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Non-Strict Temporal Exploration Thomas Erlebach and Jakob T. Spooner School of Informatics, University of Leicester (Slides mostly prepared by Jakob) Algorithmic Aspects of Temporal Graphs III (Satellite Workshop of ICALP 2020) July 7, 2020


  1. Non-Strict Temporal Exploration Thomas Erlebach and Jakob T. Spooner School of Informatics, University of Leicester (Slides mostly prepared by Jakob) Algorithmic Aspects of Temporal Graphs III (Satellite Workshop of ICALP 2020) July 7, 2020

  2. Previous Work: Strict Temporal Exploration Definition (Temporal graph G ) Temporal graph G = � G 1 , ..., G L � : ◮ underlying graph G with N vertices ◮ sequence of static graphs G i ⊆ G with V ( G i ) = V ( G ) and E ( G i ) ⊆ E ( G ) ◮ time steps 1 ≤ i ≤ L , lifetime L Strict temporal walk: Traverse at most one edge per time step. Strict exploration schedule: ◮ Strict temporal walk W through G that visits all v ∈ V ( G ). ◮ Arrival time of W : time step when last vertex is reached. AATG 2020 Thomas Erlebach and Jakob T. Spooner 2

  3. Strict Temporal Exploration Problem (TEXP) Problem ( Strict Temporal Exploration ) Input: Temporal graph G , start vertex s ∈ V ( G ). Output: A strict exploration schedule starting from vertex s with earliest arrival time. Typical assumptions: ◮ Full dynamic behaviour of G is known in advance ◮ Each G i is connected, lifetime L ≥ N 2 . (Otherwise, NP-complete to decide if exploration schedule exists (Michail and Spirakis, 2014).) AATG 2020 Thomas Erlebach and Jakob T. Spooner 3

  4. Strict TEXP : Some Known Results ◮ Introduced as TEXP by Michail and Spirakis (2014): ◮ TEXP is NP-complete ◮ TEXP admits an O ( D )-approximation, where D is the temporal diameter ( D ≤ N ) ◮ E, Hoffmann and Kammer (2015): ◮ Worst-case exploration time is Θ( N 2 ) ◮ TEXP is O ( N 1 − ε )-inapproximable. ◮ E, Kammer, Luo, Sajenko and Spooner (2019): ◮ O ( d · N 1 . 75 ) steps suffice if each G i has max degree ≤ d ◮ The exploration time of various special classes of temporal graphs has also been studied. AATG 2020 Thomas Erlebach and Jakob T. Spooner 4

  5. Non-Strict Temporal Exploration AATG 2020 Thomas Erlebach and Jakob T. Spooner 5

  6. Non-Strict Temporal Graphs ◮ Allow an arbitrary number of edges to be crossed in the same time step. ◮ Observation: Only the connected components in each time step matter (for the exploration problem). Definition (Non-strict temporal graph G ) ◮ G = � G 1 , ..., G L � with vertex set V ( | V | = N ) and lifetime L ◮ Each G i is a partition { C i , 1 , ..., C i , s i } of V AATG 2020 Thomas Erlebach and Jakob T. Spooner 6

  7. Non-Strict Temporal Walks Definition (Non-strict temporal walk W ) A non-strict temporal walk W through a graph G = � G 1 , ..., G L � is a length k -sequence of components W = C 1 , j 1 , C 2 , j 2 , ..., C k , j k with k ∈ [ L ], satisfying the following properties: ◮ For all C i , j i ∈ W we have C i , j i ∈ G i . ◮ Additionally, C i , j i ∩ C i +1 , j i +1 � = ∅ for all i ∈ [ k − 1]. ◮ A walk W visits all v ∈ � k i =1 C i , j i . ◮ If � k i =1 C i , j i = V then W is a non-strict exploration schedule . AATG 2020 Thomas Erlebach and Jakob T. Spooner 7

  8. Example: Non-Strict Exploration Schedule t = 1 t = 2 t = 3 a b b e d C 1 , 1 C 3 , 1 b C 2 , 1 c d i a c d e c f C 2 , 2 C 1 , 2 f g f i C 3 , 2 g a g h C 2 , 3 h e C 1 , 3 i h AATG 2020 Thomas Erlebach and Jakob T. Spooner 8

  9. Example: Non-Strict Exploration Schedule t = 1 t = 2 t = 3 a b b e d C 1 , 1 C 3 , 1 b C 2 , 1 c d i a c d e c f C 2 , 2 C 1 , 2 f g f i C 3 , 2 g a g h C 2 , 3 h e C 1 , 3 i h AATG 2020 Thomas Erlebach and Jakob T. Spooner 8

  10. Example: Non-Strict Exploration Schedule t = 1 t = 2 t = 3 a b b e d C 1 , 1 C 3 , 1 b C 2 , 1 c d i a c d e c f C 2 , 2 C 1 , 2 f g f i C 3 , 2 g a g h C 2 , 3 h e C 1 , 3 i h AATG 2020 Thomas Erlebach and Jakob T. Spooner 8

  11. Example: Non-Strict Exploration Schedule t = 1 t = 2 t = 3 a b b e d C 1 , 1 C 3 , 1 b C 2 , 1 c d i a c d e c f C 2 , 2 C 1 , 2 f g f i C 3 , 2 g a g h C 2 , 3 h e C 1 , 3 i h AATG 2020 Thomas Erlebach and Jakob T. Spooner 8

  12. Example: Non-Strict Exploration Schedule t = 1 t = 2 t = 3 a b b e d C 1 , 1 C 3 , 1 b C 2 , 1 c d i a c d e c f C 2 , 2 C 1 , 2 f g f i C 3 , 2 g a g h C 2 , 3 h e C 1 , 3 i h AATG 2020 Thomas Erlebach and Jakob T. Spooner 8

  13. Example: Non-Strict Exploration Schedule t = 1 t = 2 t = 3 a b b e d C 1 , 1 C 3 , 1 b C 2 , 1 c d i a c d e c f C 2 , 2 C 1 , 2 f g f i C 3 , 2 g a g h C 2 , 3 h e C 1 , 3 i h AATG 2020 Thomas Erlebach and Jakob T. Spooner 8

  14. Previous work on non-strict temporal graphs ◮ Casteigts, Chaumette and Ferreira (2009) distinguish between strict/non-strict temporal journeys in the context of distributed algorithms. ◮ Barjon, Casteigts, Chaumette, Johnen and Neggaz (2014) describe algorithms for testing strict/non-strict temporal connectivity in sparse temporal graphs. ◮ Zschoche, Fluschnik, Molter and Niedermeier (2017) consider temporal ( v , u ) -separators in non-strict and strict setting. ◮ E, Kammer, Luo, Sajenko and Spooner (2019) prove arbitrary temporal graphs can be explored in O ( N 1 . 75 ) time steps when up to 2 moves per step are allowed. AATG 2020 Thomas Erlebach and Jakob T. Spooner 9

  15. Deciding Non-Strict Temporal Exploration ◮ A non-strict temporal graph G does not necessarily admit an exploration schedule Problem ( Non-Strict TEXP Decision) Input: A non-strict temporal graph G with lifetime L , and start vertex s . Output: YES if G admits a non-strict exploration schedule W starting from s , and NO otherwise. AATG 2020 Thomas Erlebach and Jakob T. Spooner 10

  16. Deciding Non-Strict Temporal Exploration (cont.) Theorem Deciding Non-Strict TEXP is NP-complete. AATG 2020 Thomas Erlebach and Jakob T. Spooner 11

  17. Deciding Non-Strict Temporal Exploration (cont.) Theorem Deciding Non-Strict TEXP is NP-complete. Proof sketch. ◮ Take arbitrary instance I of 3SAT with n variables x i ( i ∈ [ n ]) and m = O ( n ) clauses c j . ◮ W.l.o.g., assume that no c j contains both x i and ¬ x i . ◮ Reduction: Construct non-strict temporal graph G such that: G admits exploration schedule ⇐ ⇒ I is satisfiable ◮ For all i ∈ [ n ], create 2 vertices v T and v F for variable x i of I , i i m clause vertices c j (one for each clause of I ), and an additional vertex s . ◮ Let the lifetime of G be L = 2 n . AATG 2020 Thomas Erlebach and Jakob T. Spooner 11

  18. Proof of Theorem: Reducing 3SAT to NS-TEXP Arrange vertices in components as follows (all unmentioned vertices in any step t are disconnected in that step): { s , v T 1 , v F t = 1 : 1 } { v T { v F 1 } ∪ { c j : x 1 = 1 satisfies c j } 1 } ∪ { c j : x 1 = 0 satisfies c j } t = 2 : ... t = 2 i − 1 : { v T i − 1 , v F i − 1 , v T i , v F i } ( i ∈ [2 , n ]) t = 2 i : { v T { v F i } ∪ { c j : x i = 1 satisfies c j } i } ∪ { c j : x i = 0 satisfies c j } ( i ∈ [2 , n ]) AATG 2020 Thomas Erlebach and Jakob T. Spooner 12

  19. Proof of Theorem: I satisfiable ⇐ ⇒ G explorable I satisfiable = ⇒ G admits exploration schedule W Given satisfying assignment α , move in step 2 i − 1 to v T if i α ( x i ) = 1 or to v F otherwise. In step 2 i , explore all clause vertices i satisfied by x i in α . AATG 2020 Thomas Erlebach and Jakob T. Spooner 13

  20. Proof of Theorem: I satisfiable ⇐ ⇒ G explorable I satisfiable = ⇒ G admits exploration schedule W Given satisfying assignment α , move in step 2 i − 1 to v T if i α ( x i ) = 1 or to v F otherwise. In step 2 i , explore all clause vertices i satisfied by x i in α . G admits exploration schedule W = ⇒ I is satisfiable Each c j can only be reached in a step 2 i if it is contained in the true/false component of x i . Since W visits all c j , we can set α ( x i ) = 1 or α ( x i ) = 0 depending on the component visited in step 2 i and obtain a satisfying assignment. AATG 2020 Thomas Erlebach and Jakob T. Spooner 13

  21. Proof of Theorem: Example Consider the following 3CNF formula: φ = ( x ∨ ¬ y ∨ ¬ z ) ∧ ( ¬ x ∨ y ∨ ¬ z ) ∧ ( ¬ x ∨ ¬ y ∨ ¬ z ) ∧ ( x ∨ ¬ y ∨ z ) Our reduction produces the following NS-TEXP instance: t = 1 t = 1 t = 2 t = 2 t = 3 t = 3 t = 4 t = 4 t = 5 t = 5 t = 6 t = 6 x T y T z T c 1 c 2 c 4 x T y T s c 4 x F y F x T y T z T x F y F z F x F y F z F c 1 c 1 c 2 c 3 c 2 c 3 c 4 c 3 There is a direct correspondence between the satisfying assignment x = 1 , y = 0 , z = 0 and the above exploration schedule. AATG 2020 Thomas Erlebach and Jakob T. Spooner 14

  22. Proof of Theorem: Example Consider the following 3CNF formula: φ = ( x ∨ ¬ y ∨ ¬ z ) ∧ ( ¬ x ∨ y ∨ ¬ z ) ∧ ( ¬ x ∨ ¬ y ∨ ¬ z ) ∧ ( x ∨ ¬ y ∨ z ) Our reduction produces the following NS-TEXP instance: t = 1 t = 1 t = 2 t = 2 t = 3 t = 3 t = 4 t = 4 t = 5 t = 5 t = 6 t = 6 x T y T z T c 1 c 2 c 4 x T y T s c 4 x F y F x T y T z T x F y F z F x F y F z F c 1 c 1 c 2 c 3 c 2 c 3 c 4 c 3 There is a direct correspondence between the satisfying assignment x = 1 , y = 0 , z = 0 and the above exploration schedule. AATG 2020 Thomas Erlebach and Jakob T. Spooner 14

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