How can I track the boundary

I'm going to track the boundary line ，but my photo was not very clearly about some of the point(missing point)，how can i set the missing point to NaN and replace

to Proximity value,to make the line complete ,thanks

fclose all; close all; clear all; clf; clc;
cd 'C:\Users\User\Documents\MATLAB\test1\carmer1\photo1'
a=imread('img350.png');
i2=im2double(a);
ihd=rgb2gray(i2);

cd 'C:\Users\User\Documents\MATLAB\test1\carmer1\mavg1'
b=imread('imavg1.png');
b=im2double(b);
c=ihd-b;
msim_crop = imadjust(c,[0 0.1],[0 1],1.);
msim_crop = uint8(msim_crop);
bw0 = im2bw( msim_crop,graythresh(msim_crop).*0.01 );%0.12
windsz = 7; %4
H2 = fspecial('average',windsz); )
averaged_bw1 = imfilter(bw0,H2,'replicate');

the photo will look like this

so how can i just track the white boundary only like

[KSSV]

Read about convhull, boundary.

Answer 1

[Image Analyst]

If you want to find the light blue boundary on the left, I'd scan row by row using find() to find the leftmost white pixel. Then I'd fit a sliding quadratic to it with sgolayfilt(). Then I'd subtract the fit from the actual to find the distance. Distances that are greater than some amount, I'd discard. You might be able to use rmoutliers() for that. Give it a try. Here's a start

[rows, columns] = size(binaryImage);
leftColumns = columns * ones(rows, 1);
for row = 1 : rows
    thisRow = binaryImage(row, :);
    col = find(thisRow, 1, 'first')
    if ~isempty(col)
        leftColumns(row) = col;
    end
end
smoothedColumns = sgolayfilt(leftColumns, 21, 2);
diffs = abs(smoothedColumns - leftColumns);
plot(diffs, 'b-', 'LineWidth', 2);

etc.

[HUANG YU-CHE]

sorry ，may i asked about the smoothedColumns = sgolayfilt(leftColumns, 21, 2); part，because it show that Frame length must be odd.，i have try to change some of it but it still don't work ，how to fix it，thanks

[Image Analyst]

Sorry - I reversed the polynomial order and window width. It should be sgolayfilt(leftColumns, 2, 21) or some other number than 21. A bigger number gives more smoothing.

[HUANG YU-CHE]

thank you very much