Algorithm to Make Knot Projection Identical to Unknot

Question

I found a problem in a book that went something like this(I don't have the book right now):

Given a knot projection without crossing statuses(over vs. under), come up with an algorithm to insert crossings to make the projection identical to the unknot.

I came up with this:

1. Take the first two crossings. Make them both either over or under. This separates the sections of rope on either side of the second crossing, so they are not entangled.
2. Repeat recursively until you reach the end of the crossings

So this is basically like OOUUOOUU.... or UUOOUUOO.... and I think it would make the unknot

Would this algorithm work? If not, can you give a counterexample?

Thanks