Bi- Lipschitz Bijection between the Boolean Cube and the Hamming Ball
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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 }π π΅π={π₯β {0 ,1 }π+1:|π₯|>π /2}π : β
{π₯β {0 ,1 }π+1:π₯π+1=1}
Dictator
Majority
Strings with Hamming weight
k
n
0
k
The Problem
Open Problem [LovettViola11]. Prove that for any bijection ,
.
π (π₯ )={ π₯β0|π₯|>π/2ππππ (π₯ )β1ππ‘ hπππ€ππ π
Average stretch
The βNaΓ―veβ Upper Bound.
Motivation. Related to proving lower bounds for sampling a natural distribution by circuits.
Main Theorem
Theorem. such that
1)
of
π΅πof
Main Theorem
Theorem. such that
1)
2) is computable in DLOGTIME-uniform
(Majority is -reducible to )
3) is very local
15β€πππ π‘ (π (π₯ ) ,π (π¦ ) )
πππ π‘ (π₯ , π¦ )β€4
β πβ [π ]Prπ₯
[π (π₯ )πβ π₯π ]β€π ( 1βπ )
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
πΆ π₯
πΆ π¦ 1)
The Metric Properties
xy
πΆ π₯πΆ π¦ 1)
2)
The Metric Properties
xy
πΆ π₯πΆ π¦ 1)
2)
The Metric Properties β Proof
π₯=π 0 π‘ π¦=π 1π‘
.. ..
0π1π0π+11π 0π1π+10π 1π
π>π
0π1πβπβ11π+10π+11π^ ^
πππππ‘ h=π+πβπ+π0π1π+1βπ 1π0π1π^^
πππππ‘ h=π+πβπ+π+2
π π
ππ : {0 ,1 }πβπ΅π= {π₯β {0 ,1 }π+1
:|π₯|>π /2}
0
1
1
0
1
0
1
π₯
πΆ π₯πΆ π¦
π¦
π (π1 )=π1β1π (π2)=π1β0π (π3 )=π2β1
π (π4 )=π2β0π (π5 )=π3β1π (π6 )=π3β0
π (π7 )=π4β1
π1π2
π3π4
π5π6
π7
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.
π (π’)=π’+π₯+π¦
The Hamming Ball isbi-Lipschitz Transitive
Corollary. is 20 bi-Lipschitz transitive.
π (π’)=π (πβ1 (π’)+πβ1 (π₯ )+πβ1 ( π¦ ) ) 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
π : {π₯ :π₯π+1>0 }β {π₯ :π₯1+β―+π₯π+1>0}
Does the result hold for general balanced halfspaces ?ππ : {π₯ :π₯π+1>0 }β {π₯ :π1 π₯1+β―+ππ+1π₯π+1>0 }π1 ,β¦ ,ππ+1β 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).
Goldreichβs 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!