Brain Teaser that is impossibly solvable

[Rashad]

Palpatine, you are showing that this is a traceable network, I don't think that is what we're going for here. Is the problem to show that you can 'go over' (trace) the network itself and cover every line without overlapping, like what Palpatine is doing (...I think)? Ie draw the rectangle and its accompanying segments without lifting up your writing utensil or going over the same segment twice. Or is it to prove whether or not a continuous curve (a rather insane curve) can pass through all of the small segments only once? That's what I'm trying to do...and if you say that it is impossible, that claim is insatiable without a proof.


~Rashad

 

[NateTG]

Palpatine, you are showing that this is a traceable network, I don't think that is what we're going for here. Is the problem to show that you can 'go over' (trace) the network itself and cover every line without overlapping, like what Palpatine is doing (...I think)? Ie draw the rectangle and its accompanying segments without lifting up your writing utensil or going over the same segment twice. Or is it to prove whether or not a continuous curve (a rather insane curve) can pass through all of the small segments only once? That's what I'm trying to do...and if you say that it is impossible, that claim is insatiable without a proof.

That's also impossible. There are more than two intersections where an odd number of segments meet. A similar argument as before applies; consider that, unless the path starts or ends at a particular intersection, it must go into the intersection as many times as it comes out, so the path must start or end at every intersection that connects an odd number of segments.

 

[Gokul43201]

where's Icarus, still flying ?

Didn't Icarus promise us a solution ?

 

[dave8684]

5 room puzzle solution

Quite easy if you do think outside the box:

there is a logic problem that goes like this: draw a square, divide in half horizontally, divide the top half into two equal parts with a vertical line, then divide the bottom portion into 3 equal portions with 2 vertical lines.

the task is to draw a continues line through all lines without ever crossing your own line or crossing any line two times. The problem is presented on this sight. Now according to conventional logic this problem seems impossible because line always needs an entry and exit but there are an odd number of spaces and an odd number of segments in three of them.

The real difficulty here is that an assumption is made, creating an unwritten rule. This unwritten rule, this self imposed limitation forces the problem solver to focus on the problem, NOT THE SOLUTION. By recognizing the problem (NOT FOCUSING ON IT) - a long line cannot enter and leave each space enough times without making an illegal crossing- we can find the solution.

The solution is this: use a very wide marker and cross the entire box in one diagonal line. All stated conditions are met, the problem is circumvented and the solution is found. Clearly this is not the intended answer, but it is indisputable.

 

[check]

And Dave8684's answer is verbatim from http://www.puzzles.com/PuzzleHelp/PuzzleHelpItems25_36.htm

 

[Matter]

would a spiral work!?

 

[koroljov]

Found!

I found it. First try worked. (See attachment).

edit:
You're right, I missed one.

 

[Zorodius]

I found it. First try worked. (See attachment).

Missed a spot, there. You did not cross the segment directly to the left of the center of the figure.

You will always "miss a spot", because there exist no solutions to this problem, aside from the "thinking outside the box" answers like poking holes in the paper and so on.

 

[Galileo]

There are five rooms. The wall between two given rooms must be crossed by the line.
So you have to get from one room to the next.
Give every room a vertex and connect the vertices when there is a wall between them.
This will give you the following graph (see graph.bmp).

Then the upper two rooms (vertices) must have two more lines going out, and so do the two rooms and the bottom left and bottom right.
The room in the bottom center has one more line going out.
The graph will look graph1.bmp.

Since the line must be drawn without removing your pencil from the paper, the entire graph must be connected.
However you connect the loose ends, there will be 3 odd vertices, so there is no solution to the problem.

 

[CrankFan]

I've heard of a similar problem to this one, which had no solution and was supposedly given to grade school students for extra credit. As if that isn't enough of a coincidence just like this version of the story it was rumored to have been solved by one student, so you "know" a solution has to exist!

I suspect that stories of this kind are intentional hoaxes which persist as a kind of elementary math folklore.

 

[Ethereal]

I've heard of a similar problem to this one, which had no solution and was supposedly given to grade school students for extra credit. As if that isn't enough of a coincidence just like this version of the story it was rumored to have been solved by one student, so you "know" a solution has to exist!

I suspect that stories of this kind are intentional hoaxes which persist as a kind of elementary math folklore.

That's a good one! I bet the same thing happened with all those famous conjectures which were supposedly proven by mathematicians in the past, but whose proofs were lost. Fermat comes to mind...

 

[Matter]

Is this it?

How do I post an attachment!? This is driving me maaaaad Never mind I got it!

 

[Galileo]

Dude, you missed a spot.

It can't be done!

 

[Gokul43201]

It is not possible to do this : this is an old problem known as the Bridges of Konigsberg, and was solved by Euler, nearly 300 years ago.

PROOF : There are 3 rooms with an odd number of walls. If you start inside a room with an odd number of walls, and walk through each wall once (you're a ghost), you will end up outside the room. Conversely, if you started outside, then after walking through the last wall, you will end up inside - and can't leave because you've walked through all the walls. This means that one end of your journey must always be inside an odd walled room - either the beginning or the end. Unfortunately, a journey (or a line or curve or whatever you choose to call it) has only 2 ends. But the house has 3 odd-walled rooms, and you can't have a journey with 3 ends, so it's not possible.

THE END

...unless you cheat, by traveling through vertices, (or drawing the figure on a torus), or using a really thick marker, or some other such childish ploy.

 

[Matter]

where?

Dude, you missed a spot.

It can't be done!

There are 5 rectangels with 4 walls each! 6 of these walls are shared or common walls . I went through each wall of each rectangel once ! so where did I miss one!

 

[Matter]

There are 5 rectangels with 4 walls each! 6 of these walls are shared or common walls . I went through each wall of each rectangel once ! so where did I miss one!

don't answer that I see it! Dang! I hate not having a solution!#"!

 

[Galileo]

bladibladibladibla

EDIT: Whoops. I`ve been screwing with that picture so long you posted your message in the meantime.

 

[discomb]

aaaaahhhhhhhhh.

after reading all that, I feel soooooo much better.

before, i was like this:



and now, i am like this:



cheers everybody, and particularly to Mr. Euler and his chums

 

[Myclicheisbetter]

If you approach the puzzle in three dimensions you can easily do it, in fact it only takes about 5 seconds, and the mind set of a jackass. I can attach it if you want, but I'm sure you understand what I'm getting at.

 

[Gokul43201]

Yes, the problem is insoluble only on the plane or the homotopy class of a sphere.

It has a solution, for instance, on the torus.

 

[kerryfan]

Is this the solution?

I was given this problem as a brain teaser and am wondering if I've found the solution (see attachment) or if there is a solution besides cheating, use a marker, etc. This has been driving me nuts, please help if you know the answer.
Thanks

 

[cogito²]

I was given this problem as a brain teaser and am wondering if I've found the solution (see attachment) or if there is a solution besides cheating, use a marker, etc. This has been driving me nuts, please help if you know the answer.
Thanks

If going through a corner counts as going through all three then yes this is a solution. If that is not allowed then there is no solution as has been shown like 10 times throughout this thread. It depends on what conditions you put on the problem. So you decide: is that a solution?

 

[BTruesdell07]

make it really small then draw through it with a thick marker covering the entire puzzle. You go through every line only once.

 

[Sisco424]

Hi,
I've been lookin for the answer to this on the internet because one of my teachers was offering a small cash reward to whoever could solve it and i now know that it is possible because my friend printed a picture of it completed off the net but i just got a quick glance and he took it away thinkin that we was going to steal his answer, i just want to figure it out cause its driving me crazy so if someone finds the answer tell me. It is definately possible

 

[BigStelly]

This has been seen to be impossible for several centuries ever since Euhler proved it was. This comes from the bridges of Konisberg and for centuries nobody figured out a way to cross all the bridges once and only once without crossing back over their path, Because its impossible.

 

[Data]

Sisco424, you will impress your teacher more if you take a book on Graph Theory out of the library, learn about the question, and prove that it is impossible in front of your class :)

If you want a bunch of "completed solutions" then just look through this thread: All of them either cheat or miss a side.

There is a very good reason for this.

 

[animejunkie1100]

I have worked on it for like a year almost and still can't find the answer please help!

 

[d_leet]

I have worked on it for like a year almost and still can't find the answer please help!

This is because, as many people have mentioned in this thread, that it is impossible and has been proven impossible.

 

[-Job-]

You can convert the problem to the attached graph.
The problem can be solved only if you can find a path that goes through every edge exactly once. Such a path is called a Euler path, or Euler tour. Euler showed that a graph can only contain such a path if either all or all but two nodes have an even degree. The graph corresponding to this problem has 4 nodes of odd degree and two of even degree, so it can't contain a Euler tour.

 

[ValenceE]

Hello to all !,

Guys, if we accept the fact that any corner is the junction of as many sides as are connected, then there’s a way to solve this puzzle. Actually there’s probably more than one way, but here’s what one of them looks like.

It also has a neat shape to it…


VE

edited to replace .jpg file by .bmp