The Question:

Suppose there are $n$ extremely paranoid, vulnerable mental patients at a hospital.

Each day at the lunch hour, they move around like frictionless billiard balls of radius $\rho$ metres in a square, padded cell (of side length $s$ metres), and each day, at the start of the hour, they each resume their wandering from last time as if nothing happened.

At the first instance of this strange behaviour, the patients were positioned in a specific pattern and each began traveling at a specific velocity.

Each patient is severely short-sighted and can only see $r$ metres ahead of them, all within an angle $\theta$ radians either side of their noses (with the centres in the middle of their respective bodies).

They know about each other's vision and always point their noses where they're going.

The nurses at the hospital are hopelessly inept. Whenever a patient sees a total of $m$ other patients looking at them in a given lunch hour, that patient will commit suicide that night (and thus can no longer resume their wandering the next day and are removed).

For what configurations of $(n, \rho, s, r, \theta, m)$ and positions of patients & velocities on the first day can we prevent the suicides?


This is a mathematics problem I made up in 2013 when overthinking the blue-eyed islanders puzzle. I put my question up as a Facebook status and worked on it for about a week until my undergraduate degree commitments made me leave it, so I forgot about it until I saw it in the "On This Day" feature of Facebook today.

I hope it won't be answered by appealing to how silly it is. Feel free to make simplifying assumptions.

The answer is clear if $n=1$ with $m>0$, say, or if $m\le n$, $\theta=\pi$, $r=s$, etc. I think there could be some more interesting results however.

I don't think it is something I can answer myself in any depth.

NB: I take suicide and mental health issues seriously. Please seek help if you're suffering; for instance, call Samaritans for free on 116 123 via any phone if you're in the UK. I mean no offense.

  • 1
    $\begingroup$ Can you please remove the reference to suicide from the question title? When seeing it on the front page it's hard to be sure that the question isn't a genuine call for help, and in any case it's quite distressing. $\endgroup$ – Rahul Sep 25 '18 at 18:37

OK, I'll get the ball rolling (pun intended!). Some initial assumptions are

a. Radius of patients: $0 \lt \rho \le \frac{s}{2}$

b. Number of patients: $0 \lt n \le$ max patients possible in square of side $s$

c. Max patients allowed to look: $n-1 \ge m \gt 0$.

d. Sphere of any patient: Is within the square and doesn't overlap any other patient.

Some solutions are then

  1. Blind patients

Conditions: $r \lt \rho$

The distance patients can see is less than their radius, i.e the patients are effectively blind. This means the other configuration parameters can have any values (allowed by the initial assumptions).

  1. Side by side

enter image description here

Conditions: $\theta \lt 60^{\circ}$ and all patients are initially lined up along one wall of the square and their velocity is the same and perpendicular to the wall

The patients will bounce back and forth between the initial wall and the opposite wall without ever seeing another patient.

  1. Follow the leader

Conditions: $\rho = \frac{1}{8}s$, $n=4$, $r \lt 2\rho$ and initial positions and velocities as shown below:

enter image description here

Their closest encounter with another patient (if I calculated this correctly) is when they are halfway between two walls. So they never see another patient. This configuration is extendable to $n \gt 4$.

Interesting challenge!

  • 1
    $\begingroup$ You could pack even more people into the "follow the leader" approach. We don't need to keep patients from looking at one another, we need to keep them from seeing when others are looking at them. A conga line method solves this nicely, since no one ever sees anyone who can see them (unless the field of view is more than 180 degrees or the room is too small). $\endgroup$ – Nuclear Wang Sep 25 '18 at 18:57
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    $\begingroup$ @Nuclear Wang: I know, which is why I mentioned that the "follow the leader" approach could be extended to $n \gt 4$. $\endgroup$ – Jens Sep 25 '18 at 19:03

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