while loop until x amount correct digits

Hi

Can someomene give me an example or an general way to write a while loop where the condition is that you need to have 3 correct decimals. ?

In my code I have a for loop but I need to make it more adapt.

Answer 1

[Akihumi]

Have you considered using built-in function 'round'?

https://www.mathworks.com/help/matlab/ref/round.html

Then you can just do:

if round(x,3) == round(y,3)
    ...
end

[mohamed hassan]

My original code is:

for ii = 25:50:500

N=ip;

[r y] = main(N,a,k,Ta); % it calculates u for different N. k a ,Ta are constants.

vec=[vec; y(N+1)]

end

This would give me some answers and then i can manually compare how many decimals are the same( correct) but I want to do it in a smart way.by doublig N til I get 3 correct digits. It should stop iterating then. That would mean i should use i while loop but I dont know how to do it.

Hope you understand

[mohamed hassan]

The round function dont give em the amount correct decimals ?

[Akihumi]

Let me clarify, you want to compare y with another number, is this correct?

I don't think 'round' can give you the number of 'correct decimal', because it is just doing rounding instead of comparison.

But you can use a while loop with 'round', which would look something like this:

n = 1;
while round(y,n) ~= round(x,n)  % x is the number that you want to compare with
    n = n + 1;
end

But you better set a limit, otherwise N might go into infinity. So it will look something like this

n = 1;
nLim = 10;
while n <= nLim && round(y,n) ~= round(x,n)  % x is the number that you want to compare with
    n = n + 1;
end

[Akihumi]

I see. I think in this case you want to stop the while loop when Utemputsida >= 2.5306e2, right? Then you can try something like this:

thresh = 253.06;  % which is equivalent to 1.0e+02 * 2.5306
N = 0;
NLim = 100;  % to stop the while loop if it doesn't fulfil the condition for 100 iterations
[r u] = main(N,a,k,Ta);
while N < NLim && round(u(N+1),2) < thresh
    N = N + 50;
    [r u] = main(N,a,k,Ta);
end

If you want to still use a for loop, you can actually try 'break'

thresh = 253.06;  % which is equivalent to 1.0e+02 * 2.5306
for ip = (25*2.^(0:8))
    N1=ip;
    [r u] = main(N1,a,k,Ta);
    Utemputsida=[Utemputsida; u(N1+1)] %vector with answers for different N1
    if round(Utemputsida(end),2) >= thresh
        break
    end
end

[Akihumi]

Ah I see. Then it should be

N = 0;
NLim = 1e10;  % to stop the while loop if it goes too big
decimalPlace = 2;
[r u] = main(N,a,k,Ta);
[r u2] = main(N+50,a,k,Ta);
while N < NLim && round(u(N+1),decimalPlace) ~= round(u(N+50+1),decimalPlace)
    N = N + 50;
    u = u2;
    [r u2] = main(N+50,a,k,Ta);
end

[mohamed hassan]

I need to calculate the the u untill I get 5 correct digits. In your case I only calculate two times under the while loop and in my assigment I need to double the N until I get 4 for example correct digits.

[Akihumi]

I have changed the NLim from 100 to 1e10 just now, you may check it again.

If you want to check up to 5 correct digits, try changing decimalPlace from 2 to 5.

I did a N = N + 50 in the while loop. So it will increase like what you did in your for loop (0, 50, 100,... , 400).

You can still print out the u in each loop if you want:

N = 0;
NLim = 1e10;  % to stop the while loop if it goes too big
decimalPlace = 5;
[r u] = main(N,a,k,Ta);
[r u2] = main(N+50,a,k,Ta);
disp(u)
disp(u2)
while N < NLim && round(u(N+1),decimalPlace) ~= round(u(N+50+1),decimalPlace)
    N = N + 50;
    u = u2;
    [r u2] = main(N+50,a,k,Ta);
    disp(u2)
end

[mohamed hassan]

I want to make it more general by doubling N instead of increasing with 50 until the condiction of the while loop is set.

[Akihumi]

N = 0;
NLim = 1e10;  % to stop the while loop if it goes too big
decimalPlace = 5;
[r u] = main(N,a,k,Ta);
[r u2] = main(2*N,a,k,Ta);
disp(u)
disp(u2)
while N < NLim && round(u(N+1),decimalPlace) ~= round(u(2*N+1),decimalPlace)
    N = N * 2;
    u = u2;
    [r u2] = main(2*N,a,k,Ta);
    disp(u2)
end

[Stephen23]

Rounding is not the correct approach, read these to know why:

https://www.mathworks.com/matlabcentral/answers/350303-why-0-6-is-not-equal-to-0-2-3#comment_472139

https://www.mathworks.com/matlabcentral/answers/321709-strange-behaviour-of-matlab-find-command#comment_444258

The correct way to is to compare the absolute difference against the required tolerance:

abs(A-B)<tol

[Akihumi]

@Stephen Cobeldick thank you for the lesson.

Then it should be something like this i think

N = 0;
NLim = 1e10;  % to stop the while loop if it goes too big
tol = 1e-5;
[r u] = main(N,a,k,Ta);
[r u2] = main(2*N,a,k,Ta);
disp(u(N+1))
disp(u2(2*N+1))
while N < NLim && abs(u(2*N+1)-u(N+1))>tol 
    N = N * 2;
    u = u2;
    [r u2] = main(2*N,a,k,Ta);
    disp(u2(2*N+1))
end

Answer 2

[mohamed hassan]

Thanks for the help guys, I've done the question and got it correct.