Linear Arrangement Problem A Betweenness Approach for Solving the Linear Arrangement Problem Marcus Oswald University of Heidelberg, Germany Joint work with Alberto Caprara, Emiliano Traversi, Michael Jung and Gerhard Reinelt Aussois, January 16th, 2009 Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Outline Linear Arrangement Problem 1 Lower Bounds 2 Betweenness Approach 3 Branch-and-Cut Algorithms 4 Computational Results 5 Outlook 6 Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Definition Given an undirected graph G ( V, E ) Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Definition Given an undirected graph G ( V, E ) Goal: Find a labeling π : V → { 1 , . . . , n } that minimizes � | π ( i ) − π ( j ) | . ( i,j ) ∈ E Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Definition Given an undirected graph G ( V, E ) Goal: Find a labeling π : V → { 1 , . . . , n } that minimizes � | π ( i ) − π ( j ) | . ( i,j ) ∈ E Example 2 2 1 1 3 5 3 5 4 4 We call the minimum lap ( G ) . Here lap ( G ) = 10 . Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Combinatorial Lower Bounds I Degree Lower Bound, Petit (2003) LB D = 1 � ⌊ (deg( i ) + 1) 2 / 4 ⌋ 2 i ∈ V 2 1 3 5 4 LB D = 1 2 (4 + 4 + 2 + 2 + 2) = 7 Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Combinatorial Lower Bounds II Edge Lower Bound, Petit (2003) At most ( n − 1) edges with distance 1 , ( n − 2) with distance 2 , etc. 2 1 3 5 4 LB E = 4 × 1 + 2 × 2 = 8 Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Eigenvalue Lower Bound Eigenvalue Lower Bound, Juvan, Mohar (1992) The n × n Laplacian matrix L ( G ) of G is defined as: deg( i ) i = j L ( G ) i,j := − 1 ( i, j ) ∈ E 0 otherwise. LB JM = ⌈ ( λ 2 ( n 2 − 1) / 6 ⌉ 3 − 1 − 1 − 1 0 − 1 2 0 0 − 1 L ( G ) = − 1 0 2 0 − 1 − 1 0 0 2 − 1 0 − 1 − 1 − 1 3 LB JM = ⌈ (2(5 2 − 1) / 6 ⌉ = 8 λ 2 = 2 ⇒ Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Comparison of the Bounds Name n m UB LB JM LB E LB D gd95c 62 144 506 37 250 292 gd96b 111 193 1416 42 276 702 gd96c 65 125 519 37 186 191 gd96d 180 228 2391 418 277 595 Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Linear Programming Bound Amaral, Caprara, Letchford, Salazar (2007) Introduce distance variables d ij and solve: � min d ij ( i,j ) ∈ E G ′ subgraph of G � d ij ≥ lap ( G ′ ) , s.t. ( i,j ) ∈ E ( G ′ ) d ij ≥ 1 , ( i, j ) ∈ E, where lap ( G ′ ) is known, for example for G ′ stars, cliques, etc. Note: Separation can be done in the complete graph after computing shortest paths on the d ij -values. Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Betweenness Approach Indicators For each triple i, k, j , ( i, j ) ∈ E, k ∈ V \ { i, j } and each order π χ π ikj indicates whether k lies between i and j in π : π − 1 ( i ) < π − 1 ( k ) < π − 1 ( j ) 1 or χ π π − 1 ( i ) > π − 1 ( k ) > π − 1 ( j ) ikj := 0 otherwise. Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Betweenness Approach Indicators For each triple i, k, j , ( i, j ) ∈ E, k ∈ V \ { i, j } and each order π χ π ikj indicates whether k lies between i and j in π : π − 1 ( i ) < π − 1 ( k ) < π − 1 ( j ) 1 or χ π π − 1 ( i ) > π − 1 ( k ) > π − 1 ( j ) ikj := 0 otherwise. Example 2 2 1 1 3 5 3 5 4 4 Here: χ π 132 = 0 and χ π 235 = 1 Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Reformulation of the Problem Computing the distances � | π − 1 ( i ) − π − 1 ( j ) | = 1 + χ π ikj k Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Reformulation of the Problem Computing the distances � | π − 1 ( i ) − π − 1 ( j ) | = 1 + χ π ikj k Linear Arrangement Problem � � � � χ π χ π min (1 + ikj ) = m + min ikj . π ∈ S ( n ) π ∈ S ( n ) k k ∈ V ( i,j ) ∈ E ( i,j ) ∈ E Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Branch-and-Cut based on Consecutive Ones Problem Given a 0/1 vector x . Is there a labeling π with x = χ π ? Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Branch-and-Cut based on Consecutive Ones Problem Given a 0/1 vector x . Is there a labeling π with x = χ π ? Writing x as matrix We define a matrix M ( x ) ∈ { 0 , 1 } 2 m × n like follows: 1 k = i and r = r ( i, j ) or k = j and r = r ( i, j ) + m M ( x ) r,k := 0 k = j and r = r ( i, j ) or k = i and r = r ( i, j ) + m x ikj otherwise, where r ( i, j ) denotes the edge with endnodes i and j . Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Branch-and-Cut based on Consecutive Ones Problem Given a 0/1 vector x . Is there a labeling π with x = χ π ? Writing x as matrix We define a matrix M ( x ) ∈ { 0 , 1 } 2 m × n like follows: 1 k = i and r = r ( i, j ) or k = j and r = r ( i, j ) + m M ( x ) r,k := 0 k = j and r = r ( i, j ) or k = i and r = r ( i, j ) + m x ikj otherwise, where r ( i, j ) denotes the edge with endnodes i and j . Observation x is a feasible betweenness vector if and only if M ( x ) has the consecutive ones property for rows . Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Transformation into a Consecutive Ones Problem Characterization of Tucker (1972) A 0/1 matrix M has the consecutive ones property for rows iff none of five types of forbidden matrices occur in M as submatrix. Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Transformation into a Consecutive Ones Problem Characterization of Tucker (1972) A 0/1 matrix M has the consecutive ones property for rows iff none of five types of forbidden matrices occur in M as submatrix. Integer Programming Formulation There is a set of valid inequalities that can be separated in polynomial time and cut off all forbidden matrices. Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Transformation into a Consecutive Ones Problem Characterization of Tucker (1972) A 0/1 matrix M has the consecutive ones property for rows iff none of five types of forbidden matrices occur in M as submatrix. Integer Programming Formulation There is a set of valid inequalities that can be separated in polynomial time and cut off all forbidden matrices. By solving the Weighted Consecutive Ones Problem with weights n k = i and r = r ( i, j ) or w r,k := k = j and r = r ( i, j ) − 1 otherwise, we can solve the Linear Arrangement Problem. Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Branch-and-Cut based on Betweenness-Variables Betweenness Polytope BTW = conv { χ π | π ∈ S ( n ) } P G Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Branch-and-Cut based on Betweenness-Variables Betweenness Polytope BTW = conv { χ π | π ∈ S ( n ) } P G Linear Arrangement Problem � � m + min x ikj x ∈ P G BT W k ∈ V ( i,j ) ∈ E Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Branch-and-Cut based on Betweenness-Variables Betweenness Polytope BTW = conv { χ π | π ∈ S ( n ) } P G Linear Arrangement Problem � � m + min x ikj x ∈ P G BT W k ∈ V ( i,j ) ∈ E Observations Like in the d ij -formulation: Let G ′ ⊂ G . Each valid inequality for P G ′ BTW is valid for P G BTW . For special graphs (stars, cliques, cycles, ...) we know inequalities derived from the d ij -formulation. Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Splitting distances Computing the distances | π − 1 ( i ) − π − 1 ( j ) | = 1 + � χ π ikj k Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Splitting distances Computing the distances | π − 1 ( i ) − π − 1 ( j ) | = 1 + � χ π ikj k In terms of variables � d ij = 1 + x ikj k Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
Linear Arrangement Problem Splitting distances Computing the distances | π − 1 ( i ) − π − 1 ( j ) | = 1 + � χ π ikj k In terms of variables � d ij = 1 + x ikj k 3-Star d 12 + d 13 + d 14 ≥ 4 ↓ x 132 + x 142 + x 123 + x 143 + x 124 + x 134 ≥ 1 Marcus Oswald Heidelberg A Betweenness Approach for Solving the LAP
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