Lecture-12 - Lecture-12 Face Recognition 1 Simple Approach...

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1 Lecture-12 Face Recognition
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2 Simple Approach • Recognize faces (mug shots) using gray levels (appearance) • Each image is mapped to a long vector of gray levels • Several views of each person are collected in the model-base during training • During recognition a vector corresponding to an unknown face is compared with all vectors in the model-base • The face from model-base, which is closest to the unknown face is declared as a recognized face. Problems and Solution • Problems : – Dimensionality of each face vector will be very large (250,000 for a 512X512 image!) – Raw gray levels are sensitive to noise, and lighting conditions. • Solution: – Reduce dimensionality of face space by finding principal components (eigen vectors) to span the face space – Only a few most significant eigen vectors can be used to represent a face, thus reducing the dimensionality
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3 Eigen Vectors and Eigen Values The eigen vector, x, of a matrix A is a special vector, with the following property x Ax l = Where ë is called eigen value 0 ) det( = - I A l To find eigen values of a matrix A first find the roots of: Then solve the following linear system for each eigen value to find corresponding eigen vector 0 ) ( = - x I A l Example - = 7 0 0 4 3 0 0 2 1 A 1 , 3 , 7 3 2 1 - = = = l l l = = = 0 0 1 , 0 2 1 , 4 4 1 3 2 1 x x x Eigen Values Eigen Vectors
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Eigen Values 0 ) det( = - I A l 0 ) 1 0 0 0 1 0 0 0 1 7 0 0 4 3 0 0 2 1 det( = - - l 0 ) 7 0 0 4 3 0 0 2 1 det( = - - - - l l l 7 3, , 1 0 ) 7 )( 3 )( 1 ( 0 ) 0 ) 7 )( 3 )(( 1 ( = = - = = - - - - = - - - - - l l l l l l l l l Eigen Vectors 0 ) ( = - x I A l = + - 0 0 0 ) 1 0 0 0 1 0 0 0 1 7 0 0 4 3 0 0 2 1 ( 3 2 1 x x x = 0 0 0 8 0 0 4 4 0 0 2 0 3 2 1 x x x 0 , 0 , 1
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This note was uploaded on 06/12/2011 for the course CAP 6411 taught by Professor Shah during the Spring '09 term at University of Central Florida.

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Lecture-12 - Lecture-12 Face Recognition 1 Simple Approach...

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