Integer Quadratic Programming is in NP

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1 Alberto Del Pia 1 Santanu S. Dey 2 Marco Molinaro 2 1 IBM T. J. Watson Research Center, Yorktown Heights. 2 School of Industrial and Systems Engineering, Atlanta

2 Outline

3 1

4 4 Program: Definition Definition (IQP) min s.t. x Qx + c x Ax b x Z n,

5 5 Program: Definition Definition (IQP) min s.t. x Qx + c x Ax b x Z n, We do no assume that x Qx is convex i.e., Q is not necessarily positive semi-definite.

6 6 Program: Definition Definition (IQP) min s.t. x Qx + c x Ax b x Z n, We do no assume that x Qx is convex i.e., Q is not necessarily positive semi-definite. Decision Version of IQP Does there exist x satisfying: x Qx + c x + d 0 Ax b x Z n, where we assume all the data is rational. F(Q, c, d, A, b)

7 7 Main Theorem Let n, m Z ++. Let Q Q n n, c Q n, d Q, A Q m n, b Q m.

8 8 Main Theorem Let n, m Z ++. Let Q Q n n, c Q n, d Q, A Q m n, b Q m. If F(Q, c, d, A, b) is non-empty, then there exists x 0 F(Q, c, d, A, b)

9 9 Main Theorem Let n, m Z ++. Let Q Q n n, c Q n, d Q, A Q m n, b Q m. If F(Q, c, d, A, b) is non-empty, then there exists x 0 F(Q, c, d, A, b) such that the binary encoding size of x 0 is bounded from above by a polynomial function of the size of binary encoding of Q, c, d, A, b.

10 10 Main Theorem Let n, m Z ++. Let Q Q n n, c Q n, d Q, A Q m n, b Q m. If F(Q, c, d, A, b) is non-empty, then there exists x 0 F(Q, c, d, A, b) such that the binary encoding size of x 0 is bounded from above by a polynomial function of the size of binary encoding of Q, c, d, A, b. Consequences 1.. In particular, the decision version of IQP is NP-complete. 2. Broadly speaking, this implies that these exists an algorithm to solve IQP, i.e. not undecidable.

11 11 Comparison 1: More quadratic inequalities? Undecidable! Determing the feasibility of a system with 1. Number of quadratic inequalities: 2 ( (58 2 ) ) = Number of linear inequalities: 58 ( (58 ) ) 3. Number of integer variables: = is Undecidable. 2

12 12 Comparison 1: More quadratic inequalities? Undecidable! Determing the feasibility of a system with 1. Number of quadratic inequalities: 2 ( (58 2 ) ) = Number of linear inequalities: 58 ( (58 ) ) 3. Number of integer variables: = is Undecidable. 2 Reduction from undecidability of determining the feasibility of a quartic equation in 58 non-negative integer variables. [Jones (1982)], See discussion and additional references in [Köppe (2012)].

13 13 Comparison 2: Two quadratic inequalities? Exponential size solution! Consider the system for d = 5 2n+1 : x 2 dy , x 2 + dy x, y Z.

14 14 Comparison 2: Two quadratic inequalities? Exponential size solution! Consider the system for d = 5 2n+1 : x 2 dy , x 2 + dy x, y Z. 1. The binary encoding length of smallest integer solution with minimal binary encoding length has an encoding length of: Ω(5 n ).

15 15 Comparison 2: Two quadratic inequalities? Exponential size solution! Consider the system for d = 5 2n+1 : x 2 dy , x 2 + dy x, y Z. 1. The binary encoding length of smallest integer solution with minimal binary encoding length has an encoding length of: Ω(5 n ). 2. The binary encoding length of instance: Θ(n). [Lagarias (1980)], See discussion and additional references in [Köppe (2012)]

16 16 Comparison 3: More convex quadratic inequalities? Exponential size solution! Consider the system: x 1 2 x j xj 1 2 j {2,..., n} x j Z j {1,..., n}.

17 Comparison 3: More convex quadratic inequalities? Exponential size solution! Consider the system: x 1 2 x j xj 1 2 j {2,..., n} x j Z j {1,..., n}. 1. The binary encoding length of smallest size solution is: Ω(2 n ).

18 18 Comparison 3: More convex quadratic inequalities? Exponential size solution! Consider the system: x 1 2 x j xj 1 2 j {2,..., n} x j Z j {1,..., n}. 1. The binary encoding length of smallest size solution is: Ω(2 n ). 2. The binary encoding length of instance: Θ(n).

19 19 In Conclusion In the presence of exactly one rational quadratic inequality, there exists small" poly-size feasible solutions.

20 20 In Conclusion In the presence of exactly one rational quadratic inequality, there exists small" poly-size feasible solutions. 2. With even two inequality, the binary encoding of the smallest solution may be exponential in size. 3. With many" inequalities, (a) the problem become undecidables with general quadratics, or (b) binary encoding of all solutions may be exponential in size in the convex quadratics case.

21 2

22 Overview of the proof x Qx + c x + d 0 Ax b (P) x Z n

23 23 Overview of the proof x Qx + c x + d 0 Ax b (P) x Z n Definition: Simplicial cone A simplicial cone is a cone generated by a simplex.

24 Overview of the proof x Qx + c x + d 0 Ax b (P) x Z n Definition: Simplicial cone A simplicial cone is a cone generated by a simplex. Proof Steps 1. Step 1: It is sufficient to prove the result where P is a full-dimensional simplicial cone.

25 25 Overview of the proof x Qx + c x + d 0 Ax b (P) x Z n Definition: Simplicial cone A simplicial cone is a cone generated by a simplex. Proof Steps 1. Step 1: It is sufficient to prove the result where P is a full-dimensional simplicial cone. Standard techniques to show Integer linear programming is in NP. Carathéodory Theorem. Some careful rotation using (poly-size) unimodular matrices.

26 26 Overview of the proof x Qx + c x + d 0 Ax b (P) x Z n Definition: Simplicial cone A simplicial cone is a cone generated by a simplex. Proof Steps 1. Step 1: It is sufficient to prove the result where P is a full-dimensional simplicial cone. Standard techniques to show Integer linear programming is in NP. Carathéodory Theorem. Some careful rotation using (poly-size) unimodular matrices. 2. Step 2: Verify the result for the case where P is a full-dimensional simplicial cone.

27 2.1 Step 2

28 28 Getting Started x Qx + c x + d 0 Ax (P) x Z n 1. {x Ax 0} is a simplicial cone. 2. We may assume d > 0.

29 29 P is a full-dimensional simplicial cone. "Slice " the cone P with a "carefully selected" hyperplane H P Carefully selected Hyperplane

30 30 P is a full-dimensional simplicial cone. "Slice " the cone P with a "carefully selected" hyperplane H Let x be a poly-size rational optimal solution to the problem x Qx := min x Qx s.t. x P H Poly size rational optimal solution x* P Carefully selected Hyperplane

31 31 P is a full-dimensional simplicial cone. "Slice " the cone P with a "carefully selected" hyperplane H Let x be a poly-size rational optimal solution to the problem x Qx := min x Qx s.t. x P H The quadratic problem min{x Vx x rational polytope} (where V is a rational matrix) has a rational globally optimal solution of poly-size with respect to the size of the instance. [Vavasis 1990] Poly size rational optimal solution x* P Carefully selected Hyperplane

32 32 Case analysis based on sign of x Qx x Qx := min x Qx s.t. x P H Poly size rational optimal solution x* P Carefully selected Hyperplane

33 33 Case 1: x Qx < 0 Scale and find a solution

34 34 Case 1: x Qx < 0 Scale and find a solution 1. First scale x to x so that x P Z n.

35 35 Case 1: x Qx < 0 Scale and find a solution 1. First scale x to x so that x P Z n. 2. λ = c x + d ( x) Q x 3. Then λ x P Z n and ( x) Q x ( λ x) Q( λ x) + c ( λ x) + d 0. Poly size rational optimal solution x* P Carefully selected Hyperplane

36 36 Case 2: x Qx > 0 Question: Can x Qx be arbitrarily close to zero?

37 37 Case 2: x Qx > 0 Question: Can x Qx be arbitrarily close to zero? 1. No. In fact x Qx 1 G 2 integral WLOG). where G 2 size of x (Assuming Q is

38 38 Case 2: x Qx > 0 Question: Can x Qx be arbitrarily close to zero? 1. No. In fact x Qx 1 where G 2 size of x (Assuming Q is G 2 integral WLOG). 2. Let x P and x = λ x where x belongs to slice of P. x Qx + c x + d = λ 2 x Q x + λc x + d

39 39 Case 2: x Qx > 0 Question: Can x Qx be arbitrarily close to zero? 1. No. In fact x Qx 1 where G 2 size of x (Assuming Q is G 2 integral WLOG). 2. Let x P and x = λ x where x belongs to slice of P. x Qx + c x + d = λ 2 x Q x + λc x + d λ 2 1 G 2 λ c M + d

40 40 Case 2: x Qx > 0 Question: Can x Qx be arbitrarily close to zero? 1. No. In fact x Qx 1 where G 2 size of x (Assuming Q is G 2 integral WLOG). 2. Let x P and x = λ x where x belongs to slice of P. x Qx + c x + d = λ 2 x Q x + λc x + d λ 2 1 G 2 λ c M + d > 0 for sufficiently large λ

41 Case 2: x Qx > 0 Question: Can x Qx be arbitrarily close to zero? 1. No. In fact x Qx 1 where G 2 size of x (Assuming Q is G 2 integral WLOG). 2. Let x P and x = λ x where x belongs to slice of P. x Qx + c x + d = λ 2 x Q x + λc x + d λ 2 1 G 2 λ c M + d > 0 for sufficiently large λ Bound size of all potential solution If x > x c (R) 3 and x P, then x Qx + c x + d > 0 (R is the size of the largest extreme ray of P.).

42 42 Case 3: x Qx = 0 x* such that x* T Qx* = 0 P Carefully selected Hyperplane

43 43 Looking around x... (x Qx + c x + d 0) c T x < 0 x T Qx > 0 c T x 0 c T x= 0

44 Looking around x... (x Qx + c x + d 0) c T x 0 c T x < 0 x T Qx > 0 c T x= 0 1. Not easy to bound size of feasible solution. 2. Not easy to find feasible solution of small size.

45 45 So far we have If 0 = min x T Qx s.t. x slice of P then we need to tread more cautiously...

46 46 So far we have If 0 = min x T Qx s.t. x slice of P then we need to tread more cautiously... Lets see one possible approach.

47 47 Decomposing P further Lemma Let P be a full-dimensional simplicial cone such that x Qx 0 for every x C. Let H be a hyperplane such that P H is a simplex. Then there exist a finite family of full-dimensional simple cones C i, i I such that (a) i I Ci = P, (b) for every i I, if a face F of C i satisfies min{x Hx : x F H} = 0, then there exists an extreme ray v of F with v Hv = 0, (c) for every i I, the size of C i is polynomial in the size of P.

48 48 Illustration of Lemma

49 49 Illustration of Lemma

50 50 Working with one of these cones C 1. C has n extreme rays r 1,..., r n.

51 51 Working with one of these cones C 1. C has n extreme rays r 1,..., r n. 2. Let J {1,..., n} such that (r j ) Qr j = 0 for all j J.

52 52 Working with one of these cones C 1. C has n extreme rays r 1,..., r n. 2. Let J {1,..., n} such that (r j ) Qr j = 0 for all j J. 3. x C Z n, then x = x }{{} 0 + polysize integer point n r j y j, y j Z + j {1,..., n}. j=1

53 53 Working with one of these cones C 1. C has n extreme rays r 1,..., r n. 2. Let J {1,..., n} such that (r j ) Qr j = 0 for all j J. 3. x C Z n, then x = x }{{} 0 + polysize integer point n r j y j, y j Z + j {1,..., n}. j=1 Now for each small x 0 :

54 54 Working with one of these cones C 1. C has n extreme rays r 1,..., r n. 2. Let J {1,..., n} such that (r j ) Qr j = 0 for all j J. 3. x C Z n, then x = x }{{} 0 + polysize integer point n r j y j, y j Z + j {1,..., n}. j=1 Now for each small x 0 : 4. Using (2) and (3) above:

55 55 Working with one of these cones C 1. C has n extreme rays r 1,..., r n. 2. Let J {1,..., n} such that (r j ) Qr j = 0 for all j J. 3. x C Z n, then x = x }{{} 0 + polysize integer point n r j y j, y j Z + j {1,..., n}. j=1 Now for each small x 0 : 4. Using (2) and (3) above: For each j J either 4.1 We can directly construct a small size solution or [y j > 0 in this solution] 4.2 If a solution exists, then there exist another solution with y j = 0.

56 56 Working with one of these cones C 1. C has n extreme rays r 1,..., r n. 2. Let J {1,..., n} such that (r j ) Qr j = 0 for all j J. 3. x C Z n, then x = x }{{} 0 + polysize integer point n r j y j, y j Z + j {1,..., n}. j=1 Now for each small x 0 : 4. Using (2) and (3) above: For each j J either 4.1 We can directly construct a small size solution or [y j > 0 in this solution] 4.2 If a solution exists, then there exist another solution with y j = If we are not in case (4.1) above for all j J then: If x C Z and x Qx + c x + d 0, then there exists x C Z such that x = x 0 + n r j y j, y j Z + j {1,..., n} \ J. j=1

57 57 Working with one of these cones C 1. C has n extreme rays r 1,..., r n. 2. Let J {1,..., n} such that (r j ) Qr j = 0 for all j J. 3. x C Z n, then x = x }{{} 0 + polysize integer point n r j y j, y j Z + j {1,..., n}. j=1 Now for each small x 0 : 4. Using (2) and (3) above: For each j J either 4.1 We can directly construct a small size solution or [y j > 0 in this solution] 4.2 If a solution exists, then there exist another solution with y j = If we are not in case (4.1) above for all j J then: If x C Z and x Qx + c x + d 0, then there exists x C Z such that x = x 0 + n r j y j, y j Z + j {1,..., n} \ J. j=1 6. Finally, using the structure of the cone, (5) implies all the solutions are bounded, where the bound is polynomial size.

58 58 Open Problem Is Programming in P for fixed dimension?

59 59 Thank You!

Minimizing Cubic and Homogeneous Polynomials Over Integer Points in the Plane

Minimizing Cubic and Homogeneous Polynomials Over Integer Points in the Plane Robert Hildebrand IFOR - ETH Zürich Joint work with Alberto Del Pia, Robert Weismantel and Kevin Zemmer Robert Hildebrand (ETH)