Séminaire IRCCyN
6 Janvier 2 0 00
Christian VIARD-GAUDIN
[email protected]
Institut de Recherche en
Communications et Cybernétique de Nantes
1
Multimédia et
Réseaux
fy (cycles/d°)
28.2 cy/d°
4
5
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V
Vidéo et
Multimédia
3
6
14.2 cy/d°
IV
4
5
5.7 cy/d°
III
1.5 cy/d°
3
2
6
4
2
3
2
1
1
1
fx (cycles/d°)
Modélisation
psychovisuelle
Écrit et
Documents
0.1
Angular selectivity
III : 45°
IV : 30°
V : 30°
0.35
0.15
0.25
0.7
0.8
0.3
0.6
0.4
0.2
Équipe Image et Vidéo
Communications
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Modèle z
2
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SOMMAIRE

TUTORIAL SUR LES MODÈLES MARKOV
CACHÉS

APPLICATION A LA RECONNAISSANCE DE
L’ÉCRITURE MANUSCRITE
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Première Partie : Modèles de Markov
Cachés (HMM’s)
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Origins and Scopes

Late 1960’s (BAUM, 1967) :
– Basic theory

Late 1980’s :
– Large widespread understanding
– Application to speech recognition

Used to model the behavior of
A real-world
signal
speech, characters,
Temperature, stock market, …
By mean of a statistical approach
Parametric random process
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Hidden Markov Model (1)

An HMM is a double stochastic process
1) an underlying stochastic process
generates a sequence of states :
q1, q2, … , qt, ... qT,
Where t : discrete time, regularly spaced
T : length of the sequence
qt  Q = {q1, q2, ... qN}
N : the number of possible states
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• Markov Chain Hypotheses
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1) First order : probabilistic description is truncated to
just the current state and the predecessor state :
P[qt=qj|qt-1=qi, qt-2= qk, …] = P[qt=qj|qt-1=qi]
2) Stationarity : probabilities are time invariant :
1  i, j  N
P[qt=qj|qt-1=qi] = aij
This defines a square NxN matrix, A = {aij}
(state transition probability matrix)
N
where aij  0 and  aij  1
j 1
3) an initial state distribution  = {i} should also be defined:
i = P[q1 =qi] with 1  i  N
where i  0 and
N

i 1
i
1
•The value of each state is unobservable, but ...6
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Hidden Markov Model (2)
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
An HMM is a double stochastic process
1) an underlying stochastic process
generates a sequence of states :
q1, q2, … , qt, ... qT,
But ...
2) each state emits an observation according
to a second stochastic process :
xt  X = {x1, x2, ... xM}
qt
xi : a discrete symbol
M : number of symbols
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• Observation Hypothesis
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1) The observation xt depends only of the present state qt :
P[xt = xj | qt = qi] = bij
This defines a NxM matrix B: observation probability matrix
B = {bij}
where bij  0 and
M
b
j 1
ij
1
A complete specification of an HMM (l) requires :
• Two model parameters N and M
• A specification of the symbols to
observe
• Three probability measures : , A, B
l =Copyright
(, A,
B)
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Example 1 : (Not Hidden, Just a Discrete
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Markov Model)

Model of the weather in Nantes
–N=M=3
– State = observation :
Q = {q1=rain, q2=cloudy, q3=sunny}
– t is sampled every day, at noon for instance
– ANantes = {aij} =
0.4 0.3 0.3


0
.
2
0
.
6
0
.
2


 01
. 01
. 0.8
Let us play with this model now !
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NANTES Weather Model
Given that the weather today (t=1) is
sunny (q1 = q3)
Answer these questions

 What will be the weather tomorrow (t = 2) ?
 What is the probability of rain for the day
after tomorrow (t = 3) ?
 And for the day after (t = 4) ?
 What is the probability the coming week will
be “sun-sun-rain-rain-sun-cloudy-sun”
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NANTES Weather Model
Two more questions :
 What is the probability of rain for d
consecutive days ? (ex. d = 3),
 What is the average number of
consecutive sunny days ? Cloudy days ?
Rainy days ?
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11
R
NANTES Weather Model
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
A=
Answers
C
S
0.4 0.3 0.3 R


0.2 0.6 0.2 C
 01
. 01
. 0.8 S
 What will be the weather tomorrow ?
 What is the probability of rain for the day after
tomorrow ?
Use of a trellis :
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NANTES Weather Model
A=
0.4 0.3 0.3
 0 .2 0 .6 0 .2 


 0.1 0.1 0.8
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 And for the day after ?
Just extend the trellis, from the previous values
q1: rain
q2: cloudy
q3: sunny
P(q4=q1) =
 What is the probability the coming week will be
“sun-sun-rain-rain-sun-cloudy-sun”
P(q3,q3,q1,q1,q3,q2,q3)
=
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NANTES Weather Model
 What is the probability that rain lasts for d
consecutive days ? (ex. d = 3),
More generally :
P(qi, qi, …qi,qij) =
hence P(q1, q1, q1, qj  1) =
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NANTES Weather Model
 What is the average number of
consecutive sunny days ? Cloudy days ?
Rainy days ?


d   d  p d    d  aii 1  a 
i
d 1
i
d 1
d 1
ii
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NANTES Weather Model
A=
0.4 0.3 0.3


0.2 0.6 0.2
 01
. 01
. 0.8
Topological graphic representation :
0.3
0.4
q1 =
rain
0.6
q2 =
cloudy
0.2
0.3
0.1
0.2
0.1
q3 =
sunny
0.8
This is an ergodic model : from one state, all other
states are reachable
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Example 2 : Extension to HMM

Coin tossing experiment :
We do not see how the tossing is performed :
one coin, multiple coins, biased coins ?
We only have access to the result, a sequence of tosses,
In that case, M = 2 and X = {x1 = Head, x2 = Tail}
Observation sequence : (x1,x2, …. xT) = (H,H,T,T,T,H, …H)
The problem are :

How to build a model to explain the observed sequence ?

What are the states ?

How many states ?
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Coin tossing experiment :
model 1 : assume only one biased coin

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2 states (N = 2) Q = {q1 = H, q2 = T}
Model topology :
1-P(H)
P(H)
1-P(H)
q2 =
Tail
q1 =
Head
P(H)
Only one parameter is needed : P(H), it defines matrix A
model 2 : assume two biased coins
- 2 states (N = 2)
Q = {q1 = Coin1, q2 = Coin2}
- 2 different observations (M = 2) X = {x1 = H, x2 = T}
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Model topology :
1-a11
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a11
a22
q2 =
Coin2
q1 =
Coin1
1-a22
Transition state
probabilities :
A=
 a11
1  a
22

1  a11 
a 22 
Observation symbol
probabilities
B=
 P1  H  1  P1  H  
  



P
H
1

P
H
2
 2

4 parameters are required to define this model
(A : 2, B : 2)
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Coin tossing experiment :
model 3 : assume three biased coins

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3 states (N = 3) Q = {q1 = Coin1, q2 = Coin2, q3 = Coin3}
X = {x1 = H, x2 = T}
2 different observations (M = 2)
Model topology :
a12
a11
Observation symbol
probabilities :
B=
 P1  H  1  P1  H  






P
H
1

P
H
2
2


 P3  H  1  P3  H  
q1 =
Coin1
a22
a21
a13
q2 =
Coin2
a32
a23
a31
q3 =
Coin3
9 independent parameters are required
to define this model (A : 6, B :3)
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a33
20
Coin tossing experiment :
model 3 : assume three biased coins

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3 states (N = 3) Q = {q1 = Coin1, q2 = Coin2, q3 = Coin3}
Matrix A
Consider the following example :
- State transition probabilities : 1/3
- Initial state probabilities : 1/3
Vector 
- Observation probabilities
t
t-1
q1
q2
q3
q1
q2
q3
1/3
1/3
1/3
1/3
1/3
1/3
1/3
1/3
1/3
State Values
q1
q2
q3
Initial state
probability
P(q1)
1/3
1/3
1/3
P(H)
State 1 :
q1
0.5
State 2 :
q2
0.75
State 3 :
q3
0.25
P(T)
0.5
0.25
0.75
Matrix B
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21
Coin tossing experiment :
model 3 : assume three biased coins

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3 states (N = 3) Q = {q1 = Coin1, q2 = Coin2, q3 = Coin3}
1. You observe X = (H,H,H,H,T,H,T,T,T,T)
Which state sequence G, generates most likely X ?
What is the joint probability, P(X, G | l) of the
observation sequence and the state sequence ?
2. What is the probability that the observation sequence came
entirely from state q1 ?
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1. You observe X = (H,H,H,H,T,H,T,T,T,T)
2. What is the probability that the observation sequence came entirely
from state q1 ?
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Example of non ergodic model
(left-right model)
0.7
qs
0.2
q1
0.1
0.1
0.8
q2
0.6
0.1
q3
0.4
0.9
qf
0.1
Three states + one starting state qs + one final state qf
qs and qf are non emitting states.
Assume there are 2 symbols to observe X = {x1=a,x2=b}
0.2 
0.7 
 
0
 
 0.1
Initial state
probabilities
. 01
. 
0 0.8 01
0 0 0.6 0.4

A
0 0 01
. 0.9


0
1
0 0
Transition state
probabilities
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 0.8 0.2


B  0.4 0.6
 01
. 0.9
P(a|q1)
P(b|q3)
Observation symbol
probabilities
24
The most probable state sequence with this model is : q2, q3
resulting in the symbol sequence “bb”.
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But this sequence can also be generated by other state
sequences, such as q1, q2.
• Computation of the likelihood of an observation sequence :
Given X = “aaa” compute the likelihood for this model : P(aaa | l)
The likelihood P(X | l) is given by the sum over all possible ways
to generate X.
State
sequence
Init
q 1q 2q 3
0.2
0.8
0.8
0.4
0.6
0.1
0.9
0.0027648
q 1q 3q 3
0.2
0.8
0.1
0.1
0.1
0.1
0.9
0.0000144
q 2q 3q 3
0.7
0.4
0.6
0.1
0.1
0.1
0.9
0.0001512
P(aaa|l) =
0.0029304
0.7
qs
0.2
q1
0.1
0.8
Obs a Trans Obs a Trans Obs a Trans
q2
0.4
0.1
0.9
0.6
3
0.1
0.1
q
Joint
probability
qf
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Using HMM for pattern recognition consists in computing the
model li among a set of K models which maximizes the
likehood for an observation to have been generated by this
model :
lmax = arg max P(X|li)
li
for i = 1, … K
Character recognition :
• Small lexicon : as many HMModels as words
• Otherwise, letters are individually modeled by an HMM
which can be concatenated to form word models.
lP
lA
lR
lI
lS
Word model for « PARIS »
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The three basic problems for HMMs
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
Problem 1 : Recognition
Given X = (x1,x2, … xT) and the various models li
How to efficiently compute P(X|l) ?
Forward-Backward algorithm

Problem 2 : Analysis
Given X = (x1,x2, … xT) and a model l, find the optimal state
sequence G.
How can we undiscovered the sequence of states corresponding to
a given observation ?
Viterbi algorithm

Problem 3 : Learning
Given X = (x1,x2, … xT), estimate model parameters l = (, A, B)
that maximize P(X| l)
How do we adjust the model parameters
l = (, A, B) ?
Baum-Welch algorithm
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
Problem 1 : Recognition
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How to efficiently compute P(X|l) ?
X = x1x2 …xt … xT : observation sequence
It exists several paths (G) which allow to obtain X :
P(X|l) = SG P( X,G |l) = SG P(X | G, l) x P(G|l)
Joint
probability
Depends only on state
Depends only on observation
transition probabilities : A
probabilities : B matrix
•
P(X | G, l)
The path G is defined by a sequence of states : q1q2 …qt … qT
P(X | G, l) = P(x1x2 …xt … xT | q1q2 …qt … qT , l)
= P(x1| q1…qT , l)xP(x2| x1,q1… qT , l)…P(xT| xT-1 … x1,q1… qT , l)
= P(x1| q1 , l)xP(x2| q2, l)…P(xT|qT, l)
as xt depends only of qt
  P x t | q t , l 
T
t 1
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
Problem 1 : Recognition (2)
How to efficiently compute P(X|l) ?
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P(X|l) = SG P( X,G | l) = SG P(X | G, l) x P(G|l)
•
P(G|l)
The path G is defined by a sequence of states : q1q2 …qt … qT
P(G | l)
= P(q1q2 …qt … qT | l)
Joint probability
= P(q1|l) x P(q2|q1, l)…P(qT| qT-1 … q1, l)
= P(q1|l) x P(q2| q1, l)…P(qT|qT-1, l)
 Pq1 | l    Pq t | q t 1 , l 
T
as we assume a first
order HMM
t 2
At last, by replacing :
P X | l     P xt | qt , l   Pq1 | l    Pqt | qt 1 , l 
G
T
T
t 1
t 2
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What about the computational complexity ?
P X | l     P xt | qt , l   Pq1 | l    Pqt | qt 1 , l 
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G
T
T
t 1
t 2
Number of multiplications for one path G :
Number of paths
(T-1)+1+1+(T-2) =2T-1
NT
X:
Total number of multiplications
:
(2T-1)NT
Total number of additions
:
(NT-1)
How long does it take ?
Assume N = 23, T = 15 (word check application)
Number of operations  2T.NT = 30 . 2315  1022
Assume 1 Gops  Number of seconds = 1022/109 = 1013
 Number of days = 1013/3600x24 = 108
 Number of years = 108/365 = 105
FIND SOMETHING ELSE !!!
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Forward-Backward algorithm
Use a trellis structure to carry out the computations :
at each node of trellis, store the forward variable ati
with ati = P(x1x2 … xt, qt = qi | l)
which is the probability of a partial observation
sequence up to time t and of being in the state qi at
that same time
Algorithm in 3 steps :
1. Initialization a1i = P(x1, q1 = qi | l) = P(x1| q1 = qi,l) x P(q1 = qi|l)
2. Recursion
a
j
t 1
N i

j
i
  at Pqt 1  q | qt  q , l   P xt 1 | q j , l 
 i 1

with 1  j  N, and 1  t  T-1
N
3. Termination P X | l    aTi
i 1
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Forward-Backward algorithm
atN
qN
.
qi
.
qj
.
.
q1
ati
aNj
aij
at1
1
...
at j1
a1j
t
t+1
...
N


j
i
j
i
at 1   at Pqt 1  q | qt  q , l   P xt 1| q j , l 
 i 1

with 1  j  N, and 1  t  T-1
Total number of multiplications
:
Total number of additions
:
Assume N = 23, T = 15
Number of operations
:
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
Problem 2 : Analysis
– How can we undiscovered the sequence of states
corresponding to a given observation X ?
– Choose the most likely path : Find the path (q1,q2,…,qT) that
maximizes the probability = P(q1,q2,…,qT| X,l)
– Solution by Dynamic Programming :
Viterbi algorithm
– It is an inductive algorithm that keep the best possible state
sequence ending in state qi at time t.
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
VITERBI Algorithm
– Define dt(i) = max
P(q1,q2,…,qt=qi , x1,x2,…xt | l)
– dt(i) is the highest probability path ending in state qi
– By induction, we have :
 dt+1(k) = max [dt(i) aik] . bk(xt+1),with 1 k N
1iN
STATES
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q1,q2,…,qt-1
 Memorize also t+1(k) = arg max(dt(i) aik)
N
aNk
dt+1(k)
k
t+1(k) = j
j
aik
i
dt(i)
2
1
Tracing
back the
optimal
state
sequence
a1k
1
2
x1
x2
t-1
t
t+1
TIME
xt-1
xt
xt+1
OBSERVATION
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max [dT(i)]
T-1
T
xT-1
xT
1 i N
34
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
VITERBI Algorithm
1. Initialization
For 1 i N
d1(i) = i x bi(x1);
1(i) = 0;
2. Recursive computation
For 2 t T
For 1 j N
dt(j) = max [dt-1(i) aij] . bj(xt);
1iN
t(j) = arg max(dt-1(i) aij);
1iN
3. Termination
P* = max[dt(i)];
1iN
q*T = arg max[dT(i)];
1iN
4. Backtracking For t=T-1 down to 1
q*t = t(q*t+1);
// or -ln(i)-ln(bi(x1))
// or dt(j) = min[dt-1(i)-ln( aij)]-ln(bj(xt));
// or t(j) = arg min[dt-1(i)-ln(aij));
// or P* = min[dt(i)];
// or q*T = arg min[dT(i)];
Hence P* (or exp(-P*) ) gives the required state-optimized
probability,and G* = (q1*,q2*, …, qT*) is the optimal state sequence.
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
Problem 3 : Learning
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How do we adjust the model parameters l = (, A, B) ?
Baum-Welch algorithm
1. Let initial model be l0,
2. Compute new model l based on l0 and on observation X,
3. If log P(X|l) - log(P(X|l0) < Delta stop,
4. Else set l0  l and goto step 2.
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Joint Probability
P(a, b) = P(a | b) x P(b)
= P(b | a) x P(a)
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Copyright © IRCCyN/CVG
SECONDE PARTIE : RECONNAISSANCE DE
L’ECRITURE
MANUSCRITE
ON-LINE
OFF-LINE
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Copyright © IRCCyN/CVG
OFF-LINE VERSUS ON-LINE
Retracé
Intersection
Posé de stylo
Lever de stylo
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LA RECONNAISSANCE
Image d’un mot
1D
2D
HMM
Modélisation
Graphèmes
Lettres
Colonnes
de graphe
pixels :
Ordonnancement off-line
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Copyright © IRCCyN/CVG
UN MODELE APPROPRIE
POUR UNE OPTIMISATION GLOBALE
1
DEFINITION D’UN NOEUD
2
GRAPHE ORIGINAL
R1
2
R2
Problème du voyageur de commerce
recherche d’un cycle hamiltonien
1
3
7
5
6 R3
Modèle
2
4
8
1
3
1
4
lien inter
lien intra
7
2
3
7
6
4
5
6
5
8
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8
41
Copyright © IRCCyN/CVG
UN MODELE APPROPRIE
POUR UNE OPTIMISATION GLOBALE
1
DEFINITION D’UN NOEUD
2
GRAPHE ORIGINAL
3
GRAPHE COMPLET
4
GRAPHE FINAL
5
VALUATION DU GRAPHE
6
CHEMIN HAMILTONIEN
2
3
7
2
1
5
6
4
8
1
3
1
4
2
3
4
5
P
Lien
Lien
Lien
Lien
7
inter
intra
de complétude
de départ et d’arrivée
7
6
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6
5
8
8
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Synoptique du Système de
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RECONNAISSANCE
Observations : 1D
sequence of symbols
Feature
Vectors
Feature
Symbol
Extraction
Mapping
HMM
- State transition
prob.
- Observations
symbol prob.
- Initial state prob.
rugby : -2.15
……… : ……
Result
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Copyright © IRCCyN/CVG
R
E
C
O
S
N
Y
N
S
D A
T
E I
E
S
M
S
E
A
N
C
E
Image
d’un mot
Segmentation
Fichier
on-line
Ensemble
de
segments
Normalisation
Lignes de
référence
Graphe
Ordonnancement on-line
Ordonnancement off-line
Séquence de segments orientés
et de levers-posés de stylo
Extraction de caractéristiques
Quantification vectorielle
125 32 368 56 98 225 118 …
Séquence de symboles
HMM
Vraisemblances pour chaque
mot du dictionnaire
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un
deux
frs
44
EXAMPLES of CLUSTERS
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Core line
Core line
Core line
Core line
Core line
Number of clusters : 300
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RESULTATS
IRONOFF
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Une base de données duales on-line et off-line
25 451 caractères isolés
31 346 mots cursifs
environ 700 scripteurs différents
(64% d’hommes et 34% de femmes)
âge moyen d’un scripteur : 26 ans et demi
Dictionnaire :
197 mots
Apprentissage : 20 898 mots
Test :
10 448 mots
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Copyright © IRCCyN/CVG
IRONOFF Construction
Online data
Step 1 : online acquisition
Gray level image
Step 2 : offline acquisition
Step 3 :
matching process
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Copyright © IRCCyN/CVG
IRONOFF Samples
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96
96
Top(1)
Top(1)
94
94
OnLineSeg
on-line
92
92
Ord.on
on
Ord.
90
90
Ord.off
off
Ord.
88
88
Ord.on&off
on&off
Ord.
Ord.On
On++Ord.
Ord.off
off
Ord.
86
86
OffLineSeg
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COMPARAISON OffLineSeg et OnLineSeg
84
84
00
20
20
40
40
N
N
60
60
80
80
APPROCHE « ORDREC »
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100
100
99
99
Top(K)
Top(K)
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COMPARAISON OnLineSeg et OnLinePt
98
98
OnLinePt
OnLinePt
97
97
OnLineSeg
OnLineSeg
96
96
95
95
94
94
11
2
33
4
55
6
7
7
8
99
10
K
K
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Copyright © IRCCyN/CVG
EXEMPLES
rugby
: -2.15
surgelées : -2.24
dégâts
: -2.29
whisky
Liberty
blâmez
: -1.74
: -2.10
: -2.27
Top(1) correct
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EXEMPLES
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éclabousser : -2.01
déposerait : -2.23
moissonner : -2.26
père-noël
: -1.82
pivoteras
: -2.22
pédalerions : -2.24
Hôpital : -2.01
Maître : -2.24
autre
: -2.25
Top(1) correct
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EXEMPLES
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pas : -2.26
plus : -2.34
puis : -2.37
aïeux : -2.26
une
: -2.30
onze : -2.32
Erreurs dues à un pré-traitement : normalisation du mot
Occident : -2.11
accident : -2.19
Maître
: -2.43
North
: -2.43
Neptune
: -2.46
Littérature : -2.54
Erreurs d ’annotation : mot inconnu dans une certaine casse
treize : -1.77
seize : -1.90
fonça : -2.06
dix : -1.82
six : -2.13
deux : -2.20
Erreurs dues aux limitations du modèle
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EXEMPLES
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pas : -2.26
plus : -2.34
puis : -2.37
aïeux : -2.26
une
: -2.30
onze : -2.32
Erreurs dues à un pré-traitement : normalisation du mot
Occident : -2.11
accident : -2.19
Maître
: -2.43
North
: -2.43
Neptune
: -2.46
Littérature : -2.54
Erreurs d ’annotation : mot inconnu dans une certaine casse
treize : -1.77
seize : -1.90
fonça : -2.06
dix : -1.82
six : -2.13
deux : -2.20
Erreurs dues aux limitations du modèle
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FIN
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CONCATENATION DE MODELES LETTRES
POUR DEFINIR DES MODELES MOTS
c
état initial
e
états non émetteurs
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t
état final
56
MODELE MOT : APPROCHE « LEGO »
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dégât
qD
e
d
g
a
dia
dia
qF
t
dia
dia
dégât
15
d é div_0 dia g div_+ dia â div_0 dia divDia_0 t div_F dia divDia_F
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
2
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
2
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
2
0
2
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
2
0
2
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
1
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APPROCHE
ORDREC »»
APPROCHE «« ORDREC
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Graphe des
segments
Graphe des traits




 

 
Posés et levers de stylo
Levers de stylo

  Posés de stylo et ordre des traits
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Segment features
• Dimensions of the segment enclosing rectangle : Lx(s) et Ly(s),
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• Relative Ordinate to the baseline of the center of the enclosing rectangle : y(s),
• Segment orientation : Dx(s) et Dy(s),
• Left and right extensions of the segment : Og(s) et Od(s).
Lx
Lx
d
Ly
d
Dy
Ly
Dy
Dx
Dx
y
y
y = 0 (ligne de base du mot)
Pen-up/pen-down features
• Ordinate of the vector center of the pen-up/pen-down : y(lp),
• Direction of the vector of the pen-up/pen-down : Dx(lp) et Dy(lp).
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