Saddening Goat: Artefactual coattail (Part 3)

The third and (likely) second-last installment for artefactual coattail. If all this is deterring you, fret not - I'll be posting some non-math musings soon. Anyway, onto Generalization #3 and #4

Generalization 3:

This is both trickier and simpler than Generalization 1. On the con side, because of the cyclic-ity, our symmetry is dead. But, instead of

rules, we only have

rules - this will help us a lot.
Why does this help us a lot? Let's begin a quick foray into linear algebra. We have

rules and

chemicals. This sounds suspiciously like

equations for

unknowns - however the rules are actually unknowns, not equations. How so? Well, what is unknown about each rule? The only think unknown about or relevant to each rule is how much it is applied. From the argument in part 2, (or anyway, it should be pretty evident), it doesn't matter in what order rules are applied. So let us assume rule

(the rule that subtracts the same amount from

) is used

amount (so

is subtracted in total from each of those

's). Now, the rules that affect a given

are rules

$. We can then check whether we can get 0 excess by assuming each of the

decrease by

- illustration time:

system:

In total the contribution is:

.

For zero excess, the total contribution must be

. Equating, we can say

where

and

. Of course, right now it is not in our interest to go around inverting huge

matrices and then checking what terms are negative when multiplied by our

vector. But knowing this fact will help us later on in our discussion.

For now put this aside. Let's start by looking at

systems. Now, the claim here will be slightly more bizarre than even the general claim for Generalization 1:

Claim: In a

system, if

is even there can be no excess only if

and

for all

, and if

is odd there can be no excess only if

for all

(taking

and

).

Proof:

The claim is divided into two cases, so let's divide our proof into 2 cases. First, n is even. We'll just let

for now. Then, after we apply rule 1, we have:

In order to remove the rest of

(that is, now

), we need to apply rule 2 with that:

And so on:

Eventually we'll get to:

It's crucial to note that there is a

term instead of a

term because

is even. Now, in order to cancel the remaining two terms, it's necessary that:

, or

, as desired.

(EDIT: Okay, above is the way that I saw it first. However, a much, much simpler way is just noting that

is invariant, and it is 0 at the end. This is why I should write my argument on paper first...)

So we've seen that this condition is necessary for

even. Is it sufficient? A look at a case like 1 3 1 1 2 0 should convince you otherwise. The restriction

for all

is necessary regardless of whether

is even or odd. A simple contradiction argument shows this - basically, you can't remove more than

, the two neighboring y's, from

.

This does prove the "necessary" condition asked for. It would be a lot nicer to have necessary and sufficient conditions, though. Are these two conditions taken together sufficient?

I believe it is. I haven't found a counter-example. And I have sort of intuitively reasoned why - but of course, that means nothing. Ordinarily in a situation like this, I'd offer $10 or so to someone who could provide a proof or counter-example, but unfortunately I doubt anyone that reads this has a chance... in any case, the offer is still up.

Given the fact that I cannot even solve

systems successfully, there's not much I can do with

systems in general. However, if you paid attention the the matrix thing above, you'll notice interesting things happen when

- particularly, the determinant of the matrix becomes 0, so the matrix is uninvertible. In fact, I have a rather cool theorem (which I have proved - it's actually quite the same as the short proof for the alternating sum above) that outlines a necessary condition in this case:

If we have a

system where

, there is no excess only if: