Sandwich problems on orientations Zolt an Szigeti Laboratoire - PowerPoint PPT Presentation

Matroids Definition A set system M = ( V , M ) is called a matroid if M satisfies : 1 ∅ ∈ M , 2 if F ∈ M and F ′ ⊆ F , then F ′ ∈ M , 3 if F , F ′ ∈ M and | F | > | F ′ | , then ∃ f ∈ F \ F ′ : F ′ ∪ f ∈ M . The rank of M is the maximum size of a set in M . Examples 1 Forests of a graph, 2 Linearly independent vectors of a vector space. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 5 / 21

Matroids Definition A set system M = ( V , M ) is called a matroid if M satisfies : 1 ∅ ∈ M , 2 if F ∈ M and F ′ ⊆ F , then F ′ ∈ M , 3 if F , F ′ ∈ M and | F | > | F ′ | , then ∃ f ∈ F \ F ′ : F ′ ∪ f ∈ M . The rank of M is the maximum size of a set in M . Algorithmic aspects 1 Matroid is given by an oracle that answers if F ∈ M . 2 Greedy algorithm finds a set of M of maximum size, 3 more generally, given a matroid M , F 1 ∈ M and | F 1 | ≤ k ≤ rank of M , it finds F ∈ M that contains F 1 and that has size k . Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 5 / 21

Generalized Polymatroids Definition 1 A pair ( p , b ) of set functions on V is a strong pair if p is supermodular, b is submodular, they are compliant : for all X , Y ⊂ V , p ( X ) − p ( X \ Y ) ≤ b ( Y ) − b ( Y \ X ) . 2 If ( p , b ) is a strong pair then the polyhedron Q ( p , b ) = { z ∈ R V : p ( X ) ≤ z ( X ) ≤ b ( X ) ∀ X ⊆ V } is called a generalized polymatroid. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 6 / 21

Generalized Polymatroids Definition 1 A pair ( p , b ) of set functions on V is a strong pair if p is supermodular, b is submodular, they are compliant : for all X , Y ⊂ V , p ( X ) − p ( X \ Y ) ≤ b ( Y ) − b ( Y \ X ) . 2 If ( p , b ) is a strong pair then the polyhedron Q ( p , b ) = { z ∈ R V : p ( X ) ≤ z ( X ) ≤ b ( X ) ∀ X ⊆ V } is called a generalized polymatroid. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 6 / 21

Generalized Polymatroids Definition 1 A pair ( p , b ) of set functions on V is a strong pair if p is supermodular, b is submodular, they are compliant : for all X , Y ⊂ V , p ( X ) − p ( X \ Y ) ≤ b ( Y ) − b ( Y \ X ) . 2 If ( p , b ) is a strong pair then the polyhedron Q ( p , b ) = { z ∈ R V : p ( X ) ≤ z ( X ) ≤ b ( X ) ∀ X ⊆ V } is called a generalized polymatroid. Remarks 1 A pair ( m 1 , m 2 ) of modular functions is a strong pair if and only if m 1 ( v ) ≤ m 2 ( v ) ∀ v ∈ V . 2 The pair ( i G , e G ) is a strong pair. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 6 / 21

Generalized Polymatroids Definition 1 A pair ( p , b ) of set functions on V is a strong pair if p is supermodular, b is submodular, they are compliant : for all X , Y ⊂ V , p ( X ) − p ( X \ Y ) ≤ b ( Y ) − b ( Y \ X ) . 2 If ( p , b ) is a strong pair then the polyhedron Q ( p , b ) = { z ∈ R V : p ( X ) ≤ z ( X ) ≤ b ( X ) ∀ X ⊆ V } is called a generalized polymatroid. Remarks 1 A pair ( m 1 , m 2 ) of modular functions is a strong pair if and only if m 1 ( v ) ≤ m 2 ( v ) ∀ v ∈ V . 2 The pair ( i G , e G ) is a strong pair. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 6 / 21

Generalized Polymatroids Definition 1 A pair ( p , b ) of set functions on V is a strong pair if p is supermodular, b is submodular, they are compliant : for all X , Y ⊂ V , p ( X ) − p ( X \ Y ) ≤ b ( Y ) − b ( Y \ X ) . 2 If ( p , b ) is a strong pair then the polyhedron Q ( p , b ) = { z ∈ R V : p ( X ) ≤ z ( X ) ≤ b ( X ) ∀ X ⊆ V } is called a generalized polymatroid. Remarks 1 A pair ( m 1 , m 2 ) of modular functions is a strong pair if and only if m 1 ( v ) ≤ m 2 ( v ) ∀ v ∈ V . 2 The pair ( i G , e G ) is a strong pair. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 6 / 21

Generalized Polymatroid Intersection Theorem Q ( p , b ) = { z ∈ R V : p ( X ) ≤ z ( X ) ≤ b ( X ) ∀ X ⊆ V } Theorem (Frank, Tardos ’88) 1 The g-polymatroid Q ( p , b ) is 2 The intersection of two g-polymatroids Q ( p 1 , b 1 ) and Q ( p 2 , b 2 ) is Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 7 / 21

Generalized Polymatroid Intersection Theorem Q ( p , b ) = { z ∈ R V : p ( X ) ≤ z ( X ) ≤ b ( X ) ∀ X ⊆ V } Theorem (Frank, Tardos ’88) 1 The g-polymatroid Q ( p , b ) is non-empty, 1 an integral polyhedron if p and b are integral functions. 2 2 The intersection of two g-polymatroids Q ( p 1 , b 1 ) and Q ( p 2 , b 2 ) is Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 7 / 21

Generalized Polymatroid Intersection Theorem Q ( p , b ) = { z ∈ R V : p ( X ) ≤ z ( X ) ≤ b ( X ) ∀ X ⊆ V } Theorem (Frank, Tardos ’88) 1 The g-polymatroid Q ( p , b ) is non-empty, 1 an integral polyhedron if p and b are integral functions. 2 2 The intersection of two g-polymatroids Q ( p 1 , b 1 ) and Q ( p 2 , b 2 ) is Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 7 / 21

Generalized Polymatroid Intersection Theorem Q ( p , b ) = { z ∈ R V : p ( X ) ≤ z ( X ) ≤ b ( X ) ∀ X ⊆ V } Theorem (Frank, Tardos ’88) 1 The g-polymatroid Q ( p , b ) is non-empty, 1 an integral polyhedron if p and b are integral functions. 2 2 The intersection of two g-polymatroids Q ( p 1 , b 1 ) and Q ( p 2 , b 2 ) is Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 7 / 21

Generalized Polymatroid Intersection Theorem Q ( p , b ) = { z ∈ R V : p ( X ) ≤ z ( X ) ≤ b ( X ) ∀ X ⊆ V } Theorem (Frank, Tardos ’88) 1 The g-polymatroid Q ( p , b ) is non-empty, 1 an integral polyhedron if p and b are integral functions. 2 2 The intersection of two g-polymatroids Q ( p 1 , b 1 ) and Q ( p 2 , b 2 ) is non-empty if and only if p 1 ≤ b 2 and p 2 ≤ b 1 , 1 an integral polyhedron if p 1 , p 2 and b 1 , b 2 are integral functions. 2 Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 7 / 21

Generalized Polymatroid Intersection Theorem Q ( p , b ) = { z ∈ R V : p ( X ) ≤ z ( X ) ≤ b ( X ) ∀ X ⊆ V } Theorem (Frank, Tardos ’88) 1 The g-polymatroid Q ( p , b ) is non-empty, 1 an integral polyhedron if p and b are integral functions. 2 2 The intersection of two g-polymatroids Q ( p 1 , b 1 ) and Q ( p 2 , b 2 ) is non-empty if and only if p 1 ≤ b 2 and p 2 ≤ b 1 , 1 an integral polyhedron if p 1 , p 2 and b 1 , b 2 are integral functions. 2 Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 7 / 21

In-degree constrained orientation : Characterization m -orientation Problem Instance : Given a graph G = ( V , E ) and m : V → Z + . 2 1 0 1 2 Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 8 / 21

In-degree constrained orientation : Characterization m -orientation Problem Instance : Given a graph G = ( V , E ) and m : V → Z + . Question : Does there exist an orientation � G whose in-degree vector is m that is d − G ( v ) = m ( v ) ∀ v ∈ V ? � 2 1 0 1 2 Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 8 / 21

In-degree constrained orientation : Characterization m -orientation Problem Instance : Given a graph G = ( V , E ) and m : V → Z + . Question : Does there exist an orientation � G whose in-degree vector is m that is d − G ( v ) = m ( v ) ∀ v ∈ V ? � Theorem (Hakimi’65) The answer is Yes if and only if m ( X ) ≥ i G ( X ) ∀ X ⊆ V , m ( V ) = | E | . 2 1 0 X 1 2 Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 8 / 21

In-degree constrained orientation : Applications Applications Eulerian orientation of an undirected graph (Euler), Eulerian orientation of a mixed graph (Ford-Fulkerson), Perfect matching in a bipartite graph (Hall, Frobenius), f -factor in a bipartite graph (Ore, Tutte). G Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 9 / 21

In-degree constrained orientation : Applications Applications Eulerian orientation of an undirected graph (Euler), Eulerian orientation of a mixed graph (Ford-Fulkerson), Perfect matching in a bipartite graph (Hall, Frobenius), f -factor in a bipartite graph (Ore, Tutte). G m ( v ) = dG ( v ) ∀ v ∈ V 2 Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 9 / 21

In-degree constrained orientation : Applications Applications Eulerian orientation of an undirected graph (Euler), Eulerian orientation of a mixed graph (Ford-Fulkerson), Perfect matching in a bipartite graph (Hall, Frobenius), f -factor in a bipartite graph (Ore, Tutte). G = ( V , E ∪ A ) Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 9 / 21

In-degree constrained orientation : Applications Applications Eulerian orientation of an undirected graph (Euler), Eulerian orientation of a mixed graph (Ford-Fulkerson), Perfect matching in a bipartite graph (Hall, Frobenius), f -factor in a bipartite graph (Ore, Tutte). � G = ( V , � E ∪ A ) dE ( v )+ d + A ( v )+ d − A ( v ) m ( v ) = − d − A ( v ) ∀ v ∈ V 2 Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 9 / 21

In-degree constrained orientation : Applications Applications Eulerian orientation of an undirected graph (Euler), Eulerian orientation of a mixed graph (Ford-Fulkerson), Perfect matching in a bipartite graph (Hall, Frobenius), f -factor in a bipartite graph (Ore, Tutte). V U G = ( U ∪ V ; E ) Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 9 / 21

In-degree constrained orientation : Applications Applications Eulerian orientation of an undirected graph (Euler), Eulerian orientation of a mixed graph (Ford-Fulkerson), Perfect matching in a bipartite graph (Hall, Frobenius), f -factor in a bipartite graph (Ore, Tutte). V U G = ( U ∪ V ; E ) Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 9 / 21

In-degree constrained orientation : Applications Applications Eulerian orientation of an undirected graph (Euler), Eulerian orientation of a mixed graph (Ford-Fulkerson), Perfect matching in a bipartite graph (Hall, Frobenius), f -factor in a bipartite graph (Ore, Tutte). V U G = ( U ∪ V ; E ) m ( u ) = 1 ∀ u ∈ U m ( v ) = d ( v ) − 1 ∀ v ∈ V Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 9 / 21

In-degree constrained orientation : Applications Applications Eulerian orientation of an undirected graph (Euler), Eulerian orientation of a mixed graph (Ford-Fulkerson), Perfect matching in a bipartite graph (Hall, Frobenius), f -factor in a bipartite graph (Ore, Tutte). 1 3 2 1 V U 2 1 3 1 G = ( U ∪ V ; E ) , f : U ∪ V → Z + Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 9 / 21

In-degree constrained orientation : Applications Applications Eulerian orientation of an undirected graph (Euler), Eulerian orientation of a mixed graph (Ford-Fulkerson), Perfect matching in a bipartite graph (Hall, Frobenius), f -factor in a bipartite graph (Ore, Tutte). 1 3 2 1 V U 2 1 3 1 G = ( U ∪ V ; E ), f -factor Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 9 / 21

In-degree constrained orientation : Applications Applications Eulerian orientation of an undirected graph (Euler), Eulerian orientation of a mixed graph (Ford-Fulkerson), Perfect matching in a bipartite graph (Hall, Frobenius), f -factor in a bipartite graph (Ore, Tutte). 1 3 2 1 V U 2 1 3 1 G = ( U ∪ V ; E ), f -factor m ( u ) = f ( u ) ∀ u ∈ U m ( v ) = d ( v ) − f ( v ) ∀ v ∈ V Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 9 / 21

In-degree constrained orientation : Algorithm Algorithm 1 The in-degree constrained orientation problem is in P because it is equivalent to the f -factor problem in a bipartite graph. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 10 / 21

In-degree constrained orientation : Algorithm Algorithm 1 The in-degree constrained orientation problem is in P because it is equivalent to the f -factor problem in a bipartite graph. 3 1 0 2 G , m Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 10 / 21

In-degree constrained orientation : Algorithm Algorithm 1 The in-degree constrained orientation problem is in P because it is equivalent to the f -factor problem in a bipartite graph. 3 3 1 1 1 1 1 1 1 1 0 2 0 2 G , m H , f Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 10 / 21

In-degree constrained orientation : Algorithm Algorithm 1 The in-degree constrained orientation problem is in P because it is equivalent to the f -factor problem in a bipartite graph. 3 3 1 1 1 1 1 1 1 1 0 2 0 2 G , m H , f , F Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 10 / 21

In-degree constrained orientation : Algorithm Algorithm 1 The in-degree constrained orientation problem is in P because it is equivalent to the f -factor problem in a bipartite graph. 3 3 1 1 1 1 1 1 1 1 0 2 0 2 � G , m = d − H , f , F � G Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 10 / 21

In-degree constrained orientation : Algorithm Algorithm 2 1 Take an arbitrary orientation � G of G . 2 If d − G ( v ) ≤ m ( v ) ∀ v , then it is an m -orientation, Stop. � 3 Otherwise, take a big vertex v : d − G ( v ) > m ( v ). � 4 Let X be the set of vertices u from which there exists a path P u to v . 5 Take a small vertex u ∈ X : d − G ( u ) < m ( u ). � G ′ be obtained from � 6 Let � G by reorienting P u . Go to Step 2. 7 This algorithm finds an m -orientation in polynomial time. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 10 / 21

In-degree constrained orientation : Algorithm Algorithm 2 1 Take an arbitrary orientation � G of G . 2 If d − G ( v ) ≤ m ( v ) ∀ v , then it is an m -orientation, Stop. � 3 Otherwise, take a big vertex v : d − G ( v ) > m ( v ). � 4 Let X be the set of vertices u from which there exists a path P u to v . 5 Take a small vertex u ∈ X : d − G ( u ) < m ( u ). � G ′ be obtained from � 6 Let � G by reorienting P u . Go to Step 2. 7 This algorithm finds an m -orientation in polynomial time. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 10 / 21

In-degree constrained orientation : Algorithm Algorithm 2 1 Take an arbitrary orientation � G of G . 2 If d − G ( v ) ≤ m ( v ) ∀ v , then it is an m -orientation, Stop. � v ∈ V d − (Indeed, | A | = � G ( v ) ≤ � v ∈ V m ( v ) = m ( V ) = | E | = | A | . ) � 3 Otherwise, take a big vertex v : d − G ( v ) > m ( v ). � 4 Let X be the set of vertices u from which there exists a path P u to v . 5 Take a small vertex u ∈ X : d − G ( u ) < m ( u ). � G ′ be obtained from � 6 Let � G by reorienting P u . Go to Step 2. 7 This algorithm finds an m -orientation in polynomial time. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 10 / 21

In-degree constrained orientation : Algorithm Algorithm 2 1 Take an arbitrary orientation � G of G . 2 If d − G ( v ) ≤ m ( v ) ∀ v , then it is an m -orientation, Stop. � 3 Otherwise, take a big vertex v : d − G ( v ) > m ( v ). � 4 Let X be the set of vertices u from which there exists a path P u to v . 5 Take a small vertex u ∈ X : d − G ( u ) < m ( u ). � G ′ be obtained from � 6 Let � G by reorienting P u . Go to Step 2. 7 This algorithm finds an m -orientation in polynomial time. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 10 / 21

In-degree constrained orientation : Algorithm Algorithm 2 1 Take an arbitrary orientation � G of G . 2 If d − G ( v ) ≤ m ( v ) ∀ v , then it is an m -orientation, Stop. � 3 Otherwise, take a big vertex v : d − G ( v ) > m ( v ). � 4 Let X be the set of vertices u from which there exists a path P u to v . 5 Take a small vertex u ∈ X : d − G ( u ) < m ( u ). � G ′ be obtained from � 6 Let � G by reorienting P u . Go to Step 2. 7 This algorithm finds an m -orientation in polynomial time. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 10 / 21

In-degree constrained orientation : Algorithm Algorithm 2 1 Take an arbitrary orientation � G of G . 2 If d − G ( v ) ≤ m ( v ) ∀ v , then it is an m -orientation, Stop. � 3 Otherwise, take a big vertex v : d − G ( v ) > m ( v ). � 4 Let X be the set of vertices u from which there exists a path P u to v . 5 Take a small vertex u ∈ X : d − G ( u ) < m ( u ). � G ′ be obtained from � 6 Let � G by reorienting P u . Go to Step 2. 7 This algorithm finds an m -orientation in polynomial time. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 10 / 21

In-degree constrained orientation : Algorithm Algorithm 2 1 Take an arbitrary orientation � G of G . 2 If d − G ( v ) ≤ m ( v ) ∀ v , then it is an m -orientation, Stop. � 3 Otherwise, take a big vertex v : d − G ( v ) > m ( v ). � 4 Let X be the set of vertices u from which there exists a path P u to v . 5 Take a small vertex u ∈ X : d − G ( u ) < m ( u ). (It exists because � x ∈ X m ( x ) = m ( X ) ≥ i G ( X ) = i G ( X ) + d − x ∈ X d − � G ( X ) = � G ( x ) . ) � � G ′ be obtained from � 6 Let � G by reorienting P u . Go to Step 2. 7 This algorithm finds an m -orientation in polynomial time. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 10 / 21

In-degree constrained orientation : Algorithm Algorithm 2 1 Take an arbitrary orientation � G of G . 2 If d − G ( v ) ≤ m ( v ) ∀ v , then it is an m -orientation, Stop. � 3 Otherwise, take a big vertex v : d − G ( v ) > m ( v ). � 4 Let X be the set of vertices u from which there exists a path P u to v . 5 Take a small vertex u ∈ X : d − G ( u ) < m ( u ). � G ′ be obtained from � 6 Let � G by reorienting P u . Go to Step 2. 7 This algorithm finds an m -orientation in polynomial time. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 10 / 21

In-degree constrained orientation : Algorithm Algorithm 2 1 Take an arbitrary orientation � G of G . 2 If d − G ( v ) ≤ m ( v ) ∀ v , then it is an m -orientation, Stop. � 3 Otherwise, take a big vertex v : d − G ( v ) > m ( v ). � 4 Let X be the set of vertices u from which there exists a path P u to v . 5 Take a small vertex u ∈ X : d − G ( u ) < m ( u ). � G ′ be obtained from � 6 Let � G by reorienting P u . Go to Step 2. w ∈ V | d − w ∈ V | d − (It is better : � G ′ ( w ) − m ( w ) | = � G ( w ) − m ( w ) | − 2.) � � 7 This algorithm finds an m -orientation in polynomial time. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 10 / 21

In-degree constrained orientation : Algorithm Algorithm 2 1 Take an arbitrary orientation � G of G . 2 If d − G ( v ) ≤ m ( v ) ∀ v , then it is an m -orientation, Stop. � 3 Otherwise, take a big vertex v : d − G ( v ) > m ( v ). � 4 Let X be the set of vertices u from which there exists a path P u to v . 5 Take a small vertex u ∈ X : d − G ( u ) < m ( u ). � G ′ be obtained from � 6 Let � G by reorienting P u . Go to Step 2. 7 This algorithm finds an m -orientation in polynomial time. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 10 / 21

In-degree constrained orientation : Algorithm Algorithm 2 1 Take an arbitrary orientation � G of G . 2 If d − G ( v ) ≤ m ( v ) ∀ v , then it is an m -orientation, Stop. � 3 Otherwise, take a big vertex v : d − G ( v ) > m ( v ). � 4 Let X be the set of vertices u from which there exists a path P u to v . 5 Take a small vertex u ∈ X : d − G ( u ) < m ( u ). � G ′ be obtained from � 6 Let � G by reorienting P u . Go to Step 2. 7 This algorithm finds an m -orientation in polynomial time. (0 ≤ � w ∈ V | d − G ( w ) − m ( w ) | ≤ 2 | E | . ) � Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 10 / 21

In-degree constrained orientation : Algorithm Algorithm 2 1 Take an arbitrary orientation � G of G . 2 If d − G ( v ) ≤ m ( v ) ∀ v , then it is an m -orientation, Stop. � 3 Otherwise, take a big vertex v : d − G ( v ) > m ( v ). � 4 Let X be the set of vertices u from which there exists a path P u to v . 5 Take a small vertex u ∈ X : d − G ( u ) < m ( u ). � G ′ be obtained from � 6 Let � G by reorienting P u . Go to Step 2. 7 This algorithm finds an m -orientation in polynomial time. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 10 / 21

Sandwich problems Graph Sandwich Problem for Property Π Instance : Given graphs G 1 = ( V , E 1 ) and G 2 = ( V , E 2 ) with E 1 ⊂ E 2 . Question : Does there exist E 1 ⊆ E ⊆ E 2 such that the graph G = ( V , E ) satisfies property Π ? Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 11 / 21

Sandwich problems Graph Sandwich Problem for Property Π Instance : Given graphs G 1 = ( V , E 1 ) and G 2 = ( V , E 2 ) with E 1 ⊂ E 2 . Question : Does there exist E 1 ⊆ E ⊆ E 2 such that the graph G = ( V , E ) satisfies property Π ? Golumbic, Kaplan, Shamir ’95 Split graphs (in P), [V=C+I] Cographs (in P), [no induced P 4 ] Eulerian graphs, Comparability graphs (NP-complete), [has a transitive orientation] Permutation graphs (NP-complete), [intersection graph of the chords of a permutation diagram] Interval graphs (NP-complete). [intersection graph of a family of intervals on the real line] Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 11 / 21

Degree Constrained Sandwich Problems Undirected case G 1 , G 2 undirected graphs, Π = { d G ( v ) = m ( v ) ∀ v ∈ V } ( m : V → Z + ). Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 12 / 21

Degree Constrained Sandwich Problems Undirected case G 1 , G 2 undirected graphs, Π = { d G ( v ) = m ( v ) ∀ v ∈ V } ( m : V → Z + ). Remark It is equivalent to the f -factor problem. The answer is Yes if and only if there exists an ( m ( v ) − d G 1 ( v ))-factor in the graph G 0 = ( V , E 2 \ E 1 ) . Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 12 / 21

Degree Constrained Sandwich Problems Undirected case G 1 , G 2 undirected graphs, Π = { d G ( v ) = m ( v ) ∀ v ∈ V } ( m : V → Z + ). Remark It is equivalent to the f -factor problem. The answer is Yes if and only if there exists an ( m ( v ) − d G 1 ( v ))-factor in the graph G 0 = ( V , E 2 \ E 1 ) . Directed case D 1 , D 2 directed graphs and Π = { d − D ( v ) = m ( v ) ∀ v ∈ V } ( m : V → Z + ). Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 12 / 21

Degree Constrained Sandwich Problems Undirected case G 1 , G 2 undirected graphs, Π = { d G ( v ) = m ( v ) ∀ v ∈ V } ( m : V → Z + ). Remark It is equivalent to the f -factor problem. The answer is Yes if and only if there exists an ( m ( v ) − d G 1 ( v ))-factor in the graph G 0 = ( V , E 2 \ E 1 ) . Directed case D 1 , D 2 directed graphs and Π = { d − D ( v ) = m ( v ) ∀ v ∈ V } ( m : V → Z + ). Exercise The answer is Yes if and only if d − D 2 ( v ) ≥ m ( v ) ≥ d − D 1 ( v ) ∀ v ∈ V . Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 12 / 21

m -orientation Sandwich Problem 1 Undirected Graphs : G 1 , G 2 undirected graphs, Π = G has an m -orientation ( m : V → Z + ) . Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 13 / 21

m -orientation Sandwich Problem 1 m -orientation Sandwich Problem for Undirected Graphs : Instance : Given undirected graphs G 1 = ( V , E 1 ) and G 2 = ( V , E 2 ) with E 1 ⊆ E 2 and a non-negative integer vector m on V . Question : Does there exist a sandwich graph G = ( V , E ) ( E 1 ⊆ E ⊆ E 2 ) that has an orientation � G whose in-degree vector is m that is d − G ( v ) = m ( v ) ∀ v ∈ V ? � Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 13 / 21

m -orientation Sandwich Problem 1 m -orientation Sandwich Problem for Undirected Graphs : Instance : Given undirected graphs G 1 = ( V , E 1 ) and G 2 = ( V , E 2 ) with E 1 ⊆ E 2 and a non-negative integer vector m on V . Question : Does there exist a sandwich graph G = ( V , E ) ( E 1 ⊆ E ⊆ E 2 ) that has an orientation � G whose in-degree vector is m that is d − G ( v ) = m ( v ) ∀ v ∈ V ? � Theorem (de Gevigney, Klein, Nguyen, Szigeti 2010) The answer is Yes if and only if i E 1 ( X ) ≤ m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V . Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 13 / 21

m -orientation Sandwich Problem 1 m -orientation Sandwich Problem for Undirected Graphs : Instance : Given undirected graphs G 1 = ( V , E 1 ) and G 2 = ( V , E 2 ) with E 1 ⊆ E 2 and a non-negative integer vector m on V . Question : Does there exist a sandwich graph G = ( V , E ) ( E 1 ⊆ E ⊆ E 2 ) that has an orientation � G whose in-degree vector is m that is d − G ( v ) = m ( v ) ∀ v ∈ V ? � Theorem (de Gevigney, Klein, Nguyen, Szigeti 2010) The answer is Yes if and only if i E 1 ( X ) ≤ m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V . Remark E 1 = E 2 : equivalent to Hakimi’s Theorem. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 13 / 21

Proof Theorem (de Gevigney, Klein, Nguyen, Szigeti 2010) The answer is Yes if and only if i E 1 ( X ) ≤ m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V . 1 Necessity : if sandwich graph G that has an m -orientation exists Each edge that contributes to i E 1 ( X ) must contribute to m ( X ) and 1 only the edges that contributes to e E 2 ( X ) may contribute to m ( X ). 2 2 Sufficiency : Let M = { F ⊆ E 2 : m ( X ) ≥ i F ( X ) ∀ X ⊆ V } . 1 M is a matroid of rank min { m ( V ( F )) + | E 2 \ F | : F ⊆ E 2 } . 2 By i E 1 ( X ) ≤ m ( X ) ∀ X ⊆ V , E 1 ∈ M . 3 For all F ⊆ E 2 , by m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V , applied for V \ V ( F ), and 4 by 2, rank of M is ≥ m ( V ). By 3 and 4, there exists E ∈ M that contains E 1 , of size m ( V ). 5 By 5, G = ( V , E ) is a sandwich graph that has, by Hakimi’s Theorem, 6 an m -orientation. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 14 / 21

Proof Theorem (de Gevigney, Klein, Nguyen, Szigeti 2010) The answer is Yes if and only if i E 1 ( X ) ≤ m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V . 1 Necessity : if sandwich graph G that has an m -orientation exists Each edge that contributes to i E 1 ( X ) must contribute to m ( X ) and 1 only the edges that contributes to e E 2 ( X ) may contribute to m ( X ). 2 2 Sufficiency : Let M = { F ⊆ E 2 : m ( X ) ≥ i F ( X ) ∀ X ⊆ V } . 1 M is a matroid of rank min { m ( V ( F )) + | E 2 \ F | : F ⊆ E 2 } . 2 By i E 1 ( X ) ≤ m ( X ) ∀ X ⊆ V , E 1 ∈ M . 3 For all F ⊆ E 2 , by m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V , applied for V \ V ( F ), and 4 by 2, rank of M is ≥ m ( V ). By 3 and 4, there exists E ∈ M that contains E 1 , of size m ( V ). 5 By 5, G = ( V , E ) is a sandwich graph that has, by Hakimi’s Theorem, 6 an m -orientation. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 14 / 21

Proof Theorem (de Gevigney, Klein, Nguyen, Szigeti 2010) The answer is Yes if and only if i E 1 ( X ) ≤ m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V . 1 Necessity : if sandwich graph G that has an m -orientation exists Each edge that contributes to i E 1 ( X ) must contribute to m ( X ) and 1 only the edges that contributes to e E 2 ( X ) may contribute to m ( X ). 2 2 Sufficiency : Let M = { F ⊆ E 2 : m ( X ) ≥ i F ( X ) ∀ X ⊆ V } . 1 M is a matroid of rank min { m ( V ( F )) + | E 2 \ F | : F ⊆ E 2 } . 2 By i E 1 ( X ) ≤ m ( X ) ∀ X ⊆ V , E 1 ∈ M . 3 For all F ⊆ E 2 , by m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V , applied for V \ V ( F ), and 4 by 2, rank of M is ≥ m ( V ). By 3 and 4, there exists E ∈ M that contains E 1 , of size m ( V ). 5 By 5, G = ( V , E ) is a sandwich graph that has, by Hakimi’s Theorem, 6 an m -orientation. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 14 / 21

Proof Theorem (de Gevigney, Klein, Nguyen, Szigeti 2010) The answer is Yes if and only if i E 1 ( X ) ≤ m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V . 1 Necessity : if sandwich graph G that has an m -orientation exists Each edge that contributes to i E 1 ( X ) must contribute to m ( X ) and 1 only the edges that contributes to e E 2 ( X ) may contribute to m ( X ). 2 2 Sufficiency : Let M = { F ⊆ E 2 : m ( X ) ≥ i F ( X ) ∀ X ⊆ V } . 1 M is a matroid of rank min { m ( V ( F )) + | E 2 \ F | : F ⊆ E 2 } . 2 By i E 1 ( X ) ≤ m ( X ) ∀ X ⊆ V , E 1 ∈ M . 3 For all F ⊆ E 2 , by m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V , applied for V \ V ( F ), and 4 by 2, rank of M is ≥ m ( V ). By 3 and 4, there exists E ∈ M that contains E 1 , of size m ( V ). 5 By 5, G = ( V , E ) is a sandwich graph that has, by Hakimi’s Theorem, 6 an m -orientation. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 14 / 21

Proof Theorem (de Gevigney, Klein, Nguyen, Szigeti 2010) The answer is Yes if and only if i E 1 ( X ) ≤ m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V . 1 Necessity : if sandwich graph G that has an m -orientation exists Each edge that contributes to i E 1 ( X ) must contribute to m ( X ) and 1 only the edges that contributes to e E 2 ( X ) may contribute to m ( X ). 2 2 Sufficiency : Let M = { F ⊆ E 2 : m ( X ) ≥ i F ( X ) ∀ X ⊆ V } . 1 M is a matroid of rank min { m ( V ( F )) + | E 2 \ F | : F ⊆ E 2 } . 2 By i E 1 ( X ) ≤ m ( X ) ∀ X ⊆ V , E 1 ∈ M . 3 For all F ⊆ E 2 , by m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V , applied for V \ V ( F ), and 4 by 2, rank of M is ≥ m ( V ). By 3 and 4, there exists E ∈ M that contains E 1 , of size m ( V ). 5 By 5, G = ( V , E ) is a sandwich graph that has, by Hakimi’s Theorem, 6 an m -orientation. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 14 / 21

Proof Theorem (de Gevigney, Klein, Nguyen, Szigeti 2010) The answer is Yes if and only if i E 1 ( X ) ≤ m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V . 1 Necessity : if sandwich graph G that has an m -orientation exists Each edge that contributes to i E 1 ( X ) must contribute to m ( X ) and 1 only the edges that contributes to e E 2 ( X ) may contribute to m ( X ). 2 2 Sufficiency : Let M = { F ⊆ E 2 : m ( X ) ≥ i F ( X ) ∀ X ⊆ V } . 1 M is a matroid of rank min { m ( V ( F )) + | E 2 \ F | : F ⊆ E 2 } . 2 By i E 1 ( X ) ≤ m ( X ) ∀ X ⊆ V , E 1 ∈ M . 3 For all F ⊆ E 2 , by m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V , applied for V \ V ( F ), and 4 by 2, rank of M is ≥ m ( V ). By 3 and 4, there exists E ∈ M that contains E 1 , of size m ( V ). 5 By 5, G = ( V , E ) is a sandwich graph that has, by Hakimi’s Theorem, 6 an m -orientation. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 14 / 21

Proof Theorem (de Gevigney, Klein, Nguyen, Szigeti 2010) The answer is Yes if and only if i E 1 ( X ) ≤ m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V . 1 Necessity : if sandwich graph G that has an m -orientation exists Each edge that contributes to i E 1 ( X ) must contribute to m ( X ) and 1 only the edges that contributes to e E 2 ( X ) may contribute to m ( X ). 2 2 Sufficiency : Let M = { F ⊆ E 2 : m ( X ) ≥ i F ( X ) ∀ X ⊆ V } . 1 M is a matroid of rank min { m ( V ( F )) + | E 2 \ F | : F ⊆ E 2 } . 2 By i E 1 ( X ) ≤ m ( X ) ∀ X ⊆ V , E 1 ∈ M . 3 For all F ⊆ E 2 , by m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V , applied for V \ V ( F ), and 4 by 2, rank of M is ≥ m ( V ). By 3 and 4, there exists E ∈ M that contains E 1 , of size m ( V ). 5 By 5, G = ( V , E ) is a sandwich graph that has, by Hakimi’s Theorem, 6 an m -orientation. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 14 / 21

Proof Theorem (de Gevigney, Klein, Nguyen, Szigeti 2010) The answer is Yes if and only if i E 1 ( X ) ≤ m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V . 1 Necessity : if sandwich graph G that has an m -orientation exists Each edge that contributes to i E 1 ( X ) must contribute to m ( X ) and 1 only the edges that contributes to e E 2 ( X ) may contribute to m ( X ). 2 2 Sufficiency : Let M = { F ⊆ E 2 : m ( X ) ≥ i F ( X ) ∀ X ⊆ V } . 1 M is a matroid of rank min { m ( V ( F )) + | E 2 \ F | : F ⊆ E 2 } . 2 By i E 1 ( X ) ≤ m ( X ) ∀ X ⊆ V , E 1 ∈ M . 3 For all F ⊆ E 2 , by m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V , applied for V \ V ( F ), and 4 by 2, rank of M is ≥ m ( V ). By 3 and 4, there exists E ∈ M that contains E 1 , of size m ( V ). 5 By 5, G = ( V , E ) is a sandwich graph that has, by Hakimi’s Theorem, 6 an m -orientation. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 14 / 21

Proof Theorem (de Gevigney, Klein, Nguyen, Szigeti 2010) The answer is Yes if and only if i E 1 ( X ) ≤ m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V . 1 Necessity : if sandwich graph G that has an m -orientation exists Each edge that contributes to i E 1 ( X ) must contribute to m ( X ) and 1 only the edges that contributes to e E 2 ( X ) may contribute to m ( X ). 2 2 Sufficiency : Let M = { F ⊆ E 2 : m ( X ) ≥ i F ( X ) ∀ X ⊆ V } . 1 M is a matroid of rank min { m ( V ( F )) + | E 2 \ F | : F ⊆ E 2 } . 2 By i E 1 ( X ) ≤ m ( X ) ∀ X ⊆ V , E 1 ∈ M . 3 For all F ⊆ E 2 , by m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V , applied for V \ V ( F ), and 4 by 2, rank of M is ≥ m ( V ). By 3 and 4, there exists E ∈ M that contains E 1 , of size m ( V ). 5 By 5, G = ( V , E ) is a sandwich graph that has, by Hakimi’s Theorem, 6 an m -orientation. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 14 / 21

Proof Theorem (de Gevigney, Klein, Nguyen, Szigeti 2010) The answer is Yes if and only if i E 1 ( X ) ≤ m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V . 1 Necessity : if sandwich graph G that has an m -orientation exists Each edge that contributes to i E 1 ( X ) must contribute to m ( X ) and 1 only the edges that contributes to e E 2 ( X ) may contribute to m ( X ). 2 2 Sufficiency : Let M = { F ⊆ E 2 : m ( X ) ≥ i F ( X ) ∀ X ⊆ V } . 1 M is a matroid of rank min { m ( V ( F )) + | E 2 \ F | : F ⊆ E 2 } . 2 By i E 1 ( X ) ≤ m ( X ) ∀ X ⊆ V , E 1 ∈ M . 3 For all F ⊆ E 2 , by m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V , applied for V \ V ( F ), and 4 by 2, rank of M is ≥ m ( V ). By 3 and 4, there exists E ∈ M that contains E 1 , of size m ( V ). 5 By 5, G = ( V , E ) is a sandwich graph that has, by Hakimi’s Theorem, 6 an m -orientation. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 14 / 21

Algorithmic aspects Theorem (de Gevigney, Klein, Nguyen, Szigeti 2010) The answer is Yes if and only if i E 1 ( X ) ≤ m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V . Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 15 / 21

Algorithmic aspects Theorem (de Gevigney, Klein, Nguyen, Szigeti 2010) The answer is Yes if and only if i E 1 ( X ) ≤ m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V . 1 Decide : The answer is Yes if and only if both submodular functions b 1 ( X ) = m ( X ) − i E 1 ( X ) and b 2 ( X ) = e E 2 ( X ) − m ( X ) have minimum value 0 . Submodular function minimization is polynomial (Schrijver ; Fleicher, Fujishige, Iwata’2000). 2 Find : By the previous matroid property, greedy algorithm finds the sandwich graph G , and as seen, the m -orientation of G is easy to find. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 15 / 21

Algorithmic aspects Theorem (de Gevigney, Klein, Nguyen, Szigeti 2010) The answer is Yes if and only if i E 1 ( X ) ≤ m ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V . 1 Decide : The answer is Yes if and only if both submodular functions b 1 ( X ) = m ( X ) − i E 1 ( X ) and b 2 ( X ) = e E 2 ( X ) − m ( X ) have minimum value 0 . Submodular function minimization is polynomial (Schrijver ; Fleicher, Fujishige, Iwata’2000). 2 Find : By the previous matroid property, greedy algorithm finds the sandwich graph G , and as seen, the m -orientation of G is easy to find. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 15 / 21

m -orientation Sandwich Problem 2 Mixed Graphs : G 1 , G 2 mixed graphs, Π = G has an m -orientation ( m : V → Z + ) . Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 16 / 21

m -orientation Sandwich Problem 2 m -orientation Sandwich Problem for Mixed Graphs : Instance : Given mixed graphs G 1 = ( V , E 1 ∪ A 1 ) and G 2 = ( V , E 2 ∪ A 2 ) with E 1 ⊆ E 2 , A 1 ⊆ A 2 and a non-negative integer vector m on V . Question : Does there exist a sandwich mixed graph G = ( V , E ∪ A ) with G = ( V , − → E 1 ⊆ E ⊆ E 2 and A 1 ⊆ A ⊆ A 2 that has an orientation � E ∪ A ) whose in-degree vector is m that is d − G ( v ) = m ( v ) ∀ v ∈ V ? � Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 16 / 21

m -orientation Sandwich Problem 2 m -orientation Sandwich Problem for Mixed Graphs : Instance : Given mixed graphs G 1 = ( V , E 1 ∪ A 1 ) and G 2 = ( V , E 2 ∪ A 2 ) with E 1 ⊆ E 2 , A 1 ⊆ A 2 and a non-negative integer vector m on V . Question : Does there exist a sandwich mixed graph G = ( V , E ∪ A ) with G = ( V , − → E 1 ⊆ E ⊆ E 2 and A 1 ⊆ A ⊆ A 2 that has an orientation � E ∪ A ) whose in-degree vector is m that is d − G ( v ) = m ( v ) ∀ v ∈ V ? � Theorem (de Gevigney, Klein, Nguyen, Szigeti 2010) The answer is Yes if and only if i E 1 ( X ) + � v ∈ X d − A 1 ( v ) ≤ m ( X ) ≤ e E 2 ( X ) + � v ∈ X d − A 2 ( v ) ∀ X ⊆ V . Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 16 / 21

m -orientation Sandwich Problem 2 m -orientation Sandwich Problem for Mixed Graphs : Instance : Given mixed graphs G 1 = ( V , E 1 ∪ A 1 ) and G 2 = ( V , E 2 ∪ A 2 ) with E 1 ⊆ E 2 , A 1 ⊆ A 2 and a non-negative integer vector m on V . Question : Does there exist a sandwich mixed graph G = ( V , E ∪ A ) with G = ( V , − → E 1 ⊆ E ⊆ E 2 and A 1 ⊆ A ⊆ A 2 that has an orientation � E ∪ A ) whose in-degree vector is m that is d − G ( v ) = m ( v ) ∀ v ∈ V ? � Theorem (de Gevigney, Klein, Nguyen, Szigeti 2010) The answer is Yes if and only if v ∈ X d − v ∈ X d − i E 1 ( X ) + � A 1 ( v ) ≤ m ( X ) ≤ e E 2 ( X ) + � A 2 ( v ) ∀ X ⊆ V . Special cases 1 E 2 = ∅ : result on the In-degree Constrained Sandwich Problem. 2 A 2 = ∅ : result on m -orient. Sandwich Problem for Undirected Graphs. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 16 / 21

Proof 1 Suppose that E 1 ⊆ E ⊆ E 2 has been choosen and oriented with in-degree vector m 1 . 2 Then the problem is reduced to the Dir. Degree Const. Sandw. Problem with m 2 ( v ) = m ( v ) − m 1 ( v ) ∀ v ∈ V for A 1 ⊆ A 2 , 3 which has a solution if and only if d − A 1 ( v ) ≤ m 2 ( v ) ≤ d − A 2 ( v ) ∀ v ∈ V . 4 or equivalently (1) m ( v ) − d − A 2 ( v ) ≤ m 1 ( v ) ≤ m ( v ) − d − A 1 ( v ) ∀ v ∈ V . 5 The problem is reduced to the m 1 -orientation Sandwich Problem for Undirected Graphs for E 1 ⊆ E 2 , 6 which has a solution iff (2) i E 1 ( X ) ≤ m 1 ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V . 7 The Mixed m -orient. Sandwich Problem has an Yes answer if and only if there exists a function m 1 : V → Z satisfying (1) and (2). 8 By the Generalized Polymatroid Intersection Theorem, applied for v ∈ X ( m ( v ) − d − v ∈ X ( m ( v ) − d − p 1 ( X ) = � A 2 ( v )) , b 1 ( X ) = � A 1 ( v )) , p 2 ( X ) = i E 1 ( X ) , b 2 ( X ) = e E 2 ( X ) , we are done. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 17 / 21

Proof 1 Suppose that E 1 ⊆ E ⊆ E 2 has been choosen and oriented with in-degree vector m 1 . 2 Then the problem is reduced to the Dir. Degree Const. Sandw. Problem with m 2 ( v ) = m ( v ) − m 1 ( v ) ∀ v ∈ V for A 1 ⊆ A 2 , 3 which has a solution if and only if d − A 1 ( v ) ≤ m 2 ( v ) ≤ d − A 2 ( v ) ∀ v ∈ V . 4 or equivalently (1) m ( v ) − d − A 2 ( v ) ≤ m 1 ( v ) ≤ m ( v ) − d − A 1 ( v ) ∀ v ∈ V . 5 The problem is reduced to the m 1 -orientation Sandwich Problem for Undirected Graphs for E 1 ⊆ E 2 , 6 which has a solution iff (2) i E 1 ( X ) ≤ m 1 ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V . 7 The Mixed m -orient. Sandwich Problem has an Yes answer if and only if there exists a function m 1 : V → Z satisfying (1) and (2). 8 By the Generalized Polymatroid Intersection Theorem, applied for v ∈ X ( m ( v ) − d − v ∈ X ( m ( v ) − d − p 1 ( X ) = � A 2 ( v )) , b 1 ( X ) = � A 1 ( v )) , p 2 ( X ) = i E 1 ( X ) , b 2 ( X ) = e E 2 ( X ) , we are done. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 17 / 21

Proof 1 Suppose that E 1 ⊆ E ⊆ E 2 has been choosen and oriented with in-degree vector m 1 . 2 Then the problem is reduced to the Dir. Degree Const. Sandw. Problem with m 2 ( v ) = m ( v ) − m 1 ( v ) ∀ v ∈ V for A 1 ⊆ A 2 , 3 which has a solution if and only if d − A 1 ( v ) ≤ m 2 ( v ) ≤ d − A 2 ( v ) ∀ v ∈ V . 4 or equivalently (1) m ( v ) − d − A 2 ( v ) ≤ m 1 ( v ) ≤ m ( v ) − d − A 1 ( v ) ∀ v ∈ V . 5 The problem is reduced to the m 1 -orientation Sandwich Problem for Undirected Graphs for E 1 ⊆ E 2 , 6 which has a solution iff (2) i E 1 ( X ) ≤ m 1 ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V . 7 The Mixed m -orient. Sandwich Problem has an Yes answer if and only if there exists a function m 1 : V → Z satisfying (1) and (2). 8 By the Generalized Polymatroid Intersection Theorem, applied for v ∈ X ( m ( v ) − d − v ∈ X ( m ( v ) − d − p 1 ( X ) = � A 2 ( v )) , b 1 ( X ) = � A 1 ( v )) , p 2 ( X ) = i E 1 ( X ) , b 2 ( X ) = e E 2 ( X ) , we are done. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 17 / 21

Proof 1 Suppose that E 1 ⊆ E ⊆ E 2 has been choosen and oriented with in-degree vector m 1 . 2 Then the problem is reduced to the Dir. Degree Const. Sandw. Problem with m 2 ( v ) = m ( v ) − m 1 ( v ) ∀ v ∈ V for A 1 ⊆ A 2 , 3 which has a solution if and only if d − A 1 ( v ) ≤ m 2 ( v ) ≤ d − A 2 ( v ) ∀ v ∈ V . 4 or equivalently (1) m ( v ) − d − A 2 ( v ) ≤ m 1 ( v ) ≤ m ( v ) − d − A 1 ( v ) ∀ v ∈ V . 5 The problem is reduced to the m 1 -orientation Sandwich Problem for Undirected Graphs for E 1 ⊆ E 2 , 6 which has a solution iff (2) i E 1 ( X ) ≤ m 1 ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V . 7 The Mixed m -orient. Sandwich Problem has an Yes answer if and only if there exists a function m 1 : V → Z satisfying (1) and (2). 8 By the Generalized Polymatroid Intersection Theorem, applied for v ∈ X ( m ( v ) − d − v ∈ X ( m ( v ) − d − p 1 ( X ) = � A 2 ( v )) , b 1 ( X ) = � A 1 ( v )) , p 2 ( X ) = i E 1 ( X ) , b 2 ( X ) = e E 2 ( X ) , we are done. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 17 / 21

Proof 1 Suppose that E 1 ⊆ E ⊆ E 2 has been choosen and oriented with in-degree vector m 1 . 2 Then the problem is reduced to the Dir. Degree Const. Sandw. Problem with m 2 ( v ) = m ( v ) − m 1 ( v ) ∀ v ∈ V for A 1 ⊆ A 2 , 3 which has a solution if and only if d − A 1 ( v ) ≤ m 2 ( v ) ≤ d − A 2 ( v ) ∀ v ∈ V . 4 or equivalently (1) m ( v ) − d − A 2 ( v ) ≤ m 1 ( v ) ≤ m ( v ) − d − A 1 ( v ) ∀ v ∈ V . 5 The problem is reduced to the m 1 -orientation Sandwich Problem for Undirected Graphs for E 1 ⊆ E 2 , 6 which has a solution iff (2) i E 1 ( X ) ≤ m 1 ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V . 7 The Mixed m -orient. Sandwich Problem has an Yes answer if and only if there exists a function m 1 : V → Z satisfying (1) and (2). 8 By the Generalized Polymatroid Intersection Theorem, applied for v ∈ X ( m ( v ) − d − v ∈ X ( m ( v ) − d − p 1 ( X ) = � A 2 ( v )) , b 1 ( X ) = � A 1 ( v )) , p 2 ( X ) = i E 1 ( X ) , b 2 ( X ) = e E 2 ( X ) , we are done. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 17 / 21

Proof 1 Suppose that E 1 ⊆ E ⊆ E 2 has been choosen and oriented with in-degree vector m 1 . 2 Then the problem is reduced to the Dir. Degree Const. Sandw. Problem with m 2 ( v ) = m ( v ) − m 1 ( v ) ∀ v ∈ V for A 1 ⊆ A 2 , 3 which has a solution if and only if d − A 1 ( v ) ≤ m 2 ( v ) ≤ d − A 2 ( v ) ∀ v ∈ V . 4 or equivalently (1) m ( v ) − d − A 2 ( v ) ≤ m 1 ( v ) ≤ m ( v ) − d − A 1 ( v ) ∀ v ∈ V . 5 The problem is reduced to the m 1 -orientation Sandwich Problem for Undirected Graphs for E 1 ⊆ E 2 , 6 which has a solution iff (2) i E 1 ( X ) ≤ m 1 ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V . 7 The Mixed m -orient. Sandwich Problem has an Yes answer if and only if there exists a function m 1 : V → Z satisfying (1) and (2). 8 By the Generalized Polymatroid Intersection Theorem, applied for v ∈ X ( m ( v ) − d − v ∈ X ( m ( v ) − d − p 1 ( X ) = � A 2 ( v )) , b 1 ( X ) = � A 1 ( v )) , p 2 ( X ) = i E 1 ( X ) , b 2 ( X ) = e E 2 ( X ) , we are done. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 17 / 21

Proof 1 Suppose that E 1 ⊆ E ⊆ E 2 has been choosen and oriented with in-degree vector m 1 . 2 Then the problem is reduced to the Dir. Degree Const. Sandw. Problem with m 2 ( v ) = m ( v ) − m 1 ( v ) ∀ v ∈ V for A 1 ⊆ A 2 , 3 which has a solution if and only if d − A 1 ( v ) ≤ m 2 ( v ) ≤ d − A 2 ( v ) ∀ v ∈ V . 4 or equivalently (1) m ( v ) − d − A 2 ( v ) ≤ m 1 ( v ) ≤ m ( v ) − d − A 1 ( v ) ∀ v ∈ V . 5 The problem is reduced to the m 1 -orientation Sandwich Problem for Undirected Graphs for E 1 ⊆ E 2 , 6 which has a solution iff (2) i E 1 ( X ) ≤ m 1 ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V . 7 The Mixed m -orient. Sandwich Problem has an Yes answer if and only if there exists a function m 1 : V → Z satisfying (1) and (2). 8 By the Generalized Polymatroid Intersection Theorem, applied for v ∈ X ( m ( v ) − d − v ∈ X ( m ( v ) − d − p 1 ( X ) = � A 2 ( v )) , b 1 ( X ) = � A 1 ( v )) , p 2 ( X ) = i E 1 ( X ) , b 2 ( X ) = e E 2 ( X ) , we are done. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 17 / 21

Proof 1 Suppose that E 1 ⊆ E ⊆ E 2 has been choosen and oriented with in-degree vector m 1 . 2 Then the problem is reduced to the Dir. Degree Const. Sandw. Problem with m 2 ( v ) = m ( v ) − m 1 ( v ) ∀ v ∈ V for A 1 ⊆ A 2 , 3 which has a solution if and only if d − A 1 ( v ) ≤ m 2 ( v ) ≤ d − A 2 ( v ) ∀ v ∈ V . 4 or equivalently (1) m ( v ) − d − A 2 ( v ) ≤ m 1 ( v ) ≤ m ( v ) − d − A 1 ( v ) ∀ v ∈ V . 5 The problem is reduced to the m 1 -orientation Sandwich Problem for Undirected Graphs for E 1 ⊆ E 2 , 6 which has a solution iff (2) i E 1 ( X ) ≤ m 1 ( X ) ≤ e E 2 ( X ) ∀ X ⊆ V . 7 The Mixed m -orient. Sandwich Problem has an Yes answer if and only if there exists a function m 1 : V → Z satisfying (1) and (2). 8 By the Generalized Polymatroid Intersection Theorem, applied for v ∈ X ( m ( v ) − d − v ∈ X ( m ( v ) − d − p 1 ( X ) = � A 2 ( v )) , b 1 ( X ) = � A 1 ( v )) , p 2 ( X ) = i E 1 ( X ) , b 2 ( X ) = e E 2 ( X ) , we are done. Z. Szigeti (G-SCOP, Grenoble) Sandwich problems on orientations 27 janvier 2011 17 / 21