hi im having trouble creating a linear congruential generator in MatLab, the ones that I found online work quite different than mine. then im trying to print values of the m and a (relatively prime, m being a large prime obviously) and check when the cycle is full. I know all the math stuff, im getting used to matlab and its hard to implement this for me even though i should know. my program looks like this:
M = [];
for m = 100:10000;
M(m) = m;
A = [];
for a = 2:(m-1);
A(a) = a;
B = [];
R = [];
for n = 1:1000;
R(n) = n;
B(n) = A(a) * n;
K = [];
K(n)=mod(B(n),M(m));
n=n+1;
a=a+1;
m=m+1;
if K(n) == R(n)
print (m)
print (a)
print ('the cycle is done')
end
end
end
end
also im not too familiar with MatLab so im probably creating arrays the wrong way. thanks in advance.
Well you aren't really asking a question there. Here is some advice for you:
1) Pre-allocate the matrices: M = zeros(9900,1), A = zeros(9998,1), you will get much faster results when you loop, or even better M = 100:10000 works directly if the values you want to put in are as simple as that.
2) You do not need to do the a = a+1, the for loop does it automatically for you (unless it's there for another reason I'm unaware of).
Related
I recognized this is a quite hard problem for me. I asked this problem on official Matlab side but no-one could help me either so maybe someone of you can come up with an outstanding approach.
In detail my Problem consist of:
N = 100 %some number
G = 21 %random guess < N
for x = 1:N;
a = mod(G^x,N);
end
Now I want the calculation of a to stop, if a number repeats.
For example: a = 1, 2, 3, 1 -break
Seems simple but I just can't handle it right after many tries.
For instance I've put:
for x = 1:N
a = mod(G^x,N);
b = unique(a);
if a ~= b
break
end
end
but doesn't seem to work bc. it's not element wise I guess.
This approach keeps a running log of the past Results and uses the ismember() function to check if the current value of a has been previously seen.
clc;
N = 100; %some number
G = 21; %random guess < N
Results = NaN(1,N);
for x = 1:N
a = mod(G^x,N);
disp(a);
if ismember(a,Results)
disp("-break");
break
end
Results(x) = a;
end
Ran using MATLAB R2019b
How can I simulate this question using MATLAB?
Out of 100 apples, 10 are rotten. We randomly choose 5 apples without
replacement. What is the probability that there is at least one
rotten?
The Expected Answer
0.4162476
My Attempt:
r=0
for i=1:10000
for c=1:5
a = randi(1,100);
if a < 11
r=r+1;
end
end
end
r/10000
but it didn't work, so what would be a better way of doing it?
Use randperm to choose randomly without replacement:
A = false(1, 100);
A(1:10) = true;
r = 0;
for k = 1:10000
a = randperm(100, 5);
r = r + any(A(a));
end
result = r/10000;
Short answer:
Your problem follow an hypergeometric distribution (similar to a binomial distribution but without replacement), if you have the necessary toolbox you can simply use the probability density function of the hypergeometric distribution:
r = 1-hygepdf(0,100,10,5) % r = 0.4162
Since P(x>=1) = P(x=1) + P(x=2) + P(x=3) + P(x=4) + P(x=5) = 1-P(x=0)
Of course, here I calculate the exact probability, this is not an experimental result.
To get further:
Noticed that if you do not have access to hygepdf, you can easily write the function yourself by using binomial coefficient:
N = 100; K = 10;
n = 5; k = 0;
r = 1-(nchoosek(K,k)*nchoosek(N-K,n-k))/nchoosek(N,n) % r = 0.4162
You can also use the binomial probability density function, it is a bit more tricky (but also more intuitive):
r = 1-prod(binopdf(0,1,10./(100-[0:4])))
Here we compute the probability to obtain 0 rotten apple five time in a row, the probabily increase at every step since we remove 1 good juicy apple each time. And then, according to the above explaination, we take 1-P(x=0).
There are a couple of issues with your code. First of all, implicitly in what you wrote, you replace the apple after you look at it. When you generate the random number, you need to eliminate the possibility of choosing that number again.
I've rewritten your code to include better practices:
clear
n_runs = 1000;
success = zeros(n_runs, 1);
failure = zeros(n_runs, 1);
approach = zeros(n_runs, 1);
for ii = 1:n_runs
apples = 1:100;
a = randperm(100, 5);
if any(a < 11)
success(ii) = 1;
elseif a >= 11
failure(ii) = 1;
end
approach(ii) = sum(success)/(sum(success)+sum(failure));
end
figure; hold on
plot(approach)
title("r = "+ approach(end))
hold off
The results are stored in an array (called approach), rather than a single number being updated every time, which means you can see how quickly you approach the end value of r.
Another good habit is including clear at the beginning of any script, which reduces the possibility of an error occurring due to variables stored in the workspace.
I am trying to make a portion of my code run faster in MatLab, and I'd like to use parfor. When I try to, I get the following error about one of my variables D_all.
"The PARFOR loop cannot run because of the way D_all is used".
Here is a sample of my code.
M = 161;
N = 24;
P = 161;
parfor n=1:M*N*P
[j,i,k] = ind2sub([N,M,P],n);
r0 = Rw(n,1:3);
R0 = repmat(r0,M*N*P,1);
delta = sqrt(dXnd(i)^2 + dZnd(k)^2);
d = R_prime - R0;
inS = Rw_prime(find(sqrt(sum(d.^2,2))<0.8*delta),:);
if isempty(inS)
D_all(j,i,k,tj) = D_all(j,i,k,tj-1);
else
y0 = r0(2);
inC = inS(find(inS(:,2)==y0),:);
dw = sqrt(sum(d(find(sqrt(sum(d.^2,2))<0.8*delta & d(:,2)==0),:).^2,2));
V_avg = sum(dw.^(-1).*inC(:,4))/sum(dw.^(-1));
D_all(j,i,k,tj) = V_avg;
end
end
I'm not very familiar with parallel computing, and I've looked at the guides online and don't really understand how to apply them to my situation. I guess I need to "slice" D_all but I don't know how to do that.
EDIT: I think I understand that the major problem is that when using D_all I have tj and tj-1.
EDIT 2: I don't show this above, it probably would have been helpful, but I defined D_all(:,:,:,1) = V_1; where V_1 corresponds to a previous time step. I tried making multiple variables V_2, V_3, etc. for each step and replacing D_all(j,i,k,tj-1) with V_1(j,i,k). This still led to the same error I am seeing with D_all.
"Valid indices for D_all are restricted for PARFOR loops"
The goal is generating six lotto numbers, but obviously they have to be unique. This has to be written in function form though, following is the equivalent using the library:
(randsample(42,6))'
My idea was to create the vector with all possibilities, pick one out at a time through index and making it impossible to pick this one again by grabbing it out before the next one is picked.
function numbers = lottonumbers()
pool = 1:42;
numbers = zeros(1,6);
for i=1:6
for j=42-i
randIndex = round(1+j*rand);
randNumber = pool(randIndex);
numbers(i) = randNumber;
if randIndex==1
pool = pool(2:end);
else if randIndex==length(pool)
pool = pool(1:(end-1));
else
pool = [pool(1:randIndex-1), pool(randIndex+1:end)];
end
end
end
end
Since I'm pretty noob at MATLAB (just noob at programming really) and since I solved it myself while asking the question, I'm just going to leave it here and ask you guys for suggestions (better style, other algorithm...)
Lotto is based on permutations where the order does not play a role.
% p = randperm(n,k) returns a row vector containing k unique integers selected randomly from 1 to n inclusive.
randperm( 42, 6 )
should do the trick.
From the code: "This is sometimes referred to as a K-permutation of 1:N or as sampling without replacement."
Another approach is to use rejection sampling: generate the numbers independently, and if they are not all different start again. This is efficient as long as the chance of numbers not being all different is small.
N = 6;
M = 42;
done = false;
while ~done
result = randi(M,1,N); %// generate N numbers from [1,...,M]
done = all(diff(sort(result))); %// if all are different, we're done
end
I am trying to write an algorithm in MatLab which takes as its input a lower triangular matrix. The output should be the inverse of this matrix (which also should be in lower triangular form). I have almost managed to solve this, but one part of my algorithm still leaves me scratching my head. So far I have:
function AI = inverse(A)
n = length(A);
I = eye(n);
AI = zeros(n);
for k = 1:n
AI(k,k) = (I(k,k) - A(k,1:(k-1))*AI(1:(k-1),k))/A(k,k);
for i = k+1:n
AI(i,k) = (I(i,k) - (??????????????))/A(i,i);
end
end
I have marked with question marks the part I am unsure of. I have tried to find a pattern for this part of the code by writing out the procedure on paper, but I just can't seem to find a proper way to solve this part.
If anyone can help me out, I would be very grateful!
Here is my code to get the inverse of a lower triangular matrix by using row transformation:
function AI = inverse(A)
len = length(A);
I = eye(len);
M = [A I];
for row = 1:len
M(row,:) = M(row,:)/M(row,row);
for idx = 1:row-1
M(row,:) = M(row,:) - M(idx,:)*M(row,idx);
end
end
AI = M(:,len+1:end);
end
You can see how it's done on Octave's source. This seems to be implemented in different places depending on the class of the matrix. For Float type Diagonal Matrix it's on liboctave/array/fDiagMatrix.cc, for Complex Diagonal matrix it's on liboctave/array/CDiagMatrix.cc, etc...
One of the advantages of free (as in freedom) software is that you are free to study how things are implemented ;)
Thanks for all the input! I was actually able to find a very nice and easy way to solve this problem today, given that the input is a lower triangular matrix:
function AI = inverse(A)
n = length(A);
I = eye(n);
AI = zeros(n);
for k = 1:n
for i = 1:n
AI(k,i) = (I(k,i) - A(k,1:(k-1))*AI(1:(k-1),i))/A(k,k);
end
end