Introduction to Algorithms Jiafen Liu Sept. 2013

Today’s Tasks Shortest Paths –Properties of shortest paths –Dijkstra’s algorithm –Correctness –Analysis –Breadth-first search

Paths in graphs Consider a digraph G= (V, E) with edge- weight function w: E→R. The weight of path p = v 1 →v 2 →…→v k is defined to be Example: w(p)=?

Shortest paths A shortest path from u to v is a path of minimum weight from u to v. The shortest path weight from u to v is defined as δ(u, v) = min{w(p): p is a path from u to v}. In what case there is no shortest path? –Note: δ(u, v) = ∞ if no path from u to v exists. –There is a negative cycle. In other cases, usually a finite set.

Well-definedness of shortest paths If a graph G contains a negative-weight cycle, then some shortest paths may not exist. Example:

Thought of solving Dynamic Programming Greedy Algorithms Characteristics ： –Optimal Substructure –Overlapping Subproblems –Greedy choice property

Optimal substructure What is the optimal substructure for this problem? Theorem. A subpath of a shortest path is a shortest path. How to prove? –Cut and paste

What’s the subproblems ？ Shortest path going from u to v, maybe involves computing shortest path from u to some intermediate point x, and then from x to v. Different choosing of x 1,x 2 …x n consists overlapping subproblems. How many subproblems are there? |v| 2

Triangle inequality Important for us to make greedy choice Theorem: For all u, v, x ∈ V, we have Figure Illustration:

Single-source shortest paths Problem. From a given source vertex s ∈ V, find the shortest-path weights δ(s, v) for all v ∈ V. If all edge weights w(u, v) are nonnegative, all shortest-path weights must exist. IDEA: Greedy. 1.Maintain a set S of vertices whose shortest-path distances from s are known. (s ∈ S) 2.At each step add the vertex v ∈ V–S whose distance estimate from s is minimal to S. 3.Update the distance estimates of vertices adjacent to v.

Dijkstra’s algorithm

Example

Correctness —Part I Lemma. Initializing d[s]←0 and d[v]←∞ for all v ∈ V–{s} establishes d[v] ≥ δ(s,v) for all v ∈ V, and this invariant is maintained over any sequence of relaxation steps. Any suggestions about how to prove? –Induction Proof. Base case: check the initializing term Induction: suppose that this fails to hold at some point.

Correctness —Part I Suppose not. Let v be the first vertex for which d[v] < δ(s, v), and let u be the vertex that caused d[v] to change: d[v] = d[u] + w(u, v). Then how to induct a contradition? d[v]< δ(s, v) ≤ δ(s, u) + δ(u, v) ≤ δ(s,u) + w(u, v) d[v] = d[u] + w(u, v) Contradiction. supposition triangle inequality sh. path ≤ specific path v is first violation

Correctness —Part II Lemma. Let u be v’s predecessor on a shortest path from s to v: s→…u→v. if d[u]=δ(s, u) and if we relax edge(u, v), Then, we have d[v] = δ(s, v) after the relaxation. Proof. s→…u→v is the shortest path from s to v, so δ(s, v) = w(s→… →u)+ w(u, v) = δ(s, u)+ w(u, v).

Correctness —Part II Suppose that d[v] > δ(s, v) before the relaxation. –By Correctness I, d[v] ≥ δ(s,v) –Otherwise d[v] = δ(s,v), we’re done. Then test d[v] > d[u] + w(u, v) succeeds –because d[v] > δ(s, v) = δ(s, u)+ w(u, v) = d[u] + w(u, v) Then the algorithm sets d[v] = d[u]+ w(u,v) = δ(s, v).

Correctness —Part III Theorem. Dijkstra’s algorithm terminates with d[v] = δ(s, v) for all v ∈ V. Something to be addressed: –Once we add a vertex to S, we never change its weight Proof. It suffices to show that d[v] = δ(s, v) for every v ∈ V when v is added to S.

Correctness —Part III Suppose for contradiction: Suppose u is the first vertex added to S for which d[u] ≠ δ(s, u). –d[u] > δ(s, u). Let p be a shortest path from s to u. Now look at the first place here where the path p exits S

Correctness —Part III Let y be the first vertex in V–S along a shortest path from s to u, and let x be its predecessor. Since u is the first vertex violating the claimed invariant, we have d[x] = δ(s, x).

Correctness —Part III When x was added to S, the edge (x, y) was relaxed, which implies that d[y] = δ(s, y) ≤ δ(s, u) < d[u] But, d[u] ≤ d[y] by our choice of u. Contradiction.

Analysis of Dijkstra

Note: Same formula as in the analysis of Prim’s minimum spanning tree algorithm.

Unweighted graphs Suppose that w(u, v) =1 for all (u, v) ∈ E. Can Dijkstra’s algorithm be improved? –Use a simple FIFO queue instead of a priority queue. –Analysis: Time = O(V+ E).

Example

Correctness of BFS IDEA: The FIFO Q in breadth-first search mimics the priority queue Q in Dijkstra. Invariant: v comes after u in Q implies that d[v] = d[u] or d[v] = d[u] + 1.

Step Further I

Bellman-Ford algorithm

Example of Bellman-Ford

Step Further II All-pairs shortest paths Input: Digraph G= (V, E), where V= {1, 2, …, n}, with edge-weight function w: E→R. Output: n ×nmatrix of shortest-path lengths δ(i, j)for all i, j ∈ V. Floyd alglrithm

Any idea of this problem? Dynamic programming again! Consider the n×n adjacency matrix A= (a ij ) of the digraph, and define d ij (m) =weight of a shortest path from i to j that uses at most m edges Claim: We have and for m= 1, 2, …, n–1,

Proof of Claim

Matrix multiplication Compute C= A·B, where C, A, and B are n×n matrices Time = Θ(n 3 ) using the standard algorithm. What if we map “+”→“min” and “·”→“+”?