Computer Graphics : Sutherland-Hodgman Polygon Clipping Algorithm Thanks again for watching ! If you liked it pls make it a point to comment below. Keeps me going :)
I was thinking the same thing. (Also, I think you meant V2', instead of V3'). This video is wrong. If anyone is looking around, to save time, I found a video with the correct explanation of the algorithm: ruclips.net/video/Euuw72Ymu0M/видео.html
The video was an good guide for me during my semester exam i thank you for your help and i request you to upload this kind of video with updated version of items frequently EVERY ENGINEER SHOULD WATCH
I have an exam tomorrow and wasn't being able to understand this from the book. But now i clearly understood it. Thanks a lot for this informative video!!!
Another scenario should be taken into account is the case when both vertices are outside the clipping area, while the line they form *does* cross the clipping area. I guess in this case you should not discard both v1 and v2
talsveta yeah in that case you have to assume a point on the line crossing as a vertices of the polygon(hypothetically) and then break that line into the cases mentioned in the video. Thank for pointing it out though:)
Considering points as "inside" and "outside" will not work in all cases. Check this image. i.imgur.com/e6S0V20.png Segment ED is intersecting rectangle, but nither E nor D is inside the rect.
The solution is to NOT check the 4 cases according to the rectangle but according to the clipping edge I guess. If you have a closer look at the pseudocode provided by e.g. wikipedia ( en.wikipedia.org/wiki/Sutherland%E2%80%93Hodgman_algorithm#Pseudo_code ) you quickly see that one has to pick an edge of the clipping polygon. E.g.: Take the 1. Clipping Edge (e.g. left edge) -> now take as input points all points of the polygon and check the 4 cases against this edge -> i.e. if its left of the edge -> its outside, if its right of it -> its inside -> add the new intersection points like the cases tell you and you get a new list L' -> you take that list as the input list for the next iteration Take the 2. Clipping Edge (e.g. top edge) and as point input list L' from the previous iteration. -> now perform the same steps as before and get the next output list L' ... continue this procedure for the other 2 edges of your rectangle. I think following the steps like I wrote will indeed capture the edge by E and D. ( Please correct me if I am wrong because I also need to know the correct solution :D )
Thank you very much! It was great and well explained! Just one other note - I think the last edge you made between the vertices v5' and v3' is not trivial and worth to mention.. I.e. if those vertices had not been signed with a dash, you would have needed to connect them with one direct edge and not over the actually 2 edges that are formed by the surroundings of the clipping area
I have watched a few videos of this channel and they are amazing You should publish a full fledged course on Computer Graphics as there is no good source of it on youtube Videos were
How does this algorithm handle the case when one or more corners of the rectangles is part of the clipping? You just included the corner without any explanation in the example you solved. One other case that is not handled is when both v1 and v2 are outside the clipping window but the edge between them intersects the clipping window.
To solve your first query, a corner of the window is supposed to be within its boundaries. So it's treated as 'inside'. For your second question, Sutherland Hodgeman polygon clipping algorithm clips a polygon with respect to one boundary at a time. So you actually need to repeat this four times for a rectangular window. In the video, he has tried to solve it in one go which would create confusion. I suggest you solve this case considering one boundary at a time. It definitely works. However be careful that you figure out 'inside' and 'outside' vertices considering only the boundary then and not the entire polygon.
there seems to be a minor error at time 3:40 in video for rule 'd' as you ask us to save v2' but while solving the sum you consider points v1' as at 6:55 you take point v5' instead of v1' oh !! and love your videos................Good Job :) couldnt find for liang barsky algorithm though please share the link unless you havent made one if so please make one THANK YOU
This looks like something to test. What would happen if there is a hole in the polygon? Looks like O^2 problems. If v1&v2 both outside, but their segment crossing the corner?
The BootStrappers This is a good tutorial bro, no need to be faster because there is some people who dont understand english well.. Keep the good work!!
Look the video is very good but there was a problem in the sheet 3 where you did not draw the window..How are gonna know if the line is going in or out if you dont show the window ?
In the rule 4 sir u have mentioned v1 inside and v2 outside and we must save v2' in the polygon example the line from v5 to v1 ,v5 is inside and v1 is outside according to rule 4 v1' must be saved but sir y did u mentioned v5'
i think according to rules when goes from v3 to v4 we make v4' so how we make 2nd V4' when we go from v4 to v5 that should make also V4' that will make 2 v'4 ???
an entire lecture thing in just 8 min. just wow. much helpful thanks a lot .
You didn’t explain how you clipped the low left corner. V5’ would wrap to V3’ not magically magically create the corner.
I was thinking the same thing. (Also, I think you meant V2', instead of V3').
This video is wrong. If anyone is looking around, to save time, I found a video with the correct explanation of the algorithm: ruclips.net/video/Euuw72Ymu0M/видео.html
The video was an good guide for me during my semester exam i thank you for your help and i request you to upload this kind of video with updated version of items frequently EVERY ENGINEER SHOULD WATCH
Thankyou for the lecture..👍
And your voice is so beautiful🌝
Amazingly clarified..I must appreciate your way of teaching so politely . Thankyou so mch sir 😇
I have an exam tomorrow and wasn't being able to understand this from the book.
But now i clearly understood it. Thanks a lot for this informative video!!!
Thanks a lot! Also, your voice is brilliant XD
Another scenario should be taken into account is the case when both vertices are outside the clipping area, while the line they form *does* cross the clipping area. I guess in this case you should not discard both v1 and v2
talsveta yeah in that case you have to assume a point on the line crossing as a vertices of the polygon(hypothetically) and then break that line into the cases mentioned in the video. Thank for pointing it out though:)
Yeah,I was also thinking the same!
Great tutorial otherwise!
good tutorial for Sutherland-Hodgman Algorithm. thank you, you really help me a lot
+Mark Woo
Thank you :)
Thank you sir simple and easy video with kind understanding ❤😊
Brilliant video!
I really liked your lectures they are very beneficial..... I request you to pls upload more videos on computer graphics
Considering points as "inside" and "outside" will not work in all cases.
Check this image. i.imgur.com/e6S0V20.png
Segment ED is intersecting rectangle, but nither E nor D is inside the rect.
+Ostap Hrytsyshyn yup
did you get any solution of this bro???
The solution is to NOT check the 4 cases according to the rectangle
but according to the clipping edge I guess.
If you have a closer look at the pseudocode provided by e.g. wikipedia
( en.wikipedia.org/wiki/Sutherland%E2%80%93Hodgman_algorithm#Pseudo_code )
you quickly see that one has to pick an edge of the clipping polygon.
E.g.:
Take the 1. Clipping Edge (e.g. left edge)
-> now take as input points all points of the polygon and check the 4 cases against this edge
-> i.e. if its left of the edge -> its outside, if its right of it -> its inside
-> add the new intersection points like the cases tell you and you get a new list L'
-> you take that list as the input list for the next iteration
Take the 2. Clipping Edge (e.g. top edge) and as point input list L' from the previous iteration.
-> now perform the same steps as before and get the next output list L'
... continue this procedure for the other 2 edges of your rectangle.
I think following the steps like I wrote will indeed capture the edge by E and D.
( Please correct me if I am wrong because I also need to know the correct solution :D )
Thank you so much! really well explained! God bless u!
very understandable. Thank you 🙏
Just in one word "AWESOME"
+ashwani prasad just in two words "THANK YOU"
This was great, thanks.
thank you sir, very nice explanation! very useful!
Very nicely explained . very precise
prateek parab thank you brother
thank you for this video... wonderful explanation..
thanks buddy... well explained without any doubts.
would be great if you have given these in written points,but it was such an easy explanation ,thanks
I like that. Well explained!
Thank you very much sir ❤️❤️❤️❤️
Thanks a lot for the video... It was really helpful.
very nice mann
Good job!
very good explanation, thanks
Kladimir happy to help!
Nice explanation
Thank you very much! It was great and well explained!
Just one other note - I think the last edge you made between the vertices v5' and v3' is not trivial and worth to mention.. I.e. if those vertices had not been signed with a dash, you would have needed to connect them with one direct edge and not over the actually 2 edges that are formed by the surroundings of the clipping area
talsveta sure thanks !
+talsveta Yess you are right ! i think that's the small mistake done by talented buddy :D
it really helped me.. thanks a lot man
Legend
Great tutorial, very helpful
Gr8 one dude
excellent video bro. nice keep it up...
Wow , amazing sir ,, thank you 🙏🏻
very good explanation
thanks for the video. it helped a lot.
I have watched a few videos of this channel and they are amazing
You should publish a full fledged course on Computer Graphics as there is no good source of it on youtube
Videos were
Excellent Video , Thanks Bro (Y)
awsome video
How does this algorithm handle the case when one or more corners of the rectangles is part of the clipping? You just included the corner without any explanation in the example you solved.
One other case that is not handled is when both v1 and v2 are outside the clipping window but the edge between them intersects the clipping window.
To solve your first query, a corner of the window is supposed to be within its boundaries. So it's treated as 'inside'. For your second question, Sutherland Hodgeman polygon clipping algorithm clips a polygon with respect to one boundary at a time. So you actually need to repeat this four times for a rectangular window. In the video, he has tried to solve it in one go which would create confusion. I suggest you solve this case considering one boundary at a time. It definitely works. However be careful that you figure out 'inside' and 'outside' vertices considering only the boundary then and not the entire polygon.
Not the entire window* i mean
it works if you repeat the process in EACH edge-boundary, not everything at once.
@@renzocoppola4664 Yep, and you have to treat them as infinitely long
Nice explanation.. keep it up👍👍👍👍
Was useful and helpful
Tmaari awaaaz, uffff, agar natural hai to im sorry, or ye jaanke istarah ki nikaal rhe ho to, katai jeher ho tum
Amazing !!
God bless you !
thanks mahn ....helped a lot
there seems to be a minor error at time 3:40 in video for rule 'd' as you ask us to save v2'
but while solving the sum you consider points v1' as at 6:55 you take point v5' instead of v1'
oh !!
and love your videos................Good Job :)
couldnt find for liang barsky algorithm though
please share the link
unless you havent made one
if so please make one
THANK YOU
+Kshitij Makwana sahi a 22
Thankyou very much....😊
very useful, one note, what you pronounce as v1 dash is more commonly called v1 prime
best video on SHPCA...
realy helpfull for me! thanks
Good Job..
thanks brother
It's awsm bro
thanks broda!!!
Awesome
really helping tutorial... thankyou (y)
Thank you :)
Simple mistake last step V5 -V1 not V5' its (V1')
Good work bro..
please upload 2-D and 3-D transformation lecture too....its helping a lot
thank you
nice explanation...
Helpful.....
+The BootStrappers, 8:01 tho,... That v2' & v5' should be joined connected directly... Right? :\
Anyway, awesome presentation... really helped me! :) :')
The polygon you have used for clipping is a concave polygon!!
well explained.... also post liang barsky algo...pls...
Thanks
08:22 So, initially this polygon had 5 sides. After clipping, the polygon has 7 sides. Right ? or is it 6 ?
I like to point out a small mistake in in 4 th example
The clipping from inside to outside we have to save the inside point v1dash instead of v2dash
Is the algorithm is will work for all complicated shapes??? Or only works on any shapes clipped by rectangle shape??? @The BootStrappers
please upload the video of midpoint circle drawing algorithm
their is a confusion between inside to outside point. which one we have to take you didnt mention in the lecture......
how can you go vertical form v2' and then horizontal ??( 8:00)
Vertices of the square polygon have not been saved.
U have nice communication skills though... And very nicely presented... 👍👌
Thank u bro
Well understood…plz Post liang barsky line clipping
Nice
At 3:50 it should be v1 dash not v2 dash. Thnx for great lec.
This looks like something to test.
What would happen if there is a hole in the polygon? Looks like O^2 problems.
If v1&v2 both outside, but their segment crossing the corner?
It will fail if v0 ,v1,v2 (a triangle larger as the clip rectangle) all outside the 2D screen rectangle or 3D view frustum.
DJLinux007 that’s why we have another algorithms.
Concise explanation, sir. But be a bit fast.
blackaura345 sure ! Will take note of it in the upcoming videos
The BootStrappers
This is a good tutorial bro, no need to be faster because there is some people who dont understand english well.. Keep the good work!!
There is some mistake in this video!!!!! V3 to v4 there should be saved v4' but you saved v3' and there is also mistakes in v5 to v1
yes how can we save that point
?
joining the end points? you mean to reiterate the algorithm?
Uhm, no. You know about this case after checking if it's two sides off
Thanxx sir
when V1 goes from outside to inside V2, the intersection is actually V1', not V2'
very useful fr me👌👌👌
After clipping , in the final diag you didn't name the vertex between v2' and v5' What about that?
SAYAN BOSE bhaaiiii aar poris na
Can we start from any vertex of polygon? or what would be the direction to move (clockwise or anti.C)?
Look the video is very good but there was a problem in the sheet 3 where you did not draw the window..How are gonna know if the line is going in or out if you dont show the window ?
Soham Mondal sorry buddy I'll take care of it next time!
yes,plz !!
What about the case if all points are outside the window and window is completely inside the polygon?
what about if we have a non-rectangular clipping window?
how can you tell the direction of the line when the question is not given any direction at all? :/
U said for 'inside to outside' that we consider the outside point but while solving problem u saved the inside point.... So which one we hv to save???
In the rule 4 sir u have mentioned v1 inside and v2 outside and we must save v2' in the polygon example the line from v5 to v1 ,v5 is inside and v1 is outside according to rule 4 v1' must be saved but sir y did u mentioned v5'
How did you determined the inside and outside points !!!!!!say it first
i think according to rules when goes from v3 to v4 we make v4' so how we make 2nd V4' when we go from v4 to v5 that should make also V4' that will make 2 v'4 ???
its wrong the last part v2' should be connected to v5' directly
how can a line bend between to point in clipping
You have to follow LRBT. Otherwise you can't say about the v2' and v1'.
helpful..
Isn't the last point which you saved, suppose to be v1'?