# Lexicographic Bottleneck Shortest Path in Undirected Graphs

# 1 Lexicographic Bottleneck Ordering

Let X be a totally ordered set. Let l(S) be the sorted sequence of all the elements in S, where S\subset X. We can induce an total ordering on the subset of X.

S\preccurlyeq T for S,T\subset X if l(S)\leq l(T) in lexicographic ordering. \preccurlyeq is called the lexicographic bottleneck ordering.

For nonempty sets A and B, if A\preccurlyeq B , then A - min A \preccurlyeq B - min B. Also if A\preccurlyeq B then A \preccurlyeq A\cup B \preccurlyeq B.

A\preccurlyeq A', B\preccurlyeq B' then A\cup B \preccurlyeq A'\cup B'

Let C=A\cup B and C' = A'\cup B'.

Since \preccurlyeq is a total order, we can prove it by showing if C'\preccurlyeq C then C' = C.

We prove it by structural induction on C'. The base case when C' = \emptyset is trivial, since it must mean A=B=A'=B'=\emptyset.

Consider c' = \min C', c = \min C. c'\leq c in order for C'\preccurlyeq C. But we know c\leq c'. This shows c=c'. Given that c=c', A'\cup B' = A\cup B if and only if (A'-c)\cup (B'-c) = (A-c) \cup (B-c).

First, we show that A-c \preccurlyeq A'-c.

- Assume c\in A, then c\in A', so by previous lemma this is true.
- If c\not\in A and c\not \in A', then A-c=A\preccurlyeq A'=A'-c.
- If c\not \in A but c\in A', then this implies A is empty, and \emptyset \preccurlyeq A'-c.

Similarly, B-c\preccurlyeq B'-c.

Second, we need to show that (A'-c)\cup (B'-c) \preccurlyeq (A-c) \cup (B-c). This is obvious because C'\preccurlyeq C implies C'-c \preccurlyeq C-c.

By the inductive hypothesis, (A'-c)\cup (B'-c) = (A-c) \cup (B-c), thus completes the proof.

The theorem intuitively tells us how to partition a set into smaller sets.

# 2 Lexicographic Bottleneck Path

Given a undirected graph G=(V,E), and an ordering of the edges e_1,\ldots,e_m. Let w(e_i)=i.

Find a st-path that maximizes the minimum edge weight on the path.

Any st-path that maximizes the minimum edge weight over all st-paths is a st-bottleneck shortest path(BSP). We are interested in a more general version of this problem. Find a path from s to t that is maximum with respect to the lexicographic bottleneck ordering \preccurlyeq of the path.

Find a st-path P such that P'\preccurlyeq P for all st-path P'.

The unique st-path that is maximum in lexicographic bottleneck order among all st-paths is called the st-lexicographic bottleneck shortest path(LBSP).

In order to find a BSP, we can first compute the maximum spanning tree T of G, as show in Lemma 4.1 of [1].

If T is a maximum spanning tree of G(under the weight w), then the unique st-path in T is a st-BSP in G.

It's interesting this theorem actually extends to LBSP.

If T is a maximum spanning tree of G(under the weight w), then the unique st-path in T is the st-LBSP in G.

Before proving the theorem, we consider a useful lemma.

P is a st-BSP with bottleneck edge xy. If removing edge xy result a sx-LBSP and yt-LBSP, then P is a st-LBSP.

P is a bottleneck st-path implies the st-LBSP has to reach either x or y before t.

If it reaches y before x, then the subpath from s to y then from y to t using the yt-LBSP would imply xy is not in st-LBSP, a contradiction.

Thus, we must have the st-LBSP is a concatenation of 3 paths, a sx-path P_{sx}, edge xy and a yt-path P_{ty}. Using Theorem 3, we notice P is a LBSP.

We prove by induction on the distance between the two vertices on the maximum spanning tree T.

**Base Case:** If the length of a uv on T is 1, then the edge uv is a BSP, and also a LBSP.

**Inductive Step:** Consider two vertices s and t. The tree induces a st-BSP with bottleneck edge xy. By the inductive hypothesis, removing xy result a sx-LBSP and yt-LBSP in G. The previous lemma demonstrates that st-BSP in T is a st-LBSP in G.

# References

[1] A. Shapira, R. Yuster, U. Zwick, **All-pairs bottleneck paths in vertex weighted graphs**, Algorithmica. 59 (2011) 621–633 10.1007/s00453-009-9328-x.