face recognition by Euclidean space distance seeking for help

asked 2017-12-04 22:01:04 -0500

striving gravatar image

updated 2017-12-04 23:06:36 -0500

In the n dimensional Euclidean space,

If The first point is P1 [x1,x2,...,xn]

The second point is P2 [y1,y2,...,yn]

so ,the Euclidean space distance from P1 to P2 is

d(P1,P2) = sqrt ( (x1-y1)^2 + (x2-y2)^2 + ... + (xn-yn)^2 )

so , in the face recognition field , for example , in many open source , every point have x coordinate value and y coordinate value ,

I think it is ,

The first face is P1 [(x1,y1),(x2,y2),...,(xn,yn)]

The second face is P2 [(a1,b1),(a2,b2),...,(an,bn)]

so ,the Euclidean space distance from P1 to P2 is ________________________________ ?

the result is too complex , even is unsolvable .

I don't know if I have a problem with my understanding ?

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please do not post images, but TEXT,

berak gravatar imageberak ( 2017-12-04 22:14:08 -0500 )edit

I am sorry ,the effect of the formula in the text is not so good

striving gravatar imagestriving ( 2017-12-04 22:21:32 -0500 )edit

the image is unreadable, your question cannot be indexed for search, folks cannot quote you, please just comply.

berak gravatar imageberak ( 2017-12-04 22:25:49 -0500 )edit

ok , I got,should I delete or convert text in comment ?

striving gravatar imagestriving ( 2017-12-04 22:36:21 -0500 )edit

just remove the image, and make a text version of it, we can help you with the formatting

berak gravatar imageberak ( 2017-12-04 22:39:28 -0500 )edit

the pixel values are compared, not their (2d)coordinates.

berak gravatar imageberak ( 2017-12-05 01:48:56 -0500 )edit

ok , thank you very much

striving gravatar imagestriving ( 2017-12-05 02:00:27 -0500 )edit