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by: Khalil Conroy


Khalil Conroy
University of Central Florida
GPA 3.76


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Class Notes
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This 19 page Class Notes was uploaded by Khalil Conroy on Thursday October 22, 2015. The Class Notes belongs to CAP 6411 at University of Central Florida taught by Staff in Fall. Since its upload, it has received 10 views. For similar materials see /class/227217/cap-6411-university-of-central-florida in System Engineering at University of Central Florida.

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Date Created: 10/22/15
Lecture 13 Face Recognition amp Visual Lipreading Face Recognition 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 modelbase during training During recognition a vector corresponding to an unknown face is compared with all vectors in the modelbase The face from modelbase which is closest to the unknown face is declared as a recognized face Problems and Solution Problems 7 Dimensionality of each face vector will be very large 250000 for a 512X512 image 7 Raw gray levels are sensitive to noise and lighting conditions Solution 7 Reduce dimensionality of face space by nding principal components eigen vectors to span the face space 7 Only a few most signi cant eigen vectors can be used to represent a face thus reducing the dimensionality Eigen Vectors and Eigen Values The eigen vector x of a matrix A is a special vector with the following property sz x Where is called eigen value To nd eigen values of a matrix A rst nd the roots of det xll 0 Then solve the following linear system for each eigen value to nd corresponding eigen vector A mx0 Example 1 2 0 A 0 3 4 0 0 7 Eigen Values 112222211324 1 l 1 X1 4 X2 2 X3 0 Eigen Vectors 0 Eigen Values detA7210 LI 2 01 F1 0 01 det 0 3 4710100 i0 0 7 0 0 1J 0 7171 2 det 0 371 4 o o 0 771 11317100 7171371771o 171 13 17 Eigen Vectors 12 1 A Dx0 120 100 x1 0 034010 x220 007 001 x5 0 020x1 0 02x20 0 0 4 4 x2 0 0 0 8 0 04x24x30 5 008x30 x1 1 190 x0 1 X120 0 Face Recognition Collect all gray levels in a long vector u u 111 I1NI21 12N 1M1 1MNT Collect 71 samples views of each of p persons in matrix A MN X pn A u11uiu12ufufuf Form a correlation matrix L MN X MN L AAT Compute eigen vectors 1gt 2gt 3 gt 1 of L which form a bases for whole face space Face Recognition Each face u can now be represented as a linear combination of eigen vectors m u Zq z 11 Eigen vectors for a symmetric matrix are orthonormal 1 if 139 1 T Pi 1quot 0 ifiij Face Recognition m 422 m a az f ai f 2111qu 4 um a1quq a2 j i ai f an Qq Therefore a 21 Face Recognition L is a large matrix computing eigen vectors of a large matrix is time consuming Therefore compute eigen vectors of a smaller matrix C C A7 A Let Xi be eigen vectors of C then Aw are the eigen vectors of A I Cor m ATAoc Log AAT Aug 1A0 LA0cLA0c Training Create A matrix from training images Compute C matrix from A Compute eigenvectors of C gompute eigenvectors of L from eigenvectors of Select few most significant eigenvectors of L for face recognition Compute coefficient vectors corresponding to each training image For each person coefficients will form a cluster compute the mean of cluster Recognition Create a vector u for the image to be recognized Compute coefficient vector for this u Decide which person this image belongs to based on the distance from the cluster mean for each person load faces mat CA A vectorCvalueCeigC ssdiagvalueC ssiiis01tss vectorCvectorCiii vectorLAvectorC15 Coef A vectorL for 11 30 mode1i memcoe 5i115I end While 1 imagename 39nputCEnter the lename of the image to Recognize0 stop if imagename lt1 break end imagecFAC imagename vectorL d1 disp The coef cients for this image are mess1sprintf 2f 2f 2f 2f 2f imageco1imageco2imageco3imageco4 imageco5 dispmess1 to 1 for 1230 if normmode1iimageco1ltnormmode1 top imageco1 topi end mess1sprintfquot1quothe image input was a image of person number d top disp mess 1 end WACBI Wreshapeb3451 imshowbgray255 Webpage httpVismodwwwmedia mit eduVismoddemos Visual Lipreading Image Sequences 0f A to J a In E E E E E E E E E E mmmnmlmnmmmm ls mnmmmnnlmmnmmm mmnmmmlmmmmmmx mmmmmmmmmn mmmmmmmmlnmmmmw Particulars Pattern differ spatially Solution Spatial registration using SSD Articu ations Vary in length and thus in number of frames Solution Dynamic programming for temporal Warping of sequences Features should have compact representation Solution Principle Component Analysis Feature Subspace Generation Generate a lower dimension subspace onto which image sequences are projected to produce a vector of coef cients Components Sample Matrix Most Expressive Features Generating the Sample Matrix Consider 8 letters each of which has a training set of K sequences Each sequence is compose of images 112IP Collect all graylevel pixels from all images in a sequence into a vector u 11011MN2112MNP11PMN Generating the Sample Matrix For letter a collect vectors into matrix T 1 2 K Tm u u u Create sample matrix A A71T27l The eigenvectors of a matrix L AAT are defined as L i The Most Expressive Features is an orthonormal basis of the sample matrix oAny image sequence u can be represented as Q u 2 amquot pa nl Use Q most signi cant eigenvectors The linear coef cients can be computed as T an 27quot n Training Process Model Generation Warp all the training sequences to a fixed length Perform spatial registration SSD Perform PCA Select Q most signi cant eigensequences and compute coefficient vectors a Compute mean coefficient vector for each letter Warping I AIaua2apali j Bb1b2bjbJ HL 13972 d Zlaibf tat71139 g1122d11 gi lj 22dij ldij gij min gi lj l2dij gi 2j l2di ljdij Recognition Warp the unknown sequence Perform spatial registration T aix ux 39 i dw aw ax Determine best match by min a d w Compute Extracting letters from Connected Sequences Average absolute intensity difference function fngtMiNZZIIIltxygt Imy xl yl f is smoothed to obtain g Articulation intervals correspond to peaks and nonarticulation intervals correspond to valleys in g A 12722 B 26739 C 42 55 D 57767 algmu Extracting letters from Connected Sequences Detect valleys in g From valley locations in g find locations where f crosses high threshold Locate beginning and ending frames A 12722 B 26739 C 42755 MMEMEMM WW WWM EEDMM WWWWIMW nmwmmmmmm MMWE EWEM D 57767 wwwmnmwmmm wwwmmmmmmm Results I II I I A to J one speaker 10 training seqs II A to M one speaker 10 training seqs III A to Z ten speakers two training seqsletterperson Show Video Clip v m cmw u 0 Q3113 paper httpwww csucfeduNVisi0npapersshah97NDS97pdf


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