Finding the max sum of the sub arrays.

Having a vector a, that contains sub arrays of 4 elements (their amount is unknown), knowing their starting indices in a and ending indices in a in two vectors=

start=[y1 y2 y3.....]
end=[x1 x2 x3.....]

(yet, we don't know how much sub arrays we get, so the length of the start/end vector is unknown, but they are equal). We need to create the vector of the sums like following: sums=[sum(a((y1):(x1))) sum(a((y2):(x2))) ....] and then find the max sum (that's an easy part).

for example:

a=[7 2 6 14 30 15 12 0 20 21 300 67]

start=[2 8]
end=[5 11]
sums=[52 341]

How can we write the code for the general problem?

Only these functions are allowed :

min , max , sum , find , any , all , isempty , sort, length

I can't figure it out without using the matrices, which is forbidden. Any ideas?

[Jan]

Do I understand correctly: You are not allowed to use for, subsref, subsasgn, colon, horzcat?

I do not see, how a contains subarrays of 4 elements. Which matrices are forbidden to use?

[Adam]

a, start and end are matrices so if you are forbidden to use matrices I would imagine the problem is impossible.

I assume this is a homework question though as they are the only places such stupid restrictions get made so isn't the idea to work out a solution by yourself in that case?!

[Valeria]

Not allowed. All multiple dimensions arrays are forbidden, only vectors. That doesn't matter what are the requirements for the subarrays.

The question is, knowing the starting and the ending points of them, how do I sum them to the new vector?

[Valeria]

Adam, we can use arrays. Or, if you insist, the matrix with one row only. I've been working on this question for 2 days, well, it's the last and the smallest part of it, the rest is done. But this part I just can't complete.

Answer 1

[Valeria]

I managed. We create two help arrays: ind2=start+1; ind3=start+2;

So the sum vector will be: sums=start+ind2+ind3+end;

Thank you all for your assistance!!

[Image Analyst]

You managed to create an error, not solve the problem. Even if you fix the variable names, it's not right because you're not even involving the "a" in any way. Here's your code

a=[7 2 6 14 30 15 12 0 20 21 300 67]

starting=[2 8]
ending=[5 11]
desiredSums=[52 341] % What you want to get
% Now your code:
% Create two helper arrays
ind2=starting+1; 
ind3=starting+2;
% So the sum vector will be: 
sums=starting+ind2+ind3+ending

And here's the output

desiredSums =
    52   341
sums =
    14    38

As you can see, your sums do not produce the desired result.

Answer 2

[Image Analyst]

Valeria: You simply need to give an index to the sums vector and put that index into the "a" array inside the sum function.

a=[7 2 6 14 30 15 12 0 20 21 300 67]
startingIndexes = [2, 8]
endingIndexes = [5, 11]
for k = 1 : length(startingIndexes)
    sums(k) = sum(a( you do this part) ) 
end

Inside you simply need to give the starting and ending indexes as a function of k. Trivial - I think you can do that part.

[Valeria]

Unfortunately, loops are forbidden to use too. Only basic matlab functions.

Answer 3

[Andrei Bobrov]

a=[7 2 6 14 30 15 12 0 20 21 300 67];
start=[2 8];
end1 =[5 11];

a1 = cumsum(a);
out = diff(a1([start;end1]))+a(start);

[Image Analyst]

She already told Matt that she could not use the cumsum() function. And told me that she can't use loops (so that makes it almost impossible to do a cumsum manually). Apparently her professor found a trick where it is possible to get the answer using only the functions min , max , sum , find , any , all , isempty , sort, length , and no matrices (2-D arrays) and no loops of any kind . I'm stumped.

Answer 4

[Matt J]

Hint - consider the CUMSUM command.

[Valeria]

I can't use it.