Lecture16

Lecture16 - C&O 355 Lecture 16 N. Harvey Topics Review...

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Lecture 16 N. Harvey
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Topics Review of Fourier-Motzkin Elimination Linear Transformations of Polyhedra Convex Combinations Convex Hulls Polytopes & Convex Hulls
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Fourier-Motzkin Elimination Joseph Fourier Theodore Motzkin Given a polyhedron Q μ R n , we want to find the set Q’ μ R n-1 satisfying ( x 1 , ,x n-1 ) 2 Q’ , 9 x n s.t. ( x 1 , ,x n-1 , x n ) 2 Q Q’ is called the projection of Q onto first n-1 coordinates Fourier-Motzkin Elimination constructs Q’ by generating (finitely many) constraints from the constraints of Q . Corollary: Q’ is a polyhedron.
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Elimination Example x 1 x 2 Q Project Q onto coordinates {x 1 , x 2 }... x 3
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Elimination Example x 1 x 2 x 3 Q’ Project Q onto coordinates {x 1 , x 2 }... Fourier-Motzkin: Q’ is a polyhedron. Of course, the ordering of coordinates is irrevelant.
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Elimination Example x 1 x 2 x 3 Q’’ Of course, the ordering of coordinates is irrevelant. Fourier-Motzkin: Q’’ is also a polyhedron. I can also apply Elimination twice…
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Elimination Example x 1 x 2 x 3 Q’’’ Fourier-Motzkin: Q’’’ is also a polyhedron.
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Projecting a Polyhedron Onto Some of its Coordinates Lemma: Given a polyhedron Q μ R n . Let S={s 1 ,…, s k } μ {1,…,n} be any subset of the coordinates. Let Q S = { ( x s1 , , x sk ) : x 2 Q } μ R k . In other words, Q S is projection of Q onto coordinates in S. Then Q S is a polyhedron. Proof: Direct from Fourier-Motzkin Elimination. Just eliminate all coordinates not in S. ¥
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Linear Transformations of Polyhedra Lemma: Let P = { x : Ax · b } μ R n be a polyhedron. Let M be any matrix of size p x n. Let Q = { Mx : x 2 P } μ R p . Then Q is a polyhedron. x 1 x 3 P Let M = 1 0 0 -1 1 0 0 0 1 x 2
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Linear Transformations of Polyhedra Lemma: Let P = { x : Ax · b } μ R n be a polyhedron. Let M be any matrix of size p
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Lecture16 - C&O 355 Lecture 16 N. Harvey Topics Review...

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