Here is the full title I would have posted, but it happens to be too long:

Given a source node, dest node, and intermediate nodes, how does one detect if the shortest Manhattan Distance is blocked by the intermediate nodes?

I've drawn a diagram to make it more clear. On the left side, "u" is the source node and "v" is the destination node. The nodes labeled 1 through 6 are the intermediate nodes. The shortest Manhattan Distance from u -> v would be 12, but the intermediate nodes form a wall blocking it. The diagram on the right, with u' being the source, and v' being the destination, shows that the intermediate nodes 1 through 5 do not block the shortest Manhattan distance from u' to v'.

I'm trying to find an algorithm that won't require me to actually do a graph search (e.g. BFS), because the distance between u and v could potentially be very large.

If all you want to do is detect whether the shortest path (one consisting of moves that monotonically take you in the right direction) is blocked, then you are trying to check whether the blocking nodes cut the rectangle whose corners are given by the source and destination node into two different regions that are disconnected. If no shortest path from the source to the destination is possible, then every path must have some point in it that's blocked.

Let's suppose for simplicity that your start point is below and to the left of the destination point. In that case, find, in O(n), all of the other points that are obstacle points contained in the bounding box holding the start and end point. You now want to see if there is some subset of those nodes that cuts the rectangle into two pieces, one containing the bottom-left corner and one containing the top-right corner. This is possible iff there is a path of the blocking nodes from the left side to the right side, from the left side to the bottom side, from the top side to the right side, or from the top side to the bottom side. Thus we just need to check if any of these are possible.

Fortunately, this can be done efficiently by modeling the problem as a graph search in a graph that has size O(n), where n is the number of blocking points, and has nothing to do with the size of the bounding box. That is, no matter how far apart the test points are, the size of the graph to search depends solely on the number of blocking points.

The graph you want to construct has two parts. First, build a graph where each blocking point is connected to each other blocking point in the 3x3 square surrounding it. These edges link together blocking points that could be part of the same barrier, in that no path from the source to the target could pass between two blocking points joined by an edge. Now, add in four new nodes representing the top wall, left wall, right wall, and bottom wall and connect them to each node that is adjacent to the appropriate wall. That way, for example, a path from the left wall node to the right wall node would represent a series of blocking nodes that make it impossible to get from the bottom-left node to the top-right node.

This graph has size O(n), where n is the number of blocking nodes, since there are O(n) nodes and each node can have at most 12 edges - one for each of the 8 neighbors and potentially one for each of the four walls. You could construct it in at worst quadratic time by scanning over each node and, for each other node, seeing if they are adjacent. There is probably a better way to do this, but nothing comes to me at the moment.

Now that you have the graph, for each of the pairs of walls that, if connected, would disconnect the graph, run a graph search in this graph between those two wall nodes. If a path exists, report that the shortest path is blocked. If not, report that some shortest path is unblocked. These searches could be done with a simple DFS, or since you're running multiple searches and just want to know if they're connected, using a strongly connected components algorithm once and checking if any pair of important nodes are in the same SCC. Either approach takes time O(n).

Thus the time to solve this problem is at most O(n2), with the bottleneck being the time required to construct the graph.

Hope this helps!