Computing Height-Optimal Tangles Faster Oksana Firman W u PK AW - - PowerPoint PPT Presentation
Computing Height-Optimal Tangles Faster Oksana Firman W u PK AW JZ OF AR Lviv Philipp Kindermann Alexander Wolff Johannes Zink Julius-Maximilians-Universit at W urzburg, Germany Alexander Ravsky Pidstryhach Institute for
Oksana Firman Philipp Kindermann Alexander Wolff Johannes Zink
Julius-Maximilians-Universit¨ at W¨ urzburg, Germany
Alexander Ravsky
Pidstryhach Institute for Applied Problems
National Academy of Sciences of Ukraine, Lviv, Ukraine
PK AW JZ OF W¨ u AR Lviv
Given a set of n y-monotone wires
Given a set of n y-monotone wires
1 ≤ i < j ≤ n
swap ij i j
Given a set of n y-monotone wires
1 ≤ i < j ≤ n
swap ij disjoint swaps
Given a set of n y-monotone wires
1 ≤ i < j ≤ n
swap ij disjoint swaps adjacent permutations
Given a set of n y-monotone wires
1 ≤ i < j ≤ n
swap ij disjoint swaps multiple swaps adjacent permutations
Given a set of n y-monotone wires
1 ≤ i < j ≤ n
swap ij disjoint swaps multiple swaps tangle T of height h(T) π1 π2 π3 π6 adjacent permutations π4 π5
Given a set of n y-monotone wires
1 ≤ i < j ≤ n
1 2 · · · n swap ij disjoint swaps multiple swaps tangle T of height h(T) π1 π2 π3 π6 π′
1
π′
2
π′
3
π′
5
adjacent permutations π4 π5 π′
4
1 2 · · · n
Given a set of n y-monotone wires
. . . and given a list of swaps L
1 ≤ i < j ≤ n
1 2 · · · n swap ij disjoint swaps multiple swaps tangle T of height h(T) π1 π2 π3 π6 adjacent permutations π4 π5
Given a set of n y-monotone wires
as a multiset (ℓij) . . . and given a list of swaps L
1 ≤ i < j ≤ n
1 2 · · · n swap ij disjoint swaps multiple swaps tangle T of height h(T) π1 π2 π3 π6 1 3 1 1 2 adjacent permutations π4 π5
Given a set of n y-monotone wires
as a multiset (ℓij) . . . and given a list of swaps L
1 ≤ i < j ≤ n
1 2 · · · n swap ij disjoint swaps multiple swaps tangle T of height h(T) π1 π2 π3 π6 1 3 1 1 2 Tangle T(L) realizes list L. adjacent permutations π4 π5
Given a set of n y-monotone wires
as a multiset (ℓij) . . . and given a list of swaps L
1 ≤ i < j ≤ n
1 2 · · · n swap ij disjoint swaps multiple swaps tangle T of height h(T) π1 π2 π3 π6 1 3 1 1 2 Tangle T(L) realizes list L. adjacent permutations π4 π5
Given a set of n y-monotone wires
as a multiset (ℓij) . . . and given a list of swaps L
1 ≤ i < j ≤ n
1 2 · · · n swap ij disjoint swaps multiple swaps tangle T of height h(T) π1 π2 π3 π6 1 3 1 1 2 Tangle T(L) realizes list L. adjacent permutations not feasible π4 π5
Given a set of n y-monotone wires
as a multiset (ℓij) . . . and given a list of swaps L
1 ≤ i < j ≤ n
1 2 · · · n swap ij disjoint swaps multiple swaps tangle T of height h(T) π1 π2 π3 π6 A tangle T(L) is height-optimal if it has the minimum height among all tangles realizing the list L. 1 3 1 1 2 Tangle T(L) realizes list L. adjacent permutations π4 π5
GD 2018
GD 2018
GD 2018
Algorithm for finding
GD 2018
Algorithm for finding
Complexity ?
GD 2018
layout part I: Two-layer channel routing. DAC 1991 Given: initial and
final permutations
Algorithm for finding
Complexity ?
GD 2018
layout part I: Two-layer channel routing. DAC 1991 Given: initial and
final permutations
Objective: minimize
the number of bends
GD 2013 Algorithm for finding
Complexity ?
NP-hardness by reduction from 3-Partition.
asymptotically faster than [Olszewski et al., GD’18]. O
5|L|/n n
2|L|
n2 + 1
n2
2 ϕnn
Tangle-Height Minimization is NP-hard.
Theorem.
Reduction from 3-Partition Tangle-Height Minimization is NP-hard.
Theorem.
Proof.
Reduction from 3-Partition Tangle-Height Minimization is NP-hard.
Theorem.
Proof. 3-Partition Given: Multiset A of 3m positive integers. a1 a2 a3 a3m−2 a3m · · · a3m−1
Reduction from 3-Partition Tangle-Height Minimization is NP-hard.
Theorem.
Proof. 3-Partition
· · ·
Given: Multiset A of 3m positive integers. Can A be partitioned into m groups of three elements s.t. each group sums up to the same value B? Question: a1 a2 a3 a3m−2 a3m · · ·
a3m−1
Reduction from 3-Partition Tangle-Height Minimization is NP-hard.
Theorem.
Proof. 3-Partition
B 4 < ai < B 2
B is poly in m
· · ·
Given: Multiset A of 3m positive integers. Can A be partitioned into m groups of three elements s.t. each group sums up to the same value B? Question: a1 a2 a3 a3m−2 a3m · · ·
a3m−1
Reduction from 3-Partition Tangle-Height Minimization is NP-hard.
Theorem.
Proof.
B 4 < ai < B 2
B is poly in m
· · ·
Given: Multiset A of 3m positive integers. Can A be partitioned into m groups of three elements s.t. each group sums up to the same value B? Given: Instance A of 3-Partition. Question: a1 a2 a3 a3m−2 a3m · · ·
a3m−1
Reduction from 3-Partition Tangle-Height Minimization is NP-hard.
Theorem.
Proof.
B 4 < ai < B 2
B is poly in m
· · ·
Given: Multiset A of 3m positive integers. Can A be partitioned into m groups of three elements s.t. each group sums up to the same value 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 · · ·
a3m−1
Reduction from 3-Partition Tangle-Height Minimization is NP-hard.
Theorem.
Proof. 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. A
· · ·
a1 a2 a3 a3m−2 a3m−1 a3m · · ·
Reduction from 3-Partition Tangle-Height Minimization is NP-hard.
Theorem.
Proof. 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. A
· · ·
a1 a2 a3 a3m−2 a3m−1 a3m · · ·
+1
Reduction from 3-Partition Tangle-Height Minimization is NP-hard.
Theorem.
Proof. 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. A +1
· · ·
a1 a2 a3 a3m−2 a3m−1 a3m · · ·
+1
Reduction from 3-Partition Tangle-Height Minimization is NP-hard.
Theorem.
Proof. Given: Instance A of 3-Partition. A +1 Task: construct L s.t. there is T realizing L with height at most H = 2m3( A )+7m2 iff A is a yes-instance +1
· · ·
a1 a2 a3 a3m−2 a3m−1 a3m · · ·
+1
Transforming the Instance A into a List L
ω ω′ ω ω′
Transforming the Instance A into a List L
ω ω′ ω ω′
2m swaps
Transforming the Instance A into a List L
α1 ω ω′ ω ω′ α1 α′ 1 α′ 1
Transforming the Instance A into a List L
α1 ω ω′ ω ω′ α1 α′ 1 α′ 1
Ma1
M = 2m3
Transforming the Instance A into a List L
α1 ω ω′ ω ω′ α1 α′ 1 α′ 1
Ma1
M = 2m3
Transforming the Instance A into a List L
α1 ω ω′ ω ω′ α1 α′ 1 α′ 1
Ma1
M = 2m3
Transforming the Instance A into a List L
α1 ω ω′ ω ω′ α1 α′ 1 α′ 1
Ma1
M = 2m3
Transforming the Instance A into a List L
α1 ω ω′ ω ω′ α1 α′ 1 α′ 1
Ma1
M = 2m3 What is not possible? split
Transforming the Instance A into a List L
α1 ω ω′ ω ω′ α1 α′ 1 α′ 1 α2 α′ 2 α2 α′ 2
Ma1
M = 2m3
Ma2
Transforming the Instance A into a List L
α1 ω ω′ ω ω′ α1 α′ 1 α′ 1 α2 α′ 2 α2 α′ 2
Ma1
M = 2m3
Ma2
What is not possible? put it on the same level with other α-α′ swaps
Transforming the Instance A into a List L
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 Squeezed
ω ω′ ω ω′ δ2 β1δ1 β2 β2 δ2 β1 δ1
Making Sure That the “Pockets” Can’t Be Squeezed
ω ω′ ω ω′ δ2 β1δ1 β2 β2 δ2 β1 δ1
β2 δ2 β1 δ1
Making Sure That the “Pockets” Can’t Be Squeezed
ω ω′ ω ω′ δ2 β1δ1 β2 β2 δ2 β1 δ1
β2 δ2 β1 δ1 β2 δ2 β1 δ1
Making Sure That the “Pockets” Can’t Be Squeezed
ω ω′ ω ω′ δ2 β1δ1 β2 β2 δ2 β1 δ1
Making Sure That the “Pockets” Can’t Be Squeezed
ω ω′ ω ω′ δ2 β1δ1 β2 β2 δ2 β1 δ1
Making Sure That the “Pockets” Can’t Be Squeezed
ω ω′ ω ω′ δ2 β1δ1 β2 β2 δ2 β1 δ1
Making Sure That the “Pockets” Can’t Be Squeezed
ω ω′ ω ω′ δ2 β1δ1 β2 β2 δ2 β1 δ1
Making Sure That the “Pockets” Can’t Be Squeezed
ω ω′ ω ω′ δ2 β1δ1 β2 β2 δ2 β1 δ1
Making Sure That the “Pockets” Can’t Be Squeezed
ω ω′ ω ω′ δ2 β1δ1 β2 β2 δ2 β1 δ1
Making Sure That the “Pockets” Can’t Be Squeezed
ω ω′ ω ω′ γ2 γ1 δ2 β1δ1 β2 β2 γ2 γ1 δ2 β1 δ1
2 M B M B
M = 2m3
Making Sure That the “Pockets” Can’t Be Squeezed
ω ω′ ω ω′ γ′ 1 γ′ 2 δ′ 1 β′ 2 δ′ 2 β′ 1 β′ 1 γ′ 1γ′ 2 δ′ 1 β′ 2δ′ 2
2 M B M B
γ2 γ1 δ2 β1δ1 β2 β2 γ2 γ1 δ2 β1 δ1
2 M B M B
M = 2m3
Making Sure That the “Pockets” Can’t Be Squeezed
α1 ω ω′ ω ω′ γ′ 1 γ′ 2 δ′ 1 β′ 2 δ′ 2 β′ 1 β′ 1 γ′ 1γ′ 2 δ′ 1 β′ 2δ′ 2 α6 α1 · · · α′ 1 α′ 6 α6· · · α′ 1 α′ 6 · · ·
2 M B M B
· · · γ2 γ1 δ2 β1δ1 β2 β2 γ2 γ1 δ2 β1 δ1
Ma1 Ma4 Ma5 Ma2 Ma3 Ma6
2 M B M B
M = 2m3
α1 ω ω′ ω ω′ γ′ 1 γ′ 2 δ′ 1 β′ 2 δ′ 2 β′ 1 β′ 1 γ′ 1γ′ 2 δ′ 1 β′ 2δ′ 2 α6 α1 · · · α′ 1 α′ 6 α6· · · α′ 1 α′ 6 · · ·
2 M B M B
· · · γ2 γ1 δ2 β1δ1 β2 β2 γ2 γ1 δ2 β1 δ1
Ma1 Ma4 Ma5 Ma2 Ma3 Ma6
2 M B M B
M = 2m3 A is a yes-instance
H = 2m3( A) + 7m2 is the maximum allowed height for the reduction
α1 ω ω′ ω ω′ γ′ 1 γ′ 2 δ′ 1 β′ 2 δ′ 2 β′ 1 β′ 1 γ′ 1γ′ 2 δ′ 1 β′ 2δ′ 2 α6 α1 · · · α′ 1 α′ 6 α6· · · α′ 1 α′ 6 · · ·
2 M B M B
· · · γ2 γ1 δ2 β1δ1 β2 β2 γ2 γ1 δ2 β1 δ1
Ma1 Ma4 Ma5 Ma2 Ma3 Ma6
2 M B M B
M = 2m3 A is a yes-instance by construction
H = 2m3( A) + 7m2 is the maximum allowed height for the reduction
α1 ω ω′ ω ω′ γ′ 1 γ′ 2 δ′ 1 β′ 2 δ′ 2 β′ 1 β′ 1 γ′ 1γ′ 2 δ′ 1 β′ 2δ′ 2 α6 α1 · · · α′ 1 α′ 6 α6· · · α′ 1 α′ 6 · · ·
2 M B M B
· · · γ2 γ1 δ2 β1δ1 β2 β2 γ2 γ1 δ2 β1 δ1
Ma1 Ma4 Ma5 Ma2 Ma3 Ma6
2 M B M B
M = 2m3 A is a yes-instance by construction height ≤ H
H = 2m3( A) + 7m2 is the maximum allowed height for the reduction
α1 ω ω′ ω ω′ γ′ 1 γ′ 2 δ′ 1 β′ 2 δ′ 2 β′ 1 β′ 1 γ′ 1γ′ 2 δ′ 1 β′ 2δ′ 2 α6 α1 · · · α′ 1 α′ 6 α6· · · α′ 1 α′ 6 · · ·
2 M B M B
· · · γ2 γ1 δ2 β1δ1 β2 β2 γ2 γ1 δ2 β1 δ1
Ma1 Ma4 Ma5 Ma2 Ma3 Ma6
2 M B M B
M = 2m3 A is a yes-instance no
H = 2m3( A) + 7m2 is the maximum allowed height for the reduction
α1 ω ω′ ω ω′ γ′ 1 γ′ 2 δ′ 1 β′ 2 δ′ 2 β′ 1 β′ 1 γ′ 1γ′ 2 δ′ 1 β′ 2δ′ 2 α6 α1 · · · α′ 1 α′ 6 α6· · · α′ 1 α′ 6 · · ·
2 M B M B
· · · γ2 γ1 δ2 β1δ1 β2 β2 γ2 γ1 δ2 β1 δ1
Ma1 Ma4 Ma5 Ma2 Ma3 Ma6
2 M B M B
M = 2m3 A is a yes-instance no
H = 2m3( A) + 7m2 is the maximum allowed height for the reduction
minimum height 2m3( A + 1)
α1 ω ω′ ω ω′ γ′ 1 γ′ 2 δ′ 1 β′ 2 δ′ 2 β′ 1 β′ 1 γ′ 1γ′ 2 δ′ 1 β′ 2δ′ 2 α6 α1 · · · α′ 1 α′ 6 α6· · · α′ 1 α′ 6 · · ·
2 M B M B
· · · γ2 γ1 δ2 β1δ1 β2 β2 γ2 γ1 δ2 β1 δ1
Ma1 Ma4 Ma5 Ma2 Ma3 Ma6
2 M B M B
M = 2m3 A is a yes-instance no height > H
H = 2m3( A) + 7m2 is the maximum allowed height for the reduction
minimum height 2m3( A + 1)
α1 ω ω′ ω ω′ γ′ 1 γ′ 2 δ′ 1 β′ 2 δ′ 2 β′ 1 β′ 1 γ′ 1γ′ 2 δ′ 1 β′ 2δ′ 2 α6 α1 · · · α′ 1 α′ 6 α6· · · α′ 1 α′ 6 · · ·
2 M B M B
· · · γ2 γ1 δ2 β1δ1 β2 β2 γ2 γ1 δ2 β1 δ1
Ma1 Ma4 Ma5 Ma2 Ma3 Ma6
2 M B M B
M = 2m3 A is a yes-instance no height > H
H = 2m3( A) + 7m2 is the maximum allowed height for the reduction
minimum height 2m3( A + 1) Tangle-Height Minimization is NP-hard.
Theorem.
NP-hardness by reduction from 3-Partition.
asymptotically faster than [Olszewski et al., GD’18]. O
5|L|/n n
2|L|
n2 + 1
n2
2 ϕnn
Simple lists Tangle-Height Minimization can be solved in . . . General lists
Simple lists Tangle-Height Minimization can be solved in . . . General lists
[Olszewski et al., GD’18]
2O(n2)
n – number of wires
Simple lists Tangle-Height Minimization can be solved in . . . General lists
[Olszewski et al., GD’18]
2O(n2) 2O(n log n)
n – number of wires
Simple lists Tangle-Height Minimization can be solved in . . . General lists
[Olszewski et al., GD’18]
2O(n2) 2O(n log n)
[Olszewski et al., GD’18]
O
5|L|/n n
– number of wires – length of the list L (= ℓij) – golden ratio (≈ 1.618) |L| ϕ
Simple lists Tangle-Height Minimization can be solved in . . . General lists
[Olszewski et al., GD’18]
2O(n2) 2O(n log n)
[Olszewski et al., GD’18]
O
5|L|/n n
2|L|
n2 + 1
n2
2 ϕnn
n – number of wires – length of the list L (= ℓij) – golden ratio (≈ 1.618) |L| ϕ
Simple lists Tangle-Height Minimization can be solved in . . . General lists
[Olszewski et al., GD’18]
2O(n2) 2O(n log n)
[Olszewski et al., GD’18]
O
5|L|/n n
2|L|
n2 + 1
n2
2 ϕnn
polynomial in |L|
n – number of wires – length of the list L (= ℓij) – golden ratio (≈ 1.618) |L| ϕ
polynomial in |L| for fixed n
O 2|L|
n2 + 1
n2/2ϕnn
O 2|L|
n2 + 1
n2/2ϕnn
λ = # of distinct sublists of L.
L′ is a sublist of L if ℓ′
ij ≤ ℓij
O 2|L|
n2 + 1
n2/2ϕnn
λ = # of distinct sublists of L. Consider them in order of increasing length.
L′ is a sublist of L if ℓ′
ij ≤ ℓij
O 2|L|
n2 + 1
n2/2ϕnn
λ = # of distinct sublists of L. Consider them in order of increasing length. Let L′ be the next list to consider.
L′ is a sublist of L if ℓ′
ij ≤ ℓij
O 2|L|
n2 + 1
n2/2ϕnn
λ = # of distinct sublists of L. Consider them in order of increasing length. Let L′ be the next list to consider. Check its consistency.
L′ is a sublist of L if ℓ′
ij ≤ ℓij
O 2|L|
n2 + 1
n2/2ϕnn
λ = # of distinct sublists of L. Consider them in order of increasing length. Let L′ be the next list to consider. Check its consistency. i
L′ is a sublist of L if ℓ′
ij ≤ ℓij
O 2|L|
n2 + 1
n2/2ϕnn
λ = # of distinct sublists of L. Consider them in order of increasing length. Let L′ be the next list to consider. Check its consistency. i for each wire i:
// find a position where it is after applying L′
L′ is a sublist of L if ℓ′
ij ≤ ℓij
O 2|L|
n2 + 1
n2/2ϕnn
λ = # of distinct sublists of L. Consider them in order of increasing length. Let L′ be the next list to consider. Check its consistency. i for each wire i:
// find a position where it is after applying L′
i → i + |{j : j > i and ℓ′
ij is odd}| − |{j : j < i and ℓ′ ij is odd}| L′ is a sublist of L if ℓ′
ij ≤ ℓij
O 2|L|
n2 + 1
n2/2ϕnn
λ = # of distinct sublists of L. Consider them in order of increasing length. Let L′ be the next list to consider. Check its consistency. i for each wire i:
// find a position where it is after applying L′
i → i + |{j : j > i and ℓ′
ij is odd}| − |{j : j < i and ℓ′ ij is odd}| L′ is a sublist of L if ℓ′
ij ≤ ℓij
O 2|L|
n2 + 1
n2/2ϕnn
λ = # of distinct sublists of L. Consider them in order of increasing length. Let L′ be the next list to consider. Check its consistency. i for each wire i:
// find a position where it is after applying L′
i → i + |{j : j > i and ℓ′
ij is odd}| − |{j : j < i and ℓ′ ij is odd}| L′ is a sublist of L if ℓ′
ij ≤ ℓij
O 2|L|
n2 + 1
n2/2ϕnn
λ = # of distinct sublists of L. Consider them in order of increasing length. Let L′ be the next list to consider. Check its consistency. i for each wire i:
// find a position where it is after applying L′
i → i + |{j : j > i and ℓ′
ij is odd}| − |{j : j < i and ℓ′ ij is odd}| L′ is a sublist of L if ℓ′
ij ≤ ℓij
O 2|L|
n2 + 1
n2/2ϕnn
λ = # of distinct sublists of L. Consider them in order of increasing length. Let L′ be the next list to consider. Check its consistency. i for each wire i:
// find a position where it is after applying L′
i → i + |{j : j > i and ℓ′
ij is odd}| − |{j : j < i and ℓ′ ij is odd}|
check whether the map is indeed a permutation
L′ is a sublist of L if ℓ′
ij ≤ ℓij
O 2|L|
n2 + 1
n2/2ϕnn
λ = # of distinct sublists of L. Consider them in order of increasing length. Let L′ be the next list to consider. Check its consistency. idn L′
L′ is a sublist of L if ℓ′
ij ≤ ℓij
Compute the final permutation idn L′.
O 2|L|
n2 + 1
n2/2ϕnn
λ = # of distinct sublists of L. Consider them in order of increasing length. Let L′ be the next list to consider. Check its consistency. π1 π2 . . . πh idn L′ Choose the shortest tangle T(L′′).
L′ is a sublist of L if ℓ′
ij ≤ ℓij
Compute the final permutation idn L′. πh and idn L′ are adjacent L′′∪ add. swaps = L′
O 2|L|
n2 + 1
n2/2ϕnn
λ = # of distinct sublists of L. Consider them in order of increasing length. Let L′ be the next list to consider. Check its consistency. π1 π2 . . . πh idn L′ Choose the shortest tangle T(L′′).
L′ is a sublist of L if ℓ′
ij ≤ ℓij
Compute the final permutation idn L′. πh and idn L′ are adjacent L′′∪ add. swaps = L′
O 2|L|
n2 + 1
n2/2ϕnn
λ = # of distinct sublists of L. Consider them in order of increasing length. Let L′ be the next list to consider. Check its consistency. π1 π2 . . . πh Add the final permutation to its end. idn L′ Choose the shortest tangle T(L′′).
L′ is a sublist of L if ℓ′
ij ≤ ℓij
Compute the final permutation idn L′.
O 2|L|
n2 + 1
n2/2ϕnn
λ = # of distinct sublists of L. Consider them in order of increasing length. Let L′ be the next list to consider. Check its consistency. π1 π2 . . . πh Add the final permutation to its end. idn L′ Running time O(λ · (Fn+1 − 1) · n) ≤ Choose the shortest tangle T(L′′).
L′ is a sublist of L if ℓ′
ij ≤ ℓij
Compute the final permutation idn L′.
O 2|L|
n2 + 1
n2/2ϕnn
λ = # of distinct sublists of L. Consider them in order of increasing length. Let L′ be the next list to consider. Check its consistency. π1 π2 . . . πh Add the final permutation to its end. idn L′ Running time O(λ · (Fn+1 − 1) · n) ≤ Choose the shortest tangle T(L′′).
L′ is a sublist of L if ℓ′
ij ≤ ℓij
Compute the final permutation idn L′.
O 2|L|
n2 + 1
n2/2ϕnn
λ = # of distinct sublists of L. Consider them in order of increasing length. Let L′ be the next list to consider. Check its consistency. π1 π2 . . . πh Add the final permutation to its end. idn L′ Running time O(λ · (Fn+1 − 1) · n) ≤ Choose the shortest tangle T(L′′).
Fn is the n-th Fibonacci number L′ is a sublist of L if ℓ′
ij ≤ ℓij
Compute the final permutation idn L′.
O 2|L|
n2 + 1
n2/2ϕnn
λ = # of distinct sublists of L. Consider them in order of increasing length. Let L′ be the next list to consider. Check its consistency. π1 π2 . . . πh Add the final permutation to its end. idn L′ Running time O(λ · (Fn+1 − 1) · n) ≤ Choose the shortest tangle T(L′′).
L′ is a sublist of L if ℓ′
ij ≤ ℓij
Compute the final permutation idn L′.
λ =
i<j
(ℓij + 1) ≤
n2 + 1
n2/2 Fn ∈ O(ϕn)
O 2|L|
n2 + 1
n2/2ϕnn
λ = # of distinct sublists of L. Consider them in order of increasing length. Let L′ be the next list to consider. Check its consistency. π1 π2 . . . πh Add the final permutation to its end. idn L′ Running time O(λ · (Fn+1 − 1) · n) ≤ Choose the shortest tangle T(L′′).
L′ is a sublist of L if ℓ′
ij ≤ ℓij
Compute the final permutation idn L′.
λ =
i<j
(ℓij + 1) ≤
n2 + 1
n2/2 Fn ∈ O(ϕn)
NP-hardness by reduction from 3-Partition.
asymptotically faster than [Olszewski et al., GD’18]. O
5|L|/n n
2|L|
n2 + 1
n2
2 ϕnn
[Olszewski et al., GD’18] O
5|L|/n n
2|L|
n2 + 1
n2
2 ϕnn
runtime [sec] length |L| of the list L
0 10 20 30 40 50 0.0001 0.01 1 100 3600
0 10 20 30 40 0 10 20 30 40
Problem 1 Is it NP-hard to test the feasibility of a given (non-simple) list?
Problem 1 Is it NP-hard to test the feasibility of a given (non-simple) list? Problem 2 Can we decide a feasibility of a list faster than finding its
Problem 1 Is it NP-hard to test the feasibility of a given (non-simple) list? Problem 2 Can we decide a feasibility of a list faster than finding its
Problem 3 A list (ℓij) is non-separable if ∀i<k<j:
i k j
Problem 1 Is it NP-hard to test the feasibility of a given (non-simple) list? Problem 2 Can we decide a feasibility of a list faster than finding its
Problem 3 A list (ℓij) is non-separable if ∀i<k<j:
i k j necessary
Problem 1 Is it NP-hard to test the feasibility of a given (non-simple) list? Problem 2 Can we decide a feasibility of a list faster than finding its
Problem 3 For lists where all entries are even, is this sufficient? A list (ℓij) is non-separable if ∀i<k<j:
i k j necessary
Problem 1 Is it NP-hard to test the feasibility of a given (non-simple) list? Problem 2 Can we decide a feasibility of a list faster than finding its
Problem 3 For lists where all entries are even, is this sufficient? A list (ℓij) is non-separable if ∀i<k<j:
i k j necessary