ALTERNATING PATHS AND CYCLES OF MINIMUM LENGTH William Evans - - PowerPoint PPT Presentation

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ALTERNATING PATHS AND CYCLES OF MINIMUM LENGTH William Evans - - PowerPoint PPT Presentation

Feb 03, 2023 228 likes •821 views

ALTERNATING PATHS AND CYCLES OF MINIMUM LENGTH William Evans University of British Columbia, Canada Giuseppe Liotta Universit a degli Studi di Perugia, Italy Henk Meijer U. C. Roosevelt, the Netherlands Stephen Wismath University of

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ALTERNATING PATHS AND CYCLES OF MINIMUM LENGTH William Evans Giuseppe Liotta Henk Meijer Stephen Wismath

University of British Columbia, Canada Universit´ a degli Studi di Perugia, Italy

  • U. C. Roosevelt, the Netherlands

University of Lethbridge, Canada

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Motivating problem Draw planar graph on given point set to minimize total edge length [Chan et al. GD13]

  • vertex maps to point of the same color
  • each vertex has distinct color
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Motivating problem Draw planar graph on given point set to minimize total edge length [Chan et al. GD13]

  • vertex maps to point of the same color

This is NP-hard [Bastert and Fekete 96]

  • each vertex has distinct color
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Our problem

  • vertex maps to point of the same color
  • each vertex has distinct color

Draw planar graph on given point set to minimize total edge length

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Our problem

  • vertex maps to point of the same color
  • each vertex has distinct color

Draw planar graph on given point set to minimize total edge length alternating path/cycle

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Our problem is NP-hard Draw planar alternating path/cycle on given point set to minimize total edge length Idea: Reduce from EXACT COVER

Use (modified) reduction to Euclidean TSP [Papadimitriou 77]

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Our problem Draw planar alternating path/cycle on given colinear point set to minimize total edge length

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Cycles: A lower bound

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Cycles: A lower bound Minimum number of edges crossing this gap? (for any alternating cycle)

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Cycles: A lower bound Minimum number of edges crossing this gap? (for any alternating cycle) 2

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Cycles: A lower bound Minimum number of edges crossing this gap? (for any alternating cycle) 2

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Cycles: A lower bound Minimum number of edges crossing this gap? (for any alternating cycle) 2 2

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Cycles: A lower bound Minimum number of edges crossing this gap? (for any alternating cycle) 2 2

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Cycles: A lower bound Minimum number of edges crossing this gap? (for any alternating cycle) 2 2 2

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Cycles: A lower bound Minimum number of edges crossing this gap? (for any alternating cycle) 2 2 2

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Cycles: A lower bound Minimum number of edges crossing this gap? (for any alternating cycle) 2 2 2 4

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Cycles: A lower bound 2 2 2 4 ci = 2 max{|ri − bi|, 1} ri = red, bi = blue points before gapi 6 4 Lemma 1. Minimum number of edges crossing gapi is 6 8 10 8 6 4 2

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Cycles: A lower bound 2 2 2 4 ci = 2 max{|ri − bi|, 1} ri = red, bi = blue points before gapi 6 4 Lemma 1. Minimum number of edges crossing gapi is Proof: Each cycle component to the left of gapi has the same number of red and blue points ±1. 6 8 10 8 6 4 2

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Cycles: A lower bound 2 2 2 4 ci = 2 max{|ri − bi|, 1} ri = red, bi = blue points before gapi 6 4 Cycle length ≥

i ci|gapi|

Lemma 1. Minimum number of edges crossing gapi is 6 8 10 8 6 4 2

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Cycles: Matching the lower bound with 2 bends Drawing Rules Invariants At gapi:

  • 1. Number of red components = max{ri − bi, 0}.
  • 2. Number of blue components = max{bi − ri, 0}.
  • 3. If ri = bi, one red/blue component spans spine.
  • 4. Two closest components to spine are not both above
  • r below.
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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules

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Cycles: Matching the lower bound with 2 bends Drawing Rules Theorem 1. Exists O(n log n)-time algorithm to compute a shortest planar alternating cycle on colinear

  • points. Each edge has ≤ 2 bends.
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Paths: A lower bound 2 3 Lemma 2. Minimum number of edges crossing gapi is 5 7 9 7 6 4 2 Given path endpoints r and b. ci =    2 max{|ri−bi|, 1} if r, b same side 1+2 max{bi−ri, ri−bi−1} if r left of gapi 1+2 max{ri−bi, bi−ri−1} if b left of gapi 5 2 2 3 r b

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Paths: A lower bound 2 3 Lemma 2. Minimum number of edges crossing gapi is Proof: Same. Component with red endpoint can have

  • ne more red than blue points, but zero more blue than

red points. 5 7 9 7 6 4 2 Given path endpoints r and b. ci =    2 max{|ri−bi|, 1} if r, b same side 1+2 max{bi−ri, ri−bi−1} if r left of gapi 1+2 max{ri−bi, bi−ri−1} if b left of gapi 5 2 2 3 r b

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Paths: A lower bound 2 3 r to b Path length ≥

i ci|gapi|

Lemma 2. Minimum number of edges crossing gapi is 5 7 9 7 6 4 2 Given path endpoints r and b. ci =    2 max{|ri−bi|, 1} if r, b same side 1+2 max{bi−ri, ri−bi−1} if r left of gapi 1+2 max{ri−bi, bi−ri−1} if b left of gapi 5 2 2 3 r b

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Paths: Matching the lower bound with 2 bends Given path endpoints r and b. Use (almost) the same algorithm as for cycles to find a path whose length matches the lower bound.

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Paths: Matching the lower bound with 2 bends Given path endpoints r and b. Calculate the r to b Path length lower bound for all r and b and pick the minimum. Use (almost) the same algorithm as for cycles to find a path whose length matches the lower bound. O(n2) time

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Paths: Matching the lower bound with 2 bends Given path endpoints r and b. Calculate the r to b Path length lower bound for all r and b and pick the minimum. Use (almost) the same algorithm as for cycles to find a path whose length matches the lower bound. O(n2) time Theorem 2. Exists O(n2)-time algorithm to compute a shortest planar alternating path on colinear points. Each edge has ≤ 2 bends.

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Extending to more than two colors ci = 2 max{|ri − gi|, |gi − bi|, |bi − ri|, 1} ri = red, gi = green, bi = blue points before gapi Lemma 3. Minimum number of edges crossing gapi is Similar algorithm achieves lower bound. (O(n) bends)

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Extending to more than two colors ci = 2 max{|ri − gi|, |gi − bi|, |bi − ri|, 1} ri = red, gi = green, bi = blue points before gapi Lemma 3. Minimum number of edges crossing gapi is Similar algorithm achieves lower bound. (O(n) bends) Four colors Lower bound cannot be achieved.

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Open Problems

  • Shortest alternating path in o(n2) time.
  • Shortest 3-color path/cycle with o(n)

bends.

  • Shortest 4-color path/cycle.
  • Shortest arbitrary (not alternating) 2-color

path/cycle.

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Open Problems

  • Shortest alternating path in o(n2) time.
  • Shortest 3-color path/cycle with o(n)

bends.

  • Shortest 4-color path/cycle.
  • Shortest arbitrary (not alternating) 2-color

path/cycle.

Thank you