Methodus: OK. For those the total number of duration units would be 18 and 22. The average of these two would be 20. That seems a bit high, given that the range is 12 through 24.
John: Well...no. For over three-quarters of the duration mappings the total number of duration units will be 20, 22 or 24.
Methodus: Oh...right...because there are many more ways to partition 10, 11 and 12 into 6 parts as compared to 6, 7, 8 or 9.
John: Right. The average total is 315 (three-hundred-fifteen) times 2, divided by 30. That's 21.
Methodus: Some of the pairs that you assigned do not seem to be too balanced. For example, what about the edges for which crescendo is the characteristic? You assigned 221111 and 211113 to these. For those the total number of duration units is 16 and 18.
John: Yes, I did that deliberately. I wanted crescendo phrases to be relatively fast. So I assigned duration mappings with a relatively small total number of duration units. I did the same thing for decrescendo phrases. For those I used the duration mappings 111411 and 112222.
Methodus: Here's another case that is unbalanced. You assigned 322122 and 142212 to the edges for which the characteristic is exaggerate-frequency-range. For both of those the total number of duration units is 24. It's the same for 121215 and 111441, which you assigned to edges for which the characteristic is shift-attack-time.
John: Yes...well...when I assigned these duration mappings, the characteristics were something else. I mean, at the time at which I made this decision, I planned on having these characteristics be trill and repeat-note.
Methodus: You mean, the characteristics exaggerate-frequency-range and shift-attack-time were trill and repeat-note, originally?
John: Yes, early in the process of composing this piece, I planned to have these other characteristics. Then later I changed trill to exaggerate-frequency-range. And I changed repeat-note to shift-attack-time.
Methodus: So you assigned these duration mappings thinking that the characteristics would be trill and repeat-note?
John: Yes. I suspect that I may have assigned duration mappings with long durations to the trill characteristic in order to have notes that would be sufficiently long to perform and perceive a trill.
Methodus: What about repeat-notes? Why did you use long durations for that?
John: Well...at the time that I assigned these duration mappings I thought there were only 29, rather than 30. I had accidentally omitted the partition 1+1+1+1+4+4. Initially, I assigned the duration mapping 121215 to /. By itself, it wasn't too far out of balance. It was the only duration mapping that I assigned to this characteristic, which at the time was repeat-note. As it turned out, I changed this characteristic to exaggerate-frequency-range. Later, I discovered that I had omitted one partition. I considered assigning the duration mapping for this to edge a.
Methodus: You mean the first edge of the stream?
John: Yes.
Methodus: But then it would never be used...because no stream can end on its first edge.
John: Right. But, to some extent it was satisfying to have a one-to-one correspondence between duration mappings and edges. And, after all, I had forgotten this partition. So it made some sense to assign it in such a way that it would never be used. In the end, I opted to divide the streams that end on edge / into two subclasses based on the direction of the last edge. I assigned 121215 to one class and 111441 to the other. This solution had the advantage that all partitions would be used, and symmetry would be preserved relative to the first edge. I think it was a close call between these two possible solutions. If I were presented with the same problem today, I might choose the other approach instead.
Methodus: ......Why did you decide to change the trill and repeat-note characteristics to...
Interruptus: Earlier, you described how the pitch class for each note of a particular stream is determined in Dodecahedron. Which pitch of the pitch class would be used?
John: Do you mean, what would be the octave?
Interruptus: Yes. Suppose the pitch class for a given note is R. What determines the octave of the pitch? For example, under what conditions would the pitch be R3 versus R4, or some other pitch in the pitch class R?
John: Well...for a main note the octave will depend upon the pitch of the preceding main note.
Methodus: How so?
John: Usually, the octave of a given main note is such that the interval from the preceding main note to the main note will be minimized.
Methodus: You mean, if you had a sequence of two main notes, the pitch of the second one would be selected so that the distance between the two pitches would be minimized?
John: Yes. For example, let's suppose the pitch of one main note is Q4. Say the pitch class of the subsequent main note is Z. That would mean that the pitch of the subsequent note would be Z3.
Methodus: You said usually this is the way it goes. Are there exceptions?
John: Yes. In some cases, the next closest pitch would be used instead.
Methodus: You mean for this example that you just gave, sometimes the pitch of the subsequent note would be Z4 instead of Z3?
John: Exactly.
Methodus: Are there any other possibilities? I mean would Z2 or Z5 be used sometimes?
John: No. In all cases either the closest pitch or next nearest pitch would be used.
Methodus: What determines which pitch would be used in a particular case?
John: Well...for each stream type there is what I refer to as a contour face. I call it that because it is used to determine the melodic contour of the sequence of main notes of the stream.
Methodus: How is the contour face used?
John: The contour face number represents a particular orientation of a dodecahedral die for which the contour face is the topmost face.
Methodus: So if the number of the contour face is 4, the die would be oriented so that it is sitting on a flat surface with face 4 being the top face?
John: Yes that's right. Now, let's suppose the contour face number is C. We may partition the 30 edges of the die into 5 sets. First there are the 5 edges on which the contour face is incident. Let's call that set N, for north. Next there are 5 edges of the face that is diametrically opposed to the contour face.
Methodus: You mean the bottom face?
John: Yes. The number of that face will always be 13 minus C. Let's call this set of edges S, for south. Next we have the 5 edges that are connected to the contour face, but not part of it.
Methodus: So if the contour face is 4, one of these edges would be that which is incident on faces 7 and 10, right?
John: Yes. And the other four edges would be those which are incident on faces 10 and 12, 12 and 8, 8 and 11, and 11 and 7. Let's call this set of edges LN (l-sub-n). Here I'm using L for longitudinal. The subscript N is for north.
Methodus: I think I can see where you are headed. Would the other two sets be LS (l-sub-s) and E?
John: Yes. LS would be the set of 5 edges that are connected to face 13 minus C, but not part of it. These are the longitudinal edges on the south side of the die. E would be the equator, which consists of the 10 remaining edges that are not in N, S, LN or LS.
Methodus: Do these 5 sets play some role in determining the melodic contour?
John: Yes......as a particular stream travels along the edges of the die it will pass through some or all of these sets. Along the way, it will produce a sequence of main notes. There will be one main note for each simple edge and a sequence of two main notes for each pivot edge. In general the melodic direction to a given note from the previous note, be it up in pitch or down, will be dictated by the constraint that the distance between consecutive notes should be minimized. However, this will not be the case for the first note that is produced by a stream edge that is in LN or LS.
Methodus: You're saying first note here. By that you mean the note that is produced by a simple edge, or the first note of the two that are produced by a pivot edge, right?
John: Yes. The exceptional cases are notes that are produced by simple stream edges in LN or LS, and the first note that is produced by any pivot edge that is in LN or LS.
Methodus: So what happens for these notes? I mean what determines whether the melodic motion to these notes is upward or downward?
John: Well...the first note of a stream edge that is in LN is approached from below if the edge is traversed from south to north.
Methodus: By south to north, you mean the stream travels along a longitudinal edge from a vertex on the equator to a vertex on the contour face. Right?
John: Exactly.
Methodus: So in this case the pitch of the main note must be higher than that of the preceding main note?
John: Yes, that's correct. Most importantly, that must be the case regardless of whether or not this particular pitch is the closest pitch to the preceding main note.
Methodus: In some cases it would be the closest. Right?
John: Yes, but in other cases it would be the pitch that is one octave lower that would be closest to the preceding main note.
Methodus: OK. That makes sense......What happens if the direction along an edge such as this is downward? I mean from north to south.
John: Then the note would be approached from above.
Methodus: So, the first note produced by a stream edge that is in LN would be approached from above if the edge is traversed from north to south?
John: Yes.
Methodus: OK. Then what happens for a stream edge that is in LS?
John: It is the same. If the direction by which the edge is traversed is from north to south then the first note for that edge would be approached from above. Otherwise, the note would be approached from below.
Methodus: All right. You've been saying from below or from above. What if the pitch classes of the main note and the preceding main note are the same? What would be the direction of melodic motion in that case?
John: In that case the motion would be horizontal.
Methodus: You mean the pitch of the main note would match that of the preceding main note?
John: Yes, they would be at the same pitch.
Methodus: ...There is another special case that's been bothering me.
John: What's that?
Methodus: What happens if the interval between a given main note and the preceding main note is 6? In general the pitch of the main note would be such that it minimizes the difference in pitch between these two notes. In this case, there would be more than one solution. I mean, suppose the pitch of the preceding note is O4. Say the pitch class of the main note that follows it is U. The distance from U3 to O4 would be the same as the distance for O4 to U4. What would...
Interruptus: Do you believe that there is such a thing as free will?
John: You mean do I believe that people have the freedom to make choices?
Interruptus: Yes. Exactly.
John: Well...I believe that people make choices. And I believe that they believe they have freedom to make choices.
Methodus: But do you believe that they actually have this freedom?
John: If by freedom, you mean that one has the power to make a choice then I would say yes. By that I mean I believe that one has the power to act. We breathe air and eat food. This gives us energy to act.
Methodus: Yes, but does one have the freedom to act in a different way?
John: Do you mean act in a way that is different from how they have already acted? Are you talking about something along the lines of a time machine?
Methodus: No......Let's suppose I am pondering a decision for which the result will be to act in one of two possible ways. Let's call these action A and action B. Suppose I choose action A. Do you believe that I had the freedom to choose action A or action B?
John: Well...I believe you had the freedom to consider this decision. And, I believe you have the freedom to carry out action A.
Methodus: But did I have the freedom to choose B instead?
John: Do you mean, could you have selected B instead?
Methodus: Yes.
John: No. I don't think so.
Methodus: You don't think that I could have selected B?
John: No. I believe that if we could repeat the experiment, with all things being the same as before, you would choose A every time.
Methodus: So you don't believe I actually had a choice.
John: No. I believe you had a choice. But I believe that you would make the exact same choice if it were possible to rerun the experiment.
Methodus: So, if I were to have to make this decision again at some point in the future you think I would always pick A?
John: No. That would be a different decision. You and the rest of the Universe would be different at that point in time. In that situation, you might pick B instead.
Methodus: But if we could rerun the experiment so that everything in the Universe including me were the same, you believe I would pick A again.
John: Yes.
Methodus: Well...with that, I'm not sure if I have much freedom...if I can never choose B.
John: I believe you could choose B. But I don't believe you would.
Methodus: OK. So, you believe that my decision is predictable.
John: Only, if I had witnessed one run of the experiment.
Methodus: So you think my decision is completely determined by the state of the Universe, self included, at that point in time.
John: Yes.
Methodus: Can we extend this? I mean, do you believe that the state of the Universe at some future time is completely determined by its current...
Interruptus: For Dodecahedron, how did you go about assigning a particular self-complementary pitch-class set to a particular stream type?
John: Well...first I partitioned the various stream types into 9 equivalence classes. One of the classes consists of all stream types for which the last edge is g, f, b or e, relative to the first edge. I'll refer to that class as gfbe (g f b e). Similarly, the other classes are nmcd, olhj, vi, wuqt, -+rs, *zxy, pk and /.
Methodus: These are related to the classes that you used to assign characteristics, right?
John: Yes, very much so. For example, I assigned the characteristics crescendo and decrescendo to stream types in the class gfbe. More specifically, you could speak of the subclasses gf (g f) and be (b e). The subclass gf would be the set of all stream types for which the last edge is g or f. I assigned crescendo to stream types in the subclass gf. I assigned decrescendo to stream types in be.
Methodus: Earlier you said that you used 9 different self-complementary pitch-class sets. Did you assign one of these to each of the 9 equivalence classes of stream types?
John: Yes, that is exactly what I did.
Methodus: How did you do that?
John: I sorted the elements in each of these sets of 9 elements in a particular way. I mean, I arranged them in a particular order. Then, I assigned the first self-complementary pitch-class set to the first equivalence class of stream types; the second to the second; and so on.
Methodus: How did you order the self-complementary pitch-class sets?
John: First I listed the six self-complementary pitch-class sets that contain the all-combinatorial hexachords. Then I listed the other three. The order that I used was: 66, 411411, 3333, 2111121111, 222222, 111111111111, 5115, 411114 and 312312.
Methodus: I can see that for each list you have arranged the elements in descending order by the size of the largest non-subscripted number. You have 6, 4, 3, 2, 2 and 1 followed by 5, 4 and 3. Was that the basis for this particular ordering?
John: Yes...In terms of bracelets, for each list, the self-complementary pitch-class sets are ordered using the size of the longest string of black beads.
Methodus: Why is 2111121111 before 222222?
John: Several years ago, I compiled a list of the self-complementary pitch-class sets. I listed them in a particular order. In that list 2111121111 was before 222222. That's why it is so.
Methodus: Is there some rationale for the ordering that you used for this list of self-complementary pitch-class sets that you compiled?
John: I'm not sure. It is possible that I may have been using some strategy to make sure I was able to find all 20 possibilities. I can't say for sure why I listed 2111121111 before 222222.
Methodus: What about the 9 equivalence classes of stream types. What order did you use for those?
John: For each of the 29 edges at which a stream type may end, I determined the length of the shortest stream type that ends at that edge. Then I ordered the six equivalence classes that have four-letter names.
Methodus: You mean gfbe, nmcd, olhj, wuqt, -+rs and *zxy?
John: Yes. In fact that is the order that I used.
Methodus: And to get this, you used the lengths of the shortest stream types?
John: Yes...For each of these I added the four numbers that are the lengths of the shortest stream type for each of the edges listed in the class name. For example, let's take the class gfbe. The length of the shortest stream that ends on edge g is 6. The same may be said of edge b. The length of the shortest stream that ends on edge f or e is 9. So, for this case I calculated the sum 6+9+6+9 (six plus nine plus six plus nine), which is 30. I did that for the other 5 classes too.
Methodus: What would be the length of the shortest stream for the other cases?
John: For nmcd, olhj, wuqt, -+rs and *zxy you would have 7+8+7+8, 7+9+7+9, 8+10+8+10, 9+9+9+9 and 9+10+9+10. The totals would be 30, 32, 36, 36 and 38.
Methodus: So you sorted these by using these 6 totals?
John: Yes.
Methodus: Why is gfbe before nmcd? The total is 30 for both of these classes.
John: For that case, I listed the class with the smaller minimum component first. The minimum component for gfbe is 6. For nmcd it is 7. So gfbe is before nmcd. Similarly, wuqt is before -+rs because 8 is the minimum component of wuqt, whereas the minimum component of -+rs is 9.
Methodus: What about the other 3 classes?...How did you sort the classes vi, pk and /?
John: I listed those after the other 6 classes. For those I used the average of the components. For vi, pk and /, you have 8+8, 8+10 and 10, respectively. The averages are 8, 9 and 10.
Methodus: You mean you found the length of the shortest stream type that ends on edge v...and then you did the same for edges i, p, k and /?
John: Yes. Those would be 8, 8, 8, 10 and 10, respectively.
Methodus: Then, for vi you used the average of 8 and 8, for pk you used the average of 8 and 10, and for / you used 10. Is that right?
John: That's right. So the order is 8, 9 and 10; or vi, pk and /.
Methodus: So that would mean that you assigned the self-complementary pitch-class sets 66, 411411, 3333, 2111121111, 222222, 111111111111, 5115, 411114 and 312312 to the stream-type equivalence classes gfbe, nmcd, olhj, wuqt, -+rs, *zxy, vi, pk and /, respectively. Is that right?
John: Yes. That's correct.
Methodus: ......Earlier you said that for each of these self-complementary pitch-class sets, you used two different approaches to map the pitch classes to faces; a forward mapping and a backward mapping. How do these come into play?
John: For each equivalence class of stream types, the forward mapping is used for some members, while the backward mapping is used for the others.
Methodus: Oh, OK. How are these equivalence classes subdivided? I mean...let's take the equivalence class gfbe for example. Which of the stream types in the class gfbe would use the forward mapping?
John: The forward mapping would be used for the subclass gf.
Methodus: You mean if the stream type ends on edge g or f, the forward mapping is used?
John: Yes.
Methodus: So the backward mapping is used for the subclass be (b e)?
John: Right.
Methodus: What about the other 8 equivalence classes of stream types? How are they subdivided?
John: For nmcd, olhj, wuqt, -+rs, *zxy and vi, it would be similar. You divide these names into two halves to get the names of two subclasses. For the first half the forward mapping is used. For the other half, the backward mapping is used.
Methodus: So the forward mapping is used for the subclasses nm, ol, wu, -+, *z and v. And the backward mapping is used for the subclasses cd, hj, qt, rs, xy and i. Is that right?
John: Yes. That's correct.
Methodus: What about the equivalence classes pk and /?
John: For those two cases, the direction of the last edge is used. For streams that end on p or k we have this notion of left to right, or right to left. Do you remember that?
Methodus: Yes. And for the edge /, you said there is a forward direction and a backward direction.
John: That's right. Now, suppose we have a stream type that ends on edge p or k. If the direction of the last edge is left to right, the forward mapping will be used.
Methodus: Would the backward mapping be used if the direction is from right to left?
John: Yes, that's how it works.
Methodus: OK...what if the last edge is /? Would the forward mapping be used if the direction is forward?
John: Yes. Otherwise, [knock knock knock] the backward [knock knock]...Hello. Who's there?
Complementum: I am the Chair of the search committee for the
assistant professor position in
composition for which you applied.
John: ...Uh...I don't remember applying for that.
Complementum: Oh, I should explain. You applied to this position two years ago. Our response to applicants was delayed because it took some time for us to get approval for this position from the University.
John: Oh, I see. Come on in.
Complementum: I have been looking through your application. It seems to be incomplete. We have not received any of your letters of recommendation yet.
John: ...I don't plan to do that.
Complementum: I'm not sure I understand...are you still interested in this position?
John: Well...yes, I think there is some possibility that it might be interesting.
Complementum: OK. The hiring policy at the University requires that you submit three letters of recommendation. We cannot even consider your application without these letters in hand.
John: Oh.
Complementum: Actually...let's skip that for the moment. In your CV, you have not listed any commissions that you have received as a composer. You really should include those. We will need to see a list of your commissions.
John: Well...actually I did list them...it's right here.
Complementum: What's that?
John: That's the symbol for the empty set.
Complementum: Look John...there are a lot of composers out there who would kill for a chance to get this job...if you're not dead serious, then...
John: I'm serious.
Complementum: OK......In this job, you will be required to teach composition students. You do not have a degree in music. Is that correct?
John: Yes.
Complementum: Well...that is not a requirement for this position. But it does concern me. I wonder how well you will be able to teach our students. Especially, when you consider that most of your recent compositions are so closely related to mathematics.
John: Is the mathematics a problem?
Complementum: Well...we are teaching students of music here, not mathematicians. Many of our students are not very comfortable with mathematics. Some of them have not taken a math course since high school. I'm not sure they would be very interested in your work...I'm not sure you would be able to communicate with them very well. For the most part, their approach to composition is quite different from yours.
John: Yes...
Complementum: John, I would love to talk more about this with you. But, I have to run. In a few months we will be calling some candidates in for a formal interview. At that time we will send a letter to each of the applicants to let them know where they stand.
John: OK. Thanks for stopping...
Interruptus: Why did you select the particular array of tempi that you used for Dodecahedron?
John: Well...as I said earlier, tempo was determined by the difference between the left face number and right face number of a particular stream edge. There are 18 possible differences of this type. I grouped these in pairs as 10 and -2, 9 and -3, 8 and -4, 7 and -5, 6 and -6, 5 and -7, 4 and -8, 3 and -9, and 2 and -10.
Methodus: And you mapped each of the pairs to a particular tempo.
John: That's right. So, I wanted a total of 9 different tempi.
Methodus: And you chose to select these from the metronome scale?
John: Yes. As I said before, I used the following tempi: 52, 58, 66, 76, 88, 100 (one-hundred), 112 (one-twelve), 126 and 144.
Methodus: Why did you select these particular tempi?
John: First, I determined the slowest and fastest tempi that I might use. For that I selected 48 and 144.
Methodus: How did you select these?
John: ...By listening to a metronome beating time at 48 or 144.
Methodus: But you didn't actually use 48.
John: No. I toyed with the idea of selecting 48, 144, and 7 other tempi from the metronome scale for which the ratio between consecutive tempi would be roughly the same.
Methodus: How did you go about doing that?
John: I generated a sequence of tempi using the ratio eighth root of 3.
Methodus: The eighth root of 3?
John: Yes. We have 3 because 144 is 3 times 48. If we were to construct a tempo scale with eight steps from 48 to 144 for which the ratio between any two consecutive tempi would be the same, this ratio would have to be the eighth root of 3.
Methodus: OK. So you would multiply 48 by the eighth root of 3. Then you would multiply the result by the eighth root of 3. That would give you a sequence of tempi. How did you round these values to the whole numbers that are in the metronome scale?
John: Well...I didn't use the approach that you just described.
Methodus: No? Then how did you use this ratio?
John: First, I multiplied 48 by the eighth root of 3. The eighth root of 3 is roughly 1.14720269 (one-point-one-four-seven two-oh-two-six-nine). So that gives us about 55.066. Then I found the point that is closest to this in the metronome scale.
Methodus: 55.066 is between 54 and 56. So you selected 56 because it is closer. I mean 56 minus 55.066 is 0.934, which is less than 55.066 minus 54.
John: Well...yes and no. I chose 56. But that's not how I calculated distances. I didn't use the difference. I used ratios. Here we have 55.066, which is between 54 and 56. To determine whether 55.066 is closer to 54 or 56, I calculated the ratios 55.066/54 (fifty-five-point-oh-six-six over fifty-four) and 56/55.066. Those are roughly 1.020 and 1.017, respectively. I selected 56 because 1.017 is less than 1.020.
Methodus: Oh OK. I see...So what did you do after you selected 56?
John: I multiplied 56 by the eighth root of 3. That would be roughly 64.243. This falls between the points 63 and 66 on the metronome scale. So, I selected 63. By continuing this process, I selected the tempi, 72, 84, 96, 112 and 126. Finally, I multiplied 126 by the eighth root of 3 to get 144.548 (one-forty-four-point-five-four-eight), which is between 144 and 152. I selected 144.
Methodus: Doesn't this produce the same sequence of tempi that you would get if you followed the approach that I proposed?
John: You mean where we would multiply 48 by powers of the eighth root of 3...and then select the closest point on the metronome scale for each of these?
Methodus: Yes.
John: No. That yields a different result. It's the same except for one point. It would give you 108 (one-oh-eight) in place of 112. 48 times the sixth power of the eighth root of 3 would be roughly 109.416. That's closer to 108 than 112. With the approach that I followed you have 96 times the eighth root of 3, which is roughly 110.131. That's closer to 112 than 108.
Methodus: Oh. OK......So we get the sequence 48, 56, 63, 72, 84, 96, 112, 126 and 144. How did you get from this sequence to the one that you actually used?
John: Well...I didn't like the properties of this sequence. For my purposes it has too many simple ratios.
Methodus: What do you mean?
John: You've got 96, which is twice 48; 112 which is twice 56; 126 which is twice 63; and 144 which is twice 72 and three times 48.
Methodus: Right. There's also 84, which is three-halves of 56.
John: Also, the sequence of step sizes is a bit uneven.
Methodus: What do you mean by step sizes?
John: I mean the sequence of differences between consecutive tempi. For 48, 56, 63, 72, 84, 96, 112, 126 and 144, you get 8, 7, 9, 12, 12, 16, 14 and 18.
Methodus: So, how did you come up with the sequence 52, 58, 66, 76, 88, 100, 112, 126 and 144?
John: First I selected 144. Then I proceeded through the sequence of tempi in the metronome scale, in descending order from 144. I skipped 138 and 132, and selected 126. Then I skipped 120 and 116, and selected 112. I continued in this way.
Methodus: Oh I see, it would be skip 108, skip 104, select 100, skip 96, skip 92, select 88, skip 84, skip 80, select 76, skip 72, skip 69, select 66, skip 63, skip 60, select 58, skip 56, skip 54, and select 52.
John: That's right, I took every third tempi from the metronome scale.
Methodus: I can see that the sequence of step sizes for this is non-decreasing.
John: Yes. That's a consequence of the fact that the metronome scale has this property. You have 6, 8, 10, 12, 12, 12, 14 and 18.
Methodus: ...And the ratios between different tempi in this sequence are relatively complex. The simplest ones that I can find are 88/66 (eighty-eight over sixty-six), which is 4/3 (four over three); 126/112 (one-twenty-six over one-twelve) which is 9/8 (nine over eight); 144/112 which is 9/7; and 144/126 which is 8/7. None of the other ratios can be reduced to a fraction that has a single-digit numerator or denominator. Is that right?
John: Yes. The factors of these tempi are fairly large primes. 52 is 4 times 13; 58 is 2 times 29; 66 is 6 times 11; 76 is 4 times 19.
Methodus: Right. And all these primes are...
Interruptus: Could you describe the characteristics trill and repeat-note that you did not use in Dodecahedron?
John: Sure. Actually, the trill characteristic should be called trill/tremolo (trill-slash-tremolo). The interval between the two pitches might not be small...I think of a trill as being a continuous alternation between two pitches that are relatively close. I believe the word tremolo would be used when the interval between the two pitches is large.
Methodus: Yes, I think you're right......Where would a trill occur?
John: Well...suppose you have the sequence of main notes that are produced by a stream. There would be a trill between a pair of consecutive main notes.
Methodus: You mean you would alternate between the pitches of two consecutive notes?
John: Yes, exactly.
Methodus: Would more than one pair of notes be affected in this way?
John: No, only one.
Methodus: What would be the duration of the trill?
John: It would be an unmeasured trill or tremolo that alternates between the two pitches as fast as possible. The total duration of the trill would be equal to the sum of the durations of the two notes.
Methodus: So, in effect, the trill would replace the two notes. Right?
John: Yes. Instead of playing the sequence of two notes, one would alternate between the pitches of notes as fast as possible, for the total duration of the two notes that were replaced.
Methodus: Which pair of main notes would be affected in this way?
John: Do you remember what I said earlier about the other edge-face difference?
Methodus: No...that doesn't ring a bell. What do you mean?
John: You know...the part where the difference between the numbers of the two faces on which an edge is incident is calculated for the first four edges of a stream.
Methodus: Oh right. You labeled these differences as T, D, R and O because one determines the tempo, one determines the dynamic level, one is for the register, and the fourth determines some other thing.
John: Right. Well...for the trill characteristic, this other difference would be used to select the pair of notes that would be replaced by a trill.
Methodus: How so?
John: Let's suppose a given stream produces a sequence of 7 main notes. Let's call them a through g. So we have abcdefg (a b c d e f g). Now, we could number the adjacent pairs. We have ab as 1, bc as 2, cd as 3, and so on up through fg as 6.
Methodus: OK.
John: ...All right, with that, here is how it would work. If the difference O were positive you would take O minus 1, modulo 6, plus 1. That would be the number of the pair that would be affected.
Methodus: So, let's say O is 11. In that case...
John: Wait a minute, the difference between adjacent faces cannot be 11. Face number 1 and 12 are on opposite sides of the die. The largest possible value for O is 10.
Methodus: Oh...right......So let's say it's 10. Then I would take one from this to get 9. I would calculate 9 modulo 6 to get 3. Then I would add one to this to get 4. Is that correct?
John: Yes, that's right. In this case it would be the fourth pair of main notes that would be affected.
Methodus: What happens when O is negative?
John: In that case you would enumerate the pairs from the right. For the sequence abcdefg, you would have fg as pair 1, ef as pair 2, de as 3, and so on up through ab as pair 6. And you would use the absolute value of O in your calculations.
Methodus: So if O were -10, cd would be the pair that would be affected, right?
John: Yes.
Methodus: I suppose you could keep the pair numbers the same as before. I mean you could have ab be 1, bc be 2, and so on. For the negative case, the pair that would be affected is the complement of the positive case with respect to 7. I mean you would have 7 minus the quantity: the absolute value of O, minus 1, modulo 6, plus 1. That would be the same thing, wouldn't it?
John: Yes, you could look at it that way...More generally, 7 would be the number of main notes, and 6 would be one less than that.
Methodus: What about the characteristic called repeat-notes? How would that work?
John: Exactly one of the main notes would be affected.
Methodus: You mean only one of the main notes would be repeated?
John: Yes. A new note would be inserted after the note that is to be repeated, and before what would have been the subsequent main note.
Methodus: Would the second note be identical to the original? I mean would the note that is inserted be the same as the previous note, except for the time at which it starts?
John: Yes, except for its duration. The second note would be shorter than the original.
Methodus: How much shorter?
John: The duration of the second note would be such that the ratio between the duration of these two notes would be approximately equal to the golden ratio.
Methodus: You mean if the number of duration units in the original note were 1, then the duration of the second note would be about 0.618 (point-six-one-eight) duration units?
John: No. The duration of the second note would be some whole number of duration units.
Methodus: So let's say the number of duration units for the original note is d1 (d one) and that of the second note is d2. Are you saying that d2 would be selected so that the difference between d1 divided by d2, and the golden ratio is minimized?...I mean the absolute value of this difference.
John: Yes, that's right. d2 would be the integer in the interval from 1 through d1 that would minimize the absolute value of this difference.
Methodus: OK...So if d1 is 1, d2 must be 1. If d1 is 2, d2 will still be 1. d2 will be 2 if d1 is 3 or 4. It will be 3 if d1 is 5. And if d1 is 6 or 7, d2 will be 4. Did I get that right?
John: Yes. That is correct.
Methodus: ...Why did you want the duration of the second note to be shorter than the first?
John: I felt that this would be more interesting. By truncating the duration of the second note of the pair, the note that follows the second note occurs a bit before it is expected. I think this would help to keep things moving forward, aggressively.
Methodus: Oh, OK......So which note would be repeated?
John: As with the trill characteristic, that would depend on the other edge difference O. Here we number the main notes consecutively, beginning with number 1. Suppose the total number of main notes is M. If O is positive, the number of the note that would be repeated would be O minus 1, modulo M, plus 1. If O is negative, it would be M plus 1 minus this quantity...except in this case you would use the absolute value of O, in place of O.
Methodus: So if O is negative, it would be M minus the quantity: the absolute value of O, minus 1, modulo M. Right?
John: Yes.
Methodus: OK, I get it...Now, why didn't you use the characteristics trill and repeat-note in Dodecahedron?
John: I did that because both of these characteristics involve repetition. Repetition is something that I chose to avoid.
Methodus: Why didn't you want...
Interruptus: Why have you labeled the two directions for edge / as forward and backward? It seems like you have these reversed. I mean, if they were reversed then forward would be parallel to and in the same direction as the first edge.
John: Well...for each edge, I have defined a forward direction......You could consider each of these directed edges as the beginning of a directed ray in 3-dimensional space. Then you could translate all of these rays so that they have a common origin. For example, you could move the origin of each ray to be at the centroid of the dodecahedron. Then of the two possible directions for a given edge, you could label that which is closest in direction to the ray for the first edge, as forward. And you could label the opposite direction as backward.
Methodus: But then that would mean that you have reversed the directions for edge /. Why is that?
John: No. They're not reversed. I didn't use this approach to label the directed edges.
Methodus: Oh...then how did you label them?
John: Well...as I said earlier I named the 30 edges of the dodecahedron as a through z, and +, -, * and /. For all stream types the first edge is a. The direction of the first edge is from the vertex at which the edges a, e and f meet, to the vertex at which edges a, b and g meet. For edge a, I defined this to be the forward direction.
Methodus: What about the other edges?
John: Let's suppose we have some given edge E. Of the two directions along edge E, I defined the forward direction to be that which may be connected to the forward direction of edge a with the shortest directed path.
Methodus: Oh, OK. So for example, for edge h the forward direction would be away from the vertex at which edges b, c and h meet. Right?
John: Yes. In that case, the shortest directed path to this directed edge would consist of the edges abh (a b h). The shortest directed path to the opposite direction along edge h would be agoph. Since the path abh is shorter than the path agoph, the direction by which edge h is traversed in the path abh would be the forward direction of edge h.
Methodus: For some of the edges, the shortest path would be the same for either direction. For example, for edge r you would get the following two shortest paths: abcir and abhqr. Which would be the forward direction in that case?
John: If there is a tie, the forward direction would be that which is associated with the path that has the most balance between left and right turns.
Methodus: You mean for each of the shortest paths you would calculate the absolute value of the difference between the number of left turns and the number of right turns?
John: Yes. For example for the path abcir the turns are RRLL. For the path abhqr they would be RLRR. So the forward direction for edge r would be the direction by which edge r is traversed in the path abcir.
Methodus: Right. For that case, you would have the absolute value of 2 minus 2, or 0 (zero). But for the path abhqr you would have the absolute value of 1 minus 3, or 2......What happens if you still had a tie? I mean...is that possible?
John: Yes, for edges p and k you need yet another condition to define direction. Take edge p for example. You could get to it by the paths abhp or agop. Both of these paths have the same length. And for these paths the turns would be RLL and LRR, respectively. They are equally balanced in terms of left and right turns.
Methodus: So how would the forward direction be determined for an edge such as this?
John: The forward direction is that which is associated with the path that has the most right turns.
Methodus: So for edge p, the forward direction would be away from the vertex on which the edges o, p and w are incident. Is that right?
John: Yes.
Methodus: All right...so for the edge /, the shortest path would be abhqx/.
John: Well...there's also agow*/. But that would be...
Interruptus: For Dodecahedron, what is the contour face for each stream type?
John: Well...all stream types that end on a particular edge, relative to the first edge, have the same contour face.
Methodus: So for example, all the stream types that end on edge b would have the same contour face?
John: Yes, that's right. For that class of stream types the number of the contour face would be 4.
Methodus: Why is it 4?
John: I rolled a dodecahedral die 29 times. The topmost face of the first roll was 4. I used that for the contour face of the stream types that end on edge b.
Methodus: How did you use the results of the other 28 rolls?
John: The second roll yielded a 5. I used that for the contour face of stream types that end on edge c. The rest of the rolls were 5, 8, 9, 11, 8, 1, 10, 4, 10, 10, 12, 10, 1, 9, 8, 1, 1, 6, 3, 3, 5, 9, 5, 1, 2, 5 and 6. I assigned these to stream types for which the last edge is c through z, +, -, * and /, respectively.
Methodus: So for example, you mean you used 12 as the contour face of stream types that end on edge n?
John: Yes.
Methodus: ......Are there any other elements of Dodecahedron that you determined by rolling...
Interruptus: For Dodecahedron, how is the edge-face difference O used?
John: Well...it's used to determine some property of each characteristic...except for senza-vibrato. The difference O is not used in any way for streams for which the characteristic is senza-vibrato.
Methodus: How is it used if the characteristic is accelerando?
John: It dictates the type of transition that is to occur.
Methodus: What do you mean by type of transition?
John: I mean the trajectory or path that is followed in progressing from the initial tempo to the final tempo.
Methodus: What are the possible types of transitions for accelerando?
John: It could be either sinusoidal or geometric.
Methodus: For what values of O would it be sinusoidal?
John: If O is 10, 9, 8,...
Interruptus: What do you mean by a sinusoidal transition?
John: It's a transition that is calculated by using a sine wave function.
Methodus: You mean if you graphed the tempo over time it would be a sine wave?
John: Well...no, actually it's not the tempo that follows this trajectory. It is the sequence of durations between time-scale points.
Methodus: I think you may have lost me. I thought this was a sinusoidal transition in tempo.
John: Let's take an example. Suppose that for a given sinusoidal accelerando the initial and final tempi are 58 and 100 (one-hundred), respectively.
Methodus: OK. So for this there would be 58 4-notes per minute when the phrase begins. Then at the end of the phrase the tempo should be 100 4-notes per minute.
John: Right. Now, you can think of the time scale as consisting of a sequence of non-overlapping 1-notes, where each 1-note starts at one time-scale point and ends at the subsequent time-scale point......I'm calling these things notes, but they aren't like other notes. They aren't sounded as part of the phrase. You can think of them as being silent notes.
Methodus: OK. So if there are five time-scale points shown in the score for a given phrase, there would be four 1-notes.
John: Yes...it's also helpful to think of this time scale as being extended before time-scale point 0 (zero) and after the last time-scale point listed in the score. The time-scale points that precede time-scale point 0 are numbered -1 (minus-one), -2, -3, and so on.
Methodus: What about the time-scale points after the end of the phrase? Would those just be numbered consecutively?
John: Yes. Let's say the number of 1-notes that are shown in the score is L. That would mean the number of the last time-scale point that is given in the score would be L-1 (l minus one). The time-scale points after this, which are not shown would be L, L+1 (l plus one), L+2, and so on.
Methodus: So from time-scale point 0 (zero) through L-1, the tempo would vary sinusoidally from 58 through 100?
John: Again, it's not the tempo that varies sinusoidally. It is the durations of the 1-notes that form the time scale that vary in this way.
Methodus: Could you explain that a bit more?
John: Sure. First, I think it would be more convenient for us to convert these tempi into the number of seconds per 1-note. The tempo 58 would be 4 times 58 1-notes per minute, or 1 over 4 times 58, times 60 seconds per 1-note. That would be roughly 0.259 seconds per 1-note. The tempo 100 would be exactly 0.150 seconds per 1-note.
Methodus: OK. So would the duration of the 1-note that starts at time-scale point 0 be 0.259 seconds, roughly?
John: Yes. And for the infinite sequence of 1-notes that precede this particular 1-note, the duration would also be 0.259 seconds. These are the 1-notes that begin at the time-scale points -1, -2, and so on.
Methodus: Would the duration of the last 1-note that is shown in the score be 0.150 seconds?
John: No. That would be the duration of the 1-note that begins at time-scale point L.
Methodus: You mean the 1-note that immediately follows the last 1-note that is shown in the score?
John: Yes, that's right. Also...the duration of each of the 1-notes that follow this one would be 0.150 seconds as well.
Methodus: What would be the duration of the 1-notes that begin at time-scale points 1 through L-1?
John: A sine wave function is used to calculate those durations.
Methodus: How so?
John: Well...imagine yourself walking around a unit circle that is centered at the origin of the xy-plane.
Methodus: Clockwise or counterclockwise?
John: Counterclockwise.
Methodus: OK. Where do I begin?
John: You will begin at the point x-equals-zero, y-equals-minus-one. Then you will walk halfway around the circle until you get to the point x-equals-zero, y-equals-one.
Methodus: All right.
John: Before you do that, construct a second scale along the y-axis. This should be a linear scale of 1-note durations that progresses from the initial duration of 0.259 at y-equals-minus-one to the final duration of 0.150 at y-equals-one.
Methodus: OK. So, for a given value of y in the interval -1 to 1, I would have a duration of 0.259, minus 0.109 times the quantity y plus 1, over 2. Is that right?
John: ...Right. Now, for this walk you will take a total of L+1 steps, including the step that you take to land on your starting point.
Methodus: You mean my first step will be to step onto the point x-equals-zero, y-equals-minus-one?
John: Yes. We will call that step number 0 (zero).
Methodus: So step number L would be the one that I take to land on x-equals-zero, y-equals-one?
John: Yes, that's right. Now, each of these L+1 steps should be of the same size.
Methodus: You mean the arc length between any two consecutive steps should be the same?
John: Yes...it should be pi divided by L......As you walk along your semicircular path the y-coordinate of your position will vary continuously from -1 to 1.
Methodus: Right. If we graphed my y-coordinate over time we would get a sine function.
John: Exactly. That is why I refer to this type of transition as sinusoidal......OK, here's another thing. As you walk, your speed should change.
Methodus: How?
John: When you arrive at a given step, you will determine the y-coordinate of your location. From that you should calculate the 1-note duration for that value of y. This duration is the amount of time that you should take to progress to the next step.
Methodus: In walking from one step to the next, should my speed be constant?
John: Yes. But when you arrive at the next step, you will need to adjust your speed instantaneously because the duration for that step will be shorter than that of the previous step.
Methodus: OK, I get it. But how is this related to a sinusoidal transition from the tempo 58 to 100?
John: The duration of step k will equal the duration of the kth 1-note.
Methodus: Oh...I see......Suppose this were to be a sinusoidal decelerando from 100 to 58. Would we just use a different scale of durations along the y-axis?
John: Yes, that's right. In that case, we would have a linear scale that runs from 0.150 through 0.259.
Methodus: ...Now, what if this were a geometric transition? How would...
Interruptus: Do you believe that anything in the Universe is constant?
John: Yes, I believe the truth-value of the following statement is constant: The Universe exists.
Methodus: Right, earlier you said you believe that the Universe always exists. So you believe this statement is always true......What about other things in the Universe?
John: Do you have some particular thing in mind?
Methodus: Well...what about the speed of light? Do you believe that that is constant?
John: No, I don't think so.
Methodus: You think it varies?
John: Yes. To me, it is counterintuitive to believe that the speed of light is constant. I believe that it is more likely that over time the speed of light changes.
Methodus: What do you mean, over time?
John: I mean over long periods of time, the change in the speed of light would be measurable.
Methodus: Do you believe that it is constantly changing?
John: Yes, but to a degree that might be immeasurable over relatively short durations of time.
Methodus: What about other physical constants?
John: I don't believe there are any physical things that are constant. Again, it seems counterintuitive to believe that the value of some physical property of a concrete thing could be constant.
Methodus: What about abstract things?
John: Yes. I believe ideas are...
Interruptus: I wonder if it is possible to cover all the edges of a dodecahedron with some combination of non-overlapping streams.
John: ......I suspect that you would not be able to do that.
Methodus: Why not?
John: I don't have a proof. It's just a feeling I have.
Methodus: Is there some basis for this feeling?
John: ...Here, let's try to cover the edges with various streams of type abhpog. Let's begin with the stream abhpog. That leaves us with 24 edges to be covered.
Methodus: 24 divided by 6 is 4...We might be able to do it with 4 more streams of this type.
John: Let's see. Of the 24 edges, four of them are c, q, w and n. I believe you will have to arrange these 4 streams so that they begin on these edges. Otherwise, I don't think you will be able to cover these edges.
Methodus: Right. I can't think of any other way to cover edge c. A different stream cannot traverse edges b or h because they are already covered by the first stream. No stream could travel down edge c toward the vertex at which edges b, h and c are incident and end at this vertex, if the stream has not traversed edges b and h already.
John: Let's suppose you add a second stream cdjtsi.
Methodus: I can see there's going to be a problem...When you add the third stream qrsy/x, this will overlap with the second stream on edge s.
John: I think that type of thing might always happen. There might be enough here to formulate a proof.
Methodus: You mean you think there is always going to be some edge s where the streams must overlap?
John: Yes. Suppose we select any one of the 7800 possible stream types and use that as our first stream. I believe you can show that there will be two consecutive vertices traversed by this stream for which one of the incident edges will not be traversed.
Methodus: You mean two consecutive vertices through which the stream passes that have not been visited before, and will not be visited again?
John: Yes, because the only vertices that a stream can revisit would be the first and last vertex of the stream. And since the shortest stream is of length 6, all streams consist of a sequence of at least 7 vertices.
Methodus: OK, I think I see where you're going with this. For each of these two vertices there will be exactly one edge that was not traversed by the first stream.
John: Right. These would be analogous to the edges c and q in the example that we did. Without loss of generality, we may assume that they are the edges c and q.
Methodus: Yeah. We could always start the first stream from some other edge so that that would be the case.
John: We might have to reflect the stream...I mean, we might have to replace all the left turns with right turns, and vice versa. I'm not sure...either way, it wouldn't matter. I think we could show that if we cannot cover the dodecahedron by beginning with one particular stream, we would not be able to cover it by using the reflected version of that stream.
Methodus: I think you may be right.
John: OK, we have arranged it so that the first stream passes through the edges b, h and p, but not c and q. So, we're going to need to use at least two more streams: one that begins at the vertex bch (b c h) and travels down edge c to the vertex cdi (c d i); another that begins at vertex hpq and travels down edge q to vertex qrx.
Methodus: OK, that makes sense.
John: Let's call the stream that traverses edge c, stream number 2. Let's call the one that traverses edge q, stream number 3. Now, I think you will be able to show that stream number 2 must be the stream that traverses edges d and i.
Methodus: Wait...I can see why stream 2 would have to traverse one of these edges. But why both? Couldn't stream 2 traverse edge d and then go off somewhere else? Couldn't we cover edge i with another stream that begins at the vertex dci?
John: ......Yeah, I think you're right......maybe the key thing is that stream 3 cannot traverse edges d or i. I think we might be able to prove that.
Methodus: I agree.
John: I think we could show that some stream other than stream 3 must traverse edge i in the direction from vertex cdi to vertex isr.
Methodus: Right. And that would be either stream 2 or a fourth stream that begins at vertex cdi.
John: Now, I think you can show that one of these streams must traverse edge r.
Methodus: I'll buy that.
John: Let's suppose stream 3 traverses edge r. Then the stream that traversed edge i cannot go down edge r. It must go down edge s. But then we will not be able to continue stream 3 past edge r, because edges i and s will have been traversed already.
Methodus: Right. And if it were stream 2 or 4 that had traversed edge r, that would mean that stream 2 would have to traverse edge x. So we would not be able to continue stream...
Interruptus: What if we allow the streams to overlap?
John: We could do that.
Methodus: Yeah. Then you might try to find a way to cover all the edges while minimizing the number of edges where streams overlap.
John: Or, for each edge, you could have a penalty of 1 if two streams contain the edge. Maybe the penalty would be 2 if three streams traverse a given edge. If exactly one stream traverses an edge, there would be no penalty for that edge. Then you would add up the 30 penalty values.
Methodus: Hey, this could be a game. Each player takes a turn at placing a stream. When a player places a stream, they would get one penalty point for each edge of their stream that covers an edge that has already been covered by a previous stream. The game ends when all edges are covered. The winner would be the player with the minimum number of penalty points.
John: ...Here's a variation. You might make it so that all players must use the same stream type throughout the game.
Methodus: Yeah. You could give the first player the freedom to decide the stream type to be used.
John: Or maybe each player has the freedom to choose the stream type that they will use throughout the game. In other words, the first player would place a stream of a given type on the dodecahedron. From then on, every stream that the first player places must be of this type. After the first player places his stream, the second player selects a stream type that he will use throughout the game. The second player would not be required to use the same stream type as the first player.
Methodus: Maybe a point should be awarded for being the first player to cover a given edge.
John: Yeah...that sounds interesting. Then you might have something like this: After a player places a stream, you would update their score in the following way: For each edge of their stream that covers an edge that was not already covered you would add 1 point to that player's score; For each edge of their stream that covers an edge that was already covered by a previous stream, you would deduct 1 point from their score......Each player would begin the game with a score of zero points.
Methodus: Maybe that wouldn't be very good. I mean if I were the first player, I could just place a stream of maximum length. That would give me 21 points immediately.
John: Yes. But then you would have to use this long stream type for the rest of the game. That would probably force you to take a substantial number of deductions on subsequent turns.
Methodus: I think the original version of the game might be more interesting, or at least more complex. It might be more difficult to play, even for a computer program.
John: You're probably right. For that we would allow each player to place any type of stream they wish on their turn. Each player would begin with a score of zero. And they would get one penalty point for each edge of their stream that overlaps an edge that was covered by a previous stream. It's nice because it is a relatively simple game, in terms of rules. But with each turn a player could choose from among 7800 times 60 possible streams.
Methodus: Well...that's true. But for some of these streams the set of edges would be the same. For example, the streams abhpog and agophb would be two different streams that cover the same edges.
John: Right. So, it would be less than 7800 times 60. Still, I think it would be a fairly large number of possibilities.
Methodus: ......What would we call this game?
John: How about...
Interruptus: Earlier when we were discussing contour faces, you said that for a given main note that is not the first main note of an edge in LN or LS, the direction by which the note is approached is such that the difference in pitch between the main note and the preceding main note is minimized. Is that right?
John: Yes...and if the main note is the first main note of an edge in LN or LS, it would be approached from above if the edge is traversed from north to south, and below if it is traversed from south to north.
Methodus: Right. And you also said that for this type of main note, the direction of motion would be horizontal if the pitch classes of the main note and the preceding main note are the same.
John: ......Wait, that doesn't make sense.
Methodus: What do you mean?
John: It's impossible for the pitch class of the first main note of any edge to be the same as the pitch class of the preceding main note.
Methodus: ......Yeah, I think I can see why...For any main note that is the first main note of an edge, the current face of that note must be the same as the current face of the preceding main note.
John: That's right.
Methodus: ...The neighbor face of each of these notes must be different. Or more importantly, these two neighbor face numbers cannot be equivalent, modulo 12.
John: Exactly. The pitch classes of these two main notes would be calculated by displacing the pitch class of the current face by different amounts.
Methodus: ......So are there any cases when the pitch class of two consecutive main notes can be the same?
John: Yes. This can happen. The first and second main note of a pivot edge can have the same pitch class.
Methodus: How would that happen?
John: Here's an example. Let's suppose we have a stream for which the edges are 96 (nine six), 92 (nine two), 52 (five two), 5A (five a), 7A, 74, B4, 84, C4 and A4.
Methodus: Wait. What is this notation that you are using for stream edges?
John: Oh, I'm sorry. I thought we had discussed that already.
Methodus: No, I don't think we did.
John: Oh, OK. Here, I can name each directed edge of a dodecahedral die by using a sequence of two symbols. The first symbol is the number of the face that is on the left side of the edge. The second symbol is the number of the face on the right side.
Methodus: I just want to make sure I understand what you mean by left and right side in this context. Suppose I am holding the die in front of me so that the particular stream edge in question is nearest to me and vertical. Let's say the stream traverses this edge in an upward direction. Of the two faces upon which this edge is incident, would the left face be that which is on my left?
John: Yes. And the face that is on your right would be the right face for this directed edge. You can think of it this way. Suppose a boat is traveling forward down a given stream. For each edge of the stream, the left face will be on the port side. The right face will be on the starboard side.
Methodus: OK......In these edges that you listed...are you using A, B and C to represent the face numbers 10 (ten), 11 and 12?
John: Yes. It's like the notation for hexadecimal digits.
Methodus: All right...So the first three edges of this stream are 96 (nine six), 92 (nine two) and 52. Edge 96 is a simple edge. Edge 92 would be a pivot edge. Right?
John: Yes, that's right. Now, suppose the pitch classes that are assigned to faces 9 and 2 are Q and P, respectively.
Methodus: OK...In that case, the pitch class of the first main note of the pivot edge 92 would be the pitch class that is an interval of 2 above Q.
John: Right, because the current face of this note is face number 9. The pitch class for this face is Q. The stream is traveling in a counterclockwise direction around the current face. And the neighbor face for this note is 2.
Methodus: So the pitch class for this note would be S.
John: Yes...Now, let's look at the pitch class of the second main note of this pivot edge. For that note the current face is 2 and the neighbor face is 9. The stream is traveling clockwise around face 2. So, in this case we get the pitch class that is an interval of 9 below the pitch class P.
Methodus: That would be S.
John: Exactly...So here we have a case where the two main notes that are produced by a particular pivot edge have the same pitch class.
Methodus: ...Would the octave of these two notes be the same too?
John: Yes, the octave of the second note would match that of the first. The second note of a pivot edge cannot be the first main note of an edge in LN or LS. So, the second note will be approached in such a way that the difference between the pitches of the two main notes of the pivot edge will be minimized. In this particular example, both notes have the same pitch class. So the difference is minimized by having both pitches be at the same octave. In other words, for this example the direction of melodic motion will be horizontal.
Methodus: All right...that makes sense...If a given main note is not the first main note of an edge in LN or LS, the pitch of the note is chosen so as to minimize the difference in pitch between the note and the preceding main note......What happens when the interval between the pitch classes of the note and the preceding main note is 6? That is possible, isn't it?
John: Sure. For example, suppose the first two edges of a stream are AC and 4C. Since the difference between A and 4 is 6, the interval between the first two main notes of this stream will be 6.
Methodus: OK. Let's suppose the pitch class for face C is U. We would be traveling around face C in the clockwise direction, so we need to find the pitch classes that are an interval of 10 (ten) and 4 below U......All right, so the pitch classes of the first two notes for this stream would be W and Q. What if the pitch of the first note were W4. Would the pitch of the second note be Q4 or Q5? Either way, the distance between these two notes would be the same.
John: In this case, I audition the two possibilities and choose the one that I like the most.
Methodus: You mean, first you listen to the phrase that you would get if the pitch of the second note were Q4. Then you listen to the phrase with the pitch of this note at Q5. And then you pick the one you like?
John: Yes...I wouldn't necessarily audition them in that order. But that is the basic idea.
Methodus: So these two phrases would be identical except for one note?
John: No. Not necessarily. The pitches of subsequent notes would be affected by the particular octave that is chosen for the second note.
Methodus: How's that?
John: Well...suppose that for a given stream there is exactly one pair of consecutive main notes for which the interval between the pitch classes of the notes is 6. The octave of the second note of this pair would be optional. We could pick from one of two possible octaves. The pitches of all subsequent notes would be related to the pitch of this note. I mean they would be positioned relative to the pitch...
Interruptus: Have you heard of String Theory?
John: Yes...
Interruptus: I wonder if there is a relationship between strings and streams.
John: ......Maybe there is. I mean, maybe strings are shaped like streams.
Methodus: You mean on a dodecahedron?
John: No. Not necessarily. You could have streams on any polyhedron. For example, maybe the different forms that strings can take are related to the possible streams on the Platonic...
Interruptus: Many of the elements of Dodecahedron were determined by the value of the difference between the left and right faces of a stream edge. For each possible difference of this type, how many directed edges would there be?
John: You mean, how many directed edges are there for which the difference would be a particular value?
Interruptus: Yes.
John: Well...first we can say that for any value d, the number of directed edges for which the face difference is d will be the same as the number of directed edges for which the face difference is negative d.
Methodus: Why is that?
John: You could partition the 60 directed edges into three sets in the following way. Put all the directed edges for which the face difference is positive in one set. Call that set P. In a second set, put all the directed edges for which the face difference is negative. Call that set N. Then put the rest of the directed edges in a set called Z, for zero.
Methodus: OK...Wouldn't Z be empty? I mean, no two faces have the same number.
John: Yes, that's right. Z would be empty...Now suppose the face difference for a particular directed edge is d. Consider the directed edge that is on the same edge of the die, but in the opposite direction. The face difference for that directed edge will be negative d.
Methodus: Oh, I see. With this you can establish a one-to-one correspondence between the elements of P and those of N.
John: Exactly.
Methodus: So for 30 of the directed edges, the face difference will be negative. For the other 30 the face difference will be positive.
John: Right. And the number of elements in P for which the face difference is d will be the same as the number of elements in N for which the face difference is negative d.
Methodus: So, really we would only need to find out how many elements of P have a particular face difference.
John: Yes...Here's what you get. There are two directed edges for which the face difference is 5. The same may be said of the face differences 7 and 10. There are four directed edges for which the face difference is 2. The same may be said of the face differences 6 and 8. There are six directed edges for which the face difference is 3. The same may be said of the face difference 4.
Methodus: Wait a minute...there's no way to get a face difference of 9?
John: No.
Methodus: Before, you said that the possible values for the face difference range from -10 through 10, excluding -1, 0 and 1...You said there were 18 possible values. From what you just said, I only get 16. There's 2 through 8, and 10, and -2 through -8 and -10.
John: Oh...I see what happened. I have arranged the various face differences in pairs. There are nine pairs. I have paired the face difference d with -1 times the quantity 12 minus d. So we have the pairs 10 and -2, 9 and -3, 8 and -4, and so on through 2 and -10. That's how we got 18.
Methodus: Yeah. And you assigned various tempi, dynamic levels and registers to these pairs. For example, you assigned a tempo of 52 to the pair 10 and -2.
John: That's right......One thing that I considered when making these assignments was the total number of directed edges for which the face difference is one of the values of a particular pair. For example, for the pair 10 and -2, there are a total of 6 directed edges.
Methodus: Right. There would be 2 edges for 10 and 4 edges for -2......What about the other pairs?
John: There are 6 edges for the pair 9 and -3; 10 edges for the pair 8 and -4; 4 edges for 7 and -5; 8 edges for 6 and -6; 4 edges for 5 and -7; 10 edges for 4 and -8; 6 for 3 and -9; and 6 for 2 and -10.
Methodus: That's a symmetric distribution of directed edges. I mean, you have 6, 6, 10, 4, 8, 4, 10, 6 and 6.
John: Yes, because of the way that I have formed these pairs that would have to be the case. For example, the total number of directed edges for the pair 10 and -2 must be the same as that for the pair 2 and -10.
Methodus: Oh...right...Because 10 and -10 have the same number, and so do 2 and -2......Did you do that deliberately...so there would be this symmetry?
John: Yes. But symmetry wasn't the only issue. There were other factors that played a role in my decision to arrange the 16 possible face differences in this way.
Methodus: What factors?
John: Well...I did this while I was considering various ways to map face differences to dynamic levels. I wanted the number of dynamic levels to be 7, 9 or 11.
Methodus: You chose 9. Right?
John: Yes. But I also looked at mappings for which there would be 7 or 11. To get 7, I would eliminate triple piano and triple forte. To get 11, I would add quadruple piano and quadruple forte.
Methodus: So you had to map 16 face differences to 7, 9 or 11 dynamic levels. That means some face differences would have to map to the same dynamic level. I mean, the mapping could not be one-to-one.
John: Right. Initially, I was planning to use the absolute value of the face difference. I was going to use the direction of some particular turn of the stream to determine whether the level should be loud versus soft. I was thinking about having the absolute value of the face difference determine the degree of loudness, or softness.
Methodus: You mean, the larger the absolute value, the greater the degree of the effect.
John: Yes.
Methodus: Why did you abandon that idea?
John: Because, there would be no neutral or normal level.
Methodus: But you could have mapped some absolute value to the normal dynamic level. Couldn't you?
John: Yes, I considered that. First, I determined the number of occurrences for each possible face difference.
Methodus: You mean you determined how many directed edges have a particular face difference?
John: Yes. And with this in mind, I mapped absolute values to dynamic levels. At that point, I was still thinking of using the direction of a particular turn as a determining factor as well. Here is what I proposed initially:
10: 
(quadruple
forte) or 
(quadruple piano)
7: 
(triple
forte) or 
(triple piano)
5: 
(fortissimo)
or 
(pianissimo)
2: 
(forte)
or 
(piano)
3: 
(mezzo
forte) or 
(mezzo piano)
4: 
(normal)
or (normal)
6: (mezzo
piano) or
(mezzo forte)
8: (piano)
or (forte)
The first dynamic level on each line is that which would be used if the turn were to the left. The second level would be used if the turn were to the right. That would give us the following number of directed edges for each level:
(quadruple
forte): 2
(triple
forte): 2
(fortissimo):
2
(forte):
4+4 = 8 (four plus four equals eight)
(mezzo
forte): 6+4 = 10
(normal):
6+6 = 12
(mezzo
piano): 4+6 = 10
(piano):
4+4 = 8
(pianissimo):
2
(triple
piano): 2
(quadruple
piano): 2
Methodus: But you didn't like that?
John: No. I didn't like the fact that there would be only 2 directed edges for each of the 6 extreme dynamic levels.
Methodus: You could have consolidated these by eliminating some levels.
John: Yes, I considered eliminating the levels triple forte, quadruple forte, triple piano and quadruple piano. So the face differences 5, 7 and 10 would all map to fortissimo and pianissimo, depending upon the direction of the turn.
Methodus: Why didn't you do that?
John: I'm not sure. I think it may have been because I wanted more than 7 dynamic levels......The next thing I considered was eliminating the turn as a determining factor. I replaced this with the direction along which the edge was traversed. In effect, I decided to use the face difference rather than the absolute value of the face difference. I proposed using positive face differences in place of the left turn and negative face differences in place of the right turn. That gave me the following mapping:
10: (quadruple
forte)
7: (triple
forte)
5: (fortissimo)
2 or -8:
(forte)
3 or -6:
(mezzo forte)
4 or -4:
(normal)
6 or -3:
(mezzo piano)
8 or -2:
(piano)
-5: (pianissimo)
-7: (triple
piano)
-10:
(quadruple piano)
Methodus: Ok. But that's not the mapping that you used. What was wrong with this one?
John: Well...in developing this mapping, I had considered the number of directed edges that there would be for each dynamic level. I was trying to make these numbers be somewhat balanced over the range of dynamic levels. But in doing so, I had created a mapping for which the dynamic level was not an increasing function of the face difference. That was dissatisfying.
Methodus: Where did you go from there?
John: I rearranged the face differences in this mapping so that the dynamic level would be an increasing function of positive face differences, and an increasing function of negative face differences. By doing that I got the following mapping:
10: (2)
(ten: quadruple forte)
8: (4)
(eight: triple forte)
7: (2)
6 or -2:
(4+4=8)
5 or -3:
(2+6=8)
4 or -4:
(6+6=12)
3 or -5:
(6+2=8)
2 or -6:
(4+4=8)
-7: (2)
-8: (4)
-10:
(2)
Here I have given the number of directed edges for each dynamic level in parentheses. At this point, I was still considering collapsing quadruple forte and triple forte into fortissimo, and quadruple piano and triple piano into pianissimo. That would have yielded 7 dynamic levels with the following distribution of direction edges: 8, 8, 8, 12, 8, 8, 8.
Methodus: But you didn't use this mapping.
John: No. I took this idea a bit further. I inserted the missing face difference 9 so the scale would be 10, 9, 8, and so on. That gave me the following mapping:
10: (2)
9: (0)
8 or -2:
(4+4=8)
7 or -3:
(2+6=8)
6 or -4:
(4+6=10)
5 or -5: (2+2=4)
4 or -6:
(6+4=10)
3 or -7:
(6+2=8)
2 or -8:
(4+4=8)
-9: (0)
-10:
(2)
Methodus: But then triple forte and triple piano would never be used.
John: Yes, and for that reason I compressed this scale. I decided to map 2 to triple piano, and -2 to triple forte, and in the process, I eliminated the levels quadruple forte and quadruple piano. That gave me the following mapping:
10 or -2:
(2+4=6)
9 or -3:
(0+6=6)
8 or -4:
(4+6=10)
7 or -5:
(2+2=4)
6 or -6:
(4+4=8)
5 or -7:
(2+2=4)
4 or -8:
(6+4=10)
3 or -9:
(6+0=6)
2 or -10:
(4+2=6)
Methodus: That's the one that you used, right?
John: Yes.
Methodus: Why did you accept this as the final version?
John: It addressed all my concerns. First there is an adequate number of dynamic levels for my purposes. Second, the distribution of the 60 possible directed edges is fairly even among the 9 dynamic levels. And this distribution is symmetric about the normal dynamic level. Also the fact that the most common dynamic levels would likely be forte, normal and piano was pleasing to me. Third, the dynamic level is an increasing function of positive face differences, and it is an increasing function of negative...
Interruptus: Do you believe that the future state of the Universe is completely determined by its current state?
John: ...Yes.
Methodus: So, you believe that the future state of the Universe is predictable?
John: Yes and no. I mean, if you had a complete understanding of how the Universe works, I think you could predict its future state.
Methodus: ...You don't believe that anything in the Universe occurs randomly?
John: What do you mean by random?
Methodus: I guess I mean unpredictable.
John: Well...I believe that there are things that appear to be random.
Methodus: But they're not?
John: No. I think they appear to be random only because we do not understand how they...
Interruptus: What do you think is the greatest thing that America has achieved?
John: ...Putting a man on the Moon. I think the picture of the Earth from the perspective of the Moon is a very significant image.
Methodus: Do you think that that actually happened?
John: Well...it is quite an amazing accomplishment. Especially when you consider that it was done almost 40 years ago...I can imagine that it never happened.
Methodus: You mean you think it was a hoax?
John: I didn't say that. I said that it would not surprise me if it were a hoax. I mean, I believe that it is possible that it was a hoax.
Methodus: But you believe that it did happen, right?
John: I guess so. But it doesn't really matter to me one way or another.
Methodus: But what about the image of the Earth? Some artist could have created that.
John: For me, even if I were certain that that is how it was created, it would not diminish its significance.
Methodus: Would you be surprised if you found out that it was hoax?
John: No, not really...I would probably just say, I thought that might be a hoax.
Methodus: Wouldn't it affect the faith that you have in your country?
John: Wait a minute...now you're putting words in my mouth. When did I say that I have faith in the country in which I live?
Methodus: Oh...sorry. Are you saying that you don't have faith in America?
John: Faith. I don't think faith is very useful.
Methodus: Do you have faith in anything?
John: Let's see what The American Heritage Dictionary says about faith. Here...meaning number 1 is "Confident belief in the truth, value, or trustworthiness of a person, idea, or thing.". Meaning number 2 is "Belief that does not rest on logical proof or material evidence.". Number 3 is "Loyalty to a person or thing; allegiance". Is that what you mean by faith?
Methodus: Yes.
John: Well...here, I have faith in the following idea. Faith is an obstacle to truth.
Methodus: What do you mean?
John: I believe that if you wish to seek the truth about something, then faith and loyalty will be of no help to you.
Methodus: If you have faith that this idea is true, wouldn't that faith be an obstacle to you in determining the truth-value of this statement?
John: Good point. Let me rephrase that...I believe that the following statement is true: Faith is an obstacle to truth.
Methodus: What's the difference? You just changed your feeling from faith to belief.
John: Yes. And in so doing I have removed confidence and loyalty......And as a result, now I feel more open to the possibility that somehow faith and loyalty might be useful to someone who wishes to seek the truth.
Methodus: But you still believe that faith is an obstacle to truth.
John: Yes. But it wouldn't surprise me if the opposite were true.
Methodus: I guess if you don't have faith then it's difficult to be surprised......What about the Universe? Do you have faith that the Universe will always exist?
John: This is something that I believe. But I do not have faith in the Universe. I do not have faith that it will always exist.
Methodus: ...The second meaning for faith in the dictionary is "Belief that does not rest on logical proof or material evidence.". Isn't your belief that the Universe will always exist, a belief of this type?
John: No. For one thing, my belief rests on the fact that I believe that the Universe has existed persistently since I was born.
Methodus: Wait a minute...you said you believe that the Universe has existed persistently since you were born. Why have you qualified that with I believe? Do you think it is possible that the Universe has not existed persistently since you were born?
John: Well...I can't say much about what goes on around here while I'm asleep.
Methodus: Would you be surprised if it turned out that the Universe ceased to exist at midnight?
John: How could I be...
Interruptus: How is the edge-face difference O used for the characteristic shift-attack-time in Dodecahedron?
John: For shift-attack-time, both O and the absolute value of O are used. These numbers determine which time-scale point will be displaced and the direction of the displacement.
Methodus: How so?
John: Well...suppose we have a particular stream. There will be a sequence of main notes produced by that stream. We can partition this sequence into contiguous subsequences for which all of the notes in a given subsequence will have the same current face.
Methodus: OK.
John: For the characteristic shift-attack-time, the time at which exactly one particular time-scale point occurs will be shifted in time.
Methodus: Which time-scale point would be affected in this way?
John: Suppose the edge-face difference O is positive. In that case we would assign a number to each of the main notes from beginning to end, starting with number 0 (zero). Except, we would skip the last note of each face.
Methodus: You mean you would skip the last note in each contiguous subsequence?
John: Yes.
Methodus: So for example, if the sequence of main notes were of the form (NNN) (NNNNNN) (N) (n-n-n, n-n-n-n-n-n, n), you would number these as (01-) (23456-) (-) (zero-one-dash two-three-four-five-six-dash dash).
John: Well...yes. That's the idea. But your example is impossible.
Methodus: What do you mean?
John: The number of notes in each subsequence must be greater than or equal to 2 and less than or equal to 5.
Methodus: Oh...right. It's a stream...So it has to leave the current face if it has traveled all the way around it.
John: And the current face of any particular note must be the same as the current face of the note that it follows or precedes.
Methodus: OK. So would this be right? If the sequence of main notes were of the form (NNN) (NNNNN) (NN), you would number these as (01-) (2345-) (6-).
John: Again, you've got the right idea. But this example is impossible as well.
Methodus: Why?
John: You cannot construct a stream like this. You're saying that there would be two simple edges, followed by a pivot edge, followed by three simple edges, followed by a pivot edge, followed by a simple edge. For example, you might have the following sequence of turns: LLRRRR. That would give you two simple edges, followed by a pivot edge, followed by four simple edges. The stream has to end there. You can't convert the last of these four simple edges into a pivot edge because you cannot go any further.
Methodus: Oh...I see. How about this one (NNN) (NN) (NNNNN)?
John: Yes, you could have a stream like that. For example, that would be what you would get for a stream that has the following sequence of turns: LLRLLLL.
Methodus: So in this case you would number the notes as (01-) (2-) (3456-). Right?
John: Exactly...That is, if the edge-face difference O is positive.
Methodus: Would it be different if it were negative?
John: Yes. Then the notes would be numbered from end to beginning, again skipping the last note of each current face.
Methodus: So for that case you would have (65-) (4-) (3210-)?
John: That's right.
Methodus: OK...But which time-scale point would be shifted?
John: Suppose there are a total of N main notes. Further, suppose the number of faces that the stream visits is F.
Methodus: You mean there are F subsequences?
John: Yes...To N minus F notes, we would assign numbers 0 (zero) through N minus F, minus 1.
Methodus: Right. In this case, we would be skipping a total of F notes.
John: Now,
find the note for which the number is the absolute value of O, modulo N
minus F.