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

Asked by jose sanchez

jose sanchez (view profile)

on 27 Jan 2016
Latest activity Commented on by Stephen Cobeldick

Stephen Cobeldick (view profile)

on 27 Jan 2016
Accepted Answer by Ingrid

Ingrid (view profile)

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.

Stephen Cobeldick

Stephen Cobeldick (view profile)

on 27 Jan 2016
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

Ingrid (view profile)

on 27 Jan 2016
Stephen - how would you check with this vectorized solution if the precision is found sooner?
Stephen Cobeldick

Stephen Cobeldick (view profile)

on 27 Jan 2016
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.

on 27 Jan 2016
Edited by Ingrid

Ingrid (view profile)

on 27 Jan 2016

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)