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CSC 263H               Tutorial Outline for Week 12             Winter 2004
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BFS
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For any two vertices u,v in a graph, let delta(u,v) represent the number of
edges on a shortest path from u to v (i.e., a path with the smallest number
of edges).
Prove that at the end of BFS (started from vertex s), d[v] = delta(s,v),
the number of edges on a shortest path from s to v.


First, we show that d[v] >= delta(s,v), by induction on the number of
vertices that have been dequeued during the algorithm's execution:

  - Initially, d[s] = 0 = delta(s,s), and for v =/= s, d[v] = oo >=
    delta(s,v).

  - d[v] only changes once: when v is "discovered" from some other vertex
    u.  Since u was placed in the queue before v, d[u] >= delta(s,u) by the
    induction hypothesis, so

        d[v] =  d[u] + 1 >= delta(s,u) + 1 >= delta(s,v)

    (since delta(s,v) <= delta(s,u) + 1 for every edge (u,v) in E.

    The above claim can be easily shown. If u is reachable from s, then
    v must also be reachable from s. Thus the shortest path from s to v
    cannot be longer than the shortest path from s to u followed by
    the edge (u,v). If u is not reachable from s, delta(s,u) = infinity
    and the inequality holds.)

Next, we have to show that d[v] <= delta(s,v).  First, we show that if
u_1, u_2, ..., u_k are the vertices of G in the order in which they were
enqueued (so s = u_1), then d[u_1] <= d[u_2] <= ... <= d[u_k].

  - For a contradiction, assume that this is not the case and let j be the
    first index such that d[u_{j-1}] > d[u_j].  Also, let u_i = p[u_j] (so
    i < j) and u_i' = p[u_{j-1}] (so i' < j-1).

  - Then, d[u_i] = d[u_j] - 1 < d[u_{j-1}] - 1 = d[u_i'], by definition of
    j, so it must be the case that i < i' (by definition of j again).  But
    then, u_i was enqueued before u_i', which means that u_i was dequeued
    before u_i', which means that u_j must have been enqueued before
    u_{j-1}, which is a contradiction.

This implies that if u is enqueued before v, d[u] <= d[v].

Now, it is possible to show that for every vertex v, d[v] <= delta(s,v).

For a contradiction, assume that d[v] > delta(s,v) for some vertex v.
Let v' be the vertex with the minimum value of delta(s,v) such that
d[v'] > delta(s,v'). The vertex v' must be reachable from s.
(If this were not the case, delta(s,v') = infinity >= d[v'].)

Consider the shortest path from s to v', and let u be the vertex on this
path that is directly before v'.
delta(s,v') = delta(s,u) + 1

Thus delta(s,u) < delta(s,v'), and by our choice of v', delta(s,u) <= d[u].
From the first part, we know that this implies that delta(s,u) = d[u].

Thus we have,
d[v'] > delta(s,v') = delta(s,u) + 1 = d[u] + 1. (*)

Now consider the time when u was dequeued.

Assume that v' was white when u was dequeued. Since there exists an edge
(u,v'), we update d[v'] = d[u] + 1. This contradicts (*). So v' could not
be white.

Now assume that v' was black when u was dequeued. Then v' was previously
dequeued from the queue. In other words, v' was enqueued before u. Thus
d[v] <= d[u]. (We showed this earlier in this proof.) This contradicts (*).
So v' could not be balck.

Now assume that v' was gray when u was dequeued. Then there was another
vertex, say w, that was responsible for colouring v' gray. We know that
(w,v') is an edge in the graph, and w was dequeued before u.
Thus d[w] <= d[u], and d[v'] = d[w] + 1
This implies that d[v'] <= d[u] + 1. This contradicts (*). So v' could not
be gray.

This provides a contradiction, since v' must be either white, black, or
gray.

Thus for every vertex v, d[v] <= delta(s,v).

Now we know that delta(s,v) <= d[v] and d[v] <= delta(s,v) for every vertex
v, i.e., delta(s,v) = d[v] for every vertex v.