My teacher gave this problem to my class, for any one who was failing, only one person was able to solve it correctly. It has been driving me crazy for almost two years. Alot of peolpe have said that it is impossible, but I assure you it is not. It was solved by a 17yr old student. here is the best I can explain it, with out drawing it actually. Imagine a rectangle that has a split line horizontaly through the middle. On the top half of the rectangle is a line that is verticle, in the center, connecting the first and second horizontal lines. Now on the bottom half of the rectangle, there are two verticle lines, connecting the second and third horizontal lines, one vertical line on either side of the top verticle. So in all, it should appear to be a large rectangle with two small rectanles on top of three smaller rectangles. Simply five in one. Now you must draw a line through each line, or sides, of all reactangles. There are 16 sides in all. The line cannot break, fork, or overlap. The line cannot pass through the same side twice. Please email me if this confusses you and i can email you the picture back. Thank you for your help!!
Um, as you describe it, it is mathematically impossible. Now if when you say you must draw a line through each side, you really mean draw acontinuous curve then it is possible. You really need to be more clear on definitions with these kinds of things, ambiguities like that will make it impossible. Now i'm off to solve it on the assumption you meant draw a continuous curve.
im sorry but I do not understand the meaning of an acontinuous curve, or how an acontinuous curve would make it possible if a line makes it impossible.
It's not possible. The top center rectangle, and the bottom rectangles all have an odd number of sides (5). Now, if a line goes into a rectangle, then it must come out, so unless the path starts or ends in a rectangle, only an even number of sides can be crossed by the line. Now, since the path only has two ends, it cannot start or end in all three of those rectangles. That means that you cannot make a squiggle go through all of the segments. ...unless you think outside of the box. If you punch holes in the paper inside some of the squares and then put the squiggle through, for example, you can do it.
I was given the same assignment in 6th grade. It has to pass through all the barriers, and it can not overlap. I've never been able to solve it.
twas a typo. I meant a continuous curve. A line is by definition linear, its direction cannot change. Just stop posting till it actually works. Everyone of those has missed a side. Not trying to be nasty, its just a lot of links to click. Been trying for a while and haven't been able to solve it yet. I'll prolly have better luck proving whether or not its possible. NateTG may be right in principle, but his post lacks in rigor...off to try and construct rigorous proff in either direction.
Guys: This is exactly how the puzzle should look, and the # of sides are numbered, 16 total. I am pretty sure this is unsolvable.
For a Euler path to exist (each edge traversed exactly once) two vertices need to have odd degree and the rest must have even degree. The graph has vertices with degrees, 5,5,5,9,4,4 So it looks like a no-go. The last drawing i posted had 5,4,4,4,4,9. So that at least checks out.
Assuming NateG got the right picture, and I've no reason to doubt that, then his proof is perfectly rigorous (and assuming one can't pass through corners claiming to intersect both sides simultaneously).
WEll i've been at it fairly consistently, and no luck, there is always one side i cannot get. I think arachymo and palpatine are right here, it is unsolvable. matt: palaptine's post was more what i was looking for when i said rigor, logically its right, but without the basis on things already proven to be true i was concerned that there might be a small loophole or something. but yes it seems he was right.
No offense to Palpatine, but I should point out that my argument was more rigorous than Palpatine's. (For example, he doesn't show how to construct the graph he's making Euler paths on.) I elected not to bring graph theory or the results of graph theory into it because, among other things, I expect that many of the people here have not been exposed to it. The theorem that There are, by the bye, many different 'out of the box' solutions. On a blackboard you can wipe out one of the lines after crossing it (use your left hand if you can't put the chalk down), and the gap will allow you finish the rest of the lines.