The Laser Gun

LasergunFront

I was casually browsing YouTube for nice puzzles (like one does) and stumbled upon this amazing point-line geometry puzzle that I had never seen before, so I couldn’t resist writing it up.

The full video introduces a series of nice puzzles designed to challenge common intuition and this was one of them.

You can see the introduction to the problem I am about to discuss in the clip below:

 
Fortunately (or unfortunately depending on your preference) the presenter leaves out the full explanation, although he does give the answer and some good hints on where to begin for the proof.
I thought I’d fill in the gaps to complete the argument of this excellent puzzle.

In case you did not want to bother watching, here is the problem statement from the video:

Problem (The Laser Gun):

You are standing at a fixed point in a large, rectangular room with mirrored walls.

TheLaserGunStatement

Elsewhere in the room is your enemy(red), with a laser gun.

Your only defense is that you may summon bodyguards and position them in the room so as to block all possible shots.

How many bodyguards do you need?

(I should remark that all entities in this problem are considered to be points, so they have no “thickness” although it appears that way in the picture).

 

As the presenter mentions, you could clearly do it with an uncountable continuum of points.
A circle is such a set and no matter how close the enemy is to us, we can center ourselves in a circle of bodyguards with radius half the distance between ourselves and our enemy.
(If you know some Topology you trivially recognize that this comes from the fact that \mathbb{R}^2 with the Euclidean topology is Hausdorff, and in particular a Kolmogorov space).

The presenter however tells us that only 16 bodyguards are necessary and sufficient to block all possible shots!
What is counter-intuitive about this problem is that there are infinitely many dense trajectories in the rectangle, meaning we can get hit from an infinite number of angles, and yet the claim is that only 16 bodyguards suffice to block all shots – amazing, no?

In case you wish to attempt the problem yourself here is the customary SPOILER ALERT!
 
 
 
 

Solution (The Laser Gun):

On a first look this problem seems increasingly difficult to analyze, how do you understand if an arbitrary trajectory is ever going to hit you or not?

The key to this problem was revealed in the video by the presenter.
The presenter mentions it is not too hard convincing yourself that only a countably infinite number of body guards are needed by considering reflections of the rectangle in the plane.

So what does he mean by this?
A countable infinity is in some sense smaller than an uncountable infinity (think of \mathbb{Z} vs \mathbb{R}).
This would be good because it means there are only finitely many trajectories that can hit us with the property that it may only bounce n times.
According to the presenter one can see this by looking at reflections of the rectangle.
Following his suggestion we can see that instead of bouncing, we can think of the laser as continuing straight through the mirror into another room which is the mirror image of the current room.
Each wall crossed by the laser represents a bounce in the actual trajectory.
See picture below:

LaserGunMirror

This completely removes the complexity of having piecewise linear trajectories (i.e bouncing) since we can just look at straight lines in the infinite plane. For the enemy to hit us, he now needs to hit one of our countably infinite reflected copies in any of the virtually mirrored rooms using a straight laser shot.
See picture below:

LaserGunTrajectories3

This is good, but there are still infinitely many target points.
It turns out this is fine, because one guard can block an infinite number of trajectories towards a target.
Why? Because not only are we reflected in each mirrored room, so are our bodyguards.
It therefore seems conceivable that a finite number might do given that there are only finitely many relative positions of ourselves in the rectangle across all reflected rooms (inspect above diagram and you will see).

Let’s add some mathematical precision to this argument.

Suppose the rectangle has dimensions a,b and the origin is set at the lower left corner.

LaserGunCoordinates
 
It follows that our reflections appear at the following points in the plane: (ma \pm x_1, nb \pm y_1) where m,n \in \mathbb{Z}

The equations of the straight lines between the enemy at (x_2,y_2) and our reflections are given by (y-y_2) = \frac{(nb \pm y_1) - y_2}{(ma \pm x_1) - x_2}(x - x_2) in two point form.

We now need to choose points in the rectangle in such a way that the points along with their infinite number of reflections are situated on each of the above lines.

The laser beam directly hitting us must pass through the mid-line between ourselves and our enemy so we may initially consider x = \frac{x_1+x_2}{2}.
Plugging this into the beams line equation (i.e setting m,n=0) we get y = \frac{y_1+y_2}{2}.
The point ( \frac{x_1+x_2}{2},  \frac{y_1+y_2}{2}) lies in the room so we may place a bodyguard there.
Incidentally one sees that all points of the form (ma + \frac{x_1+x_2}{2}, nb + \frac{y_1+y_2}{2}) lie on corresponding lines (y-y_2) = \frac{(nb + y_1) - y_2}{(ma + x_1) - x_2}(x - x_2).
However only targets of the form (3ma + x_1, 2nb + y_1) are situated in the line of fire protected by this guard (or some reflection of it).

Proceeding in similar fashion using symmetry, we can list all points covering required targets as:
(x,y) = ((ma \pm \frac{x_1}{2}) + \frac{x_2}{2}, (nb \pm \frac{y_1}{2}) + \frac{y_2}{2} )

Assuming x_1 < x_2, y_1 < y_2 it is enough to place guards at below 16 points since these are the points of the form above which have coordinates in the actual room (with remaining reflected points being collinear with one of them).
(\frac{x_2}{2} \pm \frac{x_1}{2}, \frac{y_2}{2} \pm \frac{y_1}{2} )
(\frac{x_2}{2} \pm \frac{x_1}{2}, b - \frac{y_2}{2} \pm \frac{y_1}{2} )
(a - \frac{x_2}{2} \pm \frac{x_1}{2}, \frac{y_2}{2} \pm \frac{y_1}{2} )
(a - \frac{x_2}{2} \pm \frac{x_1}{2}, b - \frac{y_2}{2} \pm \frac{y_1}{2} )

The other cases differ only by switching the roles of x_1,x_2 and y_1,y_2.

LaserGunGuards

 

It might be interesting to look at other shapes and try to answer the same question.

Here is an interesting example I came to think of:

The Laser Gun – Ellipse Version:

By definition an ellipse is the locus of points such that the sum of the distances to two fixed points (the foci) is a constant.

Ellipse

It follows by miracle of geometry that any trajectory passing through one of the foci must reflect into the other focus.
(btw if you want to construct the worlds easiest mini-golf course this may be it, you virtually can’t miss provided your putt has appropriate force)

Thus if the enemy is positioned at one of the foci and you on the other, then the enemy can shoot in any given direction he wishes and still be able to hit you. Therefore an uncountably infinite number of bodyguards are needed in this case to protect you. If instead your position is anywhere else and the enemy is at the focus, then at most two bodyguards are needed. All you have to do is position one bodyguard to block the direct shot and place the second bodyguard at the other focus to block every reflection.

LaserGunEllipseFoci1

 

Perhaps even more interesting would be to look at other n-sided polygons.
In fact this could make an interesting mini-project.

Clearly there is something funny going on when angles are not rational multiples of 2\pi.
I will let you work out what it is.

LaserGunIrregularAngle

It may therefore be easiest to start with regular n-gons.
For example in a room shaped like an isosceles triangle we would get a picture that looks something like the below:

LaserGunTriangle

Perhaps it may be tempting to draw some kind of conclusion from this, but inaccurate pictures have fooled many!
This needs a mathematical argument.

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One Response to The Laser Gun

  1. Pingback: Puzzles With Handicap | Mathematical Mélange

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