Slide 1

Lecture 2-3-4

ASSOCIATIONS RULES AND MACHINES

CONCEPT OF AN E-MACHINEsimulating symbolic readwrite memory by changing dynamical attributes of data in a long-term memory

Victor Eliashberg

Consulting professor Stanford University Department of Electrical Engineering

BDW

Human-like robot (DB)External world W

External system (WD)Sensorimotor devices D

Computing system B simulatingthe work of human nervous system

Slide 2

ldquoWhen you have eliminated the impossible whatever remains however improbable must be the truthrdquo (Sherlock Holmes)

SCIENTIFIC EGINEERING APPROACH

ZERO-APPROXIMATION MODEL

Slide 3

s(ν+1)

s(ν)

BIOLOGICAL INTERPRETATION

Slide 4

Working memory episodic memory and mental imagery

ASAM

Motor control

PROBLEM 1 LEARNING TO SIMULATE the Teacher This problem is simple system AM needs to learn a manageable number of fixed rules

Slide 5

NMy

Teacher

AM

symbol read

X11

X12

0y

current state of mind next state of mind

movetype symbol

X y

sel

NM1

PROBLEM 2 LEARNING TO SIMULATE EXTERNAL SYSTEM This problem is hard the number of fixed rules needed to represent a RAM with n locations explodes exponentially with n

Slide 6

y

1

2

NS

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

Programmable logic array (PLA) a logic implementation of a local associative memory (solves problem 1 from slide 5)

Slide 7

BASIC CONCEPTS FROM THE AREA OF ARTIFICIAL NEURAL NETWORKS

Slide 8

Typical neuron

Neuron is a very specialized cell There are several types of neurons with different shapes and different types of membrane proteins Biological neuron is a complex functional unit However it is helpful to start with a simple artificial neuron (next slide)

Slide 9

Neuron as the first-order linear threshold element

x1

xk xmgkg1 gm

y

u

Inputs xk RrsquoOutput y Rrsquo

Parameters g1hellip gm Rrsquo

y=L( u )

(1)

(2)

L( u) = u if u gt 00 otherwise

(3)

Equations

dudt + u =

mΣ gkxk k=1

τ

where

u0

y=L( u )

Rrsquo is the set of real non-negative numbers

u

x1xk

xm

g1

gk

gm

y

s

A more convenient notation

xk is the k-th component of input vectorgk is the gain (weight) of the k-th synapses =

Σ gkxk

m

k=1

is the total postsynaptic current

u is the postsynaptic potentialy is the neuron outputτ is the time constant of the neuron

τ

Slide 10

Input synaptic matrix input long-term memory (ILTM) and DECODING

si =

Σ gxikxk

m

k=1

i=1hellipn

(1) fdec X times Gx S (2)

An abstract representation of (1)

x1xkxm

s1 si sn

gx1

k DECODING (computing similarity)

x

s1 si sn

ILTMgxik gx

nk

Notation

x=(x1 xm) are the signals from input neurons (not shown)

gx = (gxik) i=1hellipn k=1hellipm is the matrix of synaptic gains -- we

postulate that this matrix represents input long-term memory (ILTM)s=(s1 sn) is the similarity function

Slide 11

Layer with inhibitory connections as the mechanism of the winner-take-all (WTA) choice

Slide 12

Note Small white and black circles represent excitatory and inhibitory synapses respectively

s1

d1

α

β

si

di

α

β

sn

dn

α

β

uiu1un

xinhq

τττ Equations

(1)

(2)

(3)

iwin i si=max sj gt 0

( j )

if (i == iwin) di=1 else di=0

(4)

(5)

Procedural representationRANDOM CHOICE

s1 si sn

iwin

ldquo ldquo denotes random equally probable choice

Output synaptic matrix output long-term memory (OLTM) and ENCODING

y1ykyp gy

ki

d1 di dn

gykngy

k

1

NOTATION

d=(d1 dm) signals from the WTA layer (see previous slide) gy = (gy

ki) i=1hellipn k=1hellipm is the matrix of synaptic gains -- we postulate that this matrix represents output long-term memory (OLTM)y=(y1 yp) output vector

OLTM

ENCODING (data retrieval)

y

d1 di dn

yk =

Σ gykidi

i=1k=1hellipp (1) fenc D times Gy Y (2)

An abstract representation of (1)n

Slide 13

A neural implementation of a local associative memory (solves problem 1 from slide 5) (WTAEXE)

Slide 14

DECODING

ENCODING

RANDOM CHOICE

Input long-term memory (ILTM)

Output long-term memory (OLTM)

addressing by content

retrieval

S21(Ij)

N1(j)

S21(ij)

A functional model of the previous network [7][8][11]

(WTAEXE)

(1)

(2)

(3)

(4)

(5)

Slide 15

Slide 16

HOW CAN WE SOLVE THE HARD PROBLEM 2 from slide 6

Slide 17

External system as a generalized RAM

Slide 18

Concept of a generalized RAM (GRAM)

Slide 18Slide 19

Slide 20

Representation of local associative memory in terms of three ldquoone-steprdquo procedures DECODING CHOICE ENCODING

INTERPRETATION PROCEDURE

Slide 21

At the stage of training sel=1 at the stage of examination sel=0System AS simply ldquotape-recordsrdquo its experience (x1x2xy)(0ν)

Slide 22

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

y

1

2

NS

GRAM

EXPERIMENT 1 Fixed rules and variable rules

Slide 23

EXPERIMENT 1 (continued 1)

Slide 24

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

BDW

Human-like robot (DB)External world W

External system (WD)Sensorimotor devices D

Computing system B simulatingthe work of human nervous system

Slide 2

ldquoWhen you have eliminated the impossible whatever remains however improbable must be the truthrdquo (Sherlock Holmes)

SCIENTIFIC EGINEERING APPROACH

ZERO-APPROXIMATION MODEL

Slide 3

s(ν+1)

s(ν)

BIOLOGICAL INTERPRETATION

Slide 4

Working memory episodic memory and mental imagery

ASAM

Motor control

PROBLEM 1 LEARNING TO SIMULATE the Teacher This problem is simple system AM needs to learn a manageable number of fixed rules

Slide 5

NMy

Teacher

AM

symbol read

X11

X12

0y

current state of mind next state of mind

movetype symbol

X y

sel

NM1

PROBLEM 2 LEARNING TO SIMULATE EXTERNAL SYSTEM This problem is hard the number of fixed rules needed to represent a RAM with n locations explodes exponentially with n

Slide 6

y

1

2

NS

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

Programmable logic array (PLA) a logic implementation of a local associative memory (solves problem 1 from slide 5)

Slide 7

BASIC CONCEPTS FROM THE AREA OF ARTIFICIAL NEURAL NETWORKS

Slide 8

Typical neuron

Neuron is a very specialized cell There are several types of neurons with different shapes and different types of membrane proteins Biological neuron is a complex functional unit However it is helpful to start with a simple artificial neuron (next slide)

Slide 9

Neuron as the first-order linear threshold element

x1

xk xmgkg1 gm

y

u

Inputs xk RrsquoOutput y Rrsquo

Parameters g1hellip gm Rrsquo

y=L( u )

(1)

(2)

L( u) = u if u gt 00 otherwise

(3)

Equations

dudt + u =

mΣ gkxk k=1

τ

where

u0

y=L( u )

Rrsquo is the set of real non-negative numbers

u

x1xk

xm

g1

gk

gm

y

s

A more convenient notation

xk is the k-th component of input vectorgk is the gain (weight) of the k-th synapses =

Σ gkxk

m

k=1

is the total postsynaptic current

u is the postsynaptic potentialy is the neuron outputτ is the time constant of the neuron

τ

Slide 10

Input synaptic matrix input long-term memory (ILTM) and DECODING

si =

Σ gxikxk

m

k=1

i=1hellipn

(1) fdec X times Gx S (2)

An abstract representation of (1)

x1xkxm

s1 si sn

gx1

k DECODING (computing similarity)

x

s1 si sn

ILTMgxik gx

nk

Notation

x=(x1 xm) are the signals from input neurons (not shown)

gx = (gxik) i=1hellipn k=1hellipm is the matrix of synaptic gains -- we

postulate that this matrix represents input long-term memory (ILTM)s=(s1 sn) is the similarity function

Slide 11

Layer with inhibitory connections as the mechanism of the winner-take-all (WTA) choice

Slide 12

Note Small white and black circles represent excitatory and inhibitory synapses respectively

s1

d1

α

β

si

di

α

β

sn

dn

α

β

uiu1un

xinhq

τττ Equations

(1)

(2)

(3)

iwin i si=max sj gt 0

( j )

if (i == iwin) di=1 else di=0

(4)

(5)

Procedural representationRANDOM CHOICE

s1 si sn

iwin

ldquo ldquo denotes random equally probable choice

Output synaptic matrix output long-term memory (OLTM) and ENCODING

y1ykyp gy

ki

d1 di dn

gykngy

k

1

NOTATION

d=(d1 dm) signals from the WTA layer (see previous slide) gy = (gy

ki) i=1hellipn k=1hellipm is the matrix of synaptic gains -- we postulate that this matrix represents output long-term memory (OLTM)y=(y1 yp) output vector

OLTM

ENCODING (data retrieval)

y

d1 di dn

yk =

Σ gykidi

i=1k=1hellipp (1) fenc D times Gy Y (2)

An abstract representation of (1)n

Slide 13

A neural implementation of a local associative memory (solves problem 1 from slide 5) (WTAEXE)

Slide 14

DECODING

ENCODING

RANDOM CHOICE

Input long-term memory (ILTM)

Output long-term memory (OLTM)

addressing by content

retrieval

S21(Ij)

N1(j)

S21(ij)

A functional model of the previous network [7][8][11]

(WTAEXE)

(1)

(2)

(3)

(4)

(5)

Slide 15

Slide 16

HOW CAN WE SOLVE THE HARD PROBLEM 2 from slide 6

Slide 17

External system as a generalized RAM

Slide 18

Concept of a generalized RAM (GRAM)

Slide 18Slide 19

Slide 20

Representation of local associative memory in terms of three ldquoone-steprdquo procedures DECODING CHOICE ENCODING

INTERPRETATION PROCEDURE

Slide 21

At the stage of training sel=1 at the stage of examination sel=0System AS simply ldquotape-recordsrdquo its experience (x1x2xy)(0ν)

Slide 22

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

y

1

2

NS

GRAM

EXPERIMENT 1 Fixed rules and variable rules

Slide 23

EXPERIMENT 1 (continued 1)

Slide 24

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

ZERO-APPROXIMATION MODEL

Slide 3

s(ν+1)

s(ν)

BIOLOGICAL INTERPRETATION

Slide 4

Working memory episodic memory and mental imagery

ASAM

Motor control

PROBLEM 1 LEARNING TO SIMULATE the Teacher This problem is simple system AM needs to learn a manageable number of fixed rules

Slide 5

NMy

Teacher

AM

symbol read

X11

X12

0y

current state of mind next state of mind

movetype symbol

X y

sel

NM1

PROBLEM 2 LEARNING TO SIMULATE EXTERNAL SYSTEM This problem is hard the number of fixed rules needed to represent a RAM with n locations explodes exponentially with n

Slide 6

y

1

2

NS

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

Programmable logic array (PLA) a logic implementation of a local associative memory (solves problem 1 from slide 5)

Slide 7

BASIC CONCEPTS FROM THE AREA OF ARTIFICIAL NEURAL NETWORKS

Slide 8

Typical neuron

Neuron is a very specialized cell There are several types of neurons with different shapes and different types of membrane proteins Biological neuron is a complex functional unit However it is helpful to start with a simple artificial neuron (next slide)

Slide 9

Neuron as the first-order linear threshold element

x1

xk xmgkg1 gm

y

u

Inputs xk RrsquoOutput y Rrsquo

Parameters g1hellip gm Rrsquo

y=L( u )

(1)

(2)

L( u) = u if u gt 00 otherwise

(3)

Equations

dudt + u =

mΣ gkxk k=1

τ

where

u0

y=L( u )

Rrsquo is the set of real non-negative numbers

u

x1xk

xm

g1

gk

gm

y

s

A more convenient notation

xk is the k-th component of input vectorgk is the gain (weight) of the k-th synapses =

Σ gkxk

m

k=1

is the total postsynaptic current

u is the postsynaptic potentialy is the neuron outputτ is the time constant of the neuron

τ

Slide 10

Input synaptic matrix input long-term memory (ILTM) and DECODING

si =

Σ gxikxk

m

k=1

i=1hellipn

(1) fdec X times Gx S (2)

An abstract representation of (1)

x1xkxm

s1 si sn

gx1

k DECODING (computing similarity)

x

s1 si sn

ILTMgxik gx

nk

Notation

x=(x1 xm) are the signals from input neurons (not shown)

gx = (gxik) i=1hellipn k=1hellipm is the matrix of synaptic gains -- we

postulate that this matrix represents input long-term memory (ILTM)s=(s1 sn) is the similarity function

Slide 11

Layer with inhibitory connections as the mechanism of the winner-take-all (WTA) choice

Slide 12

Note Small white and black circles represent excitatory and inhibitory synapses respectively

s1

d1

α

β

si

di

α

β

sn

dn

α

β

uiu1un

xinhq

τττ Equations

(1)

(2)

(3)

iwin i si=max sj gt 0

( j )

if (i == iwin) di=1 else di=0

(4)

(5)

Procedural representationRANDOM CHOICE

s1 si sn

iwin

ldquo ldquo denotes random equally probable choice

Output synaptic matrix output long-term memory (OLTM) and ENCODING

y1ykyp gy

ki

d1 di dn

gykngy

k

1

NOTATION

d=(d1 dm) signals from the WTA layer (see previous slide) gy = (gy

ki) i=1hellipn k=1hellipm is the matrix of synaptic gains -- we postulate that this matrix represents output long-term memory (OLTM)y=(y1 yp) output vector

OLTM

ENCODING (data retrieval)

y

d1 di dn

yk =

Σ gykidi

i=1k=1hellipp (1) fenc D times Gy Y (2)

An abstract representation of (1)n

Slide 13

A neural implementation of a local associative memory (solves problem 1 from slide 5) (WTAEXE)

Slide 14

DECODING

ENCODING

RANDOM CHOICE

Input long-term memory (ILTM)

Output long-term memory (OLTM)

addressing by content

retrieval

S21(Ij)

N1(j)

S21(ij)

A functional model of the previous network [7][8][11]

(WTAEXE)

(1)

(2)

(3)

(4)

(5)

Slide 15

Slide 16

HOW CAN WE SOLVE THE HARD PROBLEM 2 from slide 6

Slide 17

External system as a generalized RAM

Slide 18

Concept of a generalized RAM (GRAM)

Slide 18Slide 19

Slide 20

Representation of local associative memory in terms of three ldquoone-steprdquo procedures DECODING CHOICE ENCODING

INTERPRETATION PROCEDURE

Slide 21

At the stage of training sel=1 at the stage of examination sel=0System AS simply ldquotape-recordsrdquo its experience (x1x2xy)(0ν)

Slide 22

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

y

1

2

NS

GRAM

EXPERIMENT 1 Fixed rules and variable rules

Slide 23

EXPERIMENT 1 (continued 1)

Slide 24

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

BIOLOGICAL INTERPRETATION

Slide 4

Working memory episodic memory and mental imagery

ASAM

Motor control

PROBLEM 1 LEARNING TO SIMULATE the Teacher This problem is simple system AM needs to learn a manageable number of fixed rules

Slide 5

NMy

Teacher

AM

symbol read

X11

X12

0y

current state of mind next state of mind

movetype symbol

X y

sel

NM1

PROBLEM 2 LEARNING TO SIMULATE EXTERNAL SYSTEM This problem is hard the number of fixed rules needed to represent a RAM with n locations explodes exponentially with n

Slide 6

y

1

2

NS

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

Programmable logic array (PLA) a logic implementation of a local associative memory (solves problem 1 from slide 5)

Slide 7

BASIC CONCEPTS FROM THE AREA OF ARTIFICIAL NEURAL NETWORKS

Slide 8

Typical neuron

Neuron is a very specialized cell There are several types of neurons with different shapes and different types of membrane proteins Biological neuron is a complex functional unit However it is helpful to start with a simple artificial neuron (next slide)

Slide 9

Neuron as the first-order linear threshold element

x1

xk xmgkg1 gm

y

u

Inputs xk RrsquoOutput y Rrsquo

Parameters g1hellip gm Rrsquo

y=L( u )

(1)

(2)

L( u) = u if u gt 00 otherwise

(3)

Equations

dudt + u =

mΣ gkxk k=1

τ

where

u0

y=L( u )

Rrsquo is the set of real non-negative numbers

u

x1xk

xm

g1

gk

gm

y

s

A more convenient notation

xk is the k-th component of input vectorgk is the gain (weight) of the k-th synapses =

Σ gkxk

m

k=1

is the total postsynaptic current

u is the postsynaptic potentialy is the neuron outputτ is the time constant of the neuron

τ

Slide 10

Input synaptic matrix input long-term memory (ILTM) and DECODING

si =

Σ gxikxk

m

k=1

i=1hellipn

(1) fdec X times Gx S (2)

An abstract representation of (1)

x1xkxm

s1 si sn

gx1

k DECODING (computing similarity)

x

s1 si sn

ILTMgxik gx

nk

Notation

x=(x1 xm) are the signals from input neurons (not shown)

gx = (gxik) i=1hellipn k=1hellipm is the matrix of synaptic gains -- we

postulate that this matrix represents input long-term memory (ILTM)s=(s1 sn) is the similarity function

Slide 11

Layer with inhibitory connections as the mechanism of the winner-take-all (WTA) choice

Slide 12

Note Small white and black circles represent excitatory and inhibitory synapses respectively

s1

d1

α

β

si

di

α

β

sn

dn

α

β

uiu1un

xinhq

τττ Equations

(1)

(2)

(3)

iwin i si=max sj gt 0

( j )

if (i == iwin) di=1 else di=0

(4)

(5)

Procedural representationRANDOM CHOICE

s1 si sn

iwin

ldquo ldquo denotes random equally probable choice

Output synaptic matrix output long-term memory (OLTM) and ENCODING

y1ykyp gy

ki

d1 di dn

gykngy

k

1

NOTATION

d=(d1 dm) signals from the WTA layer (see previous slide) gy = (gy

ki) i=1hellipn k=1hellipm is the matrix of synaptic gains -- we postulate that this matrix represents output long-term memory (OLTM)y=(y1 yp) output vector

OLTM

ENCODING (data retrieval)

y

d1 di dn

yk =

Σ gykidi

i=1k=1hellipp (1) fenc D times Gy Y (2)

An abstract representation of (1)n

Slide 13

A neural implementation of a local associative memory (solves problem 1 from slide 5) (WTAEXE)

Slide 14

DECODING

ENCODING

RANDOM CHOICE

Input long-term memory (ILTM)

Output long-term memory (OLTM)

addressing by content

retrieval

S21(Ij)

N1(j)

S21(ij)

A functional model of the previous network [7][8][11]

(WTAEXE)

(1)

(2)

(3)

(4)

(5)

Slide 15

Slide 16

HOW CAN WE SOLVE THE HARD PROBLEM 2 from slide 6

Slide 17

External system as a generalized RAM

Slide 18

Concept of a generalized RAM (GRAM)

Slide 18Slide 19

Slide 20

Representation of local associative memory in terms of three ldquoone-steprdquo procedures DECODING CHOICE ENCODING

INTERPRETATION PROCEDURE

Slide 21

At the stage of training sel=1 at the stage of examination sel=0System AS simply ldquotape-recordsrdquo its experience (x1x2xy)(0ν)

Slide 22

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

y

1

2

NS

GRAM

EXPERIMENT 1 Fixed rules and variable rules

Slide 23

EXPERIMENT 1 (continued 1)

Slide 24

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

PROBLEM 1 LEARNING TO SIMULATE the Teacher This problem is simple system AM needs to learn a manageable number of fixed rules

Slide 5

NMy

Teacher

AM

symbol read

X11

X12

0y

current state of mind next state of mind

movetype symbol

X y

sel

NM1

PROBLEM 2 LEARNING TO SIMULATE EXTERNAL SYSTEM This problem is hard the number of fixed rules needed to represent a RAM with n locations explodes exponentially with n

Slide 6

y

1

2

NS

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

Programmable logic array (PLA) a logic implementation of a local associative memory (solves problem 1 from slide 5)

Slide 7

BASIC CONCEPTS FROM THE AREA OF ARTIFICIAL NEURAL NETWORKS

Slide 8

Typical neuron

Neuron is a very specialized cell There are several types of neurons with different shapes and different types of membrane proteins Biological neuron is a complex functional unit However it is helpful to start with a simple artificial neuron (next slide)

Slide 9

Neuron as the first-order linear threshold element

x1

xk xmgkg1 gm

y

u

Inputs xk RrsquoOutput y Rrsquo

Parameters g1hellip gm Rrsquo

y=L( u )

(1)

(2)

L( u) = u if u gt 00 otherwise

(3)

Equations

dudt + u =

mΣ gkxk k=1

τ

where

u0

y=L( u )

Rrsquo is the set of real non-negative numbers

u

x1xk

xm

g1

gk

gm

y

s

A more convenient notation

xk is the k-th component of input vectorgk is the gain (weight) of the k-th synapses =

Σ gkxk

m

k=1

is the total postsynaptic current

u is the postsynaptic potentialy is the neuron outputτ is the time constant of the neuron

τ

Slide 10

Input synaptic matrix input long-term memory (ILTM) and DECODING

si =

Σ gxikxk

m

k=1

i=1hellipn

(1) fdec X times Gx S (2)

An abstract representation of (1)

x1xkxm

s1 si sn

gx1

k DECODING (computing similarity)

x

s1 si sn

ILTMgxik gx

nk

Notation

x=(x1 xm) are the signals from input neurons (not shown)

gx = (gxik) i=1hellipn k=1hellipm is the matrix of synaptic gains -- we

postulate that this matrix represents input long-term memory (ILTM)s=(s1 sn) is the similarity function

Slide 11

Layer with inhibitory connections as the mechanism of the winner-take-all (WTA) choice

Slide 12

Note Small white and black circles represent excitatory and inhibitory synapses respectively

s1

d1

α

β

si

di

α

β

sn

dn

α

β

uiu1un

xinhq

τττ Equations

(1)

(2)

(3)

iwin i si=max sj gt 0

( j )

if (i == iwin) di=1 else di=0

(4)

(5)

Procedural representationRANDOM CHOICE

s1 si sn

iwin

ldquo ldquo denotes random equally probable choice

Output synaptic matrix output long-term memory (OLTM) and ENCODING

y1ykyp gy

ki

d1 di dn

gykngy

k

1

NOTATION

d=(d1 dm) signals from the WTA layer (see previous slide) gy = (gy

ki) i=1hellipn k=1hellipm is the matrix of synaptic gains -- we postulate that this matrix represents output long-term memory (OLTM)y=(y1 yp) output vector

OLTM

ENCODING (data retrieval)

y

d1 di dn

yk =

Σ gykidi

i=1k=1hellipp (1) fenc D times Gy Y (2)

An abstract representation of (1)n

Slide 13

A neural implementation of a local associative memory (solves problem 1 from slide 5) (WTAEXE)

Slide 14

DECODING

ENCODING

RANDOM CHOICE

Input long-term memory (ILTM)

Output long-term memory (OLTM)

addressing by content

retrieval

S21(Ij)

N1(j)

S21(ij)

A functional model of the previous network [7][8][11]

(WTAEXE)

(1)

(2)

(3)

(4)

(5)

Slide 15

Slide 16

HOW CAN WE SOLVE THE HARD PROBLEM 2 from slide 6

Slide 17

External system as a generalized RAM

Slide 18

Concept of a generalized RAM (GRAM)

Slide 18Slide 19

Slide 20

Representation of local associative memory in terms of three ldquoone-steprdquo procedures DECODING CHOICE ENCODING

INTERPRETATION PROCEDURE

Slide 21

At the stage of training sel=1 at the stage of examination sel=0System AS simply ldquotape-recordsrdquo its experience (x1x2xy)(0ν)

Slide 22

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

y

1

2

NS

GRAM

EXPERIMENT 1 Fixed rules and variable rules

Slide 23

EXPERIMENT 1 (continued 1)

Slide 24

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

PROBLEM 2 LEARNING TO SIMULATE EXTERNAL SYSTEM This problem is hard the number of fixed rules needed to represent a RAM with n locations explodes exponentially with n

Slide 6

y

1

2

NS

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

Programmable logic array (PLA) a logic implementation of a local associative memory (solves problem 1 from slide 5)

Slide 7

BASIC CONCEPTS FROM THE AREA OF ARTIFICIAL NEURAL NETWORKS

Slide 8

Typical neuron

Neuron is a very specialized cell There are several types of neurons with different shapes and different types of membrane proteins Biological neuron is a complex functional unit However it is helpful to start with a simple artificial neuron (next slide)

Slide 9

Neuron as the first-order linear threshold element

x1

xk xmgkg1 gm

y

u

Inputs xk RrsquoOutput y Rrsquo

Parameters g1hellip gm Rrsquo

y=L( u )

(1)

(2)

L( u) = u if u gt 00 otherwise

(3)

Equations

dudt + u =

mΣ gkxk k=1

τ

where

u0

y=L( u )

Rrsquo is the set of real non-negative numbers

u

x1xk

xm

g1

gk

gm

y

s

A more convenient notation

xk is the k-th component of input vectorgk is the gain (weight) of the k-th synapses =

Σ gkxk

m

k=1

is the total postsynaptic current

u is the postsynaptic potentialy is the neuron outputτ is the time constant of the neuron

τ

Slide 10

Input synaptic matrix input long-term memory (ILTM) and DECODING

si =

Σ gxikxk

m

k=1

i=1hellipn

(1) fdec X times Gx S (2)

An abstract representation of (1)

x1xkxm

s1 si sn

gx1

k DECODING (computing similarity)

x

s1 si sn

ILTMgxik gx

nk

Notation

x=(x1 xm) are the signals from input neurons (not shown)

gx = (gxik) i=1hellipn k=1hellipm is the matrix of synaptic gains -- we

postulate that this matrix represents input long-term memory (ILTM)s=(s1 sn) is the similarity function

Slide 11

Layer with inhibitory connections as the mechanism of the winner-take-all (WTA) choice

Slide 12

Note Small white and black circles represent excitatory and inhibitory synapses respectively

s1

d1

α

β

si

di

α

β

sn

dn

α

β

uiu1un

xinhq

τττ Equations

(1)

(2)

(3)

iwin i si=max sj gt 0

( j )

if (i == iwin) di=1 else di=0

(4)

(5)

Procedural representationRANDOM CHOICE

s1 si sn

iwin

ldquo ldquo denotes random equally probable choice

Output synaptic matrix output long-term memory (OLTM) and ENCODING

y1ykyp gy

ki

d1 di dn

gykngy

k

1

NOTATION

d=(d1 dm) signals from the WTA layer (see previous slide) gy = (gy

ki) i=1hellipn k=1hellipm is the matrix of synaptic gains -- we postulate that this matrix represents output long-term memory (OLTM)y=(y1 yp) output vector

OLTM

ENCODING (data retrieval)

y

d1 di dn

yk =

Σ gykidi

i=1k=1hellipp (1) fenc D times Gy Y (2)

An abstract representation of (1)n

Slide 13

A neural implementation of a local associative memory (solves problem 1 from slide 5) (WTAEXE)

Slide 14

DECODING

ENCODING

RANDOM CHOICE

Input long-term memory (ILTM)

Output long-term memory (OLTM)

addressing by content

retrieval

S21(Ij)

N1(j)

S21(ij)

A functional model of the previous network [7][8][11]

(WTAEXE)

(1)

(2)

(3)

(4)

(5)

Slide 15

Slide 16

HOW CAN WE SOLVE THE HARD PROBLEM 2 from slide 6

Slide 17

External system as a generalized RAM

Slide 18

Concept of a generalized RAM (GRAM)

Slide 18Slide 19

Slide 20

Representation of local associative memory in terms of three ldquoone-steprdquo procedures DECODING CHOICE ENCODING

INTERPRETATION PROCEDURE

Slide 21

At the stage of training sel=1 at the stage of examination sel=0System AS simply ldquotape-recordsrdquo its experience (x1x2xy)(0ν)

Slide 22

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

y

1

2

NS

GRAM

EXPERIMENT 1 Fixed rules and variable rules

Slide 23

EXPERIMENT 1 (continued 1)

Slide 24

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

Programmable logic array (PLA) a logic implementation of a local associative memory (solves problem 1 from slide 5)

Slide 7

BASIC CONCEPTS FROM THE AREA OF ARTIFICIAL NEURAL NETWORKS

Slide 8

Typical neuron

Neuron is a very specialized cell There are several types of neurons with different shapes and different types of membrane proteins Biological neuron is a complex functional unit However it is helpful to start with a simple artificial neuron (next slide)

Slide 9

Neuron as the first-order linear threshold element

x1

xk xmgkg1 gm

y

u

Inputs xk RrsquoOutput y Rrsquo

Parameters g1hellip gm Rrsquo

y=L( u )

(1)

(2)

L( u) = u if u gt 00 otherwise

(3)

Equations

dudt + u =

mΣ gkxk k=1

τ

where

u0

y=L( u )

Rrsquo is the set of real non-negative numbers

u

x1xk

xm

g1

gk

gm

y

s

A more convenient notation

xk is the k-th component of input vectorgk is the gain (weight) of the k-th synapses =

Σ gkxk

m

k=1

is the total postsynaptic current

u is the postsynaptic potentialy is the neuron outputτ is the time constant of the neuron

τ

Slide 10

Input synaptic matrix input long-term memory (ILTM) and DECODING

si =

Σ gxikxk

m

k=1

i=1hellipn

(1) fdec X times Gx S (2)

An abstract representation of (1)

x1xkxm

s1 si sn

gx1

k DECODING (computing similarity)

x

s1 si sn

ILTMgxik gx

nk

Notation

x=(x1 xm) are the signals from input neurons (not shown)

gx = (gxik) i=1hellipn k=1hellipm is the matrix of synaptic gains -- we

postulate that this matrix represents input long-term memory (ILTM)s=(s1 sn) is the similarity function

Slide 11

Layer with inhibitory connections as the mechanism of the winner-take-all (WTA) choice

Slide 12

Note Small white and black circles represent excitatory and inhibitory synapses respectively

s1

d1

α

β

si

di

α

β

sn

dn

α

β

uiu1un

xinhq

τττ Equations

(1)

(2)

(3)

iwin i si=max sj gt 0

( j )

if (i == iwin) di=1 else di=0

(4)

(5)

Procedural representationRANDOM CHOICE

s1 si sn

iwin

ldquo ldquo denotes random equally probable choice

Output synaptic matrix output long-term memory (OLTM) and ENCODING

y1ykyp gy

ki

d1 di dn

gykngy

k

1

NOTATION

d=(d1 dm) signals from the WTA layer (see previous slide) gy = (gy

ki) i=1hellipn k=1hellipm is the matrix of synaptic gains -- we postulate that this matrix represents output long-term memory (OLTM)y=(y1 yp) output vector

OLTM

ENCODING (data retrieval)

y

d1 di dn

yk =

Σ gykidi

i=1k=1hellipp (1) fenc D times Gy Y (2)

An abstract representation of (1)n

Slide 13

A neural implementation of a local associative memory (solves problem 1 from slide 5) (WTAEXE)

Slide 14

DECODING

ENCODING

RANDOM CHOICE

Input long-term memory (ILTM)

Output long-term memory (OLTM)

addressing by content

retrieval

S21(Ij)

N1(j)

S21(ij)

A functional model of the previous network [7][8][11]

(WTAEXE)

(1)

(2)

(3)

(4)

(5)

Slide 15

Slide 16

HOW CAN WE SOLVE THE HARD PROBLEM 2 from slide 6

Slide 17

External system as a generalized RAM

Slide 18

Concept of a generalized RAM (GRAM)

Slide 18Slide 19

Slide 20

Representation of local associative memory in terms of three ldquoone-steprdquo procedures DECODING CHOICE ENCODING

INTERPRETATION PROCEDURE

Slide 21

At the stage of training sel=1 at the stage of examination sel=0System AS simply ldquotape-recordsrdquo its experience (x1x2xy)(0ν)

Slide 22

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

y

1

2

NS

GRAM

EXPERIMENT 1 Fixed rules and variable rules

Slide 23

EXPERIMENT 1 (continued 1)

Slide 24

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

BASIC CONCEPTS FROM THE AREA OF ARTIFICIAL NEURAL NETWORKS

Slide 8

Typical neuron

Neuron is a very specialized cell There are several types of neurons with different shapes and different types of membrane proteins Biological neuron is a complex functional unit However it is helpful to start with a simple artificial neuron (next slide)

Slide 9

Neuron as the first-order linear threshold element

x1

xk xmgkg1 gm

y

u

Inputs xk RrsquoOutput y Rrsquo

Parameters g1hellip gm Rrsquo

y=L( u )

(1)

(2)

L( u) = u if u gt 00 otherwise

(3)

Equations

dudt + u =

mΣ gkxk k=1

τ

where

u0

y=L( u )

Rrsquo is the set of real non-negative numbers

u

x1xk

xm

g1

gk

gm

y

s

A more convenient notation

xk is the k-th component of input vectorgk is the gain (weight) of the k-th synapses =

Σ gkxk

m

k=1

is the total postsynaptic current

u is the postsynaptic potentialy is the neuron outputτ is the time constant of the neuron

τ

Slide 10

Input synaptic matrix input long-term memory (ILTM) and DECODING

si =

Σ gxikxk

m

k=1

i=1hellipn

(1) fdec X times Gx S (2)

An abstract representation of (1)

x1xkxm

s1 si sn

gx1

k DECODING (computing similarity)

x

s1 si sn

ILTMgxik gx

nk

Notation

x=(x1 xm) are the signals from input neurons (not shown)

gx = (gxik) i=1hellipn k=1hellipm is the matrix of synaptic gains -- we

postulate that this matrix represents input long-term memory (ILTM)s=(s1 sn) is the similarity function

Slide 11

Layer with inhibitory connections as the mechanism of the winner-take-all (WTA) choice

Slide 12

Note Small white and black circles represent excitatory and inhibitory synapses respectively

s1

d1

α

β

si

di

α

β

sn

dn

α

β

uiu1un

xinhq

τττ Equations

(1)

(2)

(3)

iwin i si=max sj gt 0

( j )

if (i == iwin) di=1 else di=0

(4)

(5)

Procedural representationRANDOM CHOICE

s1 si sn

iwin

ldquo ldquo denotes random equally probable choice

Output synaptic matrix output long-term memory (OLTM) and ENCODING

y1ykyp gy

ki

d1 di dn

gykngy

k

1

NOTATION

d=(d1 dm) signals from the WTA layer (see previous slide) gy = (gy

ki) i=1hellipn k=1hellipm is the matrix of synaptic gains -- we postulate that this matrix represents output long-term memory (OLTM)y=(y1 yp) output vector

OLTM

ENCODING (data retrieval)

y

d1 di dn

yk =

Σ gykidi

i=1k=1hellipp (1) fenc D times Gy Y (2)

An abstract representation of (1)n

Slide 13

A neural implementation of a local associative memory (solves problem 1 from slide 5) (WTAEXE)

Slide 14

DECODING

ENCODING

RANDOM CHOICE

Input long-term memory (ILTM)

Output long-term memory (OLTM)

addressing by content

retrieval

S21(Ij)

N1(j)

S21(ij)

A functional model of the previous network [7][8][11]

(WTAEXE)

(1)

(2)

(3)

(4)

(5)

Slide 15

Slide 16

HOW CAN WE SOLVE THE HARD PROBLEM 2 from slide 6

Slide 17

External system as a generalized RAM

Slide 18

Concept of a generalized RAM (GRAM)

Slide 18Slide 19

Slide 20

Representation of local associative memory in terms of three ldquoone-steprdquo procedures DECODING CHOICE ENCODING

INTERPRETATION PROCEDURE

Slide 21

At the stage of training sel=1 at the stage of examination sel=0System AS simply ldquotape-recordsrdquo its experience (x1x2xy)(0ν)

Slide 22

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

y

1

2

NS

GRAM

EXPERIMENT 1 Fixed rules and variable rules

Slide 23

EXPERIMENT 1 (continued 1)

Slide 24

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

Typical neuron

Neuron is a very specialized cell There are several types of neurons with different shapes and different types of membrane proteins Biological neuron is a complex functional unit However it is helpful to start with a simple artificial neuron (next slide)

Slide 9

Neuron as the first-order linear threshold element

x1

xk xmgkg1 gm

y

u

Inputs xk RrsquoOutput y Rrsquo

Parameters g1hellip gm Rrsquo

y=L( u )

(1)

(2)

L( u) = u if u gt 00 otherwise

(3)

Equations

dudt + u =

mΣ gkxk k=1

τ

where

u0

y=L( u )

Rrsquo is the set of real non-negative numbers

u

x1xk

xm

g1

gk

gm

y

s

A more convenient notation

xk is the k-th component of input vectorgk is the gain (weight) of the k-th synapses =

Σ gkxk

m

k=1

is the total postsynaptic current

u is the postsynaptic potentialy is the neuron outputτ is the time constant of the neuron

τ

Slide 10

Input synaptic matrix input long-term memory (ILTM) and DECODING

si =

Σ gxikxk

m

k=1

i=1hellipn

(1) fdec X times Gx S (2)

An abstract representation of (1)

x1xkxm

s1 si sn

gx1

k DECODING (computing similarity)

x

s1 si sn

ILTMgxik gx

nk

Notation

x=(x1 xm) are the signals from input neurons (not shown)

gx = (gxik) i=1hellipn k=1hellipm is the matrix of synaptic gains -- we

postulate that this matrix represents input long-term memory (ILTM)s=(s1 sn) is the similarity function

Slide 11

Layer with inhibitory connections as the mechanism of the winner-take-all (WTA) choice

Slide 12

Note Small white and black circles represent excitatory and inhibitory synapses respectively

s1

d1

α

β

si

di

α

β

sn

dn

α

β

uiu1un

xinhq

τττ Equations

(1)

(2)

(3)

iwin i si=max sj gt 0

( j )

if (i == iwin) di=1 else di=0

(4)

(5)

Procedural representationRANDOM CHOICE

s1 si sn

iwin

ldquo ldquo denotes random equally probable choice

Output synaptic matrix output long-term memory (OLTM) and ENCODING

y1ykyp gy

ki

d1 di dn

gykngy

k

1

NOTATION

d=(d1 dm) signals from the WTA layer (see previous slide) gy = (gy

ki) i=1hellipn k=1hellipm is the matrix of synaptic gains -- we postulate that this matrix represents output long-term memory (OLTM)y=(y1 yp) output vector

OLTM

ENCODING (data retrieval)

y

d1 di dn

yk =

Σ gykidi

i=1k=1hellipp (1) fenc D times Gy Y (2)

An abstract representation of (1)n

Slide 13

A neural implementation of a local associative memory (solves problem 1 from slide 5) (WTAEXE)

Slide 14

DECODING

ENCODING

RANDOM CHOICE

Input long-term memory (ILTM)

Output long-term memory (OLTM)

addressing by content

retrieval

S21(Ij)

N1(j)

S21(ij)

A functional model of the previous network [7][8][11]

(WTAEXE)

(1)

(2)

(3)

(4)

(5)

Slide 15

Slide 16

HOW CAN WE SOLVE THE HARD PROBLEM 2 from slide 6

Slide 17

External system as a generalized RAM

Slide 18

Concept of a generalized RAM (GRAM)

Slide 18Slide 19

Slide 20

Representation of local associative memory in terms of three ldquoone-steprdquo procedures DECODING CHOICE ENCODING

INTERPRETATION PROCEDURE

Slide 21

At the stage of training sel=1 at the stage of examination sel=0System AS simply ldquotape-recordsrdquo its experience (x1x2xy)(0ν)

Slide 22

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

y

1

2

NS

GRAM

EXPERIMENT 1 Fixed rules and variable rules

Slide 23

EXPERIMENT 1 (continued 1)

Slide 24

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

Neuron as the first-order linear threshold element

x1

xk xmgkg1 gm

y

u

Inputs xk RrsquoOutput y Rrsquo

Parameters g1hellip gm Rrsquo

y=L( u )

(1)

(2)

L( u) = u if u gt 00 otherwise

(3)

Equations

dudt + u =

mΣ gkxk k=1

τ

where

u0

y=L( u )

Rrsquo is the set of real non-negative numbers

u

x1xk

xm

g1

gk

gm

y

s

A more convenient notation

xk is the k-th component of input vectorgk is the gain (weight) of the k-th synapses =

Σ gkxk

m

k=1

is the total postsynaptic current

u is the postsynaptic potentialy is the neuron outputτ is the time constant of the neuron

τ

Slide 10

Input synaptic matrix input long-term memory (ILTM) and DECODING

si =

Σ gxikxk

m

k=1

i=1hellipn

(1) fdec X times Gx S (2)

An abstract representation of (1)

x1xkxm

s1 si sn

gx1

k DECODING (computing similarity)

x

s1 si sn

ILTMgxik gx

nk

Notation

x=(x1 xm) are the signals from input neurons (not shown)

gx = (gxik) i=1hellipn k=1hellipm is the matrix of synaptic gains -- we

postulate that this matrix represents input long-term memory (ILTM)s=(s1 sn) is the similarity function

Slide 11

Layer with inhibitory connections as the mechanism of the winner-take-all (WTA) choice

Slide 12

Note Small white and black circles represent excitatory and inhibitory synapses respectively

s1

d1

α

β

si

di

α

β

sn

dn

α

β

uiu1un

xinhq

τττ Equations

(1)

(2)

(3)

iwin i si=max sj gt 0

( j )

if (i == iwin) di=1 else di=0

(4)

(5)

Procedural representationRANDOM CHOICE

s1 si sn

iwin

ldquo ldquo denotes random equally probable choice

Output synaptic matrix output long-term memory (OLTM) and ENCODING

y1ykyp gy

ki

d1 di dn

gykngy

k

1

NOTATION

d=(d1 dm) signals from the WTA layer (see previous slide) gy = (gy

ki) i=1hellipn k=1hellipm is the matrix of synaptic gains -- we postulate that this matrix represents output long-term memory (OLTM)y=(y1 yp) output vector

OLTM

ENCODING (data retrieval)

y

d1 di dn

yk =

Σ gykidi

i=1k=1hellipp (1) fenc D times Gy Y (2)

An abstract representation of (1)n

Slide 13

A neural implementation of a local associative memory (solves problem 1 from slide 5) (WTAEXE)

Slide 14

DECODING

ENCODING

RANDOM CHOICE

Input long-term memory (ILTM)

Output long-term memory (OLTM)

addressing by content

retrieval

S21(Ij)

N1(j)

S21(ij)

A functional model of the previous network [7][8][11]

(WTAEXE)

(1)

(2)

(3)

(4)

(5)

Slide 15

Slide 16

HOW CAN WE SOLVE THE HARD PROBLEM 2 from slide 6

Slide 17

External system as a generalized RAM

Slide 18

Concept of a generalized RAM (GRAM)

Slide 18Slide 19

Slide 20

Representation of local associative memory in terms of three ldquoone-steprdquo procedures DECODING CHOICE ENCODING

INTERPRETATION PROCEDURE

Slide 21

At the stage of training sel=1 at the stage of examination sel=0System AS simply ldquotape-recordsrdquo its experience (x1x2xy)(0ν)

Slide 22

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

y

1

2

NS

GRAM

EXPERIMENT 1 Fixed rules and variable rules

Slide 23

EXPERIMENT 1 (continued 1)

Slide 24

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

Input synaptic matrix input long-term memory (ILTM) and DECODING

si =

Σ gxikxk

m

k=1

i=1hellipn

(1) fdec X times Gx S (2)

An abstract representation of (1)

x1xkxm

s1 si sn

gx1

k DECODING (computing similarity)

x

s1 si sn

ILTMgxik gx

nk

Notation

x=(x1 xm) are the signals from input neurons (not shown)

gx = (gxik) i=1hellipn k=1hellipm is the matrix of synaptic gains -- we

postulate that this matrix represents input long-term memory (ILTM)s=(s1 sn) is the similarity function

Slide 11

Layer with inhibitory connections as the mechanism of the winner-take-all (WTA) choice

Slide 12

Note Small white and black circles represent excitatory and inhibitory synapses respectively

s1

d1

α

β

si

di

α

β

sn

dn

α

β

uiu1un

xinhq

τττ Equations

(1)

(2)

(3)

iwin i si=max sj gt 0

( j )

if (i == iwin) di=1 else di=0

(4)

(5)

Procedural representationRANDOM CHOICE

s1 si sn

iwin

ldquo ldquo denotes random equally probable choice

Output synaptic matrix output long-term memory (OLTM) and ENCODING

y1ykyp gy

ki

d1 di dn

gykngy

k

1

NOTATION

d=(d1 dm) signals from the WTA layer (see previous slide) gy = (gy

ki) i=1hellipn k=1hellipm is the matrix of synaptic gains -- we postulate that this matrix represents output long-term memory (OLTM)y=(y1 yp) output vector

OLTM

ENCODING (data retrieval)

y

d1 di dn

yk =

Σ gykidi

i=1k=1hellipp (1) fenc D times Gy Y (2)

An abstract representation of (1)n

Slide 13

A neural implementation of a local associative memory (solves problem 1 from slide 5) (WTAEXE)

Slide 14

DECODING

ENCODING

RANDOM CHOICE

Input long-term memory (ILTM)

Output long-term memory (OLTM)

addressing by content

retrieval

S21(Ij)

N1(j)

S21(ij)

A functional model of the previous network [7][8][11]

(WTAEXE)

(1)

(2)

(3)

(4)

(5)

Slide 15

Slide 16

HOW CAN WE SOLVE THE HARD PROBLEM 2 from slide 6

Slide 17

External system as a generalized RAM

Slide 18

Concept of a generalized RAM (GRAM)

Slide 18Slide 19

Slide 20

Representation of local associative memory in terms of three ldquoone-steprdquo procedures DECODING CHOICE ENCODING

INTERPRETATION PROCEDURE

Slide 21

At the stage of training sel=1 at the stage of examination sel=0System AS simply ldquotape-recordsrdquo its experience (x1x2xy)(0ν)

Slide 22

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

y

1

2

NS

GRAM

EXPERIMENT 1 Fixed rules and variable rules

Slide 23

EXPERIMENT 1 (continued 1)

Slide 24

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

Layer with inhibitory connections as the mechanism of the winner-take-all (WTA) choice

Slide 12

Note Small white and black circles represent excitatory and inhibitory synapses respectively

s1

d1

α

β

si

di

α

β

sn

dn

α

β

uiu1un

xinhq

τττ Equations

(1)

(2)

(3)

iwin i si=max sj gt 0

( j )

if (i == iwin) di=1 else di=0

(4)

(5)

Procedural representationRANDOM CHOICE

s1 si sn

iwin

ldquo ldquo denotes random equally probable choice

Output synaptic matrix output long-term memory (OLTM) and ENCODING

y1ykyp gy

ki

d1 di dn

gykngy

k

1

NOTATION

d=(d1 dm) signals from the WTA layer (see previous slide) gy = (gy

ki) i=1hellipn k=1hellipm is the matrix of synaptic gains -- we postulate that this matrix represents output long-term memory (OLTM)y=(y1 yp) output vector

OLTM

ENCODING (data retrieval)

y

d1 di dn

yk =

Σ gykidi

i=1k=1hellipp (1) fenc D times Gy Y (2)

An abstract representation of (1)n

Slide 13

A neural implementation of a local associative memory (solves problem 1 from slide 5) (WTAEXE)

Slide 14

DECODING

ENCODING

RANDOM CHOICE

Input long-term memory (ILTM)

Output long-term memory (OLTM)

addressing by content

retrieval

S21(Ij)

N1(j)

S21(ij)

A functional model of the previous network [7][8][11]

(WTAEXE)

(1)

(2)

(3)

(4)

(5)

Slide 15

Slide 16

HOW CAN WE SOLVE THE HARD PROBLEM 2 from slide 6

Slide 17

External system as a generalized RAM

Slide 18

Concept of a generalized RAM (GRAM)

Slide 18Slide 19

Slide 20

Representation of local associative memory in terms of three ldquoone-steprdquo procedures DECODING CHOICE ENCODING

INTERPRETATION PROCEDURE

Slide 21

At the stage of training sel=1 at the stage of examination sel=0System AS simply ldquotape-recordsrdquo its experience (x1x2xy)(0ν)

Slide 22

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

y

1

2

NS

GRAM

EXPERIMENT 1 Fixed rules and variable rules

Slide 23

EXPERIMENT 1 (continued 1)

Slide 24

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

Output synaptic matrix output long-term memory (OLTM) and ENCODING

y1ykyp gy

ki

d1 di dn

gykngy

k

1

NOTATION

d=(d1 dm) signals from the WTA layer (see previous slide) gy = (gy

ki) i=1hellipn k=1hellipm is the matrix of synaptic gains -- we postulate that this matrix represents output long-term memory (OLTM)y=(y1 yp) output vector

OLTM

ENCODING (data retrieval)

y

d1 di dn

yk =

Σ gykidi

i=1k=1hellipp (1) fenc D times Gy Y (2)

An abstract representation of (1)n

Slide 13

A neural implementation of a local associative memory (solves problem 1 from slide 5) (WTAEXE)

Slide 14

DECODING

ENCODING

RANDOM CHOICE

Input long-term memory (ILTM)

Output long-term memory (OLTM)

addressing by content

retrieval

S21(Ij)

N1(j)

S21(ij)

A functional model of the previous network [7][8][11]

(WTAEXE)

(1)

(2)

(3)

(4)

(5)

Slide 15

Slide 16

HOW CAN WE SOLVE THE HARD PROBLEM 2 from slide 6

Slide 17

External system as a generalized RAM

Slide 18

Concept of a generalized RAM (GRAM)

Slide 18Slide 19

Slide 20

Representation of local associative memory in terms of three ldquoone-steprdquo procedures DECODING CHOICE ENCODING

INTERPRETATION PROCEDURE

Slide 21

At the stage of training sel=1 at the stage of examination sel=0System AS simply ldquotape-recordsrdquo its experience (x1x2xy)(0ν)

Slide 22

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

y

1

2

NS

GRAM

EXPERIMENT 1 Fixed rules and variable rules

Slide 23

EXPERIMENT 1 (continued 1)

Slide 24

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

A neural implementation of a local associative memory (solves problem 1 from slide 5) (WTAEXE)

Slide 14

DECODING

ENCODING

RANDOM CHOICE

Input long-term memory (ILTM)

Output long-term memory (OLTM)

addressing by content

retrieval

S21(Ij)

N1(j)

S21(ij)

A functional model of the previous network [7][8][11]

(WTAEXE)

(1)

(2)

(3)

(4)

(5)

Slide 15

Slide 16

HOW CAN WE SOLVE THE HARD PROBLEM 2 from slide 6

Slide 17

External system as a generalized RAM

Slide 18

Concept of a generalized RAM (GRAM)

Slide 18Slide 19

Slide 20

Representation of local associative memory in terms of three ldquoone-steprdquo procedures DECODING CHOICE ENCODING

INTERPRETATION PROCEDURE

Slide 21

At the stage of training sel=1 at the stage of examination sel=0System AS simply ldquotape-recordsrdquo its experience (x1x2xy)(0ν)

Slide 22

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

y

1

2

NS

GRAM

EXPERIMENT 1 Fixed rules and variable rules

Slide 23

EXPERIMENT 1 (continued 1)

Slide 24

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

A functional model of the previous network [7][8][11]

(WTAEXE)

(1)

(2)

(3)

(4)

(5)

Slide 15

Slide 16

HOW CAN WE SOLVE THE HARD PROBLEM 2 from slide 6

Slide 17

External system as a generalized RAM

Slide 18

Concept of a generalized RAM (GRAM)

Slide 18Slide 19

Slide 20

Representation of local associative memory in terms of three ldquoone-steprdquo procedures DECODING CHOICE ENCODING

INTERPRETATION PROCEDURE

Slide 21

At the stage of training sel=1 at the stage of examination sel=0System AS simply ldquotape-recordsrdquo its experience (x1x2xy)(0ν)

Slide 22

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

y

1

2

NS

GRAM

EXPERIMENT 1 Fixed rules and variable rules

Slide 23

EXPERIMENT 1 (continued 1)

Slide 24

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

Slide 16

HOW CAN WE SOLVE THE HARD PROBLEM 2 from slide 6

Slide 17

External system as a generalized RAM

Slide 18

Concept of a generalized RAM (GRAM)

Slide 18Slide 19

Slide 20

Representation of local associative memory in terms of three ldquoone-steprdquo procedures DECODING CHOICE ENCODING

INTERPRETATION PROCEDURE

Slide 21

At the stage of training sel=1 at the stage of examination sel=0System AS simply ldquotape-recordsrdquo its experience (x1x2xy)(0ν)

Slide 22

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

y

1

2

NS

GRAM

EXPERIMENT 1 Fixed rules and variable rules

Slide 23

EXPERIMENT 1 (continued 1)

Slide 24

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

Slide 17

External system as a generalized RAM

Slide 18

Concept of a generalized RAM (GRAM)

Slide 18Slide 19

Slide 20

Representation of local associative memory in terms of three ldquoone-steprdquo procedures DECODING CHOICE ENCODING

INTERPRETATION PROCEDURE

Slide 21

At the stage of training sel=1 at the stage of examination sel=0System AS simply ldquotape-recordsrdquo its experience (x1x2xy)(0ν)

Slide 22

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

y

1

2

NS

GRAM

EXPERIMENT 1 Fixed rules and variable rules

Slide 23

EXPERIMENT 1 (continued 1)

Slide 24

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

Slide 18

Concept of a generalized RAM (GRAM)

Slide 18Slide 19

Slide 20

Representation of local associative memory in terms of three ldquoone-steprdquo procedures DECODING CHOICE ENCODING

INTERPRETATION PROCEDURE

Slide 21

At the stage of training sel=1 at the stage of examination sel=0System AS simply ldquotape-recordsrdquo its experience (x1x2xy)(0ν)

Slide 22

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

y

1

2

NS

GRAM

EXPERIMENT 1 Fixed rules and variable rules

Slide 23

EXPERIMENT 1 (continued 1)

Slide 24

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

Slide 18Slide 19

Slide 20

Representation of local associative memory in terms of three ldquoone-steprdquo procedures DECODING CHOICE ENCODING

INTERPRETATION PROCEDURE

Slide 21

At the stage of training sel=1 at the stage of examination sel=0System AS simply ldquotape-recordsrdquo its experience (x1x2xy)(0ν)

Slide 22

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

y

1

2

NS

GRAM

EXPERIMENT 1 Fixed rules and variable rules

Slide 23

EXPERIMENT 1 (continued 1)

Slide 24

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

Slide 20

Representation of local associative memory in terms of three ldquoone-steprdquo procedures DECODING CHOICE ENCODING

INTERPRETATION PROCEDURE

Slide 21

At the stage of training sel=1 at the stage of examination sel=0System AS simply ldquotape-recordsrdquo its experience (x1x2xy)(0ν)

Slide 22

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

y

1

2

NS

GRAM

EXPERIMENT 1 Fixed rules and variable rules

Slide 23

EXPERIMENT 1 (continued 1)

Slide 24

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

INTERPRETATION PROCEDURE

Slide 21

At the stage of training sel=1 at the stage of examination sel=0System AS simply ldquotape-recordsrdquo its experience (x1x2xy)(0ν)

Slide 22

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

y

1

2

NS

GRAM

EXPERIMENT 1 Fixed rules and variable rules

Slide 23

EXPERIMENT 1 (continued 1)

Slide 24

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

At the stage of training sel=1 at the stage of examination sel=0System AS simply ldquotape-recordsrdquo its experience (x1x2xy)(0ν)

Slide 22

NOTE System (WD) shown in slide 3 has the properties of a random access memory (RAM)

y

1

2

NS

GRAM

EXPERIMENT 1 Fixed rules and variable rules

Slide 23

EXPERIMENT 1 (continued 1)

Slide 24

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

EXPERIMENT 1 Fixed rules and variable rules

Slide 23

EXPERIMENT 1 (continued 1)

Slide 24

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

EXPERIMENT 1 (continued 1)

Slide 24

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

EXPERIMENT 1 (continued 2)

Slide 25

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

A COMPLETE MEMORY MACHINE (CMM) SOLVES PROBLEM 2 but this solution can be easily falsified

Slide 26

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

Slide 27

GRAM as a state machine combinatorial explosion of the number of fixed rules

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

Slide 28

Concept of a primitive E-machine

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

Slide 29

s(i) gt c

(αlt 5)

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

Effect of a RAM wo a RAM buffer

abc

abc

abc

abc

1 2 3 4

b a c b

1 2 3 4

abc

abc

abc

abc

1 2 3 4

c b a c

1 2 3 4

a b c a b c a b c a b c

1 2 3 4

G-state

E-state

abc

abc

abc

abc

1 2 3 4

a c b a

1 2 3 4

Slide 30

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

EFFECT OF ldquoMANY MACHINES IN ONErdquo

Slide 31

0

00

1

00

0

10

1

10

1 2 3 4

0

01

1

01

0

11

1

11

5 6 7 8

X(1)

X(2)

y(1)

AND

OR

XOR

NAND

NOR

N=2m2

A table with n=2 m+1

represents

different m-input 1-output Boolean functions

Let m=10 Then n=2048

and N=21024

G-state

E-state

n=8 locations of LTM

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

Simulation of GRAM with A=12 and D=abε

Slide 32

a

a1

b

b1

b

b2

a

a2

1 2 3 4

b

1

a

2

5 6 7

addr

din

dout

i

s(i)

ν = 5

e(i)

se(i)se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

dout=b is read from i=2 that has se(i)=max(se)

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

Slide 33

i

s(i)

e(i)

se(i)

a

a1

b

b1

b

b2

a

a2

1 2 3 4

addr

din

dout

ν = 5se(i) = s(i) ( 1+a e(i) ) (alt5)

if ( s(i)gte(i) ) e(i)(ν+1) = s(i)(ν)

else e(i)(ν+1) = c e(i)(ν) τ=1(1-c)

s (i) is the number of matches in the first two rows Input (addrdin) = (1ε) produces s(i)=1 for i=1 and i=2

iwin i se(i)=max(se)gt0

Assume that the E-machine starts with the state of LTM shown in the table and doesnrsquot learn more so this state remains the same What changes is the E-state e(1)hellipe(4) Assume that at ν=1 e(1)=e(4)=0 Let us send the input sequence (addrdin)(15) = (1a) (1b)(2a)(2b)(1ε) As can be verified at ν = 5 the state e(i) and functions s(i) and se(i) for i=14 are as shown below Accordingly iwin=2 and dout=b

y =gy(iwin) (alt5)

gx(114)

gx(214)

gy(114)

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34

What can be efficiently computed in this ldquononclassicalrdquo symbolicdynamical computational paradigm (call it the E-machine paradigm)

What computational resources are available in the brain -- especially in the neocortex -- for the implementation of this paradigm

How can dynamical equations (such as the last equation in slide 29) be efficiently implemented in biologically plausible neural network models

Slide 34

• Slide 1
• Slide 2
• Slide 3
• Slide 4
• Slide 5
• Slide 6
• Slide 7
• Slide 8
• Slide 9
• Slide 10
• Slide 11
• Slide 12
• Slide 13
• Slide 14
• Slide 15
• Slide 16
• Slide 17
• Slide 18
• Slide 19
• Slide 20
• Slide 21
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Slide 28
• Slide 29
• Slide 30
• Slide 31
• Slide 32
• Slide 33
• Slide 34