# Puzzle

Drew
Posers and Puzzles 29 Oct '07 23:17
1. 29 Oct '07 23:17
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Try to draw a continuous line through every segment of this box. There are 15 segments. My Calculus teacher gave it to my class. I always find myself one segment short. I don't even know if it is possible at this point.

Good Luck
2. 29 Oct '07 23:40
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I assume (because of the number of segments you mentioned) that the line underneath the 'x' in my diagram is a single line and not two separate ones - ie you only go through that line once. If this is the case then it _is_ possible (hint: start in one of the bottom boxes). If this line is actually two lines then this isn't possible.
Next step is for you to figure out why!
[Euler and those pesky Bridges of Konigsburg might help...]
3. 29 Oct '07 23:58
Originally posted by Diapason
I assume (because of the number of segments you mentioned) that the line underneath the 'x' in my diagram is a single line and not two separate ones - ie you only go through that line once. If this is the case then it _is_ possible (hint: start in one of the bottom boxes). If this line is actually two lines then this isn't possible.
Next step is for you to figure out why!
[Euler and those pesky Bridges of Konigsburg might help...]
Sorry, I wasn't clear. The line doesn't have to be straight. It will weave back and forth between segments.
4. 30 Oct '07 01:11
Originally posted by Drew L
Sorry, I wasn't clear. The line doesn't have to be straight. It will weave back and forth between segments.
Now it's my turn not to be clear! I was talking about the diagram you drew. There is potentially some confusion about whether there is a single line at the base of the box in the middle of the top row (and this single line has a vertical line coming up to hit it at its halfway mark) or whether the vertical line does cut this horizontal line into two separate sections. I assume, because you mentioned 15 sections, that you mean the former. When I drew it out, though, the diagram looked a bit weird when I acted on that assumption. Hence my question!
If the 15 sections part is correct, then it can be solved (with a single meandering line) and you can do it most easily by starting in one of the boxes in the bottom row.

I hope this helps.
5. 30 Oct '07 02:27
Originally posted by Diapason
Now it's my turn not to be clear! I was talking about the diagram you drew. There is potentially some confusion about whether there is a single line at the base of the box in the middle of the top row (and this single line has a vertical line coming up to hit it at its halfway mark) or whether the vertical line does cut this horizontal line into two separ ...[text shortened]... can do it most easily by starting in one of the boxes in the bottom row.

I hope this helps.
Two seperate segments. Thus a total of 16 segments.

Sorry for the delay,
Drew
6. 30 Oct '07 02:54
Originally posted by Drew L
Two seperate segments. Thus a total of 16 segments.
In that case it is impossible!
Euler famously solve a problem related to this in Konigsburg.
7. 30 Oct '07 03:34
Originally posted by Diapason
In that case it is impossible!
Euler famously solve a problem related to this in Konigsburg.
That's what I thought. Thanks for your time!
8. 05 Nov '07 01:391 edit
I managed to draw a line through each segment (assuming you mean line). However, i can't start and finish in the same place (as is the more popular form of this riddle).

EDIT: Wait, no I skipped one.
lol
9. 05 Nov '07 02:25
Originally posted by Drew L
Two seperate segments. Thus a total of 16 segments.

Sorry for the delay,
Drew
Your original post indicates 15 segments... are you sure that you didn't mean 15, which is easily possible and provable using net-theory.
10. genius
Wayward Soul
05 Nov '07 13:04
Originally posted by Gastel
Your original post indicates 15 segments... are you sure that you didn't mean 15, which is easily possible and provable using net-theory.
My understanding of the problem is,
"draw the diagram exactly without taking your pencil from the paper, and without drawing the same line twice".

If so, it isn't possible - there are 8 "vertexes" which have three lines attached to them. If you enter a vertex you must also exit it (apart from the first and last vertexes*) which implies that the number of lines attached to every vertex must be even. If they are odd then there is a line that enters but cannot exit.

*the first one has even weight+1, as an edge exits from it. the last has even weight+1, as an edge enters it. If you required the first and last vertexes to be the same vertex (draw the diagram starting and finishing at the same point) then all the weights must be even-every vertex must have an odd number of lines connected to it.