# Go back to initializing values using for loop when condition is not met

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Mounisha Ganesan on 10 Jul 2019
Commented: infinity on 10 Jul 2019
I have a for loop where I initiialize random values to variables and I check for certain condition, if my condition is not met I have to reinitialize the values for the same iteration. How do I do so?
Here's a sample
n = 100;
for i = 1:n
x(i) = rand
y(i) = rand
if(x(i)< 0 || y(i)<0)
disp(' I need to reinitialize values for the same i and my n count should be 100')
end
end
So I have to run the loop until I reach my 'n' value and meet my conditions within the loop. Is there a way to do it or should I use different loop like while?

infinity on 10 Jul 2019
Edited: infinity on 10 Jul 2019
Hello,
There are several ways, one approach that you can use as follows
clear
n = 100;
x = zeros(1,n);
y = zeros(1,n);
for i = 1:n
xt = rand;
yt = rand;
while (xt*yt<0)
xt = rand;
yt = rand;
end
x(i) = xt;
y(i) = yt;
end

infinity on 10 Jul 2019
Hello,
"I would like to know why do we have to use * operator instead of || operator. It means when one value is less than zero the product automatically becomes less than zero? Am I right? "
I can use * operator is because it is equivalent with your condition x < 0 || y < 0. If one of them (x and y) less than 0 then x * y < 0 and inverse.
Also, as you can see that x*y < 0 is shorter than (x< 0 || y < 0). But, of course, you can use your original condition.
"And what should I do if I have to check two conditions say I have to new variables x2t and y2t which depensds on xt and yt are always between (0 and 1) in this code given below
Is my code right?"
Actually, what I have done before are not necessary since function "rand" generate a real random number between 0 and 1. Therefore, now you don't need to use condition to on xt and yt.
But for your current problem, x2 and y2 can be negative and can be greater than 1. Therefore, you should implement conditions and the code could be like
a = 0;
b= 1;
l = 0.2;
for i = 1:n
x1(i) = a + (b-a)*rand;
y1(i) = a + (b-a)*rand;
thetax = 2*pi*rand;
thetay = 2*pi*rand;
x2next = x1(i) + li(i)*cos(thetax);
y2next = y1(i) + li(i)*sin(thetay);
while (x2next < 0 || x2next > 1)
thetax = 2*pi*rand;
x2next = x1(i) + li(i)*cos(thetax);
end
while (y2next < 0 || y2next > 1)
thetay = 2*pi*rand;
y2next = y1(i) + li(i)*sin(thetay);
end
x2(i) = x2next;
y2(i) = y2next;
end
I have put a, b and l out side the loop since they are constants.
More details of function "rand" can be found by typing
help rand
Mounisha Ganesan on 10 Jul 2019
YES ! I did this already. That was my original code. But since x2 is( X2 = x1+ some value ) it might exceed value 1 or become less than 0. So I have to check for the x2 and y2 everytime so that they are between 0-1 and initialize x1 and y1 again for the same iteration until I make sure x2 and y2 are within (0-1)
infinity on 10 Jul 2019
Hello,
I got your point. You can see my comment above, which is just modified.