How to find nearest two points each other ?

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Ender Rencuzogullari
回答： Semion 2020-4-14
Hi Everyone,
I have
X = (Yt.*sin(W))-(Xt.*cos(W));
Y = (Yt.*cos(W))+(Xt.*sin(W)); which give the coordinates X and Y.
X_inv = R.*sin(B_involute);
Y_inv = R.*cos(B_involute); which give the coordinates X_inv and Y_inv.
I need to find the nearest two points between X,Y and X_inv,Y_inv.
lots of thanks from now.
  2 个评论
Image Analyst
Image Analyst 2015-11-29
What is W? What is B_involute? Can you post a diagram or plot of this and point out X, Y, X_inv, Y_inv what two points on there you'd like to have it return - which are the "two points"? And it's ambiguous. What's nearest to what? Two points nearest to each other? Two points nearest to either X,Y or X_inv,Yinv? For example if you have 2,6,7,50, then 6 and 7 are nearest to each other but 6 and 7 are closer to 2 than they are to 50, so would you return 6 and 7, or return 2???
Ender Rencuzogullari
Ender Rencuzogullari 2015-11-29
编辑：Ender Rencuzogullari 2015-11-29
sorry for confusion. I have program to plot graph of two curves by specifying the coodinates of points. And, those curves intersect at an unknown point. I want to find out this point but If can find the nearest points between both, that will be my solution. X,Y specify a point while loop, and X_inv, Y_inv specify other point's coordinate. In fact, A=[X Y] and B=[X_inv Y_inv]. to make more clarify, I attached plot image of those graphs. the blue marks on 1st curve show the coordinates of point depends on X_inv (x axis) and Y_inv (y axis). it is the same with second curve depends on X and Y.Now, I want to find out the nearest point on second curve to point on first curve. I hope I could use english to explain enough.

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采纳的回答

Star Strider
Star Strider 2015-11-29
If you have the Statistics Toolbox, use the pdist2 function.
  4 个评论
显示 2更早的评论
Star Strider
Star Strider 2015-11-29
编辑：Star Strider 2015-11-29
My pleasure!
You would simply use your ‘X’, ‘Y’, ‘X_inv’ and ‘Y_inv’ in my code, eliminating the lines I used to create them to test it.
You would use only:
for k1 = 1:length(X)
for k2 = 1:length(X_inv)
DE(k1,k2) = hypot(X(k1)-X_inv(k2), Y(k1)-Y_inv(k2)); % Euclidean Distance
end
end
[DEmin,ix] = min(DE(:));
[K1,K2] = ind2sub(size(DE),ix);
fprintf(1,'\nNearest Points:\n\tDistance = %.3f\n', DEmin)
fprintf(1,'\t\tX(%d), Y(%d) \t\t\t= %.2f, %.2f\n', K1, K1, X(K1), Y(K1))
fprintf(1,'\t\tX_inv(%d), Y_inv(%d) \t= %.2f, %.2f\n', K2, K2, X_inv(K2), Y_inv(K2))

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更多回答（2 个）

Image Analyst
Image Analyst 2015-11-29
Use a nested for loop and the sqrt() function, then sort() and find() to find the 8 closest distances at the two points where your curves intersect.
Or maybe you could use roots(curve1-curve2).
Semion
Semion 2020-4-14
Hi. Could you explain, how does method "dsearchn" select an index of multi closest points with the same distance to target point? BW, the method "dnsearch" with and without triangulation produce different results.

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