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In JPEG and MPEG, why is 8x8 matrix chosen for Discrete Cosine Transform? Why not any other, say 64x64?

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2 Answers 2

The DCT treats the block as if it were periodic and has to reconstruct the resulting jump at the boundaries. If you take 64x64 blocks, you'll most likely have a huge jump at the boundaries, and you'll need lots of high-frequency components to reconstruct that to a satisfactory precision. The $8$ results from a tradeoff that can't be theoretically optimized, but seems to work well for typical images, where the variation across 8x8 blocks is typically small enough that blocking artifacts can be avoided without having to encode too much high-frequency information.

Also $8$ is a power of $2$, so even if the optimal tradeoff from the information-content point of view had been, say, 10, one might still have chosen $8$ because the transform is much simpler and quicker to perform.

Edit in response to the comment: Compression is always a tradeoff. You can always get sharper images by keeping more of the information -- you can get sharp images with 4x4, 8x8 or 64x64 blocks simply by keeping the entire high-frequency information. The question is how to reduce the amount of information as much as possible with as little visible difference as possible. Experience shows that in 8x8 blocks you can drop a lot of the information without creating unacceptable blocking artifacts. Certainly for 4x4 blocks the boundary jump would be even less, but there would also be less opportunity for compression. Consider an 8x8 block made up of four 4x4 blocks: If you transform each of the 4x4 blocks separately, you have to store their averages (zero-frequency components) with the same (high) precision for all four of them. If you transform them together as an 8x8 block, instead of those four averages, you get one average and three oscillating components that you can store with lower precision. All this can't be put into a single precise formula; one has to look at how data in real-world images are actually distributed and then make the required tradeoffs.

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In that case 4 would be better as it would lead to sharper images? –  user9476 Apr 12 '11 at 8:25

From the MPEG FAQ:

Q22. Why was the 8x8 DCT size chosen?

A Experiments showed little compaction gains could be acheived with larger sizes, especially when considering the increased implementation complexity. A fast DCT algorithm will require roughly double the arithmetic operations per sample when the linear transform point size is doubled. Naturally, the best compaction efficiency has been demonstrated using locally adaptive block sizes (e.g. 16x16, 16x8, 8x8, 8x4, and 4x4) (See Gary Sullivan and Rich Baker 'Efficient Quadtree Coding of Images and Video,' ICASSP 91, pp 2661-2664.). Inevitably, this introduces additional side information overhead and forces the decoder to implement programmable or hardwired recursive DCT algorithms. If the DCT size becomes too large, then more edges (local discontinuities) and the like become absorbed into the transform block, resulting in wider propagation of Gibbs (ringing) and other phenomena. Finally, with larger transform sizes, the DC term is even more critically sensitive to quantization noise.

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