The document discusses connected dominating sets and short cycles. It begins by explaining that excluding longer cycles makes related problems easier to solve. Specifically, it shows that on graphs with girth at least five, high degree vertices must be in any minimum dominating set. However, this does not hold for connected dominating sets, since connectivity must also be maintained. It then describes how to obtain fixed-parameter tractable algorithms for connected dominating set problems by guessing the minimum dominating set and extending it. It also shows that these problems do not admit polynomial kernels by providing a reduction from Fair Connected Colors, which is W-hard.

1. Connected Dominating Set and
Short Cycles

2. Connected Dominating Set and
Short Cycles
Joint work with Geevarghese Philip, Venkatesh Raman
and Saket Saurabh
The Institute of Mathematical Sciences, Chennai

3. Polynomial Kernels
No Polynomial Kernels, but FPT
W-hard
e impact of excluding short cycles.
Talk Outline

4. Polynomial Kernels
No Polynomial Kernels, but FPT
W-hard
e impact of excluding short cycles.
Talk Outline

5. Polynomial Kernels
No Polynomial Kernels, but FPT
W-hard
Girth: at least seven
Talk Outline

6. Polynomial Kernels ( 7)
No Polynomial Kernels, but FPT
W-hard
Girth: five or six
Talk Outline

7. Polynomial Kernels ( 7)
No Polynomial Kernels, but FPT (5, 6)
W-hard
Girth: at most four
Talk Outline

8. Polynomial Kernels ( 7)
No Polynomial Kernels, but FPT (5, 6)
W-hard ( 4)
Girth: at least seven
Talk Outline

9. Polynomial Kernels ( 7)
No Polynomial Kernels, but FPT (5, 6)
W-hard ( 4)
e longer the cycles we exclude, the easier the problem becomes to solve.
Talk Outline

10. Polynomial Kernels ( 7)
No Polynomial Kernels, but FPT (5, 6)
W-hard ( 4)
e longer the cycles we exclude, the easier the problem becomes to solve.
Talk Outline

11. Part 1
Why Its Useful
to not have short cycles

12. Consider the problem of domination:

13. Consider the problem of domination:
Find a subset of at most k vertices S such that every v ∈ V is either in S or
has a neighbor in S.

14. Consider the problem of domination:
Find a subset of at most k vertices S such that every v ∈ V is either in S or
has a neighbor in S.
Let us consider graphs of girth at least five.

15. Consider the problem of domination:
Find a subset of at most k vertices S such that every v ∈ V is either in S or
has a neighbor in S.
Let us consider graphs of girth at least five.
We begin by examining vertices of degree more than k.

16. Does there exist a dominating set of size at most k that does not include
the red vertex?

17. Who dominates the neighbors of the red vertex?
Does there exist a dominating set of size at most k that does not include
the red vertex?

18. Who dominates the neighbors of the red vertex?
Clearly, it is not the case that each one dominates itself.

19. Who dominates the neighbors of the red vertex?
Clearly, it is not the case that each one dominates itself.

20. Either a vertex from within the neighborhood dominates at least two of
vertices...

23. or a vertex from “outside” the neighborhood dominates at least two of the
vertices...

26. High Degree Vertices
On graphs that have girth at least five:
Any vertex of degree more than k belongs to any dominating set of size at
most k.

27. Notice that the number of reds is at most k and the number of blues is at
most k2 .

28. Greens in the neighborhood of a blue vertex cannot have a common red
neighbor.

29. us, for every blue vertex, there are at most k green vertices.

30. vertices that have been dominated that have
no edges into blue are irrelevant.

31. So now all is bounded, and we have a happy ending: a kernel on at most
O(k + k2 + k3 )
vertices.

32. Part 2
FPT Algorithms
and Polynomial Kernels

33. In the connected dominating set situation, we have to backtrack a small
way in the story so far to see what fails, and how badly.

34. High Degree Vertices
On graphs that have girth at least five:
Any vertex of degree more than k belongs to any dominating set of size at
most k.

35. High Degree Vertices
On graphs that have girth at least five:
connected
Any vertex of degree more than k belongs to any dominating set of size at
most k.

36. Notice that the number of reds is at most k and the number of blues is
again at most k2 .

37. Greens in the neighborhood of a blue vertex cannot have a common red
neighbor.

38. us, for every blue vertex, there are at most k green vertices.

39. vertices that have been dominated that have
no edges into blue are irrelevant.

40. vertices that have been dominated that have
no edges into blue are irrelevant.
Not any more...

41. any connected dom set contains a minimal dom set
that resides in the bounded part of the graph.

42. Now we know what to do:

43. Now we know what to do:
Guess the minimal dominating set.

44. Now we know what to do:
Guess the minimal dominating set.
Extend it to a connected dominating set by an application of a Steiner
Tree algorithm.

45. Now we know what to do:
Guess the minimal dominating set.
Extend it to a connected dominating set by an application of a Steiner
Tree algorithm.
is is evidently FPT.

46. Now we know what to do:
Guess the minimal dominating set.
Extend it to a connected dominating set by an application of a Steiner
Tree algorithm.
is is evidently FPT. What about kernels?

47. It turns out that if G did not admit cycles of length six and less, then the
number of green vertices can be bounded.

48. It turns out that if G did not admit cycles of length six and less, then the
number of green vertices can be bounded.
And we will show (later) that on graphs of girth five, connected
dominating set is unlikely to admit polynomial kernels.

49. Low Degree Vertices
If there is blue pendant vertex, make its neighbor red.
If there is a green vertex that is pendant, delete it.
If there is a blue vertex that is isolated, say no.

53. no pair 'sees' more than one vertex.

56. So now the green vertices are bounded:
• At most O(k3 ) with at least one neighbor in the blues,
• At most O(k2 ) with only red neighbors,
• At most O(k2 ) with no blue neighbors and at least one white
neighbor.

57. So now the green vertices are bounded:
• At most O(k3 ) with at least one neighbor in the blues,
• At most O(k2 ) with only red neighbors,
• At most O(k2 ) with no blue neighbors and at least one white
neighbor.
is gives us a O(k3 ) vertex kernel.

58. The Hard Part
Infeasibility of Polynomial Kernelization
and W-hardness

59. Introducing...

60. Introducing...
F C C

61. Introducing...
F C C
NP-complete FPT Compositional

62. Introducing...
F C C
NP-complete FPT Compositional
Ready for reduction!

63. Promise: Every vertex has degree at most one into
any color class, and every color class is independent.
Question: Does there exist a colorful tree?

64. Add global vertices (with guards) to each of the color classes. subdivide
the edges,
add a global vertex for the newly added vertices (with guards).

65. Between a pair of color classes:

66. Between a pair of color classes:
• subdivide the edges,

67. Between a pair of color classes:
• subdivide the edges,
• add a global vertex for the newly added vertices

68. Between a pair of color classes:
• subdivide the edges,
• add a global vertex for the newly added vertices (with guards).

69. For vertices that don’t look inside the neighboring color class, add a path of
length two to the global vertex. add a global vertex for the newly added
vertices (with guards).

70. Here is (1/k2 )th of the full picture. subdivide the edges,
add a global vertex for the newly added vertices (with guards).

71. Notice that we have ended up with a bipartite graph... subdivide the edges,
add a global vertex for the newly added vertices (with guards).

72. ....with a cycle of length six. subdivide the edges,
add a global vertex for the newly added vertices (with guards).

73. Suppose we have a colorful tree in the source instance.

74. Consider the same subset.

75. (along with the subdivided vertices along the edges of the tree)

76. is is a connected subset, but not dominating.

77. To fix this, add all the global vertices to this set.

78. (k)
ere are k + 2 such vertices.

79. Now we have a dominating set, but it’s not connected any more!

80. Let’s look more closely.

81. e vertices global to color classes are connected to the tree.

82. Some of the other vertices are connected too!

83. Each of the rest corresponds to a non-edge in the tree,

84. for which we add the path of length two to the picture.

85. e size of the connected dominating set thus obtained is:

86. e size of the connected dominating set thus obtained is:
(k)
2 2 + 2k

87. e size of the connected dominating set thus obtained is:
(k)
2 2 + 2k
Two vertices for every pair of original vertices in the tree.
(A global vertex, and a neighbor.)

88. e size of the connected dominating set thus obtained is:
(k)
2 2 + 2k
Two vertices for every original vertex in the tree:
itself, and the corresponding global vertex.

89. And that works out to...
(k)
2 2 + 2k
= k2 + k

90. Let S be a connected dominating set of size at most:
( ) ( )
k k
+ +k+k
2 2

91. Let S be a connected dominating set of size at most:
( ) ( )
k k
+ +k+k
2 2
Of course, global vertices are forced in S, because of the guard vertices.

92. Let S be a connected dominating set of size at most:
( ) ( )
k k
+ +k+k
2 2
Since S is a connected subset, each of these vertices have at least one
neighbor in S.

93. Let S be a connected dominating set of size at most:
( ) ( )
k k
+ +k+k
2 2
Since S is a connected subset, each of these vertices have at least one
neighbor in S.
Because of the budget, each of them have...

94. Let S be a connected dominating set of size at most:
( ) ( )
k k
+ +k+k
2 2
Since S is a connected subset, each of these vertices have exactly one
neighbor in S.
Because of the budget, each of them have...

95. us, the dominating set picks exactly one vertex from each color class.

96. us, the dominating set picks exactly one vertex from each color class.
• Neglect the global vertices and “subdivision vertices” of degree one, to
be left with a connected subtree with subdivided edges.

97. us, the dominating set picks exactly one vertex from each color class.
• Neglect the global vertices and “subdivision vertices” of degree one, to
be left with a connected subtree with subdivided edges.
• is can be easily pulled back to a colorful tree of the original graph.

98. e W-hardness result.

99. A dominating set
instance
G

100. We begin by
making a copy of the graph.
G

101. G G

102. en, for every edge in G,
we stretch it across the copies.

103. en, for every edge in G,
we stretch it across the copies.

104. Finally, add a new vertex to each copy,
with one of them global to the other copy.

105. Any old dominating set + new global vertex
= new connected dominating set.

106. Any new dominating set contains the newly added
global vertex WLOG.

107. Consider now the vertices of the
new connected dominating set.

108. Observe that they form a
dominating set of G.

109. Takk