matrix sorting different way

for example I have [ 1 9 5 7 8 4 6 2 3]

I want to sort by becoming [max, function , min]

where this function is basically taking the list without the first max and min and do it again [max, function,min]

so it become like [max,[max,[max,[max, function,min],min],min],min]

[9 8 7 6 5 4 3 2 1]

without using the sort(matrix)

[KALYAN ACHARJYA]

Let me repeat the question:

You have [ 1 9 5 7 8 4 6 2 3]

Result expected: [9 8 7 6 5 4 3 2 1] any way- but without using sort function

Confirm?

or we must follow the following-

I want to sort by becoming [max, function , min]

where this function is basically taking the list without the first max and min and do it again [max, function,min]

so it become like [max,[max,[max,[max, function,min],min],min],min]

[Walter Roberson]

Recursion.

max() with two outputs and delete the indicated location from the matrix.

min() with two outputs and delete the indicated locaiton on the matrix.

Output [the_max, recurse on matrix, the_min]

The reason to not do max() and min() both and recurse on the matrix without those two elements -- the reason to do a deletion before the min() -- is that if all of the non-nan elements are equal, the location of the max and min might be the same.

[Lucy Hannah]

im confused on how to do sorting without using the sort function thats my idea but maybe im wrong

[Walter Roberson]

Consider

A = [1 9 5 7 8 4 6 2 3]

Now suppose you cannot use sort(), but you can use max():

Result = [];
while ~isempty(A)
  [maxval, maxidx] = max(A);
  Result = [Result, maxval];
  A(maxidx) = [];
end

The first step initialized Result to empty.

We then test, is A not empty? Is is not empty, so we do the steps. We find the maximum value of A, which is 9, and assign that to maxval, and we assign the index of that maximum inside A, 2, to maxidx. We then append the maximum value we just got, 9, to Result, giving us Result = [9] . We then delete A(2), giving use A = [1 5 7 8 4 6 2 3]. We loop back.

A is still not empty, so max(A) gives us maxval = 8, maxidx = 4, and we append the 8 to Result giving us Result = [9 8], and we remove A(4) giving us A = [1 5 7 4 6 2 3]. We loop back

A is still not empty, so max(A) gives us maxval = 7, maxidx = 3, and we append the 7 to Result giving [9 8 7], and we remove A(3) giving us A = [1 5 4 6 2 3]. We loop back

Next, 6 @ 4, Result = [9 8 7 6], A = [1 5 4 2 3]

Next, 5 @ 2, Result = [9 8 7 6 5], A = [1 4 2 3]

Next, 4 @ 2, Result = [9 8 7 6 5 4], A = [1 2 3]

Next, 3 @ 3, Result = [9 8 7 6 5 4 3], A = [1 2]

Next, 2 @ 2, Result = [9 8 7 6 5 4 3 2], A = [1]

Next, 1 @ 1, Result = [9 8 7 6 5 4 3 2 1], A = []. We loop back

A is now empty so we stop looping.

Now Result is the original A sorted in reverse order, but we never used sort(). Instead, at each step, we found the maximum and removed it from the array, and then processed what was left of the array.