bitrate and blocks

[cheyrn]

I'm reading compression for great video and audio by bill waggoner and he talks about H264 supporting 4x4 blocks saying that this requires edges to pass through fewer blocks and require less bitrate and I realize I have no clue how bitrate relates to video blocks.

Can you describe how physically a block (e.g. an 8x8 or 4x4 grid of coefficients) is constructed from a video signal and how bitrate relates to the blocks being constructed? I hope that's a clear enough question. Let me know if it's not clear, please.

[poisondeathray]

You can think of it sort of like a "Jigsaw puzzle" . If the encoder has the option to choose adaptive 4x4 and 8x4 blocks , you're able to achieve a better "fit" for different types of content because of a better subdivision - smaller details are better represented. Imagine you have a contour, and if all you have are 8x8 and 16x16 blocks - you won't be able to represent a smooth curve as well

If you have better prediction (e.g. more accurate seach algorithms, and variable block sizes to "fit" such as 4x4 blocks ) , you end up saving bitrate. The reason is what goes into those P and B frames is known as the "residual". It's the difference between the actual and what is predicted. If you only have 16x16 and 8x8 blocks available to use, the representation is less accurate => thus bigger differences => thus larger residuals => thus higher bitrate required for a certain level of "quality". HEVC takes this a step farther with the options of using 64x64 CTB's . It's especially advantageous on UHD/4K content.

[cheyrn]

So, when the book says edge's have to pass through fewer blocks, it means less change in an edge passes through a block? An edge passing through a smaller block is closer to being a single value for the entire block?

Is the content of a single block for the same video signal inherently different given one bitrate vs. another? Or, if it works this way, does more bitrate mean there are more blocks over time, for the same signal?

[poisondeathray]

So, when the book says edge's have to pass through fewer blocks, it means less change in an edge passes through a block? An edge passing through a smaller block is closer to being a single value for the entire block?

Is the content of a single block for the same video signal inherently different given one bitrate vs. another? Or, if it works this way, does more bitrate mean there are more blocks over time, for the same signal?

I'm not sure what he means by that. I don't have the book. You can ask BEN (not BILL btw ) directly at Doom9 forum .

A thin edge passing though a smaller block would be more accurate than if it were a 16x16 block. (i.e your motion vector would be more accurate if the encoder used smaller blocks) . When you divide up into smaller blocks, both intra (within the same frame, adjacent mb's) and inter (temporal, mb's on adjacent frames) predictive motion vectors benefit - you have more ability to "catch" the edge because each partition has a motion vector. "Catching" the edge means you successfully predict it, thus the residual will be lower. On the other hand, a lower quality motion vector search algorithm, or shorter ME range, or large blocks might "miss" the edge, thus larger differences, thus larger residuals, thus higher bitrates required for a certain level of "quality"

Although larger blocks "cost" less in general, they are less accurate - so like all things in compression it's trade off. If you divided using only 16x16 blocks it wouldn't necessarily be good, and the same for 4x4 only. This is what efficiency is all about, the "jigsaw puzzle" - that's the bottom line. If you have more accurate prediction (that follows from the option of using better, adaptive partitions), then you end up with smaller residuals

Is the content of a single block for the same video signal inherently different given one bitrate vs. another? Or, if it works this way, does more bitrate mean there are more blocks over time, for the same signal?

I don't understand that 1st part

"blocks" are determined by the encoder and motion estimation algorithm . It can choose to partition differently depending on the algorithm used . It might choose to divide a 16x16 "block" into 4 * 4x4 partitions, or some other combinations. That decision can be influenced number of things, including bitrates. If you have only low bitrates allowed, it might choose to allocate 16x16 instead of 4*4x4 . So yes, more available bitrates usually mean it can choose smaller blocks

[cheyrn]

Let's say 1 second of a checkered flag waving is recorded. If there was only one size of block available to the encoder, would a high bitrate recording mean that more blocks would be created for the 1 second than if low bitrate was used? If so, is it more blocks spatially or spread out over time? If not, and the blocks have different content, what type of difference would there be in the content?

[poisondeathray]

Let's say 1 second of a checkered flag waving is recorded. If there was only one size of block available to the encoder, would a high bitrate recording mean that more blocks would be created for the 1 second than if low bitrate was used? If so, is it more blocks spatially or spread out over time? If not, and the blocks have different content, what type of difference would there be in the content?

If you only had 1 size of block available it would be identical in number regardless of bitrate. ie. Each frame would have exactly the same number of blocks, regardless of bitrate , or encoding settings, or content, or anything else - because that is the smallest AND largest partition available - there are no other choices. And if everything else was the same, the quality would be lower at a lower bitrate than a higher bitrate - as expected