Bi- Lipschitz Bijection between the Boolean Cube and the Hamming Ball

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!