unsupervised learning the hebb rule – neurons that fire together wire together. pca rf development...
TRANSCRIPT
Unsupervised learning
• The Hebb rule – Neurons that fire together wire together.
• PCA
• RF development with PCA
Classical Conditioning and Hebb’s rule
“When an axon in cell A is near enough to excite cell B and repeatedly and persistently takes part in firing it, some growth process or metabolic change takes place in one or both cells such that A’s efficacy in firing B is increased”
Ear
Tongue
Nose
A
B
D. O. Hebb (1949)
The generalized Hebb rule:
where xi are the inputs
and y the output is assumed linear:
Results in 2D
jj
j
ii
xwy
yxdt
dw
-2 -1 0 1 2-2
-1
0
1
2
x1
x 2 m
=/3
Example of Hebb in 2D
w
(Note: here inputs have a mean of zero)
On the board:
• Solve simple linear first order ODE • Fixed points and their stability for non linear ODE.
In the simplest case, the change in synaptic weight w is:
where x are input vectors and y is the neural response.
Assume for simplicity a linear neuron:
So we get:
Now take an average with respect to the distribution of inputs, get:
€
Δwi =ηx iy
€
y = w j
j
∑ x j
€
Δwi = η xixjwjj
∑ ⎛
⎝ ⎜
⎞
⎠ ⎟
jj
ijj
jjii wQwxxEwE
Δ ][][
If a small change Δw occurs over a short time Δt then:(in matrix notation)
If <x>=0 , Q is the covariance function.
What is then the solution of this simple first order linear ODE ?
(Show on board)
www
Qdt
d
t
ΔΔ
The change in synaptic weight w is:
where x are input vectors and y is the neural response.
Assume for simplicity a linear neuron:
So we get:
Mathematics of the generalized Hebb rule
))(( 00 yyxxw ii Δ
jj
jxwy
Δ
j jjjijjii xywxxxywxxw 0000
Δ
j jjjijjii xywxExxEywxxEwE 0000 ][][][][
Δ
j jjjiji xywxywQwE 0000][
ijji QxxE ][ ][ ixE
Taking an average of the the distribution of inputs
And using and
We obtain
€
E[Δw ] =η Q − x0μJ[ ] W − y0(μ − x0) =η Q− k2J[ ]w − e^
k1
In matrix form
Where J is a matrix of ones, e is a unit vector in direction (1,1,1 … 1), and
or
Where
)( 001 xyk 02 xk
1
^
][ keE Δ wQw '
JkQQ 2'
The equation therefore has the form
]ˆ[ 1' ekQ
dt
d w
w
If k1 is not zero, this has a fixed point, however it is usually not stable.
If k1=0 then have: ww 'Qdt
d
The Hebb rule is unstable – how can it be stabilized while preserving its properties?
The stabilized Hebb (Oja) rule.
2'' )(/))(()1( Δww Δ iii wtwtw
NormalizeWhere |w(t)|=1
Appoximate to first order in Δw: (show on board)
Now insert
Get:
yxw ii Δ
)()()()1( '' twwtwwtwtwj
jjiiii ΔΔ
jjjiii
jjjiiii
wxytwyxtw
tywxtwyxtwtw
'''
''''
)()(
)()()()1(
}y
))(()()1( 2'''' ytwyxtwtww iiiii Δ
Therefore
The Oja rule therefore has the form:
)( 2ywyxdt
dwii
i
k kjjkjkiikk
i
k kjkjkjikkiii
i
wwxxwxxwdt
dw
wwxxwwxxywyxdt
dw
,
,
2 )(
Average
In matrix form:
Qw)w(wQww Tdt
d
Using this rule the weight vector converges tothe eigen-vector of Q with the highest eigen-value. It is often called a principal component or PCA rule.
•The exact dynamics of the Oja rule have been solved by Wyatt and Elfaldel 1995
•Variants of networks that extract several principal components have been proposed (e.g: Sanger 1989)
Therefore a stabilized Hebb (Oja neuron) carries out Eigen-vector, or principal component analysis (PCA).
-1 -0.5 0 0.5 1-1
-0.5
0
0.5
1
x1
x 2
=/3
Another way to look at this:
Where for the Oja rule:
At the f.p: where
So the f.p is an eigen-vector of Q.
The condition means that w is normalized.
Why? Could there be other choices for β?
))(( wQww
ydt
d
2)( yy
0dt
dw
Using this rule the weight vector converges tothe eigen-vector of Q with the highest eigen-value. It is often called a principal component or PCA rule.
wQw
€
= y 2
€
=y 2
Show that the Oja rule converges to the state |w^2|=1
The Oja rule in matrix form:
What is
Bonus question for H.W: The equivalence above, why does it prove convergence to normalization.
)( 2yydt
dwx
w
€
d | w2 |
dt= 2ηy2(1− | w2 |)
€
d | w2 |
dt=
d(wT ⋅w)
dt= 2wT ⋅
dw
dt
Show that the f.p of the Oja rule is such that the largest eigen-vector with the largest eigen-value (PC) is stable while others are not (from HKP – pg 202).
Start with:
Assume w=ua+εub where ua and ub are eigen-vectors with eigen-values λa,b
Qw)w(wQww Tdt
d
dt
ud
dt
du
dt
d ba )(w
)(()()( babababa uuuuuuuu )Q(Q T
)(()()( babababa uuuuuuuudt
d )Q(Qw T
€
d(ua + εub )
dt=η λ aua + ελ bub − λ aua −ελ aub −ελ bua( ) +O(ε 2)
Get: (show on board)
Therefore:
That is stable only when λa> λb for every b.
abdt
d
Finding multiple principal components – the Sanger algorithm.
))((1
1iij
i
kkjkji
ij ywwyxydt
dw
Subtract projection onto accounted for subspace (Grahm-Schimdtt)
))((1
i
kkjkji
ij wyxydt
dw
Standard Oja Normalization
Homework 1: (due in 1/28) Implement a simple Hebb neuron with random 2D input, tilted at an angle, θ=30o
with variances 1 and 3 and mean 0. Show the synaptic weight evolution. (200 patterns at least)1b) Calculate the correlation matrix of the input data. Find the eigen-values, eigen-vectors of this matrix. Compare to 1a.1c) Repeat 1a for an Oja neuron, compare to 1b.+ bonus question above (another 25 pt)
What did we learn up to here?
Visual Pathway
Area17
LGN
Visual Cortex
Retinalight electrical signals
•Monocular•Radially Symmetric
•Binocular•Orientation Selective
Receptive fields are:
Receptive fields are:
Re s
pon
se (
s pik
e s/s
e c)
Left Right
Tuning curves
0 180 36090 270
RightLeft
Orientation Selectivity
Binocular Binocular DeprivationDeprivation
NormalNormal
Adult
Eye-opening angle angle
Res
pon
se (
spik
es/s
ec)
Res
pon
se (
spik
es/s
ec)
Eye-opening
Adult
Monocular Monocular DeprivationDeprivation
NormalNormal
Left Right
% o
f ce
lls
group group
angleangleRes
pon
se (
spik
es/s
ec)
1 2 3 4 5 6 7
10
20
1 2 3 4 5 6 7
30
15
RightLeft
Rittenhouse et. al.
First use Hebb/PCA with toy examplesthen used with more realistic examples
Aim get selective neurons using a Hebb/PCA rule
Simple example: ))(cos(1)( '' rrqrrQ
r r r r
Why?The eigen-value equation has the form:
Q can be rewritten in the equivalent form:
And a possible solution can be written as the sum:
)()()( ''1
1
' rwrwrrQdr
))sin()sin()cos()(cos(1)( ''' rrrrqrrQ
1
0 )sin()cos()(l
ll lrblraarW
Inserting, and by orthogonality get:
))sin()cos(()sin()cos(21
0110 lrblraarqbrqaa lll
So for l=0, λ=2, and for l=1, λ=q, for l>1 there is no solution.
So either w(r)= const with λ=2 or with λ=q.)sin()cos()( 11 rbrarw
Orientation selectivity from a natural environment:
The Images:
Natural Images, Noise, and Learning
image retinal activity
•present patches
•update weights
•Patches from retinal activity image
•Patches from noise
Retina
LGN
Cortex
Raw images: (fig 5.8)
Preprocessed images: (fig 5.9)
Monocular Monocular DeprivationDeprivation
NormalNormal
Left Right
% o
f ce
lls
group group
angleangleRes
pon
se (
spik
es/s
ec)
1 2 3 4 5 6 7
10
20
1 2 3 4 5 6 7
30
15
RightLeft
Rittenhouse et. al.
Binocularity – simple examples.
Q is a 2-eye correlation function.
What is the solution of the eigen-value equation:
mmQ
baba
21
21
1
1
2
1
1
1
2
1
mm
In a higher dimensional case:
Qll, Qlr etc. are now matrixes.And Qlr=Qrl.
The eigen-vectors now have the form
rrrl
lrll
QQ2Q
m
m
m
m
2
1
2
1 21 mm
In a simpler case
This implies Qll=Qrr, that is eyes are equivalent.And the cross eye correlation is a scaled version of the one eye correlation.
If: then:
with
2Q
mQm
m
m
m
m
2
1
2
1 21 mm
)1(2,1
Positive correlations (η=0.2)
Negative correlations (η=-0.2)
Hebb with lower saturation at 0
Lets now assume that Q is as above for the 1Dselectivity example.
2Q
With 2D space included
2 partially overlapping eyes using natural images
RightLeft
Orientation selectivity and Ocular Dominance
PCA
Left Eye
Right Eye
Right SynapsesLeft
Synapses
0
50
100
0
50
100
0
50
100
0
50
100
1 3 5Bin
No
. of
Cel
ls
What did we learned today?
The Hebb rule is unstable – how can it be stabilized while preserving its properties?
The stabilized Hebb (Oja) rule.
This is has a fixed point at:
Where:
The only stable fixed point is for λmax