Can anyone please tell me what if the nargin function is doing? I've been trying to learn how to use matlab on my own. //thanx
function [m_hat,s2hat ] = yatzy(n)
for m = 1:n
value = tillsfem(5);
throws(m) = value;
end
m_hat = mean(throws);
s2hat = var(throws);
throw_max = max(throws);
p = [];
A = [0 1/6 1/36 1/216 1/1296;
0 5/6 10/36 15/216 25/1296;
0 0 25/36 80/216 250/1296;
0 0 0 120/216 900/1296;
0 0 0 0 120/1296];
e1 = [1 0 0 0 0]';
e5 = [0 0 0 0 1]';
p = zeros(throw_max, n);
for m = 1:throw_max
pm = e1'*A^m*e5;
p(m) = pm;
end
if nargin < 2
k = 1:throw_max;
p = p*n;
else
p = p*n*throw_max/k;
end
figure(1); clf;
hist(throws,k);
hold on
stem(1:throw_max,p,'r')
title('number of throws to get 5 equal dice')
legend('Numerical','Analytical')
xlabel('Number of throws')
ylabel('Frequency')
hold off
fprintf(['Expected number of throws to get: %4.2f '...
'(analytical)\n' ...
' %4.2f '...
'(numerical)\n' ...
'Expected variance: %4.2f '...
'(analytical)\n' ...
' %4.2f '...
'(numerical)\n'], ...
191283/17248, m_hat, 12125651655/297493504, s2hat);
I just need help with explaining
if nargin < 2
k = 1:throw_max;
p = p*n;
else
p = p*n*throw_max/k;

1 Comment

Kishor
Kishor on 20 Aug 2011
see MATLAB help examples on nargin.it is number input argument to your functions(count).

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 Accepted Answer

Sean de Wolski
Sean de Wolski on 19 Aug 2011

3 votes

nargin means "number of arguments in". In the above code it looks pretty useless:
if nargin < 2
%do stuff
end
nargin in this case can only be 0 or 1, since yatzy(n) only allows for one input argument, (n), or no input arguments, yatzy. If you call yatzy with two input arguments, e.g.
yatzy(4,5)
It will error out with something along the lines of "Error using yatzy, more input arguments than expected"
Now as for future reference, since you're learning: nargin is typically used for error checking and so you can have optional arguments, e.g:
let's write a program f that takes one or two input arguments with the 2nd being optional:
function out = f(x,y);
if nargin==0
%user called >>f
error('Not enough input arguments, 1 is required');
elseif nargin==1
%user called f(x)
y = pi;
end %if neither, user called f(x,y)
Here we've made sure the person enters at least one argument, x, if they entered a second one great! if not, the second one, y, = pi.
Good luck!

More Answers (4)

lasse
lasse on 19 Aug 2011

0 votes

hi thank you for the answer, but could you explain the rest of the peace for me k = 1:throw_max; p = p*n; else p = p*n*throw_max/k; what k,n,p is? and what happens later i got this from a friend so is kinda hard for me to understand everything :)

6 Comments

Sean de Wolski
Sean de Wolski on 19 Aug 2011
Why don't you ask him (her)? It's his (her) logic.
lasse
lasse on 19 Aug 2011
hes is on vacation :/ school begins soon so i really need to understand this
if you understand, please help me thanx :)
Sean de Wolski
Sean de Wolski on 19 Aug 2011
I understand it, yes, but that's not helping you. For you to understand it you have to figure it out yourself. Use the debugger, put a break point on the first line and work through it step by step and see what's happening.
lasse
lasse on 19 Aug 2011
i can tell you that ive tried to understand it for days now, but the problem is i dont have the matlab at home and this need to be done before school start. So if you can help i would be very thankful.
Sean de Wolski
Sean de Wolski on 19 Aug 2011
Then finding a computer that has ML should be your first step. Perhaps at the school's library/computer cluster etc.?
lasse
lasse on 19 Aug 2011
thats the problem,i dont have it and i live far away from school. I have wrote a whole yatzy game code and this is the last part, so if you dont mind explaining either wise, thank you anyway. :)

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Fangjun Jiang
Fangjun Jiang on 19 Aug 2011

0 votes

It is likely a bad code. The function has only one input argument so nargin<2 is always true. If you try to provide more than one input arguments like yatzy(3,4), it will cause an error saying too many input arguments.

9 Comments

lasse
lasse on 19 Aug 2011
okej i can change that but could help me explaining
k = 1:throw_max;
p = p*n;
else p = p*n*throw_max/k;
//thanx
Sean de Wolski
Sean de Wolski on 19 Aug 2011
And, assuming it's the MB game, it's misspelled!
Walter Roberson
Walter Roberson on 19 Aug 2011
Since you know that the condition will always be true, you can remove the "if" and the "else" statement, reducing it down to
k = 1:throw_max;
p = p*n;
As to what that means in the context of the code... Good luck understanding it, as it involves stochastic transition matrices which is a topic in Mathematics, not in MATLAB.
Sean de Wolski
Sean de Wolski on 19 Aug 2011
And the assumption that n is a number and not an OOP object.
Fangjun Jiang
Fangjun Jiang on 19 Aug 2011
@Sean, you got keyboard latency too?!
Fangjun Jiang
Fangjun Jiang on 19 Aug 2011
I am simply explaining the syntax of the code here.
Check the value of throw_max, k=1:throw_max is to create a vector of 1,2,3,4,...,throw_max
p=p*n, p is a vector, it multiple every element of p with n. Try an example separately in command window. p=1:5;n=3;p=p*n;
Oleg Komarov
Oleg Komarov on 19 Aug 2011
The OP doesn't have matlab but yet he wrote the code of his friend.
lasse
lasse on 19 Aug 2011
i wrote this code before summer vacation at school, just went home from my real vacation thats why its kinda hard for me to understand everything now.
Oleg Komarov
Oleg Komarov on 19 Aug 2011
Then go to the getting started guide, chapter about matrix manipulation: http://www.mathworks.com/help/techdoc/learn_matlab/f2-8955.html. You'll find the answer in 5-15 minutes and you'll understand why your story is just a story.
Friends, homework, assignments it's for your benefit. Help yourself and others will help. If you asked WHERE could you find the explanation you would have saved yourself lots of time.
You could find useful: http://www.mathworks.com/matlabcentral/answers/6200-tutorial-how-to-ask-a-question-on-answers-and-get-a-fast-answer
Give a look to the other links as well.

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Daniel Shub
Daniel Shub on 20 Aug 2011

0 votes

You do not need MATLAB to figure out what is going on. Just work through your code line by line with a pencil and paper.
function [m_hat,s2hat ] = yatzy(n)
At this point the only thing in memory is n.
for m = 1:n
now we do something n times. What do we do, well that is:
value = tillsfem(5);
Well is is not so clear since you haven't told us what tillsfem does. It looks like value could be anything. Oh wait, look,
throws(m) = value;
this means value must be a scalar. So we could actually write
throws(m) = tillsfem(5);
So now we have n and throws which is a nx1 array of something.
m_hat = mean(throws);
hmmm, I wonder what that does. Do you know what the command mean does? You can go to the online documentation and find out that mean takes the mean of an array and returns a scalar. So now you have n and m_hat which are a scalars and throws which is a nx1 array.
It is as easy as that. The nice about MATLAB is it does exactly what you tell it to do.

2 Comments

hanisah bt azhar
hanisah bt azhar on 29 Nov 2018
Edited: hanisah bt azhar on 29 Nov 2018
why it said that support for 'nargin' in the script has been removed in my matlab?
Steven Lord
Steven Lord on 29 Nov 2018
Because it has been. You can no longer call nargin, nargout, or inputname from within a script as stated in the Release Notes for release R2016b. You can still call them from within a function or a class method.
Scripts can't have inputs so they can't have a number of inputs.
Scripts can't have inputs so those (non-existent) inputs can't have names.
Scripts can't return outputs so they can't have a number of outputs.

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Aldo Díaz
Aldo Díaz on 28 Apr 2020

0 votes

nargin works fine except for nargin < 3, it seems there is a bug, at least on R2018b!!!!

1 Comment

Walter Roberson
Walter Roberson on 29 Apr 2020
Could you give us an example of nargin failing for you in R2018b ?

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