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Two problems for me to solve over the weekend:
1. Here's a problem I've been trying to solve. I haven't figured it out yet. You start with this:
The goal is to go through each line segment once. You can't go through the same one twice, and you may start wherever you like. One continuous line.Here's an example with the wrong part circled
2. I'm back in my hometown on break from college. My friends attending community college are bored as shit. None of us are 21, and I wanna do something with them to shake things up. Nothing illegal unless its hard to get caught. Open to ideas
my thoughts so far are ranging from: triple-date w/ random girls golfing off of various roofs prank war with somebody
but its cold as shit and rainy outside, so outside ideas are gonna have to be cancelled. We don't know enough people in town to have a good party, and we don't know anybody over 21 who we could reasonably ask to get us anything. After my taser incident I'm not sure I wanna drink either
open to ideas. anything and everything will be considered except obvious trolling
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Also rule to #1: the line cannot cross over itself
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pretty sure #1 is impossible
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On October 15 2009 15:26 Caller wrote: pretty easy
start from the center and draw lines squiggly through all the middle segments then draw parabolic curves touching each segment at one sole point (i.e. tangent)
i thought of this too. my sister is going crazy trying to solve this and she said that doesnt count, it actually has to go through
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motbob
United States12546 Posts
On October 15 2009 15:28 KurtistheTurtle wrote:Show nested quote +On October 15 2009 15:26 Caller wrote: pretty easy
start from the center and draw lines squiggly through all the middle segments then draw parabolic curves touching each segment at one sole point (i.e. tangent)
i thought of this too. my sister is going crazy trying to solve this and she said that doesnt count, it actually has to go through Why would you go crazy trying to solve something which is probably impossible?
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On October 15 2009 15:30 Ryan307 wrote:I think I got it~ you missed one, middle line second horizontal segment from the left
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oh fuck you're right lol.
then I give up.
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Yeah that's a graph theory problem. Yeah that's not doable.
Just look at this basic example
Take the top left square, there are 5 edges that you must cross. The only way to do this without crossing an edge twice is to start from INSIDE the square. So that's doable, but now you must start from outside the upper right square, and cross all the edges without crossing one twice. You can't do it.
Also i feel bad for your sister because this only takes a minute tops of reasoning it out instead of trying random paths.
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Yeah read the wiki article - it pretty much explains it there.
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@Divinek: If it is a well known problem for which a new type of problem solving was developed to solve it...then it isn't just 'a minute tops of reasoning it out' that most people take to solve it.
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It doesn't seem possible: unless I miscounted, there are 4 edges of odd degree, which means there is no Eulerian walk for this problem.
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So simple and so close lol
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On October 15 2009 16:14 Divinek wrote: Yeah that's a graph theory problem. Yeah that's not doable.
Just look at this basic example
Take the top left square, there are 5 edges that you must cross. The only way to do this without crossing an edge twice is to start from INSIDE the square. So that's doable, but now you must start from outside the upper right square, and cross all the edges without crossing one twice. You can't do it.
Also i feel bad for your sister because this only takes a minute tops of reasoning it out instead of trying random paths.
That is not true. Leave the top left square, but change the top right square so that it has only 2 interior edges, while not changing the rest of the squares (somehow), and the problem is now solveable. It's not that the top left square has 5 edges that is the problem, because then you have exactly two edges of odd degree (note that there are 11 possible edge destinations from each corner edge that faces the outside white space). Then, since the top right also has two edges of odd degree, you end up with 4 total edges of odd degree, which at THAT point makes the problem impossible.
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On October 15 2009 16:18 Lemonwalrus wrote: @Divinek: If it is a well known problem for which a new type of problem solving was developed to solve it...then it isn't just 'a minute tops of reasoning it out' that most people take to solve it.
seems pretty obvious that there cant be a solution just from what i said. You cant even get past that part so there's no part even fiddling with the rest.
On October 15 2009 16:23 EtherealDeath wrote:Show nested quote +On October 15 2009 16:14 Divinek wrote: Yeah that's a graph theory problem. Yeah that's not doable.
Just look at this basic example
Take the top left square, there are 5 edges that you must cross. The only way to do this without crossing an edge twice is to start from INSIDE the square. So that's doable, but now you must start from outside the upper right square, and cross all the edges without crossing one twice. You can't do it.
Also i feel bad for your sister because this only takes a minute tops of reasoning it out instead of trying random paths. That is not true. Leave the top left square, but change the top right square so that it has only 2 interior edges, while not changing the rest of the squares (somehow), and the problem is now solveable. It's not that the top left square has 5 edges that is the problem, because then you have exactly two edges of odd degree (note that there are 11 possible edge destinations from each corner edge that faces the outside white space). Then, since the top right also has two edges of odd degree, you end up with 4 total edges of odd degree, which at THAT point makes the problem impossible.
That's what i said v_v. If you use them in combination like that. Though the way that paragraph started it did seem i was talking about only the one thing isolated.
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On October 15 2009 16:24 Divinek wrote:Show nested quote +On October 15 2009 16:18 Lemonwalrus wrote: @Divinek: If it is a well known problem for which a new type of problem solving was developed to solve it...then it isn't just 'a minute tops of reasoning it out' that most people take to solve it. seems pretty obvious that there cant be a solution just from what i said. You cant even get past that part so there's no part even fiddling with the rest. I'm just saying implying someone is stupid for not immediately realizing the solution to a problem that is so troublesome it lead to the development of a new type of problem solving is kinda lame.
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On October 15 2009 16:26 Lemonwalrus wrote:Show nested quote +On October 15 2009 16:24 Divinek wrote:On October 15 2009 16:18 Lemonwalrus wrote: @Divinek: If it is a well known problem for which a new type of problem solving was developed to solve it...then it isn't just 'a minute tops of reasoning it out' that most people take to solve it. seems pretty obvious that there cant be a solution just from what i said. You cant even get past that part so there's no part even fiddling with the rest. I'm just saying implying someone is stupid for not immediately realizing the solution to a problem that is so troublesome it lead to the development of a new type of problem solving is kinda lame.
I didn't imply she's stupid, i just felt bad that the problem was driving her nuts. I mean the way she tried it is by far way way funner. I'm sure she could reason it out the same way if she didn't try a brute force method.
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On October 15 2009 16:24 Divinek wrote:Show nested quote +On October 15 2009 16:18 Lemonwalrus wrote: @Divinek: If it is a well known problem for which a new type of problem solving was developed to solve it...then it isn't just 'a minute tops of reasoning it out' that most people take to solve it. seems pretty obvious that there cant be a solution just from what i said. You cant even get past that part so there's no part even fiddling with the rest. Show nested quote +On October 15 2009 16:23 EtherealDeath wrote:On October 15 2009 16:14 Divinek wrote: Yeah that's a graph theory problem. Yeah that's not doable.
Just look at this basic example
Take the top left square, there are 5 edges that you must cross. The only way to do this without crossing an edge twice is to start from INSIDE the square. So that's doable, but now you must start from outside the upper right square, and cross all the edges without crossing one twice. You can't do it.
Also i feel bad for your sister because this only takes a minute tops of reasoning it out instead of trying random paths. That is not true. Leave the top left square, but change the top right square so that it has only 2 interior edges, while not changing the rest of the squares (somehow), and the problem is now solveable. It's not that the top left square has 5 edges that is the problem, because then you have exactly two edges of odd degree (note that there are 11 possible edge destinations from each corner edge that faces the outside white space). Then, since the top right also has two edges of odd degree, you end up with 4 total edges of odd degree, which at THAT point makes the problem impossible. That's what i said v_v. If you use them in combination like that. Though the way that paragraph started it did mean i was talking about only the one thing isolated.
Oh haha misread, interpreted what you typed the wrong way for some reason. 3:27 am ftl.
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Could you maybe go inside the walls. This way you wont cross the wall but can use it for transport.
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