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© Leo Burnett
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2 “Moe”: April 23 “Moe” : June 6
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3 Population size estimation Behavioral studies Phenotype analysis
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4 Tracking devices: Require anesthetization Expensive Unreliable Broad coverage all but impossible
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5 1.Moe93% 2.Alice5% 3.Bob2% The “ALGORITHM” Database of zebra pictures Input – “the query” Output – “zebra ranking”
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6 Face recognition algorithms ▪ Early algorithms too rigid ▪ Modern algorithm could work, but are black boxes Fingerprint recognition ▪ Generally look for well-defined features ▪ Rarely deal with occlusion, perspective skew, varying distance to camera
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7 © Colchester Zoo y x 1.Shape contour tracing 2.Spline function fitting 3.Query and retrieve splines
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9 How much information is there in the data?
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10 How much information is there in the data?
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11 Hypothesis: width and spacing of stripes are distinctive when measured finely
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12 Zebras are/have: seldom two-dimensional frequently obscured non uniform stripes high tendency towards pregnancy and violence afraid of barcode scanners © Barcodeman
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13 Build a solution by eliminating the problems!
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14 Closer to camera, more pixels for the bodyFurther away, fewer pixels for the body 10 Megapixel camera = 3648 pixels across, 2736 pixels down 1080p HD TV = 1920 pixels across, 1080 pixels down 15” MacBook Pro screen= 1440 pixels across, 900 pixels down
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15 Solution: measure widths relative to the previous stripe 52466466638875104
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16 Solution: measure widths relative to the previous stripe 524664666388751.38
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17 Solution: measure widths relative to the previous stripe 5246646663880.851.38
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18 Solution: measure widths relative to the previous stripe 52.81.41.0.951.390.851.38
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20 Shear transformation flattens small amount of perspective skew [1]. Original imageWith shear transformation
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21 Shear transformation flattens small amount of perspective skew [1]. Shear transformation is a special case of affine transformation. Affine transformation: Ratios of distances along a line are preserved
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22 Solution: measure widths relative to the previous stripe 52.81.41.0.951.390.851.38
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23 Solution: measure widths relative to the previous stripe 52.81.41.0.951.390.851.38 0.81.41.00.951.390.851.38 A “strip” of stripes
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24 1.41.00.951.390.851.380.7 From original photograph in database: 0.831.390.851.380.7 Zebra occluded from the left side: Missed the rightmost black stripe: 1.41.00.951.390.851.98 Extremely oblique viewing angle: 1.51.11.21.71.011.661.3
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25 Solution: Dynamic programming Align two strips to minimize “errors” a.k.a. Spell-check, DNA sequence alignment, Needleman-Wunsch algorithm, Smith-Waterman algorithm, edit distance, dynamic time warping, etc. Stripe-alignment! occlusion = indel cost image processing errors = indel + matching cost stripe distortion = matching cost strong perspective skew= matching cost Low alignment “cost” = fewer differences in strips = zebras are very similar
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26 1.41.00.951.390.851.380.7 COST = 0 1.41.00.951.390.851.380.7 1.41.00.981.440.851.380.7 1.41.00.951.390.851.380.7 COST = (0.98-0.95) + (1.44-1.39) = 0.08
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27 For a new picture (the “query”): Read a strip off the body at a known location Align against all the zebra strips in the database, also from the same location Rank zebras in the database by the alignment cost of their strips
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28 One click per zebra Analogous to a barcode scanner Handles occlusion, minor perspective skew Can be applied to any part of the body Computationally efficient
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29 20 zebras, ~6 photos per zebra = 109 pictures. “Transcription” errors
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30 Photos were manually identified by Rosemary at Ol’Pejeta Conservancy. Manually coded stripes along the shoulder For each photo, rank the closest matches using dynamic programming. Metric: rank of the correct zebra in the list of closest matches.
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31 Proportion of queries at or below rank Average rank = 1.5
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32 No love from Zebra #3 – “01_700” Photo 87 Zebra 3 Flank R PicID 8159 CORRECT_RANK 4 Photo 65 Zebra 3 Flank L PicID 8142 CORRECT_RANK 9 Photo 63 Zebra 3 Flank R PicID 8170 CORRECT_RANK 5 Photo 52 Zebra 3 Flank R PicID 8166 CORRECT_RANK 2 81598142 81708166
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33 Worst performance on this picture:
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34 Time to search database of 108 pictures: 0.023 seconds-- my ageing 2006 laptop If the number of stripes on a zebra is O(1), then the time complexity of a single search is linear in the number of photographs. 1,000 pictures~ 0.2 seconds 10,000 pictures~ 3 seconds
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36 Future work: 1. Build an effective user interface 2. Get field biologists to discover better ways to use it! 3. Run tests on 3,000+ pictures from January Kenya trip 4. Tweak image processing algorithms “So you see! There’s no end to the thing you might know, depending how far beyond Zebra you go.” - Dr. Seuss, Beyond Zebra References: [1] Hutchison and Barrett. Fourier-Mellin registration of line-detained tabular document images. Intl. J. Doc. Analysis 8(2):87-110, 2006. mlahiri@gmail.com
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