# Posts Tagged ‘hex crossing’

## Magic figures from crossed stick configurations

July 1st, 2015

When I was working on finding configurations to use in crossed stick puzzles, it occurred to me that the same configurations would make good bases for magic figures. Since each piece is a line segment with the same number of intersections, you could put numbers at intersection sites, trying to arrange them so that the numbers on each segment add up to the same magic sum. Since I had already figured out how to use Numberjack to solve magic figure problems in the context of magic dice, it was simple enough to adapt the code to solve other magic figures.

The puzzle that I’m selling as Grand Hex can form two different figures. Since they each have 24 intersection points, it seemed reasonable that one could use the numbers from 1 to 24 on the intersections, and get a magic figure with a magic sum of 50. And here’s one for each configuration:

The gold lines aren’t part of the original puzzle configurations, but they were the obvious places to add a few constraints, since the solver otherwise finds a very large number of solutions for the figures with just the black lines.

## A Ten Piece Triangular Grid Puzzle Configuration

June 24th, 2013

Here’s a configuration that I found while doodling a few weeks ago:

At the time, I was thinking of the 12 piece puzzles I found using the triangular grid; this configuration could be an easier challenge for that puzzle set that wouldn’t use all of the pieces.

But it could also make a good puzzle in its own right. We can use exactly the same slot scheme here as in the decagram puzzle. That puzzle used deep/shallow slots, (pointing the same direction) at the inner slot positions, and up/down slots at the outer positions. There are ten such pieces, which can be flipped horizontally but not vertically. Outer slots always connect with outers, and inners with inners, and the inner connections are bipartite, so it ought to work.

## Ups and Downs (and Deeps and Shallows) revisited

June 21st, 2013

I would like to be able to report that I have produced a successful prototype of the crossed stick puzzle I described in a previous post. But I cannot. This photo is a lie. There are two configurations for this puzzle, for which I intended to use two copies of each of the six possible pieces with up/down binary slots:

I say “intended to” because the puzzle appears to not be possible to solve. Specifically, the `0110` piece is rather intransigent for the configuration on the left, and it doesn’t seem possible to fit two of them in without changing the piece set. Which is indeed what I did to put together the solution shown at top. For the configuration on the right, `0101` is the problem piece.

Problem #41: Are there sets of pieces that can be assembled into both of the above figures? It would be ideal if all 6 possible pieces are present in at least one copy. (I’d have explored this manually, but for the other reason this prototype was a failure: the material was too thick for the width of holes I used, and a number of pieces snapped in the process of trying to assemble the above. So I am not able to test out whether I can make two different configurations with any piece set without running out of intact pieces.)

In crossed stick puzzles, we like for solutions to be “assemblable,” meaning that there is no cycle of pieces where each has a downward facing slot connected to the next piece and an upward facing slot connected to the previous piece. (If there were, there would be no piece you could place last.) The assemblability condition implies the existence of a partial ordering among the pieces, where the pieces are ordered according to which ones are “on top” of other pieces. It follows that that there must be at least one piece with all slots pointing up and one piece with all slots pointing down.

It turns out that the Star of David puzzle using a single set of these pieces, which I proved was unsolvable due to a clever parity argument, is also unsolvable because it has only one `0000` piece. (Likewise, the dual Star of David puzzle using two sets of pieces, which I proposed in order to get around the parity problem, is also unsolvable, because that would require four `0000` pieces.)

An alternative to the up and down slot scheme is the deep and shallow slot scheme, where all slots point the same direction on each piece. This scheme bypasses assemblability issues entirely, but it has another problem: because upward and downward pointing pieces may only connect to each other, the configuration they form must be bipartite. It turns out that a lot of interesting configurations aren’t! Here are the configurations possible on a triangular grid with 12 pieces with three equally spaced slots:

(Did I miss any?)

Only two of these configurations are bipartite. They can be made with a double set of 6 pieces with a deep/shallow slot scheme. Here are a couple of prototype lasercut puzzles in these configurations:

But what about the rest? Is there some way we can change our piece set to be able to make them? If we allow the middle slot to take all four possible combinations of up/down and deep/shallow, while the outer slots are deep/shallow and, as before, point the same direction, the resulting set should be more flexible in the configurations that can solve. (Conveniently, there are twelve such pieces.) Any subconfiguration with a cycle formed by an odd number of pieces that is connected only by outer slots is still impossible to solve. Therefore the configuration with the triangle shown in brown is still impossible. But the others should still be solvable.

(As I was finishing off this post, another set of pieces to use with these configurations occurred to me. We can make a set of pieces where each piece has exactly one deep, one middle, and one shallow shot, and the slots can occur in either direction and any order. As before, there are twelve such pieces. Can they be assembled into all of the above configurations?)

## Three More Crossed Stick Puzzles

March 28th, 2013

Since I last posted about crossed stick puzzles, I’ve come up with three new ones. Here’s one of them:

This one nicely fills one of the spots in the binary word table. There are eight pieces with three slots apiece; if we use binary shallow/deep slots, there are exactly eight possible pieces, because the pieces won’t be flippable vertically or horizontally within the puzzle.

The principle behind this piece configuration can be used to make other configurations. You can take any line segment, together with a center point, reflect the segment across an axis passing through the center, and create images of the original and reflected segments over successive rotations around the center to make configurations with rotational (indeed, dihedral) symmetry. However, because the pieces wouldn’t generally be horizontally flippable, (are there exceptions?) this only produces puzzles with combinatorially complete piece sets when the number of pieces is a power of two, and only 8 and 16 make reasonable numbers of puzzle pieces. In fact, I’d argue that the ideal range of numbers of pieces for a puzzle of this type is between 8 and 12, with a penumbra extending to about 6 and 16 in which puzzles will still be interesting to some solvers, but not as many. This necessarily constrains the quantity of viable puzzle configurations to a number that is finite, and may indeed be quite small.

But if the math keeps the number of possible puzzles down, inventing new ways to skirt the constraints we’ve set for ourselves can create new possibilities. Witness the following configuration:

Here the slots are ternary: intersection points can have slots that are ⅔ of the width of the piece (pointing up or down) or they can have two slots ⅓ of the width of the piece, pointing both up and down. Where three pieces intersect, one of each of these three slot types must be present. Three piece intersections are harder to work with than regular intersections. I’m pretty sure there is no finite configuration on the triangular grid containing all pieces of the same size with all intersections being three piece intersections.

So we have to fudge. This configuration looks like it has two kinds of three slot pieces, because three of the three slot pieces intersect at a two piece intersection. The trick is that the slot on the two slot pieces at the two slot intersection has a special, single ⅓ width slot, so that it can match the regular slots on the three slot pieces. Nicely, there are exactly three possible two slot pieces with one slot of this type and one regular slot.

In this scheme there are ten three slot ternary pieces, when both horizontal and vertical flipping is allowed. Since there are only nine three slot pieces in the configuration, one piece must be excluded. In fact, our hand is forced: the excluded piece must be the one that has all three slots of the double ⅓ width type. This is because the total number of double ⅓ width slots must equal the number of three piece intersections, which is nine. Before excluding that piece, there is one double ⅓ width slot in the two slot pieces, and 11 such slots in the three slot pieces. Well, if we can’t have a combinatiorially complete set of pieces, the next best thing is a combinatorially complete set minus the most exceptional member, which we could argue that this piece is, on account of it being the only piece with both horizontal and vertical reflection symmetry.

Finally, two configurations for the same piece set:

The configuration on the left was one that I implied in this post. If we use shallow/deep binary slots, we have pieces that can be flipped horizontally but not vertically. There are six of these and twelve pieces in the configuration, so we can use a double set of these as our piece set. As an extra bit of good fortune, there is a second configuration that uses these pieces.

I’m currently getting prototypes of the last two of these lasercut, so I hope to be able to show off the results soon. (I came up with the first one after I put in the order, so it will be a while before I get around to making a prototype of it.)

## Symmetry Variations on Binary Words

December 18th, 2012

In exploring the space of crossed stick puzzles, I’ve gone on a tangent into the world of binary words and their symmetries. A binary word is just a string where the “letters” can be one of two symbols; for convenience we’ll use 0 and 1. (The distinction between a “word” and a number in binary is that a word can have any number of leading 0’s, and these are significant.)

The usefulness of binary words in crossed stick puzzle design comes from our ability to encode them in the types of slots used in a piece of a puzzle. For example, my decagram puzzle uses pieces with outer slots that could point down or up, and inner slots that always point up, but could be deep or shallow. We can assign 0 to down and 1 to up for the outer slots, and 0 to deep and 1 to shallow for the outer slots. Then, reading across the slots, we get a binary word corresponding to a given piece. For example, down-shallow-deep-up is `0101`.

One benefit of using binary words is that we can enumerate the possible pieces in a puzzle by counting the binary words of a given length, (in this case four). But there is a snag: The piece we called `0101` can be flipped horizontally, after which it would read as `1010`. It seems that rather than plain binary words, what we want are the equivalence classes of binary words formed by some symmetry action, in this case, reversing the word. Examining these gives us four symmetrical words that map to themselves upon flipping, and six pairs of words that flip to each other, for a total of ten classes of length four.

Instead of horizontal flipping, we could, if our slots are only using an up vs. down slot scheme, flip them vertically. In terms of the effect on the pieces, this will invert all of the 0’s to 1 and vice versa. This gives us eight pairs of words that flip to each other. We could also allow both horizontal and vertical flipping. In this case there are six classes of words of length four.

At this point, you might want a table with the number of binary words of each type. That way, when you come up with a crossed stick puzzle idea you can look at the table on the row with the number of slots in the pieces, and see if there’s an entry that gives you the number of pieces you’d like. Or you could go the other direction, and look at an entry in the table and try to come up with a good puzzle for that entry. (This is how I designed my Decagram puzzle.)

 Length Fixed Invertible Reversible Inv. & Rev. 1 2 1 2 1 2 4 2 3 2 3 8 4 6 3 4 16 8 10 6 5 32 16 20 10 6 64 32 36 20 7 128 64 72 36

Let’s give it a try using the piece configuration from my last post:

Applying the table to this, there are 12 pieces with 4 intersections (and thus, slots) in that configuration. There are no entries of 12 in that row, so we’re going to have to get creative. There’s a 6 in the Inv. & Rev. column; we can make our (well, my) puzzle have two copies of each piece corresponding to one of those. The intersection points are conveniently symmetrical, which gives us reversibility, and if we use up/down slots, we get invertibility. So now we have a puzzle design.

Of course, there’s one step left, which is making sure that the puzzle can be solved. (Ideally, by a human being of reasonable intelligence devoting no more than a reasonable amount of time.) It is quite possible to develop a design for a puzzle of this type that can’t be solved.

For example, this star of David puzzle, using one set of the invertible & reversible 4 slot pieces, is unsolvable. To see why, look at the relevant binary words, and consider the first and last digits, corresponding to the slots at the six points of the star.

```0000 0001 0010 0011 0101 0110```

There are three 1’s in the external slots. Inverting a word may change that number by two, or keep it the same. Reversing a word will not change the number of 1’s. Therefore, the number of 1’s must always be odd. But since each intersection in a solved puzzle must have one up and one down slot, there must be exactly six 1’s in external slots in a solved puzzle. Thus, the puzzle is unsolvable. (However, if we had two copies of each piece, we could possibly make two stars of David by swapping pieces between the sets. This gives us a second challenge for the set above. It’s starting to look like a keeper!)

Stepping away from crossed stick puzzles, it turns out that all of the fun things that you can do with binary words can be done with these symmetry variants. And when I think of fun things you can do with binary words, I think of de Bruijn sequences and Gray Codes! But this post has gotten long enough already; that will have to wait for another post.