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.