[Guest]

can you draw this without lifting your pencil or repeating any lines?

the 2nd second drawing shows my first attempt. i'm missing 1 line

drawing.pdf

[Guest]

can you draw this without lifting your pencil or repeating any lines?

the 2nd second drawing shows my first attempt. i'm missing 1 line

Four of the vertices have an odd number of lines emanating from them. It is impossible for there to be more than two. Please see the original Konigsberg Bridge problem (Euler I think.)

Edited September 13, 2009 by jerbil

[Guest]

Sorry, that was a bit brief, kineticg. THe point is that if there is any vertex with an odd number of lines emanating therefrom, then it must be either a starting point or a terminating point for your pencil. Were there a third or more vertices with odd numbers of lines emanating therefrom, then it or they would have to be both visited and exited, so that an odd number of lines to and fro would remain, which would leave at least one line not drawn.

[Guest]

Network theory states that, for a netork to be traversible (i.e. you can start at one point and cover all the lines, or arcs as they are known) there must be either zero or two odd nodes. An odd node is one with an odd number od arcs meeting there. Simple logical thinking shows that this is true. If you have an odd number of lines meeting at a point, that point must be a start or finish point. There is nowhere else to go once you have used the last line.

Donjar

[Guest]

I filled quite a few pieces of paper with this one when I was first presented with it. You are correct that it is impossible using ordinary methods, but there is a way to do it.

Those who are convinced that it is impossible will probably think the solution is cheating, but there is a solution that fits the OP.

It is easier to solve with a dull pencil or marker.

[Gyvven]

I made the square and added the arcs, when that part was done I made one diagonal line, folded the paper to where my pencil was drew on that sheet to the opposite corner of the square unfolded the paper and drew my second diagonal line. Simple!

[Guest]

I truly believe that the original post was genuine, that is to say not involving trickery, as was made evident by the writer's evidence of his best "failed" attempt. Note that his best "failure" makes a figure with only two vertices which have an odd number of emanating lines.