Minimizing and Computing Inverse Geodesic Length on Trees Thesis B Presentation Joshua Lau Supervisor: A/Prof. Serge Gaspers UNSW Sydney October 9, 2018 Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 1 / 24
Outline Introduction 1 Minimizing IGL on Trees 2 Computing IGL on Treelike Graphs 3 Concluding Remarks 4 Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 2 / 24
Definitions Definition (Inverse Geodesic Length) Let G = ( V , E ) be an undirected graph, with edges of unit length. We define the Inverse Geodesic Length of G as 1 � IGL ( G ) = d ( u , v ) { u , v }⊆ V where d ( u , v ) is the length of a shortest path from u to v . 1 Note: If u and v are disconnected then d ( u , v ) = ∞ so we take d ( u , v ) = 0. Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 3 / 24
Definitions Problem ( MinIGL ) A graph G, an integer k ≤ | V | and a target IGL T Input: Question: Does there exist X ⊆ V such that | X | ≤ k and IGL ( G − X ) ≤ T? When our budget k is 0, the problem is equivalent to computing the IGL. When our target T is 0, the problem is equivalent to Vertex Cover. TreeMinIGL is MinIGL where G is a tree. Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 4 / 24
Example Consider the following graph G when k = 2. Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 5 / 24
Example Consider the following graph G when k = 2. IGL ( G ) = 9 × 1 1 + 5 × 1 2 + 5 × 1 3 + 2 × 1 4 = 13 2 3 Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 5 / 24
Example Delete these k = 2 vertices to obtain G ′ . Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 5 / 24
Example Delete these k = 2 vertices to obtain G ′ . IGL ( G ′ ) = 3 × 1 1 + 1 × 1 2 = 3 1 2 Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 5 / 24
Outline Introduction 1 Minimizing IGL on Trees 2 Computing IGL on Treelike Graphs 3 Concluding Remarks 4 Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 6 / 24
Algorithm Key Result We can solve TreeMinIGL in subexponential time, and polynomial space. Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 7 / 24
Algorithm Key Result We can solve TreeMinIGL in subexponential time, and polynomial space. Idea : combine two different algorithms, each using polynomial space. Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 7 / 24
Algorithm Key Result We can solve TreeMinIGL in subexponential time, and polynomial space. Idea : combine two different algorithms, each using polynomial space. � = O ( n k ) subsets and APSP on Naive O ( n k +2 ) = 2 O ( k log n ) time: � n k a tree in O ( n 2 ). Useful when k is (somewhat) small. Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 7 / 24
Algorithm Key Result We can solve TreeMinIGL in subexponential time, and polynomial space. Idea : combine two different algorithms, each using polynomial space. � = O ( n k ) subsets and APSP on Naive O ( n k +2 ) = 2 O ( k log n ) time: � n k a tree in O ( n 2 ). Useful when k is (somewhat) small. We give an algorithm that runs in time O ∗ � 2 poly( n k ) � . Useful when k is relatively large compared to n . This is FPT for parameter n k . Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 7 / 24
Algorithm Key Result We can solve TreeMinIGL in subexponential time, and polynomial space. Idea : combine two different algorithms, each using polynomial space. � = O ( n k ) subsets and APSP on Naive O ( n k +2 ) = 2 O ( k log n ) time: � n k a tree in O ( n 2 ). Useful when k is (somewhat) small. We give an algorithm that runs in time O ∗ � 2 poly( n k ) � . Useful when k is relatively large compared to n . This is FPT for parameter n k . Combining these gives a subexponential (2 o ( n ) ) time solution. Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 7 / 24
Exploiting large k Define an optimal subset of an instance to be a set of vertices which minimizes the IGL. Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 8 / 24
Exploiting large k Define an optimal subset of an instance to be a set of vertices which minimizes the IGL. Observation : when k is large compared to n , we expect remaining components (trees) to be small in an optimal solution. Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 8 / 24
Exploiting large k Define an optimal subset of an instance to be a set of vertices which minimizes the IGL. Observation : when k is large compared to n , we expect remaining components (trees) to be small in an optimal solution. Theorem If S is an optimal subset of a TreeMinIGL instance ( T , k ) , then every � ( n / k ) 5 � vertices. remaining component has at most L ( n , k ) = O Proof. Omitted from presentation (see thesis report). Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 8 / 24
Exploiting large k Root the tree and use DP to determine the optimal structure of remaining components. State is (subtree, budget of deletions). Recurrence: either subtree root is deleted or it is not. If deleted, then consider children’s subtrees separately. Optimally dictate how the budget is to be shared, using DP. Otherwise, root is in some remaining component. We determine this by finding its structure and mapping (diagram next slide). Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 9 / 24
Exploiting large k Recurrence (continued): T Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 10 / 24
Exploiting large k Recurrence (continued): Try every structure : an ordered tree T ′ with no more than � 4 L � L = L ( n , k ) vertices. There are O of these, by bounding sum √ L L of Catalan numbers (Topley, 2016). T ′ T 1 2 3 4 5 6 Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 10 / 24
Exploiting large k Recurrence (continued): Try every structure : an ordered tree T ′ with no more than � 4 L � L = L ( n , k ) vertices. There are O of these, by bounding sum √ L L of Catalan numbers (Topley, 2016). T ′ T mapping 1 1 2 3 2 3 4 5 6 4 5 6 For each, find optimal mapping : assignment of vertices in T ′ to distinct vertices in T . This forces some vertices to be deleted. Recurse on the remaining subtrees, dictating how the remaining budget should be shared, using DP once again. Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 10 / 24
Exploiting large k Recurrence (continued): Try every structure : an ordered tree T ′ with no more than � 4 L � L = L ( n , k ) vertices. There are O of these, by bounding sum √ L L of Catalan numbers (Topley, 2016). T ′ T mapping 1 1 2 3 2 3 4 5 6 4 5 6 For each, find optimal mapping : assignment of vertices in T ′ to distinct vertices in T . This forces some vertices to be deleted. Recurse on the remaining subtrees, dictating how the remaining budget should be shared, using DP once again. Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 10 / 24
Exploiting large k For a fixed T ′ , we can determine the optimal mapping in polynomial time and space. We can enumerate all T ′ in polynomial space, so our algorithm uses polynomial space. � 4 L L n 2 � = 2 O (( n / k ) 5 +log n ) . This is FPT for Our algorithm takes time O √ parameter n k . Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 11 / 24
Exploiting large k For a fixed T ′ , we can determine the optimal mapping in polynomial time and space. We can enumerate all T ′ in polynomial space, so our algorithm uses polynomial space. � 4 L L n 2 � = 2 O (( n / k ) 5 +log n ) . This is FPT for Our algorithm takes time O √ parameter n k . Recall the brute-force algorithm takes time 2 O ( k log n ) . Choosing brute-force whenever k ≤ n 5 / 6 log − 1 / 6 n , and our DP algorithm otherwise, gives a solution in time 2 O (( n log n ) 5 / 6 ) = 2 o ( n ) . Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 11 / 24
A generalisation Let f be a function mapping forests to reals, such that f ( F 1 ⊕ F 2 ) = f ( F 1 ) + f ( F 2 ) for any vertex-disjoint forests F 1 , F 2 . Consider the Min- f on Trees problem of choosing a set S of k vertices to delete from a tree T of n vertices, such that f ( T − S ) is minimised. Joshua Lau (UNSW Sydney) Minimizing and Computing IGL on Trees October 9, 2018 12 / 24
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