29 Oct 07
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|___|___|___|
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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
29 Oct 07
____________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.
|___|_x_|___|
|_____|_____|
Next step is for you to figure out why!
[Euler and those pesky Bridges of Konigsburg might help...]
29 Oct 07
Originally posted by DiapasonSorry, I wasn't clear. The line doesn't have to be straight. It will weave back and forth between segments.
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...]
30 Oct 07
Originally posted by Drew LNow 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!
Sorry, I wasn't clear. The line doesn't have to be straight. It will weave back and forth between segments.
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.
30 Oct 07
Originally posted by DiapasonTwo seperate segments. Thus a total of 16 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 separ ...[text shortened]... can do it most easily by starting in one of the boxes in the bottom row.
I hope this helps.
Sorry for the delay,
Drew
05 Nov 07
Originally posted by GastelMy understanding of the problem is,
Your original post indicates 15 segments... are you sure that you didn't mean 15, which is easily possible and provable using net-theory.
"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.