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public:cs:rasterization_on_larrabee [2015/09/22 20:59] – [Intra-tile rasterization: 16x16 blocks] oakfirepublic:cs:rasterization_on_larrabee [2021/01/06 17:31] (当前版本) – [Tile assignment] oakfire
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 Once we've set up the equation for an edge (by calculating B and C, as discussed when we looked at Figure 1), the first thing we do is calculate its value at the trivial reject corner of each tile. The trivial reject corner is the corner at which an edge's equation is most negative within a tile; the selection of the trivial reject corner for a given edge is based on its slope, as we'll see shortly. We set things up so that negative means inside in order to allow us to generate masks directly from the sign bit, so you can think of the trivial reject corner as the point in the tile that's most inside the edge. If this point isn't inside the edge, no point in the tile can be inside the edge, and therefore the whole triangle can be ignored for that tile. Once we've set up the equation for an edge (by calculating B and C, as discussed when we looked at Figure 1), the first thing we do is calculate its value at the trivial reject corner of each tile. The trivial reject corner is the corner at which an edge's equation is most negative within a tile; the selection of the trivial reject corner for a given edge is based on its slope, as we'll see shortly. We set things up so that negative means inside in order to allow us to generate masks directly from the sign bit, so you can think of the trivial reject corner as the point in the tile that's most inside the edge. If this point isn't inside the edge, no point in the tile can be inside the edge, and therefore the whole triangle can be ignored for that tile.
  
-当我们建立三角形一条边的方程式后( 计算 B 和 C 系数, 如图例 1 我们所讨论的), 我们做的第一件事是计算每个瓦片的平凡拒绝角:?:的值. 平凡拒绝角就是瓦片里该边方程式值最小的角; 该边的平凡拒绝角的选择基于边的斜率, 我们很快就会看到. 前面我们已经确立了负值意味着在内部以方便直接遮罩标志位, 所以你可以认为一个瓦片的平凡拒绝角是该瓦片在该条边上最内部的点. 如果这个点不在该边内部, 那么就该瓦片就没有点在该边内部, 因为该瓦片就可以忽略整个三角形.+当我们建立三角形一条边的方程式后( 计算 B 和 C 系数, 如图例 1 我们所讨论的), 我们做的第一件事是计算每个瓦片的平凡拒绝角:?:的值. 平凡拒绝角就是瓦片里该边方程式值最小的角; 该边的平凡拒绝角的选择基于边的斜率, 我们很快就会看到. 前面我们已经确立了负值意味着在内部以方便直接遮罩标志位, 所以你可以认为一个瓦片的平凡拒绝角是该瓦片在该条边上最内部的点. 如果这个点不在该边内部, 那么就该瓦片就没有点在该边内部, 因此该瓦片就可以忽略整个三角形.
  
 Figure 13 shows the trivial reject test in action. Tile 0 is trivially rejected for the black edge and can be ignored, because its trivial reject corner is positive, and therefore the whole tile must be positive and must lie outside the triangle, while the other three tiles must be investigated further. You can see here how the trivial reject corner is the corner of each tile most inside the black edge; that is, the point with the most negative value in the tile. Figure 13 shows the trivial reject test in action. Tile 0 is trivially rejected for the black edge and can be ignored, because its trivial reject corner is positive, and therefore the whole tile must be positive and must lie outside the triangle, while the other three tiles must be investigated further. You can see here how the trivial reject corner is the corner of each tile most inside the black edge; that is, the point with the most negative value in the tile.
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 All this comes down to just setting up the right values, then doing one vector add and one vector compare. Remember that the edge equation is of the form Bx + Cy; therefore, to step a distance across the tile, we just set x and y to the horizontal and vertical components of that distance, evaluate the equation, and add that to the starting value. So all we're doing in Figure 19 is adding the 16 values that step the edge equation to the 16 trivial reject corners. For example, to get the edge equation value at the trivial reject corner of the upper-left block, we'd start with the value at the tile trivial reject corner, and add the amount that the edge equation changes for a step of -48 pixels in x and -48 pixels in y, as shown by the yellow arrow in Figure 20. To get the edge equation value at the trivial reject corner of the lower-left block, we'd instead add the amount that the edge equation changes for a step of -48 pixels in x only, as shown by the purple arrow. And that's really all there is to the Larrabee rasterizer - it's just a matter of stepping the edge equation values around the tile so as to determine what blocks and pixels are inside and outside the edges. All this comes down to just setting up the right values, then doing one vector add and one vector compare. Remember that the edge equation is of the form Bx + Cy; therefore, to step a distance across the tile, we just set x and y to the horizontal and vertical components of that distance, evaluate the equation, and add that to the starting value. So all we're doing in Figure 19 is adding the 16 values that step the edge equation to the 16 trivial reject corners. For example, to get the edge equation value at the trivial reject corner of the upper-left block, we'd start with the value at the tile trivial reject corner, and add the amount that the edge equation changes for a step of -48 pixels in x and -48 pixels in y, as shown by the yellow arrow in Figure 20. To get the edge equation value at the trivial reject corner of the lower-left block, we'd instead add the amount that the edge equation changes for a step of -48 pixels in x only, as shown by the purple arrow. And that's really all there is to the Larrabee rasterizer - it's just a matter of stepping the edge equation values around the tile so as to determine what blocks and pixels are inside and outside the edges.
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 +所有这些归结为只要建立对应的值, 然后做一次向量加法与一次向量比较. 记得边方程式的形式是 Bx + Cy; 因此, 在瓦片上的步进, 可以把 x 和 y 设为横竖相对距离, 然后求值, 并加上初始值. 所以图例 19 中我们所做的就是把 16 个步进值加到 16 个平凡拒绝角. 比如, 为了得到左上角块的平凡拒绝角的边方程式值, 我们会从瓦片的平凡拒绝角的值开始, 加上 x 为 -48 及 y 为 -48 的边方程式的值, 如图例 20 中的黄色箭头. 为了得到左下角块的平凡拒绝角的值, 我们则加上 x 为 -48 及 y 为 0 的边方程式值, 如紫色箭头所示. 这就是 Larrabee 光栅化真正所有的 - 它就是步进这些边方程式值来决定哪些块和像素点是在边的外部还是内部.
  
 {{:public:cs:fig_20.jpg?direct|}} {{:public:cs:fig_20.jpg?direct|}}
  
 **Figure 20**: Examples of stepping the edge equation Bx + Cy. **Figure 20**: Examples of stepping the edge equation Bx + Cy.
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 +图例 20 :　边方程式 Bx + Cy 的步进例子.
  
 Once again, we'll do trivial accept tests as well as trivial reject tests. In Figure 21 we've calculated the trivial accept values, and determined that 6 of the 16x16 blocks are trivially accepted for the black edge, and 10 of them are not trivially accepted. We know this because the values of the equation of the black edge at the trivial accept corners of the 6 pink blocks are negative, so those blocks are entirely inside the edge, while the values at the trivial accept corners of the other 10 blocks are positive, so those blocks are not entirely inside the black edge. The trivial accept values for the blocks can be calculated by stepping in any of several different ways: from the tile trivial accept corner, from the tile trivial reject corner, or from the 16 trivial reject values for the blocks. Regardless, it again takes only one vector instruction to step and one vector instruction to test the results. Combined with the results of the trivial reject test, we also know that 5 blocks are partially accepted. Once again, we'll do trivial accept tests as well as trivial reject tests. In Figure 21 we've calculated the trivial accept values, and determined that 6 of the 16x16 blocks are trivially accepted for the black edge, and 10 of them are not trivially accepted. We know this because the values of the equation of the black edge at the trivial accept corners of the 6 pink blocks are negative, so those blocks are entirely inside the edge, while the values at the trivial accept corners of the other 10 blocks are positive, so those blocks are not entirely inside the black edge. The trivial accept values for the blocks can be calculated by stepping in any of several different ways: from the tile trivial accept corner, from the tile trivial reject corner, or from the 16 trivial reject values for the blocks. Regardless, it again takes only one vector instruction to step and one vector instruction to test the results. Combined with the results of the trivial reject test, we also know that 5 blocks are partially accepted.
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 +再次, 我们会像平凡拒绝角那样来做平凡接受角测验. 在图例 21 中我们已经计算出这些平凡接受角的值, 并判定 其中 6 个 16x16 的块是被黑边平凡接受的, 另外 10 个则不被平凡接受. 我们知道这是因为那 6 个粉丝块的黑边平凡接受角的值是负的, 那么这些块整体都在该边的内部; 而另外 10 个块的平凡接受角值是正的, 那么它们不是整体处于黑边内部. 这些块的平凡接受角的值的计算可以使用几种方法来步进:　从瓦片的平凡接受角来步进, 从瓦片的平凡拒绝角来步进, 或从 16 个平凡拒绝角的值来步进. 不管怎样, 这再次只耗费一次向量结构来步进与一次向量结构来测验. 结合平凡接受角的测验, 我们也知道有 5 个块被部分接受.
  
 {{:public:cs:fig_21.jpg?direct|}} {{:public:cs:fig_21.jpg?direct|}}
  
 **Figure 21**: The trivial accept tests for the 16 16x16 blocks in the tile. **Figure 21**: The trivial accept tests for the 16 16x16 blocks in the tile.
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 +图例 21 : 瓦片上 16 个 16x16 的块的平凡接受角测验.
  
 The 16 trivial reject and trivial accept values can be calculated with a total of just two vector adds per edge, using tables generated at triangle set-up time, and can be tested with two vector compares, which generate mask registers describing which blocks are trivially rejected and which are trivially accepted. We do this for the three edges, ANDing the results to create masks for the triangle, do some bit manipulation on the masks so they describe trivial and partial accept, and bit-scan through the results to find the trivially and partially accepted blocks. Each 16x16 that's trivially accepted against all three edges becomes one bin command; again, no further rasterization is needed for pixels in trivially accepted blocks. The 16 trivial reject and trivial accept values can be calculated with a total of just two vector adds per edge, using tables generated at triangle set-up time, and can be tested with two vector compares, which generate mask registers describing which blocks are trivially rejected and which are trivially accepted. We do this for the three edges, ANDing the results to create masks for the triangle, do some bit manipulation on the masks so they describe trivial and partial accept, and bit-scan through the results to find the trivially and partially accepted blocks. Each 16x16 that's trivially accepted against all three edges becomes one bin command; again, no further rasterization is needed for pixels in trivially accepted blocks.
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 +通过三角形建立时生成的表, 每个边的 16 个平凡拒绝和平凡接受的值可以仅由两次向量加法来计算, 并由两次向量比较来测验, 生成标记寄存器来标识哪些块被平凡拒绝哪些被平凡接受. 我们执行三条边, 生成建立三角形的标识, 在这些标识上做些位操作来表示被平凡拒绝或平凡接受, 之后通过结果的位扫描可以找出被平凡拒绝和平凡接受的块. 每个 16x16 的被三条边平凡接受的会转为一个二进制命令:?: 再次, 被平凡接受的块不需要进一步光栅化.
  
 This is not obvious stuff, so let's take a moment to visualize the process. First, let's just look at one edge and trivial accept. This is not obvious stuff, so let's take a moment to visualize the process. First, let's just look at one edge and trivial accept.
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 +这并不怎么明显, 所以让我们花一些时间来对这过程有个印象. 首先, 我们只看一条边和平凡接受.
  
 For a given edge, say edge number 1, we take the edge equation value at the tile trivial accept corner, broadcast it out to a vector, and vector add it to the precalculated values of the 16 steps to the trivial accept corners of the 16x16 blocks. This gives us the edge 1 values at the trivial accept corners of the 16 blocks, as shown in Figure 22. (The values shown are illustrative, and are not taken from a real triangle.) For a given edge, say edge number 1, we take the edge equation value at the tile trivial accept corner, broadcast it out to a vector, and vector add it to the precalculated values of the 16 steps to the trivial accept corners of the 16x16 blocks. This gives us the edge 1 values at the trivial accept corners of the 16 blocks, as shown in Figure 22. (The values shown are illustrative, and are not taken from a real triangle.)
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 +对给定边, 称为边 1, 我们取得瓦片平凡接受角的该边方程式的值, 广播给一个向量处理器(阵列处理器),阵列处理器把它加到预先计算好的 16x16 块的步进值. 这就给了我们 16 个块的边 1 平凡接受角的值. 如图 22. ( 显示的值是用过说明的, 所以不是取自真实三角形)
  
 {{:public:cs:fig_22.jpg?direct|}} {{:public:cs:fig_22.jpg?direct|}}
  
 **Figure 22**: Edge 1 trivial accept tests for the 16 16x16 blocks. **Figure 22**: Edge 1 trivial accept tests for the 16 16x16 blocks.
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 +图 22 : 16 个 16x16 块的边 1 的平凡接受测验.
  
 The step values shown on the second line in Figure 22 are computed when the triangle is set up, using a vector multiply and a vector multiply-add. At the top level of the rasterizer - testing the 16x16 blocks that make up the tile, as shown in Figure 22 - those set-up instructions are just direct additional rasterization costs, since the top level only gets executed once per triangle, so it would be accurate to add 6 instructions to the cost of the 16x16 code we'll look at shortly in Listings 1 and 2. However, as the hierarchy is descended, the tables for the lower levels (16x16-to-4x4 and 4x4-to-mask) get reused multiple times. For example, when descending from 16x16 to 4x4 blocks, the same table is used for all partial 16x16 blocks in the tile. Likewise, there is only one table for generating masks for partial 4x4 blocks, so the additional cost per iteration in Listing 3 due to table set-up would be 2 instructions divided by the number of partial 4x4 blocks in the tile. This is generally much less than 1 instruction per 4x4 block per edge, although it gets higher the smaller the triangle is. The step values shown on the second line in Figure 22 are computed when the triangle is set up, using a vector multiply and a vector multiply-add. At the top level of the rasterizer - testing the 16x16 blocks that make up the tile, as shown in Figure 22 - those set-up instructions are just direct additional rasterization costs, since the top level only gets executed once per triangle, so it would be accurate to add 6 instructions to the cost of the 16x16 code we'll look at shortly in Listings 1 and 2. However, as the hierarchy is descended, the tables for the lower levels (16x16-to-4x4 and 4x4-to-mask) get reused multiple times. For example, when descending from 16x16 to 4x4 blocks, the same table is used for all partial 16x16 blocks in the tile. Likewise, there is only one table for generating masks for partial 4x4 blocks, so the additional cost per iteration in Listing 3 due to table set-up would be 2 instructions divided by the number of partial 4x4 blocks in the tile. This is generally much less than 1 instruction per 4x4 block per edge, although it gets higher the smaller the triangle is.
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  • 最后更改: 2015/09/22 20:59
  • 由 oakfire