Considering the strong correlation between my 20-month blog silence and the amount of time I've spent playing with baby toys in the interim, it seems appropriate to start back with a post about those baby toys.
To start with there are a whole slew of these broken colored rings lying around, presumably intended to be linked together in a chain.
I noticed that the oblong shape allows them to be assembled into Borromean rings.
Then there are all of these stereotypical wooden cubes.
It's probably pretty common to probe the limits of top-heavy structures. It occurred to me to try a single block with two blocks atop it, three blocks on the next layer, and so on as high as possible.
Each successive step seemed equally dubious.
After eight layers I ran out of the standard style of blocks and had to fall back on the alternate.
I tried adding an eleventh layer but everything started falling over at that point.
An unsuspecting friend recently reported that her five-year-old son is a fan of big numbers in general, and one number in particular. His favorite number is One Googol and One Hundred (or 10100 + 100), which I will here on out refer to as the number “Samuel” to keep my sentences to a reasonable length. He wanted to be able to count to Samuel in the same way that you might count to other numbers by skipping a regular amount, such as counting to 1000 by tens.
As I started thinking through the various options, I realized that the answer depended on the prime factorization of Samuel, since any number by which you could count to Samuel would have to be a divisor of Samuel. One obvious divisor of Samuel is 100. Counting to Samuel by 100’s is certainly not feasible, but counting to Samuel by “Samuel divided by one hundred”s (or 1098 + 1 or one hundred untrigintillion and one) is, and that was my first suggestion. But then I wondered if the full factorization of Samuel would give other options. Factoring 100-digit numbers isn’t something that I know how to get my computer to do, but fortunately Wolfram Alpha was up to the task.
The result revealed that there are actually quite a handful of options for counting to Samuel in a reasonable amount of time, so I thought I would present them here, just for the joy of looking at really big numbers written out in words. These are all the numbers you can count to Samuel by in fewer than a thousand steps, sorted by and labeled by number of steps:
Once again Algebraic Topology compelled me to write code. This time it was the Ham Sandwich Theorem, when I realized that it wouldn’t be too hard to demonstrate the two dimensional version of the theorem with another interactive doohickey.
The interpretation from which the Ham Sandwich Theorem gets its name says that if you have two slices of bread and a slice of ham between them, then no matter what their respective shapes or volumes or how they are positioned relative to each other, it’s always possible to make a single straight slice through the sandwich resulting in all three pieces being exactly bisected.
The Ham Sandwich Theorem applies to any N solids in N-dimensional space, so the 2D version says that any two sets of geometric shapes can be simultaneously bisected by a line. I said “two sets” rather than “two shapes” because the theorem actually doesn’t rely on the shapes even being connected. So we can talk about a collection of red and blue circles, as in the image below:
No matter how the circles are arranged, and no matter how many of them there are, there will always be at least one line that bisects both sets by area simultaneously. The interactive version lets you add, remove, and rearrange the circles.
I’ve been spending some of my free time the last week and a half watching Norman Wildberger’s Lectures on Algebraic Topology. One of the things I’ve learned is that an m-by-n rectangle with integer dimensions can be partitioned into 2×m×n triangles with area ½ and corners at integral points. This fact by itself is rather obvious, given the most regular partitioning shown below:
The part that surprised me is that the sides of the triangles can be arbitrarily long, allowing a variety of different shapes, and the whole thing still works out. Here’s another way to partition the 5-by-5 rectangle:
The fact that they all have the same area is a consequence of Pick’s Theorem. I wanted to see what sorts of partitionings you get by starting with a bare rectangle and adding lines at random. And that’s exactly where the image above came from.
Here are a couple more, just for fun.
I’ve been doing some academic research lately that involves computing the probability that a network that grows by adding links in a particular way will end up with certain properties. I went about this by generating all graphs up to a particular order, and so I used unlabeled graphs, which seemed good enough and meant I had far fewer graphs to deal with.
Eventually I decided that it would be more realistic to at least pretend that the computation was being done on labeled graphs (by making sure that I got the same answer either way). This led me to a discovery that wasn’t too surprising by itself, but I was surprised that I hadn’t thought of it earlier: there are (unlabeled) graphs for which adding two completely different edges will produce the same graph. For illustration I found the smallest example:
The first thing to notice about this graph is that it’s completely asymmetric: The pink and orange vertices have the distinction of being the only degree-4 and degree-2 vertices respectively. The blue vertex is distinguished as being the only vertex adjacent to both the pink and orange. The green vertex is the only degree-1 vertex adjacent to the pink, and the yellow and blue vertices are clearly distinct from each other as they have different degrees. So there is no symmetry in this graph whatsoever, which means that any edge we could conceivably add to it would be “different” from any other edge we could add.
So there are two different edges we can add to this graph and end up with graphs that are isomorphic to each other. Specifically, by adding either edge we will end up with this graph:
Another way to say it is that our original colored graph is a subgraph of this graph in two different ways. This can be easily illustrated by repositioning the vertices so they match the supergraph in two different ways:
You should be able to confirm visually that these two are the same as the original graph, and that they can each be transformed into the supergraph by adding a single edge, which is drawn dashed.
So this is the smallest pair of graphs that has this property, and in fact the only example on six vertices. I’m not sure how common they are. It is not a straightforward thing to google.
Another cool thing is that, apparently just by coincidence, the smaller graph and the larger graph are complements of each other. So if we take the original graph and reverse all the links like so:
we end up with the same graph that we got by adding an edge: