for 1:2:11 rest of code
Why not try it? You can see the behavior from one line of code.
1:4:11 ans = 1 5 9
So if the increment takes you past the endpoint, you do not get 11, or 13. You get up to the step that did not exceed the endpoint.
In fact, if the start point is itself above the end point, with a positive step, you get an empty. So the for loop would not iterate at all.
2:1:0 ans = 1×0 empty double row vector
As you can see, running this code never goes through the loop even once.
for i = 2:1:0 'sdfghwrhrt' end