# How can I check in an arbitrary vector that the first index is greater than the next index and so on?

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Omer Badreldin on 26 Jun 2019
Edited: Jan on 26 Jun 2019
I'm thinking that I can make a for lp that runs through the length of the input vector and checks if the max is greater than the minimum for each index. Then make a new vector with only the inversions and finally take the length of that vector. I have this written this code so far:
function inv = inversions(board)
% Input vector
BR = [15 2 1 12 8 5 6 11 4 9 10 7 3 14 13];
L = length(BR);
for i = 1:L
if max(BR(i)) > min(BR(i))
inv
end

#### 1 Comment

You need to check each value against the preceeding value. BR(i) is a scalar so its min and max are both equal to the value itself and will not tell you anything about the sequence of numbers.
You also need to keep a running counter of cases where there is an inversion, which you would initialise to 0 before any loop.

Jan on 26 Jun 2019
Edited: Jan on 26 Jun 2019
BR = [15 2 1 12 8 5 6 11 4 9 10 7 3 14 13]
Now you want to determine, if an element is larger or smaller than the former one:
BR(2:end) > BR(1:end-1)
Or alternatively:
diff(BR) > 0
Now you get a sequence of TRUE and FALSE values, which are treated as 1 and 0 such that a sum command counts them.

Omer Badreldin on 26 Jun 2019
Thanks guys! I worked it out :-)
I did this:
function inv = inversions(board)
% Input vector (test case)
% board = [15 2 1 12 8 5 6 11 4 9 10 7 3 14 13];
L = length(board);
i = 0;
inv = 0;
% L-1 because board(i+1) would give an index that exceeds the number of indices in board.
for i = 1:L-1
if board(i) > board(i+1) % if condition true, inv will be updated.
inv = inv+1;
end
i=i+1;
end
end

#### 1 Comment

Jan on 26 Jun 2019
Do not increase the loop counter i manually inside the loop: omit the useless lines i=0 and i=i+1 .
More Matlab'ish solutions without a loop:
inv = sum(board(1:end-1) > board(2:end))
or
inv = sum(diff(board) < 0)