Proof for Floyd-Warshall's Shortest Path Derivation Algorithm Also Demonstrates the Hierarchical Path Construction Process

(THIS BLOG WAS ORIGINALLY WRTITTEN IN CHINESE WITH LINK: http://www.cnblogs.com/waytofall/p/3732920.html)

Foreword: Floyd-Warshall is a classical dynamical programming algorithm for deriving shortest paths between each pair of nodes on a graph. It has n iterations (n for the number of nodes). During each iteration k,  the shortest paths for each pair of nodes with intermediate nodes numbered no more than k are derived. Since inductively, we can assume that before iteration k, the shortest paths for each pair of nodes with intermediate nodes numbered no more than k-1 are derived, the way to derive the paths with intermediate nodes numbered no more than k can be derived as:

In order to derive all the intermediate nodes on a shortest path, the algorithm maintains a predecessor matrix Π, with its element  π_ij denoting the node before j in the shortest path from i to j. The matrix is defined as follows:

​For more detailed specification, please refer to Introduction to Algorithms.

My problem:  for two nodes i, j of graph, there is a maximal k, such that during the kth iteration, the shortest paths dynamically programmed by the algorithm was combined by shortest paths i ~ k, k ~ j, whose intermediate nodes are numbered less than k. And for the shortest paths i ~ k, k ~ j, we can also have such a iteration. Therefore, the path can be visualised as follows:

​In the tree displayed above, each parent is separated by the "middle node" k, which is the maximal number of iteration that combines two paths. The shortest paths derived from the predecessor matrix are easily proved correct, but the problem is, whether the paths derived by such methods are identical to the paths displayed by the figure above? In another word, whether the paths derived demonstrate the hierarchical path construction process of Floyd-Warshall? The answer is yes. And I'm gonna prove it.

Lemma 1: If, during the kth iteration, the shortest path from i to j was derived by combining paths i ~ k, k ~ j, and we have m = π_ij^(k) and m ≠ k; then the shortest path constructed by the algorithm from i to m also are derived from combining i ~ k,  k ~ m.

Proof: Since π_kj^(k-1) = π_ij^(k)= m, that is, in the path from k to j with intermediate nodes numbered no more than k, the predecessor of j is m, so we have:

                               d_kj^(k-1) = d_km^(k-1) + w_mj 

w_mj is the weight for edge(m, j).

Since there is an edge from m to j, from the Triangle Inequality, we have:

d_ij^(k-1) ≤ d_im^(k-1) + w_mj                                                                          (1)

which is:

d_im^(k-1)  ≥  d_ij^(k-1) - w_mj                                                                         (2)

From the condition of the lemma, we have:

                                d_ij^(k-1) > d_ik^(k-1) + d_kj^(k-1)                                                                      (3)

Subtract w_mj from both sides, we have:

                                d_ij^(k-1)  - w_mj  > d_ik^(k-1) + d_kj^(k-1)  - w_mj =  d_ik^(k-1) + d_km^(k-1)                 (4)

Which can be transformed to:

d_ij^(k-1)  - w_mj  > d_ik^(k-1) + d_km^(k-1)                                                          (5)

Combining (2) and (5), we have:

                                d_im^(k-1) > d_ik^(k-1) + d_km^(k-1)                                                                   (6)

Done.

Then we are going to prove our conclusion.

Proof:  We use mathematical induction. Before the first iteration, the paths derived from the Π⁽⁰⁾ matrix certainly meet the property.

Then, supposing that path derived from Π^(k-1) meet the property, then in the kth iteration:

(1). If the shortest path from i to j does not contain node k, we have  d_ij^(k) = d_ij^(k-1)，π_ij^(k) = π_ij^(k-1) = m. Applying the shortest path construction function, we can have a nodes sequence:

π_ij^(k) = m

π_im^(k) = n

...

π_io^(k) = p

π_ip^(k) = i

For any  π_ix^(k) (x belongs to {m,n,...,p,i}), π_ix^(k) =π_ix^(k-1). So we have:

π_ij^(k) = π_ij^(k-1) = m

π_im^(k) = π_im^(k-1) = n

...

π_io^(k) = π_io^(k-1) = p

π_ip^(k) = π_ip^(k-1) = i

That is, the shortest paths derived from Π^(k) is identical to the shortest paths derived from Π^(k-1). Since Π^(k) has the propery, the path also has the property.

(2). If the shortest path from i to j contains node k, we have:

           d_ij^(k) = d_ik^(k-1) + d_kj^(k-1)，π_ij^(k) = π_kj^(k-1)= m

and:

π_ij^(k) = π_kj^(k-1) = m

π_im^(k) = π_km^(k-1) = n

...

π_io^(k) = π_ko^(k-1) = p

π_ip^(k) = π_kp^(k-1) = k

That is, we enumerate the predecessor of j until we get to k. Now, all the intermediate nodes of the path from i to k are identical to the path derived from Π^(k-1), so the subpath meet the property. The same for path from k to j, so the shortest paths from i to j derived from Π^(k) meet the property.

Done.