John: The time-scale point at which the subsequent note starts is the one that will be shifted.

Methodus: By subsequent note, do you mean the one that has the next higher number, or do you mean the one that follows the note that we have found, in the sequence of notes produced by the stream?

John: It would be the latter.

Methodus: OK. Suppose we have the sequence of notes (NNN) (NN) (NNNNN). Say the edge-face difference O is -4 (minus-four). Because O is negative, we would number the notes beginning with the end of the sequence. We would get (65-) (4-) (3210-)......The absolute value of O, modulo N minus F would be 4 modulo 10 minus 3, or 4. The note for which the number is 4 would be the first note of the second current face. So in this case, are you saying that the time-scale point at which the second note of the second current face starts would be the one that would be shifted?

John: Yes. That's right.

Methodus: In what direction would it be shifted?

John: If O is positive, this time-scale point will be a bit late. If O is negative, it will be a bit early.

Methodus: All right, I think I understand that......Why did you skip the last note of each face?

John: I didn't want to shift the time-scale point at which the first note of a face starts.

Methodus: Why?

John: The effect of shifting a time-scale point is to shorten one note and lengthen the subsequent note. Time is borrowed from one note and given to another...as you would do in a swing rhythm.

Methodus: Yeah...but why didn't you want to do this with a pair of notes that are from different faces?

John: For the textures melodic-each-face and homophonic, the instrument that plays the last note of a given current face will be different from that which plays the first note of the subsequent current face. I think I felt that the effect would be difficult to execute and perceive if the pair of notes...

Interruptus: What would be the structure of a geometric transition from one tempo to another?

John: Well...it's similar to what I said earlier about a sinusoidal transition.

Methodus: How so?

John: Again, you can think of the time scale for a given phrase as being composed of a sequence of silent 1-notes. Suppose there are L 1-notes shown in the score for a phrase. The first of these would start at time-scale point 0 (zero). The last would start at time-scale point L-1 (l minus one).

Methodus: Could we discuss this in terms of a concrete example? I mean, let's suppose this is going to be a decelerando from an initial tempo of 144 to a final tempo of 52.

John: OK. In that case, the initial tempo would be 144 times 4 1-notes per minute; or 1 over 576 (five-seventy-six), times 60 1-notes per second. That would be roughly 0.104 seconds per 1-note......The final tempo would be roughly 0.288 seconds per 1-note.

Methodus: Right. Now you said this would be like a sinusoidal transition. So would the duration of the 1-note that begins at time-scale point 0 be 0.288 seconds?

John: Yes. And the same may be said of all the 1-notes that precede this 1-note.

Methodus: What about at the end of the phrase? Would the duration of the 1-note that begins at time-scale point L be 0.288 seconds? And would the same be true of all subsequent 1-notes?

John: Yes, and yes. The 1-notes for which there is a difference between the sinusoidal case and the geometric case would be those that begin at time-scale points 1 through L-1.

Methodus: How would those be calculated?

John: It is very much like an equal tempered scale of 1-note durations from the initial duration to the final duration.

Methodus: How's that?

John: Well...what you would do is calculate the ratio between the duration of the 1-note that begins at the time-scale point L and that of the 1-note that begins at time-scale point 0.

Methodus: So, in this case we would have 0.288 divided by 0.104, or about 2.769.

John: That's right. Now, in this equal tempered scale of durations there must be L steps to get from the initial duration to the final duration.

Methodus: OK. So to get the duration for the 1-note that begins at time-scale point k, would you multiply the duration of the 1-note that begins at time-scale point 0...by the kth power of the L-th root of 2.769?

John: Exactly.

Methodus: All right. So...let's see. Let's say L is 9. The 9th root of 2.769 is about 1.120.

John: Right. In that case, the duration of the 1-note that begins at time-scale point 1 would be 0.104 times 1.120, or 0.116.

Methodus: Actually, I think it would be 0.117 if you had used the exact values instead of 1.104 and 1.120.

John: Oh, OK. Yes, we should do that......The durations for the subsequent 1-notes would be: 0.131, 0.146, 0.164, 0.183, 0.205, 0.230, 0.258 and 0.288.

Methodus: How does this type of transition sound?

John: To my ears, it is relatively nondescript. The change in duration with each successive 1-note is very predictable. Each 1-note seems to start exactly when one would expect it. To some degree, it is like a change in tempo that is not even occurring. It could almost go unnoticed.

Methodus: How does that compare with a sinusoidal transition?

John: To me, a sinusoidal transition has a whip-like or wave-like feel to it. The 1-notes occur at a time that is unexpected. For example in an accelerando, the 1-notes after the first few seem to occur earlier than one would predict. Similarly, 1-notes near the end of the transition seem to occur too late. As a result, this type of transition conveys the fact that an accelerando is occurring more clearly than a geometric transition.

Methodus: Have you experimented with any other type of transition?

John: Yes. I considered using a cubic function. I mean a cubic polynomial for which the slope is 0 (zero) at the initial and final durations. That type of function is mentioned in David Epstein's book Shaping Time.

Methodus: How does that sound?

John: It yields almost the same result as the sinusoidal transition that I have used. The wave-like effect might be a bit more pronounced with the sinusoidal function. At least, I think the relative shapes of these two functions would suggest that this would be the case.

Methodus: Have you tried any other functions?

John: Yes, I tried a linear combination of a geometric function with a cubic function.

Methodus: What type of linear combination?

John: Something like x times the geometric function plus the quantity one minus x, times the cubic function, where x is some real number in the interval 0 to 1.

Methodus: How did that come out?

John: I listened to it. But I didn't care for it. It sounded a bit confused and too unpredictable for my...

Interruptus: What do you think is the difference between music and mathematics?

John: ...I think there might be more similarities than differences.

Methodus: What are the similarities?

John: First, I believe that anything can declare that it is a mathematician just as anything can declare that it is a composer.

Methodus: Any thing? Don't you mean any body?

John: No. I wouldn't want to say that only humans can be mathematicians and composers.

Methodus: OK...So you're saying that a declaration is sufficient to make a given thing be a mathematician or composer?

John: In my mind, it's necessary and sufficient. Also, I believe that a given mathematician A can declare that something is a mathematical work by A. Similarly, I believe that a given composer A can declare that something is a work of art by A that is musical.

Methodus: So here again you're saying that you think that a declaration is sufficient to make a thing be a work of mathematics or a work of music?

John: Yes...again both necessary and sufficient.

Methodus: So then any given thing could be a mathematician or composer and any given thing could be a work of mathematics or a work of music?

John: Yes, I believe that...provided the sufficient conditions are met.

Methodus: Well...I guess on that level, mathematicians are similar to composers, and mathematics is similar to music. But wouldn't you agree that the things that composers declare to be music seem to be quite different from the things that mathematicians declare to be mathematics?

John: Yes......I think one fundamental difference might be that mathematicians try to be correct, while composers try to be interesting. I guess now I am speaking only about my own work as a mathematician or composer. I mean, I believe that most mathematicians try to be correct. But I'm not sure if the goal of most composers is to be interesting.

Methodus: Well...OK. Then let's focus on your work. When you are doing mathematics, you say you are trying to be correct. What do you mean by that?

John: I mean, I am trying to find the truth-value about something; some proposition.

Methodus: And when you are acting as a composer you are trying to be interesting?

John: Yes.........but maybe I am trying to seek truth then as well.

Methodus: What do you mean? Are you saying that you think a composition is a proposition? How can a composition be either true or false?

John: Well...as I said earlier, I use living systems as a model for my work. Or you might say that it is the other way around. I mean, you could think of Dodecahedron as a model of living systems. Perhaps it is a primitive model of the Universe......Maybe by writing compositions in this way, I am seeking the truth about the Universe.

Methodus: ...Are you?

John: Yes, I think so......there's another thing that's part of that. I am seeking the truth about me.

Methodus: How so?

John: I am trying to write music that is interesting to me. In the process, I am discovering what is interesting to...

Interruptus: Of the 7800 stream types on a dodecahedron, I wonder how many cover a distinct set of edges.

John: For this, would you consider the stream type abcdefmng to be equivalent to agnmfedcb?

Interruptus: Yes, two stream types would be equivalent if the set of edges that are contained in one is the same as those that are contained in the other, regardless of the order or direction in which the edges are traversed.

John: So, you are asking how many equivalence classes there would be?

Interruptus: Yes.

John: I don't know.

Methodus: Do you know of what conditions must be satisfied in order for two stream types to be equivalent in this way?

John: No, I haven't given that much thought...maybe we should look at a few more examples.

Methodus: Here's another one that I mentioned earlier: abhpog and agophb.

John: Yes......maybe it has something to do with reflection. Suppose the reflection of a given stream type covers the same edges as the original.

Methodus: By reflection, you mean you replace all left turns with right turns, and vice versa.

John: Exactly. Two stream types that are related in this way that cover the same set of edges would be in the same equivalence class. I wonder if this would be a necessary condition.

Methodus: ......No, I don't think so. Here's a counterexample: abhqric and abcirqh. These cover the same edges. So they would be in the same equivalence class. But one is not the reflection of the other.

John: Right......I wonder if every equivalence class has exactly two members.

Methodus: I think you might be onto something there.

John: Yeah...let's say I have a given stream type S. From this, I could find a second stream type E that would be equivalent to S. At the vertex at which S ends, I would take the other path to form E.

Methodus: Right. For example, suppose we have the stream type abcistjeflk. The last edge k ends at the vertex kjt. So, when we get to this vertex the first time, we could have traversed edge k instead of j.

John: Yeah. That would give us abcistklfej. From that I think we would be able to show that every equivalence class must have at least two members.

Methodus: Is it possible to have more than two members in one of these equivalence classes?

John: ...I don't think so. Suppose you give me a stream type; say abcdjtsi. To construct another stream type that covers exactly these edges, I must begin at edge a. The next edge must be b or g. Since g is not in the given stream type, I must choose b. Next I come to the vertex bch. I can't traverse h because it is not in the given stream type. So I have to traverse c. So far I have edges abc.

Methodus: Right. Next you would come to the vertex cdi. At that point, you could traverse edge d or i. If you take edge d, you would ultimately regenerate the given stream type. On the other hand, if you traverse edge i, you would eventually get the stream type abcistjd.

John: In general, I think the only vertex at which you will have a choice will be at the last vertex in the given stream. All your other decisions would be forced.

Methodus: So there would be only two members in each equivalence class. And they would both end at the same vertex.

John: Yeah. To get one from the other, you could partition the stream type into two segments. The first segment would be the sequence of edges that begins at edge a and ends when we encounter the last vertex for the first time. Then the second segment would be a loop. For the equivalent stream types, the first segment would be the same. The second segment would be reversed. I mean we would traverse the loop in the opposite direction.

Methodus: Yeah, I think that's right. For example, suppose you gave me the stream type of length 21 that we talked about before. That is, agnmfedchpw-+uktsrx/z. This stream ends on edge z. Let's see...the last vertex must be either zu+ or zy/......It never traverses edge y, so the last vertex must be zu+. So I would split the stream into two segments at the vertex between edges + and u. That would give me agnmfedchpw-+ and uktsrx/z. Then I would reverse the order of the second segment to get z/xrstku. If I put this together with the first segment I get agnmfedchpw-+z/xrstku.

John: Well...it looks like the set of 7800 stream types can be partitioned into 3900 (thirty-nine-hundred) equivalence classes, where each class has exactly two elements. And the two stream types in each class cover the exact same edges.

Methodus: What if we did this for streams instead of stream types? I mean what if we allowed the first directed edge to be...

Interruptus: In Dodecahedron, for the characteristic you call exaggerate-frequency-range, is it the edge-face difference O that determines which frequency range should be exaggerated?

John: Yes, if O is greater than or equal to 4, the high overtones will be exaggerated. If it is less than or equal to -4, the fundamental will be exaggerated. Otherwise, the middle overtones will be exaggerated.

Methodus: Why did you partition the range of possible values for O in this way?

John: Well...for larger values of O, you would exaggerate a higher range of frequencies.

Methodus: No, I mean why did you set the breakpoints at 4 and -4?

John: Oh. I did it that way because then there are an equal number of directed edges in each range. There are 20 directed edges for which the face difference is greater than or equal to 4, 20 directed edges with a face difference that is less than -4, and 20 for which the face difference is less than 4 and greater than...

Interruptus: Who is your favorite painter?

John: ......I like Pollock.

Methodus: Do you have any [knock knock knock knock] other [knock knock knock knock knock knock]...

John: Hello. Who is it?

Complementum: I am a professional composer and professor of composition. I overheard what you were saying. I have a few thoughts that I would like to share with you.

John: Come on in.

Complementum: I had a chance to hear some of this stuff that you have written that you are calling music.

John: Yes.

Complementum: Well...it's no surprise to me that you list Pollock as one of your favorite painters. Your approach to composing music is just as ridiculous as his approach to painting. And in both cases the result is a mess.

John: Oh. Are you saying that you don't hear any organization when you listen to my music?

Complementum: Well, I know there is all this mathematics crap at the heart of it. But I don't think it comes across. To me this music is unintelligible......I can see you're entertained by all these mathematical ideas. But you can't expect to automatically convert this mathematics into music. It just doesn't work that way. There is much more to music...There is much more to composing than that. You have to consider the performers, the conductor and the audience.

John: I think...

Complementum: I know all about what you think. I have been in the next room listening to you drone on about this stuff endlessly......Oh my God. You're writing all this stuff down. What, are planning to publish this interview?

John: Well...yes. Initially, I was thinking it might be an interview. Then I thought it might be a novel. Now, I am thinking of it as a play.

Complementum: A play? Are you serious? You mean you actually think that someone would perform this?

John: ...

Complementum: Look, I gotta go.

John: OK. Bye.

Methodus: Were you really thinking of publishing this as a play?

John: Yes.

Methodus: Would there be a different actor for each part?

John: I'm not sure. Maybe it would be best to leave that up to the director.

Methodus: Maybe your lines could be spoken by an actor on stage. And all the other parts could be spoken by actors who are offstage.

John: Yeah......The offstage actors could be mic'ed. There could be speakers positioned in various places for each of those characters.

Methodus: Where would the different speakers be positioned?

John: If I were doing it, I might put your speaker stage right. Interruptus's speaker could be stage left.

Methodus: What about Complementum?

John: That one could be behind the audience.

Methodus: Would they all be male voices?

John: Yes, I think so. That's the way I hear...

Interruptus: In Dodecahedron, for a stream for which the characteristic is staccato, which particular notes would be shortened?

John: There are four different patterns of staccato......Let's say you have the sequence of main notes that are produced by a given stream. As I said earlier, you can partition this sequence into subsequences in such a way that all notes in a given subsequence will have the same current face. For one of the staccato patterns, the last note of each face would be shortened.

Methodus: I think it would help me to look at a specific example.

John: OK. Let's say we have a stream for which the edges are 71 (seven one), 51 (five one), 91, 93, 63, 83, 8B, 84, C4, A4, 74 and B4.

Methodus: All right. For that, the first current face would be 1.

John: Right. Then the second current face would be 9. After that the stream would visit faces 3, 8 and 4...Now, the first two edges are simple edges, so each of these would produce one note.

Methodus: OK. Then we have edge 91. That's a pivot edge so it would give us two notes. The first of these would be grouped with the first two notes of the stream. The second one would be part of the second subsequence.

John: That's right. The same thing happens for edge 93. That's a pivot edge too.

Methodus: All right. So far we have (NNN) (NN) (N.

John: 63 (six three) is a simple edge, so that gives you another N at the end. So you would have (NNN) (NN) (NN.

Methodus: 83 (eight three) is a pivot edge. So that would give us (NNN) (NN) (NNN) (N.

John: Right. 84 (eight four) is a pivot edge too. So that gives us (NNN) (NN) (NNN) (NN) (N.

Methodus: Wait...you skipped 8B. It should be (NNN) (NN) (NNN) (NNN) (N.

John: You're right.

Methodus: Now, the last four edges are simple. The stream ends by going completely around face number 4. So we get (NNN) (NN) (NNN) (NNN) (NNNNN).

John: Yes. I think that's correct.

Methodus: So for this staccato pattern, where the last note of each face is shortened...would that be (NNn) (Nn) (NNn) (NNn) (NNNNn) (n n little-n, n little-n, n n little-n, n n little-n, n n n n little-n)? I'm using lowercase n for the notes that are staccato. Is this right?

John: Yes. But, I would probably write it this way:
||. |. ||. ||. ||||. (bar-bar-dot bar-dot bar-bar-dot bar-bar-dot bar-bar-bar-bar-dot)
Here I have used a vertical line for notes of normal duration and a period for notes that would be shorter than normal.

Methodus: OK......What would the other three staccato patterns look like?

John: Well...let's call the first pattern last-note-each-face. The other patterns would be last-note-first-last-face, last-notes-first-last-face and last-notes-each-face.

Methodus: For this example, would last-note-first-last-face be:
||. || ||| ||| ||||.

John: Exactly. And last-notes-first-last-face would be:
|.. || ||| ||| |....

Methodus: ...So last-notes-each-face would be:
|.. |. |.. |.. |....

Interruptus: It looks like Morse code. Why not use a hyphen for the longer notes?

John: Well...we've got three possible durations instead of 2. There's shorter than normal for staccato, normal, and longer than normal for tenuto. I would rather use a hyphen for tenuto.

Methodus: What about tenuto? Do you have patterns for that as well?

John: Yes. There are four patterns for tenuto. You've got first-note-each-face, first-notes-each-face, not-last-note-first-last-face and not-last-notes-first-last-face.

Methodus: For this example, would first-note-each-face look like this:
-|| -| -|| -|| -|||| (dash-bar-bar dash-bar dash-bar-bar dash-bar-bar dash-bar-bar-bar-bar)

John: Exactly.

Methodus: So first-notes-each-face would be:
--| -| --| --| ----|

John: That's right.

Methodus: What would not-last-note-first-last-face be?

John: For that one, you would have:
--| -- --- --- ----|

Methodus: So, would not-last-notes-first-last-face be:
-|| -- --- --- -||||

John: Yes. That's correct.

Methodus: ......For a given stream, what determines which pattern would be used for staccato and tenuto?

John: The edge-face difference O.

Methodus: You mean, different patterns would be used for different values of O?

John: Yes. For example, if O is 10 or -2, 9 or -3, or 8 or -4, the staccato pattern would be last-note-first-last-face. For a stream for which the characteristic is tenuto, the pattern would be first-note-each-face.

Methodus: So for this you have grouped the possible values of O in pairs?

John: Yes, I used the same pairs here as I did for dynamic level, tempo and register.

Methodus: What about the other pairs?

John: If O is 4 or -8, 3 or -9, or 2 or -10, the staccato pattern would be last-note-each-face. For a tenuto stream, the pattern would be not-last-notes-first-last-face. If O is 7 or -5, or 5 or -7, the staccato pattern would be last-notes-first-last-face. For tenuto it would be first-notes-each-face. Otherwise, if O is 6 or -6 the staccato pattern would be last-notes-each-face. For tenuto it would be not-last-note-first-last-face.

Methodus: ......These are the patterns that would be used for main notes. What about neighbor notes? Would they be affected as well?

John: Yes in the homophonic texture, for a given simple edge, the neighbor note would be affected in the same way as the main note. So, if the main note is to be shorter than normal, the neighbor note will be shortened as well. Similarly, if the main note is to be longer than...

Interruptus: Maybe this discussion that we are having could be a screenplay.

John: Yeah, I could see it being used that way. I could imagine this as a film.

Methodus: It could be like the stage version, but you would be on film instead of on stage.

John: Right. And there could be quadraphonic sound.

Methodus: Yeah, there could be one speaker for your voice, one for me, one for Interruptus, and one for Complementum.

John: Maybe the voices should be all the same.

Methodus: What do you mean?

John: Maybe all the lines should be spoken by one man.

Methodus: Wouldn't that get confusing?

John: Well...you could differentiate the different characters by spatial position.

Methodus: Oh right, your speaker would be at center stage, mine would be stage right, Interruptus's would be stage left, and Complementum would be behind the audience.

John: Yes...But we could also have the voices be processed in some way so it would be easier to tell one from another.

Methodus: You mean some type of filtering?

John: Yeah...One could be bassier. Another could have more high end. Or different signal processing effects like reverb or chorus could be applied to each voice......Maybe each character could speak at a different rate of...

Interruptus: For these sinusoidal and geometric transitions that you have used in Dodecahedron, where you have the sequence of durations determined by some function...would you have gotten the same result if you had applied these functions to tempo instead of durations?

John: By that, do you mean the tempo would be varying continuously?

Interruptus: Yes. For example, over the duration of the phrase the tempo would vary continuously like a sine wave function.

John: ......I don't think that would yield the same result.

Methodus: Maybe we should look at an example.

John: Good idea. Let's look at a linear function.

Methodus: But you didn't use a linear function.

John: I know...But it's a simpler function. If we study this case maybe it will give us some insight into the more complex functions that I used.

Methodus: OK......Here's an example. Suppose this is to be an accelerando from 58 4-notes per minute to 100 (one-hundred) 4-notes per minute.

John: All right. Let's set this up like the sinusoidal or geometric transitions that I used. First you would calculate the duration of the first 1-note. That would be one over 4 times 58, times 60 seconds per 1-note. That's roughly 0.259 seconds.

Methodus: OK. Then we would calculate the duration of the 1-note that follows the last 1-note given in the score for this particular phrase. That would be one over 4 times 100, times 60 seconds per 1-note. That's exactly 0.150......What about the duration for the other 1-notes?

John: That would depend on how many there are. Let's suppose there are two 1-notes given in the score for this phrase.

Methodus: So the duration of the 1-note that starts at time-scale point 2 would be 0.150 seconds.

John: Right. And the duration of the 1-note that starts at time-scale point 0 would be 0.259 seconds.

Methodus: How would we calculate the duration for the 1-note that starts at time-scale point 1?

John: Well...this is to be a linear transition. So the duration of that 1-note should be halfway between 0.150 and 0.259. That would be 0.2045...The exact value for the duration of the 1-note that starts at time-scale point 2 is more like 0.258621. If we use that instead, the midpoint would be 0.20431. Or roughly 0.204.

Methodus: OK. Now, what would we get if we allowed the tempo to vary continuously in a linear manner from the initial tempo to the final tempo?

John: In that case, you could draw a graph that is a straight line. The independent variable of the graph would be time in seconds. Time zero would be the time at which time-scale point 0 occurs. The dependent variable would be tempo. That would be 4 times 58 1-notes per second at time 0. Then at the time at which time-scale point 2 occurs the tempo would be 4 times 100 1-notes per second.

Methodus: Wait...Shouldn't that be minutes instead of seconds?

John: Oh...right. Did I say seconds? It should be minutes.

Methodus: ...All right. So the graph of tempo over time would be a straight line from 232 (two-hundred-thirty-two) 1-notes per minute to 400 1-notes per minute......When would time-scale point 2 occur?

John: That's what we need to find out. Let's call that time T......This is like what we do in physics where we have distance equals velocity times time.

Methodus: Right. Tempo is like velocity. I guess here distance would be measured in units of 1-notes...But for this case the velocity or tempo is varying continuously.

John: That's right. So we're going to have to find the area under the curve.

Methodus: Oh...right. Distance would be the integral of velocity over time.

John: Or in this case, the number of 1-notes would be the integral of tempo over time.

Methodus: So the area under this linear function would be the number of 1-notes that occur in time T.

John: Let's figure out what this area would be......You can break this up into two parts. There's a triangle sitting on top of a rectangle.

Methodus: OK. The area of the rectangle would be 232 (two-thirty-two) times T. The area of the triangle would be one-half the base which is T, times the height which is 400 minus 232. That would be 84 times T. So the total area would be 316 (three-hundred-sixteen) times T.

John: I think that's correct. The units are 1-notes. So it would be 316 (three-hundred-sixteen) times T 1-notes.

Methodus: All right. But how do we determine the value of T?

John: Well...we said there should be 2 1-notes in the score for this phrase. So 316 (three-sixteen) times T 1-notes must equal 2 1-notes. T must be 2 over 316 seconds...I mean minutes. The units of the time axis has to be minutes.

Methodus: So if we convert to seconds we would get a value of 120 (one-twenty) over 316 (three-sixteen) seconds. That's about 0.380 seconds......So that's the time at which time-scale point 2 must occur...How does that compare with the other approach?

John: For that we would have to add the duration of the 1-note that starts at time-scale point 0 with that of the 1-note that starts at time-scale point 1. That would be 0.259 plus 0.204, or 0.463 seconds.

Methodus: So with the first approach time-scale point 2 would occur later...What about time-scale point 1? With the first approach, that would occur at 0.259 seconds. When would it occur with the second approach?

John: Well......for that you would have to find the time x at which the area under the curve from time 0 through time x equals the area under the curve from time x through time T.

Methodus: Why is that?

John: Because, then each of these two areas would equal one 1-note.

Methodus: Oh...OK. Let's call these areas A₀ (a zero) and A₁ (a one). Say L is the tempo at time x. Then A₀ would be 232 (two-thirty-two) times x, plus one-half of x times the quantity L minus 232. A₁ would be L times the quantity T minus x, plus...one-half the quantity T minus x, times 400 minus L.

John: All right. And L would be 232, plus x times the quantity 400 minus 232, divided by T. I think if we set A₀ equal to A₁ we're going to get a quadratic polynomial in x.

Methodus: We could use the quadratic formula.

John: I know. But this algebra is getting a bit messy. Let's solve this iteratively. If I leave it in terms of x and L, I get the equation 316 times x, plus x times L, minus T times L over 2, minus 200 times T, equals 0.

Methodus: Let me set that up in a spreadsheet.

John: OK. Let's try an initial guess of x equal to half of T.

Methodus: All right. That would be about 0.190...That gives us 316.056 (three-sixteen-point-oh-five-six) for L and the quadratic polynomial would be roughly -16.

John: How about 0.259? That's what we got using the other approach.

Methodus: For that I get L equal to 346.582. The value of the quadratic is about 30.

John: Oh...that's way too high. Let's try 0.220.

Methodus: That gives us 3.49.

John: We're getting closer. How about 0.215?

Methodus: For that I get 0.210. Still too high.

John: Yeah, but it's pretty close. How about 0.214?

Methodus: That gives us -0.444 (negative-point-four-four-four).

John: Well...that's close enough. The answer must be somewhere between 0.214 and 0.215. What do you get for 0.2145?

Methodus: That gives us -0.117 (negative-point-one-one-seven).

John: OK, so it's closer to 0.215.

Methodus: All right. That means that for this approach where we're using a linear tempo function, the time-scale points 0, 1 and 2 would occur at 0 seconds, 0.215 seconds, and 0.380 seconds. For the other approach for which the sequence of durations varies linearly, the times would be 0 seconds, 0.259 seconds and 0.463 seconds.

John: Or, you could say it this way. When we use a linear tempo function we get the following sequence of durations for the first three 1-notes: 0.215 seconds, 0.165 seconds and 0.150 seconds. For the approach that I used, where the durations vary linearly, the durations of the first three 1-notes would be 0.259, 0.204 and 0.150.

Methodus: ...So when a linear tempo function is used the transition happens more quickly. I mean the durations in that case are shorter.

John: Yeah...well that makes sense. For that case the tempo continues to increase during the duration of the first 1-note. But for the other case, the tempo stays constant at 58 4-notes per minute throughout the first 1-note. I mean, that's why the duration of the first 1-note is 0.259 seconds.

Methodus: Then what happens to the tempo?

John: Well...I think it increases in a stepwise way, rather than a continuous way. I think when you hear the end of the second 1-note, which is the time at which time-scale point 2 occurs, that is the first cue that you have that the tempo has changed. At that point in time the tempo changes abruptly to something else.

Methodus: What would be the new tempo?

John: The duration of the second 1-note is 0.204 seconds. So that would be one 4-note per 4 times 0.204 seconds, or 0.817 seconds if we use the exact duration for this 1-note. That would be roughly 73.418 4-notes per minute.

Methodus: ...We did all of this for a linear function. What do you think we would get if we did this for a geometric or sinusoidal transition?

John: I think the results would be similar. I mean I think you would see a difference between these two approaches with those types of transitions as well.

Methodus: I think you may be right......Why did you decide to apply these functions to the sequences of durations rather than to the tempo?

John: Actually, I don't recall considering the alternative. I guess it seemed like a better idea to work with modifying durations instead of tempo. Based on my experiments with geometric transitions, I think we might be able to perceive the ratio between two consecutive durations of different length. I constructed various sequences of 1-notes for which the durations were progressively shorter or progressively longer, and the ratio between the duration of any two consecutive notes was constant throughout the sequence. As I said earlier, I found such sequences to be extremely uniform and predictable.

Methodus: As you were listening to a sequence of this type, do you think you were estimating this ratio and using it to make a prediction about the time at which the next note would start?

John: Yes. And I think that may be why the sinusoidal transition produces a whip-like effect. In that case the ratio between the duration of successive 1-notes is not constant throughout the sequence. The ratios become progressively larger until the inflection point of the sine wave. After that they become progressively smaller.

Methodus: ......I think there's as a disadvantage to the approach that you have used to implement transitions in tempo.

John: What's that?

Methodus: You can't have an accelerando that goes from a tempo of 0 to 100 (a-hundred).

John: You mean from zero 4-notes per minute to 100 (a-hundred) 4-notes per minute?

Methodus: Yes. Then the duration of the first 1-note would be infinite. Wouldn't it?

John: You're absolutely...

Interruptus: What do you mean by strongly sharp? I mean, how sharp is strongly sharp?

John: Suppose the frequency of a given named pitch is f. If that pitch is to be strongly sharp, the frequency should be f times the eighteenth root of 2. That's how I have defined it for Dodecahedron.

Methodus: What do you mean by a named pitch?

John: That would be one of the frequencies for which a name has been assigned.

Methodus: By name, do you mean for example O4, P3, Z5?

John: Yes.

Methodus: Earlier you said that X4 is the name of the pitch that we call concert A in a traditional pitch-naming convention. Would the frequency of X4 be 440 (four-forty) hertz?

John: Yes, that would be the normal frequency of X4, provided that is the tuning standard that you have chosen to follow. I mean if you choose to base your tuning system on 432 (four-hundred-thirty-two) hertz instead, then that would be the normal frequency of X4.

Methodus: What would slightly sharp be?

John: For that you would multiply the normal frequency by the thirty-sixth root of 2.

Methodus: Oh...I see. In effect, you've got a 36-tone equal tempered scale.

John: Yes, you could think of it that way.

Methodus: What about slightly flat and strongly flat? Would you use the same ratios there as well?

John: Yes, but for that you would divide by these factors. For example, if f is the normal frequency of X4, then the frequency of X4-slightly-flat would be f divided by the thirty-sixth root of 2.

Methodus: And for strongly flat you would divide by the eighteenth root of 2?

John: Right.

Methodus: ......For a given stream for which the characteristic is sharp, what determines whether it is slightly sharp or strongly sharp?

John: That depends on the edge-face difference O. If O is 10 or -2, 9 or -3, or 8 or -4, then it will be strongly sharp. Otherwise, it will be slightly sharp.

Methodus: What about for a stream for which the characteristic is flat? Would O be used in that case as well?

John: Yes. In that case, if O is 10 or -2, 9 or -3, or 8 or -4, then it will be strongly flat. Otherwise, it will be slightly flat.

Methodus: Is there any particular reason why you partitioned the values of O in this way? I mean why did you select these particular values for O to be mapped to strongly sharp and strongly flat?

John: Well...I listed the pairs in the following order: 10, -2; 9, -3; 8, -4; 7, -5; 6, -6; 5, -7; 4, -8; 3, -9; and 2, -10. For each of these pairs, I determined the number of edges for which the face difference is equal to one of the numbers in the pair.

Methodus: You mean, like what you did when you were trying to determine how to map edge-face differences to dynamic levels.

John: Exactly. So I got the sequence 6, 6, 10, 4, 8, 4, 10, 6 and 6. That's 6 edges for which the face difference is 10 or -2, 6 edges for 9 or -3, 10 edges for 8 or -4, and so on. I used the golden ratio to determine where to partition this sequence.

Methodus: What do you mean?

John: If you add all the numbers in this sequence you get 60.

Methodus: That makes sense. It's just the number of directed edges.

John: If you add the first three terms you get 6 plus 6 plus 10, or 22. The sum of the last six terms is 38. 38 divided by 60 is roughly 0.633. That is fairly close to the inverse of the golden ratio.

Methodus: Oh, I see...Did you do something similar for streams for which the characteristic is sharp-or-flat-to-natural or natural-to-sharp-or-flat?

John: Yes, for those streams the value of O determines the amount by which the pitch should be sharp or flat at the beginning or end of the phrase.

Methodus: Did you use the same mapping as for sharp and flat streams?

John: No. When the characteristic is sharp-or-flat-to-natural, the base pitch of the tuning system can be either a bit sharp or a bit flat, initially. So the range of the mapping has to include both sharp and flat possibilities.

Methodus: So what mapping did you use for these streams?

John: If O is 10, 9, 8 or 7, the pitch will be strongly sharp...I mean, if the characteristic is sharp-or-flat-to-natural, initially the pitch will be strongly sharp. On the other hand, if the characteristic is natural-to-sharp-or-flat, the pitch will be strongly sharp at the end of the phrase. If O is 6, 5, 4, 3 or 2, it will be slightly sharp. The negative values of O are used for slightly flat and strongly flat. If O is -10 (minus-ten), -9, -8 or -7, the pitch will be strongly flat. If O is -6, -5, -4, -3 or -2, it will be slightly flat.

Methodus: Did you use the golden ratio here as well?

John: No. Not in this case.

Methodus: Then what was your rationale for partitioning the possible values of O in this way?

John: The larger the magnitude of O, the stronger the effect. Also, there is symmetry here. You end up with 8 directed edges for which the face difference maps to strongly sharp and 8 for strongly flat. There are 22 for slightly sharp and 22 for slightly flat......I can't say for sure why I partitioned it as 8 and 22. Since the effect of strongly sharp is twice that of slightly sharp, maybe I way trying to have there be roughly half as many edges for strongly sharp as for slightly sharp. But I'm just guessing here...It's not clear from the notes that I wrote when I was designing this piece.

Methodus: What about the streams for which the characteristic is pitch-bend-up-or-down? Would this mapping be used to determine the direction and amount by which the pitch of a particular note is bent?

John: No...I think I may have mentioned this before. The characteristic will be pitch-bend-up-or-down if the stream type ends on edge p or k. The direction of the bend would be determined by the direction by which the edge p or k is traversed.

Methodus: Oh...right. I remember. You said that if the stream goes from left to right along this last edge, the bend will be upward, or sharp. Otherwise the bend will be flat.

John: That's right.

Methodus: What about the degree of the bend? I mean what determines whether it is slight or strong?

John: For that, the value of O would be used. In this case, the same mapping would be used as for streams for which the characteristic is sharp or flat. I mean, if the bend is to be upward, then by the end of the duration of the note, the pitch will be strongly sharp if O is 10 or -2, 9 or -3, or 8 or -4. For all the other possible values of O, it will be slightly sharp. If the bend is to be downward, it will be strongly flat if O is 10 or -2, 9 or -3, or 8 or -4. Otherwise, it will be slightly...

Interruptus: What do you think time is?

John: I think it is an idea.

Methodus: You don't think it is a concrete thing?

John: No. I think it is a concept that we use to help us predict the future.

Methodus: Do you think it is based on any physical things?

John: Yes, I think it's based on our observation of a sequence of two or more similar events.

Methodus: You mean like the sun rising?

John: Yes. We see the sun rise once. Then we see the sun rise again.

Methodus: ...And from that comes the concept of a day?

John: Yes, I think that's how it works. I mean, I don't think that there is any concrete thing called a day. I think a day is just an idea.

Methodus: Isn't a day the amount of time between two sunrise events?

John: No. I think it is a name that we use for the sequence of sunrise events.

Methodus: How would that be different from a year? Isn't a year related to this sequence of sunrise events? Isn't it just some number of sunrise events?

John: No, I think we observe the azimuth of the sunrise. Then we observe the azimuth of the next sunrise. We notice that it is different. With each successive sunrise the azimuth moves to the right, then to the left, then back to the right.

Methodus: The motion of the azimuth is like that of a pendulum.

John: Right. Now, there will be a particular sunrise that is at our extreme left. By that I mean, the azimuth of the sunrises that precede and follow this particular extreme sunrise are more to our right. Then after many sunrise events, there will be another sunrise that is at our extreme left.

Methodus: ...And you would say that our name for this sequence of extreme sunrises is a year?

John: Exactly. More specifically, the name would be an Earth year.

Methodus: ...But don't these sequences of events exist in time? I mean don't they occur in some particular thing called a time?

John: I don't think that any event occurs at a particular time. For example, when I watch the sun rise, I would not be able to tell you what the precise moment is at which the sunrise occurs.

Methodus: Are you saying that you believe that it is impossible to say exactly when any particular event occurs?

John: Yes. And I believe the notion that an event occurs at a particular time is just a concept......In fact, I think an event is just a concept too.

Methodus: ......Earlier you said that you believe that the Universe exists for all time. What do you mean when you say all time?

John: Well...we have this concept called time. I believe that all units of time are names for particular quasi-periodic sequences of events.

Methodus: Why do you say quasi-periodic instead of periodic?

John: I don't think that anything in the Universe is periodic.

Methodus: Oh, OK. I guess that would be consistent with your belief that nothing in the Universe is constant.

John: ...Except for the truth-value of the statement: The Universe exists.

Methodus: Right. You believe that it is always true that the Universe exists. What do you mean by always?

John: ......Maybe I should just say that I believe the Universe exists...and leave it at that.

Methodus: Are you saying you would dispose of the notion of always? Or time?

John: Well...maybe that isn't such a good idea......Let's get back to that idea of time as a concept that we use to predict the future.

Methodus: You're saying future here. So you believe there is a future? I mean do you believe there is this thing called time that is separate from the units that we use to measure it?

John: Yes. I guess so. I believe that past, present and future exists. Like I said, I don't think that it is possible to say exactly when an event occurs. But I do think it is possible for us to say that we have yet to observe that a particular event has occurred. And I believe it is possible for us to say that we have observed that a particular event has occurred.

Methodus: So you believe that it is possible to establish moments in time that are before and after a given event occurs?

John: Well...at least before and after we perceive that a given event occurs......I guess by saying that the Universe exists always, I am saying that the sequence of events is infinite.

Methodus: Yes, but if an event happens in the Universe, and no one is there to perceive it, does it still happen? I mean you said that you believe an event is just a concept.

John: Good point......You know, I'm not sure if time is of any importance at all when discussing the Universe.

Methodus: What do you mean?

John: Maybe the Universe is just one big now. I mean an infinite now.

Methodus: Are you saying that some things don't happen before other things? I mean, are you saying that you think everything in the Universe happens at the same time?

John: ...I'm not sure. I think I might be saying that there is no such thing as time......Maybe everything in the Universe does happen at the same time.

Methodus: It doesn't seem that way to me.

John: Same here. But maybe it does. Maybe we are just so small compared to the size of the Universe that it seems to us, from our perspective that the Universe is unfolding or existing over time. When actually, from the perspective of the Universe, everything is happening at once.

Methodus: So, for us time appears to exist. Are you saying that maybe it makes no sense to speak of the existence of the Universe as something that occurs over time?

John: Yes, I believe the existence of the Universe is a constant. When something is constant how can it possibly be used to measure time? Perhaps for the Universe, time is constant. So from the perspective of the Universe, time stands...

Interruptus: In Dodecahedron, what type of transitions did you use for a crescendo and decrescendo?

John: Geometric and sinusoidal...I experimented with linear transitions but I didn't care for the sound of those, so I chose not to use them.

Methodus: When you say geometric and sinusoidal transitions, do you mean transitions like those that you used for accelerando and decelerando?

John: Yes. It's very much the same. Except here we're calculating the dynamic level of each note rather than its duration.

Methodus: So for a crescendo phrase you would have some initial dynamic level and some final dynamic level. And then you would use one of these functions to determine how to get from the initial level to the final level. Is that right?

John: Yes, that's right. Again, we may think of the time scale as being composed of a sequence of non-overlapping 1-notes, each of which begins at a time-scale point. A dynamic level may be calculated for each 1-note.

Methodus: Would the dynamic level of the 1-note that starts at time-scale point 0 be equal to the initial dynamic level?

John: Yes.

Methodus: I am guessing that the dynamic level of the 1-note that starts after the last 1-note that is given in the score would be equal to the final dynamic level?

John: Exactly.

Methodus: OK. What about the 1-notes that are between these two?

John: Well...let's suppose there are L 1-notes shown in the score for this phrase. I mean let's suppose the last time-scale point in the score is L-1 (l minus one). For a geometric transition, the dynamic level of the kth 1-note would be equal to the initial dynamic level times the kth power of the L-th root of the final dynamic level divided by the initial dynamic level.

Methodus: That made sense when we were talking about tempi and durations. There we had numbers. We had some number of seconds. For dynamic levels, you haven't specified any numbers. I mean, the dynamic levels that you have indicated by these symbols in the score do not have some precise numerical value on some particular loudness scale, do they?

John: No.

Methodus: Then how is it possible to calculate values for the dynamic levels between the initial and final levels?

John: Well...you select some loudness scale. On this scale, you select points that will be called triple piano, pianissimo, piano, mezzo piano, normal, mezzo forte, forte, fortissimo and triple forte. Then you calculate the intervening dynamic levels for a particular transition.

Methodus: What if the loudness scale that I use is nonlinear? I mean what if it is a logarithmic scale?

John: What you want to achieve is a transition that is perceived as being an equal tempered scale of dynamic levels from the initial level to the final level. I mean, the ratio between the apparent dynamic level between any two consecutive 1-notes should be constant throughout the phrase.

Methodus: Oh...I see......What would happen for a sinusoidal transition?

John: It would be the same idea as the geometric transition, except you would use a sine wave. To calculate the dynamic level for the kth 1-note, first you would calculate an angle in radians that would be k times pi over L, plus pi over 2. Then you would add 1 to the sine of that angle. Next you would take half that quantity and multiply it by the initial dynamic level, minus the final dynamic level. Finally, you would add that result to the final dynamic level.

Methodus: OK. That gives us the dynamic level of each 1-note. But these are just the 1-notes of the time scale. They aren't sounded notes. What about the actual notes of the phrase? How would we calculate the dynamic level for those?

John: Well...each note of a given phrase starts at the same time as one of these 1-notes of the time scale. Suppose we have a given note that starts at time-scale point k. The dynamic level for this note will be the same as that which was calculated for the 1-note of the time scale that starts at time-scale point k.

Methodus: ......Earlier you said that the dynamic level of a note will remain constant throughout its duration. Is that right?

John: Yes. That would be true. It's true of the 1-notes of the time scale as well.

Methodus: OK. So let's say there's a 7-note that begins at time-scale point 2. You're saying that the dynamic level of this 7-note will be constant and equal to that of the 1-note of the time scale that begins at time-scale point 2. Right?

John: Yes, that's how it works.

Methodus: OK, that makes sense......What determines whether a crescendo or decrescendo should be geometric or sinusoidal?

John: The value of the edge-face difference O. If O is 10 or -2, 9 or -3, or 8 or -4, the transition will be sinusoidal. For all other values of O it will be geometric.

Methodus: That's the same breakpoint that you set between strongly sharp and slightly sharp for streams for which the characteristic is sharp. Right?

John: Yes, I used the golden ratio here as well. You will have 38 out of 60 directed edges for which the transition will be geometric. For the other 22 directed edges, it will be sinusoidal.

Methodus: ......Did you do this for accelerando and decelerando too? I mean, is the transition type for these determined in the same way?

John: Yes. It is the same for those. For an accelerando or decelerando, the transition will be geometric if O is 10 or -2, 9 or -3, or 8 or -4. Otherwise it will be sinusoidal.

Methodus: You mean the other way around, don't you?

John: ......Right. It's sinusoidal if O is 10 or -2, 9 or -3, or 8 or -4. Otherwise it's...

Interruptus: Why did you do that?

John: What do you mean?

Interruptus: Why did you turn that book upside down? I think that's the third time you've done that since we've been talking.

John: Oh...I don't like to look at the faces.

Methodus: What faces?

John: The faces of the people on the covers.

Methodus: You mean like this picture of James Joyce?

John: Yeah.

Methodus: Why don't you want to see these pictures of people?

John: For me, they represent limitation.

Methodus: What kind of limitation?

John: All kinds.

Methodus: Could you explain what you mean by that?

John: Sure. For one thing, they are limited.

Methodus: How so?

John: They are limited in what they can do.

Methodus: Why do you think they are limited?

John: Well...for the most part they are not that bright. I mean their intellectual capacity is relatively low...They seem to have difficulty with focusing or concentrating......I think it might be because they are afraid. They're insecure.

Methodus: Do you see yourself as being different?

John: I don't identify with them. I mean I don't see much of myself in them. I don't think that the things that drive me are the same as what drives them.

Methodus: Does it bother you to see these pictures?

John: It's a distraction. When I am trying to think freely and openly I don't want to have some picture starring at me reminding me of a lack of possibility, and restraint.

Methodus: Do you feel that they try to limit you?

John: Absolutely. But not just me. I think they try to control everything in their environment because of their own limitations.

Methodus: What do you mean?

John: For example, they try to limit what other people can do.

Methodus: How do they do that?

John: By any means. For example, they band together. Ironically, some of these groups call themselves the avant-garde. I think a better name might be avant-guard; that's g-u-a-r-d. They become mafia-like. Loyalty to the cause becomes the most important ingredient in one's success. They become intolerant of anything that does not further the focal point of the cause.

Methodus: What are they guarding?

John: Their own territory; an intellectual territory.

Methodus: Are they successful at this?

John: Not really. Their ability to control others is limited in both space and time. The further you are away from them in both space and time, the less effect they have.

Methodus: Do you think what other people do is useful to you?

John: Yes. I didn't mean to imply that I ignore them entirely. I think that it can be stimulating to see what people have tried to accomplish. And in some rare cases the quality, quantity and depth of their work can be an inspiration.

Methodus: But you would rather not think about them all the time?

John: Exactly, I like to keep my distance......In my work, I pretend that I am an alien being.

Methodus: ...Do you believe that you are an alien?

John: Well...let's just say that if I found out that I am an alien, it wouldn't surprise me. I would probably just shrug my shoulders and say: That makes sense.

Methodus: So, as an alien, what is your purpose?

John: I think of myself as a visitor who is trying to write music for the people of my planet. I have access to all the information that man has written, in all fields. However, as much as possible, or at least to the extent that it serves my purposes, I like to pretend that I am not living in man's sociological systems. Therefore, I am not constrained by them. Then, I go about my business to create music that I believe will be interesting to my friends back home.

Methodus: So as such, you would be an explorer of Earth?

John: Yes.

Methodus: Why would the people of this alien planet have sent you here for this purpose?

John: I pretend to believe that my fellow planetmen have not done much work in the area of music. For them, music is an uncharted intellectual territory. However, they have become aware of some of the musical creations of Earth man. So they sponsored me to visit Earth to study these works, as well as those of other fields, in order to lay the groundwork for their...

Interruptus: In Dodecahedron, suppose we have a stream for which the characteristic is pitch-bend-up-or-down. What determines which notes will be affected in this way?

John: You mean for which notes will the pitch be bent?

Interruptus: Yes.

John: Only certain notes that are relatively long would be affected.

Methodus: What do you mean by relatively long?

John: I mean the longest notes of the phrase. For each duration mapping, there is a maximum possible duration. For example, for the duration mapping 411311 (four one one three one one), the maximum duration would be 4. For 333111, it would be 3.

Methodus: Oh, OK. Let's say the duration mapping for a particular stream is 112511. You're saying that only notes for which the duration is 5 would be affected?

John: Yes.

Methodus: Would all such notes be affected?

John: No. A particular note would have to satisfy other conditions as well.

Methodus: What would those be?

John: Well...suppose we have the sequence of main notes that are produced by a given stream. We may partition this sequence into subsequences by using the current face for each note.

Methodus: You mean, like we discussed earlier...where all the notes in a given subsequence have the same current face?

John: Right. There would be one such subsequence for each of the faces that the stream visits......We may assign a zero-based index to each of these subsequences. I'll refer to that as a current-face index.

Methodus: Can we discuss this in terms of an example?

John: Sure. Let's suppose the edges of a given stream are 83 (eight three), 8B, 84, 8C, 6C, 2C, 2A, 5A, 57, 51, 91, 93, 96 and 26.

Methodus: OK. In this case we would have the following subsequences: (NNNN) (NNN) (NN) (NN) (NNN) (NN) (NNN) (NN). Right?

John: Exactly. So here we have 8 subsequences, or equivalently, a sequence of 8 current faces. The first current face is 8, the second one is C, and so on. We could assign a current-face index from 0 through 7, to each of these.

Methodus: All right. What's next?

John: We can assign index values to the notes of each current face. For example, for the first current face there are 4 notes. We could label these as 0, 1, 2 and 3.

Methodus: Would the indexes of the 3 notes of the second face be 4, 5 and 6?

John: Those would be the note indexes when the notes are considered to be in the sequence of main notes. But here we're considering the note to be in a subsequence. We start again from 0 for each subsequence. So for the second current face, the indexes of the 3 notes would be 0, 1 and 2. I'll refer to this type of index as a current-face note index.

Methodus: Are the current-face index and current-face note index used to determine whether or not the pitch of a particular note should be bent?

John: Yes. But for a given note, we don't use the current-face index and current-face note index for that note. Instead, we use these indexes for the note from which the duration was derived.

Methodus: What was that? I think you lost me.

John: Do you remember how the durations for the main notes are dovetailed?

Methodus: Yeah. The duration for the last note of a subsequence is swapped with that of the first note of the subsequent subsequence. Right?

John: That's right. Now, in general for a given note we would use the current-face index and the current-face note index. But for these notes for which the duration was obtained by swapping we do something different. Specifically, for a note that is the last note of a subsequence that is followed by another subsequence, we use the current-face index and current-face note index of the first note of the subsequent subsequence. And, for a note that is the first note of a subsequence that is preceded by another subsequence, we would use the current-face index and current-face note index of the last note of the previous subsequence.

Methodus: OK. I get it. Like you said, you use the current-face index and current-face note index of the note from which the duration was obtained.

John: Exactly. For a given note, let's label these index values as F and N. And let's say L is the number of notes for the current face.

Methodus: Would L be determined like the other indexes? I mean would that be the number of notes on the face for which the index is F?

John: Yes. For that you would use the current face of the note from which the duration was derived.

Methodus: So how are these values used?

John: There is one other thing.

Methodus: What's that?

John: We will be using the absolute value of the edge-face difference O.

Methodus: OK.

John: Here's what we do. Suppose we have a given note for which the duration equals the maximum duration that is possible for the duration mapping. The pitch of such a note will be bent if and only if N equals the absolute value of O times the quantity F plus 1, modulo L.

Methodus: I think I would get a better understanding of this if we went back to the example that we were discussing.

John: OK. For that case, the stream type would be agnmluz/xridjk. The last edge is k, so the duration mapping would be 333111.

Methodus: All right. So that would mean the duration of faces 1, 2, 3, A, B and C would be 3. For the other faces it would be 1.

John: Right. Now, the sequence of current faces is 8, C, 2, A, 5, 1, 9 and 6.

Methodus: OK. So we would have the following sequence of durations: (1111) (333) (33) (33) (111) (33) (111) (11) (one-one-one-one, three-three-three, three-three, three-three, one-one-one, three-three, one-one-one, one-one).

John: That's right. Then if we swapped durations we would get: (1113) (133) (33) (31) (313) (11) (311) (11).

Methodus: What would be the value of O in this case?

John: ...The first 3 turns are LLL, so we would use the edge-face difference of the fourth edge of the stream. That would be 8C. So O would be 8 minus 12, or -4.

Methodus: OK. Let me see if I understand this. The duration of the first 3 main notes is 1. The maximum duration for this duration mapping is 3. So none of these notes would be affected. The duration of the fourth main note is 3. So, the pitch of this note might be bent. Am I right so far?

John: Yes. You're headed in the right direction.

Methodus: All right. For this note we will use the current-face index and current-face note index of the first note of the second face. So F, N and L would be C, 0 (zero) and 4, respectively.

John: No. F would be 1. We want to use the current-face index of the second face, not the face number of the second current face. And L should be 3. That's the number of notes for the second face.

Methodus: Oh, right......Now what do we do?

John: We add 1 to F to get 2. We multiply that by the absolute value of O to get 8. We calculate 8 modulo L to get 2. Then we compare 2 with N. In this case, N, which is 0 (zero) is not equal to 2. Therefore, the pitch of this particular note would not be bent.

Methodus: ......OK. So the fourth main note would not be affected...The fifth main note would not be affected because its duration is 1. So the next possibility would be the sixth main note.

John: Right. For that note F, N and L would be 1, 1 and 3, respectively......You know...I think it might be easier to forget about N for now. First, let's consider each current face and calculate the absolute value of O times the quantity F plus 1, modulo L.

Methodus: Why is that easier?

John: I am getting confused by all these variables...This way we will be able to calculate one index for each current face. Then we will just need to check if the note for that index has a duration of 3. I think it's easier to understand this way.

Methodus: OK. There are 8 current faces. For the first one we would get 4 times the quantity 0 (zero) plus 1, modulo 4. That would be 0 (zero). Right?

John: Yes. In general it would be 4 times the quantity k plus 1, modulo L, where k is in the interval 0 through 7, and L is the number of notes for the current face with index k.

Methodus: All right. So for the sequence of 8 current faces we would get: 4 mod 4, 8 mod 3, 12 mod 2, 16 mod 2, 20 mod 3, 24 mod 2, 28 mod 3, and 32 mod 2. That would be 0, 2, 0, 0, 2, 0, 1 and 0.

John: All right. Now we can mark these positions in the subsequences (NNNN) (NNN) (NN) (NN) (NNN) (NN) (NNN) (NN). Let's change N to P for the notes for which the pitch might be bent.

Methodus: OK. For the first subsequence, we would change (NNNN) to (PNNN) because for that face, the absolute value of O times F plus 1, modulo L is 0 (zero). Is that what you mean?

John: Yes, that's it.

Methodus: So we would get (PNNN) (NNP) (PN) (PN) (NNP) (PN) (NPN) (PN).

John: That's right. Now we need to swap the last position of each subsequence with that of the subsequent subsequence.

Methodus: OK. That would give us (PNNN) (NNP) (PP) (NN) (NNP) (PN) (NPP) (NN).

John: ...Right......Now we need to check the durations. For any note that we marked with a P, we need to change the P back to an N if the duration of that note is not 3.

Methodus: OK. For the durations we have (1113) (133) (33) (31) (313) (11) (311) (11)......So then we would get (NNNN) (NNP) (PP) (NN) (NNP) (NN) (NNN) (NN). So for this example, there would be four notes for which the pitch is bent?

John: ...Yes. We would bend the pitch of the 7th, 8th, 9th and 14th main note.

Methodus: ......What about neighbor notes?

John: You mean in the homophonic texture?

Methodus: Yes. What determines whether the pitch of a...

Interruptus: In Dodecahedron, suppose we partition the sequence of main notes for a given stream into a sequence of contiguous, non-overlapping subsequences for which all the notes in each subsequence have the same current face. We know that the length of each subsequence has to be in the closed interval from 2 through 5. Would any sequence of subsequence lengths be possible? I mean for example, could you have this: (NNN) (NNNNN) (NN)?

John: ......No. That particular one wouldn't be possible. I think you will find that any subsequence for which the length is 5 must be either the first or the last subsequence.

Methodus: Why would that be the case?

John: Well...suppose we have a subsequence of length 5 that is not the first subsequence. I think you can show that a subsequence of length 5 will be generated if and only if the stream goes completely around a particular face. So at some point, the stream arrives at a vertex V, and then traverses all 5 edges around a face upon which V is incident. After doing this the stream returns to vertex V. V cannot be the first vertex. Otherwise, this subsequence of length 5 would have to be the first subsequence. Since V is not the first vertex of the stream, the stream must end at vertex V.

Methodus: That's not a rigorous proof.

John: No. This is all loosely speaking. But it's the basis for...

Interruptus: Would it be possible to have a stream type for which the lengths of all its subsequences are the same? I mean, could you have this: (NN) (NN) (NN) ... (NN) (n-n n-n n-n, dot-dot-dot, n-n)?

John: You might have something there. Let's look at streams that I used for Dodecahedron......Oh...here's one. For phrase number 105 (one-oh-five) the structure is (NNNNN) (NNNNN).

Methodus: Is that the only one?

John: No, phrase 126 (one-twenty-six) has the same structure.

Methodus: Are there any that have a different structure than this?

John: Let me check......There's 304. But that's the same as the others......There's 457. But that's the same too......514 has the same structure...535 is another one like that...and 592...and 595.

Methodus: Any other ones?

John: ......No. I think that's it.

Methodus: Are all these streams of the same type?

John: Well...let's see. The stream type for phrase 105 is agnmfejkl. For 126 it's abcdeflkj.

Methodus: These two are reflections of one another.

John: Yeah. They make a figure 8 around two adjacent faces.

Methodus: Are the other ones like this too?

John: Yes. 304, 457, 592 and 595 are like 105. 514 and 535 are like 126.

Methodus: So there are only two stream types for which all the subsequences have the same length.

John: Well...we can't say that for sure. I'm fairly certain that it is true for the 659 (six-hundred-fifty-nine) streams that I used for Dodecahedron. But I'm not sure about the other stream types that I did not use...But given what we have seen so far, I would say it has a reasonable chance of being...

Interruptus: What is the largest number of subsequences that a stream type can have?

John: I'm not sure......While I was looking through the 659 streams of Dodecahedron I noticed an interesting one that is fairly long. All of its subsequences were of length 2 except for the first one......It's phrase 454 (four-fifty-four). That one traverses one edge. Then it goes all the way around an equator.

Methodus: You mean the turns are RRLRLRLRLR?

John: No, it's the reflection of that. It's LLRLRLRLRL.

Methodus: How many subsequences does that one have?

John: It visits 9 of the 12 faces. It doesn't go around the top or bottom faces, and another face that is adjacent to the equator.

Methodus: You could extend this one backwards a bit so that it would visit another face. Look, you could have the turns LRLLRLRLRLRL.

John: Right. You could also change the end a bit so that it would visit the remaining two faces. Like this: LRLLRLRLRLRRLR.

Methodus: This sequence of turns is a little hard to follow. I keep losing my place...Are you sure it visits all 12 faces?

John: Maybe it would be easier if we looked at a particular stream of this type...Let's say we start with directed edge 13 (one three). That would give us the following sequence of edges: 13 (one three), 19 (one nine), 59, 52, 5A, 7A, 74, B4, B8, 38, 36, 96, 26, 2C and AC. This stream visits the faces 1, 9, 5, A, 7, 4, B, 8, 3, 6, 2 and C......I wonder if there is a shorter stream type that visits all...

Interruptus: Can a stream have 13 (thirteen) subsequences?

John: ...I don't know. Let me search through the streams of Dodecahedron......Phrases 482 (four-eighty-two), 505 and 575 have 12. Here's one. Phrase 659 (six-fifty-nine) has 13 subsequences. The stream for that phrase traverses the following edges: 4B, 7B, 1B, 13, 93, 96, 26, C6, 86, 83, 8B, 84, C4, CA, 2A, 25, 95, 15, 75, 7A and 4A. Its length is 21.

Methodus: So that one visits the faces B, 1, 3, 9, 6, 8, 4, C, A, 2, 5, 7 and A......It visits A twice?

John: Yes.

Methodus: ......Is it possible to have a stream that visits a given face three times?

John: No. I don't think that's possible.

Methodus: Why not?

John: Well...each time a stream visits a given face, it must traverse at least 2 edges of the face. Since each face has only 5...

Interruptus: Earlier you said that for the homophonic texture that you used in Dodecahedron, it is possible for the neighbor note of a simple edge or the first note of a pivot edge to end after the start of a subsequent note that is produced by a subsequent edge. Is that right?

John: Yes. That can happen.

Interruptus: Wouldn't it be possible that two notes that overlap in this way would have to be played on the same instrument?

John: Yes. That doesn't happen too often. But it is possible.

Methodus: Could you give an example of a stream for which this happens?

John: ...Sure. Here's one. Suppose the stream edges are 2C, AC, 4C, 8C, 6C, 62, 92, 52, 5A, 57, 51 and 59.

Methodus: What is the stream type for this one?

John: It's...

Interruptus: It looks like a pair of eyeglasses.

John: Yes, I suppose it does. First we go all the way around face C, or 12 (twelve). That would be the right lens. Then we form a bridge with edges 62 (six two) and 92 (nine two), which takes us to face 5. Then we go all the way around face 5 for the left lens......The stream type name would be abcdefmnow-v.

Methodus: ...So the last edge is v. That means the duration mapping would be 161111.

John: That's right.

Methodus: OK. So the duration for faces 2 and 11 is 6. And it's 1 for all the other faces. That means that for the first five edges, the instrument assigned to face 12 would play a sequence of 5 1-notes.

John: Yes. That's correct. Those notes would be played by bass clarinet 2......Now, the texture is homophonic. So at the same time at which the bass clarinet plays its first note, the instrument assigned to face number 2 would begin playing a note for which the duration would be 6.

Methodus: Right...Because the duration for face 2 is 6. That would be oboe 2, right?

John: Yes......Now, both of these notes begin at time-scale point 0. The first note of bass clarinet 2 will end at time-scale point 1. But the first note of oboe 2 will not end until time-scale point 6.

Methodus: ...OK. Then at time-scale point 1, bass clarinet 2 starts its second 1-note.

John: Right. And at the same time, the instrument assigned to face 10 plays a 1-note. That would be horn 2.

Methodus: All right. Then we get to time-scale point 2 when bass clarinet 2 starts its 3rd 1-note. At the same time, the instrument assigned to face 4 would also play a 1-note. That would be flute 2.

John: That's right. Next we have time-scale point 3. At that time, bass clarinet 2 starts its 4th 1-note. Simultaneously, the instrument assigned to face 8 plays a 1-note. That would be clarinet 2.

Methodus: OK. The fifth edge is a pivot edge. So the durations would be swapped for those notes.

John: Yes. But in this case, it won't matter because the duration is the same for faces 6 and 12. They're both 1.

Methodus: ...Right. So at time-scale point 4 we have bass clarinet 2 starting its 5th 1-note. And at the same time, the instrument assigned to face 6 would play a 1-note. That would be bassoon 2.

John: Right...Now we get to time-scale point 5. The fifth edge was a pivot edge. So the notes of the sixth edge must begin at the time at which the second main note of the fifth edge ends.

Methodus: The second main note of edge 5 is the 1-note that was played at time-scale point 4 by bassoon 2. Right?

John: Yes. So the two notes for the sixth edge must start at time-scale point 5. One of these notes will be played by the instrument that is assigned to face 6. That's bassoon 2. The other note must be played by the instrument that is assigned to face 2. That would be oboe 2......That's where we get a conflict.

Methodus: Why?

John: Oboe 2 started playing a note at time-scale point 0 that cannot end until time-scale point 6. If we are going to treat oboe 2 as a monophonic instrument, it cannot play a note that begins at time-scale point 5.

Methodus: ......Well...even if it weren't monophonic, there still might be a problem here.

John: How's that?

Methodus: What if these two notes that overlap have the same pitch? That would be a problem, even for most polyphonic instruments.

John: You're right.

Methodus: So what do you do in this case?

John: When I selected the particular stream to be used for a given stream type, generally I had several candidates from which to choose. I automatically rejected any streams for which the notes overlap in this way.

Methodus: So you never selected a stream that has this property?

John: Right.

Methodus: What if it were necessary for you to reject all candidates for...

Interruptus: Do you think there is anything in the Universe for which the tempo is zero?

John: ...My guess would be no......If there were such a thing, how would you know?

Methodus: What do you mean?

John: How could you detect that a particular thing has a tempo of zero? I mean, let's suppose human beings behaved in the following way. Two years after being born, a human begins saying the word hello, repeatedly...And they continue doing this until they die.

Methodus: OK.

John: We could define the tempo of a particular human to be the number of times they say hello divided by the number of seconds in their lifetime, minus 2 years' worth of seconds.

Methodus: All right.

John: ...Now, suppose one given human says hello at age two, and then never says hello again.

Methodus: You mean they only say hello once. And then they die, eventually?

John: Yes......Would it ever be possible for you to say that the tempo of that particular human being is zero?

Methodus: ......No, I guess not. If they said hello at least once, then their tempo would always be nonzero. The longer they lived, the smaller their tempo would be. But it would never be zero......What if they told you that they were never going to say hello again?

John: What if they lied? They might surprise you.

Methodus: ......What if they had never said hello at all? Wouldn't their tempo be zero in that case?

John: ......You mean they lived their entire life without ever saying hello? And now they have died?

Methodus: Exactly.

John: .........I think you may be right. By definition, their tempo would be zero...Perhaps we should revise this definition......Let's define the tempo of a particular human in the following way. Suppose a given human has said hello n times, where n is some integer greater than or equal to 2. Let T_k (t k) be the time in seconds at which they said hello for the kth time. Let t be a time that is greater than or equal to T₂ (t two). For some k greater than or equal to 2, t will be greater than or equal to T_k and less than T_k+1 (t k-plus-one). The tempo at time t would be 1 over the quantity T_k+1 minus T_k, hellos per second.

Methodus: What would the tempo be for a time t that is less than T₂?

John: For those times we would say that the tempo is undefined.

Methodus: What if a human has yet to say hello more than once? Would their tempo be undefined?

John: Yes.

Methodus: ...So by this definition, the tempo of a human could never be zero.

John: Right.

Methodus: ......I'm not sure if I like this definition.

John: Why not?

Methodus: Well...let's suppose a particular human says hello for the first time at time 0 seconds. Then suppose they say hello for the second time at time 2 seconds. Now suppose their next two hellos are at times 7 and 9 seconds, respectively.

John: OK.

Methodus: Now, let's suppose this pattern continues. That is, after each odd-numbered hello there are 2 seconds before their next hello. And after each even-numbered hello there are 5 seconds before their next hello.

John: All right.

Methodus: ...By this definition of tempo, their tempo would jump alternately between 0.5 (point-five) hellos per second and 0.2 hellos per second. Right?

John: Yes. That's correct.

Methodus: I think if I were listening to such a person, I would say that their tempo is 1 beat per 7 seconds.

John: I see your point......Do you think you would perceive the downbeats as being the times at which the even-numbered hellos occur?

Methodus: Yes. Provided all other things are equal, such as dynamic level, pitch and spatial position.

John: I think after a while, I might begin to hear the tempo as being 1 beat per second.

Methodus: You mean you would hear beats in between the hellos?

John: Yes. In order to try to predict when the next odd-numbered hello is to occur, I would use the shorter duration of 2 seconds as a metric. I think I would find that there are a total of 2 and one-half of these durations after each even-numbered hello.

Methodus: So then you would know that the ratio between the long and short durations is 5 to 2.

John: Exactly. Then I would find the greatest common divisor of 5 and 2, which is 1. With this I would imagine an underlying constant tempo of 1 beat per second that would enable me to predict when the hellos would occur, more easily.

Methodus: So would you say that the tempo is 1 beat per second in this case?

John: ......No...I think I would call this an implicit tempo, to distinguish it from the concept of tempo that we have already...

Interruptus: In Dodecahedron, what kind of transitions have you used for a note-level crescendo or decrescendo?

John: Geometric and sinusoidal.

Methodus: Under what conditions would the transition be geometric?

John: If the other edge-face difference O is 10 or -2, 9 or -3, or 8 or -4, the transition would be sinusoidal. For all other values of O it would be geometric...It's the same as the mapping that is used for crescendo, decrescendo, accelerando and decelerando.

Methodus: OK......Which notes would be affected?

John: The approach that is used here is the same as that which is used for phrases for which the characteristic is pitch-bend-up-or-down.

Methodus: You mean for each current face of the stream you would calculate a current-face note index that is the absolute value of O times the quantity current-face index plus 1, modulo the number of notes for that face?

John: Exactly......If the duration of the note with this index is the longest possible duration for the duration mapping, then that note would be affected. I mean there would be a note-level crescendo or decrescendo on that note......I think it might simplify this if we used the current-face number instead of the index.

Methodus: What's the difference?

John: By current-face number, I mean the 1-based number of the current face. The current-face number of the first face would be 1. That of the second face would be 2. And so on.

Methodus: So the current-face number would be the current-face index plus 1?

John: Right......If we use that instead, we could say that one calculates the current-face note index as the current-face number times the absolute value of O, modulo the number of notes for that face.

Methodus: ......Why did you use the current-face number here instead of the current-face index?

John: What do you mean?

Methodus: Why didn't you use the current-face index times the absolute value of O instead of the current-face number times the absolute value of O?

John: Oh...because otherwise for the first face, the current-face note index would always be zero. You would have zero times the absolute value of O, modulo the number of notes for the first face. I didn't think it would be very interesting to have it this way. It would have caused there to be many phrases with a note-level crescendo or decrescendo on the first note of the phrase......This was a form of repetition that I wanted to avoid.

Methodus: Oh, I see......When you use the current-face number, what is the chance that the first note of the first face will be affected?

John: Well...let's suppose all the durations of the duration mapping are the same. In that case, all the notes would have a duration that is the maximum possible duration. Let's say there are two notes for the first face.

Methodus: OK. In that case, we would have the absolute value of O, mod 2. So it would depend on the value of O, alone......How many directed edges have a face-difference that is even?

John: For 40 of the directed edges, the absolute value of the face difference is even. For the other 20 directed edges it is odd.

Methodus: All right. So here we would have 40 cases when the first note would be affected and 20 cases when the second note would be affected......What would be the distribution if there are three notes for the first face?

John: Well...in that case we would be calculating the absolute value of O, mod 3. For this, we get an even distribution of directed edges across the three possibilities. There are 20 directed edges for which the absolute value of O mod 3 is zero, 20 for which it is 1, and 20 for which it is 2.

Methodus: OK. What about the other cases? I mean what if there are 4 notes or 5 notes?

John: For 4 notes, you would calculate the absolute value of O, mod 4. For that, the number of directed edges would be 20, 4, 40 and 16 for the possible values of 0, 1, 2 and 3 respectively. For 5 notes, the distribution would be 8, 12, 20, 12 and, 8 for the values 0, 1, 2, 3 and 4, respectively.

Methodus: .........Are any neighbor notes affected? I mean when the texture is homophonic would there be a note-level crescendo or decrescendo on a neighbor note?

John: No. None of the neighbor notes would be affected.

Methodus: ...Would that be the same when the characteristic is pitch-bend-up-or-down? I mean, would the pitch of a neighbor note ever be bent?

John: No. It's the same there as well. Neighbor notes would not be...

Interruptus: In Dodecahedron, how is the octave of the pitch for a particular note determined?

John: ...There are a couple of things that affect the octave. One would be the melodic contour.

Methodus: By contour, do you mean the direction of melodic motion from one note to the next?

John: Yes, as I said earlier there are various rules that govern whether the motion to a particular note should be upward, downward or horizontal...And in some cases, the direction is optional.

Methodus: I think I may have forgotten some of the details of this. Could we look at a particular example? I mean could we take a particular stream and determine the pitches for the notes of that stream?

John: Sure. Let's look at the stream that was used for phrase 16 of Dodecahedron. That one is relatively simple. For this stream the directed edges are 96 (nine six), 26 (two six), 2C, AC, 4C, 8C and 6C. The stream type name is abhqric.

Methodus: OK. The last edge of the stream is c. So that would mean that the pitch-class mapping would be based on the self-complementary pitch-class set 4₁1₄1₁. Right?

John: Yes, that's right.

Methodus: All right. You said that for edge c we need to use the backward mapping...The forward mapping for the pitch-class set 4₁1₄1₁ is: 3 5 7 9 ₁₂ 11 _{10 8 6 4} 1 ₂ => QRST_UV_WXYZO_P. So the backward mapping would be: 9 7 5 3 ₂ 1 _{4 6 8 10} 11 ₁₂ => QRST_UV_WXYZO_P. Is that right?

John: Yes. Exactly.

Methodus: Now, you said that we have to transpose these pitch classes in order to make the pitch class of the current face of the first edge of this stream match the pitch class that was assigned to this face in the parent stream. What is the parent stream for this stream?

John: That would be the stream that was used for phrase 15.

Methodus: The current face of the first edge is face number 6. What pitch class was assigned to this face in the parent?

John: Well...for phrase 15, the stream edges are 96 (nine six), 92, 95, 15, 17, B7, 47, A7 and 57. The stream type name is agnv+z/*-......The last edge is -, so the self-complementary pitch-class set is 2₂2₂2₂. In this case we use the forward mapping. That would be: 1 3 _{12 10} 5 7 _{8 6} 9 11 _{4 2} => OP_QRST_UVWX_YZ. But these pitch classes need to be transposed. The first current face of phrase 15 is 9. The parent of phrase 15 is phrase 13...According to my notes, the pitch class assigned to face 9 in phrase 13 is Y.

Methodus: So the pitch classes for phrase 15 would have to be transposed up by an interval of 2...I mean, for phrase 15, face 9 maps to W. So we need to transpose this up by an interval of 2 to get Y. Is that right?

John: Yes.

Methodus: All right. So, if we go back to phrase 16, we need the pitch class that was assigned to face number 6 in phrase 15. That would be V. Right?

John: Wait...you forgot to transpose that up by an interval of 2.

Methodus: Oh...right. So it would be X.

John: That's right.

Methodus: ...For phrase 16, we already have X assigned to face 6 by the backward mapping......So...we don't need to transpose the pitch classes of phrase 16...Right?

John: Yes. For phrase 16, the transposition interval would be zero.

Methodus: OK. Then for phrase 16, the pitches for each face would be: 9 7 5 3 ₂ 1 _{4 6 8 10} 11 ₁₂ => QRST_UV_WXYZO_P.

John: Right.

Methodus: What is the texture for this stream?

John: It's melodic-each-face.

Methodus: OK. So we just need to determine the sequence of main notes......The first directed edge of the stream is 96 (nine six). For the note that is produced by this edge, the current face is 6 and the neighbor face is 9. The current face is on the right, so we're traveling in a clockwise direction around the current face. So the pitch class for this note would be an interval of 9 below the pitch class that is assigned to face 6. That would be O. Right?

John: Yes. That's right.

Methodus: What would be the octave?

John: That would depend upon the instrument on which the note is to be played.

Methodus: ...How so?