Okay ... so it's impossible!
But I put it up in good faith, the guy that sent it to me said it could be done!!!

(although, not everyone agrees - this from Scott Hammond)

The reason it's impossible ... well I'll let others explain:

Think of the graph as the floor plan of a seven roomed house.
Put 'doors' in each 'wall' (line segment).
Take a walk through the house .
Note that for a room with an even number of doors it is possible to
take a walk that passes through each door that begins and ends inside
the room or that begins and ends outside the room.
For a room with an odd number of doors this is not so, if the beginning
of the walk is inside the room and each door is used only once, the end
of the walk will be outside the room and vice versa; if the beginning of
the walk is outside the room and each door is used only once the end
will be inside the room.
The 'seven roomed house' has three 'rooms' with an odd number of doors.
A walk has only one end point , so that even if you start inside one of
them ,outside of the other two, you must end up inside both of the
others which is impossible.

thank's to L.V.Wilson for that

You cannot resolve this problem. This problem is in close to Emmannuil
Kant problem about Ko:ningsberg`s bridges.
The figure you have on your picture is equal in topological case to the
following graph (see pic1.gif). If this statement isn`t clear to your,
see pic2.gif...

pic1	pic2

Take a notice on numbers in circles on pic1 and in white areas in pic2...

So, imagine that you have the point (island) with odd number of bounds
(bridges)
If you begin your path form this point you should end your path exactly
in this point not to walk the same bridge twice.
Otherwise, you can walk ower all the bridges and leave this island.
If there are even number of bridges on the initial island, you can leave
it, walk through all the bridges but not to end your path here.
If there is another even-bridges island, you can finalize your trip on it.
And if there are more than two even-bridged islands, you CAN`T walk
through all the bridges not to walk through some of them twice.

thank's to Valence for that one

Imagine that you have a rope and you have to take it through
small holes in the walls. If you have one hole in a box and
you push a rope through it, the other end has to stay inside
the box and the other end outside. Ok? On the other hand,
if you have two holes, you can pull the rope out from the
second hole and voila -- you end up with both ends either
inside or outside, depending on where you started from.
In a similar fashion you will see that whenever you have
a EVEN number of holes in the box, you will end up having the
two ends of the rope on the same side (either in- or outside)
of the box, but if you have an ODD number of holes, the other
end stays always in and the other end out.

Now in the line puzzle you have three boxes (or rooms)
with an odd number of holes (or doors, or walls) (five of them),
and two boxes (or rooms) with four holes (doors/walls).
The problem is that for each of the three rooms you need to
leave an end of the rope inside. But since you aren't allowed
to cut the rope (or to lift your pencil from the paper),
you will always have only two ends. So one door or wall
will always be left not crossed! The problem becomes very easy
if you just left one wall out or allow the rope to cross a
single wall twice.

thank's to Simo Huotari for that one.

A vertex is where one or more lines begin. For example, + has a vertex at
the center, where the four lines join, and four vertices where each line
begins, making 5 vertices in total.
An even vertex is a vertex with an even number of lines coming out of
it. An odd vertex is a point with an odd number of lines coming out of it.
So + would have 4 odd vertices (the ones where the lines begin), and 1 even
vertex, where the lines meet.

When you trace over a vertex, you touch it twice, by arriving at it, and
leaving at it. If you start there, you touch it by leaving it; when you
end, you touch it by arriving.

A network can have any number of even vertices, and, as far as I know, an
even number of odd vertices. With me so far? Good.

According to the rules of Topology, you must start at an odd vertex, and end
at one to successfully trace a network. If there are NO odd vertices, then
the puzzle is a snap. You can start at any point you please. and trace the
network.

If there are two odd vertices, them you must start at one, and end at the
other, since you must touch those vertices an odd number of times; which
means you must leave more often than you arrive, or vice versa.

But what if there are more than two odd vertices? + has four. You would
need a line with four ends. A line has only 2. Therefore, the simple
network + cannot be traced.

and thank's to John-David Kraaikamp for that

You can tell almost immediately whether a design can be drawn with one continuous line or not by counting the nodes (area where 2 or more line segments come together). Nodes are either even (even number of line segments that form the node) or odd (odd number of line segments that form the node).

If a design has zero, one or two odd nodes, the design can be drawn in one continuous line. If a design has three or more odd nodes, it can't.

Thank's to Buttmunch for that (sorry, that's the only name I have)

This is an example of the Kongsberg Bridge Problem, solved by the famous mathematician Euler ( pronounced Oiler ) Where lines meet are called nodes. The shape has one 4 node and four 5 nodes. For a diagram to be unicursal ( draw in one continous line or sweep of the pen without going over the same line twice ) it must have either two or no odd numbered nodes. You can have any number of even nodes. You can only have an odd node at either the start or end of your sweep of the pen - try it. Any crossover points will be even nodes. As this figure has FOUR odd nodes it is impossible.

Type Konigsberg Bridge Problem into Google to see ( a very long winded ) solution to a world famous mathematical problem.

Thank's to Richard from Carlisle for that one

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