Whymme

I agree, Alex ~ yet sending her on a years-long futile search
(for a solution for a problem known to have no solution)
could be considered a "waste".

I believe it was the Star Trek episode "Wolf in the Fold" in
which Spock ordered the possessed computer to calculate the
exact value of pi ~ thus occupying the computer so thoroughly
that our heroes could save the crew, the Enterprise, the United
Federation of Planets, and Life As We Know It.

~~~~~~~~~~

An "even" vertex is one which has an even number of line
segments emanting from it.
The center of the X has 4 segments; it is even.
The peak of a "roof" has 2 segments; it is even.

Elementary graph theory states: a figure is "traversable" if
(1) all vertices are even, or
(2) there are no more than two odd vertices.

Consider: When you pass through a point, you have made it an
even vertex ~ one path in, one path out. A vertex is odd if
(1) you start at that vertex, or (2) you end at that vertex.
(It may take some sketching to understand these statements.)

If there is one odd vertex, we must start at that vertex.
And the rest of the figure (all even) can be traversed.

If there are two odd vertices, we must start at one of them
and end at the other.

With more than two odd vertices, the traversing is impossible.
(And the given problem has four odd vertices.)

(Naturally, someone is bound to suggest folding the paper,
or drawing the figure on a torus, etc. ~ but I remind you,
this problem was given a child.)

1. Is suggesting to search for "Euler path" at my site implies "sending her on a years-long futile search"? Have you tried?
   
2. The problem has a solution, albeit negative, as you demonstrate below.
   

>An "even" vertex is one which has an even number of line
>segments emanting from it.
>The center of the X has 4 segments; it is even.
>The peak of a "roof" has 2 segments; it is even.
>
>Elementary graph theory states: a figure is "traversable" if
>(1) all vertices are even, or
>(2) there are no more than two odd vertices.

Is making a reference any worse than typing the same thing anew? A reference to a general approach gives one a fighting chance to use one's brain. What goal is achieved by giving the solution away?

I seem to have angered you, Alex - certainly not my intention.

I gave the solution to Martin to do as he wishes. If he wants
his daughter to use her brain, he can withold the truth. If
he truly feels he wasted valuable time in a futile endeavor,
he should be given the opportunity to decide for his daughter.

But I'll practice more descretion in future positings - promise!

Visually it will be different: a semicircle instead of a triangular roof.

The answer will remain the same, nonetheless. The question belongs to graph theory where a particular shape of an edge joining two vertices is of no consequence.

This is my first post...and it is a pleasure to frustrate someone already frustrated. If you can comb out all the tangles in the webs that have grown inside your head, perhaps it might just add to your frustration and give you a little more reason to be happier as you will be more frustrated.
Perhaps looking for the Konningen Bridge Problem, or the Travelling Salesman Problem, or something of the sort that involves tracing nodes on a graph with edges without repetition might just lessen your frustration, so i am not suggesting any of what I just said.

Q.E.D.
:wq

Well, frustrated, I'll save you the trouble of looking through the _whole_entire_thread_(who knows how long that might take)by quoting select tidbits of info.

Soroban wrote:
>With more than two odd vertices, the traversing is impossible.
>(And the given problem has four odd vertices.)

Alex B. wrote:
>Search the Web or this site for "Euler path" or "Unicursal Drawings."

I write: we're still quoting each other.

this made me laugh, thanks haha

Seems clear so far. Like laying out a string, it has no odd vertices if the ends are tied together.

>If there is one odd vertex, we must start at that vertex.
>And the rest of the figure (all even) can be traversed.

Huh? Just curious, how is one odd vertex possible? I can't seem to visualize it.

its actually really simple..
c _ b
draw this | point a
(start at the bottom) d
then make the top "roof" thing /\e

then go down diagonally to the bottom left
d
/\
c_/__\_1b 2e
/ |
/ | a
f

then go across horizontally to your beginning point (the bottom right corner)
you should have this.
d
/\
c _/__\_
/ |1b& 2e
/ |
_/___|point 1a 2g
f

sry if the letters confuse you... ignore them if they do..

the numbers mean which point they are first.

kay so then you go up diagonally to the top left corner.

d
/\
1c 2h/__\
\/ |1b& 2e
/\ |
_/__\|
f point 1a 2g

then finish it off by drawing a line downwards

ta daaa

d
/\
1c.2h /__\
| \/ |1b.2e
| /\ |
|/__\|
1f.2i point 1a.2g

and sorry to you very smart talking people if i did not use an
adequate supply of mind challenging terms that the average person cannot comprehend

make a line like _ moving from right to left
then make the line / moving from bottom to top
then make the line - from right to left, this yields the shape z
from the top of the z where your pencil should be draw ^ across the top of z
where those 2 points intersect draw | from top to bottom (or just simply down)
now draw \ through the center of your figure from right to left
now draw | from top to bottom

now send me money

I will quote only the two relevant statements:

Elementary graph theory states: a figure is "traversable" if
(1) all vertices are even, or
(2) there are no more than two odd vertices.

So the first puzzle asked about, namely a square with its diagonals and a two-line "roof" on one side (wlog say the top) IS drawable: It has FIVE vertices: the point at the top of the roof has two lines, so it's even.

Each upper corner of the square (the corners which "support" the roof) is even: from it emanate one roof line, two sides of the square, and one line of the X, a total of four lines.

And each lower corner is odd, being the intersection of two sides of the square and one line of the X.

Thus there are exactly two odd vertices, and so by rule (2) above the figure is traversable (drawable, "w/o lifting your pencil or tracing a line.")

When we change the figure by adding three more rooves, one on each of the other three sides, each of the FOUR corners of the square is then the intersection of two sides of the square, one side of each of two rooves (making four lines so far) and one line of the X, making a total of FIVE lines; thus each corner, of which there are FOUR, is an odd vertex.

Therefore, by rule (2) above, that figure is not traversable and so cannot be drawn in the prescribed manner.

And turning the four rooves into a circle (as in the posted figure) doesn't change anything. In graph theory all that matters is whether or not two points are connected by a path: the "shape" of the path is literally meaningless in that theory.

The equivalence can be verified by examining in the diagram the intersections at each corner of the square, seeing that they're the same ones I named above for the four-roofed square, and so realizing that only the endpoints matter.

I am truly interested if anyone know the solution to Janna's puzzle in message #16.