Picture Frames and Loops
Here we'll look at some topological puzzles originally cast in terms of hanging a picture frame.
Setup
This activity is inspired by a brain teaser I first heard from my colleague Micah Miller while we were teaching at Texas State. Suppose you have a picture frame with some string acting as the hanging wire. Typically you would simply hang the picture from a single nail:
Given two nails, you could hang it from both. Here's one way to do it:
In the above configuration, pulling either nail still leaves the picture hanging from the other. On the other hand, if we hang it as we have below, then pulling the red nail leaves the picture hanging from the blue nail while pulling the blue nail results in the picture crashing to the ground!
Can you find a way to hang the picture from two nails so that pulling either one results in the picture falling? (In other words, both nails support it together, but neither supports it alone.)
Before we dig into this and related puzzles, let's look at some manipulative options. If you have a disposable painting and a wall you don't mind puncturing several times, then that's one way to go! Here are some alternatives for the rest of us:
- If you construct a small cardboard wall, you can play this with pins and a loop of string. (I don't recommend thumb tacks, since the back can press the string against the wall, trapping it with friction when it should glide freely.) For better effect, I've tied some weight like a washer to the string loop to help it fall. A washer can be adorned with a small painting that the kids design to add an arts and crafts component and to make this feel a little closer to the original telling.
For a less craft-intensive option, peg boards can work. A classroom set of peg solitaire puzzles can be purchased for around $30-40. You'll still need to make a weighted loop of string, but the setup is to place two pegs in the peg board and then hang the weighted loop from them over the edge of the table so that the loop has some distance to fall.
To keep things strictly on the tabletop, you can also cast it as "can the loop be pulled free?" instead of "will the loop fall?" and students try to pull the peg board and the loop apart horizontally. I haven't tried it, but a rubber band could make an interesting loop for the tabletop version and eliminate some of the need to test by tugging, but the danger of things devolving into a rubber band war is high.
Hefty objects or magnets on a metallic sheet can also serve as pegs.
- For a more kinesthetic group experience, you can have some number of students hold out their arms horizontally as nails and then use rope to model the picture frame string, having a student pull their arm out to remove the nail and giving the rope a light downward tug to test if it's been released.
While the original puzzle was in terms of picture frames, it can be reskinned in cute ways. Here are a couple, but I'd love to hear of other ways to tell this story:
- MoMath at one point had a traveling exhibit based on this puzzle where a dog was leashed with a loop to two posts. A sequence of puzzles had visitors feed the leash around the posts in different configurations and then decide which of the posts should be removed so that the dog could run free. I liked this telling because the dog getting free seemed like a positive goal versus a picture frame crashing to the ground. The dogs were also very cute.
- One of my retellings of this puzzle has been in terms of lassoes and bulls. In each scenario, a rancher has lassoed a group of bulls. The questions are then framed around what happens if a particular bull were to slip free. Would any of the other bulls remain lassoed and, if so, which ones? Students also look for ways to lasso the bulls so that certain interesting things might happen, such as with the original picture frame brainteaser.
To keep this telling as generic as possible, we'll just describe it in terms of a loop and some pegs standing vertically on a tabletop, where the loop can be stuck (cannot be moved horizontally away from the pegs) or free (can be moved horizontally away from the pegs).
Initial Questions
- Is there a way to loop around two pegs so that the loop is initially stuck but becomes free if either peg is removed?
- We've seen a way to place the loop so that it is still stuck if the red peg is removed but free if the blue peg is removed. Can you find a way to place the loop so that it remains stuck when the blue peg is removed but becomes free when the red peg is removed?
- Are there multiple ways to create a given scenario? (For example, are there two distinct ways to place the loop so that it is stuck but becomes free when either peg is removed?)
Explorations
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Here is a relatively simple way of solving the original puzzle:
Ditching the frame, here is that solution in terms of a loop and some pegs:
It feels a little like an optical illusion in the sense that you can really only see it one way at a time, but you can interpret the above arrangement as a loop with two of its ends crossed over each other in two different ways:
Focusing on those loop ends gives a nice way to see why removing either peg would free the loop.
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In general, there will be multiple loop configurations that satisfy any given condition concerning the removal of some pegs freeing the loop. For example, here are two more configurations where removing either peg will leave the loop stuck on the remaining peg:
A great question to ponder is what these two configurations might have in common that is lacking in configurations where removing a peg frees the loop.
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One style of puzzle worth exploring is assigning students the task of finding a way to loop around some preset pegs so that some condition is met. A nice set of puzzles is collected by the Demaines et al, recast here in terms of loops and pegs:
- Place a loop around 3 pegs so that removing any 1 peg frees the loop.
- Place a loop around 3 pegs so that removing any 2 peg frees the loop but removing any 1 leaves the loop stuck.
- Place a loop around 1 red and 2 blue pegs so that removing 1 red or 2 blues frees the loop but removing only 1 blue leaves it stuck.
- Place a loop around 4 pegs so that removing any 1 peg frees the loop.
- Place a loop around 4 pegs so that removing any 2 peg frees the loop but removing only 1 leaves it stuck.
- Place a loop around 4 pegs so that removing any 3 peg frees the loop but removing 2 or fewer leaves it stuck.
- Place a loop around 2 red and 2 blue pegs so that removing 2 reds or 2 blues frees the loop but removing 1 red and 1 blue leaves it stuck.
- Place a loop around 2 red and 2 blue pegs so that removing 1 red or 2 blues frees the loop but removing only 1 blue leaves it stuck.
- Place a loop around 3 red and 3 blue pegs so that removing 1 red or 3 blues frees the loop but removing 2 or fewer blues leaves it stuck.
- Place a loop around 3 red and 3 blue pegs so that removing 1 red or 2 blues frees the loop but removing only 1 blue leaves it stuck.
- Place a loop around 2 red, 2 blue, and 2 yellow pegs so that removing 2 pegs of different colors frees the loop but removing 2 of the same color leaves it stuck.
These point to a general question of how to loop around \( n \) pegs so that the removal of any one peg frees the loop, how to loop around \( n \) pegs so that the removal of any two pegs frees it while the removal of any single peg leaves it stuck, and so on. Students can look for general algorithms for how to feed their loop around the pegs.
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The other sort of puzzle is perhaps the easier ask. Students are given a set of instructions showing how to feed the loop around the pegs and then are asked to determine the effect of removing certain pegs. This can be used as a primer before introducing the previous sorts of puzzles or as a stand-alone series of brain teasers. Consider the loop configuration below:
We can color code chunks of the loop so that each corresponds to roughly one orbit from the center of the two pegs around one of the two pegs and back to the center. (If you picture pushing the loop around so that the places where two colors meet are at the midpoint between the two pegs, it should be clear you can create exactly these orbits if you want to.) Note we've also put arrows on the loop to indicate directionality, but which of the two directions we chose was arbitrary:
This allows us to give instructions on how to wind the loop around each peg in sequence to create the full loop. I've started with the big dark blue loop on the far right so that later layers lie on top of previous layers up until that final connection of green back to blue:
I've put together some puzzle instruction sets, so check there for a fuller list. Here are a few examples one 2-4 pegs:
Does pulling the red peg or blue peg free the loop?
Does pulling the red peg or blue peg free the loop?
Does pulling the red, blue, or yellow peg free the loop? What combos of them would?
Does pulling the red, blue, yellow, or green peg free the loop? What combos of them would?
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Students will likely grasp the idea that counterclockwise around a given peg "undoes" clockwise around that peg and vice versa. However, doing some action around a second peg between these might somehow create a barrier to this "undoing." For example, in our four-step example on two pegs, the second and fourth steps want to cancel each other out, but the third step prevents this from happening. Removing the red peg effectively deletes steps 1 and 3, allowing steps 2 and 4 to cancel out.
The mathematical idea lurking behind this is the idea of homotopy -- in this case, continuously morphing a curve within a surface. The instructions in the last example almost give the game away, since they give us a straightforward reading of the loop as an ordered list of pegs together with an orientation (clockwise or counterclockwise). Let's label the colored pegs according to their initials \( R \), \( B \), \( Y \), and \( G \) (reading clockwise starting from the upper left). We'll label a step around peg \( X \) as \( x \) if the orientation is clockwise and \( x^{-1} \) if it's counterclockwise. In this language, our four-step example on two pegs can be represented as
\[ r b r b^{-1} \]
since we first travel clockwise around red, next clockwise around blue, then clockwise around red, and finally counterclockwise around blue. The removal of the red peg corresponds to deleting all \( r \) and \( r^{-1} \) symbols in this algebra, leaving
\[ b b^{-1} = 1, \]which we can interpret as a freed loop. On the other hand, removing the blue peg leaves
\[ r r = r^2, \]which we can interpret as the loop doubly-wound around the red peg.
This algebra can be useful for analyzing a loop wound around pegs, as we've seen above, assuming we can correctly interpret it step-by-step one peg at a time with orientation. On the other hand, playing with the symbols can help us create the sorts of scenarios asked for in the more advanced puzzles. For example, if we wish to come up with a scenario with two red pegs and two blue pegs where removing both red or both blue frees the loop but removing one of each color leaves it stuck, we could name these pegs \( R_1, R_2, B_1, B_2 \) and then play around with strings of the symbols in
\[ \{r_1, r_1^{-1}, r_2, r_2^{-1}, b_1, b_1^{-1}, b_2, b_2^{-1}\} \]trying to engineer one that cancels to \( 1 \) when we delete all of \( \{r_1, r_1^{-1}, r_2, r_2^{-1} \} \) or all of \( \{b_1, b_1^{-1}, b_2, b_2^{-1} \} \) but not when we delete all of \( \{r_i, r_i^{-1}, b_j, b_j^{-1} \} \) for any \( (i,j) \in \{1,2\}^2 \). This string would then translate to instructions for threading the loop. You can check that
\[ r_1 b_1 r_1^{-1} b_1^{-1} r_1 b_2 r_1^{-1} b_2^{-1} r_2 b_1 r_2^{-1} b_1^{-1} r_2 b_2 r_2^{-1} b_2^{-1} \]gets the job done, providing 16 steps for threading the loop around the four pegs.
Wrap-Up Questions
- Can you confidently predict whether removing a given peg will free the loop? What sorts of features of the loop arrangement make this prediction easier or harder?
- Once you've found an arrangement that meets a condition (e.g., pulling the red peg frees the loop but pulling the blue peg does not), are there ways to come up with a simpler arrangement that meets the same condition? Can you use a complicated arrangement to help find a simpler one or do you have to start from scratch?
- Can different instructions lead to the same loop arrangement?
Extensions
- Given a condition (e.g., pulling the red peg frees the loop but pulling the blue peg does not), what's the shortest list of instructions that can create a loop arrangement satisfying that condition? The Demaines et al explore this in a few cases, but there are some open questions relating to this problem.
- How do things change if we instead play this on the surface of a sphere? (For instance, putting your pegs in an orange.) Or the surface of a torus? (For instance, putting your pegs in a bagel.) Does our simple system of indicating a peg and then clockwise or counterclockwise still make sense in these cases, or do we have to specify other parameters to describe all possible instruction sets we might want?
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