Almost Optimal Distribution-free Junta Testing Nader H. Bshouty Technion
π -Junta A Boolean function π: 0,1 π β {0,1} is called π β junta if it depends on at most π variables/coordinates. Examples: π π¦ 1 , π¦ 2 , β¦ , π¦ 10000 = π¦ 1111 β§ π¦ 9992 β π¦ 3001 is 3 - junta Relevant variables/coordinates is 4-junta 1-junta is {π¦ π , ΰ΄₯ π¦ π , 0,1} 0-junta is {0,1} π β Junta is the class of all π β juntas.
Model of Testing 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 A distribution-free testing algorithm π΅ for π β Junta 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 β accept β 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 β reject β with probability at least 2/3. β β β π β πΎπ£ππ’π π¦~π π π¦ β β π¦ Pr β₯ π
Model of Testing π π -Junta π -Junta π΅ outputs β accept β with probability at least 2/3.
Model of Testing π π -Junta π -Junta π΅ outputs β reject β with probability at least 2/3.
Model of Testing π -Junta π -Junta π π΅ halts and outputs either β accept β or β reject β .
Results in the uniform distribution Model Result ΰ·© π·/ΰ·© π Reference Ada./NonAda Lo./Up. Lower bounds Adaptive Upper bounds NonAdaptive For the number of queries ΰ·¨ π π = π ππππ§(log π) ΰ·© Ξ© π = π /ππππ§(log π) 1 Time ππππ§ π, π
Results in the uniform distribution Model Result ΰ·© π·/ΰ·© π Reference Ada./NonAda Lo./Up. π/π Blais STOC 2009 Adaptive Upper
Results in the uniform distribution Model Result ΰ·© π·/ΰ·© π Reference Ada./NonAda Lo./Up. π/π Blais STOC 2009 Adaptive Upper π Saglam FOCS 2018 Adaptive Lower
Results in the uniform distribution Model Result ΰ·© π·/ΰ·© π Reference Ada./NonAda Lo./Up. π/π Blais STOC 2009 Adaptive Upper π Saglam FOCS 2018 Adaptive Lower 3 Blais APPROX 2008 NonAdaptive Upper π 2 /π
Results in the uniform distribution Model Result ΰ·© π·/ΰ·© π Reference Ada./NonAda Lo./Up. π/π Blais STOC 2009 Adaptive Upper π Saglam FOCS 2018 Adaptive Lower 3 Blais APPROX 2008 NonAdaptive Upper π 2 /π 3 Chen et al. CCC 2017 NonAdaptive Lower π 2 /π
Results in the distribution-free model Result ΰ·© π·/ΰ·© π Reference Ada./NonAda Lo./Up.
Results in the distribution-free model Result ΰ·© π·/ΰ·© π Reference Ada./NonAda Lo./Up. 2 π /π Halevy et al. APPROX 03 NonAdaptive Upper
Results in the distribution-free model Result ΰ·© π·/ΰ·© π Reference Ada./NonAda Lo./Up. 2 π /π Halevy et al. APPROX 03 NonAdaptive Upper 2 π/3 Chen et al. STOC 2018 NonAdaptive Lower
Results in the distribution-free model Result ΰ·© π·/ΰ·© π Reference Ada./NonAda Lo./Up. 2 π /π Halevy et al. APPROX 03 NonAdaptive Upper 2 π/3 Chen et al. STOC 2018 NonAdaptive Lower π 2 /π Chen et al. STOC 2018 Adaptive Upper
Results in the distribution-free model Result ΰ·© π·/ΰ·© π Reference Ada./NonAda Lo./Up. 2 π /π Halevy et al. APPROX 03 NonAdaptive Upper 2 π/3 Chen et al. STOC 2018 NonAdaptive Lower π 2 /π Chen et al. STOC 2018 Adaptive Upper π Saglam FOCS 2018 Adaptive Lower
Results in the distribution-free model Result ΰ·© π·/ΰ·© π Reference Ada./NonAda Lo./Up. 2 π /π Halevy et al. APPROX 03 NonAdaptive Upper 2 π/3 Chen et al. STOC 2018 NonAdaptive Lower π 2 /π Chen et al. STOC 2018 Adaptive Upper π Saglam FOCS 2018 Adaptive Lower π/π Ours Adaptive Upper
The algorithm π(π¦ 1 , β¦ , π¦ π ) Choose a random uniform partition π 1 , π 2 , β¦ , π π of π where π = 2π 2 π(π¦ π 1 β π¦ π 2 β β― β π¦ π π ) Why 2π 2 ? If π is k β junta, whp each π π contains at most one relevant coordinate Find relevant sets Find relevant sets π π 1 , π π 2 , β¦ , π π πβ² π π€ β π π€ π π β 0 π π π π 1 , π π 2 , β¦ , π π π π = π π 1 βͺ π π 2 βͺ β― βͺ π π π π π£ π β 0 ΰ΄€ π ? = π(π£) π£~π
Find relevant sets Find relevant sets π π 1 , π π 2 , β¦ , π π πβ² π π 1 , π π 2 , β¦ , π π π π = π π 1 βͺ π π 2 βͺ β― βͺ π π π π π£ π β 0 ΰ΄€ π ? = π(π£) π£~π π π£ π β 0 ΰ΄€ π β π(π£) Find a new relevant set π π π+1 = π β π π£ π β π£ π 1 β 0 π 2 ΰ΄€ π = π 1 βͺ π 2
Find relevant sets Find relevant sets π π 1 , π π 2 , β¦ , π π πβ² π π 1 , π π 2 , β¦ , π π π π = π π 1 βͺ π π 2 βͺ β― βͺ π π π π π£ π β 0 ΰ΄€ π ? = π(π£) π£~π π π£ π β 0 ΰ΄€ π β π(π£) log π = π log π Find a new relevant sets π π π+1 = π β If this is the (π + 1) -th relevant set then β reject ββ We also get a witness for π β β β 0 π β π π€ (β) β π π€ π β 1 For ΰ·¨ π π£ π β 0 ΰ΄€ π = π(π£) π π values of π£~π π ΰ·¨ π π queries π¦~π π π¦ π β 0 ΰ΄€ Pr π β π π¦ β€ π/3
The algorithm π π 1 , π π 2 , β¦ , π π πβ² , π β² β€ π Each π π π contains at least one relevant variable β€ π π = π π 1 βͺ π π 2 βͺ β― βͺ π π πβ² π¦~π π π¦ π β 0 ΰ΄€ Pr π β π π¦ 3 π β β π(π¦ π β 0 ΰ΄€ π ) is 3 β close to π with respect to π π π β: = π(π¦ π β 0 ΰ΄€ π ) π -Junta π -Junta β: = π(π¦ π β 0 ΰ΄€ π )
π π 1 , π π 2 , β¦ , π π πβ² , π β² β€ π Each π π π contains at least one relevant coordinte π = π π 1 βͺ π π 2 βͺ β― βͺ π π πβ² π(π¦) is π β far from every π β Junta with π is π β junta π ΰ·¨ π π queries respect to π β: = π(π¦ π β 0 ΰ΄€ π ) is π β junta 2π β: = π(π¦ π β 0 ΰ΄€ π ) is 3 β far Whp each π π π contains from every π β Junta with exactly one relevant coordinate respect to π We also have witness for each relevant set π π π β that is, π β 0 π ππ π π€ π β π π€ π ππ
π(π¦) is π β far from every π β Junta with π is π β junta respect to π β: = π(π¦ π β 0 ΰ΄€ π ) is π β junta 2π β: = π(π¦ π β 0 ΰ΄€ π ) is 3 β far Whp each π π π contains from every π β Junta with exactly one relevant variable respect to π π β 0 π ππ π π€ π β π π€ π ππ π β π¦ π ππ ΰ·¨ π π€ π ππ is equal to π¦ π π or π¦ π π π π queries π(log π) 1 β Junta\ 0 β Junta π β π¦ π ππ π π€ π ππ is (1/15)-close to a literal in {π¦ π π , π¦ π π } ΰ·¨ π 1 queries according to the uniform distribution
Ξ = {π(1), π(2), β¦ , π(π β² )} β: = π(π¦ π β 0 ΰ΄€ π ) is either π β junta that depends on 2π π π 1 , π π 2 , β¦ , π π π β² 3 β far from every π β junta w.r.t. π Or Fix any π§ β 0,1 π π = β(π¦ Ξ β π§ ΰ΄₯ Ξ ) is π -junta 2π β(π¦) is 3 β far from any π -junta with respect to π β(π¦) is π β junta β₯ 2π β π¦ = β(π¦ Ξ β π§ ΰ΄₯ Ξ ) π£~π β π£ β β π£ Ξ β π§ ΰ΄₯ Pr Ξ 3 β₯ 2π π£~π ,π§~π β π£ β β π§ |π§ Ξ = π£ Ξ = 0 Pr π£~π ,π§~π β π£ β β π£ Ξ β π§ ΰ΄₯ Pr Ξ 3 π ΰ·¨ π£~π ,π§~π β π£ β β π§ |π§ Ξ = π£ Ξ β₯ 2π π π queries 1 Pr ΰ·¨ π π queries 3 Given π£ ? ΰ·¨ π π queries How to draw a random uniform π§ such that π§ Ξ = π£ Ξ ?
Ξ = {π 1 , π 2 , β¦ , π(π β² )} β: = π(π¦ π β 0 ΰ΄€ π ) Given π£ ? How to draw a random uniform π§ such that π§ Ξ = π£ Ξ ? π β π¦ π ππ π π€ π ππ is (1/15)-close to a literal in {π¦ π π , π¦ π π } wrt uniform dist. β A procedure that given π£ finds π£ π π with high probability ΰ·¨ π 1 queries π§ = π§ π π1 β π§ π π2 β β― β π§ π ππβ² β π§ ΰ΄€ π Draw a random uniform π§ π π π ΰ·¨ π π queries If π§ π π = π£ π π then output( π§ π ππ ) Chen, Liu, Servedio, Sheng, Xie 2018 If π§ π π β π£ π π then output( π§ π ππ )
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