The Traveling Salesman Problem Under Squared Euclidean Distances - - PowerPoint PPT Presentation

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The Traveling Salesman Problem Under Squared Euclidean Distances - - PowerPoint PPT Presentation

Nov 24, 2023 233 likes •1.4k views

The Traveling Salesman Problem Under Squared Euclidean Distances Mark de Berg Fred van Nijnatten Gerhard Woeginger TU Eindhoven Ren e Sitters Vrije Universiteit Amsterdam Alexander Wolff Universit at W urzburg Whats the

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SLIDE 1

The Traveling Salesman Problem Under Squared Euclidean Distances

Mark de Berg Fred van Nijnatten Gerhard Woeginger TU Eindhoven Ren´ e Sitters Vrije Universiteit Amsterdam Alexander Wolff Universit¨ at W¨ urzburg

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SLIDE 2

What’s the Problem?

Notation. For points p = (p1, . . . , pd), q = (q1, . . . , qd) ∈ Rd, denote by |pq| = d

i=1(pi − qi)2 their

Euclidean distance. p q |pq|

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SLIDE 3

What’s the Problem?

Notation. For points p = (p1, . . . , pd), q = (q1, . . . , qd) ∈ Rd, denote by |pq| = d

i=1(pi − qi)2 their

Euclidean distance. Given a finite set S ⊂ Rd, find a tour π through all points in S such that π has minimum length among all tours through S w.r.t. | · | . Problem. Euclidean TSP p q |pq|

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SLIDE 4

What’s the Problem?

Notation. For points p = (p1, . . . , pd), q = (q1, . . . , qd) ∈ Rd, denote by |pq| = d

i=1(pi − qi)2 their

Euclidean distance. Given a finite set S ⊂ Rd, find a tour π through all points in S such that π has minimum length among all tours through S w.r.t. | · | . Problem. Euclidean TSP p q |pq|

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SLIDE 5

What’s the Problem?

Notation. For points p = (p1, . . . , pd), q = (q1, . . . , qd) ∈ Rd, denote by |pq| = d

i=1(pi − qi)2 their

Euclidean distance. Given a finite set S ⊂ Rd, find a tour π through all points in S such that π has minimum length among all tours through S w.r.t. | · | . Problem. Euclidean TSP p q |pq|

2 2 2

squared

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SLIDE 6

What’s the Problem?

Notation. For points p = (p1, . . . , pd), q = (q1, . . . , qd) ∈ Rd, denote by |pq| = d

i=1(pi − qi)2 their

Euclidean distance. Given a finite set S ⊂ Rd, find a tour π through all points in S such that π has minimum length among all tours through S w.r.t. | · | . Problem. Euclidean TSP p q |pq|

2 2 2

squared (d, 2)

2

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SLIDE 7

What’s the Problem?

Notation. For points p = (p1, . . . , pd), q = (q1, . . . , qd) ∈ Rd, denote by |pq| = d

i=1(pi − qi)2 their

Euclidean distance. Given a finite set S ⊂ Rd, find a tour π through all points in S such that π has minimum length among all tours through S w.r.t. | · | . Problem. Euclidean TSP p q |pq|

2 2 2

squared (d, 2)

2

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SLIDE 8

What’s the Problem?

Notation. For points p = (p1, . . . , pd), q = (q1, . . . , qd) ∈ Rd, denote by |pq| = d

i=1(pi − qi)2 their

Euclidean distance. Given a finite set S ⊂ Rd, find a tour π through all points in S such that π has minimum length among all tours through S w.r.t. | · | . Problem. Euclidean TSP p q |pq|

α α α

power-α (d, α)

α

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SLIDE 9

The Metric/Euclidean Case (α = 1)

  • Theorem. [folklore]

The MST yields a 2-approximation for metric TSP.

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SLIDE 10

The Metric/Euclidean Case (α = 1)

  • Theorem. [folklore]

The MST yields a 2-approximation for metric TSP.

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SLIDE 11

The Metric/Euclidean Case (α = 1)

  • Theorem. [folklore]

The MST yields a 2-approximation for metric TSP.

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SLIDE 12

The Metric/Euclidean Case (α = 1)

  • Theorem. [folklore]

The MST yields a 2-approximation for metric TSP.

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SLIDE 13

The Metric/Euclidean Case (α = 1)

  • Theorem. [folklore]

The MST yields a 2-approximation for metric TSP.

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SLIDE 14

The Metric/Euclidean Case (α = 1)

  • Theorem. [folklore]

The MST yields a 2-approximation for metric TSP.

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SLIDE 15

The Metric/Euclidean Case (α = 1)

  • Theorem. [folklore]

The MST yields a 2-approximation for metric TSP.

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SLIDE 16

The Metric/Euclidean Case (α = 1)

  • Theorem. [folklore]

The MST yields a 2-approximation for metric TSP.

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SLIDE 17

The Metric/Euclidean Case (α = 1)

  • Theorem. [folklore]

The MST yields a 2-approximation for metric TSP.

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SLIDE 18

The Metric/Euclidean Case (α = 1)

  • Theorem. [folklore]

The MST yields a 2-approximation for metric TSP.

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SLIDE 19

The Metric/Euclidean Case (α = 1)

  • Theorem. [folklore]

The MST yields a 2-approximation for metric TSP.

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SLIDE 20

The Metric/Euclidean Case (α = 1)

  • Theorem. [folklore]

The MST yields a 2-approximation for metric TSP.

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SLIDE 21

The Metric/Euclidean Case (α = 1)

  • Theorem. [folklore]

The MST yields a 2-approximation for metric TSP.

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SLIDE 22

The Metric/Euclidean Case (α = 1)

  • Theorem. [folklore]

The MST yields a 2-approximation for metric TSP.

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SLIDE 23

The Metric/Euclidean Case (α = 1)

  • Theorem. [folklore]

The MST yields a 2-approximation for metric TSP.

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SLIDE 24

The Metric/Euclidean Case (α = 1)

  • Theorem. [folklore]

The MST yields a 2-approximation for metric TSP.

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SLIDE 25

The Metric/Euclidean Case (α = 1)

  • Theorem. [folklore]

The MST yields a 2-approximation for metric TSP.

  • Theorem. [Christofides’76]

There is a 3/2-approximation for metric TSP.

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SLIDE 26

The Metric/Euclidean Case (α = 1)

  • Theorem. [folklore]

The MST yields a 2-approximation for metric TSP.

  • Theorem. [Papadimitriou’77]

Euclidean TSP is NP-hard in any dimension d ≥ 1

  • Theorem. [Christofides’76]

There is a 3/2-approximation for metric TSP.

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SLIDE 27

The Metric/Euclidean Case (α = 1)

  • Theorem. [folklore]

The MST yields a 2-approximation for metric TSP.

  • Theorem. [Papadimitriou’77]

Euclidean TSP is NP-hard in any dimension d ≥ 1

  • Theorem. [Christofides’76]

There is a 3/2-approximation for metric TSP.

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SLIDE 28

The Metric/Euclidean Case (α = 1)

  • Theorem. [folklore]

The MST yields a 2-approximation for metric TSP.

  • Theorem. [Papadimitriou’77]

Euclidean TSP is NP-hard in any dimension d ≥ 1

  • Theorem. [Arora’96, Mitchell’96, RaoSmith’98]

Euclidean TSP admits a PTAS for any fixed d ≥ 1.

  • Theorem. [Christofides’76]

There is a 3/2-approximation for metric TSP.

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SLIDE 29

The Metric/Euclidean Case (α = 1)

  • Theorem. [folklore]

The MST yields a 2-approximation for metric TSP.

  • Theorem. [Papadimitriou’77]

Euclidean TSP is NP-hard in any dimension d ≥ 1

  • Theorem. [Arora’96, Mitchell’96, RaoSmith’98]

Euclidean TSP admits a PTAS for any fixed d ≥ 1.

But what about TSP(d, α) for α = 1?

  • Theorem. [Christofides’76]

There is a 3/2-approximation for metric TSP.

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SLIDE 30

Motivation

  • 1. Range assignment for wireless networks
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SLIDE 31

Motivation

  • 1. Range assignment for wireless networks
  • transmission range depends on power
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SLIDE 32

Motivation

  • 1. Range assignment for wireless networks
  • transmission range depends on power
  • energy consumption ∼ dα

for some α ∈ [2, 6] (“distance-power gradient”) d

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SLIDE 33

Motivation

  • 1. Range assignment for wireless networks
  • transmission range depends on power
  • energy consumption ∼ dα

for some α ∈ [2, 6] (“distance-power gradient”) d

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SLIDE 34

Motivation

  • 1. Range assignment for wireless networks
  • transmission range depends on power
  • energy consumption ∼ dα

for some α ∈ [2, 6] (“distance-power gradient”) d

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SLIDE 35

Motivation

  • 1. Range assignment for wireless networks
  • transmission range depends on power
  • energy consumption ∼ dα

for some α ∈ [2, 6] (“distance-power gradient”) d

  • range assignment ρ induces
  • dir. communication graph Gρ
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SLIDE 36

Motivation

  • 1. Range assignment for wireless networks
  • transmission range depends on power
  • energy consumption ∼ dα

for some α ∈ [2, 6] (“distance-power gradient”) d

  • range assignment ρ induces
  • dir. communication graph Gρ
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SLIDE 37

Motivation

  • 1. Range assignment for wireless networks
  • transmission range depends on power
  • energy consumption ∼ dα

for some α ∈ [2, 6] (“distance-power gradient”) d

  • range assignment ρ induces
  • dir. communication graph Gρ
  • optimization problem:

compute min-energy range assignment ρ s.t. Gρ . . .

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SLIDE 38

Motivation

  • 1. Range assignment for wireless networks
  • transmission range depends on power
  • energy consumption ∼ dα

for some α ∈ [2, 6] (“distance-power gradient”) d

  • range assignment ρ induces
  • dir. communication graph Gρ
  • optimization problem:

compute min-energy range assignment ρ s.t. Gρ . . . – is strongly connected – contains broadcast tree – contains tour [Funke...’08]

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SLIDE 39

Motivation

  • 2. Directional antennas with circular sectors

[Caragiannis...’08]

γ

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SLIDE 40

Motivation

  • 2. Directional antennas with circular sectors

[Caragiannis...’08]

γ γ

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SLIDE 41

Motivation

  • 2. Directional antennas with circular sectors

[Caragiannis...’08]

γ γ Gρ strongly connected

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SLIDE 42

Motivation

  • 2. Directional antennas with circular sectors

[Caragiannis...’08]

γ γ Gρ strongly connected γ → 0

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SLIDE 43

Motivation

  • 2. Directional antennas with circular sectors

[Caragiannis...’08]

γ γ Gρ strongly connected γ → 0 Gρ tour

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SLIDE 44

Motivation

  • 2. Directional antennas with circular sectors

[Caragiannis...’08]

γ γ Gρ strongly connected γ → 0 Gρ tour

  • 3. Complexity

Are things becoming simpler or harder?

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SLIDE 45

Motivation

  • 2. Directional antennas with circular sectors

[Caragiannis...’08]

γ γ Gρ strongly connected γ → 0 Gρ tour

  • 3. Complexity

Are things becoming simpler or harder? Is Arora’s PTAS for Euclidean TSP a “lucky coincidence”?

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SLIDE 46

Motivation

  • 2. Directional antennas with circular sectors

[Caragiannis...’08]

γ γ Gρ strongly connected γ → 0 Gρ tour

  • 3. Complexity

Are things becoming simpler or harder? Is Arora’s PTAS for Euclidean TSP a “lucky coincidence”? If it is, how well can we approximate, say, TSP(2, 2)?

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SLIDE 47

Previous Work

Definition. dist(·, ·) fulfills the τ-relaxed triangle inequality if any three points p, q, r satisfy dist(p, r) ≤ τ · (dist(p, q) + dist(q, r)). p q r

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SLIDE 48

Previous Work

Lemma. | · | fulfills the

  • relaxed triangle inequality.

Definition. dist(·, ·) fulfills the τ-relaxed triangle inequality if any three points p, q, r satisfy dist(p, r) ≤ τ · (dist(p, q) + dist(q, r)). [Funke...’08] p q r

2

2

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SLIDE 49

Previous Work

Lemma. | · | fulfills the

  • relaxed triangle inequality.

Definition. dist(·, ·) fulfills the τ-relaxed triangle inequality if any three points p, q, r satisfy dist(p, r) ≤ τ · (dist(p, q) + dist(q, r)). [Funke...’08] For α ≥ 1, p q r

α

2α−1

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SLIDE 50

Previous Work

Lemma. | · | fulfills the

  • relaxed triangle inequality.

Definition. dist(·, ·) fulfills the τ-relaxed triangle inequality if any three points p, q, r satisfy dist(p, r) ≤ τ · (dist(p, q) + dist(q, r)). [Funke...’08] For α ≥ 1, p q r

α

2α−1 Good news. Can apply algorithms for ∆τ-TSP to TSP( · , α)!

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SLIDE 51

Previous Work

Lemma. | · | fulfills the

  • relaxed triangle inequality.

Definition. dist(·, ·) fulfills the τ-relaxed triangle inequality if any three points p, q, r satisfy dist(p, r) ≤ τ · (dist(p, q) + dist(q, r)). [Funke...’08] For α ≥ 1, p q r

α

2α−1 Good news. Can apply algorithms for ∆τ-TSP to TSP( · , α)! ∆τ TSP( · , α) τ 2 + τ 4α−1 + 2α−1 [Andreae’01] 4τ 2α+1 [BenderChekuri’00] — 3 · 2α−1 [Funke...’08]

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SLIDE 52

Previous Work

Lemma. | · | fulfills the

  • relaxed triangle inequality.

Definition. dist(·, ·) fulfills the τ-relaxed triangle inequality if any three points p, q, r satisfy dist(p, r) ≤ τ · (dist(p, q) + dist(q, r)). [Funke...’08] For α ≥ 1, p q r

α

2α−1 Good news. Can apply algorithms for ∆τ-TSP to TSP( · , α)! ∆τ TSP( · , α) τ 2 + τ 4α−1 + 2α−1 [Andreae’01] 4τ 2α+1 [BenderChekuri’00] — 3 · 2α−1 [Funke...’08]

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SLIDE 53

Our Results

5 10 15 20 25 30 2 2.2 2.4 2.6 2.8 3 3.2 3.4 2α+1

α

[BenderChekuri’00] 4α−1 + 2α−1 [Andreae’01] 2 · 3α−1 [Funke...’08]

≈ 2.71

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SLIDE 54

Our Results

5 10 15 20 25 30 2 2.2 2.4 2.6 2.8 3 3.2 3.4 2α+1

α

[BenderChekuri’00] 4α−1 + 2α−1 [Andreae’01] 2 · 3α−1 [Funke...’08]

≈ 2.71

[AndreaeBandelt’95] T3-algorithm

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SLIDE 55

Our Results

5 10 15 20 25 30 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3α−1 + √ 6

α/3

2α+1

α

[BenderChekuri’00] 4α−1 + 2α−1 [Andreae’01] 2 · 3α−1 [Funke...’08]

≈ 2.71 ≈ 3.41

[AndreaeBandelt’95] T3-algorithm geometric T3-algorithm

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SLIDE 56

Our Results

5 10 15 20 25 30 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3α−1 + √ 6

α/3

2α+1

α

[BenderChekuri’00] 4α−1 + 2α−1 [Andreae’01] 2 · 3α−1 [Funke...’08]

≈ 2.71 ≈ 3.41

[AndreaeBandelt’95] T3-algorithm geometric T3-algorithm in R2 in any metric space

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SLIDE 57

Our Results

5 10 15 20 25 30 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3α−1 + √ 6

α/3

2α+1

6α+3α+2α−1 3·2α−1

α

[BenderChekuri’00] lower bound for MST-based approx. 4α−1 + 2α−1 [Andreae’01] 2 · 3α−1 [Funke...’08]

≈ 2.71 ≈ 3.41

[AndreaeBandelt’95] T3-algorithm geometric T3-algorithm in R2 in any metric space

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SLIDE 58

The T3-Algorithm

CycleInCube(T, e = u1u2) for i ← 1 to 2 do Ti ← component of T − e that contains ui if |Ti| = 1 then Πi ← ∅; wi ← ui else pick an edge ei = uiwi incident to ui in Ti if |Ti| = 2 then Πi ← ei else Πi ← CycleInCube(Ti, ei) − ei return Π1 + e + Π2 + w1w2 u1 u2 e

[Sekanina’60, AndreaeBandelt’95]

T

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SLIDE 59

The T3-Algorithm

CycleInCube(T, e = u1u2) for i ← 1 to 2 do Ti ← component of T − e that contains ui if |Ti| = 1 then Πi ← ∅; wi ← ui else pick an edge ei = uiwi incident to ui in Ti if |Ti| = 2 then Πi ← ei else Πi ← CycleInCube(Ti, ei) − ei return Π1 + e + Π2 + w1w2 u1 u2 e

[Sekanina’60, AndreaeBandelt’95]

T

Take MST of given point set!

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SLIDE 60

The T3-Algorithm

CycleInCube(T, e = u1u2) for i ← 1 to 2 do Ti ← component of T − e that contains ui if |Ti| = 1 then Πi ← ∅; wi ← ui else pick an edge ei = uiwi incident to ui in Ti if |Ti| = 2 then Πi ← ei else Πi ← CycleInCube(Ti, ei) − ei return Π1 + e + Π2 + w1w2 u1 u2 T1 T2 e

[Sekanina’60, AndreaeBandelt’95]

T

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SLIDE 61

The T3-Algorithm

CycleInCube(T, e = u1u2) for i ← 1 to 2 do Ti ← component of T − e that contains ui if |Ti| = 1 then Πi ← ∅; wi ← ui else pick an edge ei = uiwi incident to ui in Ti if |Ti| = 2 then Πi ← ei else Πi ← CycleInCube(Ti, ei) − ei return Π1 + e + Π2 + w1w2 u1 u2 T1 T2 e e1 e2

[Sekanina’60, AndreaeBandelt’95]

T

w1 w2

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SLIDE 62

The T3-Algorithm

CycleInCube(T, e = u1u2) for i ← 1 to 2 do Ti ← component of T − e that contains ui if |Ti| = 1 then Πi ← ∅; wi ← ui else pick an edge ei = uiwi incident to ui in Ti if |Ti| = 2 then Πi ← ei else Πi ← CycleInCube(Ti, ei) − ei return Π1 + e + Π2 + w1w2 u1 u2 T1 T2 e e1 e2

[Sekanina’60, AndreaeBandelt’95]

T

w1 w2 Π1 Π2

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SLIDE 63

The T3-Algorithm

CycleInCube(T, e = u1u2) for i ← 1 to 2 do Ti ← component of T − e that contains ui if |Ti| = 1 then Πi ← ∅; wi ← ui else pick an edge ei = uiwi incident to ui in Ti if |Ti| = 2 then Πi ← ei else Πi ← CycleInCube(Ti, ei) − ei return Π1 + e + Π2 + w1w2 u1 u2 T1 T2 e e1 e2

[Sekanina’60, AndreaeBandelt’95]

T

w1 w2 Π1 Π2

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SLIDE 64

The T3-Algorithm

CycleInCube(T, e = u1u2) for i ← 1 to 2 do Ti ← component of T − e that contains ui if |Ti| = 1 then Πi ← ∅; wi ← ui else pick an edge ei = uiwi incident to ui in Ti if |Ti| = 2 then Πi ← ei else Πi ← CycleInCube(Ti, ei) − ei return Π1 + e + Π2 + w1w2 u1 u2 T1 T2 e e1 e2

[Sekanina’60, AndreaeBandelt’95]

T

w1 w2 Π1 Π2 3-shortcut

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SLIDE 65

The T3-Algorithm

CycleInCube(T, e = u1u2) for i ← 1 to 2 do Ti ← component of T − e that contains ui if |Ti| = 1 then Πi ← ∅; wi ← ui else pick an edge ei = uiwi incident to ui in Ti if |Ti| = 2 then Πi ← ei else Πi ← CycleInCube(Ti, ei) − ei return Π1 + e + Π2 + w1w2 u1 u2 T1 T2 e e1 e2

[Sekanina’60, AndreaeBandelt’95]

T

w1 w2 Π1 Π2 3-shortcut

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SLIDE 66

The T3-Algorithm

CycleInCube(T, e = u1u2) for i ← 1 to 2 do Ti ← component of T − e that contains ui if |Ti| = 1 then Πi ← ∅; wi ← ui else pick an edge ei = uiwi incident to ui in Ti if |Ti| = 2 then Πi ← ei else Πi ← CycleInCube(Ti, ei) − ei return Π1 + e + Π2 + w1w2 u1 u2 T1 T2 e e1 e2

[Sekanina’60, AndreaeBandelt’95]

T

w1 w2 Π1 Π2 3-shortcut

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SLIDE 67

The T3-Algorithm

CycleInCube(T, e = u1u2) for i ← 1 to 2 do Ti ← component of T − e that contains ui if |Ti| = 1 then Πi ← ∅; wi ← ui else pick an edge ei = uiwi incident to ui in Ti if |Ti| = 2 then Πi ← ei else Πi ← CycleInCube(Ti, ei) − ei return Π1 + e + Π2 + w1w2 u1 u2 T1 T2 e e1 e2

[Sekanina’60, AndreaeBandelt’95]

T

w1 w2 Π1 Π2 3-shortcut (2- or)

slide-68
SLIDE 68

The T3-Algorithm

CycleInCube(T, e = u1u2) for i ← 1 to 2 do Ti ← component of T − e that contains ui if |Ti| = 1 then Πi ← ∅; wi ← ui else pick an edge ei = uiwi incident to ui in Ti if |Ti| = 2 then Πi ← ei else Πi ← CycleInCube(Ti, ei) − ei return Π1 + e + Π2 + w1w2 u1 u2 T1 T2 e e1 e2

[Sekanina’60, AndreaeBandelt’95]

T

w1 w2 Π1 Π2 3-shortcut uses edges e, e1, and e2 (2- or)

slide-69
SLIDE 69

The T3-Algorithm

CycleInCube(T, e = u1u2) for i ← 1 to 2 do Ti ← component of T − e that contains ui if |Ti| = 1 then Πi ← ∅; wi ← ui else pick an edge ei = uiwi incident to ui in Ti if |Ti| = 2 then Πi ← ei else Πi ← CycleInCube(Ti, ei) − ei return Π1 + e + Π2 + w1w2 u1 u2 T1 T2 e e1 e2

[Sekanina’60, AndreaeBandelt’95]

T

w1 w2 Π1 Π2 3-shortcut uses edges e, e1, and e2 Observation. Every edge is used at most ...? (2- or)

slide-70
SLIDE 70

The T3-Algorithm

CycleInCube(T, e = u1u2) for i ← 1 to 2 do Ti ← component of T − e that contains ui if |Ti| = 1 then Πi ← ∅; wi ← ui else pick an edge ei = uiwi incident to ui in Ti if |Ti| = 2 then Πi ← ei else Πi ← CycleInCube(Ti, ei) − ei return Π1 + e + Π2 + w1w2 u1 u2 T1 T2 e e1 e2

[Sekanina’60, AndreaeBandelt’95]

T

w1 w2 Π1 Π2 3-shortcut uses edges e, e1, and e2 Observation. Every edge is used at most twice. (2- or)

slide-71
SLIDE 71

Result #1

Observation. Every edge is used at most twice.

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SLIDE 72

Result #1

Observation. Every edge is used at most twice. Lemma. Let e be a 3-shortcut using a, b, c. Let α ≥ 1. Then |e|α ≤ 3α−1 |a|α + |b|α + |c|α .

slide-73
SLIDE 73

Result #1

Observation. Every edge is used at most twice. Lemma. Let e be a 3-shortcut using a, b, c. Let α ≥ 1. Then |e|α ≤ 3α−1 |a|α + |b|α + |c|α . Corollary. For α ≥ 2, the T3-algorithm yields a (2 · 3α−1)- approximation for TSP( · , α).

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SLIDE 74

Result #1

Observation. Every edge is used at most twice. Lemma. Let e be a 3-shortcut using a, b, c. Let α ≥ 1. Then |e|α ≤ 3α−1 |a|α + |b|α + |c|α . Corollary. For α ≥ 2, the T3-algorithm yields a (2 · 3α−1)- approximation for TSP( · , α). Proof. Every edge c of the MST (w.r.t. |·|α) contributes at most twice at most 3α−1|c|α to the T3-tour. c e′ e

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SLIDE 75

Result #1

Observation. Every edge is used at most twice. Lemma. Let e be a 3-shortcut using a, b, c. Let α ≥ 1. Then |e|α ≤ 3α−1 |a|α + |b|α + |c|α . Corollary. For α ≥ 2, the T3-algorithm yields a (2 · 3α−1)- approximation for TSP( · , α). Proof. Every edge c of the MST (w.r.t. |·|α) contributes at most twice at most 3α−1|c|α to the T3-tour. c e′ e

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SLIDE 76

Result #1

Observation. Every edge is used at most twice. Lemma. Let e be a 3-shortcut using a, b, c. Let α ≥ 1. Then |e|α ≤ 3α−1 |a|α + |b|α + |c|α . Corollary. For α ≥ 2, the T3-algorithm yields a (2 · 3α−1)- approximation for TSP( · , α). Proof. Every edge c of the MST (w.r.t. |·|α) contributes at most twice at most 3α−1|c|α to the T3-tour. c e′ e

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SLIDE 77

Result #1

Observation. Every edge is used at most twice. Lemma. Let e be a 3-shortcut using a, b, c. Let α ≥ 1. Then |e|α ≤ 3α−1 |a|α + |b|α + |c|α . Corollary. For α ≥ 2, the T3-algorithm yields a (2 · 3α−1)- approximation for TSP( · , α). Proof. Every edge c of the MST (w.r.t. |·|α) contributes at most twice at most 3α−1|c|α to the T3-tour. c e′ e add up!

slide-78
SLIDE 78

Result #1

Observation. Every edge is used at most twice. Lemma. Let e be a 3-shortcut using a, b, c. Let α ≥ 1. Then |e|α ≤ 3α−1 |a|α + |b|α + |c|α . Corollary. For α ≥ 2, the T3-algorithm yields a (2 · 3α−1)- approximation for TSP( · , α). Proof. Every edge c of the MST (w.r.t. |·|α) contributes at most twice at most 3α−1|c|α to the T3-tour. c e′ e add up! 3α−1 |MST|α

slide-79
SLIDE 79

Result #1

Observation. Every edge is used at most twice. Lemma. Let e be a 3-shortcut using a, b, c. Let α ≥ 1. Then |e|α ≤ 3α−1 |a|α + |b|α + |c|α . Corollary. For α ≥ 2, the T3-algorithm yields a (2 · 3α−1)- approximation for TSP( · , α). Proof. Every edge c of the MST (w.r.t. |·|α) contributes at most twice at most 3α−1|c|α to the T3-tour. c e′ e add up! 2· 3α−1 |MST|α

slide-80
SLIDE 80

Result #1

Observation. Every edge is used at most twice. Lemma. Let e be a 3-shortcut using a, b, c. Let α ≥ 1. Then |e|α ≤ 3α−1 |a|α + |b|α + |c|α . Corollary. For α ≥ 2, the T3-algorithm yields a (2 · 3α−1)- approximation for TSP( · , α). Proof. Every edge c of the MST (w.r.t. |·|α) contributes at most twice at most 3α−1|c|α to the T3-tour. c e′ e add up! ⇒ |T3-tour|α ≤ 2· 3α−1 |MST|α

slide-81
SLIDE 81

Result #1

Observation. Every edge is used at most twice. Lemma. Let e be a 3-shortcut using a, b, c. Let α ≥ 1. Then |e|α ≤ 3α−1 |a|α + |b|α + |c|α . Corollary. For α ≥ 2, the T3-algorithm yields a (2 · 3α−1)- approximation for TSP( · , α). Proof. Every edge c of the MST (w.r.t. |·|α) contributes at most twice at most 3α−1|c|α to the T3-tour. c e′ e ≤ 2 · 3α−1·OPT add up! ⇒ |T3-tour|α ≤ 2· 3α−1 |MST|α

slide-82
SLIDE 82

Result #1

Observation. Every edge is used at most twice. Lemma. Let e be a 3-shortcut using a, b, c. Let α ≥ 1. Then |e|α ≤ 3α−1 |a|α + |b|α + |c|α . Corollary. For α ≥ 2, the T3-algorithm yields a (2 · 3α−1)- approximation for TSP( · , α). Proof. Every edge c of the MST (w.r.t. |·|α) contributes at most twice at most 3α−1|c|α to the T3-tour. c e′ e ≤ 2 · 3α−1·OPT add up!

NO GEOMETRY!

⇒ |T3-tour|α ≤ 2· 3α−1 |MST|α

slide-83
SLIDE 83

Result #2

Corollary. The T3-algorithm yields a (2·3α−1)-approximation for TSP( · , α) if α ≥ 2.

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SLIDE 84

Result #2

Theorem. For α ≥ 2, the geometric T3-algorithm yields a

  • 3α−1 +

√ 6

α/3

  • approximation for TSP(2, α).

Corollary. The T3-algorithm yields a (2·3α−1)-approximation for TSP( · , α) if α ≥ 2.

slide-85
SLIDE 85

Result #2

Theorem. For α ≥ 2, the geometric T3-algorithm yields a

  • 3α−1 +

√ 6

α/3

  • approximation for TSP(2, α).

Corollary. The T3-algorithm yields a (2·3α−1)-approximation for TSP( · , α) if α ≥ 2. 5 10 15 20 25 30 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3α−1 + √ 6

α/3

α

2 · 3α−1

slide-86
SLIDE 86

Result #2

Theorem. For α ≥ 2, the geometric T3-algorithm yields a

  • 3α−1 +

√ 6

α/3

  • approximation for TSP(2, α).

Corollary. The T3-algorithm yields a (2·3α−1)-approximation for TSP( · , α) if α ≥ 2. GeometricT3(tree T, e = u1u2 of T) for i ← 1 to 2 do Ti ← component of T − e that contains ui ... pick an edge ei = uiwi incident to ui in Ti ... Πi ← GeometricT3(Ti, ei) − ei return Π1 + e + Π2 + w1w2

slide-87
SLIDE 87

Result #2

Theorem. For α ≥ 2, the geometric T3-algorithm yields a

  • 3α−1 +

√ 6

α/3

  • approximation for TSP(2, α).

Corollary. The T3-algorithm yields a (2·3α−1)-approximation for TSP( · , α) if α ≥ 2. GeometricT3(tree T, e = u1u2 of T) for i ← 1 to 2 do Ti ← component of T − e that contains ui ... pick an edge ei = uiwi incident to ui in Ti ... Πi ← GeometricT3(Ti, ei) − ei return Π1 + e + Π2 + w1w2 MST w.r.t. | · |α

slide-88
SLIDE 88

Result #2

Theorem. For α ≥ 2, the geometric T3-algorithm yields a

  • 3α−1 +

√ 6

α/3

  • approximation for TSP(2, α).

Corollary. The T3-algorithm yields a (2·3α−1)-approximation for TSP( · , α) if α ≥ 2. GeometricT3(tree T, e = u1u2 of T) for i ← 1 to 2 do Ti ← component of T − e that contains ui ... pick an edge ei = uiwi incident to ui in Ti ... Πi ← GeometricT3(Ti, ei) − ei return Π1 + e + Π2 + w1w2 s.t. the angle ∠eei is min!

  • MST w.r.t. | · |α
slide-89
SLIDE 89

Result #2

Theorem. For α ≥ 2, the geometric T3-algorithm yields a

  • 3α−1 +

√ 6

α/3

  • approximation for TSP(2, α).

Corollary. The T3-algorithm yields a (2·3α−1)-approximation for TSP( · , α) if α ≥ 2. GeometricT3(tree T, e = u1u2 of T) for i ← 1 to 2 do Ti ← component of T − e that contains ui ... pick an edge ei = uiwi incident to ui in Ti ... Πi ← GeometricT3(Ti, ei) − ei return Π1 + e + Π2 + w1w2 s.t. the angle ∠eei is min!

  • u1

u2 e e1 e2 w1 w2 γ1 γ2 MST w.r.t. | · |α

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SLIDE 90

Main Insight

Why can we bound γ1 (and γ2) from above?

slide-91
SLIDE 91

Main Insight

Why can we bound γ1 (and γ2) from above?

  • If deg u1 ≤ 2, then we can’t.

u1 e γ1

slide-92
SLIDE 92

Main Insight

Why can we bound γ1 (and γ2) from above?

  • If deg u1 ≤ 2, then we can’t.

But then e is used by at most one 3-shortcut and maybe a 2-shortcut u1 e γ1

slide-93
SLIDE 93

Main Insight

Why can we bound γ1 (and γ2) from above?

  • If deg u1 ≤ 2, then we can’t.

But then e is used by at most one 3-shortcut and maybe a 2-shortcut—but those are cheap. u1 e γ1

slide-94
SLIDE 94

Main Insight

Why can we bound γ1 (and γ2) from above?

  • If deg u1 ≤ 2, then we can’t.

But then e is used by at most one 3-shortcut and maybe a 2-shortcut—but those are cheap.

  • Otherwise recall that in the MST (w.r.t. | · | and w.r.t. | · |α)

edges incident to the same vertex make angles ≥ 60◦. u1 e γ1

slide-95
SLIDE 95

Main Insight

Why can we bound γ1 (and γ2) from above?

  • If deg u1 ≤ 2, then we can’t.

But then e is used by at most one 3-shortcut and maybe a 2-shortcut—but those are cheap.

  • Otherwise recall that in the MST (w.r.t. | · | and w.r.t. | · |α)

edges incident to the same vertex make angles ≥ 60◦. u1 e u1 e

≥ 60◦

γ1

slide-96
SLIDE 96

Main Insight

Why can we bound γ1 (and γ2) from above?

  • If deg u1 ≤ 2, then we can’t.

But then e is used by at most one 3-shortcut and maybe a 2-shortcut—but those are cheap.

  • Otherwise recall that in the MST (w.r.t. | · | and w.r.t. | · |α)

edges incident to the same vertex make angles ≥ 60◦. u1 e u1 e

≥ 60◦ γ1 ≤ 150◦

γ1

slide-97
SLIDE 97

Main Insight

Why can we bound γ1 (and γ2) from above?

  • If deg u1 ≤ 2, then we can’t.

But then e is used by at most one 3-shortcut and maybe a 2-shortcut—but those are cheap.

  • Otherwise recall that in the MST (w.r.t. | · | and w.r.t. | · |α)

edges incident to the same vertex make angles ≥ 60◦. Thus, there is an edge e1 incident to u1 with ∠ee1 ≤ 150◦. u1 e u1 e

≥ 60◦ γ1 ≤ 150◦

γ1

slide-98
SLIDE 98

Conclusions

  • Geometry helps!
slide-99
SLIDE 99

Conclusions

  • Geometry helps!

We have improved approx. of TSP(2,2) from factor 6 to 5. There is a lower bound of 4 4

11 for MST-based methods.

slide-100
SLIDE 100

Conclusions

  • Geometry helps!
  • What about complexity?

We have improved approx. of TSP(2,2) from factor 6 to 5. There is a lower bound of 4 4

11 for MST-based methods.

slide-101
SLIDE 101

Conclusions

  • Geometry helps!
  • What about complexity?

TSP(d, α) is APX-hard for any d ≥ 3 and α > 1! This is in sharp contrast with Euclidean TSP. We have improved approx. of TSP(2,2) from factor 6 to 5. There is a lower bound of 4 4

11 for MST-based methods.

slide-102
SLIDE 102

Conclusions

  • Geometry helps!
  • What about complexity?

TSP(d, α) is APX-hard for any d ≥ 3 and α > 1! This is in sharp contrast with Euclidean TSP. We have improved approx. of TSP(2,2) from factor 6 to 5. There is a lower bound of 4 4

11 for MST-based methods.

  • What about allowing the salesperson to revisit cities?
slide-103
SLIDE 103

Conclusions

  • Geometry helps!
  • What about complexity?

TSP(d, α) is APX-hard for any d ≥ 3 and α > 1! This is in sharp contrast with Euclidean TSP. We have improved approx. of TSP(2,2) from factor 6 to 5. There is a lower bound of 4 4

11 for MST-based methods.

  • What about allowing the salesperson to revisit cities?

– Rev-TSP(d, α) is APX-hard for any d ≥ 3 and α > 1. – Rev-TSP(2, α) has a PTAS for any α ≥ 2. – Rev-TSP( · , α) has a quasi-PTAS for any 0 < α < 1.

slide-104
SLIDE 104

Conclusions

  • Geometry helps!
  • What about complexity?

TSP(d, α) is APX-hard for any d ≥ 3 and α > 1! This is in sharp contrast with Euclidean TSP. We have improved approx. of TSP(2,2) from factor 6 to 5. There is a lower bound of 4 4

11 for MST-based methods.

  • What about allowing the salesperson to revisit cities?

– Rev-TSP(d, α) is APX-hard for any d ≥ 3 and α > 1. – What about Rev-TSP(2, α) with 1 < α < 2? – Rev-TSP(2, α) has a PTAS for any α ≥ 2. – Rev-TSP( · , α) has a quasi-PTAS for any 0 < α < 1.

slide-105
SLIDE 105

Conclusions

  • Geometry helps!
  • What about complexity?

TSP(d, α) is APX-hard for any d ≥ 3 and α > 1! This is in sharp contrast with Euclidean TSP. We have improved approx. of TSP(2,2) from factor 6 to 5. There is a lower bound of 4 4

11 for MST-based methods.

  • What about allowing the salesperson to revisit cities?

– Rev-TSP(d, α) is APX-hard for any d ≥ 3 and α > 1. – What about Rev-TSP(2, α) with 1 < α < 2? – Rev-TSP(2, α) has a PTAS for any α ≥ 2. – Rev-TSP( · , α) has a quasi-PTAS for any 0 < α < 1. (At least as hard as TSP in weighted planar graphs!)

slide-106
SLIDE 106

Conclusions

  • Geometry helps!
  • What about complexity?

TSP(d, α) is APX-hard for any d ≥ 3 and α > 1! This is in sharp contrast with Euclidean TSP. We have improved approx. of TSP(2,2) from factor 6 to 5. There is a lower bound of 4 4

11 for MST-based methods.

  • What about allowing the salesperson to revisit cities?

– Rev-TSP(d, α) is APX-hard for any d ≥ 3 and α > 1. – What about Rev-TSP(2, α) with 1 < α < 2? – Rev-TSP(2, α) has a PTAS for any α ≥ 2. – Rev-TSP( · , α) has a quasi-PTAS for any 0 < α < 1. (At least as hard as TSP in weighted planar graphs!)

slide-107
SLIDE 107

Result #3

Theorem. TSP(d, α) and Rev-TSP(d, α) are APX-hard for any d ≥ 3 and any α > 1.

slide-108
SLIDE 108

Result #3

Theorem. TSP(d, α) and Rev-TSP(d, α) are APX-hard for any d ≥ 3 and any α > 1. Proof sketch. Reduction from {1, 2}-TSP, which is APX-hard.

slide-109
SLIDE 109

Result #3

Theorem. TSP(d, α) and Rev-TSP(d, α) are APX-hard for any d ≥ 3 and any α > 1. v1 v2 v3 v4

e1 e2 e3 e4 e5 e6

Proof sketch. Reduction from {1, 2}-TSP, which is APX-hard. weight 2 weight 1

slide-110
SLIDE 110

Result #3

Theorem. TSP(d, α) and Rev-TSP(d, α) are APX-hard for any d ≥ 3 and any α > 1. v1 v2 v3 v4

e1 e2 e3 e4 e5 e6

Proof sketch. Reduction from {1, 2}-TSP, which is APX-hard. weight 2 weight 1

slide-111
SLIDE 111

Result #3

Theorem. TSP(d, α) and Rev-TSP(d, α) are APX-hard for any d ≥ 3 and any α > 1. v1 v2 v3 v4 v1 v2 v3 v4

n n n n2 nm 2n e1 e2 e3 e4 e5 e6 e1 e2 e3 e4 e5 e6

Proof sketch. Reduction from {1, 2}-TSP, which is APX-hard. weight 2 weight 1

slide-112
SLIDE 112

Result #3

Theorem. TSP(d, α) and Rev-TSP(d, α) are APX-hard for any d ≥ 3 and any α > 1. v1 v2 v3 v4 v1 v2 v3 v4

n n n n2 nm 2n e1 e2 e3 e4 e5 e6 e1 e2 e3 e4 e5 e6

Proof sketch. Reduction from {1, 2}-TSP, which is APX-hard. weight 2 weight 1

slide-113
SLIDE 113

Result #3

Theorem. TSP(d, α) and Rev-TSP(d, α) are APX-hard for any d ≥ 3 and any α > 1. v1 v2 v3 v4 v1 v2 v3 v4

n n n n2 nm 2n e1 e2 e3 e4 e5 e6 e1 e2 e3 e4 e5 e6

Proof sketch. Reduction from {1, 2}-TSP, which is APX-hard. weight 2 weight 1

slide-114
SLIDE 114

Result #3

Theorem. TSP(d, α) and Rev-TSP(d, α) are APX-hard for any d ≥ 3 and any α > 1. v1 v2 v3 v4 v1 v2 v3 v4

n n n n2 nm 2n e1 e2 e3 e4 e5 e6 e1 e2 e3 e4 e5 e6

Proof sketch. Reduction from {1, 2}-TSP, which is APX-hard. weight 2 weight 1

slide-115
SLIDE 115

Conclusion (once more :-)

  • Geometry helps!
  • What about complexity?

TSP(d, α) is APX-hard for any d ≥ 3 and α > 1! This is in sharp contrast with Euclidean TSP. We have improved approx. of TSP(2,2) from factor 6 to 5. There is a lower bound of 4 4

11 for MST-based methods.

  • What about allowing the salesman to revisit cities?

– Rev-TSP(d, α) is APX-hard for any d ≥ 3 and α > 1. – What about Rev-TSP(2, α) with 1 < α < 2? – Rev-TSP(2, α) has a PTAS for any α ≥ 2. – Rev-TSP( · , α) has a quasi-PTAS for any 0 < α < 1. (At least as hard as TSP in weighted planar graphs!)