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ps2sol

# ps2sol - 15.083/6.859J Integer Programming Combinatorial...

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15.083/6.859J Integer Programming & Combinatorial Optimization Solutions to Problem Set 2 October 18, 2004 1. BW 2.15 As shown in BW example 2.3, the class of clique tree inequalities defined in Eq. (2.22) h t x e + x e 3 t + 2 h 1 , i =1 e δ ( H i ) i =1 e δ ( T i ) is valid when h = 1. We shall then use induction to show the validity. By multiplying the degree constraints by 1 and summing over H i , T i , we get 2 1 x e + x e = | H i i = 1 , . . . , h 2 | e E ( H i ) e δ ( H i ) 1 x e + x e = | T i i = 1 , . . . , t 2 | e E ( T i ) e δ ( T i ) Substituting them into the clique tree inequalities, we get h t ( | H i x e ) + ( | T i x e ) 3 t + 2 h 1 | − | − 2 i =1 e E ( H i ) i =1 e E ( T i ) h t h t 3 t + 2 h 1 x e + x e H i | + i =1 e E ( H i ) i =1 e E ( T i ) i =1 | i =1 | T i | − 2 Let C ( H, T ) be a clique tree where | H = h + 1, | T = t . Without loss of generality, consider | | handle H h +1 and let the teeth intersecting with it be T 1 , . . . , T k and k is odd. k ± Removing the nodes in H h +1 \ T i leave us with k clique trees C i ( H ( C i ) , T ( C i )), i = 1 , . . . , k i =1 each with at most h handles. Therefore, by the induction hypothesis, for all i 1

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= x e + x e | H + | T | − 3 | T ( C i ) + 2 H ( C i ) | | 2 | | − 1 H H ( C i ) e E ( H ) T T ( C i ) e E ( T ) H H ( C i ) T T ( C i ) k k Summing over the k inequalities, and given | H ( C i ) = h and | T ( C i ) = t | i =1 | i =1 h t h t 3 t + 2 h 1 x e + x e H i | + i =1 e E ( H i ) i =1 e E ( T i ) i =1 | i =1 | T i | − 2 From H h +1 , we have 1 x e + x e = H h +1 | 2 | e E ( H h +1 ) e δ ( H h +1 ) t ± 1 Adding x e 0 to the above inequality for all e δ ( H h +1 ) \ E ( T i ), we get 2 i =1 1 t x e + x e H h +1 | 2 | e E ( H h +1 ) i =1 e δ ( H h +1 ) E ( T i ) t Since H h +1 intersects with k teeth, x e k , we get i =1 e δ ( H h +1 ) E ( T i ) k x e | H h +1 | − 2 e E ( H h +1 ) and since k is odd, k + 1 x e | H h +1 | − 2 e E ( H h +1 ) Adding the contribution from H h +1 to the other handles, we get h +1 h +1 3 t +2 h k + k +1 i =1 e E ( H i ) x e + i t =1 e E ( T i ) x e h +1 | H i | + i t t =1 | T i | − 3 t +2( h +1) 1 i =1 2 2 i =1 | H i | + i =1 | T i | − 2. BW 3.13 The three matroids, M 1 , M 2 and M 3 are: (a) M 1 is the matroid ( A , I 2 ) as defined in the book where I 2 is the collection of acyclic subgraphs. (b) Define M 2 as the matroid ( A , I 5 ) where I 5 is the collection of arcs in which no more than one arc enters each node. (c) Define M 3 as the matroid ( A , I 6 ) where I 6 is the collection of arcs in which no more than one arc leaves each node. 2
Since we know how to represent the convex hull of the incidence vectors of a matroid as an integral polyhedron, and a Hamiltonian path in D contains V 1 arcs, a Hamiltonian path in digraph | | D can be formulated as the integer program Z IP = max x j e ∈A x e m 1 ( S ) , S ⊆ A e ∈S x e m 2 ( S ) , S ⊆ A e ∈S x e m 3 ( S ) , S ⊆ A e ∈S x e = | V | − 1 e ∈A , , x e ∈ { 0 , 1 } e ∈ A where m 1 ( · ), m 2 ( · ), and m 3 ( · ) are the rank functions corresponding to matroids M 1 , M 2 and M 3 , respectively.

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ps2sol - 15.083/6.859J Integer Programming Combinatorial...

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