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COSC6380-Digital Image Processing (Dr. Pranav Mantini) Fall 2022 - DIP, Fall 2022, Lecture - 7, Binary Image Processing: Thresholding, Otsu's Thresholding
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01:24 Okay. Uh Do you have any ? Any questions with respect of the
01:39 ? I believe it's due tonight. go on. Yeah. No you
01:59 have to push any of your They're automatically generated at the back
02:11 Yeah. So do not. Hard uh you know size of the image
02:17 the column? It which you're splitting of that has to be taken from
02:20 input arguments. Any more questions. one. Okay. Uh let's get
03:08 then. Um So we've been talking binary images if you remember the last
03:14 . Um you know we introduced what binary images and we said we'll ask
03:19 same two questions as we asked in beginning. That is how do you
03:23 information in the form of a binary ? The second question was how do
03:27 improve the information in a in an in this case in a binary
03:32 So in the last class we did how I can get how I can
03:36 a binary image. Um That is how I can capture information the form
03:40 a binary image. And today we'll with the second question and try to
03:45 how we can improve the information in binary image. Okay, once I
03:49 a binary image um how do I it. Right so going back again
03:54 said binary images or those that take value of either zero or one at
03:59 pixel location, thereby valued and in they tend to represent some feature and
04:05 image and we said we'll follow the that one. Well um one is
04:13 by black and zero is represented by . Okay. And um how do
04:20 get? How do we create binary ? How did we create binary images
04:27 where do we get binary images from ? We can get it from grayscale
04:35 . That is correct. One is there are devices that create binary images
04:39 ? Like caresses to pad Or we actually take a grayscale image and then
04:45 a binary image from that. So how do we create a binary
04:48 from a grayscale image? What is basic operation to do that? We
04:54 to do thresh holding right that is define a threshold and then you compare
04:59 pixel with that threshold. If it's than threshold you give it a
05:03 If it's greater than threshold you give up one, you get a by
05:06 , the major at that point. okay. And we said um the
05:16 itself will depend on the image. ? Not all thresholds will work for
05:20 images but every image you would have define a particular threshold or you have
05:26 estimate a threshold? Right now, did we go about doing it in
05:31 ? Um What kind of images can can we actually threshold to create the
05:39 images? Do they have to have certain character with respect to the hissed
05:46 gram? The instagram has to be model that is. When you look
05:57 the instagram there should be at least peaks between which you can establish a
06:03 . Right. If you do that we can find a threshold between them
06:06 we can use that to create a image. Right, okay, now
06:12 saw two ways to estimate this Uh do you guys remember what those
06:26 the 11 of them was arses thresh ? Right. How do we do
06:30 is thresh holding, what is the idea for artist thresh holding?
06:50 You take the hissed a gram and how do we do about the stress
07:03 ? What is the overall idea? ? That is funny. That was
07:12 first method we saw. Right. The first method was an iterative method
07:19 we said, okay, I do know the where the peaks are
07:22 If I knew the peaks, I find the threshold by averaging them
07:26 Um but at the same time to the peaks, I need to know
07:29 the threshold is. Right. So was kind of, we had to
07:33 with an iterative problem where we made assumption on the threshold, given the
07:37 , we took expectations on either side gave us the locations of the
07:42 And given those peaks, we computed threshold again and we kept doing that
07:46 the expectations did not move anymore. . This was a simple iterative algorithm
07:51 we saw to solve it right. the second method was the arts is
07:56 holding. So how did we do holding please? Mhm. The idea
08:13 that for every possible threshold it will the history Graham into two parts.
08:19 that will give you two statistical Right? And you want to compare
08:23 statistical values to decide which one is better threshold. Right. In our
08:28 we said a good a good metric compare is the weighted sum of intra
08:33 variances. Right? And we can that for every possible threshold and essentially
08:39 the one that gives you the least sum of inter class variances. And
08:43 be a threshold. Right. And also saw an example for that With
08:50 little six x 6 image that can six different intensity levels. And for
08:56 possible intensity level we computed the Okay, okay. So now given
09:04 image, you can use any of methods to create a grayscale image.
09:07 now you have, I'm sorry to a binary image. So now we
09:11 a binary image and we'll do things kind of improve that binary image.
09:19 . All right. So let's talk little more about history grams because it
09:24 keep coming up as we go through . Um One is that we already
09:28 in order to do thresh holding or create a binary image. Our image
09:33 to a grayscale image needs to have bimodal hissed a gram. Right.
09:37 we think of examples of things that use every day that may end up
09:41 by model hissed a grams or images we take or look at in
10:00 when there is a large contrast between background and the foreground, you end
10:04 having a bi modal hissed a Right? There are a large number
10:07 pixels that have one certain pixel value another set of pixels which are the
10:13 . That may have a different set pixel values. Right? So if
10:16 take an example like uh like a printed media, right? Usually you
10:21 black text on white pages right now background is white and the foreground is
10:27 . And if you look at the graham for something like that, it
10:31 like a bi modal hissed a Right. Another example is um uh
10:37 cells in a solution, right? it's an example we already talked about
10:42 um we'll also actually do an assignment off of it. But in general
10:46 solution has a lighter color compared to blood cells. And when you look
10:51 the history graham for something like it has a by model hissed a
10:54 . You can do thresh holding you know, separate the background in
10:58 foreground, reduce the cells in the . Okay. Um and more
11:03 you know sometimes your image may not end up having a bi modal hissed
11:07 gram. In which case you can an environment that will produce a bi
11:11 hissed a gram. Right? For , uh if you go to a
11:15 store and scan things right. One the things is you know, there's
11:18 little scanner and you can scan it in general it's an illuminated surface,
11:22 ? It looks like there's a light that, right? That's because when
11:24 put your object on top it's dark that light both combined will produce two
11:30 pixel values for the background and what scanning. Right? That's the
11:33 You get a bi modal instagram and you want to threshold it you can
11:36 that. Okay, Okay. Um again, like I said, you
11:47 , bimodal instagrams are good but you , often times you your image may
11:52 up having multiple modes, right? you may not be able to control
11:56 environment but you would still have to a grayscale image, a binary image
12:00 that application. So in these what happens is that certain number of
12:04 pixels may end up being foreground pixels the other way. Right, so
12:09 of the foreground pixels may end up classified as the background pixels. Now
12:14 these cases you would have to do sort of region correction or you would
12:18 to do additional operations to improve the in the image. Okay. And
12:23 look at how we can do that two different ways. Yeah. So
12:34 mean um so basically a lot of operations will allow us will actually look
12:40 will be doing logical operations. So we've talked about this and we
12:45 oh since it's by valued image it take only zeros and ones you can
12:49 look at them as true or false . If you look at them as
12:52 or false values, you can actually up applying logical operations on them.
12:57 . What are the logical operations that are common? You need at least
13:04 . Right and orange knock. Yeah, those are the ones we'll
13:08 and in addition we look at one operation. So what is an end
13:15 to input variables? What will be response from an and gate B for
13:24 to produce a one? Both the have to be one. Right.
13:27 all the inputs are one, you get a one that's an end.
13:31 , what's in order? That's Or all the inputs have to be
13:40 to produce zero, otherwise they'll produce one. Right? And then you
13:43 a simple compliment as well? A becomes a one and one becomes a
13:48 . Right. And of course there these uh these laws that we've also
13:56 . Does anyone remember these laws? are their? What is Law two
14:06 3 associative four and five. Sorry mutated fine. Well what about
14:23 7 distributed in the last two or morgans was perfect. Okay,
14:35 So apart from this uh so the basic ones are and or and not
14:39 apart from that will use the fourth which is called majority. Okay,
14:43 the first rule for majority is that number of inputs have to be
14:47 Okay, so you cannot apply it inputs, you cannot apply it on
14:51 inputs, but you can apply it the number of inputs is three or
14:56 or seven or so forth. And it's it's the majority because it
15:00 what it says right? The result basically equal to the majority of the
15:05 that are in there. Right? if you look at the truth table
15:08 this with three inputs, you will that whenever there are two ones you
15:13 a one or when there are two you get a zero or when there
15:17 three ones, you gotta one. ? So basically whichever one is the
15:21 value of the inputs, that's the you get. Okay, now this
15:26 one more additional operation, logical operation we'll use. Okay, is
15:31 Okay. Alright, now how do apply these on images? Right now
15:38 restrict ourselves to these four operations. is not an orange majority. Um
15:45 we'll try to apply them on Now, when I say applying logical
15:49 on image images, there are two to do it. One is that
15:53 can take two images right to binary and I can apply logical operation between
16:01 pixels in those two images to get output image. Right? That's one
16:06 to do it. In other ways I can take only one image and
16:10 can define a small region in that I can take all the binary values
16:15 there and apply a logical operation on values. Okay. That's the second
16:20 to do it. And we'll look doing both both those ways and we'll
16:24 why they're helpful in certain scenarios. right. So let's look at some
16:34 of applying the uh, the logical . Uh the most straightforward operation is
16:41 a north operation and for the north you only need one input image.
16:45 what it does is basically every pixel complemented. If it's a zero,
16:48 becomes a one. If it's a it becomes a zero. Okay.
16:52 on the other hand, I can take two images and I can apply
16:56 and operation on them. Okay, is the first time I was talking
17:00 you take two images and apply the between corresponding pixels in the image.
17:05 is I 100 and I 200. ? The row and the column same
17:13 both the pixels. Right? I that for all the pixels that are
17:15 the in both the images and I a result. Now, if I
17:19 to apply an and what I end getting is the overlap of black
17:24 Right? So if you remember we one is represented by black and zero
17:29 represented by white. Right? So I take image I one and I
17:33 here and if I apply an and between them, that's the result I
17:38 and it is the overlap of the regions. Now I can also apply
17:43 or operation between them. Now, I do that, what do I
17:46 up getting please? That's right. I get the overlap of white
17:59 Right, so it is an example do that. Okay, perfect.
18:07 let's look at an application where this be useful. This is the setup
18:12 have. Okay, now again, remind ourselves, I said I can
18:16 logical operations on on images in two . One is that I can apply
18:20 on two separate images, right? corresponding pixels. The second one is
18:25 can take one image and define a region and apply on the on that
18:29 region. Okay, so we're looking the first kind now where you have
18:33 images and trying to apply a logical between them. Okay, so here
18:40 an example where applying such operation can useful. So this is the setup
18:45 have. It's it's an industrial environment I have a conveyor belt on which
18:50 producing some a manufacturing something in this that bolt and what I want to
18:56 is have a setup where I can each bolt to decide if it's defective
19:02 not. Okay, so I have conveyor belt, I have the bolts
19:07 are manufactured are coming on it and have a mechanism to take images of
19:12 one of these boats as they Okay, now, apart from
19:16 what I also have is a reference or a model image. Now,
19:20 does images saying? Oh, perfectly bolt should look like this.
19:25 it's a model of what I should now, as I keep coming and
19:30 course, uh, since I can the environment, I will control it
19:33 lights to make sure I get a image and I can threshold it to
19:38 a binary image. Okay, so a result I have two binary
19:42 One is for the model which says is a perfect bolt and one is
19:48 binary image that I keep capturing as bolt comes on. Okay,
19:52 what I want to do is I to compare the image I capture with
19:56 model image and see how much they . Right, larger the deviation implies
20:02 amount of defect and I can do quality control with that. Okay,
20:08 this is the images I get. , so the one we see on
20:13 left is I model right, that the reference image I have and the
20:18 I see on the right is the that I captured. Okay, and
20:22 for this example, let's say that little bolt has shifted to the right
20:28 certain amount. Okay, now, order to identify how much it has
20:33 , I have to identify both the in the black pixels and the white
20:39 . Right? So how do I that? So if if a certain
20:48 value is zero in both, that's . If it's one in both,
20:53 fine. But if it's zero in and one in the other one,
20:57 need to identify those or the other if it is one in the model
21:01 if it's zero in the one I , I need to identify those right
21:06 . How do I do this using logical operations? We just saw and
21:11 not. Majority eggs are great. , that's the correct answer. But
21:23 restricting ourselves to and or and not of course we can create almost any
21:28 using and or not now, how I create X are using and or
21:33 not. All right. Oh, said not or X or Yeah,
21:52 north and or will give you a . That's fine. That doesn't give
21:56 an X. All right. We the solution is X or write an
22:00 produces a one. Uh when well produces zero if both inputs are zero
22:06 one and producer one? Uh if of those is one. Right?
22:17 again, that's right. Mhm. what combination of applying logical operations on
22:28 two majors will give me that So maybe we can break it into
23:07 parts, Maybe I first I can only deviations in the black part and
23:13 can implement something else to find deviation the white part and then maybe I
23:18 combine those two right now, if pick one, how do I know
23:23 there is a deviation in the black ? Mhm mm an orbiting them and
23:46 necessarily. Mm uh so the way can do this is that you would
23:57 by taking the uh okay, so way you would do this is that
24:08 would take the so this is two to it, right, you would
24:11 the complement of the input image and do an end with the model and
24:17 the second part you take the complement the model and end it with the
24:21 . Okay, and then you take or between the result of these two
24:26 and you have what is called X or gate, Okay, and
24:30 you do that, you get something looks like this, it shows all
24:34 black pixels are deviations that is, know, they don't match up.
24:40 , now, once I have this , I can compute uh you know
24:44 one value to say what amount of has occurred, for example, if
24:48 are 100 pixels that are deviated and total there needs to be 1000 black
24:54 . So I can look at that and say the deviation is 10% and
24:57 can do some quality control single. it's more than 5% Reject that
25:02 Right. All right. So that's simple example where you can actually apply
25:09 operations between two whole images uh and actually useful to do it.
25:14 But more often in the things we , we tend to apply logical operations
25:19 smaller regions uh and use that output . All right. So before we
25:32 that, let's talk about an algorithm blob coloring. And like I
25:38 you know, when you're hissed a is not by model in nature.
25:41 is if it's multi model or if flat, you would end up either
25:46 some of the background pixels as foreground some of the foreground pixels background.
25:51 . So if you look at if I classify by some of the
25:55 pixels, foreground pixels, you would up with having some noise objects like
26:01 . Okay. On the other if there are foreground pixels and fight
26:05 them as background pixels, I would up having holes like this.
26:10 now these are all imperfections. I catch I've captured some information but
26:15 is not perfect. So I would to do additional operations to improve the
26:20 in the image and often times this is called us the region correction.
26:24 , you're correcting regions. So in case the object that I'm interested is
26:29 large object in the middle. The big one in the middle is
26:33 I'm interested. But all those little surrounding it are noise and all these
26:39 in the object are also errors. ? So I need to do region
26:44 to remove the noise and also do correction to remove those holes.
26:49 And one algorithm that helps us in this is called blob coloring. And
26:55 it does is it groups pixels together belong to an object. Okay.
27:00 if I look at the image for , right. If I look at
27:03 pixel values, I have a whole of zeros. I have a whole
27:07 of zeros wherever this object is. have once that's all the information I
27:11 but I do not know if these of ones belong to one particular
27:15 Right. Similarly, I do not if all of these ones belong to
27:19 particular object. Right. What I to do is I want to come
27:23 with an algorithm where the output is it says, Okay, object one
27:28 pixel X one Y one X Y two X three, Y
27:32 So and so forth. That is the pixel locations that belong to one
27:36 object are grouped together. Okay, is okay. Do you understand what
27:42 the output for block coloring? So this case if I say there are
28:11 objects then what I need to get output is when I apply blob coloring
28:16 this image, I need to get one and this should say okay,
28:25 XR one XRYR one. Right? is a pixel that belongs to this
28:33 . Right? And I need to X. uh all this one one
28:41 12. Y 12 is another pixel belongs to this object. So on
28:47 the pixels that belong to that object I need to get our two but
28:53 all the pixels that belong to Okay, so basically I'm grouping pixels
29:11 belongs to one object together. Once I have this information I can
29:15 region correction because I know which one those who are noises and maybe I
29:19 just remove them. I know which or whole. Well, I don't
29:22 holds yet but we'll find a way do that. Okay, so now
29:27 do we do this is the question . How do we do block
29:31 And it's really important to understand this because it's more than likely to end
29:35 as an assignment. Okay, so coloring is basically a simple technique to
29:46 axles that belong to a certain object a blob blob is nothing but a
29:52 large object. Okay. All So let's start with something simple.
29:57 , in this case you have this object in the center that is with
30:01 the black pixels and right now my image basically has a bunch of zeros
30:06 the background and once for the pixel this object is. What I want
30:11 do is I want to write an where the where it will label each
30:15 of these ones as the label be it says, it belongs to the
30:20 . B Now can we write a algorithm to do that? That's
31:11 You want you? That's perfect. there's two ideas to what you've talked
31:22 . One is that there is a of iteration that I'm following. There
31:27 a direction of flowing. There is is at every point I'm making a
31:31 for each pixel. Right? So started the top left and I follow
31:35 follow a scanning mechanism for each Right? That way I can make
31:40 decision for each pixel and move on the next. That's one thing.
31:44 ? I'm doing the scanning. And second is that I'm also looking at
31:48 pixels that surround each one to decide it belongs to that or not.
31:53 . So there is also a sense neighborhood right now, neighborhoods can be
31:58 types. Right? I can define as being only the pixel on,
32:03 know, top of north south east west are the neighbors. That's a
32:06 for neighborhood or I can say not just north south east west,
32:10 can say north northeast. Um You east. You know so and so
32:15 . I get eight different pixels. I can even define bigger neighborhoods that
32:20 25 pixels. Right? So there a definition for neighborhood as well.
32:24 . So there are two things I . One is a scanning mechanism for
32:30 pixel. And as I scan, make decisions for each pixel and move
32:34 . Okay. In our case the natural scanning mechanism is that start at
32:39 top left. Keep making decisions until end up until the until you reach
32:43 end of that column, go to next row. Start scanning again and
32:46 making decisions right now. Before I before I before I work look at
32:52 neighboring pixels. First thing I have do is I need to define a
32:56 . Right? And then I can check those pixels to see if they
33:00 already labeled or not. If they labeled. I can give the same
33:04 or I can um or I can have to create a new label.
33:09 , so in this case that's what do right. In this case actually
33:13 don't even have to do that because a very simple. Uh there's only
33:17 object essentially I can scan and at location. If I see it's a
33:22 , I can just give it Well I can just give it a
33:25 . B and I'll be good. I'm gonna run into issues when I
33:29 an image that kind of looks like because this one needs to get a
33:34 label than this object right now. this case at every point if that
33:41 is a one, I need to at its neighbors and if the neighbor
33:45 a label, it has to get same label as that neighbor.
33:50 All right. Russian. Alright. let's let's try to solve this
34:01 Right? Let's try to come up an algorithm for this. So let's
34:05 I start scanning right? As the order. Is that started top
34:17 Okay. So the scanning orders that at top left and I scan every
34:22 and make a decision for every Right? Making a decision means giving
34:25 a label right now. The first is that what's the first thing I
34:29 to check. Huh? Well, need to check if that pixel location
34:40 a zero or a one. If it's a zero, I don't
34:43 for it. Right? If it's one, I'll need to make a
34:46 on it. Right? So let's a few things. First thing is
34:50 let's call this image I Right. let's say there is also what's called
34:55 regions image. Okay, This is output image. I want to
34:58 That is just as big as that . Right? 12345678. Okay.
35:46 . Alright. So I start scanning that image in the top left.
35:50 are all zeros. So I don't anything right? All of these end
35:54 getting zeros right? And the next as well they all end up getting
36:10 . Now before I actually do I need to define a neighborhood.
36:15 . So let's say for this particular my neighborhood is a simple one that
36:19 looks at four pixels. That is , south and east and west.
36:24 , so for every pixel I define neighbors north, south, east and
36:28 . Those are the only four I have to check. Okay,
36:31 in the next one I keep Right? So this is zero.
36:36 don't care. This is zero. don't care. This is zero.
36:38 don't care. And then I come a one. Okay. And I
36:41 to define a label for this location my regions image. Right. Are
36:46 guys with me? Okay. So this case, what do I do
36:53 need to give it a region name . What do I get? All
37:05 , let's call something a region a variable that keeps count of
37:10 Okay, let's say start with the . Well too, I don't wanna
37:15 confuse between zero and one. so in this case I can simply
37:20 this value what the region number Right. So I can give it
37:27 two. And basically this is the when I have When I have this
37:38 filled with one right? And a and a zero in this case what
37:43 doing is that I'm saying region X Y is simply getting whatever is the
37:52 for the current region counter. Is that okay? Yes now so
38:04 course I need to check the condition if I X. Common Y is
38:14 to one. In which case I say region X comma Y would take
38:21 value of our Okay. Now since already used are the next time around
38:27 come across a situation like this. actually now this is the region I
38:38 this is the neighborhood that I created ? Which means I need to check
38:41 four pixels. Not just to I to check down in below as well
38:45 we're not doing that. Why? . We haven't gone that far.
38:56 don't have a region number for Right? Even if I check something
39:00 the regions inmates there's nothing there Right? Despite creating your neighborhood as
39:06 , these two are actually not going give you any information. Right?
39:10 that's the reason we only need to top and left because those are the
39:14 that are already scanned. Okay. . So when I come across this
39:20 I know what to do right now do I check this condition though?
39:27 is a condition where that this particular is true. And also if I
39:37 comma y minus one, this is to zero and I x minus one
39:46 Y. It's equal to zero. ? If these two are ones then
39:50 would have already made a decision on and given them some region number.
39:54 which case I would have to give the same. Right? Only in
39:58 case I would say region X comma is equal to R. Okay.
40:05 then what I continue scanning right, go to the next pixel. How
40:12 this case look like? I know a one here. Right. And
40:28 it looks like this is also Right in this case, what do
40:32 do? Now checking this condition would that I External y is equal to
40:44 and I x Come on. Why is equal to what 1? Right
40:57 this one as well and I x o X -1. Common. Y
41:06 equal to zero. Right? This the condition that responds to this.
41:11 in this case what do I I need to give the same label
41:24 to write or whatever is the one to it because it belongs to the
41:27 block. Right? So I need say odd. X comma y is
41:35 to out of XO Y -1. that okay? Yes. Now.
41:47 this will just get up two And do that assignment. Okay? And
41:53 keep going on I got 000 I a 00. And then I have
42:01 condition here which is basically that I XO Y is equal to one.
42:13 what I of x minus one comma equal to and I have XO Y
42:25 . Equal to what? What is 0? That's right and this is
42:39 comma y minus. Oh yeah X on. Oh no that's not too
42:46 . We're talking about this pixel this this one here. Oh wait I'm
42:52 . No you're right. So this both this is zero and this is
42:56 as well but we always see in condition which is this one?
43:05 So what do I do now? need to follow the same rules and
43:13 need to assign one of the value others. Right now, ideally this
43:17 to be incremental. Every time this is run I need to say r
43:21 equal to r plus one. The time around when I come across a
43:24 blob I give it a new region . Right? So if I do
43:28 , this gets a value three. , okay, so let's move
43:36 And the next the next pixel is interesting because both of them actually have
43:44 region number. Now this is a where I have I of X comma
43:49 is equal to one and I of -1. Common wide is equal to
43:57 and I have I of X comma minus one is equal to one.
44:04 ? This is a condition where this now. What do I do?
44:25 label should that get now? If look to the top it will tell
44:31 that I need to give it a to. If I looked at the
44:37 tell me that I need to give the label. Three. Right?
44:45 label should I give it? It's to give to which is fine.
45:00 it also tells me that the pixel the left also belongs to the same
45:05 as the one about right. They belong to the same blog,
45:09 All the pixels which means I don't which number I give as long as
45:13 give the same number to all of . Right. Does that make
45:20 So if this case occurs we need check that if the if this value
45:26 the one to do this. If value and this label are the same
45:31 ? If they are same, there's much to do. I just give
45:34 the value of our left us. only problem is when they are different
45:39 I just realized that all of these as three are the same as the
45:44 labeled as too. So I need go back check all the trees and
45:49 them to two or I can check the tools and change them to
45:53 Is that okay? So in this there's another if statement I say f
46:01 of x minus one comma y is to region of X comma Y minus
46:10 . Okay, now if this is , I can simply say R of
46:15 comma Y is basically equal to out X minus one comma Y or R
46:20 X comma Y. It doesn't matter they're the same value. Right?
46:24 if this is not true then I realized that all trees are same.
46:32 need to iterate through this right? I need to check if a value
46:37 three and change it to two. you guys Okay. So and else
46:46 be affects all trees to write. I started trading from here and check
46:56 values 00000222. I don't care. three. Change start to cool.
47:06 . And of course I'll give this the same value as stuff. It's
47:14 . Now I go to the next in which case this is the same
47:20 , right? In this case this statement is actually true. So it
47:25 the same value as the one to top. So this gets it to
47:30 ? All right. How about the one? Which condition is the next
47:39 ? This one right? In which I just give the value to the
47:44 and I get her to right and are all zeros. I don't
47:50 I don't care. And again, condition is this? We don't have
47:58 condition for this Right? There is 4th condition now and that condition is
48:16 right? So the fourth condition is i X comma y is equal to
48:25 . I X comma y minus one equal to one and I X comma
48:35 . Sorry X minus one, comma is equal to one and I X
48:48 y minus one is equal to Okay in this case what do I
49:02 ? You just give the value that to the top right? It's kind
49:05 very much like this. You instead would say out of out of X
49:15 Y is basically equal to out of minus one comma wife and you would
49:25 up getting a to hear? Alright. The next one. What's
49:35 next condition? Which condition is the 1? Sorry, this is this
50:01 right? And this is true for . So you just give it the
50:04 value as one of those right? the next one. So the next
50:39 is the same as a three and one is true. So I just
50:44 the same value as one of the left. This is also the same
50:47 as three and I do that. That's right. Yeah. So this
50:55 is also the same condition and these zeros right? And then I have
51:02 and then what do I have It's the fourth condition and I just
51:11 to assign the value above them. one is the same as this one
51:17 and it's true which is fine. I get 000. Okay. And
51:24 0000 00. And there's one Which condition is this this one?
51:36 by this time March are has actually incremental too four. Right. So
51:49 arm was initially to I didn't want confuse between zeros and ones. And
51:54 first time when I got here it implemented to three. It became r
51:58 equal to three. And next time when when I hit the three it
52:03 implemented to four. Right? So time this one will get a value
52:08 . Right. Is that okay? now I can iterate through this image
52:20 I'll say okay all the pixels that a value to. I'll group them
52:24 one. All the pixels that get value three. I group them as
52:28 . There are no trees. All pixels that get a value for I'll
52:31 them as another one and I would group pixels that belong to the same
52:36 together. Okay. And also remember . This depends on your definition of
52:42 neighborhood. Right? This is one . I can also define a neighborhood
52:47 kind of looks like this A three 3 region for each one? I
52:55 to check more pixels. So in case what are the pixels that actually
52:59 me information in this case? So for that one the neighborhood I'm
53:17 is only top and the left. . Because the one on the right
53:19 the bottom don't give me any How about this one? But every
53:24 what are the pixels? I need check this is the one. I
53:32 to check this. This this So I need to write more conditions
53:41 check these. Okay, now I also depending on the application, I
53:49 also define a larger neighborhood which is . Sometimes it has maybe in your
53:54 it's common to have you know, pixel holds and things like that.
53:59 which case I define something larger. I would have to check all of
54:07 To actually make the decision for that pixel. Okay, now if I
54:13 to use this over here then I actually end up getting the same label
54:17 this pixel as the one about. . Mhm. Um it's well that
54:36 looks large but in reality it's not example if you think about megapixel images
54:44 have 1000 columns up and down. when you do been a realization of
54:49 , you may end up with a of holds objects that are actually really
54:53 that actually together men may be separated just one pixel and you may have
54:57 group them together. Right. So let's go back to what we
55:07 doing now. Again the idea is I need to do region correction.
55:13 ? And I said in order to region correction, my objective is that
55:18 need to remove the noise pixels and also have to remove the holes.
55:23 now I'm doing a blob coloring to group these pixels together because here I
55:30 the knowledge that the largest blob is one I care about and the
55:34 smaller blobs are just noise. I that knowledge, application knowledge. Now
55:39 can use that to remove the spectral noise uh objects. Right?
55:47 once I do block coloring, I what pixels belong to which blob.
55:53 ? And again, when you get chance, please go through this
55:56 it's it's really important and it is to be on the assignment and turning
56:00 into code would require a little bit time. Okay, so at the
56:08 of Blob covering this is what I for each set of pixels. I
56:13 region numbers. So essentially all those will get a region number one.
56:16 will get a region number 2345. now I can do accounting of
56:22 Right? So if I counted, turns out that little blob up there
56:27 793 pixels. Right? This big has about 11,000 pixels and the other
56:33 has 1000 1 73 112 pixels. now I know that the biggest one
56:40 the one I care about. Um I can basically straight through that image
56:46 I'll say if the blob member of region number is equal to one,
56:50 it a zero. If the blob is equal to three make it to
56:54 . So I'll make everything zero except blob too. Right? And as
56:59 result I would have removed all the regions. Is it? Okay.
57:06 , perfect. Alright so once I that, all the smaller regions are
57:29 now I still have one more one problem to solve that. I need
57:32 remove those holes. How do we that already moved the holes. I
58:06 a way to remove smaller objects or right? I do block coloring,
58:10 remove the ones that have a low of pixels. Now how do I
58:16 the whole, I can take the of the image and then do block
58:29 on it. Right. And then again do the counting and remove smaller
58:35 . Right? So if I take compliment this is how it would look
58:40 . Now all that background is one blob right now ideally when I do
58:46 covering all of these will get one number and these smaller ones will get
58:50 different region number. Again I literate it and if this region number belongs
58:55 what other small blob is remove it . Once I do that I get
59:05 that looks like this, right? effectively we have done some amount of
59:10 correction where we removed noise objects in , Right? So this is also
59:26 you would do cell counting when you for a blood blood work,
59:31 You create these binary images. Uh then for that binary images, you
59:36 do block coloring, right? And you can remove noise, noise blobs
59:40 things like that. And then if look at the count of those
59:44 those basically is your white blood count red blood count. Okay,
59:59 this is okay. Um Alright, let's go ahead. And and the
60:12 here, we can pick it up . In next class. We can
60:15 about morphological operations and things like Mm I wrote them on
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