Return Styles: Pseud0ch, Terminal, Valhalla, NES, Geocities, Blue Moon.

Pages: 1-

Puzzle - Is there a math solution...

Name: Anonymous 2007-11-16 22:51

that shows that this is impossible to solve?


|¯¯|¯¯|
|¯|¯|¯|
.¯.¯.¯


The above was a puzzle I spotted in /b/.  The instructions were to draw the image you see on a sheet of paper and using one continuous line, without crossing itself or the same line segment twice, pass through all 16 segments at least once.  I believe it is impossible - is math available to prove me right?

The top half of the box, to be solved in the same manner as the puzzle, requires 7 segments to be crossed (since the two boxes share a common side).  The lower half requires 10 segments to be crossed (since the middle box shares two of the would-be 12) giving us a total of 17 segments to cross IN ORDER to solve using one unbroken, uncrossed line, however there are only 16 segments available for the whole puzzle.

On any 15th move there is one unreachable segment left, so essentially you must use the 16th "move" to cross a segment twice, and the 17th move is the one which would complete the puzzle - illegally.

Is there anything you all would add to this or disagree with?  I'd like to see a simple math solution as to why this won't work.  Thanks for your assistance.

Name: Anonymous 2007-11-16 22:59

Fuck that didn't come out right... the middle line of the top box is not lined up with any line on the bottom.  I hope you guys understand how the box is supposed to look, if not then ask me to clarify.

Name: Anonymous 2007-11-16 23:50

It's supposed to look like this right?
`___________
|_____|_____|
|___|___|___|

Name: Anonymous 2007-11-17 0:05

>>3
yes, thanks

Name: Anonymous 2007-11-17 0:31

I just tried several times in MS Paint.  I can't do it.  I think it is impossible.

Name: Anonymous 2007-11-17 0:34

the two top and the middle bottom box all have 5 segments making them up.  an odd number of passes means after crossing them all you will be outside if you started inside and, inside if you started outside.

this means in order to corss all of the segments your path must start or end in those 3 boxes, since a single line has only two ends, you obviously cannot have ends in all 3.

Name: Anonymous 2007-11-17 0:41

>>6 here

Someone else is going to have to make that a formal, proof.  I'm way too sleepy to do that right now, but you can just make pentagon and verify the first part

Name: Anonymous 2007-11-17 1:19

>>6 restated Euler's proof inspired by the Bridges of Ko"nigsberg problem and is correct. There are >2 nodes of odd order, so you can't walk the graph without retracing your steps.

Name: Anonymous 2007-11-17 1:25


>>8

thanks, >>6

Name: Anonymous 2007-11-17 4:32

lol, moar houses with moar utilities and if you can't connect teh lines then all your nigger residents go without them

Name: Anonymous 2007-11-18 4:30

>>5 here.

I'll take a crack at a proof.

We have three boxes with 5 segments to pass through.  For each of these boxes, ee can either go I/O/I/O/I or O/I/O/I/O (I = "intersection that goes in"; O = "intersection that goes out").

Case 1: I/O/I/O/I
You better be finished becuase you can't get out of this box.  The last "in" stroke intersected the last segment in the box.

Case 2: O/I/O/I/O
Feel free to continue (or stop, if you're finished).  You're outside the box, and you've finished the segments for it.

Since there's three boxes of 5 segments each, you can start inside one of them, and make O/I/O/I/O strokes for it and finish that box.  But that leaves two boxes that require the I/O/I/O/I pattern.  If you complete this for one of the boxes, you can't get out to do the other.  Therefore, it's impossible to cover both of the remaining boxes, even if you conquer the first box of five segments.  QED?

Name: Anonymous 2007-11-19 15:08

bump for great justice.

Name: Anonymous 2007-11-19 15:32

AIDS.

Name: Anonymous 2007-11-19 19:47

Count how many lines meet at each of the various meeting place. All even? You have a winner.

Not all even but a pair are odd? Also a winner you will start on one part and end on the other. Otherwise no.

Name: Anonymous 2007-11-19 21:09

Impossible, you have 5 vertices of odd degree; to create a path which touches each edge exactly once, you can have at most 2 vertices of odd degree, since to exhaust the paths on a non-start/end point you must enter as many times as you leave, which forces an even degree.
Wikipedia entry: http://en.wikipedia.org/wiki/Eulerian_path
Similar Problem: http://en.wikipedia.org/wiki/Seven_Bridges_of_K%C3%B6nigsberg

Name: Anonymous 2007-11-19 21:30

>>11 here.

So did I win or not?

Name: Anonymous 2007-11-20 1:10

>>16
No. You failed. Miserably.

Name: Anonymous 2007-11-20 6:25

EULER CYCLE OH LORD!

>>15

Name: Anonymous 2007-11-20 18:55

>>11 here again.

>>17
How did I fail?  What was the flaw in my proof?

Name: Anonymous 2007-11-20 19:53

>>19
You asked 4chan a question.

Name: Anonymous 2007-11-21 0:26

7 bridges of konigsberg ftw

Name: Anonymous 2007-11-21 11:07

I'm >>5, >>11, >>16, and >>19.

>>20
Now wait a second.  I'm not OP.  I did my best to solve the problem.  I don't recall asking a question.

Name: Anonymous 2013-05-19 11:09


Don't change these.
Name: Email:
Entire Thread Thread List