3
$\begingroup$

A friend and I have been discussing turning a $O(n^2)$ graph problem's algorithm into $O(n\log n)$, or at least less than $O(n^2)$. And no - this is not a homework question. We've narrowed it down to the following subproblem:

Let $A,B,P,$ and $Q$ be a group of 4 nodes, with coordinates $(x_A,y_A),(x_B,y_B)$,etc... such that:

  • $x_A$ and $x_P$ are less than both $x_B$ and $x_Q$
  • $y_A$ and $y_B$ are greater than $y_P$ and $y_Q$.

Let there be an arbitrary cluster of nodes contained within the quadrilateral formed by $A,B,P,$ and $Q$. Find a pair of paths $A\rightarrow B$ and $P\rightarrow Q$ such that:

  • Every node in the cluster is hit by exactly one of the two paths
  • The two paths have the minimum possible total (Euclidean) length
  • The two paths do not intersect
  • The paths always increase in $x$. That is, if we enumerate path $A\rightarrow B$ as $\{A,(x_1,y_1),(x_2,y_2), ...(x_n,y_n),B\}$, then $x_i<x_j$ for $0<i<j\leq n$. Also, $x_A<x_1$ and $x_n<x_B$. This property must hold for the $P\rightarrow Q$ path as well.

It is also worth mentioning that no two nodes in the entire problem will have the same $x$ coordinate.

I'm positive some geometric trick will play a huge role in finding the solution. Keep in mind this is just a subproblem and a solution to this in $O(n^2)$ will still be helpful - though I'm hoping for $O(n\log n)$.

Here's an example problem for clarity:

Sample Problem

Any solutions or even helpful ideas are always appreciated. Diagrams are also very much appreciated.

Finally, for added clarity, here is what a sample solution might look like:

Sample Solution

$\endgroup$
  • $\begingroup$ To aid with intuition, suppose we look at the special case where $ABQP$ form an axis-aligned rectangle. Is it guaranteed that, for the optimal solution, every point on the $A\to B$ path will be higher (have a larger $y$-coordinate) than every point on the $P\to Q$ path? Can you find a proof or counterexample? If this is true, then there's a chance it might help find faster algorithms.... $\endgroup$ – D.W. May 15 '15 at 4:48
  • $\begingroup$ @D.W. Counterexample: Try drawing a 'W' shaped path between A and B, and assume the path is very dense with nodes (each node 'u' units apart - assume u to be very small).Now draw another path that comes up from P to just under the centerpoint of the 'W', whilst staying more than 'u' units away from the 'W', then connects down to Q. Assume the P->Q path is equally dense. The peak of the P->Q path will reach up inside the W and vertically surpass the bottom points of the 'W'. The path will be optimal b/c each node will require the absolute minimum cost possible to connect (u units). $\endgroup$ – Kyle McCormick May 15 '15 at 19:45
2
$\begingroup$

Since you said a $O(n^2)$ time algorithm would still be useful, there is a straightforward dynamic programming algorithm that runs in time $O(n^2)$.

In particular, a subproblem is specified by a pair of points $(A',P')$; the problem is to find the best pair of paths from $A' \to B$ and $P' \to Q$, using only points that are to the right of both $A'$ and $P'$. To solve each subproblem, you only have to look at the solution to two subproblems: in particular, to $(Z,P')$ and $(A',Z)$, where $Z$ is the point immediately to the left of the rightmost of $A',P'$. Therefore, each subproblem can be solved in $O(1)$ time. As there are $O(n^2)$ subproblems, the $O(n^2)$ running time follows immediately.

$\endgroup$
  • $\begingroup$ If I'm not mistaken, I believe "immediately left of the rightmost" should actually be "immediately right of the rightmost"? Because the left of the rightmost would already be contained in a path in the previous subproblem. Or am I misunderstanding? $\endgroup$ – Kyle McCormick May 15 '15 at 5:20
  • $\begingroup$ @KyleMcCormick, I think what I wrote was correct. Given the solution to subproblem $(Z,P')$, you can extend it to a solution to subproblem $(A',P')$ by adding the edge $Z\to A'$ to the end of the first path. Given a solution to subproblem $(A',Z)$, you can extend it to a solution to subproblem $(A',P')$ by adding the edge $Z \to P'$ to the end of the second path. Thus, given the solutions to subproblems $(Z,P')$ and $(A',Z)$, you can compute the best solution to subproblem $(A',P')$. This is a d.p. algorithm that goes "left-to-right". (You could also solve it "right-to-left", I guess.) $\endgroup$ – D.W. May 15 '15 at 5:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.