How to increment and/or decrement with different values in the same loop

This is the code I have but I was wondering if there is a more simple and efficient way to implement this. Basically, I am going from 15 to 1 but instead of decrementing with a single value i.e [15 14 13 12 ... 2 1] I want to reach the target by different incrementation such as in the screenshot below the code i.e [15 14 18 16 13 12 ... 4 1]

i = 1;
v = 15;
array(i) = 15;
while array(i) ~= 1
    i = i + 1;
    v = v - 1;
    array(i) = v;
    if array(i) == 1
        break;
    end
    i = i + 1;
    v = v + 4;
    array(i) = v;
    if array(i) == 1
        break;
    end
    i = i + 1;
    v = v - 2;
    array(i) = v;
    if array(i) == 1
        break;
    end
    i = i + 1;
    v = v - 3;
    array(i) = v;
end
disp(array)

Answer 1

[Benjamin Kraus]

Here is a version that is certainly shorter, and produces the same result, but I doubt it is any more efficient. The benefit I see to this version is that you can change the pattern more easily, and you don't have quite the same amount of redundant checks for whether the current value is 1.

pattern = [-1 4 -2 -3];
i = 1;
v = 15;
while v ~= 1
    array(i) = v;
    v = v + pattern(mod(i-1,numel(pattern))+1);
    i = i + 1;
end
array(i) = v;
disp(array)

[Benjamin Kraus]

This version cuts down on the number of iterations of the loop, as it does all four additions/subtractions simultaneously, but I'm not sure it would run any faster than my previous suggestion, or even your original code.

pattern = [-1 4 -2 -3];
pattern = cumsum(pattern);
i = 1;
array = 15;
while ~any(array(max(end-3,1):end) == 1)
    array = [array array(end)+pattern];
end
array = array(1:find(array == 1));
disp(array);

[Aren Kurumlian]

For the first solution, when i = 1 then mod(i-1,numel(pattern)+1) will become mod(0,5) which should give a remainder of zero. But somehow we're getting 1. How? Thank you very much

[Benjamin Kraus]

My code was this:

mod(i-1,numel(pattern))+1

Your version is this:

mod(i-1,numel(pattern)+1)

The difference is adding 1 after calling mod vs. adding one to the input to mod.

Breaking down my version when i == 1:

      i == 1
      i-1 == 0
          numel(pattern) == 4
  mod(i-1,numel(pattern))   == mod(0,4)   == 0
  mod(i-1,numel(pattern))+1 == mod(0,4)+1 == 1