Subdeterminants and Concave Integer Quadratic Programming Alberto - PowerPoint PPT Presentation

Subdeterminants and Concave Integer Quadratic Programming Alberto Del Pia, University of Wisconsin-Madison January 10, 2019 23rd Combinatorial Optimization Workshop, Aussois 2019

The problem

Separable Concave Integer Quadratic Programming k • Polynomially solvable in fixed dimension. [Hartmann ’89] • We are interested in algorithms in general dimension. 1 ∑ i + h ⊤ x min − q i x 2 i = 1 ( IQP ) s . t . Wx ≤ w x ∈ Z n . • Integral data, q ≥ 0. • NP -hard even if k = 0 (ILP).

ϵ -approximate solution

2 Definition ϵ -approximate solution Let x ∗ be an optimal solution. For ϵ ∈ [ 0 , 1 ] , x ⋄ is an ϵ -approximate solution if obj( x ⋄ ) − obj( x ∗ ) obj max − obj( x ∗ ) ≤ ϵ. • obj( · ) : objective function value. • obj max : maximum value of obj( x ) on the feasible region.

2 Definition • Definition used in the literature from the 80s. • Natural choice for unstructured problems: • Insensitive to change of basis. ϵ -approximate solution Let x ∗ be an optimal solution. For ϵ ∈ [ 0 , 1 ] , x ⋄ is an ϵ -approximate solution if obj( x ⋄ ) − obj( x ∗ ) obj max − obj( x ∗ ) ≤ ϵ. • Preserved under dilation and translation of obj .

Main results

Main results: general case k k Theorem 3 ∑ i + h ⊤ x min − q i x 2 i = 1 ( IQP ) s . t . Wx ≤ w x ∈ Z n . Denote by ∆ the largest absolute value of the subdeterminants of W . For every ϵ ∈ ( 0 , 1 ] there is an algorithm that finds an ϵ -approximate solution by solving ⌈√ ( )⌉) k ( ( 2 n ∆) 2 + 1 3 + ϵ ILPs of the same size of ( IQP ), and with a constraint matrix with subdeterminants bounded by ∆ .

Corollary 4 k k Main results: ∆ = 2 ∑ i + h ⊤ x min − q i x 2 i = 1 ( IQP ) s . t . Wx ≤ w x ∈ Z n . Using [Artmann Weismantel Zenklusen ’17]: Assume ∆ = 2. For every ϵ ∈ ( 0 , 1 ] there is an algorithm that finds an ϵ -approximate solution in a number of operations bounded by ⌈√ ( )⌉) k ( ( 4 n ) 2 + 1 3 + poly( n , m ) . ϵ • For ∆ = 2, this closes the gap between the best known algorithm for ( IQP ) and its continuous version. [Vavasis ’92]

The algorithm

Step 1. Feasibility and boundedness 5 k ∑ i + h ⊤ x min − q i x 2 i = 1 ( IQP ) s . t . Wx ≤ w x ∈ Z n . • Feasibility and boundedness of ( IQP ) can be checked in polynomial time by solving 2 k + 1 ILPs using [Del Pia ’16]. • Assume now that ( IQP ) is feasibile and bounded.

Step 2. Decomposition b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b 6 b b b b b b b b b b b b b b b b b b b b In particular, we obtain finite bounds on the integer hull: b b b b b b b b b b b b b b b b b b b b b b b b b l i := min { x i : Wx ≤ w , x ∈ Z n } , ∀ i = 1 , . . . , k u i := max { x i : Wx ≤ w , x ∈ Z n } ∀ i = 1 , . . . , k .

Step 2. Decomposition b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b 6 b b b b b b b b b b b b b b b b b b b b In particular, we obtain finite bounds on the integer hull: b b b b b b b b b b b b b b b b b b b b b b b b b l i := min { x i : Wx ≤ w , x ∈ Z n } , ∀ i = 1 , . . . , k u i := max { x i : Wx ≤ w , x ∈ Z n } ∀ i = 1 , . . . , k .

Step 2. Decomposition b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b 6 b b b b b b b b b b b b b b b b b b k Width assumption: We need the following. b b b b b b b b b b b b b b b b b b b b b b b b b ⌈√ )⌉ ( ( 2 n ∆) 2 + 1 u i − l i ≥ =: g ϵ

Step 2. Decomposition b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b 6 b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b If ∃ i ∈ { 1 , . . . , k } such that u i − l i < g , we replace problem ( IQP ) with all the subproblems obtained by fixing x i to each value l i , l i + 1 , l i + 2 , . . . , u i .

Step 2. Decomposition b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b 6 b b b b b b b b b b b b b b b b b b b b it, and then return the best one. the width assumption. b b b b b b b b b b b b b b b b b b b b b b b b b b • Recursively, we decompose ( IQP ) into subproblems that satisfy • For each subproblem, we obtain an ϵ -approximate solution for

Step 2. Decomposition b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b 6 b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b • Each subproblem has the same form of ( IQP ). • Thus, we now assume that ( IQP ) satisfies the width assumption.

Step 3. Mesh partition and linear underestimators b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b 7 b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b We partition the box into g k boxes.

Step 3. Mesh partition and linear underestimators b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b 7 b b b b b b b b b b b b b b b b b b b quadratic objective on the vertices of b b b b b b b b b b b b b b b b b b b b b b b b min µ ( x ) For each box, we construct the affjne function µ ( x ) that coincides with the s . t . Wx ≤ w x ∈ box the box, and solve the ILP: x ∈ Z n .

Step 3. Mesh partition and linear underestimators b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b 7 b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b The best solution x ⋄ among the g k obtained ones (one per box) is an ϵ -approximate solution to ( IQP ).

Correctness

Correctness of the algorithm We need two bounds: Let’s see how to construct x . 8 To prove that x ⋄ is an ϵ -approximate solution we need to show obj( x ⋄ ) − obj( x ∗ ) obj max − obj( x ∗ ) ≤ ϵ. • obj( x ⋄ ) − obj( x ∗ ) ≤ UB : x ⋄ is close to optimal. • obj max − obj( x ∗ ) ≥ LB : ∃ x + feasible far from optimal.

Correctness of the algorithm We need two bounds: 8 To prove that x ⋄ is an ϵ -approximate solution we need to show obj( x ⋄ ) − obj( x ∗ ) obj max − obj( x ∗ ) ≤ ϵ. • obj( x ⋄ ) − obj( x ∗ ) ≤ UB : x ⋄ is close to optimal. • obj max − obj( x ∗ ) ≥ LB : ∃ x + feasible far from optimal. Let’s see how to construct x + .

A solution far from optimal b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b 9 b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b Wlog x 1 is the most concave direction, i.e, max { q i ( u i − l i ) 2 : i ∈ { 1 , . . . , k }} = q 1 ( u 1 − l 1 ) 2 =: γ.

A solution far from optimal b b b b b b b b b b b b b b b b b b b b b b b b b b b x l b b b b b b b b b b b b b b b b b b b b b b b 9 b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b x u 1 = l 1 , x u 1 = u 1 . Let x l , x u be feasible with x l

A solution far from optimal b b b b b b b b b b b b b b b b b b b b b b b b b b b b x u x l b b b b b b b b b b b b b b b b b b b b b b b b 9 b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b The vector x ◦ = 1 2 x l + 1 2 x u is far from optimal: obj( x ◦ ) − obj( x ∗ ) ≥ γ 4 . x ◦

A solution far from optimal b b b b b b b b b b b b b b b b b b b b b b b b b b b x u x l b b b b b b b b b b b b b b b b b b b b b b b 9 b b b b b b b b b b b b b b b b b b b b b b Define the box b b b b b b b b b b b b b b b b b b b b b b b b b b b D := x ◦ + [ − n ∆ , n ∆] k . x ◦

A solution far from optimal b b b b b b b b b b b b b b b b b b b b b b b b b b 9 b x u x l b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b 2 b b b b b b b b b b b b b b b b b b b b b b b b b b b Using subdeterminants, there exist feasible x − , x + ∈ D with x ◦ = x − + x + . x + x ◦ x −