How to Update value of some matriks element with looping?

I have condition for my matrix. Matrix x1,x2, and y. When element of y is less than 10, the y element will update will sum of x1 and x2 until the element is 10 or more. This is my code.

clc;
clear;
x1=[1 2 3 4 5 6 7 8 9 10];
x2=[1 2 3 4 5 6 7 8 9 10];
y=[10 9 11 10 11 11 9 10 12 12];

while y<10
    y=x1+x2;
end
y

And then, if the problem obove is clear. I want to know how many sum (in this case, how many iteration) to achieve values of 10 or more of each element

[Ameer Hamza]

Can you further clearify the problem? For the given vectors x1, x2, and y. What is you expected output.

[Eddy Iswardi]

I want to make all elements of y is same 10 or more. For the example, let's see y(2). The value is 9. I will update the value from x1 and x2. x1(2) is 2 and x2(2) is 2. So for x1(2)+x2(2) =4. 4 is less than 10. So I sum again. 4+x1(2)+x2(2)=8. 8 is less than 10. So I sum again 8+x1(2)+x2(2)= 12. 12 is more then 10. So 12 is update value for y(2). And value of y from [10 9 11 10 11 11 9 10 12 12] become [10 12 11 10 11 11 9 10 12 12]. And the algorith next to elements of y less than 10

[Ameer Hamza]

You said it make all elements of y 10 or more but then you said

[10 9 11 10 11 11 9 10 12 12] become [10 12 11 10 11 11 9 10 12 12]

The second vector still has 9.

[Eddy Iswardi]

Yeah. But it's only an example. y(7) will be update to, some ways with y(2). So the vector will get [10 12 11 10 11 11 14 10 12 12]

Answer 1

[Ameer Hamza]

Try this:

x1=[1 2 3 4 5 6 7 8 9 10];
x2=[1 2 3 4 5 6 7 8 9 10];
y=[10 9 11 10 11 11 9 10 12 12];
k = 10;

x12 = x1 + x2;
mask = y < k;
y(mask) = ceil(k./x12(mask)).*x12(mask);

Result:

y =
    10    12    11    10    11    11    14    10    12    12

[Eddy Iswardi]

thank you

[Les Beckham]

For this part of your question: "I want to know how many sum (in this case, how many iteration) to achieve values of 10 or more of each element", the "iterations" required are given by this part of Ameer's excellent answer:

ceil(k./x12(mask))

If you do this:

iterations = zeros(size(x1));
iterations(mask) = ceil(k./x12(mask))

you will get a vector showing the number of sums required for every element of your original vector with zeros where no iterations (sums) were required.