subtracting a particular value until that the result becomes negative

Hi all,

how can I subtract elements of a column until the point that the result is negative?

example: A=[1 2 3 4 5 4 2 1 1 1 3]' , I have an external value let's say 6. I want to go to the 6th element lets say and do this:6-4. then take the result (in this case 2), and subtract it from the next element until the result to become negative. At this point I would like to stop the process. is there any way to do this?

Imagine that I have an array 48X365

thanks!

[Adam]

What do you mean by "the next element". There doesn't seem to be any obvious 'next element' after you take your 6th value and subtract it from 6.

[Nikolas Spiliopoulos]

ah sorry, lets say that the external value is 8!so 8-4=4, then 4-2=2, 2-1=1,1-1=0, 0-1=-1

[Jan]

If the real problem contains a matrix and not a vector, how is the solution defined then? Why do you start at the 6th element?

[Nikolas Spiliopoulos]

It's how the problem is defined, same thing will happen but with a different external value calculated by the previous column. If i know the answer for a vector i think I could do it column by column!

[Jan]

@Nikolas: The question is still not clear to me. Do you start at the 6th element, because the external value is 6? But for the value 8 you start at the same element. What is the wanted output? You've explained the process, but not, which values should be store as output.

[Nikolas Spiliopoulos]

ok, let me clarify forget the value 6, the external let's say it's 8! So I start doing these calculations 8-4=4,4-2=2, 2-1=1, 1-1=0,0-1=-1 etc. So I want to get as output [1 2 3 4 5 4 2 1 0 -1 3]. is it clear now? (let's say that i want to start from the 6th element).

Since, I do the calculations for one vector I think I can handle it fro the next ones.

thanks anyway

[Adam]

What is the logic behind doing 4-2=2 after 8-4=4? It is still very unclear, to me at least.

[Nikolas Spiliopoulos]

you subtract the external value 8 from element 6th. then you take the result, and subtract it from the next element (so 8-6=4, 4-2=2 etc.). Does it make sense now? sorry for being a pain..

[Adam]

It seems like there is surely an easier way to represent the problem than this!

You could just do it in a basic for loop though I would have thought.

[Nikolas Spiliopoulos]

good to know,

i would be grateful If you could tell me how

thanks!

Answer 1

[Jan]

And another guess:

A = [1 2 3 4 5 4 2 1 1 1 3]'
x = 8;
for index = 6:length(A)  % Start at 6th element
  x = x - A(index);
  A(index) = x;
  if x < 0
    break;
  end
end

[Nikolas Spiliopoulos]

thanks a lot!!

Answer 2

[dpb]

>> A =[1:5 4 2 1 1 1 3];
>> ix=6;         % initial location
>> v=8;          % external initial value
>> dA=v-A(ix);   % first try
>> while dA>=0
     ix=ix+1;
     dA=dA-A(ix)
   end
dA =
   2
dA =
   1
dA =
   0
dA =
  -1
>> ix
ix =
  10
>>

Don't know what you intend re: the "have an array 48X365" and, of course, there's the issue of ix running off the end of the array if the while condition isn't satisfied so either need to be certain that there is a solution or have a bounds check.

[Nikolas Spiliopoulos]

thanks for the answer, yeah I need to check the bounds

However, I get an error "Undefined function or variable 'ix' "

any ideas how to fix it?

[Bas Holten]

'ix' is de index in your array and is still undefined the first time you use it in the code above. Just state at the start* that

 ix = 1;

*Assuming you want to start at the first element of the array.

[dpb]

As Bas says, it's that initial point; was 6 in your example. I missed it in the cut'n paste. Fixed in answer.