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
Related
I am trying to write a MATLAB function that accepts non-integer, n, and then returns the factorial of it, n!. I am supposed to use a for loop. I tried with
"for n >= 0"
but this did not work. Is there a way how I can fix this?
I wrote this code over here but this doesn't give me the correct answer..
function fact = fac(n);
for fact = n
if n >=0
factorial(n)
disp(n)
elseif n < 0
disp('Cannot take negative integers')
break
end
end
Any kind of help will be highly appreciated.
You need to read the docs and I would highly recommend doing a basic tutorial. The docs state
for index = values
statements
end
So your first idea of for n >= 0 is completely wrong because a for doesn't allow for the >. That would be the way you would write a while loop.
Your next idea of for fact = n does fit the pattern of for index = values, however, your values is a single number, n, and so this loop will only have one single iteration which is obviously not what you want.
If you wanted to loop from 1 to n you need to create a vector, (i.e. the values from the docs) that contains all the numbers from 1 to n. In MATLAB you can do this easily like this: values = 1:n. Now you can call for fact = values and you will iterate all the way from 1 to n. However, it is very strange practice to use this intermediary variable values, I was just using it to illustrate what the docs are talking about. The correct standard syntax is
for fact = 1:n
Now, for a factorial (although technically you'll get the same thing), it is clearer to actually loop from n down to 1. So we can do that by declaring a step size of -1:
for fact = n:-1:1
So now we can find the factorial like so:
function output = fac(n)
output = n;
for iter = n-1:-1:2 %// note there is really no need to go to 1 since multiplying by 1 doesn't change the value. Also start at n-1 since we initialized output to be n already
output = output*iter;
end
end
Calling the builtin factorial function inside your own function really defeats the purpose of this exercise. Lastly I see that you have added a little error check to make sure you don't get negative numbers, that is good however the check should not be inside the loop!
function output = fac(n)
if n < 0
error('Input n must be greater than zero'); %// I use error rather than disp here as it gives clearer feedback to the user
else if n == 0
output = 1; %// by definition
else
output = n;
for iter = n-1:-1:2
output = output*iter;
end
end
end
I don't get the point, what you are trying to do with "for". What I think, what you want to do is:
function fact = fac(n);
if n >= 0
n = floor(n);
fact = factorial(n);
disp(fact)
elseif n < 0
disp('Cannot take negative integers')
return
end
end
Depending on your preferences you can replace floor(round towards minus infinity) by round(round towards nearest integer) or ceil(round towards plus infinity). Any round operation is necessary to ensure n is an integer.
I have to construct the following function in MATLAB and am having trouble.
Consider the function s(t) defined for t in [0,4) by
{ sin(pi*t/2) , for t in [0,1)
s(t) = { -(t-2)^3 , for t in [1,3)*
{ sin(pi*t/2) , for t in [3,4)
(i) Generate a column vector s consisting of 512 uniform
samples of this function over the interval [0,4). (This
is best done by concatenating three vectors.)
I know it has to be something of the form.
N = 512;
s = sin(5*t/N).' ;
But I need s to be the piecewise function, can someone provide assistance with this?
If I understand correctly, you're trying to create 3 vectors which calculate the specific function outputs for all t, then take slices of each and concatenate them depending on the actual value of t. This is inefficient as you're initialising 3 times as many vectors as you actually want (memory), and also making 3 times as many calculations (CPU), most of which will just be thrown away. To top it off, it'll be a bit tricky to use concatenate if your t is ever not as you expect (i.e. monotonically increasing). It might be an unlikely situation, but better to be general.
Here are two alternatives, the first is imho the nice Matlab way, the second is the more conventional way (you might be more used to that if you're coming from C++ or something, I was for a long time).
function example()
t = linspace(0,4,513); % generate your time-trajectory
t = t(1:end-1); % exclude final value which is 4
tic
traj1 = myFunc(t);
toc
tic
traj2 = classicStyle(t);
toc
end
function trajectory = myFunc(t)
trajectory = zeros(size(t)); % since you know the size of your output, generate it at the beginning. More efficient than dynamically growing this.
% you could put an assert for t>0 and t<3, otherwise you could end up with 0s wherever t is outside your expected range
% find the indices for each piecewise segment you care about
idx1 = find(t<1);
idx2 = find(t>=1 & t<3);
idx3 = find(t>=3 & t<4);
% now calculate each entry apprioriately
trajectory(idx1) = sin(pi.*t(idx1)./2);
trajectory(idx2) = -(t(idx2)-2).^3;
trajectory(idx3) = sin(pi.*t(idx3)./2);
end
function trajectory = classicStyle(t)
trajectory = zeros(size(t));
% conventional way: loop over each t, and differentiate with if-else
% works, but a lot more code and ugly
for i=1:numel(t)
if t(i)<1
trajectory(i) = sin(pi*t(i)/2);
elseif t(i)>=1 & t(i)<3
trajectory(i) = -(t(i)-2)^3;
elseif t(i)>=3 & t(i)<4
trajectory(i) = sin(pi*t(i)/2);
else
error('t is beyond bounds!')
end
end
end
Note that when I tried it, the 'conventional way' is sometimes faster for the sampling size you're working on, although the first way (myFunc) is definitely faster as you scale up really a lot. In anycase I recommend the first approach, as it is much easier to read.
Because for combinations of large numbers at times matlab replies NaN, the assignment is to write a program to compute combinations of 200 objects taken 90 at a time. Once this works we are to make it into a function y = comb(n,k).
This is what I have so far based on an example we were given of the probability that 2 people in a class have the same birthday.
This is the example:
nMax = 70; %maximum number of people in classroom
nArray = 1:nMax;
prevPnot = 1; %initialize probability
for iN = 1:nMax
Pnot = prevPnot*(365-iN+1)/365; %probability that no birthdays are the same
P(iN) = 1-Pnot; %probability that at least two birthdays are the same
prevPnot = Pnot;
end
plot(nArray, P, '.-')
xlabel('nb. of people')
ylabel('prob. that at least two have same birthday')
grid on
At this point I'm having trouble because I'm more familiar with java. This is what I have so far, and it isn't coming out at all.
k = 90;
n = 200;
nArray = 1:k;
prevPnot = 1;
for counter = 1:k
Pnot = (n-counter+1)/(prevPnot*(n-k-counter+1);
P(iN) = Pnot;
prevPnot = Pnot;
end
The point of the loop I wrote is to separate out each term
i.e. n/k*(n-k), times (n-counter)/(k-counter)*(n-k-counter), and so forth.
I'm also not entirely sure how to save a loop as a function in matlab.
To compute the number of combinations of n objects taken k at a time, you can use gammaln to compute the logarithm of the factorials in order to avoid overflow:
result = exp(gammaln(n+1)-gammaln(k+1)-gammaln(n-k+1));
Another approach is to remove terms that will cancel and then compute the result:
result = prod((n-k+1:n)./(1:k));
I'm struggling with determining the probability of occurrence of unique elements in thresh_strain matrix (which can be seen below as a 100 x 16). I was trying to use the code at the bottom to do this, but I get an equal probability of occurrence associated with each of the elements, whereas I want the probability of occurrence associated with unique elements in thresh_strain.
function [thresh_strain] = MCsolution()
no_iterations = 100;
thresh_strain = zeros(100, 16);
casechoice =input('Enter 1 for 1st Layup and 2 for 2nd layup:');
for i=1:no_iterations
for j=1:16
J = Nielsennew(casechoice);
thresh_strain(i,j) = J(1, j);
end
end
% [uniqueValues,~,uniqueIndex] = unique(thresh_strain);
% frequency = accumarray(uniqueIndex(:),1)./numel(thresh_strain);
Thanks
It is not really clear from the title and description, but I suppose you may be looking for something like this:
myUniqueValues = unique(myMatrix);
nelements = hist(myMatrix(:),myUniqueValues);
%plot(myUniqueValues,nelements)
Basically calculating how often each unique value occurs. From here getting the corresponding percentage is of course trivial.
So I have a list of 190 numbers ranging from 1:19 (each number is repeated 10 times) that I need to sample 10 at a time. Within each sample of 10, I don't want the numbers to repeat, I tried incorporating a while loop, but computation time was way too long. So far I'm at the point where I can generate the numbers and see if there are repetitions within each subset. Any ideas?
N=[];
for i=1:10
N=[N randperm(19)];
end
B=[];
for j=1:10
if length(unique(N(j*10-9:j*10)))<10
B=[B 1];
end
end
sum(B)
Below is an updated version of the code. this might be a little more clear in showing what I want. (19 targets taken 10 at a time without repetition until all 19 targets have been repeated 10 times)
nTargs = 19;
pairs = nchoosek(1:nTargs, 10);
nPairs = size(pairs, 1);
order = randperm(nPairs);
values=randsample(order,19);
targs=pairs(values,:);
Alltargs=false;
while ~Alltargs
targs=pairs(randsample(order,19),:);
B=[];
for i=1:19
G=length(find(targs==i))==10;
B=[B G];
end
if sum(B)==19
Alltargs=true;
end
end
Here are some very simple steps to do this, basically you just shuffle the vector once, and then you grab the last 10 unique values:
v = repmat(1:19,1,10);
v = v(randperm(numel(v)));
[a idx]=unique(v);
result = unique(v);
v(idx)=[];
The algorithm should be fairly efficient, if you want to do the next 10, just run the last part again and combine the results into a totalResult
You want to sample the numbers 1:19 randomly in blocks of 10 without repetitions. The Matlab function 'randsample' has an optional 'replacement' argument which you can set to 'false' if you do not want repetitions. For example:
N = [];
replacement = false;
for i = 1:19
N = [N randsample(19,10,replacement)];
end
This generates a 19 x 10 matrix of random integers in the range [1,..,19] without repetitions within each column.
Edit: Here is a solution that addresses the requirement that each of the integers [1,..,19] occurs exactly 10 times, in addition to no repetition within each column / sample:
nRange = 19; nRep = 10;
valueRep = true; % true while there are repetitions
nLoops = 0; % count the number of iterations
while valueRep
l = zeros(1,nRep);
v = [];
for m = 1:nRep
v = [v, randperm(nRange,nRange)];
end
m1 = reshape(v,nRep,nRange);
for n = 1:nRep
l(n) = length(unique(m1(:,n)));
end
if all(l == nRep)
valueRep = false;
end
nLoops = nLoops + 1;
end
result = m1;
For the parameters in the question it takes about 300 iterations to find a result.
I think you should approach this constructively.
It's easy to initially find a 19 groups that fulfill your conditions just by rearranging the series 1:19: series1 = repmat(1:19,1,10); and rearranged= reshape(series1,10,19)
then shuffle the values
I would select two random columns copy them and switch the values at two random positions
then make a test if it fulfills your condition - like: test = #(x) numel(unique(x))==10 - if yes replace your columns
just keep shuffling till your time runs out or you are happy
of course you might come up with more efficient shuffling or testing
I was given another solution through the MATLAB forum that works pretty well (Credit to Niklas Nylen over on the MATLAB forum). Computation time is pretty low too. It basically shuffles the numbers until there are no repetitions within every 10 values. Thanks all for your help.
y = repmat(1:19,1,10);
% Run enough iterations to get the output random enough, I selected 100000
for ii = 1:100000
% Select random index
index = randi(length(y)-1);
% Check if it is allowed to switch places
if y(index)~=y(min(index+10, length(y))) && y(index+1)~=y(max(1,index-9))
% Make the switch
yTmp = y(index);
y(index)=y(index+1);
y(index+1)=yTmp;
end
end