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Everyday I walk to class and I have to walk through a sequence of sprinklers. I usually watch them for a second and try to plan a path in which I never have to stop or back track and will not get wet. It would be even better if I could maintain a constant speed.

Question

If we consider sprinkler system to be an $n \times n$ lattice where the water extends a length $L$ from the lattice point, has random initial direction and each rotates at a constant angular velocity $v$, what is the smallest $L$ and $n$ such that there exists no smooth, path from $(0,0)$ (or more generally, from any $(a,b)$) to $(n,n)$ and $\frac{dx}{dt} \ge 0$, $\frac{dy}{dt} \ge 0$? Or, what conditions must we impose on this system to guarantee a solution (other than having them all rotating the same and having me sprint arbitrarily quickly)?

Example sprinkler system:

enter image description here

Observations

Any case where $L \le .5$ is trivial since there is always a solution by following the outside of the circles formed by the sprinklers. Similarly, $L<\sqrt{2}n$ for any $n$. Of course there will not always be such an $n$ and $L$ like in the $2 \times 2$ case where there is a solution for all $L$ and any initial orientation. I feel like if $n$ and $L$ are sufficiently large, there will be no solution since you will eventually be caught inside of a polygon that is shrinking, but I do not have bounds on this conjecture.

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In your figure all the sprinklers rotate in the same direction (which I think is like reality) but in your text every other one (using what ordering?-a checkerboard?) rotates oppositely. –  Ross Millikan Sep 27 '12 at 14:11
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1 Answer

Observations for $L < \sqrt{2}$:

Start square graph whose vertices are the sprinkler points connected by edges to adjacent sprinklers. At any fixed time, two adjacent sprinkler lines either intersect or they don't. If they do intersect, remove the edge from the graph. A solution exists for a sufficiently faster walker if there is a path from connecting the top and bottom parts of the graph.

If the walker moves at a finite speed we need to consider the angular rotations. If all the angular rotations are rational (or the same), then there is a finite time at which the pattern will repeat itself. In your case all the sprinklers rotate at $\omega$ or $-\omega$, so this is guaranteed. Each intersecting pair will have a time at where they are intersecting or not. This gives a set of potentially $2^n+1$ different graphs, each existing for a known lifetime. For a given walker speed it is now possible to determine if a safe path exists by traversing the (now time-dependent) cyclic set of graphs.

Side Notes:

  • I am neglecting going around the array, but this can be taken care of by adding additional points on the graph boundary with no sprinklers.
  • With a larger $L$ the same method should still hold, but I'm a bit unsure how the graph edges would be removed.
  • We are both implicitly assuming a pointwise walker.
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