Harmony of the Spheres 2: The Tinkering Begins

 

As soon as I saw that beautiful cornucopia of a graph that we talked about in the first Harmony of the Spheres problem, I wanted to tinker. What would the graph look like if it was only two simultaneous notes, or four? What if we split the octave into eleven or thirteen semitones?

There’s two major variables at play here: the number of different tones and the number of simultaneous voices. Let’s call them t and v; t for tones, and v for voices. Each combination of t and v should give us a different graph. Like before, each node is a possible set of notes; a sound the group could be making. Each edge is one voice moving up or down one note; if one combination of notes is being sung, and one person decides to move up one note, they’ve moved the sound from one node to the other, along an edge.

Ideally, we’d want to have a system in which we know what the graph looks like if we’re given t and v. So let’s try to find one. Let’s start way down low, at the basics, and build out this set of graphs. If t is one, it doesn’t matter what v is, because all of the voices have to sing the same pitch (the only pitch that exists). And the order of the voices doesn’t matter, so those graphs have exactly one node. Which is the most boring graph possible.

If v is one, then that voice can sing any of the tones in t. But, under our rules, the pitch doesn’t matter, either – what matters is the relationships between the tones. Therefore, these graphs will also only have one node.

Now, the first non-trivial case: 2 x 2. Two tones, two voices. Let’s start with both voices on the same tone. That’s our first node, and then… something interesting happens. Moving one voice up or down to find another node, no matter how we do it, seems to bring us to the same new node – the node in which the two voices are different.

A hard example could be helpful here. Imagine two people, Alice and Bob, taking a break from writing ciphers, and two notes, C and D. The way this musical scale works is that there’s only two notes in the octave, C and D. If you sing one note higher than D, it’s a C. One note down from C, it’s a D.

So we started with both Alice and Bob singing C. That’s our first node. Then, Bob decides to sing one higher while Alice stays on C, and by our rules, that’s an edge to another node. Now Alice is on C, and Bob is on D. New node. But now, no matter who changes their note, and in what direction, we’re going to end up back at the first node!

If Bob goes down, they’re both at C again. If Bob goes up, he goes up to C. If Alice goes up or down, she goes to D, the same note that Bob is on, and remember, we only care about the relationships between the notes, so D+D is the same as C+C! And if we try to go to a different node from the first node, we quickly find that C+D is the same as D+C!

I'll keep track of the ones we've done in a table, like so.

 

Not very interesting so far – we’ve just shown that if there’s two tones and two voices, they can sing either different notes or the same note, and that’s it. But as we add more voices and tones, things will get very complicated, fast!