How to iterate until a point satisfies a certain condition?

Hi everyone! I need a suggestion for this code. I have this code:

x = [1,2,2,2,1;2,3,2,1,1];
y = [1,1,1,2,2;1,1,2,2,1];
plot(x,y,'-x')
xlim([0,3])
ylim([0,3])
axis equal

Applying this procedure, I obtain that the points that are unique such as (3;1) are removed. The code is this one:

%converts to points with x,y coordinates
points = [x(1,:),x(2,:);y(1,:),y(2,:)]; 
%initialize test array
repeat_arr = zeros(size(points,2),size(points,2));
for i_points = 1:size(points,2)
    %find if a point is repeated excluding itself
    repeat_arr(i_points,[1:i_points-1,i_points+1:end]) = all(points(:,i_points)==points(:,[1:i_points-1,i_points+1:end]),1) ;
end
% now we know for each points if it is repeated
temp = any(repeat_arr,1); 
%Reverse the segment to point process so we can mask any segment that does not have a repeated point
edge_mask = all([temp(1:end/2); temp(end/2+1:end)]);
%Create X Y array of segment using out mask
x_edges = x(:,edge_mask);
y_edges = y(:,edge_mask);
plot(x_edges,y_edges,'-x')
xlim([0,3])
ylim([0,3])
axis equal

However, my question is: let us imagine that I have also a point of coordinates (2.5;1) like the one in the figure below, and that using this procedure I am able to remove the point (3;1). After this pruning, I would like the procedure to iterate until also the point (2.5;1) (that now is unique) is excluded and therefore execute as long as only the points in common to two or more segments remain. How can I do this in the most general way possible to be able to adapt it to other more complicated situations with more segments? I think that I need a while loop but, how can I set the condition?

Answer 1

[Walter Roberson]

You could simplify by using graph() objects https://www.mathworks.com/help/matlab/ref/graph.html

The points to be removed are those with incidence 0 or 1. If you use the adjacency table, the number of nonzero in the row is 0 or 1 for removal

[Loren99]

@Walter Roberson can you help me to apply this function graph() to my case, in order to understand if it works? Pls

[Walter Roberson]

It would be easier if you used explicit vertices and connectivity instead of having to deduce the connections

x = [1,2,2,2,1;2,3,2,1,1];
y = [1,1,1,2,2;1,1,2,2,1];

xy = [x(:), y(:)];
uxy = unique(xy, 'rows');
[~, idx1] = ismember([x(1,:); y(1,:)].', uxy, 'rows');
[~, idx2] = ismember([x(2,:); y(2,:)].', uxy, 'rows');

G = graph(idx1, idx2);
plot(G, 'XData', uxy(:,1), 'YData', uxy(:,2))

nodes_to_keep_mask = sum(full(adjacency(G)),2) >= 2;
nodes_to_remove = find(~nodes_to_keep_mask)

nodes_to_remove = 5

newG = rmnode(G, nodes_to_remove);

newxy = uxy(nodes_to_keep_mask, :);
plot(newG, 'XData', newxy(:,1), 'YData', newxy(:,2))

[Loren99]

@Walter Roberson thanks. Last question, how can I iterate the procedure? For example if I have a point of coordinates (2.5 , 1), how can I adapt your code in order to eliminate firstly (5, 1) as you have done and then (2.5, 1)? I have to make this code more general possible because I have to adapt it to another more complicated situation. Pls

[Walter Roberson]

You can rmnode one at a time, but you would have to figure out the order.

[Walter Roberson]

x = [1,2,2,2,1,3;2,3,2,1,1,3];
y = [1,1,1,2,2,1;1,1,2,2,1,1.5];

xy = [x(:), y(:)];
uxy = unique(xy, 'rows');
[~, idx1] = ismember([x(1,:); y(1,:)].', uxy, 'rows');
[~, idx2] = ismember([x(2,:); y(2,:)].', uxy, 'rows');

G = graph(idx1, idx2);
plot(G, 'XData', uxy(:,1), 'YData', uxy(:,2))

inG = G;
inxy = uxy;

while true
    nodes_to_keep_mask = sum(full(adjacency(inG)),2) >= 2;
    nodes_to_remove = find(~nodes_to_keep_mask)
    if isempty(nodes_to_remove); break; end
    inG = rmnode(inG, nodes_to_remove);
    
    inxy = inxy(nodes_to_keep_mask, :);
    figure
    plot(inG, 'XData', inxy(:,1), 'YData', inxy(:,2))
end

nodes_to_remove = 6

nodes_to_remove = 5

nodes_to_remove =

  0×1 empty double column vector

Answer 2

[the cyclist]

You need to code the condition into a while loop.

(I didn't fully understand the condition, so I can't be more specific than that.)