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Hello and welcome to lecture number 22 in the course Computer Graphics. We are currently discussing the 3D graphics pipeline. And the pipeline has got 5 stages. We have already discussed object representation that is the first stage. Then modelling transformations - second stage. Lighting or assigning colour - third stage. Currently, we are in the fourth stage that is viewing pipeline. As you can see, it consists of 5 sub-stages. We have already discussed few of those and continuing our discussion on the remaining ones.
So, among those sub-stages we have already discussed earlier. View transformation, projection transformation and viewport transformation.
Two more operations are there, as we have seen in the pipeline; clipping and hidden surface removal. Among them currently we are discussing clipping.
So, in the last lecture, we introduced the basic idea of clipping and also discussed 2D line clipping. So, will continue our discussion on clipping. Today, we are going to discuss fill area clipping as well as 3D clipping.
So, what is this fill area clipping? So, as we mentioned, when we talk of clipping, there is a clipping window and earlier we have discussed how to clip points and lines against this window. However, when we project objects the projection maybe in the form of a fill area such as a polygon where there is a boundary.
Now clipping a filled area is different than flipping a point or a line, as we shall see in todays lecture. In fact, such situations are quite frequent in practice where we have to clip polygons against the clipping window. So, it requires some mechanism to do that.
Now, what can be a very obvious and straightforward approach, let us try to understand the situation. Suppose this is our clipping window and we are given a polygon, something like this after projection say this triangle. So we have to keep this part which is inside the clipping window, which I am showing with shade and we have to clip out the, outside part. How we can do that?
One way can be to use the line clippers that we discussed in earlier lecture for each of the edge, like here is one edge, one edge, one edge of the field area. And then perform clipping on the edges and decide on the clipped region. However, as you can see from this example, that is not necessarily easy, efficient and going to give us a good approach. Sometimes it is even difficult to understand how it works
Instead, we require better approaches. There are in fact, many efficient algorithms proposed for the purpose. In this lecture we are going to discuss two of those approaches. One is Sutherland-Hodgeman algorithm and the other one is Weiler-Atherton algorithm. Let us try to understand these algorithms.
We will start with the Sutherland-Hodgeman algorithm, what this algorithm does? Here in this algorithm we start with 4 clippers. Now, these clippers are essentially the lines that define the window boundary. For example, if this is my window boundary, then each of these lines defining the boundary is a clipper.
So, there are 4 clippers in 2D clipping that is right, left, above and below. Now, each clipper takes as input a list of ordered pair of vertices which essentially indicate the edges, each pair of vertex indicate the edge. And from that input list it produces another list of output vertices that is the basic idea. So, there are 4 clippers, each clipper takes as input a list of ordered pair of vertices where each pair of vertices represent an edge. And then it performs some operations to produce an output list of vertices.
Now, when we perform these operations, we impose some order of checking against each clipper that can be any order. Here in this discussion will assume the order left clipper first, then right clipper, then bottom clipper, and at the end the top or above clipper.
Now, as we said we start with the left clipper. So, its input set is the original polygon vertices or in other words, the original polygon edges represented by the pair of vertices that is the input set to the first or the left clipper.
Now, to create a vertex list as output or also to provide the input vertex list, we need to follow a naming convention, whether to name the vertices in a clockwise manner or anticlockwise manner. Here again, we will assume that we will follow an anticlockwise naming of vertices. With these conventions, let us denote input vertex list to a clipper by the set V having these vertices.
Now, for each edge or the pair of vertices in the list denoted by vi, vj. We perform some checks and based on the check results, we take some action. So, what are those checks?
If vi is inside and vj is outside of the clipper then we return the intersection point of the clipper with the edge represented by the vertex pair vi, vj. If both vertices are inside the clipper, then we return vj.
If vi is outside and vj is inside of the clipper, then we return two things. One is the intersection point of the clipper with the edge represented by the pair vi, vj and also vj. Both the things we return intersection point and vj. And finally, if both vertices are outside the clipper then we do not return anything, we return null.
Now here we have use the terms inside and outside. So, how they are defined? In fact these terms are to be interpreted differently for different clipper. So, there is not a single meaning to these terms based on the clipper we define these terms.
And let us now go through this definition for each of the 4 clippers. So, for the left clipper, when you talk of insight we mean that the vertex is on the right side of the clipper and when we talk of outside we mean that it is on the left side of the clipper. For right clipper it is just the opposite.
When the vertex is on the left side, we call it inside. Otherwise it is outside. For top clipper if a vertex is below the clipper that means it is inside.
Otherwise it is outside. And for a bottom clipper, it is again just the opposite of top clipper that means inside vertex means it is above the clipper, whereas outside means it is below. And how do we determine whether a vertex is on the right side or left side or above or below, just by considering the coordinates values, by comparing the coordinated values of the vertex with respect to the particular clipper.
For example, suppose this is the top clipper, suppose it is equation is given by x = y = 4. Now suppose a point is denoted by (3, 5). Now we check here the y-value of the point that is 5, clearly 5 is greater than 4 which is the y-value of the boundary, top boundary. Then we can say that this point is outside because it is above the clipper. Similarly, we can determine the inside and outside based on comparing the x or y coordinate value of the vertex with respect to the clipper values.
If the vertex is on the clipper then it is considered inside in all the cases. So, for a left clipper inside means either it is on the right side or on the clipper, otherwise it is outside. And same is true for all other clippers.
Now, let us try to understand this algorithm in terms of an illustrative example. Let us consider this situation here we have defined one clipping window and we have a fill area. Now this fill area is defined by the vertices {1, 2, 3} as you can see here, we followed a counter clockwise or anticlockwise naming convention to list the vertices.
Our objective is to determine the clipped polygon. That is this polygon denoted by the vertices {2’, 3’, 3’’, 1’ and 2}. And we use to do that by following the Sutherland-Hodgeman algorithm.
So, at the beginning we start with the left clipper. Then we check against right clipper, then top clipper and then bottom clipper.
Let us see what happens after checking against the left clipper. So here the input vertex list is the original vertices that is {1, 2, 3} which indicates the three edges represented by the vertex pair; {1, 2}; {2, 3} and {3, 1}. So, for each pair we perform the check. For pair {1, 2} we can see that
both the vertices are on the right side of the left clipper that means both are inside. So, Vout is 2 as per the algorithm.
Similarly, after checking for {2, 3} against the left clipper, we can see that the final Vout becomes {2, 3} taking into account 2. And after checking {3, 1} the final list becomes {2, 3, 1}. In all the cases against the left clipper all the vertices are inside.
Now, let us check against right clipper. So, now the input vertex list is {1, 2, 3} same and initially Vout is NULL. So, pair of vertices has to be checked {1, 2; 2, 3} and {3, 1}, all the three edges we need to check. For 1, 2 both are inside the right clipper. We can check by comparing their coordinative values because both of them are on the left side of the right clipper. So, Vout is now 2. Then we check the pair {2, 3} here we can see that 2 is inside whereas 3 is outside. So, in that case we compute the intersection point 2’ this point and then set Vout to be {2, 2’}.
Then we check {3, 1} here vertex 3 is outside because it is on the right side of the clipper, whereas 1 is inside because it is on the left side. So, here we calculate the intersection point 3’ and then finalize the output vertex list as {2, 2’, 3’ (and) 1}; because in this case we return 1 also. So, then after checking against the right clipper, we get this output list; {2, 2’, 3’ (that means this point and), 1}.
Then we check against the top clipper. Now in this case, the Vin or the input vertex list is the output vertex list after checking against the right clipper. So that is 2, 2’, 3’, 1. So, initially Vout is NULL. And the pair of vertices we need to check are 4; {2, 2’, 2’, 3’, 3’ 1} and {1, 2}.
So, first we check {2, 2’} against the top clipper and we find that both 2 and 2 dash are inside because both of them are below the clipper. So, output list becomes 2’. Then we check the next vertex pair {2’, 3’} again {2’, 3’} both are below, so inside then Vout becomes 2’ and 3’.
Then we check 3’, 1 in this case, we see that 3’ is inside, whereas 1 is outside. Then we calculate the intersection point 3’’ here and modify our output list to be {2’, 3’, (and) 3’’}. Finally, we check {1, 2}. Here we see that 1 is outside whereas 2 is inside. Then again we calculate the intersection point 1’ and modify Vout to be {2’, 3’, 3’’, 1’ (and) 2}. So, this is our output list after checking against top clipper, and this serves as the input list to the remaining clipper to be checked that is bottom clipper.
This is the input list for the bottom clipper and output list is initially null and as we can see all these vertices 2, 2’, 3’, 3’’ and 1’ are inside because they are above the bottom clipper. So, the output list becomes the same that is {2’, 3’, 3’’, 1’, 2}. This is also the output of the algorithm, because here no more clippers are there to check against and the algorithm stops. So, at the end of the algorithm, we get this vertex list which represents the clipped region. That is how the algorithm works.
Now let us move to our next algorithm that is Weiler-Atherton algorithm. Now, the Sutherland-Hodgeman algorithm that we just discussed works well when the fill area is a convex polygon and it is to be clipped against a rectangular clipping window. So, if this condition satisfy then Sutherland-Hodgeman algorithm works well.
However that need not be the case always and that Weiler-Atherton algorithm provides a more general solution. This algorithm can be used for any polygon, either concave or convex against any polygonal clipping window. Need not be only a rectangle. Let us see how it works.
So, we will try to understand the algorithm in terms of an example rather than formal steps. Let us consider this scenario here we have a rectangular clipping window and a fill area. So, we will try to understand how the algorithm helps us identify the parts to be discarded that is this region
and the parts to be kept after clipping that is these two regions this one and this one. So, here we start with processing the fill area edges in a particular order, which is typically anticlockwise order. So, here we start with processing the fill area edges in a particular order which typically is anticlockwise order.
So, what we do in the processing, we check the edges one by one, continue along the edges till we encounter an edge that crosses to the outside of the clip window boundary. Let us start with this edge (1, 2) this edge. So, we check it whether it crosses the window boundary or not, that is our processing. It does not cross so we continue to the next stage, that is {2, 3} represented by the vertex pair {2, 3}.
Now this edge crosses to the outside of the window boundary. Note that here we are following anticlockwise order. If the edge does not cross to the outside instead if the edge is crossing into inside of the window then we just record by intersection point, whereas if the edge is crossing to the outside boundary, then we stop and perform some different action, what we do in that case.
At the intersection point, we make a detour. So, here the intersection point is 2’ this point. So, then we make a detour. We no longer continue along this direction. Instead what we do we now follow the edge of the clip window along the same direction, maintaining the traversal order.
So, now in this example so we will follow this anticlockwise direction and make a detour from here now along the window boundary, so here we will follow this order. So, essentially, how you are traversing, we initially traversed in this way then while traversing in this way found that this edge is crossing to the outside. So, then we traverse in this way instead of continuing along the edge.
Now, this along the boundary traversal, we continue till we encounter another fill area edge that crosses to the inside of the clip window. So, here as you can see, if we follow a anticlockwise traversal, then this edge is actually crosses to the inside. So, the edge is 6, 1 denoted by the vertex pair 6, 1 which crosses to the inside of the window and we encountered it while traversing along the window boundary. At that point what we do?
At that point, we resume the polygon edge traversal again along the same direction. So, we stop here and then again continue along the same direction till we encounter previously processed intersection point. So, here we continue up to point 1 because point one is already processed. So, we stop here.
So, then after this part, we see that we started from here, then traversed up to this point, determined this intersection point, then traversed along this line up to this intersection point, traversed back up to the originating point. So, there are two rules of traversal from an intersection point due to outside to inside fill area edge we should follow the polygon edges from an intersection point due to inside to outside fill area edge we should follow the window boundaries.
So, these are the rules we applied while performing the traversal. But this gives us one part of the clipped area that is this part and apparently here it stopped. So, how to get to the other part? Actually, here the algorithm does not stop. What happens next?
Before we go into that, also, you should remember that whenever we are traversing the traversal direction remains the same, irrespective of whether you are traversing along the edge or along the windows boundary. So, if you are following an anticlockwise direction, it should be anticlockwise always.
And after this traversal ends, the output is the vertex list representing a clipped area, as we have seen. So, in this case the traversal ended at 1. So, we get the vertex list {1, 2, 2’, (and) 1’} which gives us this clipped area.
But clearly here, the whole fill area is not covered. Some of the vertices are still not processed, so then what do we do? We resume traversal in case all the vertices are not processed. We resume the traversal along the polygon edges in the same direction from last intersection point of
an inside outside polygon edge. So, our last intersection point of an inside outside polygon edge is 2’ here. Remember that this 1’ is outside inside edge. So, it is not applicable. So, what is applicable is 2’. So, from there we resume our traversal till we cover the remaining vertices.
And this traversal is in a similar way that we have done before. So, here what we do, we traverse along this anticlockwise direction to the vertex here. So, we traverse this edge, then this edge. But here, as you can see, there is an outside to inside crossing. So, we do not do anything, we keep on traversing this way, this way. Now at this point we can see that one inside to outside crossing is there. In the earlier case, it was outside to inside.
Here it is, inside to outside at 6’. So, now we traverse along the edge. Then we encountered this intersection point again. This is from outside to inside. So, now we resume our traversal along edge. So, finally what we did, we traversed this direction, this direction, this direction, then this direction, then this direction, this direction. Now since already we have encountered 4 before so we stop our traversal here when we encounter 4. Then we get this remaining portion of the clipped area also just like the way we got it earlier. So, that is how Weiler-Atherton works.
So, we encountered or we discussed two algorithms; one is Sutherland-Hodgeman, one is Weiler-Atherton. Sutherland Hodgeman is simpler but it has restrictive use. It is applicable when we have a convex polygon which is clipped against a rectangular window. Whereas Weiler-
Atherton is more generic it is applicable for any fill area, polygonal fill area, either concave or convex against any polygonal clipping window.
So, so far we have discussed clipping in 2D. So, we have learned how to clip a point line and fill area. Now let us try to understand clipping in 3D, because here our main focus is 3D graphic pipeline. So, we will try to understand clipping in 3D which is essentially extension of the ideas that we have already discussed that is clipping in 2D. Let us see how these extensions are done.
Only thing we have to keep in mind is that here we are talking about clipping against normalized view volume which is usually a symmetric cute with each coordinate in the range minus 1 to 1 in the 3 directions. That is a normalized view volume we assume while developing the or performing the clipping. Now, Cohen-Sutherland we can extend the basic 2D version to 3D with some modification.
Point clipping also, we can extend, so let us first talk about point clipping. Here we check for x, y and z earlier we are checking only for x and y whether these values are within the range of the canonical volume. If that is so, then the point is to be kept. Otherwise it is to be clipped out.
In case of Cohen Sutherland line clipping algorithm, it can be easily extended to 3D clipping. However, with some modifications, core idea remains the same. That is, we divide view coordinate space into regions. Now, earlier we had 9 regions. Now since we are dealing with 3D
we have 27 regions, 3 times. Now since we have 27 regions. So, each region needs to be represented with 6 bits. Each bit for the 6 planes that define the canonical view volume. Far, near, top, bottom, right, left this is in contrast with the 4 bits earlier used to denote the 4 sides of the window.
Now for each plane, we have this 9 regions defined so there are 9 regions behind the far plane. There are 9 regions between near and far plane and there are 9 regions in front of the near plane. Together there are 27 regions and each region is represented with this 6 bit code, where bit 6 represent the far region, bit 5 is the near region, bit 4 is the top region, bit 3 is the bottom region, bit 2 represents the right region and bit 1 represents the left region. The idea remains the same with 2D only the size changes because we are now dealing with 3D. The other steps remained the same.
Now let us try to understand the extension of the algorithms for fill area clipping. So, here what we do. We first check if the bounding volume of the polyhedron that is the fill area is outside the view volume simply by comparing their maximum and minimum coordinate values in each of the x, y and y directions. If the bounding volume is outside then we clip it out and entirely. Otherwise we apply 3D extension of the Sutherland Hodgeman algorithm for clipping.
Here also the core idea of 3D Sutherland Hodgeman algorithm remains the same with 2D version with two main differences. What are those differences?
A polyhedron is made up of polygonal surfaces. So, here we take one surface at a time to perform clipping. Earlier what we were doing, we took one line at a time. Here we are taking one surface at a time. Now, usually polygons divided into triangular meshes and there are algorithms to do so which you can refer to in the reference material at the end of this lecture. So, using those algorithms, we can divide a polygon into a triangular mesh and then each triangle is processed at a time.
And the second difference is, instead of the 4 clippers that we had earlier, we now have 6 clippers. Which correspond to the 6 bounding surfaces of the normalized view volume which is a cube. So, these are the differences between the 2D version of the algorithm and the 3D version that earlier we are considering line at a time for clipping. Now we are considering a surface at a time. Now these surfaces are polygonal surfaces and we can convert these surfaces into triangular meshes.
And then we perform clipping for each triangle at a time that is one difference. Other difference is earlier we are dealing with 4 clippers, now we have 6 clippers representing the 6 bounding planes of the view volume which is a cube. So, that is in summary the major differences between 2D clipping and 3D clipping. Core ideas remain the same some minor changes are there. So, with that we come to the end of our discussion on clipping.
And our next topic will be hidden surface removal. So, here few things omitted during the discussion. For example the triangular mesh creation from given polygon. So, for these details you may refer to the material that will be mentioned in the next slide.
So, whatever I have covered today can be found in this book. You can go through chapter 7, section 7.1.3, 7.1.4 and section 7.2. For the topics that I have covered however outside this topics
also there are few interesting things that I did not discuss but you can find that in the book. So, you may like to go through those material as well. That is all for today. Thank you and goodbye.
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