Multistage Graph (Shortest Path)
A Multistage graph is a directed graph in which the nodes can be divided into a set of stages such that all edges are from a stage to next stage only (In other words there is no edge between vertices of same stage and from a vertex of current stage to previous stage).
We are given a multistage graph, a source and a destination, we need to find shortest path from source to destination. By convention, we consider source at stage 1 and destination as last stage.
Following is an example graph we will consider in this article :-
Now there are various strategies we can apply :-
- The Brute force method of finding all possible paths between Source and Destination and then finding the minimum. That’s the WORST possible strategy.
- Dijkstra’s Algorithm of Single Source shortest paths. This method will find shortest paths from source to all other nodes which is not required in this case. So it will take a lot of time and it doesn’t even use the SPECIAL feature that this MULTI-STAGE graph has.
- Simple Greedy Method – At each node, choose the shortest outgoing path. If we apply this approach to the example graph give above we get the solution as 1 + 4 + 18 = 23. But a quick look at the graph will show much shorter paths available than 23. So the greedy method fails !
- The best option is Dynamic Programming. So we need to find Optimal Sub-structure, Recursive Equations and Overlapping Sub-problems.
Optimal Substructure and Recursive Equation :-
We define the notation :- M(x, y) as the minimum cost to T(target node) from Stage x, Node y.
Shortest distance from stage 1, node 0 to destination, i.e., 7 is M(1, 0). // From 0, we can go to 1 or 2 or 3 to // reach 7. M(1, 0) = min(1 + M(2, 1), 2 + M(2, 2), 5 + M(2, 3))
This means that our problem of 0 —> 7 is now sub-divided into 3 sub-problems :-
So if we have total 'n' stages and target as T, then the stopping condition will be :- M(n-1, i) = i ---> T + M(n, T) = i ---> T
Recursion Tree and Overlapping Sub-Problems:-
So, the hierarchy of M(x, y) evaluations will look something like this :-
In M(i, j), i is stage number and j is node number M(1, 0) / | \ / | \ M(2, 1) M(2, 2) M(2, 3) / \ / \ / \ M(3, 4) M(3, 5) M(3, 4) M(3, 5) M(3, 6) M(3, 6) . . . . . . . . . . . . . . . . . .
So, here we have drawn a very small part of the Recursion Tree and we can already see Overlapping Sub-Problems. We can largely reduce the number of M(x, y) evaluations using Dynamic Programming.
The below implementation assumes that nodes are numbered from 0 to N-1 from first stage (source) to last stage (destination). We also assume that the input graph is multistage.
Time Complexity : O(n2)