Shortest Paths and Treewidth
Algorithms and Networks
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(nk2) time solve the Single Pair Shortest
Paths problem
– In O(nk2) time determine if there is a negative
cycle
– In O(n2k) time solve the All Pairs Shortest
Paths problem
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Shortest paths and treewidth
Contents
• Short background
• Nice tree decompositions and top bags
• Directed Path Consistency on nice tree
decompositions
• Main algorithm
• Conclusions
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Shortest paths and treewidth
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
•
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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 treewidth
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 Xi = Xj = Xj’.
– Introduce: node with one child j with Xi = Xj  {v} for
some vertex v
– Forget: node with one child j with Xi = Xj - {v} for
some vertex v
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Shortest paths and treewidth
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.
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Shortest paths and treewidth
Top bags
• For each vertex v, there is a exactly one bag
i with v in Xi and v not in Xparent(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
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Shortest paths and treewidth
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)
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Shortest paths and treewidth
A property of tree
decompositions
• If P is a path from v to w, then X contains at
least one vertex from P
v
X
w
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Shortest paths and treewidth
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 Xi we compute Bi(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)
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Shortest paths and treewidth
Plan
• A Dynamic Programming algorithm
computes a table Bi for all bags i, each in
O(k2) time
• Note that we must compute O(k2) values for
each bag: O(1) time per computation
• Procedure for each of the four types of
nodes
• Total time is O(nk2) for these steps
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Shortest paths and treewidth
Leaf nodes
• Let i be a leaf node.
• For all v,w in Xi: Bi (v,w) = length of arc
from v to w (and if arc does not exist)
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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 Xi:
Bi (v,w) = min (Bj (v,w) , Bj’ (v,w) )
+
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Why?
=
Shortest paths and treewidth
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 Xi.
– 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 Xi or have an edge from a vertex in
the subtree of j to a vertex in the subtree of j’
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Shortest paths and treewidth
Introduce nodes
• Let i be a node with child j,
with Xi = Xj  {v}.
• For all w,x in Xj:
v
– Bi(v,w) = length of arc (v,w)
if existing, otherwise
– Bi(w,x) = Bj(w,x)
Trivial case
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Shortest paths and treewidth
Forget nodes
• Let i be a node with child
j, with Xi = Xj - {v}.
• For all w,x in Xi:
v
v
– Bi(w,x) = min (Bj(w,x),
Bj(w,v)+Bj(v,x))
The path now can use v as internal vertex
w
x
v
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Shortest paths and treewidth
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) = BTB(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) = BTB(w)(v,w)
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Shortest paths and treewidth
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(nk2) time
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Shortest paths and treewidth
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.)
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Shortest paths and treewidth
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
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Shortest paths and treewidth
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
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Shortest paths and treewidth
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 XTB(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
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Shortest paths and treewidth
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
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Shortest paths and treewidth
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)
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Shortest paths and treewidth
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)
–
–
–
–
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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 treewidth
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)
x…
x
w
v
xw
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TB(x)
Shortest paths and treewidth
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)
x
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w
v
Shortest paths and treewidth
Proof of correctness (3)
• w is in XTB(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 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) }
•
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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 XTB(v) , w  v do
Return d
Shortest paths and treewidth
Time
• If we have a tree
decomposition of
width k:
• O(n k2) for DPC
• O(n2 k) for 2nd step
• Total: O(n2 k) time
•
•
•
• 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) }
•
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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 XTB(v) , w  v do
Return d
Shortest paths and treewidth
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
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Shortest paths and treewidth