New Hardness Results for Diophantine Approximation Friedrich - PowerPoint PPT Presentation

New Hardness Results for Diophantine Approximation Friedrich Eisenbrand & Thomas Rothvoß Institute of Mathematics EPFL, Lausanne APPROX’09

Simultaneous Diophantine Approximation (SDA) Given: ◮ α 1 , . . . , α n ∈ Q ◮ bound N ∈ N ◮ error bound ε > 0 Decide: � � � α i − Z � ≤ ε � � ∃ Q ∈ { 1 , . . . , N } : max � � Q Q i =1 ,...,n

Simultaneous Diophantine Approximation (SDA) Given: ◮ α 1 , . . . , α n ∈ Q ◮ bound N ∈ N ◮ error bound ε > 0 Decide: � � � α i − Z � ≤ ε � � ∃ Q ∈ { 1 , . . . , N } : max � � Q Q i =1 ,...,n ⇔ ∃ Q ∈ { 1 , . . . , N } : max i =1 ,...,n |⌈ Qα i ⌋ − Qα i | ≤ ε

Simultaneous Diophantine Approximation (SDA) Given: ◮ α 1 , . . . , α n ∈ Q ◮ bound N ∈ N ◮ error bound ε > 0 Decide: � � � α i − Z � ≤ ε � � ∃ Q ∈ { 1 , . . . , N } : max � � Q Q i =1 ,...,n ⇔ ∃ Q ∈ { 1 , . . . , N } : max i =1 ,...,n |⌈ Qα i ⌋ − Qα i | ≤ ε ◮ Yes, if N ≥ (1 /ε ) n [Dirichlet]

Simultaneous Diophantine Approximation (SDA) Given: ◮ α 1 , . . . , α n ∈ Q ◮ bound N ∈ N ◮ error bound ε > 0 Decide: � � � α i − Z � ≤ ε � � ∃ Q ∈ { 1 , . . . , N } : max � � Q Q i =1 ,...,n ⇔ ∃ Q ∈ { 1 , . . . , N } : max i =1 ,...,n |⌈ Qα i ⌋ − Qα i | ≤ ε ◮ Yes, if N ≥ (1 /ε ) n [Dirichlet] ◮ NP -hard [Lagarias ’85]

Simultaneous Diophantine Approximation (SDA) Given: ◮ α 1 , . . . , α n ∈ Q ◮ bound N ∈ N ◮ error bound ε > 0 Decide: � � � α i − Z � ≤ ε � � ∃ Q ∈ { 1 , . . . , N } : max � � Q Q i =1 ,...,n ⇔ ∃ Q ∈ { 1 , . . . , N } : max i =1 ,...,n |⌈ Qα i ⌋ − Qα i | ≤ ε ◮ Yes, if N ≥ (1 /ε ) n [Dirichlet] ◮ NP -hard [Lagarias ’85] ◮ 2 O ( n ) -approximation via LLL-algo [Lagarias ’85]

Simultaneous Diophantine Approximation (SDA) Given: ◮ α 1 , . . . , α n ∈ Q ◮ bound N ∈ N ◮ error bound ε > 0 Decide: � � � α i − Z � ≤ ε � � ∃ Q ∈ { 1 , . . . , N } : max � � Q Q i =1 ,...,n ⇔ ∃ Q ∈ { 1 , . . . , N } : max i =1 ,...,n |⌈ Qα i ⌋ − Qα i | ≤ ε ◮ Yes, if N ≥ (1 /ε ) n [Dirichlet] ◮ NP -hard [Lagarias ’85] ◮ 2 O ( n ) -approximation via LLL-algo [Lagarias ’85] ◮ Gap version NP -hard [R¨ ossner & Seifert ’96, Chen & Meng ’07]

Inapproximability Theorem (R¨ ossner & Seifert ’96, Chen & Meng ’07) Given α 1 , . . . , α n , N , ε > 0 it is NP -hard to distinguish ◮ ∃ Q ∈ { 1 , . . . , N } : max i =1 ,...,n |⌈ Qα i ⌋ − Qα i | ≤ ε O (1) O (1) ◮ ∄ Q ∈ { 1 , . . . , n log log n N } : max log log n ε i =1 ,...,n |⌈ Qα i ⌋ − Qα i | ≤ n

Inapproximability Theorem (R¨ ossner & Seifert ’96, Chen & Meng ’07) Given α 1 , . . . , α n , N , ε > 0 it is NP -hard to distinguish ◮ ∃ Q ∈ { 1 , . . . , N } : max i =1 ,...,n |⌈ Qα i ⌋ − Qα i | ≤ ε O (1) O (1) ◮ ∄ Q ∈ { 1 , . . . , n log log n N } : max log log n ε i =1 ,...,n |⌈ Qα i ⌋ − Qα i | ≤ n Theorem Given α 1 , . . . , α n , N , ε > 0 it is NP -hard to distinguish ◮ ∃ Q ∈ { N/ 2 , . . . , N } : max i =1 ,...,n |⌈ Qα i ⌋ − Qα i | ≤ ε O (1) ◮ ∄ Q ∈ { 1 , . . . , 2 n · N } : max log log n · ε i =1 ,...,n |⌈ Qα i ⌋ − Qα i | ≤ n even if ε small .

Reduction SDA → DDA Theorem (Directed Diophantine Approximation (DDA)) Given α 1 , . . . , α n , N , ε > 0 it is NP -hard to distinguish ◮ ∃ Q ∈ { N/ 2 , . . . , N } : max i =1 ,...,n |⌈ Qα i ⌉ − Qα i | ≤ ε O (1) i =1 ,...,n |⌈ Qα i ⌉ − Qα i | ≤ 2 n · ε ◮ ∄ Q ∈ { 1 , . . . , n log log n · N } : max even if ε small.

Reduction SDA → DDA Theorem (Directed Diophantine Approximation (DDA)) Given α 1 , . . . , α n , N , ε > 0 it is NP -hard to distinguish ◮ ∃ Q ∈ { N/ 2 , . . . , N } : max i =1 ,...,n |⌈ Qα i ⌉ − Qα i | ≤ ε O (1) i =1 ,...,n |⌈ Qα i ⌉ − Qα i | ≤ 2 n · ε ◮ ∄ Q ∈ { 1 , . . . , n log log n · N } : max even if ε small. ◮ Given SDA instance α 1 , . . . , α n , ε, N , choose δ := 2 ε N α ′ := α i − δ ∀ i = 1 , . . . , n i α ′ := − ( α i + δ ) ∀ i = 1 , . . . , n i + n ε ′ := 3 ε

Reduction SDA → DDA Theorem (Directed Diophantine Approximation (DDA)) Given α 1 , . . . , α n , N , ε > 0 it is NP -hard to distinguish ◮ ∃ Q ∈ { N/ 2 , . . . , N } : max i =1 ,...,n |⌈ Qα i ⌉ − Qα i | ≤ ε O (1) i =1 ,...,n |⌈ Qα i ⌉ − Qα i | ≤ 2 n · ε ◮ ∄ Q ∈ { 1 , . . . , n log log n · N } : max even if ε small. ◮ Given SDA instance α 1 , . . . , α n , ε, N , choose δ := 2 ε N α ′ := α i − δ ∀ i = 1 , . . . , n i α ′ := − ( α i + δ ) ∀ i = 1 , . . . , n i + n ε ′ := 3 ε ◮ Note that |⌈− x ⌉ − ( − x ) | = |⌊ x ⌋ − x |

Reduction SDA → DDA Theorem (Directed Diophantine Approximation (DDA)) Given α 1 , . . . , α n , N , ε > 0 it is NP -hard to distinguish ◮ ∃ Q ∈ { N/ 2 , . . . , N } : max i =1 ,...,n |⌈ Qα i ⌉ − Qα i | ≤ ε O (1) i =1 ,...,n |⌈ Qα i ⌉ − Qα i | ≤ 2 n · ε ◮ ∄ Q ∈ { 1 , . . . , n log log n · N } : max even if ε small. ◮ Given SDA instance α 1 , . . . , α n , ε, N , choose δ := 2 ε N α ′ := α i − δ ∀ i = 1 , . . . , n i α ′ := − ( α i + δ ) ∀ i = 1 , . . . , n i + n ε ′ := 3 ε ◮ Note that |⌈− x ⌉ − ( − x ) | = |⌊ x ⌋ − x | ◮ Assume ε < 1 6 · ( 1 2 ) n

Proof: SDA-YES ⇒ DDA-YES ◮ Let Q ∈ { N/ 2 , . . . , N } : max i =1 ,...,n |⌈ Qα i ⌋ − Qα i | ≤ ε . Z Qα i ≤ ε

Proof: SDA-YES ⇒ DDA-YES ◮ Let Q ∈ { N/ 2 , . . . , N } : max i =1 ,...,n |⌈ Qα i ⌋ − Qα i | ≤ ε . Q · ( α i − δ ) Z Qα i Q · ( α i + δ ) ≤ ε Qδ ∈ [ ε, 2 ε ] Qδ ∈ [ ε, 2 ε ]

Proof: SDA-YES ⇒ DDA-YES ◮ Let Q ∈ { N/ 2 , . . . , N } : max i =1 ,...,n |⌈ Qα i ⌋ − Qα i | ≤ ε . round down round up Q · ( α i − δ ) Z Qα i Q · ( α i + δ ) ≤ ε Qδ ∈ [ ε, 2 ε ] Qδ ∈ [ ε, 2 ε ]

Proof: SDA-YES ⇒ DDA-YES ◮ Let Q ∈ { N/ 2 , . . . , N } : max i =1 ,...,n |⌈ Qα i ⌋ − Qα i | ≤ ε . round down round up Q · ( α i − δ ) Z Qα i Q · ( α i + δ ) ≤ ε Qδ ∈ [ ε, 2 ε ] Qδ ∈ [ ε, 2 ε ] ◮ Conclusion: Q is DDA solution � |⌈ Q ( α i − δ ) ⌉ − Q ( α i − δ ) | , � j =1 ,..., 2 n : |⌈ Qα ′ j ⌉− Qα ′ max j | = max ≤ 3 ε | Q ( α i + δ ) − ⌊ Q ( α i + δ ) ⌋| i =1 ,...,n

Proof: SDA-NO ⇒ DDA-NO ◮ Show: ¬ DDA-NO ⇒ ¬ SDA-NO

Proof: SDA-NO ⇒ DDA-NO ◮ Show: ¬ DDA-NO ⇒ ¬ SDA-NO 1 ◮ Let Q ≤ n O ( log log n ) N with j ) ≤ 2 n · 3 ε < 1 / 2 j =1 ,..., 2 n ( ⌈ Qα ′ j ⌉ − Qα ′ max

Proof: SDA-NO ⇒ DDA-NO ◮ Show: ¬ DDA-NO ⇒ ¬ SDA-NO 1 ◮ Let Q ≤ n O ( log log n ) N with j ) ≤ 2 n · 3 ε < 1 / 2 j =1 ,..., 2 n ( ⌈ Qα ′ j ⌉ − Qα ′ max ◮ Suppose ∄ integer in [ Q ( α i − δ ) , Q ( α i + δ )] ∈ Z ∈ Z Q · ( α i − δ ) Q · ( α i + δ )

Proof: SDA-NO ⇒ DDA-NO ◮ Show: ¬ DDA-NO ⇒ ¬ SDA-NO 1 ◮ Let Q ≤ n O ( log log n ) N with j ) ≤ 2 n · 3 ε < 1 / 2 j =1 ,..., 2 n ( ⌈ Qα ′ j ⌉ − Qα ′ max ◮ Suppose ∄ integer in [ Q ( α i − δ ) , Q ( α i + δ )] round up round down ∈ Z ∈ Z Q · ( α i − δ ) Q · ( α i + δ )

Proof: SDA-NO ⇒ DDA-NO ◮ Show: ¬ DDA-NO ⇒ ¬ SDA-NO 1 ◮ Let Q ≤ n O ( log log n ) N with j ) ≤ 2 n · 3 ε < 1 / 2 j =1 ,..., 2 n ( ⌈ Qα ′ j ⌉ − Qα ′ max ◮ Suppose ∄ integer in [ Q ( α i − δ ) , Q ( α i + δ )] round up round down ∈ Z ∈ Z Q · ( α i − δ ) Q · ( α i + δ ) j =1 ,..., 2 n |⌈ Qα ′ j ⌉ − Qα ′ max j | ≥ 1 / 2 Contradiction! ◮

Proof: SDA-NO ⇒ DDA-NO ◮ Show: ¬ DDA-NO ⇒ ¬ SDA-NO 1 ◮ Let Q ≤ n O ( log log n ) N with j ) ≤ 2 n · 3 ε < 1 / 2 j =1 ,..., 2 n ( ⌈ Qα ′ j ⌉ − Qα ′ max ◮ Suppose ∄ integer in [ Q ( α i − δ ) , Q ( α i + δ )] ∈ Z Q · ( α i − δ ) Q · ( α i + δ ) j =1 ,..., 2 n |⌈ Qα ′ j ⌉ − Qα ′ max j | ≥ 1 / 2 Contradiction! ◮

Proof: SDA-NO ⇒ DDA-NO ◮ Show: ¬ DDA-NO ⇒ ¬ SDA-NO 1 ◮ Let Q ≤ n O ( log log n ) N with j ) ≤ 2 n · 3 ε < 1 / 2 j =1 ,..., 2 n ( ⌈ Qα ′ j ⌉ − Qα ′ max ◮ Suppose ∄ integer in [ Q ( α i − δ ) , Q ( α i + δ )] ∈ Z Q · ( α i − δ ) Qα i Q · ( α i + δ ) ≤ Qδ j =1 ,..., 2 n |⌈ Qα ′ j ⌉ − Qα ′ max j | ≥ 1 / 2 Contradiction! ◮ log log n ) · ε hence ¬ SDA-NO 1 ◮ | Qα i − Z | ≤ Qδ ≤ n O (

Mixing Set min c s s + c T y s + a i y i ≥ b i ∀ i = 1 , . . . , n R ≥ 0 s ∈ Z n y ∈

Mixing Set min c s s + c T y s + a i y i ≥ b i ∀ i = 1 , . . . , n R ≥ 0 s ∈ Z n y ∈ ◮ Poly-time if a i = 1 [G¨ unl¨ uk & Pochet ’01, Miller & Wolsey ’03]

Mixing Set min c s s + c T y s + a i y i ≥ b i ∀ i = 1 , . . . , n R ≥ 0 s ∈ Z n y ∈ ◮ Poly-time if a i = 1 [G¨ unl¨ uk & Pochet ’01, Miller & Wolsey ’03] ◮ Poly-time if a 1 | a 2 | . . . | a n [Zhao & de Farias ’08, Conforti et al. ’08]