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Published byCory Marshall Modified over 9 years ago
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Invariants to translation and scaling Normalized central moments
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Invariants to rotation M.K. Hu, 1962 - 7 invariants of 3rd order
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Hard to find, easy to prove:
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Drawbacks of the Hu’s invariants Dependence Incompleteness Insufficient number low discriminability
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Consequence of the incompleteness of the Hu’s set The images not distinguishable by the Hu’s set
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Normalized position to rotation
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Invariants to rotation M.K. Hu, 1962
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General construction of rotation invariants Complex moment in polar coordinates Complex moment
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Basic relations between moments
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Rotation property of complex moments The magnitude is preserved, the phase is shifted by (p-q)α. Invariants are constructed by phase cancellation
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Rotation invariants from complex moments Examples: How to select a complete and independent subset (basis) of the rotation invariants?
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Construction of the basis This is the basis of invariants up to the order r
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Inverse problem Is it possible to resolve this system ?
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Inverse problem - solution
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The basis of the 3rd order This is basis B 3 (contains six real elements)
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Comparing B 3 to the Hu’s set
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Drawbacks of the Hu’s invariants Dependence Incompleteness
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Comparing B 3 to the Hu’s set - Experiment The images distinguishable by B 3 but not by Hu’s set
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Difficulties with symmetric objects Many moments and many invariants are zero
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Examples of N-fold RS N = 1 N = 2 N = 3 N = 4 N = ∞
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Difficulties with symmetric objects Many moments and many invariants are zero
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Difficulties with symmetric objects The greater N, the less nontrivial invariants Particularly
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Difficulties with symmetric objects It is very important to use only non-trivial invariants The choice of appropriate invariants (basis of invariants) depends on N
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The basis for N-fold symmetric objects Generalization of the previous theorem
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Recognition of symmetric objects – Experiment 1 5 objects with N = 3
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Recognition of symmetric objects – Experiment 1 Bad choice: p 0 = 2, q 0 = 1
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Recognition of symmetric objects – Experiment 1 Optimal choice: p 0 = 3, q 0 = 0
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Recognition of symmetric objects – Experiment 2 2 objects with N = 1 2 objects with N = 2 2 objects with N = 3 1 object with N = 4 2 objects with N = ∞
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Recognition of symmetric objects – Experiment 2 Bad choice: p 0 = 2, q 0 = 1
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Recognition of symmetric objects – Experiment 2 Better choice: p 0 = 4, q 0 = 0
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Recognition of symmetric objects – Experiment 2 Theoretically optimal choice: p 0 = 12, q 0 = 0 Logarithmic scale
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Recognition of symmetric objects – Experiment 2 The best choice: mixed orders
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Recognition of circular landmarks Measurement of scoliosis progress during pregnancy
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The goal: to detect the landmark centers The method: template matching by invariants
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Normalized position to rotation
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Rotation invariants via normalization
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