tangles faster computing height-optimal · 2019. 9. 18. · tangles faster oksanafirman...

Computing Height-OptimalTangles Faster

Oksana FirmanPhilipp Kindermann

Alexander WolffJohannes Zink

Julius-Maximilians-Universitat Wurzburg,

Germany

Alexander RavskyPidstryhach Institute for Applied Problems

of Mechanics and Mathematics,National Academy of Sciences of Ukraine,

Lviv, Ukraine

PK AW JZ OF Wu AR Lviv

Introduction

Given a set of ny -monotone wires

Introduction


1 ≤ i < j ≤ n

swap i j

i j

Introduction


1 ≤ i < j ≤ n

swap i j

disjoint swaps

Introduction


1 ≤ i < j ≤ n

swap i j

disjoint swaps

adjacentpermutations

Introduction


1 ≤ i < j ≤ n

swap i j

disjoint swaps

multiple swaps


Introduction


1 ≤ i < j ≤ n

swap i j

disjoint swaps

multiple swaps

tangle T ofheight h(T )

π1

π2

π3

π6


π4

π5

Introduction


1 ≤ i < j ≤ n

1 2 · · · n

swap i j

disjoint swaps

multiple swaps


π1

π2

π3

π6

π′1

π′2

π′3

π′5


π4

π5

π′4

1 2 · · · n

Introduction


. . . and given a list ofswaps L

1 ≤ i < j ≤ n

1 2 · · · n

swap i j

disjoint swaps

multiple swaps


π1

π2

π3

π6


π4

π5

Introduction


as a multiset (`i j)


1 ≤ i < j ≤ n

1 2 · · · n

swap i j

disjoint swaps

multiple swaps


π1

π2

π3

π6

1

3112


π4

π5

Introduction




1 ≤ i < j ≤ n

1 2 · · · n

swap i j

disjoint swaps

multiple swaps


π1

π2

π3

π6

1

3112

Tangle T (L) realizes list L.


π4

π5

Introduction




1 ≤ i < j ≤ n

1 2 · · · n

swap i j

disjoint swaps

multiple swaps


π1

π2

π3

π6

1

3112



not feasible

π4

π5

Introduction




1 ≤ i < j ≤ n

1 2 · · · n

swap i j

disjoint swaps

multiple swaps


π1

π2

π3

π6

A tangle T (L) is height-optimal if it has the minimum heightamong all tangles realizing the list L.

1

3112



π4

π5

Related Work• Olszewski et al. Visualizing the template

of a chaotic attractor.GD 2018



mlist



mlist

Algorithm for findingoptimal tangles



mlist


Complexity ??



mlist

• Wang. Novel routing schemes for IClayout part I: Two-layer channel routing.DAC 1991

Given:initial andfinal permutations


Complexity ??



mlist

• Wang. Novel routing schemes for IClayout part I: Two-layer channel routing.DAC 1991

Given:initial andfinal permutations

Objective: minimizethe number of bends

• Bereg et al. Drawing Permutations with Few Corners.GD 2013


Complexity ??

Overview

• Complexity:NP-hardness byreduction from3-Partition.

• New algorithm: using dynamic programming;asymptotically faster than [Olszewski et al., GD’18].

O(ϕ2|L|

5|L|/nn)

O(( 2|L|

n2 + 1) n2

2 ϕnn)

• Experiments: comparison with [Olszewski et al., GD’18]

Complexity

Tangle-Height Minimization is NP-hard.Theorem.

Complexity

Reduction from 3-Partition


Proof.

Complexity



Proof.3-Partition

Given: Multiset A of 3m positive integers.

a1 a2 a3 a3m−2 a3m· · · a3m−1

Complexity



Proof.3-Partition

∑1 = B · · · ∑

m = B

Given: Multiset A of 3m positive integers.Can A be partitioned into m groups ofthree elements s.t. each group sums up tothe same value B?

Question:

a1 a2 a3 a3m−2 a3m· · ·∑2 = B

a3m−1

Complexity



Proof.3-Partition

B4 < ai <

B2

B is poly in m

∑1 = B · · · ∑

m = B


Question:

a1 a2 a3 a3m−2 a3m· · ·∑2 = B

a3m−1

Complexity



Proof. B4 < ai <

B2

B is poly in m

∑1 = B · · · ∑

m = B


Given: Instance A of 3-Partition.

Question:

a1 a2 a3 a3m−2 a3m· · ·∑2 = B

a3m−1

Complexity



Proof. B4 < ai <

B2

B is poly in m

∑1 = B · · · ∑

m = B


Given: Instance A of 3-Partition.Task: Construct L s.t. there is T realizing L with height at

most H = 2m3(∑

A)+7m2 iff A is a yes-instance.

Question:

a1 a2 a3 a3m−2 a3m· · ·∑2 = B

a3m−1

Complexity



Proof.


most H = 2m3(∑


∑A

∑1 = B · · · ∑

m = B

a1 a2 a3 a3m−2 a3m−1 a3m· · ·∑2 = B

Complexity



Proof.


most H = 2m3(∑


∑A

∑1 = B · · · ∑

m = B

a1 a2 a3 a3m−2 a3m−1 a3m· · ·∑2 = B +1

Complexity



Proof.


most H = 2m3(∑


∑A +1

∑1 = B · · · ∑

m = B

a1 a2 a3 a3m−2 a3m−1 a3m· · ·∑2 = B +1

Complexity



Proof.

Given: Instance A of 3-Partition.

∑A +1

Task: construct L s.t. there is T realizing L with height atmost H = 2m3(

∑A )+7m2 iff A is a yes-instance+1

∑1 = B · · · ∑

m = B

a1 a2 a3 a3m−2 a3m−1 a3m· · ·∑2 = B +1

Transforming the Instance A into a List Lω ω′

ω ω′

Transforming the Instance A into a List Lω ω′

ω ω′

2m swaps

Transforming the Instance A into a List L

α1

ω ω′

ω ω′

α1 α′1

α′1


α1

ω ω′

ω ω′

α1 α′1

α′1

Ma1M = 2m3


α1

ω ω′

ω ω′

α1 α′1

α′1

Ma1M = 2m3

What is not possible?

split


α1

ω ω′

ω ω′

α1 α′1

α′1

α2 α′2

α2 α′2

Ma1M = 2m3

Ma2


α1

ω ω′

ω ω′

α1 α′1

α′1

α2 α′2

α2 α′2

Ma1M = 2m3

Ma2

What is not possible?

put it on the same levelwith other α-α′ swaps


α1

ω ω′

ω ω′

α6 α1· · · α′1 α′6

α6· · · α′1 α′6· · ·

· · ·

Ma1

Ma4

Ma5

Ma2

Ma3

Ma6

M = 2m3

Making Sure That the “Pockets” Can’t Be Squeezedω ω′

ω ω′δ2β1δ1 β2

β2δ2 β1δ1



β2δ2 β1δ1

β2δ2 β1δ1



β2δ2 β1δ1

β2δ2 β1δ1

β2 δ2β1 δ1



β2δ2 β1δ1


ω ω′γ2γ1 δ2β1δ1 β2

β2γ2γ1 δ2 β1δ1

2MB

MB

M = 2m3


ω ω′

γ′1 γ′2δ′1 β′2δ′2β′1

β′1 γ′1γ′2δ′1β′2δ

′2

2MB

MB

γ2γ1 δ2β1δ1 β2


2MB

MB

M = 2m3

Making Sure That the “Pockets” Can’t Be Squeezed

α1

ω ω′

ω ω′

γ′1 γ′2δ′1 β′2δ′2β′1

β′1 γ′1γ′2δ′1β′2δ

′2

α6 α1· · · α′1 α′6

α6· · · α′1 α′6· · ·

2MB

MB

· · ·



Ma1

Ma4

Ma5

Ma2

Ma3

Ma6

2MB

MB

M = 2m3

Proof of Correctness

α1

ω ω′

ω ω′

γ′1 γ′2δ′1 β′2δ′2β′1

β′1 γ′1γ′2δ′1β′2δ

′2

α6 α1· · · α′1 α′6

α6· · · α′1 α′6· · ·

2MB

MB

· · ·



Ma1

Ma4

Ma5

Ma2

Ma3

Ma6

2MB

MB

M = 2m3

A is a yes-instance

H = 2m3(∑

A) + 7m2

is the maximum allowedheight for the reduction


α1

ω ω′

ω ω′

γ′1 γ′2δ′1 β′2δ′2β′1

β′1 γ′1γ′2δ′1β′2δ

′2

α6 α1· · · α′1 α′6

α6· · · α′1 α′6· · ·

2MB

MB

· · ·



Ma1

Ma4

Ma5

Ma2

Ma3

Ma6

2MB

MB

M = 2m3

A is a yes-instance

by construction

H = 2m3(∑

A) + 7m2



α1

ω ω′

ω ω′

γ′1 γ′2δ′1 β′2δ′2β′1

β′1 γ′1γ′2δ′1β′2δ

′2

α6 α1· · · α′1 α′6

α6· · · α′1 α′6· · ·

2MB

MB

· · ·



Ma1

Ma4

Ma5

Ma2

Ma3

Ma6

2MB

MB

M = 2m3

A is a yes-instance

by construction

height ≤ H

H = 2m3(∑

A) + 7m2



α1

ω ω′

ω ω′

γ′1 γ′2δ′1 β′2δ′2β′1

β′1 γ′1γ′2δ′1β′2δ

′2

α6 α1· · · α′1 α′6

α6· · · α′1 α′6· · ·

2MB

MB

· · ·



Ma1

Ma4

Ma5

Ma2

Ma3

Ma6

2MB

MB

M = 2m3

A is a yes-instanceno

H = 2m3(∑

A) + 7m2



α1

ω ω′

ω ω′

γ′1 γ′2δ′1 β′2δ′2β′1

β′1 γ′1γ′2δ′1β′2δ

′2

α6 α1· · · α′1 α′6

α6· · · α′1 α′6· · ·

2MB

MB

· · ·



Ma1

Ma4

Ma5

Ma2

Ma3

Ma6

2MB

MB

M = 2m3


H = 2m3(∑

A) + 7m2


minimum height2m3(

∑A + 1)


α1

ω ω′

ω ω′

γ′1 γ′2δ′1 β′2δ′2β′1

β′1 γ′1γ′2δ′1β′2δ

′2

α6 α1· · · α′1 α′6

α6· · · α′1 α′6· · ·

2MB

MB

· · ·



Ma1

Ma4

Ma5

Ma2

Ma3

Ma6

2MB

MB

M = 2m3


height > H

H = 2m3(∑

A) + 7m2


minimum height2m3(

∑A + 1)


α1

ω ω′

ω ω′

γ′1 γ′2δ′1 β′2δ′2β′1

β′1 γ′1γ′2δ′1β′2δ

′2

α6 α1· · · α′1 α′6

α6· · · α′1 α′6· · ·

2MB

MB

· · ·



Ma1

Ma4

Ma5

Ma2

Ma3

Ma6

2MB

MB

M = 2m3


height > H

H = 2m3(∑

A) + 7m2


minimum height2m3(

∑A + 1)


X

Overview



O(ϕ2|L|

5|L|/nn)

O(( 2|L|

n2 + 1) n2

2 ϕnn)


Improving Exact Algorithms

Simple lists

Tangle-Height Minimization can be solved in . . .

General lists


Simple lists


General lists

[Olszewski et al., GD’18]

2O(n2)

n – number of wires


Simple lists


General lists


2O(n2) 2O(n log n)

our runtime

n – number of wires


Simple lists


General lists


2O(n2) 2O(n log n)

our runtime


O(ϕ2|L|

5|L|/nn)

n – number of wires– length of the list L (=

∑`i j )

– golden ratio (≈ 1.618)|L|ϕ


Simple lists


General lists


2O(n2) 2O(n log n)

our runtime


O(ϕ2|L|

5|L|/nn)

O(( 2|L|

n2 + 1) n2

2 ϕnn)our runtime


∑`i j )



Simple lists


General lists


2O(n2) 2O(n log n)

our runtime


O(ϕ2|L|

5|L|/nn)

O(( 2|L|

n2 + 1) n2

2 ϕnn)our runtime

polynomial in |L|


∑`i j )


polynomial in |L|for fixed n

Dynamic Programming Algorithm

O(( 2|L|

n2 + 1)n2/2

ϕnn)

Let L = (`i j) be the given list of swaps.


O(( 2|L|

n2 + 1)n2/2

ϕnn)


λ = # of distinct sublists of L.L′ is a sublist of L if`′i j ≤ `i j


O(( 2|L|

n2 + 1)n2/2

ϕnn)


λ = # of distinct sublists of L.

Consider them in order of increasing length.L′ is a sublist of L if`′i j ≤ `i j


O(( 2|L|

n2 + 1)n2/2

ϕnn)



Consider them in order of increasing length.

Let L′ be the next list to consider.

L′ is a sublist of L if`′i j ≤ `i j


O(( 2|L|

n2 + 1)n2/2

ϕnn)





Check its consistency.



O(( 2|L|

n2 + 1)n2/2

ϕnn)





Check its consistency.i



O(( 2|L|

n2 + 1)n2/2

ϕnn)






for each wire i :// find a position where it is after applying L′



O(( 2|L|

n2 + 1)n2/2

ϕnn)







i 7→ i + |{j : j > i and `′i j is odd}| − |{j : j < i and `′i j is odd}|



O(( 2|L|

n2 + 1)n2/2

ϕnn)







i 7→ i + |{j : j > i and `′i j is odd}| − |{j : j < i and `′i j is odd}|check whether the map is indeed a permutation



O(( 2|L|

n2 + 1)n2/2

ϕnn)





Check its consistency.

idn L′


Compute the final permutation idn L′.


O(( 2|L|

n2 + 1)n2/2

ϕnn)





Check its consistency. π1π2...

πhidn L′

Choose the shortest tangle T (L′′).



πh and idn L′ are adjacentL′′∪ add. swaps = L′


O(( 2|L|

n2 + 1)n2/2

ϕnn)






πhAdd the final permutation to its end.idn L′





O(( 2|L|

n2 + 1)n2/2

ϕnn)







Running time

O(λ · (Fn+1 − 1) · n) ≤





O(( 2|L|

n2 + 1)n2/2

ϕnn)







Running time

O(λ · (Fn+1 − 1) · n) ≤


Fn is the n-th Fibonacci number




O(( 2|L|

n2 + 1)n2/2

ϕnn)







Running time

O(λ · (Fn+1 − 1) · n) ≤




λ =∏i<j

(`i j + 1) ≤(

2|L|n2

+ 1)n2/2

Fn ∈ O(ϕn)

Overview



O(ϕ2|L|

5|L|/nn)

O(( 2|L|

n2 + 1) n2

2 ϕnn)



O(ϕ2|L|

5|L|/nn)

O(( 2|L|

n2 + 1) n2

2 ϕnn)Our algorithm

run

tim

e[s

ec]

length |L| of the list L

0 10 20 30 40 50

0.0001

0.01

1

100

3600

Experiments

0 10 20 30 40 0 10 20 30 40

Open Problems

Problem 1

Is it NP-hard to test the feasibility of a given (non-simple) list?

Open Problems

Problem 1


Problem 2

Can we decide a feasibility of a list faster than finding itsoptimal realization?

Open Problems

Problem 1


Problem 2


Problem 3A list (ì j) is non-separableif ∀i<k<j :

(ìk = `kj = 0 implies ì j = 0

).

i k j

Open Problems

Problem 1


Problem 2


Problem 3A list (`i j) is non-separableif ∀i<k<j :


).

i k j

necessary

Open Problems

Problem 1


Problem 2


Problem 3

For lists where all entries are even, is this sufficient?

A list (`i j) is non-separableif ∀i<k<j :


).

i k j

necessary

Open Problems

Problem 1


Problem 2


Problem 3

For lists where all entries are even, is this sufficient?

A list (`i j) is non-separableif ∀i<k<j :


).

i k j

necessary

Thank you!

tangles faster computing height-optimal · 2019. 9. 18. · tangles faster oksanafirman...

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