[Martin Taylor 2017.10.29.12.58]

```
You are right that modelling helps get things correct. Your results
```

have been puzzling me, so I checked my memory of what I learned so

long ago against Wikipedia’s presentation of flip-flops, with rather

salutary results. I have been mixing two different forms of

flip-flop, analogue and digital and telling you to produce something

that was neither. That’s not helpful. So let’s walk through it

properly, avoiding reliance on my obviously faulty memory.

```
In the flip-flop described in the Wikipedia article, the equivalent
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of PCT perceptual functions are logic circuits that take inputs that

are 1 or 0 and produce outputs that are 1 or 0. It’s binary all the

way. The flip-flop that I argue creates categories is not like that.

It can take as input and produce as output any analogue value

whatever. The flip-flop behaviour whose strength varies from a

slight contrast enhancement to discrete A or B but not both as a

function of some other “modulator” variable (e.g. task stress,

context…) can be represented as a simple cusp catastrophe:

```
![cusp A-H2.jpg|712x413](upload://tMIca5ACWkEJ78oMjBCWyDDz0v4.jpeg)
A linear system cannot create such a cusp catastrophe, and I have
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been proposing to you a linear system in which the perceptions are

linear functions (sums) of their inputs and the cross-links are

simple multipliers. However, since Weber and Fechner in the mid-19th

century it has been taken for granted that most perceptions are

approximations to logarithmic functions of their inputs rather than

being linear, meaning in our context that the firing rate of a

perceptual output approximates the logarithm of a linear function of

the inputs.

```
Would changing your "A" and "B" functions to log(Aness + Wba*B) and
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log(Bness + Wba*A) produce the cusp as the W multipliers decrease*

from zero to become increasingly negative? The analysis below

suggests that it might. Obviously, the log cannot be a true

logarithm, because firing rates cannot go below zero, so one would

have to use an approximation to log(X) that does not go below zero

when X <1. Since neither the X-ness nor the A and B outputs can

go below zero, a suitable approximation might be A = log(1+min(0,

Aness+WbaB)). Since Wba*B is the inhibitory influence on the A

perceptual function, its value may be considerably negative, but it

can’t reduce the output below zero, and log(1+X) becomes ever closer

to log(X) as X grows large.

```
To see whether this nonlinearity will produce the cusp catastrophe,
```

we could simulate the operation of the circuit or we could analyze.

I trust you to do the simulation, while here I do an informal

analysis. The equations are hard to solve explicitly, so the

analysis is in terms of limits and directions of influence.

```
Let us look for equilibrium states for A and B, given values of
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Aness and Bness, There should be one equilibrium condition if

Wab*Wba is less than a critical value (the analysis below suggests

that this critical value is 1.0) and three if the product exceeds

the critical value. If we make the situation symmetrical by setting

Wab=Wba, we expect one of the equilibrium states to be where A=0,

another where B=0, and a third where A=B. If these turn out to be

equilibrium states, we then have to ask whether they are stable,

metastable, or unstable. If the first two are stable and the third

unstable, all that remains is to determine whether there might be

other equilibria in the dynamic.

```
If A=0, then Wab*A =0, B=log(1+Bness). Wba*B = Wba*log(1+Bness), the
```

amount of inhibition of A. Because A = 0, we know that Aness <

-Wba*log(1+Bness). So long as this last inequality holds, A=0,*

B=log(1+Bness). We know that this equilibrium is at least

metastable, because it holds for all Aness values less than

-Wbalog(1+Bness). Arguing from symmetry, the other equality B=0, A

= log(1+Aness) is also at least metastable.

```
But what about A=B? Is that an equilibrium state? It certainly is if
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both are zero, which could happen only if Aness and Bness are both

zero. So let’s assume that both are greater than zero, and equal,

because if they are not equal, neither will be A and B.

```
A = log(1+min(0,Aness+Wba*B)).
By assumption, B=A > 0, so A = log(1+min(0, Aness+Wba*A) =
```

log(1+Aness+Wba*A) >0

```
Exponentiating both sides,
e<sup>A</sup> = 1+ Aness +Wba*A
The RHS of this equation grows linearly with A, while the LHS grows
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exponentially, …

```
![LinearExp.jpg|781x528](upload://wQ90i3eC0KNSugpNPfN9BLvAUgz.jpeg)(Wba=2, Aness = 2)
... and since for A=0 and Aness>0 the LHS is less than the RHS,
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there exists for all values of Wba and Aness a value of A for which

the equation holds (LHS=RHS), and the symmetrical expression in B

also holds. So A=B is an equilibrium state. Is it stable? To test

this, we augment Aness by ∂Aness, without changing Bness.

```
e<sup>A</sup> = 1+ Aness + ∂Aness +Wba*A
To equate the two sides of the equation, e<sup>A</sup> must
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increase, meaning that A must increase. What about B? We can no

longer argue from symmetry, so we must examine its corresponding

expression directly.

```
B = log(1+min(0,Bness+Wab*A). We still know that B >0 (indeed,
```

before we added ∂Aness, B=A), so we can forget the “min” expression

and exponentiate as before.

```
e<sup>B</sup> = 1 + Bness + Wab*A
When Aness = Bness, A = B, but now A is greater than it was, so e<sup>B</sup>
is decreased (Wab and Wba are both negative) and so is B. The
```

reduction in B reduces the inhibition on A, increasing A and further

decreasing B because of the increased inhibition. The A=B

equilibrium is unstable unless Wab = Wba = 0 (the two sides are

unconnected), but there may be a nearby equilibrium value if the

added increase in A is less than ∂Aness, implying a loop gain less

than 1.0.

```
The loop gain consists of the product of four components, Wab, Wba,
```

and the derivatives of the perceptual function outputs as a function

of their inputs. Our perceptual functions are “output =

log(1+input)”, for which the derivative is 1/(1+input), so long as

the input is greater than zero. The maximum loop gain occurs where

the total input to a perceptual function (say Aness+Wba*B) is just*

above zero, where the gain is just under 1.0. The critical point

where the overall loop gain crosses 1.0 therefore occurs when

WabWba ≈ 1. For Wab*Wba >1, there are the three equilibrium*

states mentioned earlier, whereas for WabWba<1 there is only

one, near A=Aness, B= Bness for low values of the product,

exaggerating the contrast more and more as the product increases,

until as it increases beyond 1, the steepness of the contrast at A≈B

overlaps and the single equilibrium surface becomes the fold.

```
![bkcfefhmkjjilgga.jpg|279x201](upload://fQz9rAeZIcs0dHcM1wEVJDxHkY0.jpeg) (Modulator
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is Wab*Wba in this case, Data is Aness-Bness)

```
That is an argument from boundary conditions and qualitative
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effects. I hope you are able to simulate it and that the simulation

bears out the analysis. If we could have an explicit solution to the

equation e^{A} = 1+ Aness +Wba*A, it would be possible to

provide a proper explicit equilibrium analysis of the loop and have

no doubt.

```
Sorry for having in my senility led you astray earlier. I hope I'm
```

not doing that now.

`Martin`