Home | Contact Us | FAQ | Search & Site Map | Link to Us
Sign In | Join | Other 45 Sites in Network
Home
Discussion Groups
Mathematics
General TopicsResearchOperations ResearchStatisticsMathematical LogicNumerical AnalysisUndergraduate MathAlgebra HelpRecreational Math
Math Software
MapleMathematicaMATLABScilabSASSPSS

Math Forum / Mathematics / Recreational Math / June 2009



Tip: Looking for answers? Try searching our database.

Drawing with 5 lines

Thread view: 
Enable EMail Alerts  Start New Thread
Thread rating: 
tk_ - 28 Jun 2009 12:20 GMT
Hey guys, I'm having a very hard time to prove this
question that a teacher gave me. I have been trying to
prove it for the past 3 months whenever I get the time
but just can't seem to do it.

It's a drawing consisting of 5 cubes with X's inside them
I have 5 attempts to draw the whole thing but cannot overwrite a line I have already drawn.

Please see the attachment for how the drawing looks like

Help will be appreciated!

Attachment available at http://mathforum.org/kb/servlet/JiveServlet/download/129-1961110-6768230-560348/
square%205.jpg

tk_ - 28 Jun 2009 12:20 GMT
Please ignore the green line in the picture it means nothing. Just think of it as colour black lol
Matthew Schauer - 29 Jun 2009 00:21 GMT
> Hey guys, I'm having a very hard time to prove this
> question that a teacher gave me. I have been trying to
[quoted text clipped - 11 lines]
> Attachment available at
> http://mathforum.org/kb/servlet/JiveServlet/download/129-1961110-6768230-560348/
square%205.jpg

It

can't be done.  If there are more than two corners where an odd number
of lines meet, then you can't trace the figure without going back over
your lines or lifting your pen(cil) from the paper.
Ben Saucer - 29 Jun 2009 01:42 GMT
Notice the outer corners each have three lines connected. If a point has
an even number of lines connected, then for each line entering the point
there is a line leaving the point. If none of the points have an odd
number of lines connected, then the figure can be drawn starting at any
point and finished at the same point.

If a figure has exactly two points with an odd number of lines
connected, then the path must start at one of these two points and
finishes at the other. If more than two points have an odd number of
lines connected, then the figure can not be drawn by one continuous path.

Since the given figure has eight points with three lines connected, and
four points with seven points connected, it cannot be drawn with one
continuous path.

> Hey guys, I'm having a very hard time to prove this
> question that a teacher gave me. I have been trying to
[quoted text clipped - 9 lines]
>
> Attachment available at http://mathforum.org/kb/servlet/JiveServlet/download/129-1961110-6768230-560348/
square%205.jpg
tk_ - 29 Jun 2009 12:26 GMT
So, is this drawing impossible for only 1 continuous drawing of the picture of even with the 5 attempts of drawing it?

I am in denial lol
bobmck - 29 Jun 2009 12:27 GMT
On 6/29/09
tk wrote:

> So, is this drawing impossible for only 1 continuous
> drawing of the picture of even with the 5 attempts of
> drawing it?
>
> I am in denial lol

********************************
Other respondents correctly point out the topological considerations for cases in which you complete the line graph with only one placement of pen on the paper and subsequent removal from the paper.

But...if you define "attempt" as each time you place pen to paper until you remove it, then it does become possible...with multiple "attempts"

Label all corners of the squares in your drawing in alphabetical order, starting with A as the upper-most, left node. For clarity, include a label even at a corner where only two sides meet.  Moving left to right, and top to bottom, you should end up with the bottom-most, right node, as letter L.

"2-node": only B
"3-node": A,C,F,G,J,K,L
"6-node": only D
"7-node": E,H,I

One solution, of several possibilities:

1. A-B-E-D-A-E-I-J-E-F-J
2. F-I-H-L-K-H-E
3. K-I-L
4. I-D-C-G-H-D-G
5. C-H
tk_ - 30 Jun 2009 12:57 GMT
By stating that as one solution of several many possibilities were you stating that your sequence was an answer which was correct?

If you were, unfortunately you missed out a line which is from D to B. =(

I just can't seem to find a way possible for this, you still think its do-able?
Philippe 92 - 30 Jun 2009 14:25 GMT
tk_ wrote :
> By stating that as one solution of several many possibilities were you
> stating that your sequence was an answer which was correct?
[quoted text clipped - 3 lines]
> I just can't seem to find a way possible for this, you still think its
> do-able?

Hi,

It is not a question of "I have tried all possibilities" or
"I just haven't succeeded", but of what IS a valid proof.

However I don't agree with bobmck. He wrote :
>> Label all corners of the squares in your drawing in alphabetical
>> order, starting with A as the upper-most, left node.
>>
>> "2-node": only B

No. etc. It seems he completely discarded the X's, and even then
not counting properly the edges.

Apart with the problem of wording (your "attempts", and I can't
see a way of considering your figure as a set of "cubes" either),
the problem you state seems to be the one usually described as
"drawing a picture in one stroke".

A stroke being a line drawn without lifting the pen up,
and it is forbidden to draw a segment twice.
Defined as "continuous path" by Ben.

The result, which is _logically_ prooved (by Ben Saucer for
instance) is that it is impossible to draw a figure in one
stroke if it has more than two vertices with odd number of lines
at them.

So your figure can't be drawn in one stroke, as it has 12 odd
vertices (not just the 8 outer, but also the 4 inner corners
which have 7 lines. Only the midpoints of the X's have even number
of lines)

Now an additional result is that there are in any drawing an
even number of odd vertices (or none, but 0 is an even number)
By counting the total number of edges, it is half times the
sum of the numbers of lines at each point, as each line is counted
twice, one for each end. So this sum must be an even number,
hence an even number of odd numbers in the sum.

And then we can draw any drawing in half that number of strokes.
As we can draw a single stroke between any two odd vertices,
leaving n-2 odd vertices, the number of lines at these vertices
being reduced by an even number (every entering at that point
is followed by an exit).
Then a new stroke can be drawn between two of these remaining odd
vertices etc...

That is here you have 12 odd vertices, hence you need 12/2 = 6
strokes to draw your figure.

Hence impossible to draw it in 5 strokes (or 5 "attempts")
Effectively drawing it in 6 strokes or more is quite easy.

Regards.

Signature

Philippe Ch., mail : chephip+news@free.fr
site : http://mathafou.free.fr/   (recreational mathematics)

bobmck - 30 Jun 2009 18:06 GMT
On 6/29/09 tk wrote:

> By stating that as one solution of several many
> possibilities were you stating that your sequence was
[quoted text clipped - 5 lines]
> I just can't seem to find a way possible for this,
> you still think its do-able?

******************************

I interpreted (incorrectly) your message of Jun 29, 2009, 3:07 AM "Please ignore the green line in the picture..."
as meaning that there was no line from node B to node D.

Putting that back in, philippe is correct and a 5-stroke solution is impossible. (My not mentioning the "X's" as nodes was not an oversight: I simply sought a convenient way to describe the graph in a textual manner, with an understanding that X type nodes are irrelevant in these type problems...unless they include a prohibition of line-crossing.)
 
Sign In
Join
My Latest Posts
My Monitored Threads
My Blog
My Photo Gallery
My Profile
My Homepage

Start New Thread
Enable EMail Alerts
Rate this Thread



©2010 Advenet LLC   Privacy Policy - Terms of Use
This website includes both content owned or controlled by Advenet as well as content owned or controlled by third parties.