Minimum Spanning Tree

Weiss Ch. 9.5

MST of UNdirected graph is usual case: tree that connects all vertices of G with minimal cost. So G must be connected. MST not necessarily unique. Natural idea: how connect all nodes cheapest?

Elementary reasoning assures us a greedy approach should work.


Pick a root and then add one vertex at a time. Just add that vertex v and edge from u to v, such that u is in the tree, v is not, and Cost(u,v) is minimal amongst all such edges.

This is just like Dijkstra's A, only dv is the cost of one edge, and the update rule is simpler. Thus Weiss brings back his tables, which for instance look like this: Here we have just chosen v1 as root, and its three naighbors v2, v3, v4 have their edge costs inserted. They're not in the tree yet so they're unknown, but it's easy to see we're going to pick v4 to add next, and when we do the cost of adding v3 is going to drop to 2, and v3's pv will change from v1 to v4.

So as for D's A, order O(|V|2) without priority queues (heaps), optimal for dense graphs and with binary heaps get
O(|E| log|V|), good for sparse graphs.


Another greedy plan: select edges in order of of smallest weight, accept one if it doesn't cause a cycle.

K's A starts with a forest of single nodes, which are merged to form one final MS tree. The underlying algorithm is union-find, but we're going to want a priority queue too. We define nodes to be in the same set if they are connected in the current spanning forest.

Rather than sort the edges, heapify and deleteMin. Then if both ends (u,v) of an edge are in same set, reject it (cycle). Else union the two sets containing u and v. Worst case you have to examine all the edges (there may be one really big one); O(|E| log |E|), dominated by the heap.

Depth-First Graph Search

Not only a key AI technique but useful for computing several graph properties and problems. DFS just visits every node, then its successors, recursively. With trees this is super-simple, with graphs one needs not to get stuck in cycles so it's only simple. Weiss has 4-liner basic program, p. 400.

DFS Visualization

DFS induces a "depth-first spanning tree" on the graph, since it doesn't cycle. In AI we often talk of a search tree in a state-space graph. To relate the two one can use "back edges" that are not taken in the search, which we can imagine being generated at each newly-visited node: all its successors that have been visited already are given back edges pointing at them. An unconnected tree needs a re-start for the search process if all nodes aren't visited when the DFS ends.

Establishing Properties etc. with DFS

Biconnectivity: A graph is biconnected when it cannot be disconnected by removal of any single vertex. Clear logistical analogs--e.g. redundant routes, components. Disconnecting vertices are "articulation points". Weiss gives pseudocode and discusses some proofs and niddles.

DFS is the basis of a linear-time algorithm to find articulation points, thus answer the biconnectivity question.

  1. Do DFS, make DFS tree with backedges, number vertices Num(v) in order of visit.
  2. For each vertex v compute the lowest-numbered vertex Low(v) that can be reached from v by 0 or more tree edges and possibly a backedge (in that order)
  3. Low(v) can be efficiently computed from its definition: Low(v) is the minimum of:

    • Num(v) if no edges are taken
    • the lowest Num(v) amongst all back edges (v,w) if take no tree edges, one back edge.
    • The lowest Low(w) amongst all tree edges (v,w), a recursive defn that moves down the tree.


Since we need to compute Low(v) for all v's children before computing v's, we have postorder traversal algorithm. If Num(v)>Num(w) it's a tree edge, else a backedge. So to compute Low(v), use adjacency list to look at all nodes adjacent to v (from adj list), keep track of minimum. That's O(|E|+|V|).

Articulation points?

  1. root is articulation pt iffi it's got >1 child.
  2. v is a.p. iffi v has a child w s.t.
    Low(w) ≥ Num(v).


Defns say C and D are articulation points. D's child E only has one way to get to any node above D: go through D.

Naively implementing all this takes three passes, preorder traversal to get Num(v), postorder to get Low(v), Third pass finds articulation points. BUT! job can be done much more elegantly (Code in Figs 9.67 - 9.69).

Euler Circuits and their Ilk:

These puzzles were the subject of the first work in graph theory in 1736.

Puzzle: Draw these figures w/o lifting pencil, each line covered exactly once. And can you finish where you started?

"Euler path (tour) and circuit" problems: traverse all the edges (NOT vertices!!) just once. The Euler circuit ends where it started, Form of solution is same as path, so Weiss does circuit. Some things obvious:

Algorithm builds up sub-solutions by splicing. This graph has an Euler circuit.

If start at 5, a DFS might find 5,4,10,5, so we'd have a sub-circuit and a remaining problem:

If then start at 4 could DFS and get 4,1,3,7,4,11,10,7,9,3,4. Splicing:
5,4,1,3,7,4,11,10,7,9,3,4,10,5. Etc.

Splicing is O(1) if path is a linked list. Maintain pointer to last edge scanned in adj. list so can start there later. Begin next depth-first search at start of the splice work in vertex search is O(E) over the algorithm, so can make whole job O(|V|+|E|).


Hamiltonian cycle is simple cycle in an undirected graph that visits every vertex. Strangely, no efficient algorithm for it is known. You could get famous if....

Directed Graphs:
DFS can be extended to directed graphs and used to find "strong components", which are subsets of strongly-connected vertices (a path between any two).

Directed Graphs

Recall a graph is strongly connected if there is a path from every vertex to every other vertex. Can DFS a directed graph but from any given vertex may not visit all nodes unless graph is strongly connected. So restart search elsewhere. Process gives forward and backward edges but also cross edges.

A digraph is acyclic iffi it has no back edges.

DFS can test strong-connectedness and find strongly-connected components. Doing it with two DFS's is easier than with one, (W. 9.65).


P and NP

Last update: 7.24.13