This one is also impossible ....
Thanx due to Ali for explaining why ........


This is in reference to "Another Line Puzzle".

Each point where a line crosses another - call it a node.
At each node, count the number of lines coming out of it.
Any drawing with more than TWO nodes with an odd number of lines coming
out of it is impossible to draw.

Suppose there are 3 nodes with an odd number of lines coming out (call
an "odd node").
Starting at any node, you will obviously have to come to an odd node for
the first time.
You will then leave that odd node.  That means you will have drawn over
2 of those lines.  Keep returning to that odd node, and eventually you
will (as it is an odd node) leave the node with one un-drawn line left.
This means you must have to come back to that node and stop, as there
are no more lines to draw (in order for you to leave the node).
This means that you must finish on THIS odd node.
So every odd node is a termination point.

There is one exception:  If you start at an odd node, the last time you
visit that node will be to leave it (for example a node with 3 lines
will:  leave, return, leave again and never return).
So there is ONE STARTING point, which must be odd.

As we are assuming three odd nodes.  If you don't start on an odd node,
this means that you must visit 3 odd nodes in the course of drawing.
But the above argument shows that every odd node must eventually be a
termination point.  There are 3 of them - a problem.
Assume we start on an odd node.  This still means that there are two
termination points.
Therefore, any drawing with more than 2 odd nodes is impossible.

If there are 2 odd nodes, then you must start at one and finish at the
other.
If there is 1 odd node, you must start and finish at the same node.

If there are no odd nodes, then you can start at any point, but you must
finish at the same point.

Non-Example:  An X has 5 nodes - four of them (the ends) have one line
coming out, and one node (the centre) has 4 lines coming out.  there are
more than 2 odd nodes, therefore impossible to draw under those rules
(as is obvious)

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