Puzzle from 3rd Grade
 

I still can't figure it out, but I haven't put a lot of time into it. You have a square and another square in it. There are lines connecting the edges. How can you trace over it without picking up your pencil or retracing? The paper said it was possible.

Send your picture to Xyzzy and he will put it up on the server.

Re: Puzzle from 3rd Grade
 

[quote="clowns789"]You have a square and another square in it. There are lines connecting the edges.[/quote]

By this, I assume that he means a figure that is the projection of a wire-frame cube as viewed looking at the center of a face.

To solve this problem, you should read about Euler graphs.

Look carefully at the figure -- count the number of intersections with an odd number of line segments meeting:[code:1]|__ _|_ ___
| | etc.[/code:1]
If there are more than 2, (start at one, end at the other) then it is a trick question. The only way to solve it is to "cheat". ;)

Is there proof?

[quote="clowns789"]Is there proof?[/quote]

If you don't start or end at a particular intersection, then you must leave that intersection as many times as you enter it - hence there must be an even number of segments connected to that intersection.

Well it said it was possible. It didn't say anything about overlapping but that would be quite unintelligent. But what happens if you get four of those and form it in another square 4x bigger?

[b]clown789[/b],

I missed your reply, or I would've responded sooner. For you and any others who are interested, here is a restatement of the problem.

Trace the figure without lifting your pencil or retracing.

[img]http://www.mersenneforum.org/jpg/Puzzle-1a.jpg[/img]

I've labeled the intersections so we can discuss paths.

Remember, this is a puzzle for 3rd or 4th grade students. Yes, there is a solution, and it does not require delving into obscure network theory.

[quote="Maybeso"]Remember, this is a puzzle for 3rd or 4th grade students. Yes, there is a solution, and it does not require delving into obscure network theory.[/quote]

It's a dirty trick to play on kids. There is no "standard" solution - the points w, x, y, and z all have an odd number of segments, so they must all be starting points or ending points of a continuous curve that doesn't leave the lines shown. I've seen several "solutions" to these trick questions.

The simplest is to add another segment - say x to y, and trace the extended figure starting at w and ending at z - the directions didn't say you could trace ONLY the figure.

Most out of the box, in my view, is to put a second piece of paper on top of the first to carry the line from X to Y, then remove it when finished. You didn't pick up the pencil, and the rules didn't say no extra paper.

The mathematician's favorite solution is to fold the paper - perhaps by wrapping it around a cylinder to that w and z touch. (Except for those mathematicians who stick to the mathematical aesthetic that proving something is impossible IS a solution to the problem).

The artist's favorite solution is to use a pencil that is worn to a nub so you can lean it sideways and draw from x to y without leaving a mark on the page.

Sorry, that's not it. There's 4 lines connecting, not 2. So it's in the corners and not the sides.

[quote="clowns789"]Sorry, that's not it. There's 4 lines connecting, not 2. So it's in the corners and not the sides.[/quote]

[img]http://www.mersenneforum.org/jpg/Puzzle-1a.jpg[/img]

So do you mean that we should modify the diagram by removing lines w-x and y-z and then adding lines a-f, b-g, c-h, and d-i ?

That is exactly the diagram that I was trying to describe last month.

If you do that, instead of 4 intersections with three lines, you will have 8.

[quote]Remember, this is a puzzle for 3rd or 4th grade students. Yes, there is a solution, and it does not require delving into obscure network theory[/quote]

The theory may be "obscure" to some of you. However, it is an every-day tool for others of us. And it does prove that the only possible solution is some "trick" that meets the letter of the rules but not the obvious intent. wblipp described a number of the "tricks" quite well. Every technique that he suggested for the diagram in the posted graphic applies equally well to the one that I have described here.

[quote="clowns789"]Sorry, that's not it. There's 4 lines connecting, not 2. So it's in the corners and not the sides.[/quote]
Yes, there are dozens of variations of the puzzle, but the premise is the same - once you have more than two intersections with an odd number of connecting segments, there are only 'trick' solutions.

[b]wblipp[/b], you are correct, there is no legal solution without stretching the rules. The solution I recall from the book is a combination of two of yours - and just as lame.

[b]Figure 1.[/b] Trace as much of the figure as you can. Here I start at [b]w[/b] and end at [b]z[/b]. w-a-b-z - y-g-f-x - w-d-c-z.
. . . . [b](1)[/b] . . [img]http://www.mersenneforum.org/jpg/Puzzle-1b.jpg[/img] . . . . [b](2)[/b] . . [img]http://www.mersenneforum.org/jpg/Puzzle-1c.jpg[/img]

[b]Figure 2. [/b] Instead of using a second piece of paper, we now fold the corner over so one edge touches the pencil at [b]z[/b], and the other touches [b]y[/b].
This lets us hop over to [b]y[/b] and finish tracing the figure. y-h-i-x.

To be slightly more "pure", you could fold over the paper so the exact corner touches the pencil, move the pencil onto the corner, then slide the pencil-and-corner over to the other point. That way no extra segments were actually drawn - just a point. :? It's still a lame solution.

As to the artist's solution, why not gently lay the pencil down (you're not "lifting" it), roll it over to the other point, then carefully press the tip down onto the point. :?

What a dreadfully evil puzzle. :shock:

Is this the answer?
 

Fold the piece of paper so that the one square touches a part of the other square, and then move your pencil from where it was along that part of the perimeter onto the perimeter of the other square...

I think that this is known as topology... :banana:

Simon
(The Gladiator)

I Must Apologize
 

I was moving too quickly through this thread and missed the solution, or a variant thereof, to this puzzle.

So while I thought I had solved it, I was mistaken, so I do apologize if I have stolen the thunder that was rightfully yours.

Apologetically,

Simon
(An Innocent)

:innocent:

Here's a drawing (sort of)...

[CODE]

l--------l
l\ /l
l l----l l
l l l l
l l----l l
l/ \l
l--------l

[/CODE]

Can you solve it WITHOUT folding or overlapping? Stay on the lines!

That didn't work, and it took me a long time to do it on Paint and Photoshop but here it is...

EDIT: There are supposed to be lines on the bottom, top, and right sides.

How is this different from the puzzle that you posted at the start of this thread two years ago?

From what I can see, it is the same puzzle and the same answers still apply.

Puzzle from 3rd.Grade
 

[QUOTE=Wacky]How is this different from the puzzle that you posted at the start of this thread two years ago?

From what I can see, it is the same puzzle and the same answers still apply.[/QUOTE]
:rolleyes:
I am happy that you have revived this thread as it is very educational.
The points that have been made are worth noting
1) Its a topological projection of a cube (strictly speaking not a wire cube as
wires have thickness no matter how fine and hence drawn as two lines not the ideal geometric one), but we get the point/idea.
I would say that even on an actual 3D cube it does not work. It could work if diagonals are allowed to be drawn but there is a clause against this.
Maybe it has a 4D solution as the pen has to be lifted into another dimension .

2) The other good point made is that every 'incoming line' must have an 'outgoing' line. In other words the lines meetng at an intersection or vertex should be even. :innocent:
Mally :coffee: