Exploration of another Sol Lewitt puzzle from Barry Cipra

TL;DR

This paper documents a new Sol Lewitt-inspired puzzle introduced at MOVES 2019, providing insights and details to facilitate understanding and further exploration by puzzle enthusiasts and researchers.

Contribution

It introduces and analyzes a novel Sol Lewitt puzzle, sharing insights to encourage further engagement and development in puzzle design.

Findings

The puzzle offers unique geometric challenges.

Insights facilitate understanding and potential variations.

Encourages community involvement in puzzle exploration.

Abstract

At MOVES 2019, Barry Cipra casually introduced a new "Sol Lewitt" puzzle to fellow conference goers. Several brainstorming sessions ensued with Barry, Peter Winkler , Donna Dietz, and other attendees. This paper is to document the puzzle and some insights so others can enjoy and build on this lovely puzzle. (Look for it in an upcoming book by Peter!)

Equations

$$A+C+B^{\prime }+D^{\prime }=A^{\prime }+C^{\prime }+B+D.$$

$$A+B+C+D=A^{\prime }+B^{\prime }+C^{\prime }+D^{\prime }.$$

$$A+A^{\prime }+B+B^{\prime }=C+C^{\prime }+D+D^{\prime },and$$

$$A+A^{\prime }+D+D^{\prime }=B+B^{\prime }+C+C^{\prime }.$$

$$A+A^{\prime }=C+C^{\prime },and$$

$$B+B^{\prime }=D+D^{\prime }.$$

$$(A+C)-(A^{\prime }+C^{\prime })=(A^{\prime }+C^{\prime })-(A+C),$$

$$A+C=A^{\prime }+C^{\prime },andsimilarly$$

$$B+D=B^{\prime }+D^{\prime }.$$

$$A=A^{\prime }=C=C^{\prime },and$$

$$B=B^{\prime }=D=D^{\prime }.$$

Full text

1. Introduction

Barry Cipra111bcipra@rconnect.com has long been inspired by the artwork of Sol Lewitt, particularly his “Fifteen Etchings” work from 1973. In this piece of work a square is subdivided into 16 smaller squares, each with an etching (except for one blank spot). These etchings contain horizontal, vertical, and diagonal lines. In Barry’s first Sol Lewitt puzzle, the goal was to rearrange the smaller squares, so that no lines terminated inside the 4x4 grid. This puzzle and its solutions can be explored online at many sites including [1].

As a takeoff on this original puzzle, Barry designed 16 new square puzzle tiles filled with curved segments, such that Cartesian placements of the tiles will never result in curves termanating within the tiled region as demonstrated in Figure 1. If you follow any curve in either direction, it will either be part of a loop, or both ends will termanate at the border.

Barry brought these tiles to MOVES 2019, asking fellow attendees (including Peter Winkler222peter.winkler@dartmouth.edu and Donna Dietz333dietz@american.edu) to ponder them and tell him what they thought. In the true spirit of mathematical adventure, we encourage readers to immediately stop reading, and go play with these tiles either on paper by cutting out Figure 2 with scissors or online at [2], before returning to read about what we discovered.

2. The design

The tiles themselves can be thought of as rooms with four entrances, one each side of the tile, allowing access to the center of the room. Now also imagine curtains for each entrance which could be left open or closed. When open, they sit on either side of the entrance, but when closed will overlap. Another way to visualize this is to see that each side of each tile involves either just the two points on the side, or it involves a third point, where the curves cross. On each side, we choose whether to activate that crossing point or not. Since there are 16 ways to do this, we end up with 16 oriented tiles.

We can also focus on the curves rather than the spaces made by the curves. Each curve is made up of arcs which deflect 90 degrees as they pass though the tile. Although there are different physical lengths, we think of each arc as having length one. There are binary strings or numbers associated with each tile. The digits represent the crossing or non-crossing (Figure 6) nature of each side of the tile, starting with the left side of the tile and working clockwise.

3. A loopy question

After a bit of exploration, we agreed to place the tiles on a 4x4 grid representing a torus, and to preserve the original orientations of the tiles. (This is by no means necessary, and readers are encouraged to explore as many variations to this puzzle as they can create!)

When placed on a 4x4 torus, the far right and left edges are one single edge, just as the top and bottom edge are the same edge. (This is the same structure as a Pac-Man board or a standard doughnut has.) There are now no edges so curves cannot terminate. The only option left for any curve is to be a loop!

The loops may be strictly planar, or they may circle around the torus once or more either horizontally or vertically.

In Figure 3, Curve A could just as easily have been created on a plane as on a torus, so we will call it a planar loop. Curve B is also a planar loop, and it can be drawn as a circle when the torus is cut differently. Curve C, however, wraps around the torus in order to form the loop. We will refer to these as planar loops and torus loops respectively.

The first questions which arose between us concerned the maximal and minimal numbers of loops possible and their lengths. Experience had led to several hunches, and a few configurations stood out as particularly interesting. Figure 4 demonstrates how it is possible to have 8 torus loops on the torus, while Figure 5 demonstrates how it is possible to have 16 planar loops.

Given our numbering convention, these two configurations have something special in common. We can find them by using Gray Code. Gray Code is named after Frank Gray, and is discussed in several places online, such as [3]. Gray Code allows binary numbers to be listed one after another and returning to the starting point, from 0 to $2^{n}-1$ while only changing the value of one digit at a time. This efficiency can pay dividends in many applications. Here, it creates aesthetically pleasing patterns. Both Figure 4 and Figure 5 exhibit this property in both the horizontal and vertical directions. Other Gray Code arrangements exist for these tiles. The reader is encouraged to explore and enjoy them.

The configuartion in Figure 4 shows how the maximum number of torus loops may be attained. Since there are 64 arcs in total, and we note that it takes 8 arcs to reach around the torus in either direction, there cannot be more than 8 torus loops.

The configuration in Figure 5 shows how the maximum number of loops (all plane loops) can be attained. Since arcs deflect 90 degrees, you cannot make any loop with fewer than four arcs, thus there cannot be more than 16 loops.

These configurations, as well as the one in Figure 7 were found by manual experimentation with the puzzle pieces. Figure 7 demonstrates a loop of length 60 together with a loop of length four.

4. Parity of number of loops matches parity of crossings

The question of course was whether or not a single loop of length 64 could be formed. We suspected this was not possible, because all configurations of the tiles had resulted in an even number of loops! After several attempts at a proof, a simple, inductive case-by-case proof was found by Peter.

Theorem 1**.**

(Parity Theorem) The number of loops and the number of crossings have the same parity.

As a base case, take any configuration. However, since it is not necessary to restrict the use of tiles, tile 0000 may be used in all 16 positions so there are initially no crossings in the configuration, making it easy to visualize.

Case A) As in Figure 8, there could be a tile with non-crossed arcs coming through adjacent corners, which (perhaps unbeknownst to us) are each part of two separate loops. If we replace that set of arcs with crossed arcs using the same attachment points on the tiles, this will result in a single loop. We have decreased the loops by one and increased the crossings by one, thereby maintaining parity.

Case B) As in Figure 9, there could be a tile which starts out as a single loop, making a figure eight crossing outside of the tile, but after the switch it becomes two loops. This will become two separate loops as shown. We have increased the number of crossings by one and also increased the number of loops by one thereby maintaing parity.

Case C) This case is special, because it is a red herring. We will show that this case does not exist for this set of tiles, given the rules we have agreed upon. However it would appear that if there is a connection between the lower to attachment points which does not intersect the connection between the upper attachment points, when the non-crossed arcs are replaced with crossed arcs, we should increase only the crossings without changing the number of loops. In order to explain why Case C cannot happen, we need another theorem.

Theorem 2**.**

(Checkerboard Crossing Theorem) Once the direction is set for a loop, this induces a checkerboard pattern on all the tiles, constraining only horizontal entry (H) into the tile or only vertical entry into the tile (V).

When the loop enters a tile, that arc must come from either a horizontal or a vertical side. Since all arcs turn a quarter-circle, it must exit opposite to how it entered thereby entering the next tile the other way. Our torus is 4x4, or 8x8 if you were to count arc steps. Therefore, the entire torus (everywhere the curve may wander) has to honor this checkerboard rule. In Case C (Figure 10), in order for this case to happen, the directions of the arcs must be opposite, which is not possible. A loop is given in Figure 11 to help the reader visualize why this is true. A zoom-in of one tile is given in Figure 12 indicating the local directional structure induced on that tile as a result of choosing the curve direction.

So, although the curve direction is arbitrary, and therefore the H or V designation is arbitrary, as soon as a direction is chosen for a curve, the whole curve must abide by that directional choice.

In our original tile portfolio, we had one of each kind of tile, which means we had an even parity on the curve crossings. Therefore we have to maintain an even number of loops, making 60 the largest possible loop, as demonstrated in Figure 7.

5. Possible loop lengths

It may appear obvious that the loops must have a length divisible by four, but precisely because it seems obvious, it would seem prudent to prove this fact. We have two proofs, in fact!

The first proof is is short and sweet, and it makes use of the fact that we have agreed to use a torus made of even numbers of tiles in each direction. Since we alternate our tile crossings as vertical-horizontal-vertical-horizontal, we must alternate switching the column, then row, then column, then row. This means we have four kinds of tiles which must be visited in a specific order. In Figure 13, this color sequence is white, red, purple, blue. Since we visit all colors equally, the total path length is divisible by four.

The second proof is not very elegant, but it is true without the restriction that the torus must have even numbers of tiles in each direction.

Since arcs turn through quarter circles, it is obvious that the minimal number of arcs in a loop must be four. Also obvious, due to argments in the proof above (Theorem 2), the loop must have an even length. One possible attack would be to take existing loops and join them as follows. Take the entire torus and cover it with tile 0000, thus making 16 loops.

One by one, replace tiles with tiles containing just one more arc crossing, until the desired configuration is attained. One example of such a transition is shown in Figure 14. However, even though this makes sense intuitively, the actions on the loops as they split and join are quite subtle, so we will focus on the analysis of an existing loop rather than creating loops.

All loops can be decomposed into their components based on their topology. Each type of loop must satisfy the property that the length is a multiple of four. When joining or splitting loops, we have seen that the number of edges remains a constant.

Type I) A planar loop with some integer number of full circular turns will have a length divisible by four, provided any net-zero deflection loop also does. This is because each full turn will require four arcs.

Type II) A torus loop in either direction will also have a length divisible by four, provided any net-zero deflection loop also does. This is because the arc-height and arc-width of our torus is 8, which is divisible by four.

Type III) Any net-zero deflection loop in the plane must have length divisible by four. The simplest case for this type of loop is a figure eight. (This will be demonstrated in Theorem 3.)

Type IV) Any other loops must be decomposible into the previous types.

Theorem 3**.**

(Net zero deflection Theorem) Zero net-deflection loops in a plane have length divisible by four.

There are four types of arcs we draw on the page, but each can have two directions. Thus, we will discuss all eight arcs as if they are different. Figure 15 gives the eight arc types, along with + and - symbols indicating their x and y displacements.

The symbols A, B, C, D, A’, B’, C’, and D’ will be used not only to represent the eight arc types, but also to indicate the number of times that arc appears in a loop. Thus, arc A appears A times in our loop.

Due to the rules we’ve set for this game, not all arc types may follow each other. In fact, once an arc has been chosen, only two possibilites follow. If the arc is at a point of inflection, it will continue in the same direction but the concavity will change. Otherwise, it must move to the next direction permitted. There must be an even number of points of inflection in a loop. The permitted transitions are shown in Figure 16. Note that this is a cube and is therefore bipartite. Eight edges are directional, while four edges can be traversed in either direction. The cube’s corners represent arcs in the loop, while the edges represent points in the loop.

Since the loop must follow this bipartite graph, and since we know from Theorem 2 that there must be an even number of arcs in the loop, we have this Bipartite loop condition as follows:

[TABLE]

Since we have a net zero deflection loop, we have a Zero deflection condition:

[TABLE]

Finally, since we have to balance our + and - tallies on both x and y coordinates, we have two more conditions, the Balancing conditions:

[TABLE]

[TABLE]

Working with just the balancing conditions, subtracing them from each other gives the Simplified balancing conditions:

[TABLE]

[TABLE]

If the Bipartite loop condition is subtracted from the zero deflection condition (flipped), we get that

[TABLE]

which tells us that

[TABLE]

[TABLE]

But this reveals something interesting. Let $A+A^{\prime}=C+C^{\prime}=2x$, $A+C=A^{\prime}+C^{\prime}=2y$ which means that $x=y$. This gives the Solved Balancing Equations:

[TABLE]

[TABLE]

Since the total number of arcs in a zero deflection plane curve is $A+B+C+D+A^{\prime}+B^{\prime}+C^{\prime}+D^{\prime}$, or $4x+4y$, it is divisible by four! Figure 17 shows how this works out for one specific loop, where $A=A^{\prime}=C=C^{\prime}=4$ and $B=B^{\prime}=D=D^{\prime}=2$.

6. Right of way

As we explored this set of game tiles, it occured to us that it might be nice to indicate a three dimensionality on the tiles, so one arc would appear to be above and one below at each crossing. We decided to use right of way to determine which arc would get the upper level. So, as shown in Figure 18, as two arcs come in from the bottom of the tile, the one on the right has priority. A maximum of four crossings may occur on one tile, such as shown in Figure 19.

An example of the entire torus being marked this way is shown in Figure 20.

You may notice in Figure 20, that there is a wonderful weaving motion, over-under-over-under, on any of these loops. Why is that?

Theorem 4**.**

(Weave Theorem) Tiles marked with right-of-way bridges will always alternate over-under-over-under.

First, let us consider all possible crossings, by simply drawing all possible arcs on one single tile as in Figure 21, and let us now focus on what happens only upon entry up from the right side, on a horizontal side as in Figure 22.

In the case of exiting at point A, consider the gray curves and the black curves from Figure 23. We will have one crossing from each set of curves, thus we’ll enter the tile with right of way, cross above, then below, then retain right of way as we exit, entering the next tile also on the right.

In the cases of exiting at points B or C from the tile, we will enter the tile on the right with right of way, then use our right of way once, but leaving the tile on the left, and without right of way.

In the final case, exiting at point D, we enter and exit with right of way, like in A, but we’ve used zero crossings, which is fine, as zero is just as even as two is.

A similar analsis can be accomplished on the four remaining paths of travel through the tile. Thus, weaving will always be maintained as we traverse our loops in this puzzle.

7. Edge loops

One variation idea by Jim Propp was to not use a torus, but to put loops on all the edge pieces as shown in Figure 27. This certainly changes the puzzle in a dramatic way!

8. Magic Squares

Another variation is to consider the binary labels as binary numbers, and then look at arrangements that form pandiagonal magic squares. Figure 28 shows one such arrangement.

9. Conclusions

In conclusion, we feel this is a fun set of tiles, and worthy of more exploration, more questions, and more theorems. Perhaps we may add more restrictions or loosen them. Feel free to explore and touch base with the authors about your findings!