README ¶
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Introduction
This simulation shows how an individual neuron can act like a detector, picking out specific patterns from its inputs and responding with varying degrees of selectivity to the match between its synaptic weights and the input activity pattern.
We will see how a particular pattern of weights makes a simulated neuron respond more to some input patterns than others. By adjusting the level of excitability of the neuron, we can make the neuron respond only to the pattern that best fits its weights, or in a more graded manner to other patterns that are close to its weight pattern. This provides some insight into why the point neuron activation function works the way it does.
The Network and Input Patterns
We begin by examining the NetView
tab, showing the Detector network. The network has an Input
layer that will have patterns of activation in the shape of different digits, and these input neurons are connected to the receiving neuron (RecvNeuron
) via a set of weighted synaptic connections. We can view the pattern of weights (synaptic strengths) that this receiving unit has from the input, which should give us an idea about what this unit will detect.
- Select
r.Wt
as the value you want to display (on the left side of the 3D network view) and then click on theRecvNeuron
to view its receiving weights.
You should now see the Input
grid lit up in the pattern of an 8
. This is the weight pattern for the receiving unit for connections from the input units, with the weight value displayed in the corresponding sending (input) unit. Thus, when the input units have an activation pattern that matches this weight pattern, the receiving unit will be maximally activated. Input patterns that are close to the target 8
input will produce graded activations as a function of how close they are. Thus, this pattern of weights determines what the unit detects, as we will see. First, we will examine the patterns of inputs that will be presented to the network.
- Click on the
DigitPats
button (next to thePats
label) on the left side of the window -- this control panel contains all the main ingredients for this model.
The display that comes up shows all of the different input patterns that will be presented ("clamped") onto the Input
layer, so we can see how the receiving unit responds. Each row of the display represents a single trial that will be presented to the network. As you can see, the input data this case contains the digits from 0 to 9, represented in a simple font on a 5x7 grid of pixels (picture elements). Each pixel in a given event (digit) will drive the corresponding input unit in the network.
Running the Network
To see the receiving neuron respond to these input patterns, we will present them one-by-one, and determine why the neuron responds as it does given its weights. Thus, we need to view the activations again in the network window.
- Select
Act
in theNetView
to view activations, then clickTest Trial
in the toolbar at the top of the window.
This activates the pattern of a 0
(zero) in the Input
, and shows 20 cycles of settling process where the activation of the receiving unit is iteratively updated over a series of cycles according to the point neuron activation function (just as the unit in the neuron
simulation was updated over time). We have selected 20 cycles as enough time for the receiving neuron to fully respond to the input.
The receiving unit showed an activity value of 0 because it was not activated above threshold by the 0
input pattern. Before getting into the nitty-gritty of why the unit responded this way, let's proceed through the remaining digits and observe how it responds to other inputs.
- Press
Test Trial
for each of the other digits, until the number8
shows up.
You should have seen the receiving unit finally activated when the digit 8
was presented, with an activation of zero for all the other digits. Thus, as expected, the receiving unit acts like an 8
detector: only when the input perfectly matches the input weights is there enough excitatory input to drive the receiving neuron above its firing threshold.
-
You can use the "VCR" style buttons after the
Time
label at the bottom right of theNetView
to review each cycle of updating, to see the progression of activation over time. -
Go ahead and do one more
Test Trial
to see what happens with9
.
We can use a graph to view the pattern of receiving unit activation across the different input patterns.
- Click on the
TstTrlPlot
(test trial plot) tab.
The graph shows the activation (Act
) for the unit as a function of trial (and digit) number along the X axis. You should see a flat line with a single peak at 8.
Computing Excitatory Conductance Ge
(Net Input)
Now, let's try to understand exactly why the unit responds as it does. The key to doing so is to understand the relationship between the pattern of weights and the input pattern.
- Click the
DigitPats
button again (or find the already-open window if you had not closed it) and scroll inside so that the8
digit is visible. Now resize the whole window so it is roughly big enough to only show the8
and position the window so it is next to the main window that includes theNetview
display. The idea is that you can see both side-by-side, so you may want/need to shrink the main window to prevent occlusion of theDigitPats
window. Then doInit
in the toolbar andTest Trial
for each input digit in turn.
Question 2.8: For each digit, report the number of active
Input
units where there is also a weight of 1 according to the8
digit pattern. In other words, report the overlap between the input activity and the weight pattern. HINT: Strictly speaking, the8
display in theDigitPats
window is NOT representing the weights per se, but as we saw earlier using ther.Wt
functionality inNetView,
they are the same pattern -- and displaying the windows side-by-side just makes the counting easier.
The number of inputs having a weight of 1 that you just calculated should correspond to the total excitatory input Ge
, also called the net input, going into the receiving unit, which is a function of the average of the sending activation Act
times the weight Wt
over all the units, with a correction factor for the expected activity level in the layer, Alpha
:
Ge = (1 / Alpha) * [Sum(Act * Wt) / N]
You can click on Ge
in the NetView
to see these values as you step through the inputs.
- Do
Init
andTest Trial
to see the0
input again. If you hover over the RecvUnit with your mouse, you should see it has a value ofGe = .3352..
. To apply the above equation, you should have observed that0
has 6 units in common with8
, andN
= 35 (7*5), so that is about .1714. Next, we need to apply theAlpha
correction factor, which we set to be the activity level of the8
, which is 17 of the 35 units active. Thus, we should get:
Ge = (1 / (17 / 35)) * (6 / 35) = .3529...
That is not quite right. The problem is that our Input
activations are not quite 1, but rather .95, because it is impossible for our rate-coded neuron activations to get all the way to 1, so we clip them at a maximum of .95. If you multiply the above number by .95, you indeed get the correct Ge = .3352..
!
As a result of working through the Ge net input calculation, you should now have a detailed understanding of how the net excitatory input to the neuron reflects the degree of match between the input pattern and the weights. You have also observed how the activation value can ignore much of the graded information present in this input signal, due to the presence of the threshold. This gives you a good sense for why neurons have these thresholds: it allows them to filter out all the "sub-threshold noise" and only communicate a clear, easily-interpreted signal when it has detected what it is looking for.
Manipulating Leak
Next, we will explore how we can change how much information is conveyed by the activation signal. We will manipulate the leak current (GbarL
), which has a default value of 2, which is sufficient to oppose the strength of the excitatory inputs for all but the strongest (best fitting) input pattern (the 8).
IMPORTANT: you must press Init
for changes in GbarL
to take effect!
- Reduce the
GbarL
value from 2 to 1.8, and doInit
thenTest Trial
(you might want to changeViewUpdt
toAlphaCycle
instead ofCycle
so it only shows the final result of setting for each input). You can alternatively just hitTest All
and look at theTstTrlPlot
.
Question 2.9: What happens to the pattern of receiving neuron activity over the different digits when you change GbarL to 1.8, 1.5, and 2.3 -- which input digits does it respond to for each case? In terms of the tug-of-war model between excitatory and inhibition & leak (i.e., GbarL = leak), why does changing leak have this effect (a simple one-sentence answer is sufficient)?
Question 2.10: Why might it be beneficial for the neuron to have a lower level of leak (e.g., GbarL = 1.8 or 1.5) compared to the original default value, in terms of the overall information that this neuron can convey about the input patterns it is "seeing"?
It is clearly important how responsive the neuron is to its inputs. However, there are tradeoffs associated with different levels of responsivity. The brain solves this kind of problem by using many neurons to code each input, so that some neurons can be more "high threshold" and others can be more "low threshold" types, providing their corresponding advantages and disadvantages in specificity and generality of response. The bias weights can be an important parameter in determining this behavior. As we will see in the next chapter, our tinkering with the value of the leak current Gbar.L is also partially replaced by the inhibitory input, which plays an important role in providing a dynamically adjusted level of inhibition for counteracting the excitatory net input. This ensures that neurons are generally in the right responsivity range for conveying useful information, and it makes each neuron's responsivity dependent on other neurons, which has many important consequences as one can imagine from the above explorations.