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Shortest Paths and Treewidth Algorithms and Networks

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This lecture Given a graph G=(V,E) with a tree decomposition of width k, and with each edge a length (possibly negative), we can: –In O(nk 2 ) time solve the Single Pair Shortest Paths problem –In O(nk 2 ) time determine if there is a negative cycle –In O(n 2 k) time solve the All Pairs Shortest Paths problem Shortest paths and treewidth2

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Contents Short background Nice tree decompositions and top bags Directed Path Consistency on nice tree decompositions Main algorithm Conclusions Shortest paths and treewidth3

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Background Directed Path Consistency, by Dechter, Meiri, Pearl, 1991 Earlier results: –Chaudhuri, Zaroliagis, 2000 Slower, but also dynamic –Chleq, 1995 Single Source Shortest Paths; gives same asympthotic bound for APSP, but larger factor in O-notation Algorithm here (SNOWBALL) from Plancken, de Weerdt, van der Krogt, Computing All-Pairs Shortest Paths by Leveraging Low Treewidth, Journal of Artificial Intelligence Research 43 (2012) 353-388 Shortest paths and treewidth4

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Nice tree decompositions (variant with larger leaf bags) Rooted tree, and four types of nodes i: –Leaf: leaf of tree (note: the size 1 restriction of usual nice tree decompositions is dropped) –Join: node with two children j, j’ with X i = X j = X j’. –Introduce: node with one child j with X i = X j {v} for some vertex v –Forget: node with one child j with X i = X j {v} for some vertex v 5

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Making it nice Theorem. Given a tree decomposition of a graph G=(V,E) with n vertices, of width k and with r bags, one can build a nice tree decomposition of G of width k, with O(n) bags, in O(k(n+r)) time. Shortest paths and treewidth6

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Top bags For each vertex v, there is a exactly one bag i with v in X i and v not in X parent(i) Call this bag the top bag of v, TB(v) The parent of the top bag of v is a forget node that forgets v: different vertices have different top bags Shortest paths and treewidth7

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A numbering of the vertices Traverse the tree of the tree decomposition in pre-order Number the vertices in the order in which their top bags appear in this ordering Equivalently: number the vertices in the root bag, and then number vertices in order we see their forget nodes Claim: if the top bag of v is a descendant of the top bag of w, then nb(v) > nb(w) Shortest paths and treewidth8

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A property of tree decompositions If P is a path from v to w, then X contains at least one vertex from P Shortest paths and treewidth X v v w w 9

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Directed Path Consistency DPC Problem: –For all pairs v, w, such that there is a bag that contains v and w, compute B(v,w) = the length of a shortest path from v to w that uses (except v and w) only vertices with numbers larger than max(nb(v),nb(w)) What we do: –For each bag i, for all v,w in X i we compute B i (v,w) = the length of a shortest path from v to w that uses only vertices whose top bag is a descendant of i (i.e. only live in the subtree below i) Shortest paths and treewidth10

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Plan A Dynamic Programming algorithm computes a table B i for all bags i, each in O(k 2 ) time Note that we must compute O(k 2 ) values for each bag: O(1) time per computation Procedure for each of the four types of nodes Total time is O(nk 2 ) for these steps Shortest paths and treewidth11

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Shortest paths and treewidth Leaf nodes Let i be a leaf node. For all v,w in X i : B i (v,w) = length of arc from v to w (and if arc does not exist) 12

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Shortest paths and treewidth Join nodes Let i be a join node with children j, j ’. For all v,w in X i : B i (v,w) = min (B j (v,w), B j’ (v,w) ) + = Why? 13

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Why? Take a path P from v to w using only vertices with top bag below i Claim: P uses only nodes in the subtree of j OR P uses only nodes in the subtree of j’ –If not, it must use another node in the bag X i. –To see this: suppose not. Choose a vertex x with top bag in subtree of j from P and a vertex y with top bag in subtree of j’ –Go along P from x to y. At some point, we either visit a vertex in X i or have an edge from a vertex in the subtree of j to a vertex in the subtree of j’ Shortest paths and treewidth14

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Shortest paths and treewidth Introduce nodes Let i be a node with child j, with X i = X j {v}. For all w,x in X j : –B i (v,w) = length of arc (v,w) if existing, otherwise –B i (w,x) = B j (w,x) v Trivial case 15

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Shortest paths and treewidth Forget nodes Let i be a node with child j, with X i = X j {v}. For all w,x in X i : –B i (w,x) = min (B j (w,x), B j (w,v)+B j (v,x)) v v The path now can use v as internal vertex w x v 16

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Back to DPC For all pairs v, w, such that there is a bag that contains v and w, compute B(v,w) = the length of a shortest path from v to w that uses (except v and w) only vertices with numbers larger than max(nb(v),nb(w)) If nb(v) < nb(w): –The top bag of w is a descendant of the top bag of v –B(v,w) = B TB(v) (v,w) –Argument uses again properties of tree decompositions: induction shows you cannot leave the subtree rooted at v in the path Similar, if nb(w) < nb(v): B(v,w) = B TB(w) (v,w) Shortest paths and treewidth17

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Single pair shortest paths The shortest path from s to t: –Add (s,t) as edge to G –Make tree decomposition of G+(s,t) –Width is at most one more than width of G (e.g., add t to all bags of a tree decomposition of G) –Choose a bag that contains s and t –Make a nice tree decomposition with root bag precisely {s,t} –The distance of s to t equals B(s,t) O(nk 2 ) time Shortest paths and treewidth18

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Checking for negative cycles Claim: there is a negative cycle, if and only if there are two vertices (that appear together in a bag) v, w with B(v,w) + B(w,v) < 0 – : trivial – : take a negative cycle C. Let v and w be the two lowest numbered vertices on the cycle Take a node i of minimum distance to root bag that contains two vertices of C. All vertices of C must belong to bag i or a descendant of i. (If not, then C would have a separator of size at most 1, namely C intersected with the bag of the parent of i.) Shortest paths and treewidth19

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All Pairs Shortest Path algorithm Step 1 Compute the values B(v,w) as discussed Check if there is a negative cycle, as discussed If there is no negative cycle, we do Step 2 Shortest paths and treewidth20

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Step 2 - Scheme Scheme: For q = 1 to n do –Suppose nb(v) = q –Compute d(v,w) and d(w,v) correctly for all w with nb(w) < q Shortest paths and treewidth21

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Step 2 - Code For all v,w, initialize d(v,w) = For all v, initialize d(v,v) = 0 For q = 1 to n do –Suppose nb(v) = q –For all w in X TB(v), w v do For r = 1 to q-1 do –Suppose nb(x) = r –d(x,v) = min { d(x,v), d(x, w) + B(w,v) } –d(v,x) = min { d(v,x), B(v,w) + d(w,x) } Return d Shortest paths and treewidth22

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Correctness Induction hypothesis: –For all v, w with nb(v) < q and nb(w) < q, d(v,w) equals the length of the shortest path from v to w From standard shortest path algorithms analysis, it is clear that always –d(x,v) distance from x to v –d(v,x) distance from v to x Shortest paths and treewidth23

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To show the IH No negative cycles: d(v,v) = 0 is correct We must argue that after the round working with v with nb(v) = q, all values d(v,x) and d(x,v) for nb(x) < q are correct Our proof looks at the values d(x,v); the other case d(v,x) is similar Consider one vertex x with nb(x) < nb(v) Shortest paths and treewidth24

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Proof of correctness (1) nb(x) < nb(v) Note: top bag of x is not a descendant of top bag of v Consider the shortest path P from x to v. Claim 1: P contains another vertex ( v) from TB(v) –Say, last arc of P is (y,v) –y belongs to subtree of v –x belongs to bags not in subtree of v –TB(x) contains a vertex of part of P from x to y Shortest paths and treewidth25

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Proof of correctness 2 Look at the vertices y on P that belong to the top bag of v Let w be the last vertex of this type on P Note: TB(w) is ancestor of TB(v), so nb(w) < nb(v) Shortest paths and treewidth xwv x w TB(x) x … 26

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Proof of correctness (2) Claim 2: d(x,v) = d(x, w) + B(w,v) When we work with v, the value d(x,w) (first part of path) was already correctly computed, as nb(x) = r < q = nb(v) and nb(w) < nb(v) Second (blue) part of path uses only higher numbered vertices: by DPC-subroutine, given in B(w,v) Shortest paths and treewidth xwv 27

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Proof of correctness (3) w is in X TB(v) So, d(x,v) will be set to the min of its old value at the length of P when we consider w After this step, equal to distance For all v,w, initialize d(v,w) = For all v, initialize d(v,v) = 0 For q = 1 to n do –Suppose nb(v) = q –For all w in X TB(v), w v do For r = 1 to q-1 do –Suppose nb(x) = r –d(x,v) = min { d(x,v), d(x, w) + B(w,v) } –d(v,x) = min { d(v,x), B(v,w) + d(w,x) } Return d Shortest paths and treewidth28

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Time If we have a tree decomposition of width k: O(n k 2 ) for DPC O(n 2 k) for 2 nd step Total: O(n 2 k) time For all v,w, initialize d(v,w) = For all v, initialize d(v,v) = 0 For q = 1 to n do –Suppose nb(v) = q –For all w in X TB(v), w v do For r = 1 to q-1 do –Suppose nb(x) = r –d(x,v) = min { d(x,v), d(x, w) + B(w,v) } –d(v,x) = min { d(v,x), B(v,w) + d(w,x) } Return d Shortest paths and treewidth29

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Remarks Most results for treewidth have faster algorithms for (NP-) hard problems This one: faster for polynomial time solvable problem Experiments by Plancken show results are practical Shortest paths and treewidth30

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