Since you're getting even numbers at least 50% of the time, eventually you should go on a run.
You also benefit from hitting a y =2^n x N number because you get to greatly reduced the size of y
Further, every time you make it down to 1, you can append the values you hit on your way down to a list of seen values, as they're part of a convergence chain, meaning if you see them in a future cycle, you can stop your iteration early because you have a known outcome.
So this doesn't necessarily become a computationally heavy problem with large numbers.
Yeah, this is definitely at least a semi-decidable problem. By following the process of the video in the beginning you can generate all values that converge towards 1, and you will reach each value at some point. Now, all that is missing is the other direction. How would you test if a number doesn't converge towards 0 in an algorithmic way that always finishes?
Since you're getting even numbers at least 50% of the time, eventually you should go on a run.
You also benefit from hitting a y =2^n x N number because you get to greatly reduced the size of y
Further, every time you make it down to 1, you can append the values you hit on your way down to a list of seen values, as they're part of a convergence chain, meaning if you see them in a future cycle, you can stop your iteration early because you have a known outcome.
So this doesn't necessarily become a computationally heavy problem with large numbers.