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Impossible Problem?

Blogs > KurtistheTurtle
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1 2 3 Next All
KurtistheTurtle
Profile Blog Joined December 2008
United States1966 Posts
October 15 2009 06:16 GMT
#1
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:

[image loading]


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
[image loading]


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


“Reject your sense of injury and the injury itself disappears."
KurtistheTurtle
Profile Blog Joined December 2008
United States1966 Posts
October 15 2009 06:19 GMT
#2
Also rule to #1: the line cannot cross over itself
“Reject your sense of injury and the injury itself disappears."
SnowFantasy
Profile Blog Joined September 2006
4173 Posts
October 15 2009 06:24 GMT
#3
pretty sure #1 is impossible
Caller
Profile Blog Joined September 2007
Poland8075 Posts
October 15 2009 06:26 GMT
#4
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)

qed

http://en.wikipedia.org/wiki/Seven_Bridges_of_Königsberg
Watch me fail at Paradox: http://www.teamliquid.net/forum/viewmessage.php?topic_id=397564
KurtistheTurtle
Profile Blog Joined December 2008
United States1966 Posts
October 15 2009 06:28 GMT
#5
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
“Reject your sense of injury and the injury itself disappears."
Ryan307 :)
Profile Blog Joined January 2004
United States1289 Posts
October 15 2009 06:30 GMT
#6
[image loading]


I think I got it~
Dont let the action of factual things fracture your casual swing
motbob
Profile Blog Joined July 2008
United States12546 Posts
October 15 2009 06:30 GMT
#7
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?
ModeratorGood content always wins.
KurtistheTurtle
Profile Blog Joined December 2008
United States1966 Posts
October 15 2009 06:32 GMT
#8
On October 15 2009 15:30 Ryan307 wrote:
[image loading]


I think I got it~

you missed one, middle line second horizontal segment from the left
“Reject your sense of injury and the injury itself disappears."
Ryan307 :)
Profile Blog Joined January 2004
United States1289 Posts
October 15 2009 06:33 GMT
#9
oh fuck you're right lol.

then I give up.
Dont let the action of factual things fracture your casual swing
Divinek
Profile Blog Joined November 2006
Canada4045 Posts
Last Edited: 2009-10-15 07:16:28
October 15 2009 07:14 GMT
#10
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.
Never attribute to malice that which can be adequately explained by stupidity.
Oh goodness me, FOX tv where do you get your sight? Can't you keep track, the puck is black. That's why the ice is white.
d3_crescentia
Profile Blog Joined May 2009
United States4054 Posts
October 15 2009 07:17 GMT
#11
Yeah read the wiki article - it pretty much explains it there.
once, not long ago, there was a moon here
Lemonwalrus
Profile Blog Joined August 2006
United States5465 Posts
Last Edited: 2009-10-15 07:19:20
October 15 2009 07:18 GMT
#12
@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.
EtherealDeath
Profile Blog Joined July 2007
United States8366 Posts
October 15 2009 07:19 GMT
#13
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.
kOre
Profile Blog Joined April 2009
Canada3642 Posts
October 15 2009 07:19 GMT
#14
So simple and so close lol

[image loading]
http://www.starcraftmecca.net - Founder
EtherealDeath
Profile Blog Joined July 2007
United States8366 Posts
October 15 2009 07:23 GMT
#15
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.
Divinek
Profile Blog Joined November 2006
Canada4045 Posts
Last Edited: 2009-10-15 07:28:07
October 15 2009 07:24 GMT
#16
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.
Never attribute to malice that which can be adequately explained by stupidity.
Oh goodness me, FOX tv where do you get your sight? Can't you keep track, the puck is black. That's why the ice is white.
Lemonwalrus
Profile Blog Joined August 2006
United States5465 Posts
October 15 2009 07:26 GMT
#17
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.
Divinek
Profile Blog Joined November 2006
Canada4045 Posts
October 15 2009 07:27 GMT
#18
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.
Never attribute to malice that which can be adequately explained by stupidity.
Oh goodness me, FOX tv where do you get your sight? Can't you keep track, the puck is black. That's why the ice is white.
EtherealDeath
Profile Blog Joined July 2007
United States8366 Posts
October 15 2009 07:27 GMT
#19
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.
Ota Solgryn
Profile Blog Joined January 2008
Denmark2011 Posts
October 15 2009 07:28 GMT
#20
Could you maybe go inside the walls. This way you wont cross the wall but can use it for transport.
ihasaKAROT: "Wish people would stop wasting their lives on finding flaws in others"
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