I'm new to matlab I'm having trouble with this one.

Write a function called mypi which consumes a number that specifies the required accuracy and approximates the value of pi to that accuracy. Use the following algorithm to approximate pi:

Think about a quarter circle inside of a unit square (quarter circle has area pi/4). Pick a random point inside the square. If it is in the quarter circle, you get a "hit", otherwise it's a "miss". The approximate area of the quarter circle will be given by the number of "hits" divided by the number of points chosen.

The function should repeat the process of counting hits and misses until at least 10,000 tries have been made or pi is within the required accuracy. It should return the the estimate for pi.

[Ingrid]

what have you tried? if you do not show any attempt you probably have no problem with the matlab coding but with the problem itself

[jose sanchez]

total= hits
hit= 0
i=1:10000
for i = 1:10000
    x = rand ()
    y = rand ()
    if x^2 + y^2 <= 1
    else hit = hit + i 
    end

end

piest = 4 * (hits/total)

error = (pi-(piest))
if error<.001
   break
end
   disp(hits/total)

[Stephen23]

The basic task can be vectorized quite easily:

>> N = 1e6;
>> X = rand(N,1);
>> Y = rand(N,1);
>> I = sqrt(X.^2 + Y.^2)<=1;
>> 4*nnz(I)/N
ans =  3.1395

[Ingrid]

Stephen - how would you check with this vectorized solution if the precision is found sooner?

[Stephen23]

I did not mean to imply that this was a complete solution for the OP: I would have given it as an Answer if it was! My comment simply shows how vectorized code can be used to estimate π using this concept, i.e. the "basic task".

To meet the precision condition requires either a loop of some kind, or else calculating a large number of X&Y values and finding the first subset which meets the condition.

Answer 1

[Ingrid]

The most important mistake is adding i to hit in the loop instead of just one, i is a counter that goes from 1 to 10000 so it gets really big.

Also, you need to do the error checking with the for loop to allow for preliminary exit after the required precision has been reached.

Finally, be careful with correct naming of your variables. It matters if you write hit or hits and it is not clear were you define hits and it should definitely not be used in calculating the estimated pi as there you want to use the actual number stored in hit

The code below should work as you wanted

maxIt = 10000;
hit= 0;
for i = 1:maxIt
    x = rand();
    y = rand();
    if x^2 + y^2 <= 1
        hit = hit + 1 ;
    end
    piest = 4 * (hit/i);
    error = abs(pi-(piest));
    if error < .001
        sprintf('Number of iterations = %1.0f',i)
        break
    end

end

disp(piest)