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An MPEG file encodes video partly by using the previous frame of video as a refe

ID: 3738900 • Letter: A

Question

An MPEG file encodes video partly by using the previous frame of video as a reference. Each “macroblock” of 16x16 pixels gets one “motion vector” associated with it that indicates where to look in the previous frame for a block of 16x16 pixels that could be used as a starting point for the new 16x16 block of pixels. Suppose for this problem that we aren’t interested in the magnitude of each motion vector - we want to encode, for each frame of video, just the rough direction of each motion vector. It’s either up, up-right, right, down-right, down, down-left, left, up-left, or no motion (9 possibilities).

a. If the “no motion” possibility occurs half the time, how many bits should we use to encode this possibility in an efficient encoding of these directions?

b. Of the remaining possibilities, left movement occurs 1/8 the time, right movement occurs 1/8 the time, down and up each occur 1/16 of the time, and the diagonals each occur 1/32 of the time apiece. Calculate the minimum number of bits necessary to represent each direction, assuming we have one code per symbol.

c. What is the entropy of this directional data, using the frequencies given in the preceding steps? (Up to 3 decimal places.)

d. Suppose instead of the frequencies just mentioned, we have more active video where all 9 symbols are equally likely. Calculate the entropy of the motion directions in this scenario. (Up to 2 decimal places.)

e.  Explain why it makes sense from a qualitative point of view that the number you obtained in the preceding step is larger or smaller than the entropy that you calculated before.

Explanation / Answer

Let suppose time is t(say)

Let say no. of bits = b

answer a: occurance of no motion = t/2

bits used = b/2

answer b: occurance of left movement = t/8

  occurance of right movement = t/8

  occurance of down movement = t/16

  occurance of up movement = t/16

occurance of diagonals = 4 * t/32 = t/8

number of bits = 1/8 + 1/8 + 1/16 + 1/16 + 1/8 = b/2

  

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