SLIDE 1 Lip ipsch chitz itz an and ou d oute ter bi r bi-Lip ipschi chitz tz ex exte tendabi ndabilit lity
Yury Makarychev, TTIC Sepideh Mahabadi, Columbia β TTIC Konstantin Makarychev, Northwestern Ilya Razenshteyn, Microsoft Research
University of Notre Dame, March 21, 2018
SLIDE 2 Plan
- The Lipschitz extendability problem
- Known results and open problems
- Vertex sparsifiers
- Connection between vertex sparsifiers and the
Lipschitz extendability problem
- Outer bi-Lipschitz extendability: definition, results,
and open problems
SLIDE 3 HahnβBanach Theorem
Let π be a normed space and π β π be its linear
- subspace. Every bounded linear map
π: π β β can be extended to α π βΆ π β β so that α π = π Is there an analogue for
- Lipschitz maps?
- maps into βπ or other normed spaces?
- H. Hahn
- S. Banah
SLIDE 4
Preliminaries
π: π β π is Lipschitz if for every π£, π€ β π ππ π π£ , π π€ β€ π·ππ(π£, π€) The Lipschitz constant π πππ of π is the minimum π· s.t. that the inequality holds. π is bi-Lipschitz if for some π·1, π·2 > 0, every π£, π€ β π π·1ππ π£, π€ β€ ππ π π£ , π π€ β€ π·2ππ(π£, π€) The bi-Lipschitz constant or distortion πΈ(π) of π is the minimum of π·2/π·1 s.t. that the inequality holds.
SLIDE 5 Preliminaries
βπ
π is βπ equipped with the β
π norm:
π¦ π = ΰ· π¦π π
1/π
π¦ β = max |π¦π|
SLIDE 6 McShaneβWhitney Theorem
βnon-linear HahnβBanachβ
Let (π, π) be a metric space and π΅ β π. Every Lipschitz map π: π΅ β β can be extended to α π βΆ π β β so that α π πππ = π πππ
β π α π(π¦)
SLIDE 7 Kirszbraun Theorem
Let π΅ β β2
π. Every Lipschitz map π: π΅ β β2 π can
be extended to α π βΆ β2
π β β2 π so that
α π πππ = π πππ βΆ
β2
π
β2
π
π
SLIDE 8
Lipschitz Extension Constant
Let π be a metric space and π be a normed space. ππ(π, π) is the min π· s.t. for every π΅ β π of size β€ π and π: π΅ β π there exists an extension α π: π β π with α π πππ β€ π· π πππ McShaneβWhitney: ππ π, β = 1 Kirszbraun: ππ β2, β2 = 1
SLIDE 9 Lipschitz Extension Constant
In general, ππ π, π > 1 E.g., π3 β1
3, β2 2 β₯
3
π΅ = π, π, π π = 1,0,0 π = 0,1,0 π = 0,0,1 π = (0,0,0) π(π) π(π) π(π) β2
2
2 2 2
SLIDE 10 π β π ππ(π, π) any β β or ββ 1 McShane, Whitney β34 β2 β β2 1
Kirszbraun β34
βπ β β2 π β€ 2 β€ π·π log π
1 πβ1 2
Marcus, Pisier β84 1 < π < 2 β₯ ππ log π log log π
1 πβ1 2 Johnson, Lindenstrauss
β84 any β β2 β€ π· log π JL β84 βπ β βπ 1 < π β€ 2 β€ π β€ 24
πβ1 πβ1
Naor, Peres, Schramm, Sheffield β04
π, π β (1, β) ππ β β as π β β
SLIDE 11
Extension Results
JohnsonβLindenstruassβSchechtman β86 ππ π, π β€ π· log π LeeβNaor β03 ππ π, π β€ π· log π log log π Best lower bounds are: ππ βΏ π log π Open Problem: what is the dependence of ππ on π ?
SLIDE 12 [JL β84] Technique for proving lower bounds on ππ π, π
Prove a lower bound for linear extensions Reinterpret it as a lower bound for Lipschitz extensions
- Linear extension (βprojectionβ)
constant is up to dim π
- [JL β84] ππ π, π β₯ π
log π log log π .
π
π
SLIDE 13 Open Problems
- Can the upper bound of βΌ log π / log log π be
improved?
- Are there any π and π with ππ(π, π) β«
log π ?
- Ball β92: Is it true that ππ β2, β1 β€ π· < β ?
SLIDE 14 Graph Sparsification
Given: a huge graph π» Goal: find a βsimplerβ graph πΌ βsimilarβ to π»
- compact representation
- algorithms work faster on the new graph
- can obtain better approximation results
SLIDE 15
Bottleneck & Routing Problems
SLIDE 16 Bottleneck Problem
- graph π» = (π, πΉ) with edge capacities ππ
- set of terminals π β π
For π β π, bkπ» π is the capacity of the minimum capacity cut in π» that separates π and π β π in π».
π’1 π’2 π’3 π’4
SLIDE 17 Bottleneck Problem
[Moitra β09] Graph πΌ = (π, πΉβ²) with capacities ππ
β² is
a vertex cut sparsifier for π» with distortion πΈ β₯ 1 if bkπ» π β€ bkπΌ π β€ πΈ β
bkπ» π βπ β π Given πΌ, can easily compute bottlenecks between terminals in the network!
π’1 π’2 π’3 π’4
SLIDE 18
Network Routing Problem
Routing problem: send a certain amount of data πππ from each terminal π’π to π’π so that the total amount sent over each edge π is at most its capacity ππ. [LM β10] A vertex flow sparsifier is an analogue of a vertex cut sparsifier for the network routing problem.
SLIDE 19 Known Results
Moitra β09 and Leigthon and Moitra β10
π = π
- π· log π / log log π
existential upper bound
- π· log2 π / log log π
algorithmic upper bound
lower bound for cut sparsifiers
lower bound for flow sparsifiers
Open Questions:
- π· log π / log log π
algorithmic upper bound?
- Better lower bounds?
- Better upper bounds?
SLIDE 20
Papers on Vertex Sparsification
Charikar, Leighton, Li and Moitra β10 Englert, Gupta, Krauthgamer, RΓ€cke, Talgam and Talwar β10 Makarychev and Makarychev β10
SLIDE 21 Main Results
- Define βMetric Sparsifiersβ
- Give π· log π / log log π algorithmic upper bound
[independently, CLLM β10, EGKRTT β10]
- Establish a direct connection between Vertex
Sparsifiers and Lipschitz Extendability
π
π
ππ£π’ = ππ(β1, β1)
π
π
ππππ₯ = ππ ββ, ββ β1 β¦ β1 ββ
SLIDE 22 Lower bounds via Lipschitz Extendability
Using lower bounds for βprojection constantsβ [GrΓΌnbaum β60], we get π
π
ππππ₯ β₯ ππ(ββ, β1) β₯ π· log π/log log π
Figiel, Johnson, and Schechtman β88 implies π
π
ππ£π’ = ππ(β1, β1) β₯ π· log π
log log π
SLIDE 23 Proof Idea: π
π
ππ£π’ β€ π
β‘ ππ(β1, β1)
Consider a game: π» and {ππ} are fixed Alice: defines πΌ by providing ππ
β²
Bob: presents π1, π β π1 and (π2, π β π2) bkπ» S1 β€ bkπΌ S1 and bkπΌ S2 β€ π
bkπ» S2 Alice wins Bob wins
yes no
SLIDE 24 Proof Idea: π
π
ππ£π’ β€ π
β‘ ππ(β1, β1)
Consider a game: π» and {ππ} are fixed Bob: distribution of π1, π β π1 and (π2, π β π2) Alice: defines πΌ by providing ππ
β²
π½bkπ» S1 β€ π½bkπΌ S1 π½bkπΌ S2 β€ π
π½bkπ» S2 Alice wins Bob wins
yes no
SLIDE 25
Distribution of cuts
Distribution π of cuts (π, π β π) on π defines a map π: π β π1(Ξ©, π): π π£ = α0, if π¦ β π 1, if π¦ β π
π’1 π’2 π’3 π’4
SLIDE 26
Distribution of cuts
Distribution π of cuts (π, π β π) on π defines a map π: π β π1(Ξ©, π): π π£ = α0, if π¦ β π 1, if π¦ β π
(0,0) (1,1) (1,0) (0,1)
SLIDE 27 Distribution of cuts
π π£ = α0, if π¦ β π 1, if π¦ β π Pr π£, π€ are separated by π = π π£ β π π€
1
π½ bkπΌ π = ΰ·
π£,π€βπ
πβ² π£, π€ β
π π£ β π π€
1
π½ bkπ» π = min
α π
ΰ·
π£,π€βπ
πβ² π£, π€ β
α π π£ β α π π€
1
SLIDE 28 Bob: gives maps π
1: π β π1 and π 2: π β π1
Need: bkπ» ΰ·© π
1 β€ bkπΌ π 1
bkπΌ π
2 β€ π
bkπ» ΰ·©
π
2
π1 π1 π
1
π
2
Relate π
1π 2 β1and ΰ·©
π
1 ΰ·©
π
2 β1
π
1π 2 β1
SLIDE 29 Bob: gives maps π
1: π β π1 and π 2: π β π1
Need: bkπ» ΰ·© π
1 β€ bkπΌ π 1
bkπΌ π
2 β€ π
bkπ» ΰ·©
π
2
π1 π1 ΰ·© π
1
ΰ·© π
2
ΰ·© π
1 ΰ·©
π
2 β1
Relate π
1π 2 β1and ΰ·©
π
1 ΰ·©
π
2 β1
SLIDE 30 Ballβs Open Problem & Sparsification
[MM β10] π
π
ππ£π’ = ππ β1, β1 β€ ππ β2, β1 β
π· log π log log π
here, π· log π log log π is the distortion of the FrΓ©chet embedding of β1 into β2 by Arora, Lee, Naor β07. If ππ β2, β1 β€ π·Ball then π
π
ππ£π’ β€ π·β² log π log log π
SLIDE 31
Outer bi-Lipschitz extendibility
SLIDE 32 1
Bi-Lipschitz Kirszbraun Theorem?
Let π΅ β β2
π. Can we extend a bi-Lipschitz map π: π΅ β β2 π
to a bi-Lipschitz map α π: β2
π β β2 π ?
No!
- β2 β β. Assume β β β2. Extend f = ππβ ?
There is even no injective extension of π to β2.
- β β β. π maps 0, 1, 2 to 0, 2, 1, respectively. There is
even no continuous one-to-one extension. 2 1 2
SLIDE 33 Bi-Lipschitz Kirszbraun Theorem?
Can overcome these obstacles by using extra dimensions!
- β2 β β. Assume β β β2. Extend f = ππβ ?
Assume that target β β β2. Then α π = ππβ2ββ2
- β β β. π maps 0, 1, 2 to 0, 2, 1, respectively.
1 2
SLIDE 34 Outer bi-Lipschitz extension
Given π΅ β β2
π and bi-Lipschitz π: π΅ β β2 π .
Assume β2
π β β2 π.
α π: β2
π β β2 π is an outer bi-Lipschitz extension of π if
α π π = π(π) for every π β π΅
and α
π is bi-Lipschitz.
[MMMR β18] For every bi-Lipschitz π: π΅ β β2
π, there
exists an outer bi-Lipschitz extension with πΈ α π β€ 3πΈ π .
SLIDE 35
Outer bi-Lipschitz extension
Near isometric case. What happens when πΈ(π) = 1 + π ? If π = 0, i.e., π is an isometric map, there is an isometric extension α π Is there a bi-Lipschitz extension with πΈ α π = 1 + π(1) as π β 0 ? For π: π΅ β β, π΅ β β, yes! There is an extension with πΈ( α π) = 1 + 1 log2 1/π The bound is tight.
SLIDE 36 Outer bi-Lipschitz extension
Near isometric case. What happens when πΈ(π) = 1 + π ? If π = 0, i.e., π is an isometric map, there is an isometric extension α π Is there a bi-Lipschitz extension with πΈ α π = 1 + π(1) as π β 0 ? For a one-point extension of π: π΅ β β2
π
πΈ( α π) = 1 + π· π The bound is tight.
SLIDE 37 Summary
Characterized the optimal distortion of cut and flow vertex sparsifiers in terms of Lipschitz extension constants.
- Find ππ β1, β1 and ππ(ββ, ββ β1 β¦ β1 ββ)
- Is ππ β2, β1 < β ?
Defined outer bi-Lipschitz extension and proved an analogue of Kirzsbraun theorem for it. Partial results for nearly isometric maps.
- Understand the nearly isometric case.
Applications to a prioritized dimension reduction.