llu s a. belanche [email protected]/docencia/apren/2009-10/teoria/tema9.pdf · ejemplo del...
TRANSCRIPT
![Page 1: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/1.jpg)
Hopfield networks
Lluıs A. [email protected]
Soft Computing Research Group
Dept. de Llenguatges i Sistemes Informatics (Software department)
Universitat Politecnica de Catalunya
2010-2011
![Page 2: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/2.jpg)
Hopfield networks
Introduction
Content-addressable autoassociative memory
Deals with incomplete/erroneous keys: “Julia xxxxazar: Rxxxela”H
ENERGYFUNCTION
Basins of attraction
Attractors (memories)
A mechanical system tends to minimum-energy states (e.g. a pole)
Symmetric Hopfield networks have an analogous behaviour:
low-energy states ⇔ attractors ⇔ memories
![Page 3: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/3.jpg)
Hopfield networks
Introduction
A Hopfield network is a single-
layer recurrent neural network,
where the only layer has n neurons
(as many as the input dimension)
![Page 4: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/4.jpg)
Hopfield networks
Introduction
The data sample is a list of keys (aka patterns) L = {ξi | 1 ≤ i ≤ l}, with
ξi ∈ {0,1}n that we want to associate with themselves.
What is the meaning of autoassociativity? when a new pattern Ψ is shown to
the network, the output is the key in L closest (in Hamming distance) to Ψ.
Evidently, this can be achieved by direct programming; yet in hardware
it is quite complex (and the Hopfield net does it quite distinctly)
We therefore set up two goals:
1. Content-addressable autoassociative memory (not easy)
2. Tolerance to small errors, that will be corrected (rather hard)
![Page 5: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/5.jpg)
Hopfield networks
Intuitive analysis
The state of the network is the vector of activations e of the n
neurons. The state space is {0,1}n:
ξ1
ξ2
ξ3
.
.
.
001
000010
100
111
011
101
110
When the network evolves, it “traverses” the hypercube
![Page 6: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/6.jpg)
Hopfield networks
Formal analysis (I)
For convenience, we change from {0,1} to {−1,+1}, by s = 2e − 1.
Dynamics of a neuron:
si := sgn
n∑
j=1
wijsj − θi
, with sgn(z) =
{
+1 : z ≥ 0−1 : z < 0
Let us simplify the analysis by using θi = 0, and uncorrelated keys ξi
(from a uniform distribution).
![Page 7: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/7.jpg)
Hopfield networks
Formal analysis (II)
How are neurons chosen to update their activations?
1. Synchronous: all at once
2. Asynchronous: there is a sequential order or a fair probability dis-tribution
The memories do not change; the end attractor (the recovered me-mory) does
We will assume asynchronous updating
When does the process stop?
−→ When there are no more changes in the network state (we are ina stable state)
![Page 8: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/8.jpg)
Hopfield networks
Formal analysis (III)
PROCEDURE:
1. Formula for setting the weights wij to store the keys ξi ∈ L
2. Sanity check that the keys ξi ∈ L are stable (they are autoassociated
with themselves)
3. Check that small perturbations of the keys ξi ∈ L are corrected (they
are associated with the closest pattern in L)
![Page 9: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/9.jpg)
Hopfield networks
Formal analysis (IV)
Case l = 1 (only one pattern ξ):
Present ξ to the net: si := ξi, 1 ≤ i ≤ n
Stability:
si := sgn
n∑
j=1
wijsj
= sgn
n∑
j=1
wijξj
= ... should be ... = ξi
One way to get this is: wij = αξiξj, α > 0 ∈ R (called Hebbian learning)
sgn
n∑
j=1
wijξj
= sgn
n∑
j=1
αξiξjξj
= sgn
n∑
j=1
αξi
= sgn (αnξi) = sgn(ξi) = ξi
![Page 10: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/10.jpg)
Hopfield networks
Formal analysis (V)
Take α = 1/n. This has a normalizing effect: ‖wi‖2 = 1/√
n
Therefore Wn×n = (wij) = 1nξ × ξ, i.e. wij = 1
nξiξj
If the perturbation consists in b flipped bits, where b ≤ (n − 1) ÷ 2,
then the sign of∑n
j=1 wijsj does not change!
=⇒ if the number of correct bits exceeds that of incorrect bits, the
network is able to correct the error bits
![Page 11: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/11.jpg)
Hopfield networks
Formal analysis (VI)
Case l ≥ 1:
Superposition: wij = 1n
l∑
k=1ξkiξkj and Wn×n = 1
n
l∑
k=1ξk × ξk
Defining En×l = [ξ1, . . . , ξl], we get Wn×n = 1nEEt
Note Wn×n is a symmetric matrix
![Page 12: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/12.jpg)
Hopfield networks
Formal analysis (VII)
Stability of a pattern ξv ∈ L:
Initialize the net to si := ξvi, 1 ≤ i ≤ n
Stability: si := sgn
(n∑
j=1
wijξvj
)
= sgn(hvi)
hvi =n∑
j=1
wijξvj = 1n
n∑
j=1
l∑
k=1
ξkiξkjξvj = 1n
n∑
j=1
(l∑
k=1,k 6=v
ξkiξkjξvj
)
+ ξvi ξvjξvj︸ ︷︷ ︸
+1
= ξvi +1n
n∑
j=1
l∑
k=1,k 6=v
ξkiξkjξvj = ξvi + crosstalk(v, i)
If |crosstalk(v, i)| < 1 then sgn(hvi) = sgn(ξvi + crosstalk(v, i)) = sgn(ξvi) = ξvi
![Page 13: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/13.jpg)
Hopfield networks
Formal analysis (and VIII)
Analogously to the case l = 1, if the number of correct bits of a
pattern ξv ∈ L exceeds that of incorrect bits, the network is able to
correct the bits in error
(the pattern ξv ∈ L is indeed a memory)
When is |crosstalk(v, i)| < 1? If l << n nothing is wrong, but as we
have l closer to n, the crosstalk term is quite strong compared to ξv:
there is no stability
The million euro question: given a network of n neurons, what is the
maximum value for l? What is the capacity of the network?
![Page 14: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/14.jpg)
Hopfield networks
Capacity analysis
For random uncorrelated patterns (for convenience):
lmax =
{ n2ln(n)+lnln(n)
: perfect recall
≈ 0,138n : small errors (≈ 1,6%)
Realistic patterns will not be random, though some coding can make
them more so
An alternative setting is to have negatively correlated patterns:
ξv · ξu ≤ 0, ∀v 6= u
This condition is equivalent to stating that the number of different bits is at least
the number of equal bits
![Page 15: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/15.jpg)
Hopfield networks
Spurious states
When some pattern ξ is stored, the pattern −ξ is also stored. In
consequence, all initial configurations of ξ where the majority of bits
are incorrect will end up in −ξ!
−→ Hardly surprising: from the point of view of −ξ, there are more correct bits than
incorrect bits
The network can be made more stable (less spurious states) if we set
wii = 0 for all i
Wn×n = 1n
l∑
k=1ξk × ξk − l
nIn×n = 1n(EEt − lIn×n)
![Page 16: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/16.jpg)
Hopfield networks
Example
ξ1 = (+1 − 1 + 1), ξ2 = (−1 + 1 − 1), E3×2 =
+1 −1−1 +1+1 −1
+1−1+1
(+1 − 1 + 1) =
+1 −1 +1−1 +1 −1+1 −1 +1
−1+1−1
(−1 + 1 − 1) =
+1 −1 +1−1 +1 −1+1 −1 +1
−→ 13
0 −2 +2−2 0 −2+2 −2 0
(+1 + 1 + 1)(−1 − 1 + 1)(+1 − 1 − 1)
−→ (+1 − 1 + 1),
(+1 + 1 − 1)(−1 − 1 − 1)(−1 + 1 + 1)
−→ (−1 + 1 − 1)
The network is able to correct 1 erroneous bit
![Page 17: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/17.jpg)
Hopfield networks
Practical applications (I)4. Otros paradigmas Conexionistas 2. Memorias asociativas: 20
Convergencia y capacidad de la red de Hopfield
ENERGIA DE LA RED O FUNCION DE LYAPUNOV: para
PROPIEDAD 1:
PROPIEDAD 2: ¡acotada!
PROPIEDAD 3:
CONVERGENCIA: La red de Hopfield converge en un numero finito de iteraciones.
CAPACIDAD:
Redes Neuronales Artificiales . 2001-2002 Facultat d’Informatica – UPV: 18 de diciembre de 2001
4. Otros paradigmas Conexionistas 2. Memorias asociativas: 21
Ejemplo del comportamiento de una red de Hopfield
Una red de 120 nodos (12x10)
CONJUNTO DE OCHO PATRONES
SECUENCIA DE SALIDA PARA LA ENTRADA "3" DISTORSIONADA
0 1 2 3 4 5 6 7
Redes Neuronales Artificiales . 2001-2002 Facultat d’Informatica – UPV: 18 de diciembre de 2001
4. Otros paradigmas Conexionistas 2. Memorias asociativas: 22
Memorias heteroasociativas: memoria bidireccional asociativa
Capa Y
Capa Xw ij
1 2 m
1 2 n
APRENDIZAJE DE LOS PESOS:Dado con ,
para y
Redes Neuronales Artificiales . 2001-2002 Facultat d’Informatica – UPV: 18 de diciembre de 2001
4. Otros paradigmas Conexionistas 2. Memorias asociativas: 23
Funcionamiento sıncrono de la BAM
Entrada:
Inicializacion: , y ,
Iteracion: Propagacion de a
Propagacion de a
Convergencia: No se produzcan cambios en ni en :
Salida:
Redes Neuronales Artificiales . 2001-2002 Facultat d’Informatica – UPV: 18 de diciembre de 2001
A Hopfield network that stores and recovers several digits
![Page 18: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/18.jpg)
Hopfield networks
Practical applications (II)
Ex: pattern-recognition– Handwritting recognition
– Reconeixement d’objectius militars
Una xarxa neu-ronal que haaprès a memo-ritzar formes otrossos de for-mes dels avionsserbis, desco-breix un aviómilitar amagatsota un avió civil.
A Hopfield network that stores fighter aircraft shapes is able to reconstruct them. The
network operates on the video images supplied by a camera on board a USAF bomber
![Page 19: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/19.jpg)
Hopfield networks
The energy function (I)
If the weights are symmetric, there exists a so-called energy function:
H(s) = −n∑
i=1
n∑
j>i
wijsisj
which is a function of the state of the system.
The central property of H is that it always decreases (or remains
constant) as the system evolves
Therefore the attractors (the stored memories) have to be the local
minima of the energy function
![Page 20: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/20.jpg)
Hopfield networks
The energy function (II)
Proposition 1:
A change of state in one neuron i yields a change ∆H < 0
Let hi =n∑
j=1wijsj. If neuron i changes is because si 6= sgn(hi)
Therefore ∆H = −(−si)hi − (−sihi) = 2sihi < 0
Proposition 2:
H(s) ≥ −n∑
i=1
n∑
j=1|wij| (the energy function is lower-bounded)
![Page 21: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/21.jpg)
Hopfield networks
The energy function (III)
The energy function H is also useful to derive the proposed weights
Consider the case of only one pattern ξ: we want H to be minimum when the correlationbetween the current state s and ξ is maximum and equal to 〈ξ, s〉2 = 〈ξ, ξ〉2 = ‖ξ‖2 = n
Therefore we choose H(s) = −K 1n〈ξ, s〉2, where K > 0 is a suitable constant
For the many-pattern case, we can try to make each of the ξi into local minima of H:
H(s) = −K1
n
n∑
v=1
〈ξv, s〉2 = −K1
n
n∑
v=1
(n∑
i=1
ξvisi
)2
= −K1
n
n∑
v=1
(n∑
i=1
ξvisi
)
n∑
j=1
ξvjsj
= −K
n∑
i=1
n∑
j=1
(
1
n
n∑
v=1
ξviξvj
)
sisj = −K
n∑
i=1
n∑
j=1
wijsisj = −1
2
n∑
i=1
n∑
j>i
wijsisj
![Page 22: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/22.jpg)
Hopfield networks
The energy function (and IV)
Exercises:
1. In the previous example,
a) Check that H(s) = 23(s1s2 − s2s3 + s1s3)
b) Show that H decreases as the network evolves towards the stored
patterns
2. Prove Proposition 2
![Page 23: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/23.jpg)
Hopfield networks
Spurious states (continued)
We have not shown that the explicitly stored memories are the only
attractors of the system
We have seen that the reversed patterns −ξ are also stored (they have
the same energy than ξ)
There are stable mixture states, linear combinations of an odd num-
ber of patterns, e.g.:
ξ(mix)i = sgn
α∑
k=1
±ξvk,i
, 1 ≤ i ≤ n, α odd
![Page 24: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/24.jpg)
Hopfield networks
Dynamics (continued)
Let us change the dynamics of a neuron by using an activation with
memory:
si(t + 1) =
+1 if hi(t + 1) > 0hi(t) if hi(t + 1) = 0−1 if hi(t + 1) < 0
where hi(t + 1) =n∑
j=1wijsj(t)
![Page 25: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/25.jpg)
Hopfield networks
Application to optimization problems
Problem: find the optimum (or close to) of a scalar function subject
to a set of restrictions
e.g.: The traveling-salesman problem (TSP) or graph bipartitioning (GBP); oftenwe deal with NP-complete problems.
TSP(n) = tour n cities with the minimum total distance; TSP(60) ≈ 6,93 · 1079
possible valid tours. In general, TSP(n) = n!2n
.
GBP(n) = given a graph with an even number n of nodes, partition it in two
subgraphs of n/2 nodes each such that the number of crossing edges is minimum
Idea: setup the problem as a function to be minimized, setting the
weights so that the obtained function is the energy function of a
Hopfield network
![Page 26: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/26.jpg)
Hopfield networks
Application to graph bipartitioning
Representation: n graph nodes = n neurons, with
si =
{
+1 : node i in left part−1 : node i in right part
Connectivity matrix: cij = 1 indicates a connection between nodes i and j.
![Page 27: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/27.jpg)
Hopfield networks
Application to graph bipartitioning
H(S) = −n∑
i=1
n∑
j>i
cijSiSj
︸ ︷︷ ︸
connections between parts
+ α
n∑
i=1
Si
2
︸ ︷︷ ︸
bipartition
=⇒ H(S) = −n∑
i=1
n∑
j>i
(cij − α)SiSj ≡ −n∑
i=1
n∑
j>i
wijSiSj +n∑
i=1
θiSi
=⇒ wij = cij − α, θi = 0
(general idea: wij is the coefficient of SiSj in the energy function)
![Page 28: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/28.jpg)
Hopfield networks
Application to the traveling-salesman problem
Representation: 1 city −→ n neurons, coding for position;
hence n cities = n2 neurons.
Sn×n = (Sij) (network states); Dn×n (distances between cities, given)
with Sij = +1 if the city i is visited in position j and −1 otherwise.
Restrictions on S:
1. A city cannot be visited more than oncen∑
i=1
n∑
j=1
n∑
k>j
SijSik = 0
−→ every row i must have only one +1: R1(S)
2. A city can only be visited in sequence (no ubiquity gift
−→ every column j must have only one +1: R1(St)
3. Every city must be visited
(n∑
i=1
n∑
j=1
Sij − n
)2
= 0
−→ there must be n +1s overall: R2(S)
![Page 29: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/29.jpg)
Hopfield networks
Application to the traveling-salesman problem
4. The tour has to be the shortest possiblen∑
k=1
n∑
i=1
n∑
j>i
DijSik(Sj,k−1 + Sj,k+1)
Total sum of distances in the tour: R3(S, D)
H(S) = −n∑
i=1
n∑
j=1
n∑
k>i
n∑
l>j
wij,klSijSkl+
n∑
i=1
n∑
j>i
θijSij ≡ αR1(S)+βR1(St)+γR2(S)+εR3(S, D)
=⇒ wij,kl = −αδik(1 − δjl) − βδjl(1 − δik) − γ − εDik(δj,l+1 + δj,l−1) and θij = −γn
This is a continuous Hopfield network: Sij ∈ (−1,1) since Sij := tanhχ(hij − θij)
Problem: set up good values for α, β, γ, ε > 0, χ > 0
![Page 30: Llu s A. Belanche belanche@lsi.upcbelanche/Docencia/apren/2009-10/Teoria/tema9.pdf · Ejemplo del comportamiento de una red de Hopfield Una red de 120 nodos (12x10) CONJUNTO DE OCHO](https://reader030.vdocuments.us/reader030/viewer/2022021713/5ba410c909d3f280548c9dcf/html5/thumbnails/30.jpg)
Hopfield networks
Application to the traveling-salesman problem
Example for n = 5: 120 valid tours (12 different), 25 neurons
225 potential attractors; actual num-ber of them: 1,740 (among them, thevalid 120). ⇒ 1,740 - 120 = 1,620spurious attractors (invalid tours)
The valid tours correspond to thestates of lower energy (correct de-sign)
But ... we have a network with 30times more invalid tours than validones: a random initialization will endup in an invalid tour with probability0.966
Moreover, we aim for the best tour!A possible workaround is to use stochastic unitswith simulated annealing