Billionaire's Gates

[borschevsky]

I think a torus would work - just build room B around the hole in the torus. Then you can get 'out' of room B without using up a door.

[mathgrant]

I was thinking about wrapping the shed around something like a cylinder so that room A would touch room C and room D would touch room E.

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[This message is being edited by mathgrant (being edited right now).]

[Ghost Post]

How about putting it on a Sphere. That way all walls would be the walls of another room and there would be NO outside....

And just think of all of the contractor bills he'd avoid! The path would be :

C-E-B-D-A-B-D-E-B-C-E-D-A-C



[This message has been edited by Fink (edited 01-08-2001).]

[borschevsky]

Like so:

[happymath]

I like borschevsky's answer. But now I'm thinking that the original phrasing of the problem is a bit weird. It clearly allows for a torus-shape (for example), but that allows simple closed loops which do NOT cut the surface into two pieces. (For the plane and for spheres, the Jordan curve theorem says that a simple closed loop always cuts the surface into exactly two pieces, but that's false for other surfaces, like the torus.)

But the puzzle says
{QUOTE]Every room had a doorway to all the adjacent rooms, plus one to the outside.[/QUOTE]

So what does "the outside" mean? Normally it means the area outside all the rooms, but this might not exist on a torus. So I guess we're allowed to choose the outside to be anything we want, provided the rooms together cover the whole surface.

In that case the torus answer works, because you have a door from B (viewed as a room) to B (viewed as the outside), and TWO doors from every other room to B (one considering B as a room and the other considering it as the outside).

[ZenBeam]

If there still has to be an outside, the problem is still solvable. Just make the two ends of the torus open into rooms B and E instead of B and the outside. This could've been done on Earth with some stairs and a tunnel. Stupid Billionaire didn't have to go to an asteroid after all.

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It is too clear, and so it is hard to see.

[Coyote]

Yes indeed. In fact, given the stipulations given in the original puzzle, any number of rooms in any configuration is solvable with one simple algorithm:

Rule 1: Keep passing through 'unused' doors til you reach a room (or outside area) with no unused doors. Then dig a tunnel to any room with at least one unused door and return to the start of 'Rule 1'.

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Gravity is a harsh mistress.



[This message has been edited by Coyote (edited 01-09-2001).]

[Ghost Post]

I think mathgrant's solution is good and simple. By wrapping it around a cylinder, this forces a new door to be added from A to C, which leaves you with only one room with an odd number of doors.

[This message has been edited by LtCmdrData (edited 01-10-2001).]

[Knoque]

The only problem with mathgrant's solution is that in order to connect A and C with a door, then D and E would also be connected. This now makes three rooms with an odd number of doors. Unless it is possible to connect A and C without connecting D and E, I don't think this solution would be the correct one.

[TVaughn]

To: Knoque

But D and E are *already* connected (even on the Euclidean plane), so another door on the other "wrapped-around" wall would be redundant (and, therefore, forbidden).

The problem with a simple cylinder is that now there are *two* "outsides", one "above" the shed (connected to A, B and C) and one "below" (connected to D and E).

The torus shape can solve this, if you connect the two "outsides" through the "hole" -- picture the shed constructed around the outside of a (rather large) donut instead of a (cylindrical) soda can.

Even better for visualization (though topologically the same) would be to imagine the shed wrapped around the *inside* of the "donut hole", as the proportion of area "outside" the shed would be greater to that "inside" the shed.

So, back to the asteroid....

First, drill a tunnel completely through the asteroid from one part of its surface to another (of course, the tunnel doesn't have to pass through the exact center of the asteroid, but you can imagine that it does). Second, construct the shed around the walls of this tunnel (sort of an "inside-out" cylinder, if you will). Be sure to build a doorway on only *one* of the two walls you constructed between rooms D and E (they're adjacent on two sides, remember). You now have only one room with an odd number of doors (room B), so that's where you start and stop your path.

[Ghost Post]

TVaughn:

It's all one outside; I would choose to exit through the roof of each room to get to the outside, where all the exits to the outside would be.

Psst: If B has an odd number of doors, it's where you either start or end your path, but not both.

[This message has been edited by LtCmdrData (edited 01-11-2001).]

[Rollercoaster]

I just can't stop staring at Borschevsky's beautiful shiny donut - makes it impossible to think about the problem at hand...

[Ghost Post]

The rules state every adjacent room has a door. Every room has a door
to the outside. This implicity says that yes, there is an outside. You cannot
simply solve the problem by saying the outside space doesnt exist. It is a
given that it does exist.
My solution is to build 3 stories.
The bottom story has 2 rooms. (Rooms A & B)
The middle story has 2 rooms. (Rooms C & D)
The third story has 1 room (Room E).

Since each story is on a different plane, there are no doors
connecting floors. The path is then E-->D-->C-->B-->A (or its reverse).
Please let me know if this is not within the rules of the puzzle.
charles@zzyzx.com

[Ghost Post]

The only problem with the 3-story answer is if you argue that floors on
different floors are adjacent. However I believe that adjacency doesnt
apply across planes. Therefore the 3-story idea works.
However, My co-worker, Kim Fortenberry, came up with another idea.
Each room is enclosed within another room.
Thus, A is totally enclosed within B, and B within C, etc.
Draw a cube, then a cube around that cube, and then a cube around that outer
cube, etc.. until you have 5 cubes. Each cube has only 1 door. (a door to the
outside. No rooms are adjacent to any other room) Therefore, the path
is a simple line from the inner cube (A) to the next outter cube (B) to the
next outter cube (C) to the next outter cube (D) to the next outter cube (E) to
the outside. Note that the outside of say, Cube B is outside Cube A.


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"Data wants to be free, Information expesnive"
- Roland D. Steedlam

[Qball]

I don't think you can change the basic structure of the toolshed. Therefore, the five cube idea seems to be outside the scope of the puzzle.

I think Borshevsky has the simplest solution.

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Sorry about that, Chief -- Agent 86, Maxwell Smart

[hanno5]

The solution to the Billionaire's Gates problem is simply to identify the exterior doors of rooms A, B, and C. Start inside room A, B, or C and go outside. Since the exterior doors for these three rooms are identified, exiting any one of them is the same as exiting all three of them. Then go back inside room E and follow this path... E-C-B-E-D-B-A-D-Outside.

[Ghost Post]

Whether each room has 1 door to the outside or 2, the solution's roughly the same. We basically have to have the ability to "tunnel" exactly once from one room to another without using a door.

Take a prismatic asteroid that, in cross-section, looks like a letter C.

For the 1-door case:
Curl your shed and an apron of exterior space "lengthwise" around the C, so that one of the 3-door rooms hangs above the other. Enter one of the 3-door rooms from the outside, then jump from that one into the other. You now have exactly 2 rooms with an odd number of unused doors (3,5). The rest have an even number (including the "apron" or exterior of the shed).

For the 2-door case:
Similarly, curl your shed around the C, but rotated 90 degrees from the orientation in the case above, so that one of the 5-door rooms in the lower half of the diagram hangs above the other. Enter one of these 5-door rooms from the outside, then jump from there into the other. You now have exactly two rooms with an odd number of unused doors (both 5); all the rest have an even number.

[Ghost Post]

Quoting from the original problem: "Every room had a doorway to all the adjacent rooms, plus one to the outside."

A is adjacent to BD
B is adjacent to ACDE
C is adjacent to BE
D is adjacent to ABE
E is adjacent to BCD

A has 3 doors.
B has 5 doors.
C has 3 doors.
D has 4 doors.
E has 4 doors.

Any room with an odd number of doors must lie at either the end or beginning of the path. But there are three rooms with an odd number of doors, so such a path is impossible.

The geometry of the space that the graph lies in is immaterial -- all that matters is the connectivity between rooms. Nothing is the restated problem suggests that the connectivity between rooms changes -- they all still have only one route to the outside and are still connected to the same neighbors by a single door. There still is no path that solves the problem.

[daniel801]

[daniel801]

does the above work?

[borschevsky]

Shrike, the geometry of the space does affect the problem, because in some geometries we can move from room to room (or node to node on the graph) without using a door (or an edge).

[Bicho the Inhaler]

I agree with Coyote.

[ewan3]

I think ive solved it. join rooms a and c by bending the surface so that they have a door bettween them (becuase they are now adjacent). now join the outside at the top with the outside at the bottom. now all rooms have 4 doors except b which has 5. resulting surface looks a bit like two hoops connecting at a saddle.

[LOG]

I think that the rooms would be wrapped around half of a sphere, therefore many combinations would be reached.... One solutin would be A>D>B>C>E and returning back to A once more.

[NewHere]

http://www.driveway.com/share?sid=5df07a79.c893f&name=Pictures&view=0

If you wrap the room around a cylinder lengthwise, so that there are doors from top to bottom, but leave the side open to the outside, the route is easy.
I'm hoping this picture link works... Not certain it will...

[NewHere]

Picture link works...Could somebody who's registered kindly post the gif to this board, pretty please?

[daniel801]

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I yell because I care.

[NewHere]

Thanks Daniel801

[CrystyB]

So are the solutions borschevsky and me proposed valid? Can there NOT be an actual outside, but it being the same with one of the rooms?

Or are we to move the rooms around our surfaces? And so have only 4 rooms and one outside?