[bushindo]

Here's a picture. There are 16 connectors connecting the dots. Find a way to draw a single line so that it crosses through every single connector once and only once. You can start from any point on the image, and you can finish at any point. The line must be continuous, and crosses every connector once each.

Here's an attempt. For ease of discussion, I label the edges from 1-16. The lines goes through connectors (2, 1, 7, 5, 4, 14, 15, 13,12,10, 9, 11, 8). Note that this doesn't satisfy the requirement above because the connectors 6 and 16 haven't been crossed.

Have fun.

[Guest]

I know a way to cross every line once! The puzzle gives us a two dimensional image but never states that the line must be two dimensional as well! therefore with a three dimensional line you can go over or under the line instead of crossing it directly!

Wow i just read the post above mine... way to steal my thunder man...15 min before me..

oh well at least i drew a picture

Edited May 27, 2009 by BishopofDis

[bushindo]

I know a way to cross every line once! The puzzle gives us a two dimensional image but never states that the line must be two dimensional as well! therefore with a three dimensional line you can go over or under the line instead of crossing it directly!

Wow i just read the post above mine... way to steal my thunder man...

oh well at least i drew a picture

You guys are hilarious. I didn't think of the 3 dimensional thing, so good job.

[Guest]

Does anyone have a solution to this? It seems impossible but I'm probably just overlooking it. (And I'm not buying the 3D solution, that would be a serious middle finger to any one trying to solve this)

[CaptainEd]

Does anyone have a solution to this? It seems impossible but I'm probably just overlooking it. (And I'm not buying the 3D solution, that would be a serious middle finger to any one trying to solve this)

it's impossible, due to Euler's original graph theorem. There are 4 land masses (the outside counts as one) that have odd degree. You can't have the single path unless there are 0 or 2 landmasses of odd degree.

[Guest]

it's impossible, due to Euler's original graph theorem. There are 4 land masses (the outside counts as one) that have odd degree. You can't have the single path unless there are 0 or 2 landmasses of odd degree.

Wow that's irritating. I guess we had to literally think outside the box for this one. Which is fine but this didn't "tease" the brain much less mislead it.

[Guest]

You should count the sides of all five figures within the large figure. Three of them have an odd number of sides, two have an even number of sides. So you should start in a figure with an odd number of sides (you can't get in and out if the number is odd) and you should also finish in a figure of odd number of sides (or at least don't go out of the large figure again). And then you're stuck because you have one odd figure left...

Edited May 28, 2009 by Henoch

[Guest]

it's impossible, due to Euler's original graph theorem. There are 4 land masses (the outside counts as one) that have odd degree. You can't have the single path unless there are 0 or 2 landmasses of odd degree.

CaptainEd suggests that this is impossible due to Euler's graph theorem. Frankly I haven't googled it, but sure it's correct. I wonder why many guys waste their time to solve this. I see that my explanation is not enough to satisfy them. Let me try again:

Imagine a pentagon, it has 5 edges. Due to game's rule, you have to pass each edge once. In two pass, you enter and go out of pentagon and come to start position. In next two, you again in-out and come to start position. But there is left a last pass. After this pass now you are in opposite side of your start position. I mean if you start from outside, after 5 pass you are now inside of the pentagon. Or if you start from inside, now you are in outside of the pentagon.

There are 3 pentagons. You may start from inside of only one pentagon. You can't start from inside of other two pentagons. So if you start from outside of other one pentagon, you finish inside of it, and from now on you can not pass again and go to outside of it. You are imprisoned in that pentagon. This is true for the third pentagon, after 5 pass you have to be inside of it and can't go outside of it again, so imprisoned in that third pentagon. As a result you are imprisoned in two distict pentagons. This is a dilemma, so solution is impossible.

Simply draw a shape containing any odd number of pentagons, with no rectangles, say it 3. You can't solve this, because you will be imprisoned in two of them, but you have only one right to be imprisoned in a pentagon and this will be the end of your line.

[CaptainEd]

CaptainEd suggests that this is impossible due to Euler's graph theorem. Frankly I haven't googled it, but sure it's correct. I wonder why many guys waste their time to solve this. I see that my explanation is not enough to satisfy them. Let me try again:

Imagine a pentagon, it has 5 edges. Due to game's rule, you have to pass each edge once. In two pass, you enter and go out of pentagon and come to start position. In next two, you again in-out and come to start position. But there is left a last pass. After this pass now you are in opposite side of your start position. I mean if you start from outside, after 5 pass you are now inside of the pentagon. Or if you start from inside, now you are in outside of the pentagon.

There are 3 pentagons. You may start from inside of only one pentagon. You can't start from inside of other two pentagons. So if you start from outside of other one pentagon, you finish inside of it, and from now on you can not pass again and go to outside of it. You are imprisoned in that pentagon. This is true for the third pentagon, after 5 pass you have to be inside of it and can't go outside of it again, so imprisoned in that third pentagon. As a result you are imprisoned in two distict pentagons. This is a dilemma, so solution is impossible.

Simply draw a shape containing any odd number of pentagons, with no rectangles, say it 3. You can't solve this, because you will be imprisoned in two of them, but you have only one right to be imprisoned in a pentagon and this will be the end of your line.

Exactly, Euler got a prize for that very same reasoning. Good Job, Nobody! This wasn't a waste, it was an opportunity for a dozen people to see an interesting challenge and for several people to invent a new principle.

[bushindo]

it's impossible, due to Euler's original graph theorem. There are 4 land masses (the outside counts as one) that have odd degree. You can't have the single path unless there are 0 or 2 landmasses of odd degree.

Incidentally, when I was saw the spoiler title, I thought that CaptainEd was claiming that no one had a solution by the time he posted it. Anyways, good job to nobody. It is very impressive that you were able to solve it without prior knowledge of graph theory. If you were born before euler, there would be a theorem called "nobody's graph theorem" by now.

Edited May 28, 2009 by bushindo

[Guest]

Incidentally, when I was saw the spoiler title, I thought that CaptainEd was claiming that no one had a solution by the time he posted it. Anyways, good job to nobody. It is very impressive that you were able to solve it without prior knowledge of graph theory. If you were born before euler, there would be a theorem called "nobody's graph theorem" by now.

Technically, before Euler came along, it was known as nobody's graph theorem.

[Guest]

Note I didn't go through a line I went through a dot.

Edited May 29, 2009 by smata

[Guest]

I will have the solution to this teaser tomorrow...

Tony