Demon in Pentagon Recap

So far, we have

1) Shown that the Demon may be powerful, but he can't change the net sum of the values of the vertices. This number is always constant, and nonzero. His actions have no net effect on the total.

ie

x1+x2+x3+x4+x5 > 0, where x1..x5 are all integers
x1+x2+x3+x4+x5 = c, where c > 0.

START

Find a negative number, can be any negative number in the set

Case #1

IF all numbers in the set of 5 are positive or zero. then EXIT

Case #2

If the neg_num is bounded on both sides by any 2 positive numbers say x1, x2
such that neg_num + x1 >=0, and neg_num + x2 >=0 then the neg_num is removed.
(the middle neg_num, becomes positive, and its neighbours are both positive or zero.)

the signs of (x1, neg_num, x2) go from (+,-,+) to (+,+,+)

CASE #3

(x1+neg_num<0, and x2+neg_num>=0) or (x1+neg_num>=0,and x2+neg_num<0)

If the negative number is bounded on one side by a larger positive number or a number that is equal to that number but different in sign, and on the other by a number that becomes more negative when added to the middle negative number, then the resulting negative number travels around the pentagon in a specific direction.

signs for (x1,neg_num, x3) change from {{-,-,+)or(+,-,-)} to {(-,+,+)or(+,+,-)}

(Remember for our purposes 0, is considered a positve number)

The negative number "rounds the bases" so to speak, in the direction away from the positive number or zero behind it. and after several turns, the negative number, will either cover its tracks, until it collides with a number that is larger, and destroys it or it will meet another negative number that makes it more negative. Eventually the negative number gets canceled out since there are more positive numbers (of total higher value) "manning" the bases than negative ones. each time the negative number meets a positive number, it gets closer and closer to zero, (from the negative side of zero) or itself becomes positive.

Case 4

The only remaining case, is when the Negative number is bounded by numbers x1,x2 such that abs(neg_number) >= x1, and abs(neg_number) >=x2
or in this case x1+neg_num < 0, and x2+neg_num <0

so the numbers x1, neg_num, x2 go from either {(-,-,-)or(+,-,+)} to (-,+,-)

This split in sign leads to either (consider one of the (x1<0,x2<0) to be the new center) {case 3 (-.-.+) or (+,-,-) or case 4 (+,-,+), but eventually to case 2 (also (+,-,+). (we know this since there are an ODD number of vertices, and there is always either a single value larger than any negative number or there is a subset of positive values that are collectively larger than any single negative vertex or the sum of the subset of the negative vertices), which eventually leads to case 1. and we are done. Phew

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This is probably easier to see with triangles than pentagons, as the -numbers in
this case can never be surrounded by 2 negative numbers.

upon trying this out with squares it seems that the daemon would finish sometimes and not others. In the case where you finish with a square you would have one negative number on a corner. if you have two negative numbers on opposite corners (kitty corner) to each other, it seems that the situation arises where you always eliminate one negative number and create another, creating a sort of an endless loop that never terminates.

In the case of the odd sided figures, ie triangle, pentagon, heptagon etc., eventually the negative numbers get trapped between 2 larger numbers, or they get reduced. The moving negative numbers make the positive numbers smaller in the same way, while simultaneously and always ensuring that (positive + negative >0). Each of the positive numbers and the negative numbers wobble towards zero. when the values converge, so that they are both positive, the daemon is finished