New Hardness Results for Diophantine Approximation Friedrich - - PowerPoint PPT Presentation

▶
new hardness results for diophantine approximation
SMART_READER_LITE
LIVE PREVIEW

New Hardness Results for Diophantine Approximation Friedrich - - PowerPoint PPT Presentation

Jan 07, 2024 275 likes •876 views

New Hardness Results for Diophantine Approximation Friedrich Eisenbrand & Thomas Rothvo Institute of Mathematics EPFL, Lausanne APPROX09 Simultaneous Diophantine Approximation (SDA) Given: 1 , . . . , n Q bound N

slide-1
SLIDE 1

New Hardness Results for Diophantine Approximation

Friedrich Eisenbrand & Thomas Rothvoß

Institute of Mathematics EPFL, Lausanne

APPROX’09

slide-2
SLIDE 2

Simultaneous Diophantine Approximation (SDA)

Given:

◮ α1, . . . , αn ∈ Q ◮ bound N ∈ N ◮ error bound ε > 0

Decide: ∃Q ∈ {1, . . . , N} : max

i=1,...,n

  • αi − Z

Q

  • ≤ ε

Q

slide-3
SLIDE 3

Simultaneous Diophantine Approximation (SDA)

Given:

◮ α1, . . . , αn ∈ Q ◮ bound N ∈ N ◮ error bound ε > 0

Decide: ∃Q ∈ {1, . . . , N} : max

i=1,...,n

  • αi − Z

Q

  • ≤ ε

Q ⇔ ∃Q ∈ {1, . . . , N} : max

i=1,...,n |⌈Qαi⌋ − Qαi| ≤ ε

slide-4
SLIDE 4

Simultaneous Diophantine Approximation (SDA)

Given:

◮ α1, . . . , αn ∈ Q ◮ bound N ∈ N ◮ error bound ε > 0

Decide: ∃Q ∈ {1, . . . , N} : max

i=1,...,n

  • αi − Z

Q

  • ≤ ε

Q ⇔ ∃Q ∈ {1, . . . , N} : max

i=1,...,n |⌈Qαi⌋ − Qαi| ≤ ε ◮ Yes, if N ≥ (1/ε)n [Dirichlet]

slide-5
SLIDE 5

Simultaneous Diophantine Approximation (SDA)

Given:

◮ α1, . . . , αn ∈ Q ◮ bound N ∈ N ◮ error bound ε > 0

Decide: ∃Q ∈ {1, . . . , N} : max

i=1,...,n

  • αi − Z

Q

  • ≤ ε

Q ⇔ ∃Q ∈ {1, . . . , N} : max

i=1,...,n |⌈Qαi⌋ − Qαi| ≤ ε ◮ Yes, if N ≥ (1/ε)n [Dirichlet] ◮ NP-hard [Lagarias ’85]

slide-6
SLIDE 6

Simultaneous Diophantine Approximation (SDA)

Given:

◮ α1, . . . , αn ∈ Q ◮ bound N ∈ N ◮ error bound ε > 0

Decide: ∃Q ∈ {1, . . . , N} : max

i=1,...,n

  • αi − Z

Q

  • ≤ ε

Q ⇔ ∃Q ∈ {1, . . . , N} : max

i=1,...,n |⌈Qαi⌋ − Qαi| ≤ ε ◮ Yes, if N ≥ (1/ε)n [Dirichlet] ◮ NP-hard [Lagarias ’85] ◮ 2O(n)-approximation via LLL-algo [Lagarias ’85]

slide-7
SLIDE 7

Simultaneous Diophantine Approximation (SDA)

Given:

◮ α1, . . . , αn ∈ Q ◮ bound N ∈ N ◮ error bound ε > 0

Decide: ∃Q ∈ {1, . . . , N} : max

i=1,...,n

  • αi − Z

Q

  • ≤ ε

Q ⇔ ∃Q ∈ {1, . . . , N} : max

i=1,...,n |⌈Qαi⌋ − Qαi| ≤ ε ◮ Yes, if N ≥ (1/ε)n [Dirichlet] ◮ NP-hard [Lagarias ’85] ◮ 2O(n)-approximation via LLL-algo [Lagarias ’85] ◮ Gap version NP-hard [R¨

  • ssner & Seifert ’96,

Chen & Meng ’07]

slide-8
SLIDE 8

Inapproximability

Theorem (R¨

  • ssner & Seifert ’96, Chen & Meng ’07)

Given α1, . . . , αn, N, ε > 0 it is NP-hard to distinguish

◮ ∃Q ∈ {1, . . . , N} : max i=1,...,n |⌈Qαi⌋ − Qαi| ≤ ε ◮ ∄Q ∈ {1, . . . , n

O(1) log log n N} : max

i=1,...,n |⌈Qαi⌋ − Qαi| ≤ n

O(1) log log n ε

slide-9
SLIDE 9

Inapproximability

Theorem (R¨

  • ssner & Seifert ’96, Chen & Meng ’07)

Given α1, . . . , αn, N, ε > 0 it is NP-hard to distinguish

◮ ∃Q ∈ {1, . . . , N} : max i=1,...,n |⌈Qαi⌋ − Qαi| ≤ ε ◮ ∄Q ∈ {1, . . . , n

O(1) log log n N} : max

i=1,...,n |⌈Qαi⌋ − Qαi| ≤ n

O(1) log log n ε

Theorem

Given α1, . . . , αn, N, ε > 0 it is NP-hard to distinguish

◮ ∃Q ∈ {N/2, . . . , N} : max i=1,...,n |⌈Qαi⌋ − Qαi| ≤ ε ◮ ∄Q ∈ {1, . . . , 2n · N} : max i=1,...,n |⌈Qαi⌋ − Qαi| ≤ n

O(1) log log n · ε

even if ε small.

slide-10
SLIDE 10

Reduction SDA→ DDA

Theorem (Directed Diophantine Approximation (DDA))

Given α1, . . . , αn, N, ε > 0 it is NP-hard to distinguish

◮ ∃Q ∈ {N/2, . . . , N} : max i=1,...,n |⌈Qαi⌉ − Qαi| ≤ ε ◮ ∄Q ∈ {1, . . . , n

O(1) log log n · N} : max

i=1,...,n |⌈Qαi⌉ − Qαi| ≤ 2n · ε

even if ε small.

slide-11
SLIDE 11

Reduction SDA→ DDA

Theorem (Directed Diophantine Approximation (DDA))

Given α1, . . . , αn, N, ε > 0 it is NP-hard to distinguish

◮ ∃Q ∈ {N/2, . . . , N} : max i=1,...,n |⌈Qαi⌉ − Qαi| ≤ ε ◮ ∄Q ∈ {1, . . . , n

O(1) log log n · N} : max

i=1,...,n |⌈Qαi⌉ − Qαi| ≤ 2n · ε

even if ε small.

◮ Given SDA instance α1, . . . , αn, ε, N, choose δ := 2ε N

α′

i

:= αi − δ ∀i = 1, . . . , n α′

i+n

:= −(αi + δ) ∀i = 1, . . . , n ε′ := 3ε

slide-12
SLIDE 12

Reduction SDA→ DDA

Theorem (Directed Diophantine Approximation (DDA))

Given α1, . . . , αn, N, ε > 0 it is NP-hard to distinguish

◮ ∃Q ∈ {N/2, . . . , N} : max i=1,...,n |⌈Qαi⌉ − Qαi| ≤ ε ◮ ∄Q ∈ {1, . . . , n

O(1) log log n · N} : max

i=1,...,n |⌈Qαi⌉ − Qαi| ≤ 2n · ε

even if ε small.

◮ Given SDA instance α1, . . . , αn, ε, N, choose δ := 2ε N

α′

i

:= αi − δ ∀i = 1, . . . , n α′

i+n

:= −(αi + δ) ∀i = 1, . . . , n ε′ := 3ε

◮ Note that |⌈−x⌉ − (−x)| = |⌊x⌋ − x|

slide-13
SLIDE 13

Reduction SDA→ DDA

Theorem (Directed Diophantine Approximation (DDA))

Given α1, . . . , αn, N, ε > 0 it is NP-hard to distinguish

◮ ∃Q ∈ {N/2, . . . , N} : max i=1,...,n |⌈Qαi⌉ − Qαi| ≤ ε ◮ ∄Q ∈ {1, . . . , n

O(1) log log n · N} : max

i=1,...,n |⌈Qαi⌉ − Qαi| ≤ 2n · ε

even if ε small.

◮ Given SDA instance α1, . . . , αn, ε, N, choose δ := 2ε N

α′

i

:= αi − δ ∀i = 1, . . . , n α′

i+n

:= −(αi + δ) ∀i = 1, . . . , n ε′ := 3ε

◮ Note that |⌈−x⌉ − (−x)| = |⌊x⌋ − x| ◮ Assume ε < 1 6 · (1 2)n

slide-14
SLIDE 14

Proof: SDA-YES ⇒ DDA-YES

◮ Let Q ∈ {N/2, . . . , N} : max i=1,...,n |⌈Qαi⌋ − Qαi| ≤ ε.

Qαi Z ≤ ε

slide-15
SLIDE 15

Proof: SDA-YES ⇒ DDA-YES

◮ Let Q ∈ {N/2, . . . , N} : max i=1,...,n |⌈Qαi⌋ − Qαi| ≤ ε.

Qαi Z ≤ ε Q · (αi − δ) Q · (αi + δ) Qδ ∈ [ε, 2ε] Qδ ∈ [ε, 2ε]

slide-16
SLIDE 16

Proof: SDA-YES ⇒ DDA-YES

◮ Let Q ∈ {N/2, . . . , N} : max i=1,...,n |⌈Qαi⌋ − Qαi| ≤ ε.

Qαi Z ≤ ε Q · (αi − δ) Q · (αi + δ) Qδ ∈ [ε, 2ε] Qδ ∈ [ε, 2ε] round up round down

slide-17
SLIDE 17

Proof: SDA-YES ⇒ DDA-YES

◮ Let Q ∈ {N/2, . . . , N} : max i=1,...,n |⌈Qαi⌋ − Qαi| ≤ ε.

Qαi Z ≤ ε Q · (αi − δ) Q · (αi + δ) Qδ ∈ [ε, 2ε] Qδ ∈ [ε, 2ε] round up round down

◮ Conclusion: Q is DDA solution

max

j=1,...,2n : |⌈Qα′ j⌉−Qα′ j| = max i=1,...,n

|⌈Q(αi − δ)⌉ − Q(αi − δ)|, |Q(αi + δ) − ⌊Q(αi + δ)⌋|

  • ≤ 3ε
slide-18
SLIDE 18

Proof: SDA-NO ⇒ DDA-NO

◮ Show: ¬DDA-NO ⇒ ¬SDA-NO

slide-19
SLIDE 19

Proof: SDA-NO ⇒ DDA-NO

◮ Show: ¬DDA-NO ⇒ ¬SDA-NO ◮ Let Q ≤ nO(

1 log log n)N with

max

j=1,...,2n(⌈Qα′ j⌉ − Qα′ j) ≤ 2n · 3ε < 1/2

slide-20
SLIDE 20

Proof: SDA-NO ⇒ DDA-NO

◮ Show: ¬DDA-NO ⇒ ¬SDA-NO ◮ Let Q ≤ nO(

1 log log n)N with

max

j=1,...,2n(⌈Qα′ j⌉ − Qα′ j) ≤ 2n · 3ε < 1/2 ◮ Suppose ∄ integer in [Q(αi − δ), Q(αi + δ)]

Q · (αi − δ) Q · (αi + δ) ∈ Z ∈ Z

slide-21
SLIDE 21

Proof: SDA-NO ⇒ DDA-NO

◮ Show: ¬DDA-NO ⇒ ¬SDA-NO ◮ Let Q ≤ nO(

1 log log n)N with

max

j=1,...,2n(⌈Qα′ j⌉ − Qα′ j) ≤ 2n · 3ε < 1/2 ◮ Suppose ∄ integer in [Q(αi − δ), Q(αi + δ)]

Q · (αi − δ) Q · (αi + δ) ∈ Z ∈ Z round up round down

slide-22
SLIDE 22

Proof: SDA-NO ⇒ DDA-NO

◮ Show: ¬DDA-NO ⇒ ¬SDA-NO ◮ Let Q ≤ nO(

1 log log n)N with

max

j=1,...,2n(⌈Qα′ j⌉ − Qα′ j) ≤ 2n · 3ε < 1/2 ◮ Suppose ∄ integer in [Q(αi − δ), Q(αi + δ)]

Q · (αi − δ) Q · (αi + δ) ∈ Z ∈ Z round up round down

◮

max

j=1,...,2n |⌈Qα′ j⌉ − Qα′ j| ≥ 1/2

Contradiction!

slide-23
SLIDE 23

Proof: SDA-NO ⇒ DDA-NO

◮ Show: ¬DDA-NO ⇒ ¬SDA-NO ◮ Let Q ≤ nO(

1 log log n)N with

max

j=1,...,2n(⌈Qα′ j⌉ − Qα′ j) ≤ 2n · 3ε < 1/2 ◮ Suppose ∄ integer in [Q(αi − δ), Q(αi + δ)]

Q · (αi − δ) Q · (αi + δ) ∈ Z

◮

max

j=1,...,2n |⌈Qα′ j⌉ − Qα′ j| ≥ 1/2

Contradiction!

slide-24
SLIDE 24

Proof: SDA-NO ⇒ DDA-NO

◮ Show: ¬DDA-NO ⇒ ¬SDA-NO ◮ Let Q ≤ nO(

1 log log n)N with

max

j=1,...,2n(⌈Qα′ j⌉ − Qα′ j) ≤ 2n · 3ε < 1/2 ◮ Suppose ∄ integer in [Q(αi − δ), Q(αi + δ)]

Q · (αi − δ) Q · (αi + δ) ∈ Z Qαi ≤ Qδ

◮

max

j=1,...,2n |⌈Qα′ j⌉ − Qα′ j| ≥ 1/2

Contradiction!

◮ |Qαi − Z| ≤ Qδ ≤ nO(

1 log log n) · ε hence ¬SDA-NO

slide-25
SLIDE 25

Mixing Set

min css + cT y s + aiyi ≥ bi ∀i = 1, . . . , n s ∈ R≥0 y ∈ Zn

slide-26
SLIDE 26

Mixing Set

min css + cT y s + aiyi ≥ bi ∀i = 1, . . . , n s ∈ R≥0 y ∈ Zn

◮ Poly-time if ai = 1

[G¨ unl¨ uk & Pochet ’01, Miller & Wolsey ’03]

slide-27
SLIDE 27

Mixing Set

min css + cT y s + aiyi ≥ bi ∀i = 1, . . . , n s ∈ R≥0 y ∈ Zn

◮ Poly-time if ai = 1

[G¨ unl¨ uk & Pochet ’01, Miller & Wolsey ’03]

◮ Poly-time if a1 | a2 | . . . | an

[Zhao & de Farias ’08, Conforti et al. ’08]

slide-28
SLIDE 28

Mixing Set

min css + cT y s + aiyi ≥ bi ∀i = 1, . . . , n s ∈ R≥0 y ∈ Zn

◮ Poly-time if ai = 1

[G¨ unl¨ uk & Pochet ’01, Miller & Wolsey ’03]

◮ Poly-time if a1 | a2 | . . . | an

[Zhao & de Farias ’08, Conforti et al. ’08]

◮ Conforti et al. ’08: Poly-time?

slide-29
SLIDE 29

Mixing Set

min css + cT y s + aiyi ≥ bi ∀i = 1, . . . , n s ∈ R≥0 y ∈ Zn

◮ Poly-time if ai = 1

[G¨ unl¨ uk & Pochet ’01, Miller & Wolsey ’03]

◮ Poly-time if a1 | a2 | . . . | an

[Zhao & de Farias ’08, Conforti et al. ’08]

◮ Conforti et al. ’08: Poly-time? ◮ → NP-hard by reduction from:

∃Q ∈ {1, . . . , N} :

n

  • i=1

|Qαi − ⌊Qαi⌋| ≤ ε

slide-30
SLIDE 30

Proof

◮ ∃Q ∈ {1, . . . , N} : n

  • i=1

|Qαi − ⌊Qαi⌋| ≤ ε

slide-31
SLIDE 31

Proof

◮ ∃Q ∈ {1, . . . , N} : n

  • i=1

|Qαi − ⌊Qαi⌋| ≤ ε min

n

  • i=1

(Qαi − yi) (IP) Qαi − yi ≥ ∀i = 1, . . . , n Q ≥ 1 Q ≤ N Q ∈ Z y1, . . . , yn ∈ Z

slide-32
SLIDE 32

Proof

◮ ∃Q ∈ {1, . . . , N} : n

  • i=1

|Qαi − ⌊Qαi⌋| ≤ ε min

n

  • i=1

(Qαi − yi) (IP) Q − yi/αi ≥ ∀i = 1, . . . , n Q ≥ 1 Q ≤ N Q ∈ Z y1, . . . , yn ∈ Z

slide-33
SLIDE 33

Proof

◮ ∃Q ∈ {1, . . . , N} : n

  • i=1

|Qαi − ⌊Qαi⌋| ≤ ε min

n

  • i=1

(Qαi − yi) (IP) Q − yi/αi ≥ ∀i = 1, . . . , n Q − 0 · y0 ≥ 1 Q ≤ N Q ∈ Z y1, . . . , yn ∈ Z

slide-34
SLIDE 34

Proof

◮ ∃Q ∈ {1, . . . , N} : n

  • i=1

|Qαi − ⌊Qαi⌋| ≤ ε min

n

  • i=1

(Qαi − yi) (IP) Q − yi/αi ≥ ∀i = 1, . . . , n Q − 0 · y0 ≥ 1 Q ≤ N Q ∈ Z y1, . . . , yn ∈ Z

Theorem

Polytime optimization ⇔ Polytime optimization over any non-empty face

slide-35
SLIDE 35

Proof

Theorem

Polytime optimization ⇔ Polytime optimization over any non-empty face

slide-36
SLIDE 36

Proof

Q ≥ yn+1

Theorem

Polytime optimization ⇔ Polytime optimization over any non-empty face

slide-37
SLIDE 37

Proof

Q ≥ yn+1

b

c

Theorem

Polytime optimization ⇔ Polytime optimization over any non-empty face

slide-38
SLIDE 38

Proof

Q ≥ yn+1

b

c M× normal vector

Theorem

Polytime optimization ⇔ Polytime optimization over any non-empty face

slide-39
SLIDE 39

Proof

Q ≥ yn+1

b

c M× normal vector c

Theorem

Polytime optimization ⇔ Polytime optimization over any non-empty face

slide-40
SLIDE 40

Proof

Q ≥ yn+1

b

c M× normal vector c c′

Theorem

Polytime optimization ⇔ Polytime optimization over any non-empty face

slide-41
SLIDE 41

Proof

◮ ∃Q ∈ {1, . . . , N} : n

  • i=1

|Qαi − ⌊Qαi⌋| ≤ ε min

n

  • i=1

(Qαi − yi) (IP) Q − yi/αi ≥ ∀i = 1, . . . , n Q − 0 · y0 ≥ 1 Q ≤ N Q ∈ Z y1, . . . , yn ∈ Z

slide-42
SLIDE 42

Proof

◮ ∃Q ∈ {1, . . . , N} : n

  • i=1

|Qαi − ⌊Qαi⌋| ≤ ε min

n

  • i=1

(Qαi − yi) (IP) Q − yi/αi ≥ ∀i = 1, . . . , n Q − 0 · y0 ≥ 1 Q ≤ N Q ∈ Z y1, . . . , yn ∈ Z

slide-43
SLIDE 43

Proof

◮ ∃Q ∈ {1, . . . , N} : n

  • i=1

|Qαi − ⌊Qαi⌋| ≤ ε min

n

  • i=1

(Qαi − yi) + ε

N · (Q − N)

(IP) Q − yi/αi ≥ ∀i = 1, . . . , n Q − 0 · y0 ≥ 1 Q ∈ Z y1, . . . , yn ∈ Z

slide-44
SLIDE 44

Proof (2)

◮ ∃Q ∈ {1, . . . , N} : n

  • i=1

|Qαi − ⌊Qαi⌋

=:yi

| ≤ ε

slide-45
SLIDE 45

Proof (2)

◮ ∃Q ∈ {1, . . . , N} : n

  • i=1

|Qαi − ⌊Qαi⌋

=:yi

| ≤ ε (IP) ≤

n

  • i=1

(Qαi − yi) + ε N · (Q − N)

slide-46
SLIDE 46

Proof (2)

◮ ∃Q ∈ {1, . . . , N} : n

  • i=1

|Qαi − ⌊Qαi⌋

=:yi

| ≤ ε (IP) ≤

n

  • i=1

(Qαi − yi)

  • ≤ε

+ ε N · (Q − N)

slide-47
SLIDE 47

Proof (2)

◮ ∃Q ∈ {1, . . . , N} : n

  • i=1

|Qαi − ⌊Qαi⌋

=:yi

| ≤ ε (IP) ≤

n

  • i=1

(Qαi − yi)

  • ≤ε

+ ε N · (Q − N)

  • ≤0
slide-48
SLIDE 48

Proof (2)

◮ ∃Q ∈ {1, . . . , N} : n

  • i=1

|Qαi − ⌊Qαi⌋

=:yi

| ≤ ε (IP) ≤

n

  • i=1

(Qαi − yi)

  • ≤ε

+ ε N · (Q − N)

  • ≤0

≤ ε

slide-49
SLIDE 49

Proof (2)

◮ ∃Q ∈ {1, . . . , N} : n

  • i=1

|Qαi − ⌊Qαi⌋

=:yi

| ≤ ε (IP) ≤

n

  • i=1

(Qαi − yi)

  • ≤ε

+ ε N · (Q − N)

  • ≤0

≤ ε

◮ Let (IP) ≤ ε n

  • i=1
  • ≥0

n

  • i=1

(Qαi − yi) + ε N · (Q − N) ≤ ε

slide-50
SLIDE 50

Proof (2)

◮ ∃Q ∈ {1, . . . , N} : n

  • i=1

|Qαi − ⌊Qαi⌋

=:yi

| ≤ ε (IP) ≤

n

  • i=1

(Qαi − yi)

  • ≤ε

+ ε N · (Q − N)

  • ≤0

≤ ε

◮ Let (IP) ≤ ε n

  • i=1
  • ≥0

n

  • i=1

(Qαi − yi)

  • ≥0

+ ε N · (Q − N) ≤ ε

slide-51
SLIDE 51

Proof (2)

◮ ∃Q ∈ {1, . . . , N} : n

  • i=1

|Qαi − ⌊Qαi⌋

=:yi

| ≤ ε (IP) ≤

n

  • i=1

(Qαi − yi)

  • ≤ε

+ ε N · (Q − N)

  • ≤0

≤ ε

◮ Let (IP) ≤ ε n

  • i=1
  • ≥0

n

  • i=1

(Qαi − yi)

  • ≥0

+ ε N · (Q − N)

  • ≤ε

≤ ε

slide-52
SLIDE 52

Proof (2)

◮ ∃Q ∈ {1, . . . , N} : n

  • i=1

|Qαi − ⌊Qαi⌋

=:yi

| ≤ ε (IP) ≤

n

  • i=1

(Qαi − yi)

  • ≤ε

+ ε N · (Q − N)

  • ≤0

≤ ε

◮ Let (IP) ≤ ε n

  • i=1
  • ≥0

n

  • i=1

(Qαi − yi)

  • ≥0

+ ε N · (Q − N)

  • ≤ε

≤ ε

◮ ε N · (Q − N) ≤ ε ⇒ Q ≤ 2N

slide-53
SLIDE 53

Proof (2)

◮ ∃Q ∈ {1, . . . , N} : n

  • i=1

|Qαi − ⌊Qαi⌋

=:yi

| ≤ ε (IP) ≤

n

  • i=1

(Qαi − yi)

  • ≤ε

+ ε N · (Q − N)

  • ≤0

≤ ε

◮ Let (IP) ≤ ε n

  • i=1
  • ≥0

n

  • i=1

(Qαi − yi) + ε N · (Q − N)

  • ≥−N

≤ ε

◮ ε N · (Q − N) ≤ ε ⇒ Q ≤ 2N

slide-54
SLIDE 54

Proof (2)

◮ ∃Q ∈ {1, . . . , N} : n

  • i=1

|Qαi − ⌊Qαi⌋

=:yi

| ≤ ε (IP) ≤

n

  • i=1

(Qαi − yi)

  • ≤ε

+ ε N · (Q − N)

  • ≤0

≤ ε

◮ Let (IP) ≤ ε n

  • i=1
  • ≥0

n

  • i=1

(Qαi − yi) + ε N · (Q − N)

  • ≥−ε

≤ ε

◮ ε N · (Q − N) ≤ ε ⇒ Q ≤ 2N

slide-55
SLIDE 55

Proof (2)

◮ ∃Q ∈ {1, . . . , N} : n

  • i=1

|Qαi − ⌊Qαi⌋

=:yi

| ≤ ε (IP) ≤

n

  • i=1

(Qαi − yi)

  • ≤ε

+ ε N · (Q − N)

  • ≤0

≤ ε

◮ Let (IP) ≤ ε n

  • i=1
  • ≥0

n

  • i=1

(Qαi − yi)

  • ≤2ε

+ ε N · (Q − N)

  • ≥−ε

≤ ε

◮ ε N · (Q − N) ≤ ε ⇒ Q ≤ 2N ◮ ∃Q ≤ 2N : n i=1(Qαi − ⌊Qαi⌋) ≤ 2ε

slide-56
SLIDE 56

Proof (2)

◮ ∃Q ∈ {1, . . . , N} : n

  • i=1

|Qαi − ⌊Qαi⌋

=:yi

| ≤ ε (IP) ≤

n

  • i=1

(Qαi − yi)

  • ≤ε

+ ε N · (Q − N)

  • ≤0

≤ ε

◮ Let (IP) ≤ ε n

  • i=1
  • ≥0

n

  • i=1

(Qαi − yi)

  • ≤2ε

+ ε N · (Q − N)

  • ≥−ε

≤ ε

◮ ε N · (Q − N) ≤ ε ⇒ Q ≤ 2N ◮ ∃Q ≤ 2N : n i=1(Qαi − ⌊Qαi⌋) ≤ 2ε ◮ to good for NO-case

slide-57
SLIDE 57

Proof (2)

◮ ∃Q ∈ {1, . . . , N} : n

  • i=1

|Qαi − ⌊Qαi⌋

=:yi

| ≤ ε (IP) ≤

n

  • i=1

(Qαi − yi)

  • ≤ε

+ ε N · (Q − N)

  • ≤0

≤ ε

◮ Let (IP) ≤ ε n

  • i=1
  • ≥0

n

  • i=1

(Qαi − yi)

  • ≤2ε

+ ε N · (Q − N)

  • ≥−ε

≤ ε

◮ ε N · (Q − N) ≤ ε ⇒ Q ≤ 2N ◮ ∃Q ≤ 2N : n i=1(Qαi − ⌊Qαi⌋) ≤ 2ε ◮ to good for NO-case ⇒ YES case of DDA

slide-58
SLIDE 58

Other applications

Theorem (Submitted to SODA 2010)

Testing EDF-schedulability of n periodic tasks (Ci, Di, Pi) (Ci = running time, Di = deadline, Pi = period), i.e. ∀t ≥ 0 :

n

  • i=1

t − Di Pi

  • + 1
  • · Ci ≤ t

is coNP-hard.

slide-59
SLIDE 59

Other applications

Theorem (Submitted to SODA 2010)

Testing EDF-schedulability of n periodic tasks (Ci, Di, Pi) (Ci = running time, Di = deadline, Pi = period), i.e. ∀t ≥ 0 :

n

  • i=1

t − Di Pi

  • + 1
  • · Ci ≤ t

is coNP-hard.

Thanks for your attention