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Bi- Lipschitz Bijection between the Boolean Cube and the Hamming Ball. Gil Cohen. Joint work with Itai Benjamini and Igor Shinkar. Cube vs. Ball. Majority. Dictator. n. Strings with Hamming weight k. k. 0. The Problem. - PowerPoint PPT Presentation
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Bi-Lipschitz Bijection between the Boolean
Cube and the Hamming BallGil Cohen
Joint work with Itai Benjamini and Igor Shinkar
Cube vs. Ball
{0 ,1 }𝑛 𝐵𝑛={𝑥∈ {0 ,1 }𝑛+1:|𝑥|>𝑛 /2}𝑓 : →
{𝑥∈ {0 ,1 }𝑛+1:𝑥𝑛+1=1}
Dictator
Majority
Strings with Hamming weight
k
n
0
k
The Problem
Open Problem [LovettViola11]. Prove that for any bijection ,
.
𝑓 (𝑥 )={ 𝑥∘0|𝑥|>𝑛/2𝑓𝑙𝑖𝑝 (𝑥 )∘1𝑜𝑡 h𝑒𝑟𝑤𝑖𝑠𝑒
Average stretch
The “Naïve” Upper Bound.
Motivation. Related to proving lower bounds for sampling a natural distribution by circuits.
Main Theorem
Theorem. such that
1)
of
𝐵𝑛of
Main Theorem
Theorem. such that
1)
2) is computable in DLOGTIME-uniform
(Majority is -reducible to )
3) is very local
15≤𝑑𝑖𝑠𝑡 (𝜓 (𝑥 ) ,𝜓 (𝑦 ) )
𝑑𝑖𝑠𝑡 (𝑥 , 𝑦 )≤4
∀ 𝑖∈ [𝑛 ]Pr𝑥
[𝜓 (𝑥 )𝑖≠ 𝑥𝑖 ]≤𝑂 ( 1√𝑛 )
The BTK Partition [DeBruijnTengbergenKruyswijk51]
1 0 0 1 1 0 0 1
Partition of to symmetric chains.
A symmetric chain is a path .
1 0 0 1 1 0 0 1
^^
1 0 0 1 1 0 0 1
^^^^
1 0 0 1 1 0 0 1
^^^^^^1 0 _ 1 1 0 0 _
0 0
1 0 _ 1 1 0 0 _
1 0 _ 1 1 0 0 _
11
1 0 _ 1 1 0 0 _
10
The BTK Partition
The Metric Properties
xy
𝐶 𝑥
𝐶 𝑦 1)
The Metric Properties
xy
𝐶 𝑥𝐶 𝑦 1)
2)
The Metric Properties
xy
𝐶 𝑥𝐶 𝑦 1)
2)
The Metric Properties – Proof
𝑥=𝑠0 𝑡 𝑦=𝑠1𝑡
.. ..
0𝑎1𝑏0𝑐+11𝑑 0𝑎1𝑏+10𝑐 1𝑑
𝑏>𝑐
0𝑎1𝑏−𝑐−11𝑐+10𝑐+11𝑑^ ^
𝑙𝑒𝑛𝑔𝑡 h=𝑎+𝑏−𝑐+𝑑0𝑎1𝑏+1−𝑐 1𝑐0𝑐1𝑑^^
𝑙𝑒𝑛𝑔𝑡 h=𝑎+𝑏−𝑐+𝑑+2
𝑖 𝑖
𝝍𝜓 : {0 ,1 }𝑛→𝐵𝑛= {𝑥∈ {0 ,1 }𝑛+1
:|𝑥|>𝑛 /2}
0
1
1
0
1
0
1
𝑥
𝐶 𝑥𝐶 𝑦
𝑦
𝜓 (𝑐1 )=𝑐1∘1𝜓 (𝑐2)=𝑐1∘0𝜓 (𝑐3 )=𝑐2∘1
𝜓 (𝑐4 )=𝑐2∘0𝜓 (𝑐5 )=𝑐3∘1𝜓 (𝑐6 )=𝑐3∘0
𝜓 (𝑐7 )=𝑐4∘1
𝑐1𝑐2
𝑐3𝑐4
𝑐5𝑐6
𝑐7
The Hamming Ball isbi-Lipschitz Transitive
Defintion. A metric space M is called k bi-Lipschitz transitive if for any there is a bijection such that , and
Example. is 1 bi-Lipschitz transitive.
𝑓 (𝑢)=𝑢+𝑥+𝑦
The Hamming Ball isbi-Lipschitz Transitive
Corollary. is 20 bi-Lipschitz transitive.
𝑓 (𝑢)=𝜓 (𝜓−1 (𝑢)+𝜓−1 (𝑥 )+𝜓−1 ( 𝑦 ) ) is convex in
Open Problems
The Constants
Are the 4,5 optimal ?
We know how to improve 4 to 3 at the expense of unbounded inverse.
Does the 20 in the corollary optimal ?
General Balanced Halfspaces
Switching to notation
𝜓 : {𝑥 :𝑥𝑛+1>0 }→ {𝑥 :𝑥1+⋯+𝑥𝑛+1>0}
Does the result hold for general balanced halfspaces ?𝜓𝑎 : {𝑥 :𝑥𝑛+1>0 }→ {𝑥 :𝑎1 𝑥1+⋯+𝑎𝑛+1𝑥𝑛+1>0 }𝑎1 ,… ,𝑎𝑛+1∈ R
One possible approach: generalize BTK chains.Applications to FPTAS for counting solutions to 0-1 knapsack problem [MorrisSinclair04].
Lower Bounds for Average Stretch
Exhibit a density half subset such that any bijection has super constant average stretch.
Conjecture: monotone noise-sensitive functions like Recursive-Majority-of-Three (highly fractal) should work.
We believe a random subset of density half has a constant average stretch.
Even average stretch 2.001 ! (for 2 take XOR).
Goldreich’s Question
Is it true that for any with density, say, half there exist and , both with density half, with bi-Lipschitz bijection between them ?
Thank youfor your attention!