[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]

god your gonna get some ocd kid trying this forever

[Guest]

There are 3 pentagons. At each step (crossing an edge) you eather enter in a pentagon or go out of a pentagon. Since a pentagon has 5 edges if you start drawing in pentagon A, after 5 steps (an odd number) you'll be outside. But if you start drawing outside of pentagon A, after 5 steps, you'll be in pentagon A. There are 3 pentagons, so if you start in one of them, at the end you'll have to be in both other 2 pentagons. This is impossible, you can!t be in two other pentagons at the same time. So this is impossible. If one of the pentagons was square, it would be possible.

[Guest]

I agree - using a bir of graph theory and converting each area into a "node" you get 6 nodes (the uoter area is also a node). To cross each pathway once gives you 2 nodes with 4 paths, 3 nodes with 5 paths and 1 (the outer node) with 9 paths. I'm fairly sure that according to Euler's theorems you can't have more than two "odd" nodes to complete this journey

There are 3 pentagons. At each step (crossing an edge) you eather enter in a pentagon or go out of a pentagon. Since a pentagon has 5 edges if you start drawing in pentagon A, after 5 steps (an odd number) you'll be outside. But if you start drawing outside of pentagon A, after 5 steps, you'll be in pentagon A. There are 3 pentagons, so if you start in one of them, at the end you'll have to be in both other 2 pentagons. This is impossible, you can!t be in two other pentagons at the same time. So this is impossible. If one of the pentagons was square, it would be possible.

[Guest]

Oops! Thanks Mel....

Edited May 27, 2009 by SteveK

[Guest]

yOUR ATTEMPT DOESN'T GO THROUGH THE BOTTOM OF THE MIDDLES PENATAGON!!!!!!!!

[Guest]

hmm, I thought I had it but its wrong

Edited May 27, 2009 by Dempsey Collins

[Guest]

edit: nope! got it wrong . lol

Edited May 27, 2009 by Napsalm

[Guest]

There are tons of solutions...it only says the line has to pass through the connectors and must be continuous. It says nothing about passing through the end points...

[Guest]

There are tons of solutions...it only says the line has to pass through the connectors and must be continuous. It says nothing about passing through the end points...

This is a good idea, but if you look at OP's sample drawing you see that the line is far thicker then the end points and they are named as "dots". If OP named them "buildings" and not drew the line so thick this would make sense. But maybe OP made these errors to trick us, since there is no restriction about the thickness of line.

I still suggest that -other than this solution- in regular way, it is impossible.

[Guest]

It is simple... nm, I realized I have crossed a line twice.

Edited May 27, 2009 by Nester22

[Guest]

It is simple... nm, I realized I have crossed a line twice.

Oooops. You crossed the upper left hand connector twice.... I don't think it is possible considering Euler's theorem....

[Guest]

Edit: Ah, damnit, noticed I skipped a line....

Edited May 27, 2009 by charlatantric

[Guest]

I may be wrong but it looks to me like its one of those that looks possible but really isnt.

[Guest]

The way that I first saw this they were all squares and rectangles three equal squares on bottom two rectangles on top both equal to each other I personally have never solved this or even seen it solved. It still has the same amount of crossing lines 16. Good Luck

[plainglazed]

Didnt mention, that I could tell, whether or not the line can cross itself. If it can, am thinking this works?

[plainglazed]

yeah I see it now, crossed the bottom line twice, oops

[Guest]

I don't think there is any reason this solution doesn't work.

[Pickett]

I don't think there is any reason this solution doesn't work.

You didn't cross the center middle line. I believe, as it has been stated multiple times, that this is not possible without maybe some trickery or something that is hidden in the OP...

[Guest]

I don't think there is any reason this solution doesn't work.

It would but you missed a line

[Guest]

Wow, you're right. I guess that's what I get for trying to sneak a post in when I should be working....easy to overlook stupid mistakes.

[bushindo]

god your gonna get some ocd kid trying this forever

Hilarious. I hope that wasn't you.

[Guest]

Superman Saves the Day Once Again

[bushindo]

Superman Saves the Day Once Again

you missed the top vertical line. Looks like this is your kryptonite

[Guest]

I think i got it!!

sorry guys i'm goin overthe top line twice!! i did not get it

Edited May 27, 2009 by mumBhai

[Guest]

Bushindo, In the original description of the problem you do not stipulate that the line has to stay on the XY plain, only that the continuous line has to start on the image and that it can finish at any point ( <= doesn’t stipulate on the image).

Is drawing the line using the Z direction permitted? If so, the problem is easy. Simply draw a line that punches in or out of the 2 dimensional image at each connector line. Think of it like using a needle and thread with the thread representing the line drawn in all three dimensions, and the needle moving through the image at each connector.

I would include an example image that fits this method but its rather hard to draw an example.