Stochastic model for a line network

 

Bayesian approach

Notations :

• D : the observed image
• T : finite grid on the image
• $d_{t}, t \in T$ : the gray level of a pixel

 
Hypothesis : the line network, S, we want to extract is composed of a finite number of segments s_i

$$\begin{array}{l} S=\{s_{i},i=1,..., n \}\\ n \in N \end{array}$$ (1)

We have :

• s_i=(p_i,m_i) : a segment in the configuration
• p_i=(x_i,y_i) : the coordinates of its center
• $m_{i}=(w_{i},l_{i},\theta_{i})$ : the vector containing the width, the length and the orientation of the segment

 
 
Problem : to detect the number of the segments, their location and their parameters
 
Solution : Bayes rule

$$P(S/D)=\frac{P(D/S)P(S)}{P(D)}$$ (2)

MAP estimation :

$$\hat{S} \approx \arg max _{S}\{P(D/S)P(S)\}$$ (3)

 
Considering we have a Gibbs point process :

$$f(S/D) \propto \beta^{n}\exp(- U)=\beta^{n}\exp( -(U_{D}(S)+U_{P}(S)))$$ (4)

 
The estimator of the line network is :

$$\hat{S}=\arg min_{S}\{U_{D}(S)+U_{P}(S)-n\log\beta\}$$ (5)

 

Candy model

The parameters of a segment are independent random variables :

• $(x_{n},y_{n}) \in T$ : the coordinates of the center of the segment
• $w_{n} \sim \mathcal{U}(\{w_{min},\ldots,w_{max}\})$ : the width of the segment
• $l_{n} \sim \mathcal{U}(\{l_{min},\ldots,l_{max}\})$ : the length of the segment
• $\theta_{n} \sim \mathcal{U}([0,2\pi])$ : the orientation of the segment

$\mathcal{U}$ is a uniform law.
 
State of segment : a segment has two extremities to be connected with

• $s=s^{\emptyset}$ : no connection
• s=s^t||q : one connected extremity
• s=s^tq : two connected extremities

 
Probability density :

$$f(S) \propto \beta^{n}\prod_{s_{i} \in S} g(i) \prod_{s_{i} \diamond s_{j}, i<j} h(s_{i},s_{j})$$ (6)

 
State penalties : the short segments and the free segments are penalized

$$g(s_{i})=g_{1}(s_{i}) \times g_{2}(s_{i})$$ (7)

$$g_{1}(s_{i})=exp\left (\frac{h_{s(i)}-h_{max}}{h_{max}}\right)$$ (8)

$$g_{2}(s_{i})=\left\{ \begin{array}{l} g_{21}, \quad s_{i}=s... ... q}\\ g_{23}= 1, \quad s_{i}=s_{i}^{tq} \end{array} \right.$$ (9)

We have : $g_{21} \ll g_{22} < 1$.

 
Rejection interactions : we penalize the overlaping segments, but we enable the crossing segments.

  

Figure: Rejection region for a segment $\overline {QT}$

  

Figure 2: Rejection interactions btw segments

 
 
Attraction interactions : to form a network, the segments attract each other. We penalize the segments which are not well aligned.

  

Figure 3: Attraction region for a segment

  

Figure 4: Attraction interactions btw segments

If there is a rejection interaction btw two segments :

$$h(s_{i},s_{j})=h_{r} \leq 1$$ (10)

If there is an attraction interaction btw two segments, the function h is penalizing the orientation between segments :

$$h(s_{i},s_{j})=h_{a} \leq 1 \quad \text{if} \quad \tau > \tau_{max}$$ (11)

 

For each segment, the local prior energy is:

$$U_{P}(s_{i})= - \log g(s_{i})-\sum_{s_{i} \diamond s_{j}, i \neq j} \log h(s_{i},s_{j})$$ (12)

hence for the configuration S :

 

$$\sum_{s_{i} \in S} U_{P}(s_{i})$$ U_P(S) =

$$-\sum_{s_{i}\in S} \log g(s_{i})+\sum_{s_{i} \diamond s_{j}, i < j} \log h(s_{i},s_{j})$$ = (13)