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

Transcription

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

2 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

3 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 This is NP-hard [Bastert and Fekete 96]

4 Our problem Draw planar graph on given point set to minimize total edge length vertex maps to point of the same color each vertex has distinct color

5 Our problem alternating path/cycle Draw planar graph on given point set to minimize total edge length vertex maps to point of the same color each vertex has distinct color

6 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]

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

8 Cycles: A lower bound

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

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

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

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

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

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

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

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

17 Cycles: A lower bound Lemma 1. Minimum number of edges crossing gap i is c i = 2 max{ r i b i, 1} r i = red, b i = blue points before gap i

18 Cycles: A lower bound Lemma 1. Minimum number of edges crossing gap i is c i = 2 max{ r i b i, 1} r i = red, b i = blue points before gap i Proof: Each cycle component to the left of gap i has the same number of red and blue points ±1.

19 Cycles: A lower bound Lemma 1. Minimum number of edges crossing gap i is c i = 2 max{ r i b i, 1} r i = red, b i = blue points before gap i Cycle length i c i gap i

20 Cycles: Matching the lower bound with 2 bends Drawing Rules Invariants At gap i : 1. Number of red components = max{r i b i, 0}. 2. Number of blue components = max{b i r i, 0}. 3. If r i = b i, one red/blue component spans spine. 4. Two closest components to spine are not both above or below.

21 Cycles: Matching the lower bound with 2 bends Drawing Rules

22 Cycles: Matching the lower bound with 2 bends Drawing Rules

23 Cycles: Matching the lower bound with 2 bends Drawing Rules

24 Cycles: Matching the lower bound with 2 bends Drawing Rules

25 Cycles: Matching the lower bound with 2 bends Drawing Rules

26 Cycles: Matching the lower bound with 2 bends Drawing Rules

27 Cycles: Matching the lower bound with 2 bends Drawing Rules

28 Cycles: Matching the lower bound with 2 bends Drawing Rules

29 Cycles: Matching the lower bound with 2 bends Drawing Rules

30 Cycles: Matching the lower bound with 2 bends Drawing Rules

31 Cycles: Matching the lower bound with 2 bends Drawing Rules

32 Cycles: Matching the lower bound with 2 bends Drawing Rules

33 Cycles: Matching the lower bound with 2 bends Drawing Rules

34 Cycles: Matching the lower bound with 2 bends Drawing Rules

35 Cycles: Matching the lower bound with 2 bends Drawing Rules

36 Cycles: Matching the lower bound with 2 bends Drawing Rules

37 Cycles: Matching the lower bound with 2 bends Drawing Rules

38 Cycles: Matching the lower bound with 2 bends Drawing Rules

39 Cycles: Matching the lower bound with 2 bends Drawing Rules

40 Cycles: Matching the lower bound with 2 bends Drawing Rules

41 Cycles: Matching the lower bound with 2 bends Drawing Rules

42 Cycles: Matching the lower bound with 2 bends Drawing Rules

43 Cycles: Matching the lower bound with 2 bends Drawing Rules

44 Cycles: Matching the lower bound with 2 bends Drawing Rules

45 Cycles: Matching the lower bound with 2 bends Drawing Rules

46 Cycles: Matching the lower bound with 2 bends Drawing Rules

47 Cycles: Matching the lower bound with 2 bends Drawing Rules

48 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.

49 Paths: A lower bound Given path endpoints r and b Lemma 2. Minimum number of edges crossing gap i is 2 max{ r i b i, 1} if r, b same side c i = r 1+2 max{b i r i, r i b i 1} if r left of gap i 1+2 max{r i b i, b i r i 1} if b left of gap i b

50 Paths: A lower bound Given path endpoints r and b Lemma 2. Minimum number of edges crossing gap i is 2 max{ r i b i, 1} if r, b same side c i = r 1+2 max{b i r i, r i b i 1} if r left of gap i 1+2 max{r i b i, b i r i 1} if b left of gap i Proof: Same. Component with red endpoint can have one more red than blue points, but zero more blue than red points. b

51 Paths: A lower bound Given path endpoints r and b Lemma 2. Minimum number of edges crossing gap i is 2 max{ r i b i, 1} if r, b same side c i = r 1+2 max{b i r i, r i b i 1} if r left of gap i 1+2 max{r i b i, b i r i 1} if b left of gap i b r to b Path length i c i gap i

52 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.

53 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. O(n 2 ) time Use (almost) the same algorithm as for cycles to find a path whose length matches the lower bound.

54 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. O(n 2 ) time Use (almost) the same algorithm as for cycles to find a path whose length matches the lower bound. Theorem 2. Exists O(n 2 )-time algorithm to compute a shortest planar alternating path on colinear points. Each edge has 2 bends.

55 Extending to more than two colors Lemma 3. Minimum number of edges crossing gap i is c i = 2 max{ r i g i, g i b i, b i r i, 1} r i = red, g i = green, b i = blue points before gap i Similar algorithm achieves lower bound. (O(n) bends)

56 Extending to more than two colors Lemma 3. Minimum number of edges crossing gap i is c i = 2 max{ r i g i, g i b i, b i r i, 1} r i = red, g i = green, b i = blue points before gap i Similar algorithm achieves lower bound. (O(n) bends) Four colors Lower bound cannot be achieved.

57 Open Problems Shortest alternating path in o(n 2 ) time. Shortest 3-color path/cycle with o(n) bends. Shortest 4-color path/cycle. Shortest arbitrary (not alternating) 2-color path/cycle.

58 Open Problems Shortest alternating path in o(n 2 ) 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