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Almost Optimal Distribution-free Junta Testing Nader H. Bshouty - - PowerPoint PPT Presentation
Almost Optimal Distribution-free Junta Testing Nader H. Bshouty - - PowerPoint PPT Presentation
Almost Optimal Distribution-free Junta Testing Nader H. Bshouty Technion -Junta A Boolean function : 0,1 {0,1} is called junta if it depends on at most variables/coordinates. Examples: 1 , 2 , ,
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Model of Testing
A distribution-free testing algorithm π΅ for π βJunta is an algorithm that, given an access to the two oracles and a distance parameter π as an input , 1) if π is π-junta then π΅ outputs βacceptβ with probability at least 2/3. 2) if π is π-far from every π-junta with respect to the distribution π then π΅ outputs βrejectβ with probability at least 2/3. Given a black box that contains a Boolean function π: 0,1 π β {0,1} Given two oracles: 1) when π¦ β 0,1 π is queried, it returns π(π¦) and 2) returns π£ β 0,1 π chosen acc. to unknown distribution π β β β π β πΎπ£ππ’π Pr
π¦~π π π¦ β β π¦
β₯ π
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π-Junta π-Junta π
Model of Testing
π΅ outputs βacceptβ with probability at least 2/3.
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π-Junta π-Junta π
Model of Testing
π΅ outputs βrejectβ with probability at least 2/3.
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π-Junta π-Junta π
Model of Testing
π΅ halts and outputs either βacceptβ or βrejectβ.
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Results in the uniform distribution Model
Result ΰ·© π·/ΰ·© π Lo./Up. Ada./NonAda Reference
Adaptive NonAdaptive Lower bounds Upper bounds For the number of queries
ΰ·© Ξ© π = π /ππππ§(log π) ΰ·¨ π π = π ππππ§(log π)
Time ππππ§ π,
1 π
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Result ΰ·© π·/ΰ·© π Lo./Up. Ada./NonAda Reference π/π Upper Adaptive Blais STOC 2009
Results in the uniform distribution Model
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Result ΰ·© π·/ΰ·© π Lo./Up. Ada./NonAda Reference π/π Upper Adaptive Blais STOC 2009 π Lower Adaptive Saglam FOCS 2018
Results in the uniform distribution Model
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Result ΰ·© π·/ΰ·© π Lo./Up. Ada./NonAda Reference π/π Upper Adaptive Blais STOC 2009 π Lower Adaptive Saglam FOCS 2018 π
3 2/π
Upper NonAdaptive Blais APPROX 2008
Results in the uniform distribution Model
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Result ΰ·© π·/ΰ·© π Lo./Up. Ada./NonAda Reference π/π Upper Adaptive Blais STOC 2009 π Lower Adaptive Saglam FOCS 2018 π
3 2/π
Upper NonAdaptive Blais APPROX 2008 π
3 2/π
Lower NonAdaptive Chen et al. CCC 2017
Results in the uniform distribution Model
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Results in the distribution-free model
Result ΰ·© π·/ΰ·© π Lo./Up. Ada./NonAda Reference
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Results in the distribution-free model
Result ΰ·© π·/ΰ·© π Lo./Up. Ada./NonAda Reference 2π/π Upper NonAdaptive Halevy et al. APPROX 03
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Results in the distribution-free model
Result ΰ·© π·/ΰ·© π Lo./Up. Ada./NonAda Reference 2π/π Upper NonAdaptive Halevy et al. APPROX 03 2π/3 Lower NonAdaptive Chen et al. STOC 2018
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Results in the distribution-free model
Result ΰ·© π·/ΰ·© π Lo./Up. Ada./NonAda Reference 2π/π Upper NonAdaptive Halevy et al. APPROX 03 2π/3 Lower NonAdaptive Chen et al. STOC 2018 π2/π Upper Adaptive Chen et al. STOC 2018
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Results in the distribution-free model
Result ΰ·© π·/ΰ·© π Lo./Up. Ada./NonAda Reference 2π/π Upper NonAdaptive Halevy et al. APPROX 03 2π/3 Lower NonAdaptive Chen et al. STOC 2018 π2/π Upper Adaptive Chen et al. STOC 2018 π Lower Adaptive Saglam FOCS 2018
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Results in the distribution-free model
Result ΰ·© π·/ΰ·© π Lo./Up. Ada./NonAda Reference 2π/π Upper NonAdaptive Halevy et al. APPROX 03 2π/3 Lower NonAdaptive Chen et al. STOC 2018 π2/π Upper Adaptive Chen et al. STOC 2018 π Lower Adaptive Saglam FOCS 2018 π/π Upper Adaptive Ours
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The algorithm
Choose a random uniform partition π1, π2, β¦ , ππ of π where π = 2π2 π(π¦1, β¦ , π¦π) π(π¦π1 β π¦π2 β β― β π¦ππ ) Find relevant sets ππ1, ππ2, β¦ , πππβ² π π£π β 0 ΰ΄€
π ? = π(π£)
ππ1, ππ2, β¦ , πππ π = ππ1 βͺ ππ2 βͺ β― βͺ πππ π£~π Why 2π2? If π is kβjunta, whp each ππ contains at most one relevant coordinate
Find relevant sets
π π€ β π π€ππ β 0ππ
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Find relevant sets
Find relevant sets ππ1, ππ2, β¦ , πππβ² π π£π β 0 ΰ΄€
π ? = π(π£)
ππ1, ππ2, β¦ , πππ π = ππ1 βͺ ππ2 βͺ β― βͺ πππ Find a new relevant set πππ+1 = πβ π π£π β 0 ΰ΄€
π β π(π£)
π π£π β π£π
1 β 0π 2
π£~π ΰ΄€ π = π
1 βͺ π 2
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Find relevant sets
Find relevant sets ππ1, ππ2, β¦ , πππβ² π π£π β 0 ΰ΄€
π ? = π(π£)
ππ1, ππ2, β¦ , πππ π = ππ1 βͺ ππ2 βͺ β― βͺ πππ Find a new relevant sets πππ+1 = πβ π π£π β 0 ΰ΄€
π β π(π£)
π π£π β 0 ΰ΄€
π = π(π£)
π£~π For ΰ·¨ π
1 π values of π£~π
If this is the (π + 1)-th relevant set then βrejectββ Pr
π¦~π π π¦π β 0 ΰ΄€ π β π π¦
β€ π/3 log π = π log π ΰ·¨ π
π π queries
We also get a witness for πβ π π€(β) β π π€πβ
β β 0πβ
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The algorithm
ππ1, ππ2, β¦ , πππβ² , πβ² β€ π Pr
π¦~π π π¦π β 0 ΰ΄€ π β π π¦
β€ π 3 π = ππ1 βͺ ππ2 βͺ β― βͺ πππβ² β β π(π¦π β 0 ΰ΄€
π) is π 3 βclose to π with respect to π
Each πππ contains at least one relevant variable π-Junta π-Junta π π β: = π(π¦π β 0 ΰ΄€
π)
β: = π(π¦π β 0 ΰ΄€
π)
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ππ1, ππ2, β¦ , πππβ² , πβ² β€ π π = ππ1 βͺ ππ2 βͺ β― βͺ πππβ² Each πππ contains at least one relevant coordinte π is π βjunta π(π¦) is π βfar from every π βJunta with respect to π β: = π(π¦π β 0 ΰ΄€
π) is π βjunta
Whp each πππ contains exactly one relevant coordinate β: = π(π¦π β 0 ΰ΄€
π) is 2π 3 βfar
from every π βJunta with respect to π We also have witness for each relevant set πππβ that is, π π€ π β π π€πππ
π β 0πππ
ΰ·¨ π
π π queries
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π is π βjunta π(π¦) is π βfar from every π βJunta with respect to π β: = π(π¦π β 0 ΰ΄€
π) is π βjunta
Whp each πππ contains exactly one relevant variable β: = π(π¦π β 0 ΰ΄€
π) is 2π 3 βfar
from every π βJunta with respect to π π π€ π β π π€πππ
π β 0πππ
π π€πππ
π β π¦πππ
is equal to π¦π π or π¦π π π π€πππ
π β π¦πππ
is (1/15)-close to a literal in {π¦π π , π¦π π } according to the uniform distribution ΰ·¨ π π queries 1 βJunta\0 βJunta π(log π) ΰ·¨ π 1 queries
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Ξ = {π(1), π(2), β¦ , π(πβ²)} π = β(π¦Ξ β π§ΰ΄₯
Ξ)
β(π¦) is
2π 3 βfar from any π-junta with respect to π
Pr
π£~π β π£ β β π£Ξ β π§ΰ΄₯ Ξ
β₯ 2π 3 Pr
π£~π ,π§~π β π£ β β π£Ξ β π§ΰ΄₯ Ξ
β₯ 2π 3 β: = π(π¦π β 0 ΰ΄€
π) is either π βjunta that depends on
β(π¦) is π βjunta β π¦ = β(π¦Ξ β π§ΰ΄₯
Ξ)
Pr
π£~π ,π§~π β π£ β β π§ |π§Ξ = π£Ξ β₯ 2π
3 Given π£? How to draw a random uniform π§ such that π§Ξ = π£Ξ? Or
2π 3 βfar from every π βjunta w.r.t. π
ΰ·¨ π π queries ΰ·¨ π
1 π queries
ΰ·¨ π
π π queries
Pr
π£~π ,π§~π β π£ β β π§ |π§Ξ = π£Ξ = 0
is π-junta Fix any π§ β 0,1 π ππ1, ππ2, β¦ , πππβ²
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Ξ = {π 1 , π 2 , β¦ , π(πβ²)} β: = π(π¦π β 0 ΰ΄€
π)
Given π£? How to draw a random uniform π§ such that π§Ξ = π£Ξ? π π€πππ
π β π¦πππ
is (1/15)-close to a literal in {π¦π π , π¦π π } wrt uniform dist. βA procedure that given π£ finds π£π π with high probability Draw a random uniform π§πππ If π§π π = π£π π then output(π§πππ) If π§π π β π£π π then output(π§πππ) π§ = π§ππ1 β π§ππ2 β β― β π§πππβ² β π§ ΰ΄€
π
ΰ·¨ π π queries ΰ·¨ π 1 queries Chen, Liu, Servedio, Sheng, Xie 2018
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Results in the distribution-free model
Result ΰ·© π·/ΰ·© π Lo./Up. Ada./NonAda Reference 2π/π Upper NonAdaptive Halevy et al. APPROX 03 2π/3 Lower NonAdaptive Chen et al. STOC 2018 π/π Upper Adaptive Ours CCC 2019 π Lower Adaptive Saglam FOCS 2018
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Result ΰ·© π·/ΰ·© π Lo./Up. Ada./NonAda Reference 2π/π Upper NonAdaptive Halevy et al. APPROX 03 2π/3 Lower NonAdaptive Chen et al. STOC 2018 π/π Upper Adaptive Ours CCC 2019 π Lower Adaptive Saglam FOCS 2018
Open Problems
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Open Problems
Result ΰ·© π·/ΰ·© π Lo./Up. Ada./NonAda Reference 2π/π Upper NonAdaptive Halevy et al. APPROX 03 2π/3 Lower NonAdaptive Chen et al. STOC 2018 ? π(1)-Round π/π Upper Adaptive Ours CCC 2019 π Lower Adaptive Saglam FOCS 2018
Almost non-Adaptive
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Open Problems
Result ΰ·© π·/ΰ·© π Lo./Up. Ada./NonAda Reference 2π/π Upper NonAdaptive Halevy et al. APPROX 03 2π/3 Lower NonAdaptive Chen et al. STOC 2018 ? π(1)-Round poly(π/π) ?-Round π/π Upper Adaptive Ours CCC 2019 π Lower Adaptive Saglam FOCS 2018
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