Coffeehouse Thread
29 postsThe simplest puzzle to keep whole highschool busy for a week

Yes, that's true. This is probably the most simplest and the most trickiest puzzle and it is guaranteed to keep classroom of highschool students including their teachers busy for at least a day.
My sister (who is currently in high school) brought it to me  it's kinda funny because she said that one of her classmates came up with it and said he saw it on internet and that the puzzle is solvable. As outcome the whole class became obsessed with it, including one teacher (who reportedly did not listen to one pupil's homework report, because she was busy with solving ).
Anyway, do you think you have what it takes to solve this puzzle?
If you do, then here are the rules:
You have 16 lines connected like this.
Your objective is to draw one continuous line (that means you cannot hold up your pencil and start drawing somewhere else), which will cross all lines once and only once. The line you draw doesn't need to be straight, is allowed to cross itself any can begin or end anywhere and you don't need to finnish where you started drawing it.
Sounds pretty simple? Well, it often happens that you think you've solved it, but what really happen is that you forgot to cross one line or you crossed a line twice.
Try it yourself or show it to other people and watch how they'll waste tones of paper and go nuts with it.
Btw, I've solved it, tell me if you need any hints 
Hardly difficult. Did it in 20 seconds.

It is a hamiltonian graph, and there are rules as to its solvability.

Lloyd_Humph wrote:Hardly difficult. Did it in 20 seconds.
I don't believe you can do it so quicky. Read the rules or show what you did 
Lloyd_Humph wrote:Hardly difficult. Did it in 20 seconds.
I concurr.
See:
EDIT: Oh, I made a mistake. Gonna change it now

Rossj wrote:It is a hamiltonian graph, and there are rules as to its solvability.
++; don't spoil with solution
It's great puzzle because a guy with some knowledge of graph theory can do it in 5 min, but it kills highschool pupils (and other mortals)
PS: I've solved it with eulerian path

RoyalSchrubber wrote:I've solved it with eulerian path
How?
Eulerian paths traverse graph edges between nodes, they don't "cross" nodes, nor do they contact edges.
Whilst this sure looks like a graphtheory puzzle, since you can eliminate the four evendegree corner nodes to reduce it to a regular Euleian Circuit for K(8), but this new graph, with 8 nodes of odddegree is unsolvable.
Spills the beans.

W3bbo wrote:
RoyalSchrubber wrote:
I've solved it with eulerian path
How?
Eulerian paths traverse graph edges between nodes, they don't "cross" nodes, nor do they contact edges.
Whilst this sure looks like a graphtheory puzzle, since you can eliminate the four evendegree corner nodes to reduce it to a regular Euleian Circuit for K(8), but this new graph, with 8 nodes of odddegree is unsolvable.
Spills the beans.
I'm also stuck somehow. I tried to build partial graphs and connect them somehow, but I'm always missing one line. I have always three endpoints! 
SecretSoftware wrote:
You must not hold up your pencil when you draw the line 
littleguru wrote:
W3bbo wrote:
RoyalSchrubber wrote:
I've solved it with eulerian path
How?
Eulerian paths traverse graph edges between nodes, they don't "cross" nodes, nor do they contact edges.
Whilst this sure looks like a graphtheory puzzle, since you can eliminate the four evendegree corner nodes to reduce it to a regular Euleian Circuit for K(8), but this new graph, with 8 nodes of odddegree is unsolvable.
Spills the beans.
I'm also stuck somehow. I tried to build partial graphs and connect them somehow, but I'm always missing one line. I have always three endpoints!
Well, you have to create new graph by assigning every face its own point on the new graph, including the outer face.
The edge between two new points exist if you can directly draw line between the two faces that are on the first graph and represent points we are now connecting. The new edges can be multiple if two adjacent faces share more than one common edge that separate them  the mutiplicity is then determined by the number of shared edges of two cells. In other words, we get new graph of possible pencil movements. Now we look if we have eulerian path (we have to be careful with points attached to edges that are multiple)
I'll post tomorrow a picture explaining it (I can't do it on the computer I am using now).
Anyway, I would like some nonmathematician solve it. It's the fun that never ends ..well at least it was for my family members, who tried to beat me by 'conventional' methods. 
edit: crap, too late at night. I'm probably wrong.

AlkalineProdigy wrote:I did in like 5 mins to easy give me something harder.
Want something harder? Try solving it correctly. Notice two lines going through the top right segment.

RoyalSchrubber wrote:
PS: I've solved it with eulerian path
It can't be done. Converted to a graph there are three odd vertices, making it unsolvable. 
The problem I have come across is that when one crosses two of the sides of a shape it will leave 2 or 3 sides left. To cross either 2 or 3 sides, one must enter the shape over one side and exit over the other. This leaves the issue that one must be able to enter the shape to cross certain lines. One could enter the shape and cross the line at the same time, however this leaves one inside a rectangle from which there is not escape as all lines have been crossed. I think this is what makes this puzzle difficult.
Angus Higgins 
Thought I had it but I didnt im guessing this puzzle cant be solved.
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