The network-untangling problem: From interactions to activity timelines Polina Rozenshtein (Nordea DS Lab, Finland) Nikolaj Tatti (University of Helsinki, Finland) Aristides Gionis (Aalto University, Finland) ECML/PKDD’17 + journal extension
Temporal networks • Temporal graph ! = #, % • # – set of entities (e.g. people, sensors, locations.. ) • Edges &, ', ( ∈ % – instantaneous interactions over entities • &, ' ∈ V • ( is the time of interaction • tweets, emails, comments on social networks..
Problem setting • consider a set of entities • entities can become active or inactive • entities interact over time, forming a temporal network • each interaction is attributed to an active entity
Problem setting • consider a set of entities • entities can become active or inactive • entities interact over time, forming a temporal network • each interaction is attributed to an active entity • can we reconstruct the activity timeline that explains best the observed temporal network? • assumption: being active is more costly, thus we want to minimize total activity time
Motivating example • analyze a discussion in twitter about a topic (e.g., brexit) • entities are hashtags • two hashtags interact if they appear in the same tweet • summarize the discussion by reconstructing a timeline • pick a set of important hashtags and the time intervals they are active
Motivating example #economy #brexit #negotiations #hardbrexit #tory time
Motivating example #economy #brexit #negotiations #hardbrexit #tory time
Problem formulation • given a temporal network ! = ($, &) with & = {(), *, +)} - . = / . , 0 . – activity interval of ) ∈ $ (starts at / . and ends • at 0 . ) • find a set of activity intervals for all nodes • at most 2 per each node ) ∈ $
Problem formulation: preliminaries • given a temporal network ! = ($, &) with & = {(), *, +)} - . = / . , 0 . – activity interval of ) ∈ $ (starts at / . and ends • at 0 . ) • find a set of activity intervals for all nodes • at most 2 per each node ) ∈ $ • Activity timeline of ! is a set of activity intervals 3 = - .4 .∈5,4∈[7,8] • The timeline 3 covers temporal network ! , if for each edge ), *, + ∈ & we have + ∈ - .4 or + ∈ - :4 for some ; ∈ [1, 2] .
Problem formulation Problem 1. (Sum-Span) • Find a timeline ! = # $% $∈',%∈[*,+] that covers - and minimizes total length of ! . Problem 2. (Max-Span) • Find a timeline ! = # $% $∈',%∈[*,+] that covers - and minimizes maximum length of intervals in ! . • For the ease of analysis consider . = 1 and . > 1 separately
1-Sum-Span Problem 1-Sum-Span is NP-hard Consider subproblem Coalesce : • Assume we are also given one active time point ! " for each vertex # ∈ % . • Find an optimal activity timeline & , which contains the corresponding active time points ! " "∈' .
1-Sum-Span • Coalesce can be solved in linear time with factor 2 approximation, based on Binary LP-formulation. • Define a variable ! "# ∈ {0,1} for each vertex * ∈ + and time stamp , ∈ -(*) (moments of interactions of * ). • ! "# = 1 indicates that , is either the beginning or end of the active interval of * . • Binary LP: – Cost function min ∑ ",# |, − 7 " | ! "# – Constraints to ensure feasibility
1-Sum-Span • Relax the integrality and write the dual • Maximal solution to the dual program is a 2-approximation for Coalesce • Maximal solution can be found in one pass ( ! " , Alg. Maximal ) Iterate to solve 1-Sum-Span (Alg. Inner ): • Start with " $ = ("'( ) * + ",- ) * )/2 • Run Maximal and update " $ • Repeat until no improvement.
k-Sum-Span k-Sum-Span is are inapproximable Consider subproblem k-Coalesce : • Assume we are also given k active time points ! "# for each vertex $ ∈ & • One for each of activity intervals of $ • Find an optimal activity timeline ' , which contains the corresponding active time points ! "# "∈(,#∈[+,,] . • Similar BLP and Alg. k-Maximal , . !
k-Sum-Span Iterate to solve k-Sum-Span (Alg. k-Inner ): • Start with ! "# as centroids of a k-clustering algorithm • Run k-Maximal and update ! " • Repeat until no improvement
1-Max-Span 1-Max-Span can be solved efficiently Subproblem Budget : • Assume we are also given a set of budgets ! " "∈$ of interval durations for each vertex. • Find an optimal activity timeline % = ' " "∈$ , such that length of each activity interval ' " is at most ! " .
1-Max-Span Budget can be solved optimally in linear time Map Budget into 2-SAT: • Variable ! "# for each vertex $ and timestamp % ∈ '($) . • Clause (! "# ∨ ! +# ) – cover each edge ,, $, % . • Clause (! "/ ∨ ! "# ) – ensure budget: for each 0, % ∈ '($) , such that 0 − % > 3 " • Solution for Budget : time intervals where all boolean variables are True .
1-Max-Span Linear time: • 2-SAT is solved in linear-time of the number of clauses (Aspvall et all [1]). We have ! " # clauses. • Bottleneck: SCC decomposition !(" # + ") • algorithm by Kosaraju [2] for SCC decomposition • Use of temporal structure → perform DFS in !(") . Solve 1-Max-Span by binary search to find the optimal maximum length for intervals (Algorithm Budget, !(" log(")) ).
k-Max-Span k-Max-Span inapproximable • consider two nested subproblems Subproblem k-Partition : • Assume we are also given k-1 inactive time points ! "# for each vertex $ ∈ & • One for each of gap between the activity intervals of $ • Find an optimal activity timeline ' , which interleaves with corresponding gap points ! "( "∈),(+[-,./-]
k-Max-Span Problem k-Partition can be solved in polynomial time through • iteration of Problem k-Budget, which sets a budget for each interval. Subproblem k-Budget : Assume we are given a set of budgets ! " "∈$ of interval durations • for each vertex; k-1 inactive time points % "& for each vertex • Find an optimal activity timeline ' = ) " "∈$ , such that length of • each activity interval ) "& is at most ! "& and the gap points are interleaved k-Budget can be solved *(,) , similarly to Budget
k-Max-Span Iterate to solve k-Sum-Span (Alg. k-Budget ): • Start with ! "# as mean points of the largest intervals with no activity of node $ • Solve k-Partition : – do binary search on budgets with solving k-Budget – update ! "# • Repeat until no improvement
Summary Problem 1: Sum-Span • ! = 1 NP-hard • ! > 1 inapproximable • Subproblem (k-)Partition with inner points • 2-approximation in linear time via BLP dual for (k-)Partition
Summary Problem 2: Max-Span • ! = 1 polynomially solvable • ! > 1 inapproximable • Subproblem (k-)Budget with budgets • Exact solution in linear time via 2-SAT for (k-)Budget
Experiments: case study Tweets from Helsinki region, November 2013 • Inner algorithm ( 1-Sum-Span ) • winwin xbone yandex webdesign vision walkbase winner younited zenrobotics slush13 pureview nuijankopautus nokiaegm illuusio here kirkkonummi nokia typaikka elop nordis mtvema bestvideo bron emaazing ema2013 exo bestpop emazing worldwideactexo voteaustinmahone 3 6 9 12 15 18 21 24 27 30 Nov 1
Experiments: case study Helsinki Twitter, years 2011-2013 • k-Inner algorithm with k = 3 ( k-Max-S ) • slush2013 uutisraivaaja slushpitstop slush13 comingtoslush pelotoncamp slush12 tediili sailfish jolla digitalist aller startups garage48 slush11 startupsauna aaltoes startup padlette crowdfunding slush lumia sxsw digasell ipad supercell tivitforesight slush2012 slush2008 Jan, 11 Apr Jul Oct Jan, 12 Apr Jul Oct Jan, 13 Apr Jul Oct Jan, 14
Performance: Inner • Synthetic dataset, with planted ground truth • overlap ! is set to 0.5 • values are averaged over 100 runs. 1.00 5.5 Relative total length 0.98 5.0 4.5 0.96 Quality 4.0 0.94 3.5 0.92 3.0 P 0.90 2.5 R 0.88 2.0 F 0.86 1.5 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 iterations iterations
Performance : k-Inner • Synthetic dataset, k=10 intervals 0.86 8.5 Relative total length 8.0 0.84 7.5 7.0 Quality 0.82 6.5 0.80 6.0 P 5.5 R 0.78 5.0 F 4.5 0.76 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 iterations iterations
Performance : k-Budget • Synthetic dataset, k=10 intervals 54 0.95 Relative max length 52 0.90 50 0.85 48 Quality 46 0.80 44 0.75 P 42 R 40 0.70 F 38 1 2 3 4 5 6 7 8 9 10 0.65 iterations 1 2 3 4 5 6 7 8 9 10 iterations
Baseline comparison 0.9 0.8 Baseline: greedily ’cover’ the • 0.7 F-measure 0.6 k-Inner longest activity intervals of the k-Budget 0.5 nodes. k-Baseline 0.4 0.3 0.2 0.1 2 3 4 5 6 7 8 9 10 number of intervals 160 20 140 18 Relative max length Relative total length 16 120 k-Inner 14 100 k-Budget 12 80 k-Baseline k-Inner 10 k-Budget 60 8 k-Baseline 40 6 4 20 2 3 4 5 6 7 8 9 10 2 3 4 5 6 7 8 9 10 number of intervals number of intervals
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