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CHAPTER 16. Adaptive Resonance Theory. Objectives. There is no guarantee that, as more inputs are applied to the competitive network, the weight matrix will eventually converge . - PowerPoint PPT Presentation
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Ming-Feng Yeh 1
CHAPTER 16CHAPTER 16
AdaptiveAdaptiveResonanceResonance
TheoryTheory
Ming-Feng Yeh 2
ObjectivesObjectives
There is no guarantee that, as more inputs are applied to the competitive network, the weight matrix will eventually converge.
Present a modified type of competitive learning, called adaptive resonance theory (ART), which is designed to overcome the problem of learning stability.
Ming-Feng Yeh 3
Theory & ExamplesTheory & Examples
A key problem of the Grossberg network and the competitive network is that they do NOT always from stable clusters (or categories).
The learning instability occurs because of the network’s adaptability (or plasticity), which causes prior learning to be eroded by more recent learning.
Ming-Feng Yeh 4
Stability / PlasticityStability / Plasticity
How can a system be receptive to significant new patterns and yet remain stable in response to irrelevant patterns?
Grossberg and Carpenter developed the ART to address the stability/plasticity dilemma. The ART networks are based on the Grossberg ne
twork of Chapter 15.
Ming-Feng Yeh 5
Key InnovationKey Innovation
The key innovation of ART is the use of “expectations.” As each input is presented to the network, it is
compared with the prototype vector that is most closely matches (the expectation).
If the match between the prototype and the input vector is NOT adequate, a new prototype is selected. In this way, previous learned memories (prototypes) are not eroded by new learning.
Ming-Feng Yeh 6
OverviewOverview
Input
Layer 1(Retina)
Layer 2(Visual Cortex )
LTM(AdaptiveWeights)
STM
Normalization ConstrastEnhancement
Basic ART architecture
Grossberg competitive network
Ming-Feng Yeh 7
Grossberg NetworkGrossberg Network
The L1-L2 connections are instars, which performs a clustering (or categorization) operation. When an input pattern is presented, it is multiplied (after normalization) by the L1-L2 weight matrix.A competition is performed at Layer 2 to determine which row of the weight matrix is closest to the input vector. That row is then moved toward the input vector.After learning is complete, each row of the L1-L2 weight matrix is a prototype pattern, which represents a cluster (or a category) of input vectors.
Ming-Feng Yeh 8
ART Networks -- 1ART Networks -- 1
Learning of ART networks also occurs in a set of feedback connections from Layer 2 to Layer 1. These connections are outstars which perform pattern recall.When a node in Layer 2 is activated, this reproduces a prototype pattern (the expectation) at layer 1.Layer 1 then performs a comparison between the expectation and the input pattern.When the expectation and the input pattern are NOT closely matched, the orienting subsystem causes a reset in Layer 2.
Ming-Feng Yeh 9
ART Networks -- 2ART Networks -- 2
The reset disables the current winning neuron, and the current expectation is removed.
A new competition is then performed in Layer 2, while the previous winning neuron is disable.
The new winning neuron in Layer 2 projects a new expectation to Layer 1, through the L2-L1 connections.
This process continues until the L2-L1 expectation provides a close enough match to the input pattern.
Ming-Feng Yeh 10
ART SubsystemsART SubsystemsLayer 1 Comparison of input pattern and expectation.L1-L2 Connections (Instars)
Perform clustering operation. Each row of W1:2 is a prototype pattern.
Layer 2Competition (Contrast enhancement)
L2-L1 Connections (Outstars)Perform pattern recall (Expectation).Each column of W2:1 is a prototype pattern
Orienting SubsystemCauses a reset when expectation does not match input patternDisables current winning neuron
Ming-Feng Yeh 11
Layer 1Layer 1
Ming-Feng Yeh 12
Layer 1 OperationLayer 1 Operation
Equation of operation of Layer 1:
Output of Layer 1:
)(][)()()()()( 211121:2111
1
tttttdt
tdaWnbaWpnbn
n
Excitatory input:Input pattern + L1-L2 expectation
Inhibitory input:Gain control from L2
)( 11 nhardlima
0,00,1
)(nn
nhardlim
Ming-Feng Yeh 13
Excitatory Input to L1Excitatory Input to L1
The excitatory input:
Assume that the jth neuron in Layer 2 has won the competition, i.e.,
The excitatory input to Layer 1 is the sum of the input pattern and the L2-L1 expectation.
)(21:2 taWp
.,0,1 22 jkaa kj
1:21:21:21:22
1:21
21:2
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2 jSj wwwwwaW
1:221:2jwpaWp
Ming-Feng Yeh 14
Inhibitory Input to L1Inhibitory Input to L1
The inhibitory input – the gain control
The inhibitory input to each neuron in Layer 1 is the sum of all of the outputs of Layer 2.
The gain control to Layer 1 will be one when Layer 2 is active (one neuron has won the competition), and zero when Layer 2 is inactive (all neurons having zero output).
21][ aW
111
111111
1
W
Ming-Feng Yeh 15
Steady State Analysis -- 1Steady State Analysis -- 1
The response of neuron i in Layer 1:
Case 1: Layer 2 is inactive – each
In steady state:If thenIf thenThe output of Layer 1 is the same as the input pattern
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1
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Ming-Feng Yeh 16
Steady State Analysis -- 2Steady State Analysis -- 2
Case 2: Layer 2 is active – and
In steady state:
Layer 1 is to combine the input vector with the expectation from Layer 2. Since both the input and the expectation are binary pattern, we will use a logic AND operation to combine the two vectors. if either or is equal to 0 if both and are equal to 1
12 ja .,02 jkak
111:2,
1111
ijiiiii nbwpnbn
dt
dn
01
dt
dni1:2
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11:2,
11
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jii
jiii
wp
bwpbn
01 in ip 1:2, jiw
01 in ip 1:2, jiw 02 11 bb
011 bb
1112 bbb 1:21jwpa
Ming-Feng Yeh 17
Layer 1 ExampleLayer 1 Example
Let
Assume that Layer 2 is active and neuron 2 of Layer 2 wins the competition.
10
,1011
,5.1,1,1.0 1:211 pWbb
22
1
211
1
21:2,
1111 S
jji
S
jjjiiii
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dn
5305.03)5.1(10)1(1.0 11
111
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11 n
dt
dnnnnn
dt
dn
5405.04)5.1(11)1(1.0 11
121
212
12
12
12 n
dt
dnnnnn
dt
dn
Ming-Feng Yeh 18
Response of Layer 1Response of Layer 1
0 0.05 0.1 0.15 0.2-0.2
-0.1
0
0.1
0.2
n11
t 16---– 1 e
30t–– =
n21
t 18--- 1 e
40t–– =
11:22 1
011
10
awp
Ming-Feng Yeh 19
Layer 2Layer 2
From the orienting subsyste
m
Ming-Feng Yeh 20
Layer 2 OperationLayer 2 Operation
Equation of operation of Layer 2:
The rows of adaptive weights , after training, will represent the prototype patterns.
)(][)(
)(][)()()(
22222
12:12222222
tt
tttdt
td
nfWnb
aWnfWnbnn
on-center feedbackadaptive instar
excitatory input
off-surround feedbackinhibitory input
2:1W
otherwise,0
])[(max)(if,1 12:112:12
awaw Tj
j
Ti
ia
Ming-Feng Yeh 21
Layer 2 ExampleLayer 2 Example
Let
015.05.0
)()(
,11
,11
,1.0 2:12
2:112:122
T
T
ww
Wbb
0,,00,)(10)(
22
nnnnf
)()(1)()()(1)()(
1.0 22
221
12:11
21
221
21
21 tnftntnftntndt
tdn T aw
)()(1)()()(1)()(
1.0 21
222
12:12
22
222
22
22 tnftntnftntndt
tdn T aw
Ming-Feng Yeh 22
Response of Layer 2Response of Layer 2
0 0.05 0.1 0.15 0.2
-1
-0.5
0
0.5
1
w1:22
Ta1
n22
t
w1:21
Ta1
n12
t
t
102a
011a
Ming-Feng Yeh 23
Orienting SubsystemOrienting Subsystem
Determine if there is a sufficient match between the L2-L1 expectation (a1) and the input pattern (p)
Ming-Feng Yeh 24
Orienting Subsyst. Operat.Orienting Subsyst. Operat.
Equation of operation of the Orienting Subsystem:
excitatory input:
inhibitory input:
Whenever the excitatory input is larger than the inhibitory input, the Orienting Subsystem will be driven on.
100000000
)()()()(
aWpW tnbtnbtndt
tdn
excitatory input inhibitory input
2
1
01
pppW
S
jjp
21
1
11101
)( aaaW
S
jj ta
Ming-Feng Yeh 25
Steady State OperationSteady State Operation
Steady state:
Let , then if , or
if (vigilance)
The condition that will cause a reset of Layer 2.
210200212
21002000
1
}){(}){(0
apap
ap
bbn
nbnbn
212
21020
0
1 ap
ap
bbn
100 bb 00 n 0212 ap
00 n
2
21
p
a
Ming-Feng Yeh 26
Vigilance ParameterVigilance Parameter
. The term is called the vigilance
parameter and must fall in the range
If is close to 1, a reset will occur unless is close to
If is close to 0, need not be close to to present a reset.
, whenever Layer 2 is active.The orienting subsystem will cause a reset when there is enough of a mismatch between and
2
21
p
a
10 1a p
1a p
1:21jwpa
212ap
p 1:2jw
otherwise,0
if,1221
0 paa
Ming-Feng Yeh 27
Orienting Subsystem Ex.Orienting Subsystem Ex.
Suppose that
In this case a reset signalwill be sent to Layer 2,since is positive.
01
,11
),75.0(4,3,1.0 1ap
0 0.05 0.1 0.15 0.2-0.2
-0.1
0
0.1
0.2
t
n0
t )}(4{)(1
)}(3{)(1
)()(
1.0
12
11
021
0
00
aatn
pptn
tndt
tdn
20)(110)( 0
0
tndt
tdn
)(0 tn
Ming-Feng Yeh 28
Learning LawLearning Law
Two separate learning laws:one for the L1-L2 connections,(instar) and another for L2-L1connections (outstar).
Both L1-L2 connections and L2-L1 connections are updated at the same time.Whenever the input and theexpectation have an adequate match.
The process of matching, and subsequentadaptation, is referred to as resonance.
2:1W1:2W
Ming-Feng Yeh 29
Subset / Superset DilemmaSubset / Superset Dilemma
Suppose that ,
so that the prototype patterns are
If the output of Layer 1 isthen the input to Layer 2 will be
Both prototype vectors have the same inner product with a1, even though the 1st prototype is identical to a1 and the 2nd prototype is not.This is called subset/superset dilemma.
1110112:1W
2:12
2:112:1
2
2:11
111
011ww
w
w
T
T
T0111 a
2212:1 aW
Ming-Feng Yeh 30
Subset / Superset SolutionSubset / Superset Solution
One solution to the subset/superset dilemma is to normalize the prototype patterns.
The input to Layer 2 will then be
The first prototype has the largest inner product with a1. The first neuron in Layer 2 will be active.
313131021212:1W
32112:1 aW
Ming-Feng Yeh 31
Learning Law: L1-L2Learning Law: L1-L2
Instar learning with competition:
When neuron i of Layer 2 is active, the ith row of , , is moved in the direction of a1. The learning law is that the elements of compete, and thereforeis normalized.
)(])}[({)(][)}({)()]([ 12:112:12
2:1
tttttadt
tdiii
i aWwbaWwbw
011
101110
,
100
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0
00
,
1
11
WWbb
2:1W2:1
1w2:1
1w
Ming-Feng Yeh 32
Fast LearningFast Learning
For fast learning, we assume that the outputs of Layer 1 and Layer 2 remain constant until the weights reach steady state.
assume that and setCase 1:
Case 2:
Summary:
jk
kjijjiiji tatwtatwta
dt
tdw)()()()}(1{)(
)( 12:1,
12:1,
22:1
,
1)(2 tai 0)(2:1, dttdw ji
1)(2 ta j
2:1,
21212:1,
2:1, )1()1()1(0 jijiji www aa
121
2:1,
a
jiw
0)(2 ta j
212:1
,0 ajiw 02:1, jiw
1,1
21
12:1
,
a
ajiw
Ming-Feng Yeh 33
Learning Law: L2-L1Learning Law: L2-L1
Typical outstar learning:
If neuron j in Layer 2 is active (has won the competition), then column j of is moved toward a1.
Fast learning: assume that andColumn j of converges to the output of Layer 1, a1, which is a combination of the input pattern and the appropriate prototype pattern. The prototype pattern is modified to incorporate the current input pattern.
)()()()]([ 11:22
1:2
tttadt
tdjj
j aww
1:2W
12 ja 0w
dt
td j )]([ 1:2
11:2 aw j
1:2W
Ming-Feng Yeh 34
ART1 Algorithm SummaryART1 Algorithm Summary
0. Initialization: The initial is set to all 1’s. Every elements of the initial is set to .
1. Present an input pattern to the network.Since Layer 2 is NOT active on initialization, the output of Layer 1 is .
2. Compute the input to Layer 2, , and activate the neuron in Layer 2 with the largest input
In case of tie, the neuron with the smallest index is declared the winner.
1:2W2:1W )1( 1 S
pa 1
12:1 aW
otherwise,0
])[(max)(if,1 12:112:12
awaw Tj
j
Ti
ia
Ming-Feng Yeh 35
Algorithm Summary Cont.Algorithm Summary Cont.
3. Compute the L2-L1 expectation (assume that neuron j of Layer 2 is activated):
4. Layer 2 is active. Adjust the Layer 1 output to include the L2-L1 expectation:
5. Determine the degree of match between the input pattern and the expectation (Orienting Subsystem):
6. If , then set , inhibit it until an adequate match occurs (resonance), and return to step 1.If , then continue with step 7.
1:221:2jwaW
1:21jwpa
otherwise,0
if,1221
0 paa
10 a 02 ja
00 a
Ming-Feng Yeh 36
Algorithm Summary Cont.Algorithm Summary Cont.
7. Update row j of when resonance has occurred:
8. Update column j of :
9. Remove the input pattern, restore all inhibited neurons in Layer 2, and return to step 1.
The input patterns continue to be applied to the network until the weights stabilize (do not change).
ART1 network can only be used for binary input patterns.
2:1W
1,1
21
12:1
,
a
ajiw
1:2W11:2 aw j
Ming-Feng Yeh 37
Solved Problem: P16.5Solved Problem: P16.5
Train an ART1 network using the parameters and , and choosing (3 categories), and using the
following three input vectors:
Initial weights:
1-1: Compute the Layer 1 response:
24.0 32 S
011
,001
,010
321 ppp
5.05.05.05.05.05.05.05.05.0
,111111111
2:11:2 WW
010
11 pa
Ming-Feng Yeh 38
P16.5 ContinuedP16.5 Continued
1-2: Compute the input to Layer 2
Since all neurons have the same input, pick the first neuron as winner.
1-3: Compute the L2-L1 expectation
5.05.05.0
010
5.05.05.05.05.05.05.05.05.0
12:1 aW
T0012 a
1:21
21:2
111
001
111111111
waW
Ming-Feng Yeh 39
P16.5 ContinuedP16.5 Continued
1-4: Adjust the Layer 1 output to include the expectation
1-5: Determine the match degree: Therefore (no reset)
1-6: Since , continued with step 7.
1-7: Resonance has occurred, update row 1 of
010
111
010
1:211
1 wpa
4.0112
1
21 pa
00 a
00 a2:1W
5.05.05.05.05.05.0
010
010
12
2 2:1121
12:1
1 Waa
aw
Ming-Feng Yeh 40
P16.5 ContinuedP16.5 Continued
1-8: Update column 1 of :
2-1: Compute the new Layer 1 response
(Layer 2 inactive):
2-2: Compute the input to Layer 2:
Since neurons 2 and 3 have the same input, pick the second neuron as winner:
1:2W
110111110
010
1:211:21 Waw
001
21 pa
5.05.0
0
001
5.05.05.05.05.05.0
01012:1 aW
T0102 a
Ming-Feng Yeh 41
P16.5 ContinuedP16.5 Continued
2-3: Compute the L2-L1 expectation:
2-4: Adjust the Layer 1 output to include the expectation
2-5: Determine the match degree: Therefore (no reset)
2-6: Since , continued with step 7.
111
1:22
21:2 waW
001
111
001
1:222
1 wpa
4.0112
1
21 pa
00 a
00 a
Ming-Feng Yeh 42
P16.5 ContinuedP16.5 Continued
2-7: Resonance has occurred, update row 2 of
2-8: Update column 2 of :
3-1: Compute the new Layer 1 response:
3-2: Compute the input to Layer 2:
2:1W
5.05.05.0001010
001
12
2 2:1121
12:1
2 Waa
aw
1:2W
100101110
001
1:211:22 Waw
011
31 pa
001
111
011
5.05.05.0001010
212:1 aaW
Ming-Feng Yeh 43
P16.5 ContinuedP16.5 Continued
3-3: Compute the L2-L1 expectation:
3-4: Adjust the Layer 1 output to include the expectation
3-5: Determine the match degree: Therefore (no reset)
3-6: Since , continued with step 7.
010
1:22
21:2 waW
010
010
011
1:213
1 wpa
4.0212
3
21 pa
00 a
00 a
Ming-Feng Yeh 44
P16.5 ContinuedP16.5 Continued
3-7: Resonance has occurred, update row 1 of
3-8: Update column 2 of :
This completes the training, since if you apply any of the three patterns again they will not change the weights. These patterns have been successfully clustered.
2:1W
5.05.05.0001010
010
12
2 2:1121
12:1
1 Waa
aw
1:2W
100101110
010
1:211:22 Waw
Ming-Feng Yeh 45
Solved Problem: P16.6Solved Problem: P16.6
Repeat Problem P16.5, but change the vigilance parameter to .
The training will proceed exactly as in Problem P16.5, until pattern p3 is presented.
3-1: Compute the Layer 1 response:
3-2: Compute the input to Layer 2:
6.0
011
31 pa
001
111
011
5.05.05.0001010
212:1 aaW
Ming-Feng Yeh 46
P16.6 ContinuedP16.6 Continued
3-3: Compute the L2-L1 expectation:
3-4: Adjust the Layer 1 output to include the expectation
3-5: Determine the match degree: Therefore (reset)
3-6: Since , set , inhibit it until an adequate match occurs (resonance), and return to step 1.
010
1:22
21:2 waW
010
010
011
1:213
1 wpa
6.0212
3
21 pa
10 a
10 a 021 a
Ming-Feng Yeh 47
P16.6 ContinuedP16.6 Continued
4-1: Recompute the Layer 1 response:
(Layer 2 inactive)
4-2: Compute the input to Layer 2:
Since neuron 1 is inhibited, neuron 2 is the winner:
4-3: Compute the L2-L1 expectation:
4-4: Adjust the Layer 1 output to include the expectation
011
31 pa
111
011
5.05.05.0001010
12:1 aW
T0102 a
001
1:22
21:2 waW
001
001
011
1:223
1 wpa
Ming-Feng Yeh 48
P16.6 ContinuedP16.6 Continued
4-5: Determine the match degree: Therefore (reset)
4-6: Since , set , inhibit it until an adequate match occurs (resonance), and return to step 1.
5-1: Recompute the Layer 1 response:
5-2: Compute the input to Layer 2:
Since neurons 1 & 2 are inhibited,
neuron 3 is the winner:
6.0212
3
21 pa
10 a
10 a 022 a
011
31 pa
111
011
5.05.05.0001010
12:1 aW
T1002 a
Ming-Feng Yeh 49
P16.6 ContinuedP16.6 Continued
5-3: Compute the L2-L1 expectation:
5-4: Adjust the Layer 1 output to include the expectation
5-5: Determine the match degree: Therefore (no reset)
5-6: Since , continued with step 7.
111
1:23
21:2 waW
011
111
011
1:233
1 wpa
6.0222
3
21 pa
00 a
00 a
Ming-Feng Yeh 50
P16.6 ContinuedP16.6 Continued
5-7: Resonance has occurred, update row 3 of
5-8: Update column 2 of :
This completes the training, since if you apply any of the three patterns again they will not change the weights. These patterns have been successfully clustered.
2:1W
03232001010
03232
12
2 2:1132
21
12:1
1 Waa
aw
1:2W
000101110
011
1:211:23 Waw
Ming-Feng Yeh 51
Solved Problem: P16.7Solved Problem: P16.7
Train an ART1 network using the following input vectors.Present the vectors in the order p1-p2-p3-p1-p4. Use theparameters and , and choose (threecategories). Train the network until the weights haveconverged.
The initial matrix is an matrix of 1’s.
The initial matrix is an matrix, with equal to
blue square 1; white square 0
55 grids 25-dimensional vectors
2 6.0 32 S
1:2W 32521 SS2:1W 25312 SS
0769.0)1252(2)1( 1 S
Ming-Feng Yeh 52
P16.7 ContinuedP16.7 Continued
Training sequence: p1-p2-p3-p1-p4
: resonance
√: reset