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# for loop, if loop, while loop or combination?

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Chris Ron de on 11 Sep 2020
Commented: Chris Ron de on 14 Sep 2020
Hi,
I am stuck with a problem which I thought was quite easy, but it didn't work out so until now.
I will briefly explain the idea.
I have four numbers as a starting point and whatever change I want to make in the numbers, the total sum of the numbers always have to be the same.
With each of the four points I have to make the same calculation and add the results. When the calculation with the four points is completed, I have to change the four numbers and do the calculation again. When the new added result is higher than the former one, the new result have to be the chosen as solution.
I will try to explain it with an example:
I have the values [2,2,2,2] as a startingpoint. With formula A, I get four different values. These values add up to B.
When I take the values [2.5, 1.5, 2, 2], I can do the calculations again. The new calculated value of B is larger than the old one, so [2.5, 1.5, 2, 2] will be my new starting point.
Next I take the values [2.5, 2, 1.5, 2], but they don't give an improvement of B, so I go on with [2.5, 2, 2, 1.5] which also is not an improvement.
I can go on with [1.5, 2.5, 2, 2], [2, 2.5, 1.5, 2], [2, 2.5, 2, 1.5] etc, but I think at this point a for loop will be necessary. I have some problem finding it while I have to add and subtract and change positions.
Moreover, I don't want my numbers to be negative. So when one of the values is 0, the loop cannot subtract from 0. I don't know if I have to use I while loop for this or an if-else-elseif.
Can someone help me with this problem?
##### 4 CommentsShowHide 3 older comments
Chris Ron de on 14 Sep 2020
Thank yor for your answer Mario. Most of the topics in the Onramp were already known, but I did learn some new things.

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### Accepted Answer

Ayush Gupta on 14 Sep 2020
Refer to the following code on how to do this:
det_rate = [0.3, 0.2, 0.5, 0.1];
T = 8; % T is total searching time
delta = 0.5; % take delta from one value of t_start and add it to another value
a= 0;
b=0;
c=0;
d=0;
number_of_iterations = T/delta;
max_B = 0;
P = zeros(length(t),4);
s = zeros(length(t),1);
for i= 1:number_of_iterations
for j = 1:number_of_iterations
for k = 1:number_of_iterations
for l=1:number_of_iterations
d = T - (a+b+c);
if(d>0)
result = 1-exp(-1.*det_rate.*t(i,:));
if(sum(result) > max_B)
max_B = sum(result);
end
end
end
c = c+delta;
end
b = b+delta;
end
a = a+delta;
end
Refer to this documentation page to see how for loop works, here.
##### 2 CommentsShowHide 1 older comment
Chris Ron de on 14 Sep 2020
Hi Ayush,
I think I cheared to early. Your example choses all the different possibilities my script have to run, but I think with another heuristic aproach. What I wanted to do was making a new value of t in which one of the values in the vector is added with delta and subsequently the other values are subtracted with delta.
So the new values of t will become in order appearing:
t(first) = t(1)+delta, t(2)-delta, t(3), t(4)
t(second) = t(1)+delta, t(2), t(3)-delta, t(4)
t(third) = t(1)+delta, t(2), t(3), t(4)-delta
t(fourth) = t(1)-delta, t(2)+delta, t(3), t(4)
t(fifth) = t(1), t(2)+delta, t(3)-delta, t(4)
etc.
I have some difficulty in finding the way to do this with for-loops. Is there something I overlooked or is this not possible with Matlab in the way I want to do it.
Another possibility is that I don't understand the way you solved it in your example. In that case, sorry for that from this rookie.

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