Demon in the Pentagon (part 5)

I thought this morning on the bus about an interesting and (perhaps simpler) case.

What if there are only 2 numbers in the pentagon? IE the set of values is: {0,0,0,8,-7}?

Then there is only one negative value to concern ourselves with, and this makes the problem a little easier to digest.

Then (see the scanned sheet) something interesting happens, the negative number "splits or partitions" for the lack of a better word, the number 8 into +7, and +1
then the -7 goes around the pentagon in circular fashion. every time the negative number (in this case -7) passes the 1 (think of it like a finish line) then the -7 becomes a -6. When the -6 starts its trek around the circle it creates a new #1 behind it, thus creating a new finish line.

So we have a progression like this (add both numbers together, you always get a 1.)

(8,-7), (7,-6), (6,-5), (5,-4), (4,-3), (3,-2), (2,-1), (1, 0)

now we are done since both numbers are considered positive or 0.

In this case we have simplified the problem somewhat, but it is indeed the same problem.

The fact that (neg_numbers) + (pos_numbers) = c, c>0
forces this to eventually terminate.

This specific case is actually the same as the general case. This may or may not seem obvious, but when there are only 2 numbers, this is the same as the subset of the negative numbers, versus the subset of the positive numbers.