Improved NP-inapproximability for 2-variable linear equations Johan - - PowerPoint PPT Presentation

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Improved NP-inapproximability for 2-variable linear equations Johan - - PowerPoint PPT Presentation

May 11, 2023 241 likes •1.24k views

Improved NP-inapproximability for 2-variable linear equations Johan Hstad Sangxia Huang Rajsekar Manokaran KTH KTH KTH Ryan ODonnell John Wright CMU CMU 2Lin x 1 = x 5 x 10 = -x 3 x 61 = -x 24 ... x 48 = -x 5 (x i = -1,1) 2Lin

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SLIDE 1

Improved NP-inapproximability for 2-variable linear equations

Sangxia Huang KTH Ryan O’Donnell CMU John Wright CMU Johan Håstad KTH Rajsekar Manokaran KTH

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SLIDE 2

2Lin

x1 = x5 x10 = -x3 x61 = -x24 ... x48 = -x5

(xi = -1,1)

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SLIDE 3

2Lin

x1 = x5 x10 = -x3 x61 = -x24 ... x48 = -x5

(xi = -1,1)

2Lin(2) ∈ 2Lin(q) ≈ UniqueGames

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SLIDE 4

2Lin

x1 = x5 x10 = -x3 x61 = -x24 ... x48 = -x5

(xi = -1,1)

2Lin(2) ∈ 2Lin(q) ≈ UniqueGames (Actually, simplest case of UG)

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SLIDE 5

2Lin

x1 = x5 x10 = -x3 x61 = -x24 ... x48 = -x5

(xi = -1,1)

2Lin(2) ∈ 2Lin(q) ≈ UniqueGames (Actually, simplest case of UG) Folklore wisdom: get 2Lin(2) right and 2Lin(q) will follow.

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SLIDE 6

Known results

Suppose val(I) = α. Can we guarantee a solution of value C*α?

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SLIDE 7

Known results

Suppose val(I) = α. Can we guarantee a solution of value C*α? [GW]: .878-approx algorithm

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SLIDE 8

Known results

Suppose val(I) = α. Can we guarantee a solution of value C*α? [GW]: .878-approx algorithm [KKMO]+[MOO]: (.878+ε)-approx UG-hard

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SLIDE 9

Known results

Suppose val(I) = α. Can we guarantee a solution of value C*α? [GW]: .878-approx algorithm [KKMO]+[MOO]: (.878+ε)-approx UG-hard [Håstad]+[TSSW]: 16/17 ≈ .941-approx NP-hard

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SLIDE 10

Known results

Suppose val(I) = α. Can we guarantee a solution of value C*α? [GW]: .878-approx algorithm [KKMO]+[MOO]: (.878+ε)-approx UG-hard [Håstad]+[TSSW]: 16/17 ≈ .941-approx NP-hard seems we’re close, right?

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SLIDE 11

A different perspective...

Suppose val(I) = (1 - ε). Can we guarantee a solution of value (1 - C*ε)?

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SLIDE 12

A different perspective...

Suppose val(I) = (1 - ε). Can we guarantee a solution of value (1 - C*ε)? Def: Such an algo. gives an (ε, C*ε)-approx.

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SLIDE 13

A different perspective...

Suppose val(I) = (1 - ε). Can we guarantee a solution of value (1 - f(ε))? Def: Such an algo. gives an (ε, f(ε))-approx.

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SLIDE 14

A different perspective...

Suppose val(I) = (1 - ε). Can we guarantee a solution of value (1 - f(ε))? Def: Such an algo. gives an (ε, f(ε))-approx. Usually called “Min-2Lin(2)-Deletion”. Let me just call this 2Lin.

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SLIDE 15

Unratio state of affairs

[easy]: (ε, ε)-approx NP-hard

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SLIDE 16

Unratio state of affairs

[easy]: (ε, ε)-approx NP-hard [Håstad]+[TSSW]: (ε, ⁵/₄*ε)-approx NP-hard

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SLIDE 17

Unratio state of affairs

[easy]: (ε, ε)-approx NP-hard [Håstad]+[TSSW]: (ε, ⁵/₄*ε)-approx NP-hard [KKMO]+[MOO]: (ε, O(ε1/2))-approx UG-hard

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SLIDE 18

Unratio state of affairs

[easy]: (ε, ε)-approx NP-hard [Håstad]+[TSSW]: (ε, ⁵/₄*ε)-approx NP-hard [KKMO]+[MOO]: (ε, O(ε1/2))-approx UG-hard [GW]: (ε, O(ε1/2))-approx algorithm

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SLIDE 19

Unratio state of affairs

[easy]: (ε, ε)-approx NP-hard [Håstad]+[TSSW]: (ε, ⁵/₄*ε)-approx NP-hard [KKMO]+[MOO]: (ε, O(ε1/2))-approx UG-hard [GW]: (ε, O(ε1/2))-approx algorithm asymptotically off from the truth

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SLIDE 20

Unratio state of affairs

[easy]: (ε, ε)-approx NP-hard [Håstad]+[TSSW]: (ε, ⁵/₄*ε)-approx NP-hard [KKMO]+[MOO]: (ε, O(ε1/2))-approx UG-hard [GW]: (ε, O(ε1/2))-approx algorithm asymptotically off from the truth [Rao]: If (ε, O(f(q)*ε1/2))-approx is NP-hard for 2Lin(q), for f(q) = Ω(1), then UG is true.

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SLIDE 21

This work

[Håstad]+[TSSW]: (ε, ⁵/₄*ε)-approx NP-hard

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SLIDE 22

This work

[Håstad]+[TSSW]: (ε, ⁵/₄*ε)-approx NP-hard [Us]: (ε, 11/8*ε)-approx NP-hard

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SLIDE 23

This work

[Håstad]+[TSSW]: (ε, 1.25*ε)-approx NP-hard [Us]: (ε, 1.375*ε)-approx NP-hard

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SLIDE 24

This work

[Håstad]+[TSSW]: (ε, ⁵/₄*ε)-approx NP-hard [Us]: (ε, 11/8*ε)-approx NP-hard

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SLIDE 25

This work

[Håstad]+[TSSW]: (ε, ⁵/₄*ε)-approx NP-hard [Us]: (ε, 11/8*ε)-approx NP-hard Cons:

  • Still haven’t proven UniqueGames. 😟
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SLIDE 26

This work

[Håstad]+[TSSW]: (ε, ⁵/₄*ε)-approx NP-hard [Us]: (ε, 11/8*ε)-approx NP-hard Cons:

  • Still haven’t proven UniqueGames. 😟

Pros:

  • First improvement since 1997.
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SLIDE 27

This work

[Håstad]+[TSSW]: (ε, ⁵/₄*ε)-approx NP-hard [Us]: (ε, 11/8*ε)-approx NP-hard Cons:

  • Still haven’t proven UniqueGames. 😟

Pros:

  • First improvement since 1997.
  • Study new type of “gadget reduction”
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SLIDE 28

This work

[Håstad]+[TSSW]: (ε, ⁵/₄*ε)-approx NP-hard [Us]: (ε, 11/8*ε)-approx NP-hard (and more!) Cons:

  • Still haven’t proven UniqueGames. 😟

Pros:

  • First improvement since 1997.
  • Study new type of “gadget reduction”
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SLIDE 29

Proving (ε, ⁵/₄*ε)-hardness

Standard two-step plan.

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SLIDE 30

Proving (ε, ⁵/₄*ε)-hardness

Standard two-step plan. [Håstad]: Given 3Lin instance I, NP-hard to distinguish

  • Yes: val(I) ≥ (1 - ε)
  • No: val(I) ≤ (½ + ε)
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SLIDE 31

Proving (ε, ⁵/₄*ε)-hardness

Standard two-step plan. [Håstad]: Given 3Lin instance I, NP-hard to distinguish

  • Yes: val(I) ≥ (1 - ε)
  • No: val(I) ≤ (½ + ε)

(In our language, (ε, ½ - ε)-approxing 3Lin is NP-hard.)

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SLIDE 32

Proving (ε, ⁵/₄*ε)-hardness

Standard two-step plan. [Håstad]: Given 3Lin instance I, NP-hard to distinguish

  • Yes: val(I) ≥ (1 - ε)
  • No: val(I) ≤ (½ + ε)

(In our language, (ε, ½ - ε)-approxing 3Lin is NP-hard.) Step 2: gadget reduce 3Lin to 2Lin [TSSW]

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SLIDE 33

Proving (ε, ⁵/₄*ε)-hardness

Standard two-step plan. [Håstad]: Given 3Lin instance I, NP-hard to distinguish

  • Yes: val(I) ≥ (1 - ε)
  • No: val(I) ≤ (½ + ε)

(In our language, (ε, ½ - ε)-approxing 3Lin is NP-hard.) Step 2: gadget reduce 3Lin to 2Lin [TSSW] (see also [OW12])

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SLIDE 34

3Lin

x1x3x5 = 1 x10x16x3 = -1 ... x47x11x98 = -1

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SLIDE 35

3Lin

x1x3x5 = 1 x10x16x3 = -1 ... x47x11x98 = -1

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SLIDE 36

3Lin

x1x3x5 = 1 x10x16x3 = -1 ... x47x11x98 = -1

x10 x3 x16

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SLIDE 37

3Lin

x1x3x5 = 1 x10x16x3 = -1 ... x47x11x98 = -1

x10 x3 x16 (aux vars)

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SLIDE 38

3Lin

x1x3x5 = 1 x10x16x3 = -1 ... x47x11x98 = -1

x10 x3 x16 (aux vars)

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SLIDE 39

3Lin

x1x3x5 = 1 x10x16x3 = -1 ... x47x11x98 = -1

2Lin gadget

x10 = -x3 y61 = -y24 ... x16 = -y5

x10 x3 x16 (aux vars)

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SLIDE 40

3Lin

x1x3x5 = 1 x10x16x3 = -1 ... x47x11x98 = -1

2Lin gadget

x10 = -x3 y61 = -y24 ... x16 = -y5

x10 x3 x16 (aux vars) Final 2Lin inst: union all the gadgets

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SLIDE 41

3Lin

x1x3x5 = 1 x10x16x3 = -1 ... x47x11x98 = -1

2Lin gadget

x10 = -x3 y61 = -y24 ... x16 = -y5

x10 x3 x16 (aux vars) Final 2Lin inst: union all the gadgets The hope: - xi’s satisfy 3Lin eq’n ⇒ good assgn to yi’s

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SLIDE 42

3Lin

x1x3x5 = 1 x10x16x3 = -1 ... x47x11x98 = -1

2Lin gadget

x10 = -x3 y61 = -y24 ... x16 = -y5

x10 x3 x16 (aux vars) Final 2Lin inst: union all the gadgets The hope: - xi’s satisfy 3Lin eq’n ⇒ good assgn to yi’s

  • xi’s don’t ⇒ no good assgn to yi’s
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SLIDE 43

Gadgets

Def: A (c, s)-gadget

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SLIDE 44

Gadgets

Def: A (c, s)-gadget

  • xi’s satisfy 3Lin eq’n ⇒ an assgn to yi’s of value (1 - c)
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SLIDE 45

Gadgets

Def: A (c, s)-gadget

  • xi’s satisfy 3Lin eq’n ⇒ an assgn to yi’s of value (1 - c)
  • xi’s don’t ⇒ no assgn to yi’s beats value (1 - s)
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SLIDE 46

Gadgets

Def: A (c, s)-gadget

  • xi’s satisfy 3Lin eq’n ⇒ an assgn to yi’s of value (1 - c)
  • xi’s don’t ⇒ no assgn to yi’s beats value (1 - s)

[TSSW]: there is a 3Lin-to-2Lin (¼, ⅜)-gadget

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SLIDE 47

Gadgets

Def: A (c, s)-gadget

  • xi’s satisfy 3Lin eq’n ⇒ an assgn to yi’s of value (1 - c)
  • xi’s don’t ⇒ no assgn to yi’s beats value (1 - s)

[TSSW]: there is a 3Lin-to-2Lin (¼, ⅜)-gadget

(ε, ½ - ε)-hardess for 3Lin

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SLIDE 48

Gadgets

Def: A (c, s)-gadget

  • xi’s satisfy 3Lin eq’n ⇒ an assgn to yi’s of value (1 - c)
  • xi’s don’t ⇒ no assgn to yi’s beats value (1 - s)

[TSSW]: there is a 3Lin-to-2Lin (¼, ⅜)-gadget

(ε, ½ - ε)-hardess for 3Lin ⇒ - Yes case: ¼

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SLIDE 49

Gadgets

Def: A (c, s)-gadget

  • xi’s satisfy 3Lin eq’n ⇒ an assgn to yi’s of value (1 - c)
  • xi’s don’t ⇒ no assgn to yi’s beats value (1 - s)

[TSSW]: there is a 3Lin-to-2Lin (¼, ⅜)-gadget

(ε, ½ - ε)-hardess for 3Lin ⇒ - Yes case: ¼

  • No case: ½ * (¼ + ⅜)
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SLIDE 50

Gadgets

Def: A (c, s)-gadget

  • xi’s satisfy 3Lin eq’n ⇒ an assgn to yi’s of value (1 - c)
  • xi’s don’t ⇒ no assgn to yi’s beats value (1 - s)

[TSSW]: there is a 3Lin-to-2Lin (¼, ⅜)-gadget

(ε, ½ - ε)-hardess for 3Lin ⇒ - Yes case: ¼

  • No case: ½ * (¼ + ⅜)

= 5/16

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SLIDE 51

Gadgets

Def: A (c, s)-gadget

  • xi’s satisfy 3Lin eq’n ⇒ an assgn to yi’s of value (1 - c)
  • xi’s don’t ⇒ no assgn to yi’s beats value (1 - s)

[TSSW]: there is a 3Lin-to-2Lin (¼, ⅜)-gadget

(ε, ½ - ε)-hardess for 3Lin ⇒ - Yes case: ¼

  • No case: ½ * (¼ + ⅜)

= 5/16 = 5/4 * ¼

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SLIDE 52

How do you find gadgets?

Gadgets are just 2Lin instances, so can just monkey around with small instances.

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SLIDE 53

How do you find gadgets?

Gadgets are just 2Lin instances, so can just monkey around with small instances. More principled: [TSSW] show that the optimal gadget can be found via linear program!

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SLIDE 54

How do you find gadgets?

Gadgets are just 2Lin instances, so can just monkey around with small instances. More principled: [TSSW] show that the optimal gadget can be found via linear program!

  • key insight: one can bound # of auxiliary variables
  • can certify optimality via dual LP.
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SLIDE 55

How do you find gadgets?

Gadgets are just 2Lin instances, so can just monkey around with small instances. More principled: [TSSW] show that the optimal gadget can be found via linear program!

  • key insight: one can bound # of auxiliary variables
  • can certify optimality via dual LP.
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SLIDE 56

Our strategy

Old reduction from Håstad’s 3Lin hardness result.

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SLIDE 57

Our strategy

Old reduction from Håstad’s 3Lin hardness result. We now have better starting points.

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SLIDE 58

Our strategy

Old reduction from Håstad’s 3Lin hardness result. We now have better starting points. [Chan]: Given 3Lin instance I, NP-hard to distinguish

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SLIDE 59

Our strategy

Old reduction from Håstad’s 3Lin hardness result. We now have better starting points. [Chan]: Given 3Lin instance I, NP-hard to distinguish

  • Yes: val(I) = (1 - ε)
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SLIDE 60

Our strategy

Old reduction from Håstad’s 3Lin hardness result. We now have better starting points. [Chan]: Given 3Lin instance I, NP-hard to distinguish

  • Yes: val(I) = (1 - ε)
  • No: no matter what assignment, the variables “appear” to

be uniformly random

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SLIDE 61

Our strategy

Old reduction from Håstad’s 3Lin hardness result. We now have better starting points. [Chan]: Given 3Lin instance I, NP-hard to distinguish

  • Yes: val(I) = (1 - ε)
  • No: no matter what assignment, the variables “appear” to

be uniformly random Stronger No condition. Might help out the gadget.

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SLIDE 62

New gadgets

Def: A (c,s)-Chan-gadget

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SLIDE 63

New gadgets

Def: A (c,s)-Chan-gadget

  • xi’s satisfy 3Lin eq’n ⇒ an assgn to yi’s of value (1 - c)
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SLIDE 64

New gadgets

Def: A (c,s)-Chan-gadget

  • xi’s satisfy 3Lin eq’n ⇒ an assgn to yi’s of value (1 - c)
  • xi’s random ⇒ on avg., expected best value ≤ (1 - s)
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SLIDE 65

New gadgets

Def: A (c,s)-Chan-gadget

  • xi’s satisfy 3Lin eq’n ⇒ an assgn to yi’s of value (1 - c)
  • xi’s random ⇒ on avg., expected best value ≤ (1 - s)

Upside: can still solve using an LP

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SLIDE 66

New gadgets

Def: A (c,s)-Chan-gadget

  • xi’s satisfy 3Lin eq’n ⇒ an assgn to yi’s of value (1 - c)
  • xi’s random ⇒ on avg., expected best value ≤ (1 - s)

Upside: can still solve using an LP Downside: best 3Lin-to-2Lin gadget no better than in ’97!

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SLIDE 67

Our strategy (revised)

Old reduction from Håstad’s 3Lin hardness result. We now have better starting points. [Chan]: Given 3Lin instance I, NP-hard to distinguish

  • Yes: val(I) = (1 - ε)
  • No: no matter what assignment, the variables “appear” to

be uniformly random

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SLIDE 68

Our strategy (revised)

Old reduction from Håstad’s 3Lin hardness result. We now have better starting points. [Chan]: Given “balanced pairwise independent subgroup predicate” instance I, NP-hard to distinguish

  • Yes: val(I) = (1 - ε)
  • No: no matter what assignment, the variables “appear” to

be uniformly random

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SLIDE 69

Our strategy (revised)

Old reduction from Håstad’s 3Lin hardness result. We now have better starting points. [Chan]: Given “balanced pairwise independent subgroup predicate” instance I, NP-hard to distinguish

  • Yes: val(I) = (1 - ε)
  • No: no matter what assignment, the variables “appear” to

be uniformly random We instantiate with BPISP = Hadk, specifically k = 3.

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SLIDE 70

One of Chan’s problems

Had3(x1, x2, x3, x1,2, x1,3, x2,3, x1,2,3) = 1 iff

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SLIDE 71

One of Chan’s problems

Had3(x1, x2, x3, x1,2, x1,3, x2,3, x1,2,3) = 1 iff

  • xi’s allowed to be arbitrary
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SLIDE 72

One of Chan’s problems

Had3(x1, x2, x3, x1,2, x1,3, x2,3, x1,2,3) = 1 iff

  • xi’s allowed to be arbitrary
  • xa,b = xa∙ xb
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SLIDE 73

One of Chan’s problems

Had3(x1, x2, x3, x1,2, x1,3, x2,3, x1,2,3) = 1 iff

  • xi’s allowed to be arbitrary
  • xa,b = xa∙ xb
  • x1,2,3 = x1∙ x2∙ x3
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SLIDE 74

One of Chan’s problems

Had3(x1, x2, x3, x1,2, x1,3, x2,3, x1,2,3) = 1 iff

  • xi’s allowed to be arbitrary
  • xa,b = xa∙ xb
  • x1,2,3 = x1∙ x2∙ x3 = x1,2∙ x3
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SLIDE 75

One of Chan’s problems

Had3(x1, x2, x3, x1,2, x1,3, x2,3, x1,2,3) = 1 iff

  • xi’s allowed to be arbitrary
  • xa,b = xa∙ xb
  • x1,2,3 = x1∙ x2∙ x3 = x1,2∙ x3 = x1,3∙ x2 = x1∙ x2,3
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SLIDE 76

One of Chan’s problems

Had3(x1, x2, x3, x1,2, x1,3, x2,3, x1,2,3) = 1 iff

  • xi’s allowed to be arbitrary
  • xa∙ xb∙ xa,b= 1
  • x1,2,3 = x1∙ x2∙ x3 = x1,2∙ x3 = x1,3∙ x2 = x1∙ x2,3
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SLIDE 77

One of Chan’s problems

Had3(x1, x2, x3, x1,2, x1,3, x2,3, x1,2,3) = 1 iff

  • xi’s allowed to be arbitrary
  • xa∙ xb∙ xa,b= 1
  • xa∙ x{1,2,3}\a∙ x1,2,3 = 1 for all a ∈ {1,2,3}
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SLIDE 78

One of Chan’s problems

Had3(x1, x2, x3, x1,2, x1,3, x2,3, x1,2,3) = 1 iff

  • xi’s allowed to be arbitrary
  • xa∙ xb∙ xa,b= 1
  • xa∙ x{1,2,3}\a∙ x1,2,3 = 1 for all a ∈ {1,2,3}

Contains many simultaneous 3Lin tests. Difficult to satisfy!

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SLIDE 79

One of Chan’s problems

Had3(x1, x2, x3, x1,2, x1,3, x2,3, x1,2,3) = 1 iff

  • xi’s allowed to be arbitrary
  • xa∙ xb∙ xa,b= 1
  • xa∙ x{1,2,3}\a∙ x1,2,3 = 1 for all a ∈ {1,2,3}

Contains many simultaneous 3Lin tests. Difficult to satisfy! (not too hard to generalize to Hadk)

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SLIDE 80

One of Chan’s problems

Had3(x1, x2, x3, x1,2, x1,3, x2,3, x1,2,3) = 1 iff

  • xi’s allowed to be arbitrary
  • xa∙ xb∙ xa,b= 1
  • xa∙ x{1,2,3}\a∙ x1,2,3 = 1 for all a ∈ {1,2,3}

Contains many simultaneous 3Lin tests. Difficult to satisfy! (not too hard to generalize to Hadk) [Us]: there is a (1/8, 11/64)-Chan-gadget from Had3 to 2Lin

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SLIDE 81

One of Chan’s problems

Had3(x1, x2, x3, x1,2, x1,3, x2,3, x1,2,3) = 1 iff

  • xi’s allowed to be arbitrary
  • xa∙ xb∙ xa,b= 1
  • xa∙ x{1,2,3}\a∙ x1,2,3 = 1 for all a ∈ {1,2,3}

Contains many simultaneous 3Lin tests. Difficult to satisfy! (not too hard to generalize to Hadk) [Us]: there is a (1/8, 11/8∙

1/8)-Chan-gadget from Had3 to 2Lin

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SLIDE 82

Solving the LP

[TSSW]: optimal gadget only needs 27 = 128 variables

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SLIDE 83

Solving the LP

[TSSW]: optimal gadget only needs 27 = 128 variables (so no pictures like these)

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SLIDE 84

Solving the LP

[TSSW]: optimal gadget only needs 27 = 128 variables

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SLIDE 85

Solving the LP

[TSSW]: optimal gadget only needs 27 = 128 variables LP has to consider all possible 2128 = 3 x 1038 assignments to these variables: too large to be feasible!

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SLIDE 86

Solving the LP

[TSSW]: optimal gadget only needs 27 = 128 variables LP has to consider all possible 2128 = 3 x 1038 assignments to these variables: too large to be feasible! So…….

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SLIDE 87

Solving the LP

[TSSW]: optimal gadget only needs 27 = 128 variables LP has to consider all possible 2128 = 3 x 1038 assignments to these variables: too large to be feasible! So…….

  • lots of work by hand
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SLIDE 88

Solving the LP

[TSSW]: optimal gadget only needs 27 = 128 variables LP has to consider all possible 2128 = 3 x 1038 assignments to these variables: too large to be feasible! So…….

  • lots of work by hand
  • lots of computer simulation (/brute force searching)
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SLIDE 89

Solving the LP

[TSSW]: optimal gadget only needs 27 = 128 variables LP has to consider all possible 2128 = 3 x 1038 assignments to these variables: too large to be feasible! So…….

  • lots of work by hand
  • lots of computer simulation (/brute force searching)
  • lots more work by hand
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SLIDE 90

Solving the LP

[TSSW]: optimal gadget only needs 27 = 128 variables LP has to consider all possible 2128 = 3 x 1038 assignments to these variables: too large to be feasible! So…….

  • lots of work by hand
  • lots of computer simulation (/brute force searching)
  • lots more work by hand

but (spoiler alert) it all works out in the end.

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SLIDE 91

Full statement of our results

  • (ε, 11/8*ε)-approx NP-hard (for 2Lin and Max-Cut)
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SLIDE 92

Full statement of our results

  • (ε, 11/8*ε)-approx NP-hard (for 2Lin and Max-Cut)
  • (1/8, 11/8∙

1/8)-Chan-gadget from Had3 to 2Lin

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SLIDE 93

Full statement of our results

  • (ε, 11/8*ε)-approx NP-hard (for 2Lin and Max-Cut)
  • (1/8, 11/8∙

1/8)-Chan-gadget from Had3 to 2Lin

  • prove optimality of this gadget via dual solution
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SLIDE 94

Full statement of our results

  • (ε, 11/8*ε)-approx NP-hard (for 2Lin and Max-Cut)
  • (1/8, 11/8∙

1/8)-Chan-gadget from Had3 to 2Lin

  • prove optimality of this gadget via dual solution
  • can’t beat (ε, 2.54 ∙ ε)-hardness via a Chan gadget

starting from *any* BPISP predicate

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SLIDE 95

Open problems

We give an optimal gadget reduction from Had3 to 2Lin.

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SLIDE 96

Open problems

We give an optimal gadget reduction from Had3 to 2Lin. We give a gadget from Hadk to 2Lin, along with a Game Show Conjecture which would imply (ε, 1.5 ∙ ε)-hardness.

slide-97
SLIDE 97

Open problems

We give an optimal gadget reduction from Had3 to 2Lin. We give a gadget from Hadk to 2Lin, along with a Game Show Conjecture which would imply (ε, 1.5 ∙ ε)-hardness. We couldn’t say anything about the normal hardness ratio. Maybe you can?

slide-98
SLIDE 98

Thanks!