How many punching bags do you need to survive a bullet ricocheting in a room?

[lugita15]

Here is an interesting puzzle from Steve Landsburg's excellent blog:
http://www.thebigquestions.com/2012/05/22/tuesday-puzzle/

"You’re in a rectangular room. Elsewhere in the room is a man with a gun, who shoots a bullet in a random direction. The bullet careens around the room, bouncing off walls, until it hits either you or one of the various punching bags you’ve placed around the room for purposes of absorbing the bullet. The punching bags must be positioned before you know the random direction of the bullet (though you do know both your own location and the bad guy’s location, neither of which you can change). How many punching bags do you need to guarantee your survival?

This being a math problem, you should treat the room as two dimensional, and yourself, the bullet and the punching bags as points."

 

[Whovian]

None. As you're a point, and so is the bullet, the probability that, when shot in a random direction, the bullet hits you, for any finite amount of time, is 0.

 

[lugita15]

None. As you're a point, and so is the bullet, the probability that, when shot in a random direction, the bullet hits you, for any finite amount of time, is 0.

Of course, the problem is asking you to survive with certainty, not just with 100% probability.

 

[collinsmark]

None. As you're a point, and so is the bullet, the probability that, when shot in a random direction, the bullet hits you, for any finite amount of time, is 0.

From the comments section of the website having the original riddle:

However, if my assailant is shooting at random (in the sense in which that is normally understood), I can contrive a 100% chance of survival with 0 punching bags. Does a 100% chance of success count as a “guarantee”?

100% chance isn’t good enough. You need guaranteed survival.

That was from the author of the riddle. So apparently a probability approaching 100% isn't good enough for the answer. It needs to be 100% guaranteed.

 

[collinsmark]

My proposed answer:

Spoiler

1 punching bag. Place the bag at the exact location of the shooter. (The exact location of the shooter is one of the few things that you do know exactly.)

 

[lugita15]

That was from the author of the riddle. So apparently a probability approaching 100% isn't good enough for the answer. It needs to be 100% guaranteed.

Even a probability equalling 100% is not good enough. Certainty is required.

 

[lugita15]

My proposed answer:

Spoiler

1 punching bag. Place the bag at the exact location of the shooter. (The exact location of the shooter is one of the few things that you do know exactly.)

Both the shooter and the punching bag are points, so it will just be as if the bullet is leaving the punching bag.

 

[Topher925]

Zero. Point the shooter perpendicular to the wall so the bullet bounces back and hits him.

 

[Whovian]

But the problem did say he shot in a random direction.

 

[Topher925]

But the problem did say he shot in a random direction.

The direction can be random. It just needs to be perpendicular to a wall.

 

[Whovian]

So you know the direction beforehand, then?

And I don't understand how an angle of, say 30˚ relative to a random wall in a rectangular room could be perpendicular to the wall.

 

[Topher925]

So you know the direction beforehand, then?

And I don't understand how an angle of, say 30˚ relative to a random wall in a rectangular room could be perpendicular to the wall.

No, you don't know the direction the bullet will travel so its still random. The room is rectangular, so 4 sides. There is a random probability of 25% that the bullet will hit one of the 4 walls. Just don't stand between the wall and the bullet and you'll survive as long as the shooter is standing perpendicular to the wall.

 

[Whovian]

But what if it ricochets, bouncing off one wall, so it hits another wall, and ricochets back towards you? We were never told the bullet's direction would be perpendicular to the wall, and the only situation in which it would just bounce back and hit the shooter is if it's perpendicular to the wall it hits.

 

[lugita15]

No, you don't know the direction the bullet will travel so its still random. The room is rectangular, so 4 sides. There is a random probability of 25% that the bullet will hit one of the 4 walls. Just don't stand between the wall and the bullet and you'll survive as long as the shooter is standing perpendicular to the wall.

It doesn't matter what direction the shooter is faced in, regardless he will still shoot at a random angle, so the probability that he will fire perpendicular to a wall is zero.

 

[Borek]

No matter where I put the bag and myself, if we are points and we are not in the same place, it is always possible that the guy with the gun will stay between me and the bag and will aim exactly at me. So I don't see how it can be solved in a general case. It is not clear if putting bag and myself in the same position makes me safe or not.

 

[collinsmark]

Both the shooter and the punching bag are points, so it will just be as if the bullet is leaving the punching bag.

The riddle statement specifically said, "This being a math problem, you should treat the room as two dimensional, and yourself, the bullet and the punching bags as points."

What that implies is that the gun shooter and bullet (before the bullet is shot, if it's not already inside a bag) can be treated as a single point. The bullet stops the moment the bullet location is exactly equal to to the location of a bag or you. Placing a punching bag at the location of the shooter satisfies the mathematical requirement for stopping the bullet.

Think of it as placing the shooter, gun, and bullet inside a single punching bag, if that helps.

[Edit: or alternately, think of it as sticking a single punching bag inside the bullet (which happens to be at the exact location as the shooter and gun). In either case, the bullet and bag intercept, from a mathematical perspective. The requirement is satisfied. The bullet is stopped.]

 

[lugita15]

No matter where I put the bag and myself, if we are points and we are not in the same place, it is always possible that the guy with the gun will stay between me and the bag and will aim exactly at me. So I don't see how it can be solved in a general case.

Remember, your location and his location are fixed, and you can put the punching bags wherever you want. If you want to prevent him from shooting directly at you, just put a bag between you and him.

 

[Borek]

Remember, your location and his location are fixed, and you can put the punching bags wherever you want. If you want to prevent him from shooting directly at you, just put a bag between you and him.

OK, I misread the problem. Positions are known beforehand and fixed, somehow I missed that part [PLAIN]http://www.bpp.com.pl/IMG/grumpy_Borek.png.

Still, there is an infinite number of directions in which the gunner can shoot to hit me, so I don't see how it can solved with a finite number of bags.

But I don't have to be right.

 

[lugita15]

OK, I misread the problem. Positions are known beforehand and fixed, somehow I missed that part [PLAIN]http://www.bpp.com.pl/IMG/grumpy_Borek.png.

Still, there is an infinite number of directions in which the gunner can shoot to hit me, so I don't see how it can solved with a finite number of bags.

But I don't have to be right.

The thing is, there may be a finite set of points through which all of the infinitely many ricocheting paths go.

 

[lugita15]

The riddle statement specifically said, "This being a math problem, you should treat the room as two dimensional, and yourself, the bullet and the punching bags as points."

What that implies is that the gun shooter and bullet (before the bullet is shot, if it's not already inside a bag) can be treated as a single point. The bullet stops the moment the bullet location is exactly equal to to the location of a bag or you. Placing a punching bag at the location of the shooter satisfies the mathematical requirement for stopping the bullet.

Think of it as placing the shooter, gun, and bullet inside a single punching bag, if that helps.

[Edit: or alternately, think of it as sticking a single punching bag inside the bullet (which happens to be at the exact location as the shooter and gun). In either case, the bullet and bag intercept, from a mathematical perspective. The requirement is satisfied. The bullet is stopped.]

The bullet stops when it HITS a bag, not when it starts from or leaves a bag.

 

[Jimmy Snyder]

I will assume that the location of the shooter and my location are off limits for punching bags.

Spoiler

In that case, it is easy to prove that countably many bags suffice. If the bullet is to strike me without ricochet, then it must be aimed at me and I can stop it with a single bag. If it is to strike me after a single ricochet, then there are no more than 4 directions that I need to protect against, one for each wall. I can protect myself with 4 bags. For two ricochets, there are 12 directions, first hitting one wall, then a different wall, then me. 12 bags suffice. For each subsequent number of ricochets, multiply the number of bags by 3. That is to say, a bullet can ricochet off of any wall except the wall it just ricocheted off of, and then hit me. So for any finite number of ricochets, I need a finite number of bags. And since there can only be a countable number of ricochets, there can only be a countable number of directions I need to protect against. That is to say, a countable union of finite sets is countable.

 

[lugita15]

I will assume that the location of the shooter and my location are off limits for punching bags.

Spoiler

In that case, it is easy to prove that countably many bags suffice. If the bullet is to strike me without ricochet, then it must be aimed at me and I can stop it with a single bag. If it is to strike me after a single ricochet, then there are no more than 4 directions that I need to protect against, one for each wall. I can protect myself with 4 bags. For two ricochets, there are 12 directions, first hitting one wall, then a different wall, then me. 12 bags suffice. For each subsequent number of ricochets, multiply the number of bags by 3. That is to say, a bullet can ricochet off of any wall except the wall it just ricocheted off of, and then hit me. So for any finite number of ricochets, I need a finite number of bags. And since there can only be a countable number of ricochets, there can only be a countable number of directions I need to protect against. That is to say, a countable union of finite sets is countable.

That's clear enough, but the hard part would presumably be to prove that finitely many bags suffice.

 

[lugita15]

Moreover:


If neither I nor my assailant are touching a wall and are not both in the same place, then it is equally easy to prove that countably many bags are necessary. Since we are not in the same place, either our x or y (or both) coordinates differ. Say our x coordinates differ. Then with a single ricochet off of a wall that is parallel to the x direction I can be hit. With two ricochets, I can be hit with a bullet that hits first one, and then the other wall parallel to the x direction, then me. The bullet will make a zig-zag path. And so on with a bullet that alternates ricocheting off of the two walls parallel to the x direction and making a zig-zag path to me. Then for each n, there is a path with n zigs and zags that approaches me with the last zag at a different angle closer and closer to 90 degrees. Each of these paths requires a punching bag and so no less than countably many will suffice.

The thing is, the paths can intersect. So just because you have distinct paths doesn't mean you need distinct punching bags for each path. It may be that there is a finite set X of points such that for any path P, among the infinitely many ricocheting paths, there exists a point in X such that P passes through that point.

 

[Jimmy Snyder]

I know. That's why I deleted it. Not soon enough though.

 

[Jimmy Snyder]

I think it can be done with

Spoiler

only 7

bags.

Just let me know if I'm right and if so I'll provide the proof. Otherwise it's back to the drawing board. Literally.

 

[lugita15]

I think it can be done with

Spoiler

only 7

bags.

Just let me know if I'm right and if so I'll provide the proof. Otherwise it's back to the drawing board. Literally.

The right answer is not known yet. But it would still be nice to hear what you got.

 

[lugita15]

I have an idea, I don't know how useful it may be. In the case where the two side lengths and the cosine of the angle of shooting are all commensurable (meaning the ratio of any two of them is a ratio of whole numbers, or equivalently there is some unit relative to which all three numbers are whole-number multiples), then the path of the bullet will be finite, in the sense that it will go back and hit the shooter eventually.

 

[Jimmy Snyder]

I'm sorry, there was a flaw in my proof. However, I will prove the following.

Spoiler

If there is a finite number of bags that suffice for one placement of the shooter and yourself in the room, then the same number of bags suffices for all rooms and all placements.

Pf. Suppose that a finite number of bags suffice for a particular layout. Draw the layout and call it the base cell. Now tesselate the plane with mirror images of the base cell. Draw a line from the shooter in the base cell to the copy of yourself in each of the cells. Each of these lines when folded and reflected back into the base cell will be the trajectory of a bullet that would have hit you were it not for a bag. Therefore, there will be a bag in its path between its endpoints. For any other configuration of room and people, leave all the copies of the shooter, yourself, and the bags as is, and erase the lines that represent the walls of the room. Now redraw a rectangle to represent the room in such a way that there is exactly one copy of the shooter and one copy of yourself within it and so that the the room has the correct dimensions and the poeple have the correct positions. Then tesselate the plane with the walls of this new room. You may need to fold and reflect bags so that one copy of each bag ends up in the new room. Since no change was made to the trajectories, they still are interupted on their flight from shooter to yourself by a bag and when folded and reflected across the walls of the new rooms they still represent the trajectories of bullets.

I thought that I had found a particulary nice symmetric layout of room and people that required 7 bags. However, I was able to find a trajectory that missed all 7 of these bags so I have no proof of how many it takes. However, I think the idea of tesselating the plane makes it easy to look for symmetrical layouts that are easy to work with.

 

[lugita15]

I'm Gunna say 100 punching bag should secure your safety. Just jam them all in there.

The punching bags are all points.

 

[256bits]

Only one punching bag should suffice iif you can also choose your own location.You would place the bag between you and the shooter so he cannot aim at you directly.

My hint is that it is a case of odd/even numbers, and whether we are working with points or the same discrete size of bullet/punching bag doesn't play into the solution at all.

PS. of course the size if the bullet/bag mutilpyied by a whole number must equal the length of a wall which would also be true for the set of points.

 

[mathal]

The solution is to be provided in 2 dimensions.

Spoiler

Put the punching bag in a diagonal position in a 'corner' of the 2 dimensional room with you behind the bag.

mathal
As the puzzle is presented, an infinite number of bags is required even in a 2 dimensional room.-my spoiler is wrong.

 

[Jimmy Snyder]

Here's a related problem. How many people will post wrong answers without even reading the OP? Hopefully it's finite.

 

[lugita15]

As the puzzle is presented, an infinite number of bags is required even in a 2 dimensional room.

What is your justification for this? Just because there are infinitely many paths does not mean that there are infinitely many bags, does it? You don't need one punching bag for each path, because you can place a bag at the intersection point of multiple paths. Isn't it possible that all the infinitely many paths intersect in finitely many points? That is to say, can't there be a finite set of points such that anyone of the infinitely many paths passes through at least one of the points in the finite set?

 

[lugita15]

Only one punching bag should suffice iif you can also choose your own location.You would place the bag between you and the shooter so he cannot aim at you directly.

Are you saying that you want to be able to choose you're location given the shooter's shooting angle, or do you want to choose your location and the punching bag location BEFORE the shooter decides what angle to shoot at? If the former, then the problem becomes trivial, because you just put the bag in whatever angle the shooter is shooting at. If the latter, how will one punching bag suffice? What if the shooter shoots at you indirectly?

My hint is that it is a case of odd/even numbers, and whether we are working with points or the same discrete size of bullet/punching bag doesn't play into the solution at all.

If we had discrete bullets and punching bags, then the problem could have an entirely different solution, because as TheNewGuys said we can just jam all the bags in there.

 

[256bits]

Are you saying that you want to be able to choose you're location given the shooter's shooting angle, or do you want to choose your location and the punching bag location BEFORE the shooter decides what angle to shoot at? If the former, then the problem becomes trivial, because you just put the bag in whatever angle the shooter is shooting at. If the latter, how will one punching bag suffice? What if the shooter shoots at you indirectly?
If we had discrete bullets and punching bags, then the problem could have an entirely different solution, because as TheNewGuys said we can just jam all the bags in there.

The actual answer is infinite bags, and my one bag is for a particular solution only.

The catch is that between any 2 points on a line there is an infinite number of points.
And that angle of incidence = anle of reflection.

If X is the length of the short side, one can divide that side into N points. The other side can be M= N, N+1, N+2, ... number of points. The distance between points is dx = X/N ( or dx=X/N+1 depending on where you place your points ).

With 4 aims perpendicular to the wall the shooter shoots himself and you are safe. You also need one bag between you and the shooter to be safe. If you place the bag at a symetrical location to where the shooter is located and place yourself at a position dx/2 behind the bag then after multiple reflections the shooter hits himself or the bag, and never you. By moving to dx/2 you will have though increased the number of points on the wall to 2N.

To hit you, with one reflection, the shooter will just have to alter his aim so that it is dx/4 on the wall of the previous location where it hit the bag or himself. Note that you can move yourself to dx/4 behind the bag and the shooter alters by dx/8 to hit you so you are never safe in the dimensionless scenario. You have increased the number of points to 4N, 8N,...
ad infinituim.

I hope that makes sense. There probably is a more rigorous solution of why 2 points can never be blocked and can always be joined by an infinite series of reflected lines.

 

[BobG]

Spoiler

Five.

You don't want the shooter to shoot directly at you. Other than directly at you, there are only four directions the shooter can shoot and hit you.

Rebounds from the wall come off the wall at the same angle they approached the wall. The sum of the angle on the side wall plus the end wall has to equal 90 degrees. (In other words, if the bullet approached and left the side wall at 30 degrees, it will approach and leave the end wall at 60 degrees.

That means a shot will not cover the entire area of the rectangle no longer how many times it bounces. You could select an angle that will cover a huge percentage of the rectangle, but, it can never cover the entire rectangle and there are still only four possible directions that you can be hit from.

Essentially, you could consider the rebounds as the mirror image of an infinitely expandable rectangle and the path of the bullet could eventually be expanded into a large parrellelogram that constantly repeats.

And, because the departure from the wall has to equal the approach to the wall, there's only four possible indirect directions the shooter can shoot that will actually hit you. And four instead of two because you also have to account for your mirror image on the other side of the shooter (in other words, the bullet can actually travel two different directions along the same line).

 

[Jimmy Snyder]

But a bullet might ricochet many times without ever striking 2 adjacent walls in succession.

 

[BobG]

Wouldn't matter. All distances would be expanded by the same factor until the size of the rectangle and the distance between the two persons were large enough for the parallelogram to fit. You could alternatively calculate each of the rebounds for cross bounces, but the paths of alternate bounces have to be parallel to each other and we're only worried about direction - not which bounce hits him.

 

[Jimmy Snyder]

Wouldn't matter. All distances would be expanded by the same factor until the size of the rectangle and the distance between the two persons were large enough for the parallelogram to fit. You could alternatively calculate each of the rebounds for cross bounces, but the paths of alternate bounces have to be parallel to each other and we're only worried about direction - not which bounce hits him.

You are not allowed to expand distances. The shape of the room and the placement of shooter and yourself are fixed before the bags are placed. By ricocheting off of oppostite walls, the bullet may approach you from infinitely many different angles.

 

[BobG]

You are not allowed to expand distances. The shape of the room and the placement of shooter and yourself are fixed before the bags are placed. By ricocheting off of oppostite walls, the bullet may approach you from infinitely many different angles.

Expanding the dimensions is just to find an easy way to calculate the direction the shooter has to fire in order to hit you. Once shot, the bullet only travels in four directions over and over and over. All the parallel paths can be treated as a single path.

But six or eight might be the right answer (five definitely is wrong since I made an assumption that might not be true). You have two opposite directions along each side of the parallelogram, plus directly at you, which is the sum of the vectors forming the two sides, plus directly away from you provided the line formed by you and the shooter isn't perpendicular to the wall (the bullet actually has to hit you before it hits the shooter).

I think the difference of the two vectors might hit you, as well, with there being two different differences since subtraction isn't commutative (in other words, you can travel two different directions along the difference, as well). But the problem with the difference idea is that the shooter and you have to be at opposite corners of whatever parallelogram you form. Intuitively, I think the difference is guaranteed to miss you.

Keep in mind the possibilities are restricted by the fact that the angle of approach to a wall has to equal the angle of departure. The bullet can't approach the wall at a 30 degree angle and then bounce off at a 50 degree angle, so really aren't an infinite number of parallelograms that intersect your point. You have to toss out all of the parallelograms that can't possibly occur, and that turns out be almost all of them.

So, for example, if I had a 10x8 rectangle, and the shooter were located at coordinates 1,2 and I were located at coordinates 8,5; then any combination of vectors that added up to 7+3i or -7-3i (or some scalar multiple that gives me the same directions) would hit me. That's an infinite number of vector combinations.

For example, the bullet travels 2+5i, bounces off at 3-9i, and then bounces off the other wall at 2+7i. The bullet hits me.

The lengths of any parellel vectors can just be added together as if they were one vector, since they're traveling the same direction. Additionally, for each vector I add together I also have a vector along the other side that has to be added to the other side of the parallelogram. I don't care that adding them together gives me a fictional vector that travels beyond the rectangle. That still leaves me with an infinite number of vectors that can hit me.

For example, the bullet travels 2+5i, bounces off at 1-18i, then bounces off at 4+10i. It's the same result as if you just added the two parallel vectors together (6+15i), creating a two vector solution of (6+15i) + (1-18i).

However, only vectors where the approach angle and the departure angle will be equal will actually occur. There's an infinite number of vector combinations that don't meet that requirement, and once I've subtracted out the infinite number of vectors that don't meet the requirement, only six remain that meet both requirements.

In other words, if the bullet travels a 2+5i, it isn't going to bounce off at 3-9i or 1-18i. It's going to bounce off at 2-5i (or some multiple, such as 4-10i). And once it finally reaches the side wall, it's going to bounce off at -2+5i or -2-5i (depending on which direction the bullet was traveling when it hit the side wall). So only those vectors that can meet that requirement, plus add up to 7+3i are valid solutions. (2+5i is not a valid solution for the dimensions and locations of the shooter and me).

 

[256bits]

If off an angle of reflection of 45 degrees the bullet hits you, then the bullet will also hit you with an angle of reflection of 22.5, 11.25, ... All the angles can never be covered.

 

[BobG]

If off an angle of reflection of 45 degrees the bullet hits you, then the bullet will also hit you with an angle of reflection of 22.5, 11.25, ... All the angles can never be covered.

The numbers are off unless both the shooter and the target are against one of the walls, but the principle is correct.

After playing with it a little bit, I think this is better handled as a wave problem(except with a sawtooth wave instead of a sine wave), which is sort of what you did with your solution. But, just as with a sine wave, you could get an infinite number of angles approaching the target by varying the wavelengths and the amplitudes (and the wavelengths that aren't harmonics of the shooter-target wavelength just miss and miss and miss).

But, yes, there will be an infinite number of harmonic wavelengths.

(Too bad. That seemed so cool to realize that every solution was just a parallelogram that folded over whenever it hit a wall, which it is, I guess, just with an infinite number of parallelograms.)

 

[mathal]

What is your justification for this? Just because there are infinitely many paths does not mean that there are infinitely many bags, does it? You don't need one punching bag for each path, because you can place a bag at the intersection point of multiple paths. Isn't it possible that all the infinitely many paths intersect in finitely many points? That is to say, can't there be a finite set of points such that anyone of the infinitely many paths passes through at least one of the points in the finite set?

As the puzzle is two dimensional and all the 'pieces' are 'points', in the corner, 2 points, in the 'outside' points of a square would preclude any point of the same 'size' reaching you (exclude the diagonal as contact would necessarily be made with one or both of the other bags first.). As long as a regressive analysis of this solution to infinitely small 'points' is allowed this should suffice as a solution.
mathal

 

[lugita15]

As the puzzle is two dimensional and all the 'pieces' are 'points', in the corner, 2 points, in the 'outside' points of a square would preclude any point of the same 'size' reaching you (exclude the diagonal as contact would necessarily be made with one or both of the other bags first.).

I have no idea what you're saying. What is the "size" of a point?

 

[AlephZero]

One way to get a handle on this is to imagine the walls are perfect mirrors, and ask yourself how many images of the shooter you would see.

You know the image in a mirror is the same distance behind the mirror as the object is in front. So plot the 4 images of the shooter in the 4 mirrors, then the images of the images in the 4 mirrors, then the images of the images of the images, etc.

You can then trace the path of the bullet backwards from you an image of the shooter, and reflect it off the walls till it reaches the "real" shooter.

It's fairly obvious that a finite number of bags will only work in special cases, where a infinite number of multiple reflections all "line up" behind each other when viewed from your position.

 

[mathal]

Originally Posted by mathal
As the puzzle is two dimensional and all the 'pieces' are 'points', in the corner, 2 points, in the 'outside' points of a square would preclude any point of the same 'size' reaching you (exclude the diagonal as contact would necessarily be made with one or both of the other bags first.).
As long as a regressive analysis of this solution to infinitely small 'points' is allowed this should suffice as a solution.

I have no idea what you're saying. What is the "size" of a point?

a point can be considered to be a circle of radius 0 at a particular coordinate position. The limit approaching 0 is 1 divided by infinity. If the argument I presented for finite circles in a corner of the room is accepted then divide the radius of the point 'circles' by 2 an infinite number of times.
The 'puzzle' is more philosophical than anything else. When is a point, not a point-when it has names and -properties a point can't posess- such as absorbing a bullet.

mathal

 

[legend_b0bby]

pi, because it is the amount needed to cover the circumference of a point. or 4 because a point cannot be cut into 0.14159265358979...

 

[Whovian]

pi, because it is the amount needed to cover the circumference of a point. or 4 because a point cannot be cut into 0.14159265358979...

The "circumference" of a point (considering a point as a circle with 0 radius) is 0. Also, again, the bags are points. How would 3 punching bags cover 3 units of distance if they're pointlike?

 

[lugita15]

Steve Landsburg, who originally posted the puzzle, now posted his solution:
http://www.thebigquestions.com/2012/06/12/tuesday-solution/