Attempt to reference field of non-structure array.

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Cat 2015-2-22

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编辑： Cat 2015-2-24

So I'm trying to write a code to count the inversions in an array. So in the array [6,4,3,5,1,2] there are 12 inversions. 6 is greater than 5 of the numbers to the right of it, 4 is greater than three of the numbers to the right of it, and so on. However, I need it so that the function doesn't count zero. So if there was a zero in the place of the 3, 6 would only be greater than 4 of the numbers to the right of it. (zero counts as a space, not as a number. I got some help, and this is the code I have to far

while (a <= array.length)
  while (b <= array.length)
      if array(a)>array(b);
          count = count + 1;
      else
          count = count + 0;
      end
      b = b + 1;
  end
  a = a + 1;
end

When I try to run an example array through it and then try to display the count, it gives me the error "Attempt to reference field of non-structure array." Any help with what I'm doing wrong? Or how to add to the function so it doesn't count being greater as an inversion?

UPDATE: fixing the error of using left in place of right

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Image Analyst 2015-2-23

What's your definition of inversion? And 6 doesn't have any numbers to the left of it. 4 has 1 number to the left of it, not 3. And this isn't .Net - there is no .length method to the variables. Use the length() function.

Cat 2015-2-24

编辑：Cat 2015-2-24

So sorry, I meant numbers to the right. And yes, it has to be compared to every number to the right. It's part of a larger function. We are trying to see if a matrix is solvable as a square puzzle. http://www.cs.bham.ac.uk/~mdr/teaching/modules04/java2/TilesSolvability.html here is a link with more information on the inversions and the goal of the function if it helps

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Answer 1

Andrew Newell 2015-2-24

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Suppose we start with the example from your link:

array = [12 1 10 2 7 11 4 14 5 0 9 15 8 13 6 3];

To eliminate the zero entries, just do this:

array = array(array~=0);

A pretty good way of counting the inversions expresses them as a logical relation, e.g.,

array(2:end) > array(1)
ans =
     0     0     0     0     0     0     1     0     0     1     0     1     0     0

Then you can just add up the 1's to get the number of inversions. You can do this in a loop:

count = 0;
for ii=1:length(array)-1
  count = count + sum(array(ii+1:end)<array(ii));
end

There are more compact ways of doing this, but they're pretty cryptic.

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Answer 2

Chris McComb 2015-2-23

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I think that your problem is that array is not a structure. In order to get the length of a, use:

length(array)

instead of:

array.length

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Cat 2015-2-24

Do you have any suggestions for how to not include the zero? I should have put this in the original question. http://www.cs.bham.ac.uk/~mdr/teaching/modules04/java2/TilesSolvability.html This link contains a really good example of the inversions and how it is supposed to play into a bigger function

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