5 DP Bellman Ford Algorithm

An algorithm that works on graphs with negative weights.

From Djikstra's, we could possibly add a constant to all edges to make them all positive. However, if the shortest path (by weight) uses many edges, then the weight would have been increased for the entire path.

Negative weight graphs with cycles could cause a cycle and endlessly loop to decrease weight.

How it works¶

Build paths backward from a given node $T$. We can start by finding the shortest (least weight) path with one edge to $T$. Then we can repeat with two edges to $T$. When we repeat with three edges, if we find a better path then we update the node to $T$ with that path.

This algorithm may run for $n-1$ iterations, OR until two iterations where the shortest path does not decrease (need to be more clear).

Recursive Function¶

Cases: - The shortest $v-t$ path uses at most $i-1$ edges. - The shortest path uses $OPT(i-1,u)+w(v,u)$ for some $u\in N_{out}(v)$ - $mi{n}_{u\in N_{out}(v)}(OPT(i-1,u)+w(v,u))$ - Where $N_{out}(v)$ is the out neighbourhood of $v$

This function will end up being $O(n^3)$ time because for every node, it loops $1\to n$ edge lengths and for each edge length it will look at every edge connected to that node which in worst case is $n$.

It will be $O(n^2)$ space as each iteration of edge length, from $1\to (n-1)$, each $n$ nodes have some shortest path value.

Improving Space/Time Complexity¶

Improving Space Complexity We don't need to store every single column to make this work, just the $i-1$ iteration.

This means we only need to store: - the adjacency list of each node (in order to determine the in and out neighbours for each node) which is $m$ edges - They current column of the table, $M[v]$, and the successor $s[v]$ (the $N_{out}$ of $v$?) which in total is $O(n)$

Therefore, space complexity is reduced to $O(m+n)$

Improving Time Complexity We don't need to check $O(n)$ nodes for every entry $M(i,v)$, just the nodes which are in the out neighbourhood of $v$ ($N_{out}[v]$)

We will only compare up to $O(m)$ edges, resulting in $O(nm)$ time complexity.

Improvement: Push values ahead¶

• Notice that at each step of the algorithm we perform a fetch action to retrieve $OPT(i-1,v)$ or $OPT(i-1,u)$.
• If we can instead push these values ahead each time, we can skip the nodes whose values won’t change.
• Initially when $i=0$ only $t$ has a finite value - it is only reachable from itself. This means all paths using 1 edge must be of the form $(u,t)$.
• So the iteration when $i=0$, need only check those vertices $u\in V$ such that $t\in N_{out}(u)$, i.e., (u, t).

Basically, if we know a node $u$ takes less than $i$ edges to get to the goal node $t$, then we can just push it's weight to the next column for $i+1$ edges. In a subsequent iteration, if we find that we can reach $t$ taking a path from $u\to v\to t$, then we flag it as something we updated and have to check in a future iteration, and update it's optimal weight.

Negative Cycles¶

Since Bellman Ford works for graphs with negative edge weights, what if there exists some negative cycle?

Reasonably, that implies we could have a path longer than the number of nodes $n$ which would then be deemed the shortest/least weight path from $s\to t$. $$ \implies OPT(i,v) \text{ decreases as } i \geq n $$

All we need to do is check that $OPT(n,v)=OPT(n-1,v)$, where $n$ is the number of nodes in the graph.

Logically, this just means that if the shortest/least weight path from $s$ to $t$ contains more than $n-1$ edges, the weight will not decrease. If it does, then we have found a negative weight cycle somewhere.

Claim: There is no negative cycle on a path to $t⟺$ $OPT(n,v)=OPV(n-1,v)$ for all nodes $v$

Proof that the Bellman Ford has no cycles¶

Need to prove the iff

The version two of the proof for the forward direction

Reverse/Backward Direction of the IFF