Daemon in a Pentagon (part 3)

In the last post, I had found 2 simple cases for the demon pentagon to clearly finish. Now it seems that the problem is getting harder.

Case #3

If one vertex is surrounded by 2 negative vertices then it will become positive, but the 2 adjacent vertices will become more negative. (But remember that the sum doesn't get any worse)

ie try

{1,-3,-2,-3,10} If the Daemon picks the -2, . . .

(Hey I'm not going to argue with a Daemon, maybe they breathe fire, or they might have large teeth and hinged jaws that drool green digestive juices . . .)

then the result becomes

{1,-5,2,-5,10} which doesn't exactly finish right away, as the negative fives aren't surrounded by nice larger numbers. (at least the sum is still the same)

It seems that the quantity of negative numbers is getting smaller, I wonder if this is always the case?

It seems not as shown on my sample sheet (the scan from Daemon #1 post) There are originally 4 vertices with the negative numbers. (all -3) each step seems to decrease the number of negative values. This is true until step 4.

The Pentagon goes from having only one negative number (-12) to having 2 negative numbers again (-9,-9) so it is possible for the amount of negative values to increase. We want them to all go away.

In the sample solution, they do all go away. I get an intuitive feeling that they should all go away eventually. In the several examples it always seems to work. but how to prove it?

Progression of Negative Vertices

Step#, # of negative vertices

0,4 START
1,3
2,3
3,2
4,1
5,2
6,2
7,1
8,2
9,2
10,1
11,2
12,1
13,2
14,1
15,2
16,2
17,1
18,1
19,1
20,0 Done

So there does seem to be a progression, with the number of negative vertices decreasing quickly from 4 to 1, and then jumping from one to two and back again until settling on one and then stopping at zero.

If the demon picked other vertices than I did in each step, then the number of steps may be different. (or not? the same?)

Also the fact that this is a pentagon may be another factor. if it was an even sided figure, I'd bet that it would be less likely to finish. I wonder what would happen for a triangle or a square?

It seems to work the same for triangles and squares. So reducing the number of sides doesn't seem to help much.

The triangle and the square do seem to progress faster. I am not entirely sure that if you had an even number of sides that you couldn't get a condition where the numbers flip back and forth.

One example that I tried with a square, with 2 negative sides and 2 positive sides would never seem to finish. Proving that the square would never finish doesn't seem any easier than proving that the pentagon will finish, however, so this isn't exactly a simpler problem.