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```									Texture Mapping using Surface Flattening via
Multi-Dimensional Scaling

G.Zigelman, R.Kimmel, N.Kiryati
IEEE Transactions on Visualization and
Computer Graphics
2002
Multidimensional scaling (MDS)

   The idea: compute the pairwise geodesic distances
between the vertices of the mesh:

M   dist (xi , x j ) 
2
nn

2
   Now, find n points in R , so that their distance matrix is
as close as possible to M.
q2

q1

2
MDS – the math details

We look for X’,
 |          | 
               
X    x1       xn   R d n
 |
            | 
               
such that || M’ – M || is as small as possible, where

                              
2
M   dist (xi , x j )   xi  x j   R nn
2

            
M’ is the Euclidean distances matrix for points xi’.

3
MDS – the math details

 
Ideally, we want: M    xi  x j 
2

                                                                     M
            

 x   x , x   x  
i       j             i       M
j

|| x  ||  || x  ||  2  x , x     M
i
2
j
2
i         j

 || x  || || x  ||
 1
|| x1 || 

 || x  || || x  ||
 1
|| xn || 

       x1        |           | 
                                  
1                                      2

 || x  || || x  ||   || x 2 ||     || x  || || x  ||          || xn || 
 2

2


 1

2


                  x1        x n 
                                                                                            
                  |
             |  
xn
 || x  || || x  ||   || xn ||      || x  || || x  ||          || xn || 
 n                                
n                       1             2                                                                         

want to get rid of these                                                              X T               X
4
MDS – the math details

Trick: use the “magic matrix” J :
 1        1
n           1
n
 1                        
J  n       1    1
n      1
n
                          
 1
                         
 n             1
n      1  n n


a     a         a  J  0

b
 
J b  0
 
 
b
 
5
MDS – the math details

Cleaning the system:
 || x  || || x  ||   || x1 ||   || x1 || || x 2 ||    || xn || 
 1             1
                                   
 || x  || || x  ||   || x 2 ||   || x1 || || x 2 ||   || xn || 
J     2             2
                                    2X  X   M
T
J
                                                                     

 || x  || || x  ||                                                 
 n             n       || xn ||   || x1 || || x 2 ||
                         || xn || 


2 X T X   JMJ
X T X    1 JMJ : B
2

X T X   B

6
How to find X’

We will use the spectral decomposition of B:

|  1
T
 |                                       |                   | 
X T X   B   v1                      
v n 
                       
                                         v1                 vn 
 |                    |            n  |                   | 
                                                             
T
|   1                    1                                            T
 |     |                                                           |       |        | 
                                                                                 
 |     |     |                                                |       |        | 
X T X    v1   vd    vn           d           
d               v1     vd       vn 
                                                                                 
|                                            
 |
 |
|
| 
  d d                                    
 |
 |
|         | 
| 
        |                          n 
                 n 
           |           

nd

X T                                       X
7
How to find X’

So we find X’ by throwing away the last nd eigenvalues

        1 v1          
                       
X                        

                       

        d vd           d n

X   arg min X T X   B
X                           L2

A L2          Aij 2
i, j

8
Flattening results (Zigelman et al.)

9
Flattening results (Zigelman et al.)

10
Flattening results (Zigelman et al.)

11
The end

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