A "Find the Path" Puzzle

Question

I invented a puzzle, inspired by Simon Tatham's Portable Puzzle Collection. Is there anything that it could improve on? Was it too easy, or was it too hard? Anything I could add? What must I call it?

Is this the right site to post this? Perhaps Meta is better.

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Puzzle:

You have the following grid: $$\begin{array}{|r|c|} \hline \rm A &\rm B &\rm C \\ \hline \rm D &\rm E &\rm F\\ \hline \rm G &\rm H &\rm I\\ \hline \end{array}$$ The letters represent numbers ranging from $1$ to $4$ inclusive. They cannot all be the same number, but all the numbers from $1$ to $4$ don't have to be in the grid either... but there's a catch.

You have to draw a path from the left bottom corner of the grid to the top right corner, following the edges of the squares. That includes outer edges as well. The path cannot cross an edge more than once, but it can meet an intersection more than once (intersections are corners/vertices with four edges that join to it). The numbers in each square in the grid show how many edges of that square must be crossed with the path. Here is an example:

Example:

Find the path. $$\begin{array}{|r|c|} \hline \rm 1 &\rm 3 &\rm 2 \\ \hline \rm 1 &\rm 2 &\rm 3\\ \hline \rm 2 &\rm 1 &\rm 1\\ \hline \end{array}$$ SOLUTION:

$\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\,\,\;$

There might be more than one solution, but I hope this puzzle is fun. Try it with the following one. Remember, you can't cross an edge more than once, but you can meet an intersection more than once.

Challenge:

Find the path. $$\begin{array}{|r|c|} \hline \rm 4 &\rm 3 &\rm 3 \\ \hline \rm 3 &\rm 4 &\rm 4\\ \hline \rm 3 &\rm 4 &\rm 2\\ \hline \end{array}$$

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Edit:

Made a harder one.

Find the path. $$\begin{array}{|r|c|} \hline \rm 2 &\rm 3 &\rm 3 &\rm 3 &\rm 2\\ \hline \rm 2 &\rm 4 &\rm 3 &\rm 2 &\rm 3\\ \hline \rm 2 &\rm 1 &\rm 2 &\rm 3 &\rm 2\\ \hline \rm 3 &\rm 3 &\rm 3 &\rm 3 &\rm 1 \\ \hline \rm 3 &\rm 3 &\rm 1 &\rm 2 &\rm 3 \\ \hline\end{array}$$

Okay, apparently this was impossible, as declared by comments below, with a formal proof given by @JonMarkPerry. I must admit, my ones do look like twos (with the only thing differing them being their curvature, so my messy handwriting does not do to well). I believe in the first column, all the ones are supposed to be twos. I have fixed that now, and have tried it three times.

Answer 1

Your 5x5 has a solution now. See below:

It took about 2-3 minutes with pen and paper (about half that time copying it onto an envelope), so at that size I think it is a little too trivial to provide much interest. But a slightly bigger puzzle - maybe 8x8 - might be interesting enough to solve.

Answer 2

Fairly easy puzzle. I'd think of it as a slight generalization of Slitherlink (or Loopy, as Simon Tatham calls it). It would probably be more interesting if it were a bigger grid, but I get the sense that there would always be multiple solutions if it was scaled up. Maybe it could be improved with some additional constraints (like how many times the path bends, or some given segments) to limit the possibilities.

Answer 3

I think your $5\times5$ is:

impossible

Here is my proof:

The $4\mid1$ line is fixed, so the $3$ at $S(1,1)$ can be defined. The bottom left origin cannot have two lines coming from $P(0,0)$, so must be as given. But now the line from $P(1,1)$ to $P(1,2)$ has nowhere to go.
(Coordinates are Square(x,y) and Point(x,y))

Answer 4

In user477343's example, there is an alternative solution:

For the 3x3 case, there are 360 sets of edges which are valid paths across the grid. This assumes that it is possible to have the number zero in a square. 320 of these sets are unique.

The other 40 sets of edges form a series of 19 'groups', where all members have the same numbers. There are between 2 and 4 members of a group. For example:

If you enlarge to 4x4, then there are 38024 sets of edges, with 37428 being unique. The other 596 are in 294 'groups', where the largest group has 4 members:

A 5x5 grid has over a million solutions: I have not yet evaluated this.