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1 Storage of images for Efficient Retrieval Representing IDB as relations straightforward Representing IDB with spatial data structures represent the image as rectangles Representing IDB using image transformations can be represented by a vector of length k (k is between 100 and 200) However, different photographs of the same person may vary, depending on a variety of factors F time of the day at which the photographs were taken F the lighting conditions F the camera used F the exact position of the subject’s head and facial expression F etc.

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2 Representing IDB with Relations (1) Create a relation called images having the scheme: (Image, ObjID, XLB,XUB,YLB,YUB) F where Image is the name of the image file and (XLB,XUB,YLB,YUB) represent the rectangle in question Suppose R is a rectangle specified by (XLB, XUB,YLB,YUB) and R is in Rec(I), the there exists a tuple F (I, newid, XLB XUB,YLB,YUB) (2) For each property p, create a relation R p having the scheme: (Image, XLB, XUB,YLB,YUB, Value) F where Image is the image file and the (XLB, XUB,YLB,YUB) denote a rectangular cell in the image Properties can be F Pixel level properties (RGB values) F Object/region level properties (Name, age) F Image level properties (when image was captured, where, by whom)

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3 Querying IDB Relations Eliciting the contents of an image is done using image processing algorithms Image processing algorithms are only partially accurate This implies that tuples placed in an IDB relation by an image processing program has certain associated probabilistic attributes The probability that John Lee is the name of o 2 is 0.75 The probability that Ken Yip is the name of o 2 is 0.15 There is a 10% missing probability ObjId Name Prob o 1 Jim Hatch 0.8 o 2 John Lee 0.75 o 2 Ken Yip 0.15 o 3 John Lee 1

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4 Complex Queries Query: What is the probability that pic1.gif contains both Jim Hatch and Ken Yip? (assume o 1 and o 2 are in pic1.gif) Is it 0.8*0.15 = 0.12? In general, the answer is NO ObjId Name Prob o 1 Jim Hatch 0.8 o 2 John Lee 0.75 o 2 Ken Yip 0.15 o 3 John Lee 1

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5 Complex Queries Consider pic8.gif with o 10 and o 11 Four possibilities possibility 1: o 10 is Ken Yip and o 11 is John Hatch possibility 2: o 10 is Ken Yip and o 11 is not John Hatch possibility 3: o 10 is not Ken Yip but o 11 is John Hatch possibility 4: o 10 is not Ken Yip and o 11 is not John Hatch Suppose p i denotes the probability of possibility i p 1 + p 2 = 0.5 p 3 + p 4 = 0.5 p 1 + p 3 = 0.8 p 2 + p 4 = 0.2 p 1 + p 2 + p 3 + p 4 = 1 ObjId Name Prob o 10 Ken Yip 0.5 o 10 Jim Hatch 0.4 o 11 Jim Hatch 0.8 o 11 John Lee 0.1

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6 Interval Probability Need to solve these linear equations to determine the probability that pic8.gif contains both Ken Yip and John Hatch The result is an interval of probabilities p 1 between 0.3 and 0.5 Requires the use of Interval probabilities!!! Interval probabilities allow us to represent the margin of error of image processing algorithms in identifying the object e.g. with 3% error ObjId Name Prob(l) Prob(u) o 1 Jim Hatch 0.77 0.83 o 2 John Lee 0.72 0.78 o 2 Ken Yip 0.12 0.18 o 3 John Lee 0.97 1

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7 Representing image DBs with R-Trees Create a relation with two attributes (ImageId, ObjId) Create an R-tree that stores rectangles if the same rectangle appears in two images, then we have an overflow list Each rectangle has an associated set of fields that specifies the object/region level properties Not good for nearest neighbor queries as the tree is constructed with only two dimensions out of the n+2 dimensions of the image Generalized R-Trees use all the n+2 dimensions to construct the tree

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8 Implementations Many use object-oriented implementation Support methods such as rotate, segment, edit Most implementations assume the whole image to compare Perform feature extraction and thus represent an image as a vector of n fields an index is created on an n-dimensional vector multidimensional extension of point quadtree R-Tree To perform similarity search, they compute the Euclidian distance between the vector representing the query image and those of all images

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