Chapter 5

1. 1. multiply all the luminance matrix elements by α = 50/25 = 2

   2. , and small elements of the matrix will be 1 and larger ones become 2.

   3. α = 0, hence all the matrix elements will be 1.

2. the same as problem 1.

3. 1. 62	0	3	1	0	0	1	0
      0	1	-1	0	1	0	0	0
      0	0	0	0	1	0	0	0
      -1	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	0	0	0	0	0	0
      0	0	0	0	0	0	0	3

   2. 31	0	1	0	0	0	0	0
      0	0	0	0	0	0	0	0
      0	0	0	0	0	0	0	0
      0	0	0	0	0	0	0	0
      0	0	0	0	0	0	0	0
      0	0	0	0	0	0	0	0
      0	0	0	0	0	0	0	0
      0	0	0	0	0	0	0	1

4. for 50% quality: DIFF = 62 - 50 = 12, symbol_1 = 4; symbol_2 = 12 scanned pairs (3,1)(0,3)(0,1)(0,-1)(1,-1)(6,1)(6,1)(1,1)(1,1)(35,3) and the resultant events:

   (3,1)(0,2)(0,1)(0,1)(1,1)(6,1)(6,1)(1,1)(1,1)(15,0)(15,0)(3,3)

   for 25% quality: DIFF= 31-50 = -19, symbol_1 = 5, symbol_2 = -19-1 =-20

   scanned pairs: (4, 1)(57, 1)

   events: (4,1)(15,0)(15,0)(15,0)(9,1)

5. for DC: DIFF= -19 ⇒ CAT = 5; DIFF-1 = -20

   VLC for CAT = 5 is 110 and -20 in binary is 11101100, hence the VLC for the DC coefficient is 11001100

   for AC, using the AC VLC tables:

   for each (15,0) the VLC is 11111111001

   and for (9, 1) the VLC is 111111001

   total number of bits: 8 + 3 x 11 + 9 = 50 bits.

6. At bit-plane 6 coefficient 65 at clean-up pass. At bit-plane 5 coefficient 65 at all passes and coefficient 50 at clean-up pass.