We study the problem of finding the next-to-shortest paths in a
graph. A next-to-shortest $(u,v)$-path is a shortest $(u,v)$-path
amongst $(u,v)$-paths with length strictly greater than the length of
the shortest $(u,v)$-path. In constrast to the situation in directed
graphs, where the problem has been shown to be NP-hard, providing edges of length zero are allowed,
we prove the somewhat surprising result that there is a polynomial
time algorithm for the undirected version of the problem

Bang-Jensen

Eppstein

I. Krasikov

Jiménez

Lalgudi

S.D. Noble

Information Processing Letters

English

We study the problem of finding the next-to-shortest paths in a graph.

A next-to-shortest (u, v)-path is a shortest (u, v)-path amongst (u, v)-

paths with length strictly greater than the length of the shortest (u, v)-

path. In constrast to the situation in directed graphs, where the problem

has been shown to be NP-hard, providing edges of length zero are allowed,

we prove the somewhat surprising result that there is a polynomial time

algorithm for the undirected version of the problem

Krasikov, I.

Noble, Steven

Birkbeck Institutional Research Online

Finding next-to-shortest paths in a graph

Krasikov, I

Noble, S D

Brunel University Research Archive

Finding Next-To-Shortest Paths in a GraphI. Krasikov and S. D. NobleDepartment of Mathematical SciencesBrunel UniversityKingston LaneUxbridgeUB8 3PH∗February 16, 2008AbstractWe study the problem of finding the next-to-shortest paths in a graph.A next-to-shortest (u, v)-path is a shortest (u, v)-path amongst (u, v)-paths with length strictly greater than the length of the shortest (u, v)-path. In constrast to the situation in directed graphs, where the problemhas been shown to be NP-hard, providing edges of length zero are allowed,we prove the somewhat surprising result that there is a polynomial timealgorithm for the undirected version of the problem.Keywords: Graph algorithms, computational complexity, shortest paths.1 IntroductionSolutions to the problem of finding the shortest path between two vertices (orall pairs of vertices) in both undirected and directed graphs are well-known [1].Finding the K shortest paths between two vertices has also been well-studied,in for instance [2, 3]. Problems involving finding a path between two vertices oflength strictly greater than the length of the shortest such path have receivedmuch less attention due to the fact that in directed graphs, when we allow edgesof length zero, the problem has been shown to be NP-hard [4]. Here we considerthe problem of finding next-to-shortest paths in an undirected graph. We givea polynomial time algorithm for the undirected version of the problem whenall edge lengths are strictly positive, thus answering a problem raised in [4] andsuggesting that the undirected version of the problem is significantly easier thanthe directed version.∗{mastiik,mastsdn}@brunel.ac.uk1Our graph theoretical notation is standard and we restrict our graphs tobeing simple, that is, having no loops or multiple edges. For the most part, wewill deal with weighted graphs consisting of a graph G = (V,E) and a functionl : E → Q+ giving the length of each edge. Note that by Q+ we mean the set{x : x ∈ Q, x > 0} so we do not allow edges with length zero. Given a set, A,of edges we define l(A) in the obvious way, that isl(A) =∑e∈Al(e).We use n(G) andm(G) to denote, respectively, the number of vertices and edgesof G. Whenever the context is clear we just use n and m. Since we need tomake frequent use of walks and paths, we give their standard definition. A walkis an ordered list of vertices and edges v0, e1, v1, e2, v2, . . . , ek, vk such that for1 ≤ i ≤ k, the endpoints of ei are vi−1 and vi. Sometimes we will deal withdirected graphs in which case we require ei = (vi−1, vi). Depending on thecontext we may just specify a walk by giving either an ordered list of verticesor of edges. A path is a walk for which the vertices v0, . . . , vk are distinct. A(u, v)-path is a path for which v0 = u and vk = v. We define a (u, v)-walk insimilar way.Given a weighted graph, G, the next-to-shortest (u, v)-path, is the shortest(u, v)-path, amongst those (u, v)-paths having length strictly greater than thelength of the shortest (u, v)-path. If no such path exists, we say that the next-to-shortest (u, v)-path has length ∞.2 Main ResultThis section is devoted to a proof of our main result.Theorem 1 There is a polynomial time algorithm which inputs an undirectedgraph G = (V,E), a length function l : E → Q+ and specified vertices u and v,and finds a next-to-shortest (u, v)-path.The main idea of the proof is to consider the set of edges that occur in a(u, v)-path and the direction in which they are traversed. Given a weightedgraph G = (V,E) and specified vertices u and v, we define the shortest pathdigraph DG(u, v) to be the digraph D = (V,A), where the arc (x, y) is in A ifthere is a shortest length (u, v)-path of the form u, v1, . . . , vi, x, y, vi+3, . . . , vm, v.We begin with a simple, preliminary and probably well-known lemma.Lemma 2 For all w, any (u,w)-walk in DG(u, v) is a shortest (u,w)-path inG.Proof: Suppose not and let v0 = u, v1, . . . , vk = w be a walk in DG(u, v) withthe fewest number of arcs amongst those walks that are not shortest paths.Denote this walk byW . Since (vk−1, vk) is an arc of DG(u, v), there is a shortest(u, v)-path P passing through (vk−1, vk). Because W is a walk with the fewest2arcs that is not a shortest path, deleting the last arc from W gives a shortest(u, vk−1)-path W ′. So if we replace the portion of P between u and vk−1 byW ′, we obtain a (u, v)-walk that is no longer than P . Since all edge lengths arestrictly positive, this must be a shortest (u, v)-path and the portion of this pathbetween u and vk is exactly W . Hence W must be a shortest (u,w) path.We say that an arc (x, y) is a forward arc (of DG(u, v)) if (x, y) ∈ DG(u, v)and a backward arc if (y, x) ∈ DG(u, v). Lemma 2 implies that an arc cannotbe both forward and backward.The next lemma is a simple consequence of minimum cost flow techniquesand is presumably well-known but we give a sketch proof here since it is a keypart of the main theorem.Lemma 3 Given a weighted, undirected graph, G, and specified vertices, u, vand w, there is a polynomial time algorithm to find the shortest length (u, v)-pathpassing through w.Proof: Form a network N from G by replacing each edge by two arcs directedin opposite directions and adding a new vertex t together with the arcs (u, t)and (v, t). Assign a capacity of ∞ to all the arcs and to the vertices t and w,and one to every other vertex. Give the arcs joined to t zero cost and give eachother arc cost equal to the length in G. Finding the shortest path in G from uto v passing through w corresponds to finding a minimum cost flow of volumetwo from w to t in N . This can be done with two iterations of the augmentationstep of the Ford-Fulkerson maximum flow algorithm or equivalently two shortestpath computations and hence gives an O(n2) algorithm, [1].We now give the proof of the main theorem.Proof of Theorem: We first verify that DG(u, v) can be constructed in poly-nomial time. Let uw denote the length of the shortest (u,w)-path. UsingDijkstra’s algorithm we can compute uw for each w in time O(n2). The arc(x, y) is present in DG(u, v) if and only ifuy = ux + l({x, y}).Thus DG(u, v) can be constructed in time O(n2).Lemma 2 shows that any walk made up of forward arcs of DG(u, v) is ashortest (u, v)-path. We first find the shortest path containing an arc that isneither forward or backward. The shortest (u, v)-path passing through {x, y}(in either direction) can be found by adding vertex w to subdivide {x, y} andapplying Lemma 3. The shortest such path (if it exists) can therefore be foundby applying Lemma 3, O(m) times.The other type of (u, v)-path that needs to be considered is one containingonly forward and backward arcs but including at least one backward arc. Such a3path must contain a forward arc (x, y) followed by a backward arc (y, z). Givenforward arcs (x, y) and (z, y), form G′ from G by deleting all edges adjacent toy except {x, y} and {y, z}. Applying Lemma 3 to G′ with w = y will find theshortest path containing one of (x, y) and (z, y) as a forward arc and the otheras a backward arc. Using the above procedure for each pair of forward arcs ofthe form (x, y) and (z, y) will give the shortest path of this type and requireO(nm) applications of Lemma 3.The next-to-shortest (u, v)-path is the shortest path found in either of thetwo steps of the algorithm and can be found using O(nm) shortest path com-putations, that is in O(n3m) steps overall.3 ConclusionWe have shown that next-to-shortest paths can be found in undirected graphswith strictly positive edge lengths in time O(n3m). The complexity status ofthe next-to-shortest paths problem in directed graphs when the edge lengthsare required to be strictly positive is open, although we suspect that it is NP-complete. Similarly we do not know the complexity of the next-to-shortest pathsproblem in undirected graphs when edge lengths of zero are allowed.A further interesting open question is to consider the structure of next-to-shortest paths and whether any analogue of the Bellman optimality equationshold. The position seems far from obvious.References[1] J. Bang-Jensen and G. Gutin. Digraphs: Theory, Algorithms and Applica-tions. Springer, 2001.[2] D. Eppstein. Finding the k shortest paths. SIAM Journal of Computing,28(2):652–673, 1999.[3] V. M. Jime´nez and A. Marzal. Computing the k shortest paths: a new algo-rithm and an experimental comparison. In J. S. Vitter and C. D. Zaroliagis,editors, Algorithm Engineering: 3rd International Workshop, WAE’99, vol-ume 1668 of Lecture Notes in Computer Science, pages 15–29, Berlin, Hei-delberg, 1999. Springer-Verlag.[4] K. N. Lalgudi and M. C. Papaefthymiou. Computing strictly-second shortestpaths. Information Processing Letters, 63:177–181, 1997.4

Finding next-to-shortest paths in a graph

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