sparse image representation via combined transforms - Convex ...
sparse image representation via combined transforms - Convex ... sparse image representation via combined transforms - Convex ...
158 APPENDIX B. FAST EDGELET-LIKE TRANSFORM B.2.4 Algorithm Here we give the algorithm to compute the fast edgelet-like transform. Note when the original image is N by N, our algorithm generates an N × 2N matrix. Recall that I(i, j) denotes the image value at pixel (i, j), 1 ≤ i, j ≤ N. Computing the discrete Radon transform via Fourier transform A. Compute the first half of the DRT matrix (size is N by N): 1. For j =0to N − 1 step 1, take DFT of (j + 1)th column: [g(0,j),g(1,j),... ,g(N − 1,j)] = DFT N ([I(1,j+1),I(2,j+1),... ,I(N,j + 1)]); End; 2. For k =0to N − 1 step 1, (a) pointwise multiply row { k +1,[g(k, 0),g(k, 1),... ,g(k, N − 1)], ( ) } with complex sequence e −2π√ −1 j·s(k)+∆(k) j2 2 : j =0, 1,... ,N − 1 ; (b) convolve it with complex sequence { } e π√ −1∆(k)(t) 2 : t =0, 1,... ,N − 1 ; (c) pointwise { multiply the sequence with complex } sequence e −2π√ −1∆(k) l2 2 : l =0, 1,... ,N − 1 ; we get another row vector: { F ( k N ,s(k)+l · ∆(k)) : l =0, 1,... ,N − 1} ; End; 3. For l =0to N − 1 step 1, taper the (l + 1)th column by the raised cosine, then take inverse 1-D DFT; End; B. Compute the second half of the DRT matrix (size is N by N): 1. For i =0to N − 1 step 1, For j =0to N − 1 step 1, modulate: g(i, j) =I(i +1,j+1)e −2π√ −1 j N ;
B.3. ADJOINT OF THE FAST TRANSFORM 159 End; End; 2. For i =0to N − 1 step 1, take 1-D DFT of the (i + 1)th row: [g(i, 0),g(i, 1),... ,g(i, N − 1)] = DFT N ([I(i +1, 1),I(i +1, 2),... ,I(i +1,N)]); End; 3. flip matrix (g(i, k)) 0≤i,k≤N−1 by columns: h(i, k) =g(i, N − 1 − k); 4. For k =0to N − 1 step 1, (a) pointwise multiply (k + 1)th column, {h(0,k),h(1,k),... ,h(N − 1,k)}, { ( ) } with complex sequence e −2π√ −1 i·s(k)+∆(k) i2 2 : i =0, 1,... ,N − 1 ; (b) convolve it with complex sequence { } e π√ −1∆(k)(t) 2 : t =0, 1,... ,N − 1 ; (c) pointwise { multiply the sequence with complex } sequence e −2π√ −1∆(k) l2 2 : l =0, 1,... ,N − 1 ; we get another column vector: { F (s(k)+l · ∆(k), N − 1 − k } ):l =0, 1,... ,N − 1 ; N End; 5. take transpose; 6. For l =0to N − 1 step 1, taper the (l + 1)th column by the raised cosine function, then take inverse 1-D DFT; End; B.3 Adjoint of the Fast Transform First of all, note that the previous fast algorithm is a linear transform, and so is every step in it. It is obvious that if we take the adjoint of each step, and do them in reverse order,
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- Page 171 and 172: A.3. DETAILS 143 Ordering of Dyadic
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- Page 193 and 194: B.5. EXAMPLES 165 And so on. Note f
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- Page 199 and 200: Bibliography [1] Sensor Data Manage
- Page 201 and 202: BIBLIOGRAPHY 173 [24] C. Victor Che
- Page 203 and 204: BIBLIOGRAPHY 175 [50] David L. Dono
- Page 205 and 206: BIBLIOGRAPHY 177 [75] Vivek K. Goya
- Page 207 and 208: BIBLIOGRAPHY 179 [100] Stéphane Ma
- Page 209 and 210: BIBLIOGRAPHY 181 [127] C. E. Shanno
158 APPENDIX B. FAST EDGELET-LIKE TRANSFORM<br />
B.2.4<br />
Algorithm<br />
Here we give the algorithm to compute the fast edgelet-like transform. Note when the<br />
original <strong>image</strong> is N by N, our algorithm generates an N × 2N matrix. Recall that I(i, j)<br />
denotes the <strong>image</strong> value at pixel (i, j), 1 ≤ i, j ≤ N.<br />
Computing the discrete Radon transform <strong>via</strong> Fourier transform<br />
A. Compute the first half of the DRT matrix (size is N by N):<br />
1. For j =0to N − 1 step 1,<br />
take DFT of (j + 1)th column:<br />
[g(0,j),g(1,j),... ,g(N − 1,j)]<br />
= DFT N ([I(1,j+1),I(2,j+1),... ,I(N,j + 1)]);<br />
End;<br />
2. For k =0to N − 1 step 1,<br />
(a) pointwise multiply row<br />
{<br />
k +1,[g(k, 0),g(k, 1),... ,g(k, N − 1)],<br />
(<br />
)<br />
}<br />
with complex sequence e −2π√ −1 j·s(k)+∆(k) j2 2<br />
: j =0, 1,... ,N − 1 ;<br />
(b) convolve it with complex sequence<br />
{<br />
}<br />
e π√ −1∆(k)(t) 2 : t =0, 1,... ,N − 1 ;<br />
(c) pointwise { multiply the sequence with complex } sequence<br />
e −2π√ −1∆(k) l2 2 : l =0, 1,... ,N − 1 ;<br />
we get another row vector: { F ( k N ,s(k)+l · ∆(k)) : l =0, 1,... ,N − 1} ;<br />
End;<br />
3. For l =0to N − 1 step 1,<br />
taper the (l + 1)th column by the raised cosine, then take inverse 1-D<br />
DFT;<br />
End;<br />
B. Compute the second half of the DRT matrix (size is N by N):<br />
1. For i =0to N − 1 step 1,<br />
For j =0to N − 1 step 1,<br />
modulate: g(i, j) =I(i +1,j+1)e −2π√ −1 j N ;