Assingment Description with output test cases provided as well as class notes.
1
THE FLOYD-WARSHALL ALGORITHM
• An intermediate vertex of a simple path p = {v1, v2, …, vq}
is any vertex of p other than vi
to vq
. i.e., any vertex in the set {v2, v3, …, vq-1}
For any pair of vertices i, j V,
we consider all paths from i to j whose intermediate vertices are
all drawn from {1, 2, …, k} k<= n = |V|
Let p be a minimum weight path among them.
2
Let p be a minimum weight path among them.
All intermediate vertices in {1,2,…k-1} All intermediate vertices in {1,2,…k-1}
i k j
P: All intermediate vertices in {1,2,…k}
p1 p2
3
•If k is not an intermediate vertex of path p,
then all intermediate vertices of path p are in the set {1,2,…,k-1}.
Thus a shortest path from vertex i to vertex j with all intermediate
vertices in the set {1,2,…,k-1} is also a shortest path from i to j.
with all intermediate vertices in the set {1,2,…,k}.
•If k is an intermediate vertex of path p.
then we break p down into
p1 is a shortest path from i to k with all intermediate vertices in the set
{1,2,…,k-1} and p2 is a shortest path from k to j with all
intermediate vertices in the set {1,2,…,k-1}.
k
j
i
p1 p2
4
A recursive solution to the all-pairs shortest-paths problem.
d (k)
ij : the weight of a shortest path from vertex i to vertex j with
all intermediate vertices in the set {1,2,…,k}
Recursive Definition:
d (k)
ij = w ij if k = 0
min (d (k-1)
ij, d (k-1)
ik + d (k-1) kj ) if k > 0
The matrix D (n) = (d (n)
ij ) gives the final answer.
i.e.,
d (n)
ij = (i, j) for all i, j V
because all intermediate vertices are in the set {1,2,…,n}
5
FLOYD-WARSHALL (W)
1 n=row[D]
2 let D (0) = W
3 for k=1 to n do
4 for i=1 to n do
5 for j=1 to n do
6 d (k)
ij = min (d (k-1)
ij, d (k-1)
ik + d (k-1) kj )
7 return D (n)
The running time: (n3
)
6
•Constructing s shortest path.
•Compute from D ---- O(n2
)
OR
•Compute “on-line” just as the Floyd-Warshall algorithm
compute the matrices D (n) :
(0)
ij = NIL if i = j or w ij =
i if i j and w ij <
For k 1
(k)
ij =
(k-1)
ij if d (k-1)
ij d (k-1)
ik + d (k-1) kj
(k-1) kj if d (k-1)
ij > d (k-1)
ik + d (k-1) kj
7
8 2 1 5 4 3 3 4
-4 6 7 8 1
-5 2
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