Young David Posted October 30, 2002 Posted October 30, 2002 Okay ... s been a while since I posted a math problem. Here's one again. Your mission, should you choose to accept it, is to draw this figure in one single line. You are not allowed to take your pen off the paper. Go ahead
Rogue15 Posted October 30, 2002 Posted October 30, 2002 I did that before...well, sorta. pretty tricky.
Dagobahn Eagle Posted October 30, 2002 Posted October 30, 2002 Not again! Spare meeeeeeeee!!! Okay, here goes..
Rogue15 Posted October 30, 2002 Posted October 30, 2002 damn, this one's hard. i'm only missing by part of the square. :/
Rogue15 Posted October 30, 2002 Posted October 30, 2002 AAAAAAAAAAAAAHHHHHHH!!!!!!!!!!!!!!!!!!!!!! I ALMOST GOT A HEADACHE!!! OW OWO OWO OW OWWWWWWWWWW u are mean. this is impossible. but I'm a rogue, and specialize in impossibility. so i'll get this done by the end of the day. edit, i'm missing a line, always in the center or side.
Zygomaticus Posted October 30, 2002 Posted October 30, 2002 is this really a math problem? or is a thinking problem...if it's the latter, i wont participate...my brain heats up on these kinds of things, and i've got enough of then already... besides i come to lucasforums for fun, not to heat up my brain :D
Rogue15 Posted October 30, 2002 Posted October 30, 2002 ok forget it, i've wasted a whole sheet of paper (both sides i might add), and I can't seem to get the last line. I have a headache now. damn math, it's a miracle that i passed it.
Zygomaticus Posted October 30, 2002 Posted October 30, 2002 I have a headache now. damn math, it's a miracle that i passed it. all math isn't this hard...
Rogue15 Posted October 30, 2002 Posted October 30, 2002 Well....I looked at what my friend who is in liek 10th grade is doing for math, it's algebra, but i don't understand ANY of it. so, if you hate math, but want to pass...take a GED exam, cause there's not alot of algebra on that. when people see this piece of paper i wonder if they'll think i'm a psycho.
Darth Homer Posted October 30, 2002 Posted October 30, 2002 algebra is child's play...now take a look at Calculus and there's a true math challenge (I should know...had Calc 1 twice in college & calc 2 once bf giving up on it)
Young David Posted October 30, 2002 Author Posted October 30, 2002 Originally posted by Kylilin I did it, yay for me! Prove it
Zygomaticus Posted October 30, 2002 Posted October 30, 2002 i'll be starting calc this year, and deeper next year oh boy
Zygomaticus Posted October 30, 2002 Posted October 30, 2002 Your mission, should you choose to accept it, is to draw this figure in one single line. You are not allowed to take your pen off the paper. following just those instructions, i did it...it's pretty easy...i think one of the rules you need to add is "You may not go over a line already drawn" that's what most similar problems ask...
Young David Posted October 30, 2002 Author Posted October 30, 2002 Originally posted by krkode following just those instructions, i did it...it's pretty easy...i think one of the rules you need to add is "You may not go over a line already drawn" that's what most similar problems ask... Yeah, that rule applies
Breton Posted October 30, 2002 Posted October 30, 2002 Just one simple question Young David: Is it possible to do it at all? It certainly does not seem so.
Crazy_dog no.3 Posted October 30, 2002 Posted October 30, 2002 Heck I don't even understand the question!
Artoo Posted October 30, 2002 Posted October 30, 2002 You should know that the answer will wind up being 42 in some way. Stupid ingrate. I bet you don't even have a towel with you right now.
Zygomaticus Posted October 31, 2002 Posted October 31, 2002 I did it!! Eureka! It's 43-12=31!! That's it!! 31! I'm a geneous!!!!!!!! oops...wrong thread... And Qui-Gon, although i haven't solved it myself, i'm sure it's possible to do...it's just some weird method that some of us don't thikn of right away...and then again, since it's symetrical, if it's possible, there should be more than one way to do it...
Tie Guy Posted October 31, 2002 Posted October 31, 2002 Alright, i literally tried it for about an hour and it never worked. I found numerous ways to get it to only one side left but no matter what i did there was always that one on the far side of where i ended up. Very frustrating.
Jatt13 Posted October 31, 2002 Posted October 31, 2002 You should know that the answer will wind up being 42 in some way. Stupid ingrate. wrong question. i could probably do it, but i don't want to take the time.
Rogue15 Posted October 31, 2002 Posted October 31, 2002 Originally posted by Tie Guy Alright, i literally tried it for about an hour and it never worked. I found numerous ways to get it to only one side left but no matter what i did there was always that one on the far side of where i ended up. Very frustrating. same here.
WolfmanNCSU Posted October 31, 2002 Posted October 31, 2002 Sorry to blow this one, but it is not possible. Here is why... There are conditions for determining whether a graph contains an Eulerian cycle, or path (path or cycle through a graph that visits each edge exactly once): An undirected graph contains an Eulerian cycle iff (1) it is connected and (2) each vertex is of even degree. An undirected graph contains an Eulerian path iff (1) it is connected and (2) all but two vertices are of even degree. These two vertices should be the starting and ending points. A directed graph contains an Eulerian cycle iff (1) it is connected and (2) each vertex has the same in-degree as out-degree. A directed graph contains an Eulerian path iff (1) it is connected and (2) all but two vertices have the same in-degree as out-degree, and these two vertices have their in-degree and out-degree differ by only one. Anyway, that is enough math for me for one day. This is just a discrete math model. Pretty cool stuff and it works alot with computers and stuff. I studied this kinda thing for 2 semesters.
Zygomaticus Posted October 31, 2002 Posted October 31, 2002 Originally posted by WolfmanNCSU Sorry to blow this one, but it is not possible. Here is why... There are conditions for determining whether a graph contains an Eulerian cycle, or path (path or cycle through a graph that visits each edge exactly once): An undirected graph contains an Eulerian cycle iff (1) it is connected and (2) each vertex is of even degree. An undirected graph contains an Eulerian path iff (1) it is connected and (2) all but two vertices are of even degree. These two vertices should be the starting and ending points. A directed graph contains an Eulerian cycle iff (1) it is connected and (2) each vertex has the same in-degree as out-degree. A directed graph contains an Eulerian path iff (1) it is connected and (2) all but two vertices have the same in-degree as out-degree, and these two vertices have their in-degree and out-degree differ by only one. Anyway, that is enough math for me for one day. This is just a discrete math model. Pretty cool stuff and it works alot with computers and stuff. I studied this kinda thing for 2 semesters. hmm...that made no sense...but i trust in your knowledge, and i repeal my previous statement of this problem's possibliity...maybe one day i'll know this myself
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