Chapter 10

1. For A: c0 = c2 = c3 = 1, index = 1 + 4 + 8 = 13, the given frequency table in Appendix D is for prob(0), hence prob(0) = 29789, but since A is a 1 pixel, then its probability prob(1) = 65535 - 29789 = 35746 out of 65535.

   For B: c0 = c1 = c2 = c3 = c4 = 1, and index = 1 + 2 + 4 + 8 + 16 = 31, prob(0) = 6554. As we see this odd pixel of 0 among the 1s has a lower probability.

   For C: c1 = c2 = c3 = c4 = c5 = c7 = 1, and the index becomes 190. Like pixel A the prob(0) = 91, but its prob(1) = 65535 - 91 = 65444 out of 65535, which is as expected.

2. The chain code is 0, 1, 0, 1, 7, 7. The differential chain code will be: 0, 1, -1, 1, -2, 0 with bits 1+2 + 3 + 2 + 5 + 1 = 14 bits

3. At level 2 the indices (without swapping) will be 0 for the two blank blocks and index = 27 × 2 + 9 × 2 + 0 + 2 = 74 and index =0 + 0 + 3x2 + 2 = 8 for the two blocks. At the upper level the index is: index = 0 + 0 + 3 x 1 + 1 =4

4. The coefficients of the shape adaptive DCT will be

   127	-40	1
   10	-15
   23	-7
   -6

   confined to the top left corner, while for those of the normal DCT with padded zeros, the significant coefficients are scattered all over the 8 × 8 area.