);
%What we want
%plot(bits, mean_entropy);
%What we have
plot(1:length(mean_entropy), mean_entropy);
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I have a set of 120 pairs, which are derived from a set of 16 items (i.e. c(16,2)=120).
I want to determine how many combinations of 56 pairs can be chosen from the 120 pairs, but with the constraint that each combination has to contain exactly 7 of each of the 16 items (i.e. in each subset of 56 pairs, each of the 16 items is equally represented).
In addition to determining the number of combinations, I also need to list them. Can anyone help with how I could code this in Matlab please?
It is an interesting combinitorial question since nchoosek(120,56) gives 7.4149e+34 possible ways to select 56 pairs out of the 120. So you shouldn't solve this by a loop over all selections.
To handle the problem I introduce a representation of selected pairs given by a matrix A = (aij), where aij == 1 if and only if (i,j) oder (j,i) is a pair among the 56 selected ones. This matrix represents a selection of 56 pairs as described above if and only if row sums are 7, i.e. A*ones(16,1) = 7*ones(16,1) (symmetric!). It is easy to solve this systematically to print all possible solutions, but this results in about 1e20 candidates...
But according to this high number you can easily select one example randomly by some stupid code like the following:
%% find some random example
A = zeros(16,16);
b1 = ones(16,1);
b7 = 7*b1;
while norm(A*b1 - b7) ~= 0
A = zeros(16,16);
change = true;
while norm(A*b1 - b7) ~= 0 && change
change = false;
todo = find(A*b1<7);
idx = todo(randperm(length(todo)));
for k = 1:2:length(idx)-1
if A(idx(k),idx(k+1)) == 0
change = true;
end
A(idx(k),idx(k+1)) = 1;
A(idx(k+1),idx(k)) = 1;
end
end
end
%% print it
pIdx = 0;
for lIdx = 1:16
for cIdx = 1:lIdx-1
if A(lIdx,cIdx) == 1
fprintf(['(' num2str(cIdx) ',' num2str(lIdx) ') ']);
pIdx = pIdx + 1;
if mod(pIdx,7) == 0
fprintf('\n');
end
end
end
end
Example result:
(1,3) (2,4) (1,5) (2,5) (2,6) (3,6) (4,6)
(5,6) (1,7) (3,7) (6,7) (4,8) (5,8) (2,9)
(5,9) (6,9) (7,9) (8,9) (2,10) (3,10) (4,10)
(5,10) (7,10) (1,11) (3,11) (6,11) (7,11) (9,11)
(1,12) (4,12) (7,12) (11,12) (3,13) (8,13) (9,13)
(10,13) (1,14) (2,14) (4,14) (8,14) (12,14) (13,14)
(1,15) (3,15) (5,15) (8,15) (11,15) (12,15) (13,15)
(2,16) (4,16) (8,16) (10,16) (12,16) (13,16) (14,16)
Update: Added code to scan all possible solutions...
function main()
%% provide n-choose-k sequences (compute only once!)
global ncks maxResult;
disp('Prepare n choose k sequences...');
for n = 1:15
fprintf([num2str(n) '... ']);
for k = 1:min(7,n)
ncks{n}{k} = nchoosek_sequence(n,k);
end
end
fprintf('\n');
%% start finding solutions
disp('Scan space of solutions...')
A = zeros(16,16);
maxResult = -Inf;
tic;
findMats(A,1)
toc;
end
function findMats(A,curRow)
global ncks maxResult;
remain = 7-sum(A+A');
if curRow == 16
if remain(16) == 0
A = A + A';
if norm(sum(A)-7*ones(1,16)) ~= 0
error('AHHH');
end
%A
%pause();
maxResult = max(YOUR_FUNCTION(A),maxResult); % <- add your function here
end
return;
end
if remain(curRow) == 0
findMats(A,curRow+1);
elseif remain(curRow) > 0
remainingRows = curRow + find(remain(curRow+1:end) > 0);
if ~isempty(remainingRows) && length(remainingRows) >= remain(curRow)
for r = 1:nchoosek(length(remainingRows),remain(curRow))
row = remainingRows(ncks{length(remainingRows)}{remain(curRow)}(r,:));
Anew = A;
Anew(curRow,row) = 1;
findMats(Anew,curRow+1);
end
end
end
end
% very simple and not optimized way to get a sequence of subset-selections
% in increasing order
%
% Note: This function was easy to implement and sufficient for our
% purposes. For more efficient implementations follow e.g.
% http://stackoverflow.com/questions/127704/algorithm-to-return-all-combin
% ations-of-k-elements-from-n
function s = nchoosek_sequence(N,k)
s = zeros(nchoosek(N,k),k);
idx = 1;
for n=(2^k-1):(2^N-1)
if sum(dec2bin(n)=='1') == k
temp = dec2bin(n);
s(idx,:) = find([zeros(1,N-length(temp)), temp=='1']);
idx = idx + 1;
end
end
% just for convinience ;-)
s = s(end:-1:1,:);
end
But this can take a while... ;-)