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All right, people.

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We are now on to stage two, which is mapping.

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Our primary goal for this state is to first convert the maize extractor in the localized state into

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some kind of clever symbol map.

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The road map for achieving that goal is to simplify our means into pathways and then find interest points

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between them.

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For this purpose, we create a new file that we name as mapping.

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Now, the very first thing that we do inside this farm is perform the necessary inputs.

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This includes the CB two

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and Numpad.

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Then we declare the class that we performed, the actual task of mapping and we name it as bought mapper.

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And the very first function that we define inside this class is the and function.

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Basically what this function will be doing is simply defining the instance variable that are necessary

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for this political class at its initializing state.

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So the function that we will be defining here is the.

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Boolean variable that basically carries the state of the class or whether the mapping process has been

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completed or not.

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So because this is basically converting our means to graph, so we name it as graphic file and because

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this is the initialized state.

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So we said this two fold.

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Then we define our second function that we name as graph invite to signify.

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I mentioned to the user that this perform the process of converting our mes to a nice looking graph.

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Now the stakes in the first argument of self.

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And the next argument is extracted means that we can see from the previous stage that is the localising

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stage.

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And then we simply check the side of this function, whether the graph function has been or the mapping

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function has been done or not.

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And that can be done by simply calling instance variable of graphing flight.

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And if it is not in graph ified or the mapping process has not been completed, then we proceed on to

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perform the mapping process.

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And the very first thing that we do inside this function is to display to the user the input that we

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receive from the previous stage, and that is the extracted means.

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Then we perform or simplify our means to get the simplified pathways instead of the complete region

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of interest.

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And that can be done by using the open CV function of pinning, which is part of X image processing

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library.

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And we access its kidney function, which takes in a source argument that we provide as extracted means.

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Once we provided this this basically upwards of a ten meters or possibly so we named it as ten.

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And then once we have this output, we display to the user and then proceed on to the next step and

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before moving on, the next thing that we do is to simply first perform our deletion over the spot phase

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and then perform turning again.

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And the reason we are going to do this is because there can be a lot of interest points when we find

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in this particular image.

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So you want to simply get a lot less interest points that we might get in this case.

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So that can be done by simply deletion and then performing the thing again to simplify our means or

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pathways even further.

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So we do exactly that by using the morphology, execute to perform the dilation, by using a structural

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element and then performing thinning again.

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So once we have this simplified thing, we then move on to perform a little bit of cropping.

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And the reason are performing cropping because a lot of other things might get on the boundary outside

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of the maize, inside the extracted means variable.

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And we don't want that.

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We only want our maize inside that variable.

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So to do that, we are going to crop all the boundaries of our maize.

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So to do that, we need a new variable inside that initialization function that is the crop mark.

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So we are going to set this to five because we want to remove five pixels all around our maize.

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So let's perform the cropping now and that can be done by simply accessing ten variable calling and

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cropping the rules by writing itself, not crop amount going all the way to the end end goal and rules.

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We're writing ten does zero, which is the rules and subtracting the self or crop mount.

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So these are crop cropping for the rules, then we crop for the columns and that can be done the same

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way we're writing Crop Mount Ten.

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That ship one which is the columns and then subtracting the crop mount to crop out the column on both

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sides.

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So basically our ten.

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Variable has been cropped on all sides by five pixels.

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So we named this new variable as ten cropped.

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So once we have this up, we can then display this thin crop thin and crop pathways to the user by overlaying

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it over the original extracted myth.

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So we need to also perform the cropping over the extracted mes so that we could overlay this image over

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that one.

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Suppose we display this to the user and then perform the overlay.

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And the way this works is, is simply cropping our extracted means and then overlaying the tin crop

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over our extracted means.

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And once we have this information, we can now call the last function that we perform or complete our

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mapping.

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That is the one function inside this function.

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The only thing that we are going to do right now is to extract the interest points from our members.

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So let's call that function.

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That is the one false the sticks and the only argument of the tin group that we have just received right

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now.

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So now we need to define this particular function that we perform or extract interest points from our

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moves.

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Let's define that function.

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So we go right over here to find one path and then give it the first argument or so the next argument

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of me.

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And then the very first thing that we do here is to first create a mes authority.

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I have as Ms. is definitely not the means.

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So we try and instead of pn.

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So once we have the means now correctly, we can then move on to extracting or creating a mes bcr or

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a coloured mes.

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Now this is necessary to display the interest points that we have found out in this particular function.

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So we create that means BGR or Ms. BGR by writing or using the opens in function of convert colour.

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So this takes in the source, which is our means.

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And for the code I pass in.

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Color gray to BGR and this converts or creates our PDR color means.

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Now, once I upgraded this, the next step is to simply create a window to display this björkman images

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so that can be done by simply writing CV to window.

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And then I extract the total number of rows and columns of our mes and this wasn't necessary to simply

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loop over all the mes pixels.

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And that will be done like mentioning the theory in the form of left or right and then top to bottom.

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So we write here for loop.

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Rule in range of all the rules or total rules.

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So it goes from zero to total rule.

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And then we go from column in range of total columns.

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And then once we're done with this, we then need to check, identify the probable interest points and

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the reason the probable interest points.

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And the way that we can find out the probable interest points is by checking the means where it has

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a one value or a positive value because that in indicates a pathway that our interest point could be

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present.

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So we can write it right or find it by writing if missed rule or the current crew and column.

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If it is equal to positive value there is to 55, then we are on a probable probable interest point.

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So what we need to do to confirm whether this is an interest point or not or not just a part, we need

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to find the surrounding pixel intensities.

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So what that will be defining will be defining a new function.

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Right.

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About.

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So let's define that function right here.

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And there it is.

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The gets around fixing intensities.

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This takes in the MES and the current column and the control and then gives out the pixel intensities

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along with the pathways.

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Now to get all the pathways or the number of pathways, we need to convert this mes positive values

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from 255 to 1.

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So if we have four pathways, then four pixels around our current pixel will have one value and we can

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simply sum them up to get the total number of path around the pixel.

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So for that purpose, we'll be using a threshold function that takes in the means or the trash we provide

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in.

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Once any value that is given, then one should get squished to the max value, which is one.

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And because we are performing my binary thresholding, so we right here.

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THRUSH binary and because threshold provides to our cause and we only one the second we write one.

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So this gives me a nice looking move.

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So once we have this information, we can then proceed on to extracting the total number of rows and

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columns that are present inside.

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This means and that can be done by simply writing Miss Dos Sheep zero.

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And for the column we write means dodge sheep.

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What?

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Once we have this information, we proceed on to declare the state variables that basically address

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the boundary conditions.

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So these are the top row, bottom row, left column and right column.

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Whether our pixel is one of these boundary conditions.

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So let's check whether the pixel was the boundary condition or not.

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And that can be done by simply checking.

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If our current show equal equals zero, that indicates that we are on a boundary condition.

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There is that we are on the top row.

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So we said that true in the same way we addressed other cases and checked whether we are on a boundary

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condition or not.

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So if our control is the last row and the bottom row should be set to true, if the current column is

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the first column in the left column, which after true.

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So this is the we are we check whether we are on the boundary condition or not.

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Now that we know whether we are on the board, recognition or not mean to address these cases.

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So let's address the corner cases.

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That is if.

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If we are on the top row or the left column.

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Oh, the left column.

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Then what we need to do.

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We can.

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We can see now that the top left can simply not be accessed.

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So we need to give it a default value of zero.

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Otherwise, if that's not the case, we can access the top left value by simply writing the maze and

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giving it the current row minus one for the top and current column, minus one for the left.

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And the same way we address the rest of the cases.

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All right.

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Once we address these cases, we need to address the lost cases that are not part of the court cases,

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but the general cases.

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So, for example, if we are on the top row, if we are the top on the top row, then definitely anything

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that is above our pixel cannot be accessed because we are already on the top.

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The top will be set to zero.

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Else the top can be accessed by simply writing means.

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Right in Crunchyroll minus one fixed at the top and the same column or the current column in the same

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way we address other cases there is.

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Computing the right bottom.

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And look, once we are done with this information, we can then proceed on to finding out the total

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number of pathways around the current pixel.

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So that can be done by simply adding all the positive surrounding fixed line densities that are all

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one.

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So if we have two pathways, then we have two surrounding pixel that have both intensity of what.

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So if we have a number of part there is good in a do I want it to display to the user.

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But simply debug whether we are doing this correctly or not.

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And then we return everything that we computed.

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Once we're done with this, we can then proceed on.

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Right before what we were doing inside our one pass function.

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So we can call this function of good, strong pixel intensities and get all the following output.

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Once we have this output, we need to check whether our parts.

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Ah, of a specific number or not.

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But before that, one thing that we need to check is whether we are on the start point or an end point

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start point representing the MIS entry on the end point, representing the music.

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So first we need to find these points and that can be found by simply writing if row row equal to zero

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or row equal to.

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Rose minus one.

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Rules minus one or.

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Call them equal equals zero or.

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Going.

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Equals equals two columns minus one.

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All right.

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So inside this condition, if such a condition occurs, we checked, we are sure that it's either a

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start point or end point.

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Now we need to find whether this is a start point or not.

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So if they're equal, equal to zero, and this is a start point and we need to give it a distinct color

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inside the BGR.

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So the user could see it or identify it.

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So that can be done by simply providing the row in column for the BGR and then giving it a color of

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orange so that we could identify that this is the stock.

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Now, once I have done this, I need to address the other case that is and so earn or make exit.

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So if such a case occurs, I give it a distinct color.

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I give it a distinct colour of green so that the user could identify that this is the music.

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Now we need to display this to the user by simply writing I am show means ingress point.

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Now, once we have addressed the start and end point, we have found a our major address, other cases

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where the pathways are the one, two, three or four.

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So if the pathways is one, that means only one path leads up to our interest point.

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That means that interest point is a dead end.

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So let's address that condition.

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So if the path is equal to equal to one, that means it is a dead end.

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And we simply give it a distinct colour in the state.

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We use it the same way we address the case for whether we are doing or not.

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So what it's doing, we have a slightly different case because if the parts are to a number, then it

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could be either undoing or either a state.

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But so we want to omit all state parts and only key interest points.

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That is a done so we check there using this condition and this gives us only the tones and it's the

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straight path.

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So we give it a nice looking blue on the straight or the tones, and then we address the last four two

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conditions.

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That is, if it is either a three junction or a four junction and we give them distinct colour.

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Show them to the user.

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And this completes.

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Our one fast algorithm for finding the interest points.

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And now we can simply run this and see whether we have implemented this correctly and completed the

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first two tasks of simplifying our MES and then finding the interest points or not.

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So let's do that.

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Now that we have completed board mapping up until finding the interest points, we need to run it to

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see whether it is doing its job or not.

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But before that, a few things.

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One I missed out the declaration of draw interest points, a boolean variable there to draw interest

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points once they have been detected.

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And we said this to false.

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And second, we come out of one box calling of the one algorithm.

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For the moment, then we had to work with the myth solver after we imported the bot locus.

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Either we invoke the bot mapper from the Borton mapping file, so we input more mapper and then we head

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over to the initialization function.

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The main solver class we after we have created object for what?

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Localized.

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We create the bot mapper object.

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So once we have done this and move over to the missile and function after we have localized our robot,

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we now perform the mapping by calling bot mapper.

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And then graphic fire and passing and.

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The self-doubt, both localised.

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Localize the.

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Don't miss occupancy as the expected miss.

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This completes the calling process of the mapping module and now we can simply head over to the terminal

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when the project.

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We launched a simulation.

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As soon as a simulation starts up, we will run our Ms. solver.

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So as you can see, the main server has started up.

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This is the base occupancy that we started from the localized stage and then for press enter and as

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you can see starting from the extracted maze, this was the input that we see in the mapping module.

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This is the extracted means that we receive from the localizing state and then we perform pinning on

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it which to reduce this region of interest into pathways.

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Once we have simplified it to part ways, we simplify it even further by performing dilation and then

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performing pinning again.

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So we get tuning twice.

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So this has simplified our process to even further.

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And then we process and then we get crop out outside of the maze to remove all the extra boundary that

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was just not required.

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This gives us the ten crop.

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And then finally we overlay this pathway that we have computed or extracted from the maze over the extracted

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maze.

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And this is the final output that you got up until this moment.

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As you can see, this yellow marking represents the pathways that we have computed from our extracted

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maze.

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Now, to test out the one path algorithm of finding the interest points we head over.

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To our school where I have created a desk mapping of sorts, a pest mapping file of sorts that is used

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to test the one fast algorithm.

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And the way this works is it simply imposed a new means.

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And this means is a tiny maze.

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And it being tiny helps us to easily identify whether our interest points have been correctly identified

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or not.

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So I imploded and then passing this tiny image through the OnePlus algorithm.

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And if I run it.

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Once it starts up, you'll see that this is our myth.

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So this is very simple.

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And you can easily identify that the top pixel is easily the stock on the bottom is the lower pixel

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is easily the end point of them.

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Is this the left edge is basically the dead end because only one part is leading up to it.

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So if I press enter, you will see interest points that have been detected in the phone colors.

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So if you remember we said the orange for start point.

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So it has been correctly identified green for End Point.

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That is also correct.

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And the red is a dead end.

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So you can see only one pixel is leading up to this one.

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One part is leading up to this interest point.

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So this is a dead end.

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And the same way, if you look at the dark blue color which represents does so as you can see, these

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three pixels make 90 degree angle with each other.

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So in the utterance, because only two pathways are leading up to it and purple represents a four junction,

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one, two, three, four.

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So by this tiny image, easily able to identify that our interest points have been correctly identified.

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Now let's move over to our main.

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To our men, both mapping and then onwards we are terminal.

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We have simulation run.

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Build this particular project once more.

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But if we're building that project, we uncomment out the one false algorithm.

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And also said the interest points to true so that we can now draw the interest points so the user can

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see that.

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So.

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We build this again?

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And once this is done, we run the over.

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And this time, once we get them, is occupancy grade from the local ocean state.

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You press enter and you will see that we have a new output after the pathways being overlaid over the

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inmates.

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This new output is the main interest points, the interest point that have been identified in our midst.

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And as you can see, this red circle represents the dead ends and the three number in our maze, and

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they have been correctly identified.

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As you can see, only one path is leading up to them.

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So if you look at right here, this is what the deadline might look like for this last case.

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And as you can see, only the red right pixel is leading up to this pixel.

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So this is a dead end under the hood.

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And as it can see, this pink triangle represents a three junction and three junction, a three paths

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leading up to it.

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And this blue rectangle represents a four junction.

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The this full path leading up to it.

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So in this way, we have confirmed that our.

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What mapping process up until finding the interest points is indeed working correctly.

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This means now we need to create a graph class to store all the interest points that we are just one.

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So let's go do the.