[Martin Taylor 991222 23:30] correcting a bad mistake in

[Martin Taylor 991222 13:30] -- an addendum to

[Martin Taylor 991222 9:37]

Without any loss of generality, we can do as Rick did, and set

P2n > P2(n+1) as the perceptual function of all but one of the third

level control systems. If there are k second and third level systems, we

set (again without loss of generality) P2k > 0 as the perceptual function

of the k'th third level system.

...

With respect to the fixed second-level reference, there are three cases:

the fixed one is the k'th, the (k-1)th, or any other one.Case 1: If the fixed one is the k'th, the situation is simple. Since P3k

is (P2k>0), the value of P3k is permanently fixed to be "true" or "false"

and either will match or will not match its reference value. The rest of

the level 3 systems have no internally contradictory sets of reference

values.

This is correct, but it's not one set of reference values that is

unachievable, it is half of all the sets. That is to say, all the sets

that have a reference value of "true" for P3k when it is fixed at "false"

or vice-versa.

Case 2: If the fixed one is R2(k-1), then if it is fixed to be greater

than zero, there are four possibilities for the reference values R3(k-1)

and R3(k).R3(k-1), R3k

true, true achievable when P2k is positive but less than

the fixed value of R2(k-1))

true, false This means P2k < 0, so this is always achievable.

false, true This can be achieved if P2k is greater than the fixed value

false, false This implies P2k is greater than the fixed (positive) value

of P2(k-1) and at the same time is negative.So there is one unachievable set of reference values in this case, too.

All the other possible reference value sets can be achieved.

And the implication again is not that there is one unachievable set of

reference values, but that 1/4 of the reference value sets cannot be

achieved.

Case 3: The fixed one is R2(k-x) where x>1. One can set up a chain of

"greater than = true" or "greater than = false" that will lead to the

requirement that P2k is greater than a positive fixed value of P2(k-x)

and less than zero, or that P2k is less than a negative value of P2(k-x)

and greater than zero, but not both. The situation resolves into a

direct analogue of case 2. Again there is only one set of level 3

reference values that leads to a contradiction.

This is wrong. it doesn't resolve into a direct analogue of case 2.

Here's a table for x = 2 (three level 3 and three level 2 units involved).

We assume the fixed reference is R2(k-2) and it is fixed at some positive

value.

R3(k-2), R3(k-1), R3k

true, true, true achievable by making R2(k-1) positive but less than

the fixed value of R2(k-2)

true, true, false easier than the first case, because R2k will be negative

true, false, true achievable by making R2(k-1) negative

true, false, false achievable by making R2(k-1) more negative than R2k

false, true, true achievable by making R2(k-1) greater than R2(k-2)

false, true, false easier than the last case

false, false, true achievable by making R2(k) greater than the fixed value

of R2(k-2)

false, false, false not achievable, because R2(k) must be greater than

the (positive) fixed value of R2(k-2) and at the

same time it must be negative.

In this case, one of the eight configurations is unachievable.

You can see the pattern. If the fixed second level reference is R2k, and

P3k is (P2k > 0), half the potential sets of reference values are

unachievable. If the fixed one is R2(k-1), 1/4 are unachievable. If the

fixed one is R2(k-2) 1/8 are unachievable. And so on. There is, as I

said, one chain of reference values that makes the set unachievable, but

as the chain lengthens, there are more and more configurations involving

the control units in the chain that are achievable, so the likelihood

of the hierarchy running into problems is reduced.

This analysis holds only for the "Rick-style" spreadsheet with an added

arbitrary constant. In a symmetric hierarchy--the kind I called "pure",

there is always one set of reference values that cannot be achieved

if there are the same number of third and second level control units.

But if there is one more second-level unit, all third-level references

can be achieved--even if one of the second-level units has a "committed"

reference level. Rick's spreadsheet emulates such a set-up, with a

"ghost" seventh second-level unit that controls perfectly to a reference

value of zero.

Here, then are the probabilities that a randomly chosen set of level 3

reference values will be achievable,

Number of units per level "Rick style" "pure"

2 3/4 1/2

3 7/8 3/4

4 15/16 7/8 ....

The numbers for the "Rick-style" spreadsheet are clearly wrong, because

the, probabilities depend on just which second-level reference is fixed.

That doesn't matter for the "pure" spreadsheet, because it is symmetric.

Those numbers seem to be correct.

If that analysis is correct (and I haven't proved it),

It was wrong, and so the following is also wrong. The number could be

higher or lower.

then in the

six-unit-per-level Rick-style hierarchy that is the spreadsheet, there

should be 64 possible sets of reference values. If three of the second

level reference values are fixed (as I did), 4*3/2, or 6 of the 64 should

be unachievable. But that still leaves a 90% chance that a random set

of reference values set by a hypothetical fourth-level system should

be achievable, leaving the fourth-level system able to control perfectly

most of the time.

Sorry for adding to the confusion of an already confused debate.

Martin