Can you solve this puzzle?

[alexjohnson16]

Puzzle

You must go through each line segment that is numbered. You cannot cross back through a segment once you have gone through it. I've heard two versions: you can and can not cross over your line, but I can't solve it either way.

Can anyone do this? If not, can you explain why it is impossible?

Edit: I'd better clarify some rules - you can start inside or outside any of the squares, and you do not have to go in any specific order. The numbers are there just to show each line you must go through.

 

[TitanDiddly]

Done.

 

[alexjohnson16]

Cancel this one, accidentally hit reply when I meant to edit.

 

[StrangerGuy]

I don't understand the question.

 

[Asthmaboy]

I'm pretty sure I got it.

 

[Mo0o]

http://mathforum.org/isaac/problems/bridges1.html

 

[alexjohnson16]

Originally posted by: Mo0o
http://mathforum.org/isaac/problems/bridges1.html

I'm not catching how this relates???

Anyways, to those who think they have solved it, post it on a webhost or email me at johnson.alexREMOVE@gmail.com with the attached image and I'll see if it works.

 

[alexjohnson16]

On a roll with stupid double posts.

 

[Mo0o]

Originally posted by: alexjohnson16

Originally posted by: Mo0o
http://mathforum.org/isaac/problems/bridges1.html

I'm not catching how this relates???

Anyways, to those who think they have solved it, post it on a webhost or email me at johnson.alexREMOVE@gmail.com with the attached image and I'll see if it works.

Your problem is basically a rewording of what I've just linked. Read the link carefully. Pretend your lines are rivers

 

[Asthmaboy]

Scratch that.
I can't find any possible solutions.
Every single time I use up 4 for every box except one, where 3 is left, one line is left because you can't cross twice.

 

[DrPizza]

It's impossible.
Explanation is simple. Look at the points where segments meet.
Every time you enter a point, you have to leave the point, thus every point has to have an even number of segments that connect to it.
That is, with the exception of the starting and ending point.
Thus, at the very most, 2 points can have an odd number of segments attached to it. You have 8 such points. So, it is quite clearly impossible.

 

[DrPizza]

(I'm sure that what I just said is buried somewhere in those pages that Mo0o linked to)

 

[moshquerade]

absolutely unimpossible

 

[Midlander]

Originally posted by: DrPizza
It's impossible.
Explanation is simple. Look at the points where segments meet.
Every time you enter a point, you have to leave the point, thus every point has to have an even number of segments that connect to it.
That is, with the exception of the starting and ending point.
Thus, at the very most, 2 points can have an odd number of segments attached to it. You have 8 such points. So, it is quite clearly impossible.