Jump to content

Home

Math problem (show how smart you are)


Young David

Recommended Posts

Posted

Okay ... s been a while since I posted a math problem. Here's one again.

 

huh.JPG

 

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 :D

Posted

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. :p

 

edit, i'm missing a line, always in the center or side. :mad:

Posted

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 :o

 

:D :D :p;)

Posted

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.

Posted

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. :D

Posted
  Quote
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...

Posted
  Quote
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 :D

Posted

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. :D

Posted

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...

Posted

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.

Posted
  Quote
You should know that the answer will wind up being 42 in some way. Stupid ingrate.

 

wrong question. :D

 

i could probably do it, but i don't want to take the time.

Posted
  Quote
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.

Posted

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.

Posted
  Quote
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 :D

Archived

This topic is now archived and is closed to further replies.

×
×
  • Create New...