John: Well...I needed 9 in all. I picked the all-combinatorial hexachords because they are relatively symmetric, like Platonic solids. To select the other 3, I used a list of the 20 equivalence classes that I had constructed to enumerate them. I removed the 6 that contain the all-combinatorial hexachords. From the 14 remaining classes, I selected the first on the list that contained a string of 5 black beads. That was 5115. Then I did the same for 4 black beads and 3 black beads. That is how I selected 411114 and 312312.

Methodus: Earlier you said that in order to assign pitches, you partitioned the stream types into equivalence classes according to the position and direction of the last edge of the stream, relative to the first edge. Then, for each of these equivalence classes you defined one pitch-class mapping to be used for all stream types in the class. How did you use these 9 equivalence classes of self-complementary pitch-class sets to accomplish this?

John: In order for me to tell you about that, I think it would help if we discussed the naming convention that I used for the edges of a dodecahedron, relative to a given edge.

Methodus: OK.

John: To name the 30 edges, I used the letters a through z, along with symbols for the arithmetic operations of addition, subtraction, multiplication and division.

Methodus: How did you assign these names to edges?

John: Suppose you have selected a particular edge of the dodecahedron. Call that edge a. Next, select one of the faces that is incident upon edge a. Set the dodecahedron down on a flat surface so that the face that you selected is on top. In clockwise order, name the other four edges of this face as b, c, d and e.

Methodus: So as you look down upon the dodecahedron, the topmost edges would be a, b, c, d and e in clockwise order. Right?

John: That's right. Next, we shall name the set of 5 edges that are incident upon the topmost vertices, that we have yet to name. The edge that meets edges a and e at a vertex is called edge f. Now, while looking down on the topmost face, name the other 4 edges in this set as g, h, i and j...in clockwise order.

Methodus: So edges e, a and f meet at a vertex...and the same may be said of the edges: a, b and g; b, c and h; c, d and i; and d, e and j. OK, what's next?

John: Next, we name the 10 edges that are around what you might call the equator of the dodecahedron.

Methodus: The equator?

John: If you looked down upon the dodecahedron from a point that is directly above it, the equator would be the 10 edges at the boundary of the projection that you would see from that point of view. Or, you could think of this dodecahedron oriented as it is, as being formed from a top half and bottom half, where each half has 6 faces. The equator would be the edges where these two halves meet.

Methodus: Oh, I see...so what would be the names for these 10 edges?

John: These are named k through t, in clockwise order.

Methodus: You mean, looking down on the die, the edges of the equator would be named k, l, m, n, o, p, q, r, s and t, in clockwise order. Is that right?

John: Yes.

Methodus: Which one would be k?

John: Edges k and j are incident upon a common vertex.

Methodus: Edge t would be incident on this vertex, too...right?

John: Yes, the clockwise traversal around the equator begins and ends at this vertex.

Methodus: What would be the names of the 10 edges that are below the equator?

John: First, we name the 5 edges that connect a vertex of the equator with a vertex of the bottom face. In clockwise order, name these edges u, v, w, x and y. Make edge u be the one that meets edges k and l at a vertex.

Methodus: So, the following edges meet at a vertex: m, n and v; o, p and w; q, r and x; and s, t and y. Right?

John: That's right. Now, we have to name the 5 remaining edges that are incident upon the bottom face. In clockwise order, those would be edges z, + (plus), - (dash), * (star) and / (slash). This is done, beginning with the vertex at which edge y meets the bottom face.

Methodus: Let me make sure I have this right. If I look down upon the top face, I see the following five edges in clockwise order: abcde. If I do this for the other 11 faces, I would get: afmng, bgoph, chqri, distj, ejklf, ktyzu, mlu+v, onv-w, qpw*x, srx/y and...

Interruptus: Earlier you said that the longest possible stream on a dodecahedron is 21. Could you give an example of a stream type for which the length is 21, in terms of these edge names?

John: Sure, here is one: agnmfedchpw-+uktsrx/z.

Methodus: Are there other stream types for which the length is 21?

John: Yes, quite a few...Out of the 7800 possible stream types, there are 196 (one-hundred-ninety-six).

Methodus: ...I wonder what would be the probability of generating the particular stream type of length 21 that you have given as an example.

John: You mean, if you made each turn based on the result of flipping a fair coin?

Methodus: Yes...I know it would be one-half to the power D where D is the number of turns minus the number of forced edges...so......well, as far as I can tell, this stream has only one forced edge. Is that right?

John: Yes, for the stream type agnmfedchpw-+uktsrx/z, there is exactly one forced edge. It occurs at the turn made to go from edge f to edge e. Here, you are at the vertex upon which the edge a, f and e are incident. When you arrive at this vertex after traversing edge f, you must traverse edge e next.

Methodus: Right...because you will have already traversed edge a at that point in the stream. So, no coin toss is necessary for this decision. So the probability of generating this particular stream type would be one-half to the power 19. Right?

John: Yes. That's right.

Methodus: You know...it is surprising to me that this stream type has only one forced edge.

John: Why is that?

Methodus: I would think that a stream type that is as long and complex as this would have several forced edges. Is this stream type a special case?

John: No. As it turns out, no stream type can have more than one forced edge.

Methodus: How did you figure that out?

John: Brute force...the method of exhaustion. I wrote a program to count the number of forced edges for each stream type.

Methodus: Do you have some other way of proving this?

John: ...Yes. It is possible to show that the only vertex along a stream at which a forced edge will occur would be at the vertex at which the stream starts. That is a consequence of the fact that the valence of each vertex in the graph of a dodecahedron is 3.

Methodus: How's that?

John: Well suppose you are traveling down a stream and you arrive at a vertex that is not the first vertex of the stream. That is, it is not the vertex at which the stream starts. There are only two possibilities. You have either visited this particular vertex already, or you have not. If you have not, then either one of the two other edges that are incident upon this vertex could be the next edge of the stream. If you have visited it already, then this must be the last vertex of the stream.

Methodus: Why?

John: If you had visited this vertex then you must have approached it along some edge, because it is not the first vertex of the stream. Then, you would have departed from it along another edge, because it was not the last edge of the stream.

Methodus: So if a stream returns to the vertex from which it began, there will be a forced edge. Otherwise, there will be no forced edges. Right?

John: Yes.

Methodus: Do all 196 stream types of length 21 have exactly one forced edge?

John: Yes.

Methodus: Can you give some explanation for why that would be the case?

John: Sure. Suppose you were able to determine that the length of a stream on a dodecahedron cannot be longer than 21. Suppose you had a particular stream for which the length is 21. Now suppose this stream has no forced edges. That would imply that the stream does not revisit the vertex from which it started. So, that would mean that only one of the three edges that is incident upon the first vertex has been traversed. If that were the case, then it would be possible to extend the stream backwards from the starting vertex along one of these two edges that has not been traversed. But, that would produce a stream for which the length is greater than 21. Since that is impossible, the stream cannot have no forced edges.

Methodus: That makes sense......Are you aware of any other stream lengths for which there must be at least one forced edge?

John: No. For all possible stream lengths except 21, there exists at least one stream type that has no forced edges.

Methodus: What about stream lengths for which there can be no forced edges? Are there any of those?

John: Yes...for example if the stream length is 6, there can be no forced edges.

Methodus: I can see that. In that case there are only two stream types. As a sequence of left or right turns those would be RLLLL and LRRRR. Those are the only possibilities, right?

John: Yes. And for those, there are no forced edges.

Methodus: Are there any other lengths like this?...where there are no forced edges.

John: Yes. Stream types of length 7, 8 and 10 cannot have any forced edges.

Methodus: I can see that for 7. The only stream types of length 7 that I can imagine would be LLRRRR, LRLLLL, RLRRRR and RRLLLL.

John: You're right. Those are the only ones. The stream types of length 8 have a similar structure too. There are 8 of those. Each begins with one of the 8 possible ways to make the first three turns. Then, they proceed around a face at the end. For example, one of them is LRLRRRR.

Methodus: What about the case when the stream length is 10? That seems to be a bit more complex. Why can there be no forced edges for that case?

John: I haven't thought of a simple proof for...

Interruptus: Maybe it would be a good idea if we came up with a name for you.

John: For me?

Interruptus: Yes, we have John, Methodus, Interruptus and Complementum. Complementum is everything that is not you. If John is you, then where does that put me and Methodus? I mean are we part of Complementum or you?

John: .........Well...my first question would be: Are you part of the Universe?

Methodus: Aren't we a part of you? I mean aren't we a part of John?

John: Well...you are fictional characters that I created for this interview.

Methodus: Did you do that alone...or with our help?

John: What do you mean by that?

Methodus: I mean, are we just fictional characters that you created, or are we real entities that helped you create these characters? Are we real entities that have helped you with all your work?

John: If you had helped me with all my work, why would you have to ask questions about it? Wouldn't you already know everything there is to know about it?

Methodus: Because you asked me to ask questions about...

Interruptus: And you allowed me to join the conversation.

John: If the two of you are part of John, can't I still be all of John?

Methodus: I guess the question is: Are you?

John: What you seem to be getting at is that there are components of John called Methodus and Interruptus. Where does that leave me? Am I just another component of John or am I John?......if you were to consider me to be a component of John, what would be my name?

Methodus: I was thinking something along the lines of Director, Conductor or Manager.

John: Maybe I am more of a conduit.

Methodus: Between Interruptus and me?

John: Yes, maybe I facilitate communication.

Interruptus: Perhaps you are an arbiter. What role have you been playing in this interview?

John: I have been answering questions about John's work.

Methodus: Do you consider yourself to be an authority on this subject?

John: Well...I know a great deal about how it was created......you know, I think John might be the best name for me because it is the whole of John that I have represented throughout this interview. I would say that the two of you are a part of John rather than Complementum, but for the purpose of this interview, John has called upon you to play roles as entities of Complementum.

Methodus: So we are doing double duty. We are asking questions. But we are also involved in the process of writing John's responses, because we are part of John. Right?

John: Yes. You can think of it this way. In this interview, John is asking John questions. Through the course of this interview, John decided that this would be done by having two interviewers rather than one. Each interviewer is a distorted version of John. Methodus, in this interview, for the most part you may think of yourself as John with Interruptus removed or suppressed. Interruptus, you may think of yourself as John with Methodus removed or...

Interruptus: You know...a ladybug is like a turtle with wings.

John: See what I mean?

Methodus: Yes...they both have the shape of a hemisphere.

John: ...And when they are threatened, they retract into their...

Interruptus: For Dodecahedron, you said that you assigned a musical characteristic to each stream type. Suppose I had the particular stream type: abhpog. What would be its musical characteristic?

John: Since the last edge is g, the characteristic would be crescendo.

Methodus: And if the stream were agophb, it would also be crescendo, right?

John: No. Why do you say that? It would be decrescendo.

Methodus: Earlier you said that from a given stream type, you could generate another stream type by replacing all right turns with left turns, and all left turns with right turns. You said that you assigned the same characteristic to two stream types that are related in this way.

John: I said that?

Methodus: Yes, that's what I thought you said.

John: Oh...well that isn't right. Sorry about that. I assigned complementary characteristics to stream types that are related in this way.

Methodus: Wait...you partitioned the stream types into 29 equivalence classes. For this, two stream types are equivalent if they end on the same edge. Right?

John: Yes.

Methodus: But earlier you listed only 16 possible musical characteristics. So some of these equivalence classes must have the same characteristic. Is there another equivalence class for which the characteristic is crescendo?

John: Yes. If the stream ends on edge g or f, the characteristic will be crescendo. If the last edge is b or e, it will be decrescendo.

Methodus: I can see there is some structure to this. These are the four edges that are adjacent to edge a. Edges f and g are on one side of edge a. Edges b and e are on the other side. Would there be a similar structure to the way that you have assigned the other musical characteristics?

John: Yes, there is. More generally, you can think of it this way. Consider the following two operations that you might perform on a stream type. I will call them rotation and reflection. We have already discussed reflection. That would be where you replace the left turns with right turns, and vice versa. With rotation, you change the direction of the first edge of the stream.

Methodus: So if you perform the operation of reflection on the stream type abhpog you would get agophb. And if you rotate abhpog, you would get...aflkje. Right?

John: That's right. Now, consider the following composite operation that I shall call rotation-reflection. With this operation you can partition the set of stream types into equivalence classes. Here, two stream types are said to be equivalent under rotation-reflection if and only if one may be obtained from the other by rotation-reflection.

Methodus: OK. For example, if you rotate abhpog, you get aflkje. Now if you reflect that you would get aejklf. So abhpog and aejklf would be equivalent under rotation-reflection.

John: Yes. Now, I have assigned the same musical characteristic to any two stream types that are equivalent under rotation-reflection.

Methodus: So two different stream types will have the same characteristic if they end on the same edge, or if they are equivalent under rotation-reflection.

John: Yes. Perhaps it would be simpler to introduce the terms last-edge-equivalent and rotation-reflection-equivalent for these two different ways that stream types may be equivalent. With that we may say that two different stream types will have the same characteristic if they are last-edge-equivalent or rotation-reflection-equivalent.

Methodus: ...and if one stream type is the reflection of another the two will have complementary characteristics.

John: That's right.

Methodus: OK. Crescendo and decrescendo is one complementary pair. Among the others, I would guess that the other pairs would be: sharp-or-flat-to-natural and natural-to-sharp-or-flat; staccato and tenuto; note-level-crescendo and note-level-decrescendo; accelerando and decelerando; and sharp and flat. Are there any others?

John: Yes, I have treated exaggerate-frequency-range and senza-vibrato as complementary.

Methodus: You have assigned crescendo to streams types for which the last edge is g or f, and decrescendo if the last edge is b or e. What about the other edges?

John: If the last edge is n or m, the characteristic will be sharp-or-flat-to-natural. If it is c or d, it will be natural-to-sharp-or-flat. If it is o or l, the characteristic will be staccato. If it is h or j, it will be tenuto. For v and i, the characteristics are note-level-crescendo and note-level-decrescendo, respectively. For w and u, the characteristic is accelerando. For q and t, it is decelerando. For - and + it's sharp. For r and s it's flat. For * and z it's exaggerate-frequency-range. For x and y it's senza-vibrato. Edges p, k and / are special.

Methodus: How so?

John: The reflection of a stream type that ends on edge p, will also end on edge p.

Methodus: But it will be traveling down edge p in the opposite direction.

John: True, but here we are not concerned about direction. Now, the same may be said of edge k. For these edges, there are no complementary characteristics. If a stream type ends on edge p or k, the characteristic will be pitch-bend-up-or-down.

Methodus: You mean if it ends on p, it will be pitch-bend up. And if it ends on k it will be pitch-bend down?

John: No. You can have a pitch-bend up, or down for these edges.

Methodus: When would it be up versus down?

John: For edges p and k, if the stream is traveling from left to right along the edge, the bend will be up. Otherwise it will be down.

Methodus: What do you mean by left to right?

John: Imagine that you are holding the dodecahedron in front of you. Now, orient it so that the first edge of the stream is the edge that is nearest to your point of view. Rotate the dodecahedron so that the first edge is directed upwards.

Methodus: OK.

John: Then, edge p will be the topmost edge. Edge k will be on the lowest. While we're here, I should mention three other edges that are special.

Methodus: What would those be?

John: There is the leftmost edge, which is v; the rightmost edge, which is i; and the edge that is farthest from your point of view and diametrically opposed to edge a, which is edge /.

Methodus: OK. I can see that.

John: It is from this point of view that the sense of left and right are defined for edges p and k.

Methodus: So to travel down edges p or k from left to right means to travel along these edges from left to right, with respect to this point of view. Right?

John: Yes...while you have that in mind, there is a concept of direction for the edge /, as well. For that edge, there is forward and backward. If the direction is from top to bottom, that would be forward. Otherwise, it would be backward.

Methodus: What characteristic is assigned to edge /?

John: That would be shift-attack-time.

Methodus: You mean, shift the time at which notes start?

John: Yes. For these phrases, some notes might be early or late.

Methodus: So, would notes be early if the direction on edge / were forward, and late if it were backward?

John: No. The direction by which the start time of a note would be shifted depends on...

Interruptus: In Dodecahedron, you said that each duration mapping is based on a particular way to partition n into 6 parts, where n is some integer in the closed interval 6 through 12. Each of these parts specifies a particular number of duration units. And these are assigned to faces. How did you decide which part should be assigned to a particular face?

John: Well...as I said earlier I assigned the same duration to faces that are diametrically opposed on the die. Said another way, any two faces for which the sum of the face numbers is 13 will have the same number of duration units.

Methodus: So when you define the duration mapping for a particular partition, you only need to specify the durations that are assigned to the odd faces. Right?

John: That's right. Or you could specify the duration for each even face...or you could consider only those faces for which the face number is in the interval 1 through 6.

Methodus: Right. So for a given partition, let's say 1+1+1+1+2+3, how did you select the particular number of duration units to be assigned to a given face? I mean, in this case there would be quite a number of ways to assign the components to faces. First, you could select any 4 of the 6 faces for which a 1 would be assigned. Then, for each of these cases there would be two ways to assign the remaining components 2 and 3. That would be a total of 30 different possibilities. How did you go about selecting one of these?

John: Well...this decision wasn't made in isolation...or independently.

Methodus: What do you mean?

John: When defining a particular duration mapping, I considered the effect it would have on two statistical properties calculated over all 30 duration mappings. First, I wanted to define the duration mappings in such a way that the average of the durations that are assigned to a face would be roughly the same for every face.

Methodus: I'm not sure if I am following what you mean by this average.

John: Well...in all there are 30 different duration mappings. A duration mapping specifies the number of duration units that are assigned to each face. For any given face, you could calculate the average number of duration units that would be assigned to that face, over the 30 different mappings. That would be some total number of duration units divided by 30. You could calculate this average for each face.

Methodus: All right. Let me see if I understand this correctly. Suppose I constructed a matrix with 30 rows and 12 columns. In each row I place the numbers for one of the duration mappings. The number in the j-th column of the i-th row would be the number of duration units assigned to face number j in duration mapping number i.

John: OK. Here, column j would be identical to column 13 minus j...because two faces for which the sum of the face numbers is 13 must have the same number of duration units.

Methodus: Right...so I guess we could eliminate columns 7 through 12. Let's do that. Now, the average number of duration units that are assigned to face number 1 would be the average of the 30 numbers in the first column, right?

John: Yes.

Methodus: So you could calculate an average like this for each column. If I understand you correctly, you arranged the six components of the partition in each row so that these 6 averages would be roughly the same.

John: Yes. Exactly.

Methodus: You said that you used two statistical properties. These averages are one. What would be the other one?

John: I wanted every particular number of duration units to occur with roughly the same frequency on every face.

Methodus: So in terms of this matrix, by frequency you mean the number of times that a particular number occurs in a given column.

John: Yes. You could think of these frequencies as forming a second matrix. Call the first matrix P, for partitions. Call the second matrix F for frequency. The elements of P are in the range 1 through 7. For the matrix F, there would be 7 rows and 6 columns. The element in the i-th row of the j-th column of F would be the number of times that the number i occurs in the j-th column of matrix P.

Methodus: OK. So you tried to rearrange the elements of each row of matrix P so that in each row of matrix F, the elements would be roughly the same.

John: Correct...while trying to have the average of each column of P be roughly the same.

Methodus: How did you accomplish that?

John: I did it manually. I mean I didn't devise an algorithm to solve this problem. I used an ad hoc approach until the results that I got were adequate.

Methodus: What was the result?

John: Well...in terms of the matrix P, the 30 rows of P would be as follows:

1 1 1 1 1 1 (one one one, one one one)
1 1 1 1 1 2 (one one one, one one two)
1 1 1 1 3 1
2 2 1 1 1 1
1 1 1 4 1 1
2 1 1 1 1 3
2 1 1 2 1 2
1 1 5 1 1 1
1 2 1 1 1 4
1 1 1 1 3 3
1 1 1 2 2 3
1 1 2 2 2 2
1 6 1 1 1 1
1 1 2 5 1 1
4 1 1 3 1 1
1 1 4 2 2 1
3 1 1 3 1 2
2 3 1 1 2 2
2 2 2 2 2 1
7 1 1 1 1 1
1 2 1 1 6 1
1 3 5 1 1 1
1 2 1 2 1 5
1 1 1 4 4 1
1 2 3 1 4 1
1 4 2 2 1 2
3 3 3 1 1 1
2 1 2 3 3 1
3 2 2 1 2 2
2 2 2 2 2 2

Methodus: Is this an optimal solution?

John: I am not sure. It's not too bad, though. The sum of the elements in columns 1 through 6 is: 52, 52, 52, 54, 54 and 51, respectively. So the average element in each column is almost the same. And, in each row of the frequency matrix F, the elements are roughly the same. The seven rows of F look like this:

18 17 18 16 17 17
7 8 7 8 7 8
3 3 2 3 3 3
1 1 1 2 2 1
0 0 2 1 0 1 (zero zero two, one zero one)
0 1 0 0 1 0
1 0 0 0 0 0

Methodus: This optimization problem...I can think of a more general form.

John: What do you have in mind?

Methodus: Here...given a matrix with integer elements, rearrange the elements of each row in such a way that the average of the elements of each column is as close as possible to the average of all the elements of the matrix, and simultaneously, make the number of times which a given integer occurs in each column be roughly the same.

John: Yes, I can imagine a generalization along a different direction.

Methodus: What's that?

John: Here we were looking at a family of matrices P for which the rows are the possible ways to partition n into 6 parts, where n varies from 6 through 12. This could be generalized as the ways to partition n into k parts, where n varies from k through 2k. We have been discussing the case when k is 6. This generated a family of matrices that we could call P6 (p six). Similarly, we could consider a family of matrices Pk (p k) for which the rows are the different ways to partition n into k parts, where n varies from k through 2k.

Methodus: Right. Then you could pose the same optimization problem for Pk.

John: Yes. And again you would be rearranging elements to make the average of the elements in each column be as close a possible to the average of all the elements of Pk...I have toyed around with this a bit. From what I can see, I think there is a chance that the average element of Pk converges as k goes to infinity.

Methodus: Here you're talking about a sequence a1 (a one), a2, a3 and so on, where ak is the average of the elements of Pk, right?

John: Yes, that's the sequence.

Methodus: What is the limit of ak?

John: I'm not sure. I took a quick look at this up through a45. Maybe it converges to 2.

Methodus: What are the terms of the sequence?

John: I think the first seven are: 3/2 (three-halves), 13/8 (thirteen-eighths), 35/21 (thirty-five over twenty-one), 82/48, 164/95 (one-sixty-four over ninety-five), 315/180 (three-fifteen over one-eighty) and 555/315 (five-fifty-five over three-fifteen). The reduced terms would be 3/2, 13/8, 5/3 (five-thirds), 41/24 (forty-one over twenty-four), 164/95, 7/4 (seven-fourths) and 37/21 (thirty-seven over twenty-one). It looks like a45 might be 1.83976075 (one-point-eight-three-nine seven-six-oh seven-five)...

Interruptus: What do you think happens to a person after they die?

John: You mean, does a person still exist in some form after their body ceases to function?

Interruptus: Yes.

John: I think at a minimum, they exist as an idea.

Methodus: What does the idea consist of for a given individual?

John: It would be everything that they did while they were alive.

Methodus: What about their beliefs?

John: Yes, their beliefs would have been expressed or communicated through things that they did while they were alive.

Methodus: Do you believe that a person can cause certain things to happen after they die?

John: Well...an idea can be quite forceful. When people believe in a particular idea, it can lead them to act in certain ways. But, I don't know if any person can cause something to happen whether they are alive or dead. By that I mean, independently of other entities of the Universe.

Methodus: By cause, I don't mean independently necessarily. I just mean that something happens as a result of something a person does. For example, if you turned your cup of water over so that the water spills on the floor, I would say that you caused the water to fall on the floor.

John: Provided you didn't push me.

Methodus: Yes.

John: But, suppose it was you who caused that water to be in the cup.

Methodus: How's that?

John: Well, suppose the main reason why the water is in the cup in the first place is because you asked me so many questions, that my throat became sore. So I needed to get water.

Methodus: Yes, but I didn't spill the water on the floor.

John: I know. But it would not be on the floor, if you had not caused me to get it.

Methodus: OK. I guess to some degree I caused the water to be on the floor. Maybe I should rephrase my question. Do you believe that a person can be one of the causes for a particular event to happen after they die?

John: Yes, as I said I believe an idea can play a role in how events unfold.

Methodus: That's not exactly what I had in mind by cause. Let's try an example. Suppose that your Aunt dies on August 7th, 1995. Now suppose that on July 7th, 1996, you wake up to find exactly 27 pigeons sitting on the railing of your porch. You know that your Aunt was born on June 4th, 1927. Do you believe that it is possible that somehow your Aunt has caused these pigeons to be on your porch?

John: Yes I believe that is possible.

Methodus: Would that be a strong belief?

John: That would depend. For example, suppose my Aunt and I had a close relationship. Or, suppose I knew that my Aunt had been an avid bird watcher all her life. Or, suppose that on September 7th, 1995, there had been 27 crows circling directly over my home for several minutes. Or, suppose that I have never seen a pigeon near my home before this event. Any of these things would cause me to have a stronger belief.

Methodus: Why would a dead person do this?

John: Perhaps for no other reason than to let you know that they still exist, and that they can do such things.

Methodus: So your Aunt would be trying to communicate with you by causing these pigeons to be on your porch.

John: Yes. Perhaps. But again, the degree to which I would believe that this particular event were caused by my Aunt would depend on probabilities.

Methodus: Suppose the additional conditions that you described were true. What would be your thought process to calculate these probabilities?

John: You mean the four conditions that I said would strengthen my belief?

Methodus: Yes.

John: Well...when I saw the 27 crows, I would think...that's strange. It's exactly one month since my Aunt died. And here are these crows. 27 of them. My Aunt was born in 1927. And she loved birds. I wonder if this is a message from my Aunt. We were very close. I miss her very much. Perhaps she is trying to let me know she is OK...and that I will be OK too, because she is present.

Methodus: Then what would you think when you saw the pigeons?

John: I would say to myself...wow, look at that. That's weird. What the heck are those pigeons doing there? Hey...there are 27 of them. Like the 27 crows. What day is it? It's one month before the first anniversary of my Aunt's death. Perhaps with these 27 birds she is trying to let me know that she is present and OK.

Methodus: So you're saying that if you can establish some relationship between the event and your Aunt, your belief would be stronger.

John: Yes. Suppose one believes that there is some chance that a dead person could communicate in this way. I believe it is reasonable to assume that they would try to accomplish this by using symbols for themselves in the context of an unlikely event.

Methodus: That seems logical. If the event were not unlikely then, then it would go unnoticed and the message would be missed.

John: Right. And if a symbol were not used then you would not know that the message was from them.

Methodus: So in this case, we have two unlikely events: the crows and the pigeons. These birds serve as symbols for your Aunt.

John: And the number 27 is a symbol because she was born in 1927.

Methodus: The particular dates would be symbolic too. That is, the events occurred on dates which bear a simple relationship to the date on which your Aunt...

Interruptus: For Dodecahedron, you said you used 9 equivalence classes of self-complementary pitch-class sets. How did you use these to define pitch-class mappings?

John: Well...from each of these equivalence classes, I selected exactly one self-complementary pitch-class set.

Methodus: How did you do that?

John: For example, let's take the equivalence class that we have been calling 411411. First, I used this name to draw a bracelet with 6 black beads and 6 white beads arranged in the following order BBBBWBWWWWBW (b-b-b-b w b w-w-w-w b w), where B and W mean black and white, respectively. Then, I located the first black bead that is after the longest string of white beads. In this case, that would be the black bead that occurs after the string of four white beads. I labeled this bead as pitch class O. Then, I labeled the beads that follow it as P, Q, R, and so on around the bracelet, through Z. To form the particular pitch-class set to be used for this equivalence class, I selected the pitch-class names that were on the black beads.

Methodus: So for the equivalence class 411411, that would be...OQRSTV (o q r s t v), right?

John: Yes. That's right.

Methodus: So for the equivalence class 2111121111 you would have a bracelet with the following sequence of beads: BBWBWBWWBWBW. Here, O would be the black bead that is after the two white beads. The next bead would be P. That's a white bead. After that there is a black bead, which is Q. Then there is a white bead, which is R. After that there are two black beads. Those would be S and T. The next four beads which are white, black, white and black, would be U, V, W and X. The remaining two white beads would be Y and Z. So, here the black beads would be OQSTVX. Is that right?

John: Yes. That's correct. Similarly, for 66, you would have OPQRST.

Methodus: OK. What about 3333? There are two strings of white beads of length 3. Which of these strings do you choose to determine the black bead that would be O?...Oh...I guess it doesn't matter. We would get the same pitch class either way. I mean, regardless of which one you pick you will get the pitch-class set OPQUVW, right?

John: That's right. The same thing happens with 222222 and 111111111111. For those the pitch-class sets will be OPSTWX and OQSUWY, respectively.

Methodus: Now, what about the 3 equivalence classes that you used that do not contain all-combinatorial hexachords?

John: For 5115, 411114 and 312312 I did the same thing.

Methodus: OK. So for 5115, we get OPQRSU; for 411114 we get OPQRTV; and for 312312 we have ORSTVW. Are those right?

John: Yes, that would be correct.

Methodus: Wait a minute...there's something bothering me here.

John: What's that?

Methodus: Well...these pitch-class sets are determined from the names that you are using to describe these equivalence classes.

John: Yes...

Methodus: These names seem somewhat arbitrary to me. For example, you could have called the equivalence class 5511 instead of 5115. Had you done that, we would have ended up with the pitch class set OQRSTU instead of OPQRSU. I can see how it would not matter for the six equivalence classes that are all-combinatorial hexachords, but for these other three classes, the orientation of the bracelet makes a difference. I mean, if I flip it over I get a different pitch-class set.

John: That's true. But, suppose we do select OPQRSU. The complement of this will be TVWXYZ. And the complement will have the same structure as the alternative pitch-class set OQRSTU.

Methodus: OK...because TVWXYZ is OQRSTU, transposed up by an interval of 5. But how does that help us?

John: Well...both the pitch-class set OPQRSU and its complement TVWXYZ will be used. The pitches of the selected pitch-class set OPQRSU will be assigned to the odd faces. Those of the complementary pitch-class set TVWXYZ will be assigned to the even faces.

Methodus: Oh, OK...how did you assign the pitches to particular...

Interruptus: Earlier you described the way by which one of 16 different musical characteristics is assigned to each stream type in Dodecahedron...I guess we could call this the characteristic mapping. For example, you said that for all stream types for which the last edge is g or f, the characteristic is crescendo. Why did you make this particular class of stream types be crescendo? More generally, what was your rationale for assigning a particular characteristic to a particular class of stream types?

John: I think I may have said this before...I'm not sure......I defined this characteristic mapping before the randomly generated sequence of 697 stream types was created.

Methodus: So you made these decisions about characteristics before you knew which stream types would be used for the piece?

John: Yes.

Methodus: How did you do that?

John: My decisions were based on probabilities.

Methodus: You mean the probability that a given stream type would occur?

John: Yes...but more specifically, the probability that a randomly generated stream type would end on a given edge.

Methodus: How did you calculate these probabilities? Did you use the fact that the probability for a given stream type to occur is one-half to the power D where D is the number of turns minus the number of forced edges?

John: Yes. In order to apply that formula, I analyzed the 7800 possible stream types. For each of the 29 edges at which a stream type can end, I counted the number of stream types of a given length. Then for each case, I determined the number of stream types that have a given number of forced edges.

Methodus: Could you give an example of this for one particular edge?

John: Sure. Let's consider edge b. Of the 7800 stream types, 280 (two-hundred-eighty) end at edge b. Exactly one of these stream types has length 6. This stream type does not have a forced edge. So the probability of it occurring would be one-half to the power 5......The probability of generating a stream type that ends at edge b, has a length of 6 and has no forced edges, is 0 (zero). Because there are no such stream types.

Methodus: ...There will be 5 turns in a stream of length 6, so...OK, that makes sense. Can you give another example that is more complicated?

John: Sure. Let's continue with edge b. Let's consider the probability that a stream type of length 16 will end on edge b. There are a total of 45 stream types of length 16 that end on edge b. Of these, 27 have no forced edges. The other 18 have one forced edge......So, the probability that a stream type ends at edge b, has a length of 16 and has no forced edges would be one-half to the power 15.

Methodus: Wait...wouldn't it be 27 times that?

John: Oh...yeah, sorry, the probability would be 27 times the quantity one-half to the power 15. Now, the probability that a stream type ends at edge b, has a length of 16 and has exactly one forced edge would be 18 times the quantity one-half to the power 14.

Methodus: So, to get the total probability that a stream type of length 16 will end on edge b, you would add these two probabilities together, right?

John: Yes, that's it.

Methodus: So for each edge, each possible length, and each possible number of forced edges you counted the number of stream types. From that you calculated a probability.

John: Yes.

Methodus: Let's see, stream lengths vary from 6 through 21...the number of forced edges must be either 0 (zero) or 1......so that would be 29 possible edges times 16 possible stream lengths times 2 possible number of forced edges. That's 928 (nine-hundred-twenty-eight) probabilities. Are all of these nonzero?

John: Oh no. Many are zero. Remember there are only two stream types of length 6. One ends on edge b. The other ends on edge g. The probability that a stream of length 6 will end on any other edge is 0 (zero).

Methodus: Oh yeah, of course. And from what you said earlier, there is no chance that a stream of length 21 can have no forced edges, regardless of where it ends.

John: Right. And there is no chance that a stream of length 6, 7, 8 or 10 can have a forced edge, regardless of where it ends.

Methodus: Are there any other cases like this? I mean, are there any other classes of cases for which the probability is zero?

John: Well...yes, all stream types that end on edge e or f must have one forced edge.

Methodus: So the probability that a stream type with no forced edges will end on edge e or edge f is zero? Why is that?

John: Edges a, e and f share a common vertex. The first edge is a. The second edge is either b or g. If a stream type ends on edge e, then the second to last edge must be f. In this case, e must be a forced edge. If a stream type ends on edge f, then the second to last edge must be e. In this case, f must be a forced edge.

Methodus: OK, I'll buy that. So you have this matrix of 928 probabilities. Is that what you used to determine how to assign musical characteristics?

John: Yes. As I said, when I made these assignments I did not know precisely which stream types would occur in the randomly generated sequence. But, by using this table of probabilities, I could estimate how frequently a member of a given class of stream types would occur. And from this, I could estimate how often a given characteristic would occur.

Methodus: So for example, you could calculate the probability that a stream type would end on a particular edge by adding the 32 probabilities for each possible length and number of forced edges.

John: Yes. I calculated that probability for each edge.

Methodus: What are those probabilities?

John: The probability that a stream type will end on edge b is roughly 0.0894 (point-oh-eight-nine-four). The same is true of edge g.

Methodus: That makes sense. For every stream type that ends on edge b, there is one that ends on edge g, and vice versa. These pairs of stream types are reflections of one another.

John: Right. You could collapse this into a single probability of 0.1789 that a stream type will end on edge b or g.

Methodus: Wouldn't it be twice 0.0894 or 0.1788?

John: No, it's 0.1789. All these probabilities are approximate. They are rounded to the fourth decimal place.

Methodus: Oh, OK...what about the other edges?

John: There are a lot of pairs of edges like this for which the probability would be equal. For example, the probability that a stream type will end on edge e is the same as the probability that it will end on f. Similarly, there are the pairs c and n, d and m, h and o, j and l, i and v, q and w, t and u, r and -, s and +, x and *, and y and z.

Methodus: Yes, I can see that.

John: All right then, here are the 16 different probabilities. I'll write them down from largest to smallest. As I said, the probability that a stream type ends on edge b is 0.0894. I'll write that as P(b) = 0.0894 (p-of-b equals point-oh-eight-nine-four). Also, we have P(g) = P(b) (p-of-g equals p-of-b). These are the edges for which the probability is highest. The probabilities for the other edges would be:

P(h) = P(o) = 0.0476 (p-of-h equals p-of-o equals point-oh-four-seven-six),
P(c) = P(n) = 0.0429,
P(j) = P(l) = 0.0352,
P(p) = 0.0351,
P(i) = P(v) = 0.0343,
P(q) = P(w) = 0.0320,
P(k) = 0.0299,
P(d) = P(m) = 0.0279,
P(x) = P(*) = 0.0275,
P(r) = P(-) = 0.0273,
P(s) = P(+) = 0.0259,
P(y) = P(z) = 0.0251,
P(/) = 0.0244,
P(t) = P(u) = 0.0227 and
P(e) = P(f) = 0.0176.

Methodus: So you defined the characteristic mapping using these probabilities.

John: Yes. I assigned characteristics to edges based on these probabilities. If I wanted a particular characteristic to occur relatively often, I assigned it to an edge that would be likely to occur relatively often.

Methodus: For the particular characteristic mapping that you have used, what would be the probability for a given characteristic to occur. For example, what would be the chance of getting a stream type for which the characteristic is crescendo?

John: Well...it would be crescendo if the stream type ends on edge g or f. The probability of that would be P(g) plus P(f). That would be 0.1070 (point-one-oh-seven-oh). The probability of getting a decrescendo would be the same. That would happen if the stream type ends on edge b or e.

Methodus: What about the other characteristics?

John: The probability of staccato and the probability of tenuto would be 0.0827. For sharp-or-flat-to-natural and natural-to-sharp-or-flat it is 0.0708.

For pitch-bend it's 0.0651.
For accelerando and decelerando it's 0.0547.
For sharp and flat it's 0.0532.
For exaggerate-frequency-range and senza-vibrato it's 0.0526.
For note-level-crescendo and note-level-decrescendo it's 0.0343.
For shift-attack-time it's 0.0244.

Methodus: In addition to these aggregate probabilities, you also had it broken down by stream length too. I mean, for each edge you knew the probability that there would be a stream type of a given length that would end on that edge. Did you use that information as well?

John: Yes, I gave some consideration to that in defining the characteristic mapping. For example, I knew that if I assigned crescendo and decrescendo to edges g and b, respectively, then it would be likely that there would be a relatively large number of relatively short phrases of this type in the piece.

Methodus: Right. Because these are the only edges at which the shortest possible stream type can end.

John: And, the probability of generating these stream types is relatively high because they involve a smaller number of coin flips.

Methodus: Or equivalently, there are fewer turns so the formula for their probability has a lower power of one-half.

John: Yes.

Methodus: I guess when you generate these stream types randomly, you would expect to get more stream types of length 6, than of length 7; more stream types of length 7 than 8; and so on.

John: Wait...that isn't true. The probability of generating a stream type of a given length is not a decreasing function of length.

Methodus: ......oh, right. I forgot to account for the fact that there are more ways to get a stream type of length 7 than 6. For a length of 6, there is only...

Interruptus: For Dodecahedron, how did you assign the pitches of a given self-complementary pitch-class set to the faces of the die?

John: Well...I used two different approaches. I'll refer to these as the forward mapping and backward mapping. First, let's talk about the forward mapping. Let's take the pitch-class set ORSTVW for example. For the forward mapping, I assigned the pitch classes in this set to the odd faces, in order.

Methodus: You mean you assigned pitch class O to face 1, R to face 3, S to face 5, T to face 7, V to face 9, and W to face 11?

John: Exactly. Then I assigned the pitches of the complementary pitch-class set PQUXYZ to the even faces.

Methodus: How did you do that?

John: For that, we should go back to the name of the equivalence class of self-complementary pitch-class sets of which ORSTVW is a member.

Methodus: You mean, 312312.

John: Yes that's it. Now, do you remember how we generated the pitch-class set ORSTVW from the name 312312?

Methodus: Yes. We drew a bracelet with 6 black beads and 6 white beads in the order BBBWBBWWWBWW. Then, we found the first black bead after the longest string of white beads, and we labeled that as pitch class O. Subsequent beads were labeled with subsequent pitch classes.

John: That's right. We could write that in the following way: So we got RSTUVWXYZOPQ (r s t sub-u v w sub-x sub-y sub-z o sub-p sub-q). Here, I have written the pitch classes that are assigned to white beads as subscripted letters.

Methodus: OK. If we use this same ordering, so far we have 3 5 7 # 9 11 # # # 1 # # (three five seven pound nine eleven pound pound pound one pound pound). I mean, O is assigned to 1, R is assigned to 3, S is assigned to 5, and so on. Here, I have used the subscripted number signs as placeholders for the pitch classes that we have yet to assign to face numbers.

John: All right. For the complementary pitch-class set, we locate the first white bead that is before the longest string of black beads.

Methodus: In this case that would be the white bead that we labeled Q, right?

John: Yes, that's it. Now, we assign this pitch class to face number 2. Then, we assign the remaining pitches of the complementary pitch class to the other even faces in reverse order.

Methodus: You mean, Q would be assigned to face 2, P would be assigned to face 4, Z goes to face 6, Y goes to face 8, X goes to face 10 and U goes to face 12?

John: Yes, exactly. Now we can fill in the placeholders in your sequence of face numbers to get
3 5 7 12 9 11 10 8 6 1 4 2 (three five seven sub-twelve nine eleven sub-ten sub-eight sub-six one sub-four sub-two).

Methodus: OK. I understand that. But with some of the other pitch classes there will be more than one longest string of black beads. For example, take the pitch-class set OPSTWX which we obtained from the equivalence class 222222 (two two two two two two). The corresponding bracelet would be BBWWBBWWBBWW. For this case, the length of the longest string of black beads is 2. But there are three strings for which the length is 2. When assigning pitches to the even faces, which of these do you use?

John: I didn't mention this before when we were deriving the pitch-class set OPSTWX from the name 222222...I should have said that in that case, we find the first black bead that is after the last longest string of white beads. So for that case, we have OPQRSTUVWXYZ.

Methodus: OK, but it doesn't really matter in that case because no matter where you begin, you end up with the pitch-class set OPSTWX, right?

John: Yes, that's right. That's why I didn't mention it before.

Methodus: So, for this case the face numbers would be 1 3 # # 5 7 # # 9 11 # #. What about the even faces?

John: For those, you would locate the first white bead that is before the first longest string of black beads.

Methodus: So, in this case that would be the white bead to which the pitch class Z has been assigned, right?

John: Yes.

Methodus: So then we would have the following sequence of face numbers: 1 3 12 10 5 7 8 6 9 11 4 2. Is that right?

John: Yes. That's correct.

Methodus: All the names that we have been using for these equivalence classes of self-complementary pitch-class sets begin with the longest string of black beads. For example, you have 411411 and 2111121111. So that would mean that the first white bead that is before the first longest string of black beads would always be the last white bead.

John: Yes, I suppose you're right. In that case, I guess the pitch class that was assigned to the last white bead would always be assigned to face number 2.

Methodus: OK. Let me see if I can plow through the rest of the cases myself.

John: All right.

Methodus: So far, we have discussed the equivalence classes 312312 and 222222. For those we have:

312312 => (goes to) RSTUVWXYZOPQ  => (which goes to) 3 5 7 12 9 11 10 8 6 1 4 2.
222222 => OPQRSTUVWXYZ  => 1 3 12 10 5 7 8 6 9 11 4 2.

For the other seven equivalence classes that you used, we would get:

66 => OPQRSTUVWXYZ => 1 3 5 7 9 11 12 10 8 6 4 2.
411411 => QRSTUVWXYZOP => 3 5 7 9 12 11 10 8 6 4 1 2.
3333 => OPQRSTUVWXYZ => 1 3 5 12 10 8 7 9 11 6 4 2.
2111121111 => STUVWXYZOPQR => 5 7 12 9 10 11 8 6 1 4 3 2.
111111111111 => OPQRSTUVWXYZ => 1 12 3 10 5 8 7 6 9 4 11 2.
5115 => OPQRSTUVWXYZ => 1 3 5 7 9 12 11 10 8 6 4 2.
411114 => OPQRSTUVWXYZ => 1 3 5 7 12 9 10 11 8 6 4 2.

Is all of that correct?

John: ...Yes...that looks good.

Methodus: All right. So is that it?

John: Well...no. This is the forward mapping. There is a second type of mapping that I used, which I call the backward mapping.

Methodus: What would that be?

John: For that, the order of the odd numbers would be reversed. And the same for the even numbers. So for 312312, you would get 9 7 5 2 3 1 4 6 8 11 10 12.

Methodus: Let me see if I understand this. For 222222 you would get 11 9 2 4 7 5 6 8 3 1 10 12. Right?

John: Yes. That's correct.

Methodus: You could think of this as reversing the order. Or, you could get the backward mapping from the forward mapping by taking the complement of the odd numbers relative to 12, and the complement of the even numbers relative to 14.

John: Yes, that's another way to look at it.

Methodus: Why do you call these mappings forward and backward?

John: Because, for the forward mapping, the odd numbered faces are assigned in increasing order. Whereas for the backward mapping, these faces are assigned in decreasing order.

Methodus: Oh, OK......why have you defined two different ways to map the pitch classes to face numbers? Under what conditions would one approach be used instead of the...

Interruptus: Suppose it is true that people are able to send messages to us after they die. And suppose they can cause events to occur in order to accomplish this. How do they do this?

John: Well...as I said before, they might do this by employing some symbol of themselves in the context of an extremely unlikely event.

Interruptus: No, I mean how could they do this? I mean where are they, what are they and by what force would they be able to accomplish this?

John: Perhaps they exist in a part of the Universe that we have not yet discovered. Perhaps they exist in a form that is unknown to us. Perhaps they use a force over which we have little or no control or understanding.

Methodus: You mean they might exist in a part of the Universe that is beyond our view?

John: Yes, but not necessarily because it is too far. Perhaps it is intangible. Perhaps they are intangible. Perhaps they are abstract.

Methodus: Like an idea?

John: Yes, but I don't mean just an idea that someone may ponder. I mean an idea as a thing that exists regardless of whether or not anyone thinks about it.

Methodus: Wait...you think that there are abstract things that exist, independently of us?

John: Yes.

Methodus: Could you give an example?

John: Sure. The thing that we represent with the symbol 0 (zero) and word zero.

Methodus: You don't think that the notion of zero is just something that was created by humans?

John: No. I think the thing that we represent with the word zero has always existed and will always exist regardless of whether or not humans had developed a symbol and word for it. I believe it is part of the Universe, regardless of whether or not humans exist.

Methodus: OK, suppose that a person becomes an abstract thing after they die. What force could they use to cause a particular event to happen?

John: Maybe it is an abstract force.

Methodus: How could an abstract thing be forceful? How could it cause something to happen?

John: Do you believe that love is abstract?

Methodus: Yes, I think it is an abstract idea.

John: Do you believe that love can cause someone to act in a particular way?

Methodus: Yes......for example, take patriotism. I believe that one's love of their country can cause them to act in certain ways.

John: So would love be an abstract thing that can be forceful?

Methodus: Yes...I suppose so.

John: Well...perhaps there is a part of the Universe that is abstract...where things like love exist.

Methodus: ...and where people who have died...

Interruptus: Earlier you said that for a dodecahedron there is a concept of direction for the edge that is diametrically opposed to the first edge of a stream type. You said you refer to the two possible directions as forward and backward. I think you may have gotten these reversed?

John: You mean what I said was forward should be backward, and vice versa?

Interruptus: Yes, I think so.

John: OK, let's go over it again. I asked you to imagine a dodecahedron oriented so that the first edge of the stream is nearest to you and directed upwards. Now, the first edge of the stream will run along edge a from the vertex at which edges a, e and f are incident to the vertex at which edges a, g and b are incident.

Methodus: Wait a minute. Why must the first edge run in this direction along edge a? Why can't it begin at the vertex at which edges a, g and b are incident?

John: Haven't I said that this is the convention that I follow?

Methodus: No. I don't remember you saying that. You mean for all stream types the first edge is a, and the second edge is either g or b?

John: Yes. Are you sure I didn't say that already?

Methodus: So, all along here when you have been talking about the class of stream types that end on a particular edge, you have been talking only about stream types for which the first edge is a, and the second edge is either g or b?

John: Yes. I thought I had said that.

Methodus: I don't think you did.

John: I'm sorry about that. I meant to say that early on, when we talking about how I named the edges of the dodecahedron. I should have said then that I follow the convention that all stream type names begin with ab or ag. So for example you would have abhpog or agophb. When I say the class of streams that end on edge x, I mean the subset of streams that begin on edge a, proceed to edge b or g, and eventually end on edge x.

Methodus: But, when we were talking about rotation and reflection of stream types we were rotating streams so that they would go in the opposite direction down edge a. For example, we had discussed rotating the stream type abhpog to get the stream type aflkje.

John: Oh, yeah I remember that.

Methodus: So now you are saying there are no stream types aflkje?

John: I think what I was trying to say there is that if you want to find the other edge that has the same characteristic you can use rotation and then reflection to find it. For example, if the last edge of a stream type is g, the characteristic will be crescendo. Now, to find the other edge to which crescendo is assigned, you could select any stream type that ends on edge g. Then rotate and reflect this stream. The last edge of the resulting stream type would be the other edge to which crescendo is assigned. So, for example take the stream type abhpog. Rotate it to get aflkje. Then reflect it to get aejklf. The last edge is f. So, stream types that begin on a, proceed to b or g, and end on g or f would have the characteristic of crescendo.

Methodus: Whoa...that's not what you said before. You said that you assigned the same musical characteristic to any two stream types that are equivalent under rotation-reflection.

John: Wow...I think I may have made a mess of that. The set of stream types that I have been talking about isn't even closed under the operation of rotation. Sorry about that. I just came up with that rotation-reflection stuff on the fly while we were talking. I should have given that a bit more thought. I was just using those operations to try to explain how one can find a pair of edges for which the characteristic would be the same.

Methodus: So, would the corrected statement be as follows: All stream types begin on edge a and proceed to either edge b or g. When you speak of a stream type that ends on a particular edge, you mean a stream type that begins on edge a, proceeds to either edge b or g, and ends on the particular edge. With that, we may say that you assigned the same musical characteristic to any two stream types that end on the same edge.

John: Yes, that is correct. In addition, you could say that I assigned the same characteristic to two stream types for which the last edges are related to one another by rotation-reflection about edge a.

Methodus: ...You are focusing on the set of stream types S for which the first two edges are ab or ag. But, you could also speak of the closure of S with respect to the operation of rotation.

John: Yes. The closure of S would have twice as many elements as that of S. There would be 2 times 7800 elements in the closure of S.

Methodus: Well...you could assign musical characteristics to the stream types that are in the set which is the closure of S, minus S.

John: Yes. One could do that.

Methodus: Then, you could say that two different stream types in the closure of S with respect to rotation, will have the same characteristic if they are last-edge-equivalent or rotation-reflection-equivalent. That would be right, wouldn't it?

John: Yes. That would be a correct way of saying what I was...

Interruptus: You know...I was just thinking about these stream types on a dodecahedron. A stream type is a sequence of connected edges. Sometimes you can remove the stream type from the dodecahedron.

Methodus: What do mean by remove?

Interruptus: I mean you could imagine the stream type being made of wire. Sometimes you can lift a stream type off of the dodecahedron. For example, abhpog is like a halo. If you rolled the dodecahedron, the halo would fall off. For other stream types, the stream surrounds the dodecahedron so that it cannot be removed. I mean some stream types trap the dodecahedron. For example, that happens with the stream type of length 21 that we discussed earlier.

Methodus: There might be some interesting mathematical questions regarding this issue.

John: Here's one. What is the length of the shortest trapping stream type?

Methodus: Maybe removable would be a better word for this?

John: Yes, then you would have removable and unremovable.

Methodus: Maybe that should be removable and irremovable? Unremovable isn't in The American Heritage Dictionary, but irremovable is.

John: Oh, OK. Then the question would be: What is the length of the shortest irremovable stream type?

Methodus: You could also ask: What is the length of the...longest removable stream type?

John: Yes...and perhaps something could be said about the structure of the right and left turns of all irremovable stream types...maybe something along the lines of this: A given stream type will be irremovable if and only if the absolute value of the total number of right turns minus the total number of left turns is greater than some number.

Methodus: I don't know...that doesn't seem right. This difference would be 3 for the stream type abhpog, which is removable.

John: Yeah, maybe it should be less...

Interruptus: For Dodecahedron you used a musical characteristic called sharp. What do you mean by that?

John: Well...all of these musical characteristics that I have used may be viewed as transformations.

Methodus: What do you mean by that?

John: There is a normal phrase that would be produced by a given stream. The phrase consists of a sequence of notes. There are phrase-level properties such as the time at which the phrase is to start, the tempo of the phrase, the dynamic level of the phrase, and the texture of the phrase. And there are note-level properties. Each note has a certain pitch, duration, timbre and relative dynamic level, the latter of which is relative to the prevailing dynamic level of the phrase and is indicated with accent and breve symbols.

Methodus: What happens when a phrase is sharp?

John: All the notes of the phrase should be performed at a pitch that is sharper than normal.

Methodus: I get it. You could consider this characteristic called sharp to be a transformation that converts a normal phrase into one that is sharper than normal.

John: Yes, exactly.

Methodus: So, would you think of this as being a transformation of a phrase-level property or a note-level property, or both?

John: I consider this to be a modification of the phrase-level property called tuning, and I have notated it as such. The pitches of the notes are still written as O4, P3, et cetera. But, for a phrase that is sharp, the tuning is altered so that the frequency that is X4 is higher than normal. Similarly, for a flat phrase, the frequency of X4 will be lower than normal.

Methodus: Why are you using X4 here?

John: That is the name of the pitch in the OZ pitch-naming convention that one would conventionally use to specify the frequency upon which a tuning system is based. It is the OZ pitch name for what one would call "concert A" in a traditional pitch-naming convention.

Methodus: How have you indicated that a phrase should be sharp or flat?

John: For phrases such as this there will be an accidental in the heading of the page on which the phrase is notated.

Methodus: By accidental you mean one of the symbols: strongly flat, slightly flat, slightly sharp and strongly sharp?

John: Yes.

Methodus: What about the characteristics sharp-or-flat-to-natural and natural-to-sharp-or-flat? Would those be similar?

John: Yes. In that case the frequency of X4 would be varied over the duration of the phrase from either sharp or flat to natural, or from natural to either sharp or flat.

Methodus: And that would affect the frequency of the pitch of every note of the phrase?

John: Yes. But the pitch of each note would be constant throughout its duration. The frequency for the pitch of a given note would be determined by the current value of the frequency of X4 at the time at which the note starts.

Methodus: So the tuning system varies continuously over the duration of the phrase. Specifically, the frequency upon which the tuning is based varies toward normal or away from normal. The frequency of the pitch of a given note would depend upon the state of the tuning system at the time at which the note starts. Is that how it goes?

John: Uh-huh.

Methodus: Would the characteristics crescendo and decrescendo be analogous to these?

John: Yes. There, the prevailing dynamic level is changed continuously over the duration of the phrase. For crescendo, it becomes louder. For decrescendo it becomes softer.

Methodus: For a given note, would its dynamic level be determined relative to the current value of the prevailing dynamic level at the time at which the note starts?

John: Yes, here again it is the start time of the note that is used to determine the properties of the note. And, although the prevailing dynamic level of the phrase would continue to change after the note had started, the dynamic level of the note would remain constant.

Methodus: OK...what is the nature of the characteristic that you refer to as shift-attack-time?

John: That would be a transformation of the time scale for a phrase. Every phrase has a time scale. The notes of a given phrase start and end relative to numbered points on its time scale. For a phrase for which the characteristic is shift-attack-time, some time-scale points will be displaced from the time at which one would expect them to occur, given the prevailing tempo.

Methodus: Oh, I get it. So for example, all the notes that are to start at a given time-scale point would be affected?

John: Their start time would be affected. However, no other time-scale points would be affected. I mean, you can think of it this way. In the case when the tempo of a phrase is constant, you can view the time-scale points as being located at the tips of the teeth of a comb. You would shift a given time-scale point by bending its corresponding tooth, slightly, without bending any of the other teeth.

Methodus: Would the characteristics accelerando and decelerando be similar? I mean would they be transformations of the time scale as well?

John: Yes, but in that case all time-scale points after number 0 (zero) would be affected. The times at which they occur would be altered so that the distance in time between consecutive time-scale points would vary over the duration of the phrase.

Methodus: From what I have seen of the score, the characteristics exaggerate-frequency-range and senza-vibrato are transformations like sharp and flat. That is, they affect all notes equally. Is that right?

John: Yes. For a phrase for which the characteristic is exaggerate-frequency-range, the same frequency range would be exaggerated for each note of the phrase, to the same degree. The same may be said of senza-vibrato. There, no vibrato would be used for any note of the phrase.

Methodus: You used a few other characteristics like staccato and tenuto. Would the staccato phrases be those for which a staccato dot has been used for some notes?

John: No, not necessarily. The duration of some notes of a staccato phrase will be shortened. However, this might not be indicated with a staccato dot in all cases. I have used a staccato dot to shorten the duration of a 1-note or 2-note, only. For longer notes, an n-note would be transformed to an m-note, where m is half of n when n is even, and half of n minus one when n is odd.

Methodus: For the characteristic staccato, only a few notes are affected. Would that be the same for the characteristics tenuto, pitch-bend-up-or-down, note-level-crescendo and note-level-decrescendo?

John: Yes. For tenuto, the tenuto articulation is used for some notes. For pitch-bend-up-or-down, the pitch of only a few notes would be affected. For those, the pitch would be changed gradually from normal to either sharp or flat by some specified amount. Similarly, for note-level-crescendo and note-level-decrescendo only a few notes would be affected. For those, the dynamic level would be altered gradually from normal to some level that is either louder or softer than normal.

Methodus: For these characteristics for which only a few notes would be affected, is there some factor that determines which...

Interruptus: For Dodecahedron, why are some notes louder or softer than normal?

John: You mean, why do some notes have an accent or breve articulation?

Interruptus: Yes, why is that?

John: Well...for a given stream, the only notes that might be louder or softer than normal would be the main note produced by the first edge of the stream, and the second main note of a pivot edge. All other notes would be normal. We should discuss the second case first, because that one is simpler.

Methodus: So, under what conditions would the second main note of a pivot edge be louder than normal?

John: It will be louder than normal if the number of the current face of the second note is larger than that of the first note of the pivot edge.

Methodus: Let me see if I understand this correctly. A pivot edge produces two main notes. For each of these notes, there is a current face. Because it is a pivot edge, these two current faces will be different; they will be on opposite sides of the pivot edge. So the numbers of these two faces will be different. Now, you are saying that for a given pivot edge of a given stream, the second main note of the pivot edge will be louder than normal if the number on the current face of the second main note is greater than the number on the current face of the first main note. Is that right?

John: Yes, exactly.

Methodus: Would it be softer than normal if the second current face were less than the first?

John: Yes, that's how it goes. If the number on the face of the current edge of the second note of the pivot edge is......sorry, that isn't right. If the number on the current face of the second note of the pivot edge is less than that of the first note of the pivot edge, then the second note of the pivot edge will be softer than normal.

Methodus: OK. What about the main note that is produced by the first edge of the stream? Under what conditions would that be louder or softer than normal?

John: Well...the basic idea is this. Every stream except the first is the child of some parent stream. To determine the relative dynamic level of the first note of a stream, the number on the current face of the first edge is compared to the number on the face that is the current face of the parent when the child starts.

Methodus: I think you may have lost me. What is the current face of the parent when the child starts?

John: Do you remember what I said about entry points?

Methodus: Yes, to some degree. You said that there is an entry point for each turn of a parent stream. And from such points a child may emerge.

John: That's right. And we discussed two forms of entry points.

Methodus: Right. For one case the edge of the parent that precedes the turn is simple. For the other case it is a pivot edge.

John: Yes...well...let's take the case for which the edge is simple.

Methodus: OK.

John: For that case, when the child starts, the current face of the parent would be the current face of this simple edge of the parent.

Methodus: For this case, this simple edge of the parent and the first edge of the child coincide, right?

John: Yes. And the child stream takes the turn that the parent avoided.

Methodus: OK. So in this case the neighbor face of this parent edge would be the current face of the first edge of the child.

John: That's right.

Methodus: So, to determine the relative dynamic level of the main note that is produced by the first edge of the child, you would compare the number of the face that is the current face of the parent when the child begins, with the number of the current face of the first edge of the child.

John: Yes. If the former is greater than the latter, the main note of the first edge of the child will be louder than normal. Otherwise, the former will be less than the latter and this main note will be softer than normal.

Methodus: That would be equivalent to saying that in order to determine the relative dynamic level of the main note of the first edge of a stream, you compare the neighbor face and current face of the first edge of the stream. If the number of the current face is greater than that of the neighbor face, then the note would be louder than normal. Otherwise it would be softer than normal. Wouldn't that be right?

John: Yes, that is correct...provided you are only talking about the case when the edge that precedes the turn of the parent with which the entry point is associated, is a simple edge. It would not be true in the case when the edge that precedes the turn is a pivot edge.

Methodus: What happens in that case?

John: Well...in that case, when the child starts, the current face of the parent would be the current face of the first main note of the pivot edge.

Methodus: In this case the child starts at the vertex where the turn is made in the parent after this pivot edge, right?

John: Yes. And the child proceeds down the edge that the parent avoided at this turn.

Methodus: Right. And then, if the parent had turned right, the first turn of the child would be left; and vice versa. Right?

John: Right. I mean correct.

Methodus: So it seems like the current face of the first edge of the child must coincide with the face that was the current face of the parent when the child started. Is that right?

John: Yes, that is true. And for that reason, the relative dynamic level of the main note of the first edge of the child stream will be normal...rather than louder than normal or softer than normal.

Methodus: So for this note there will be no accent or breve symbol.

John: Yes, that's right.

Methodus: ......What happens if the child has no parent? I mean, what about the first stream. It...

Interruptus: You know...I've been thinking about the way you assigned pitch-classes to faces. It seems like the pitch class O is always assigned to face number 1.

John: Well...no, that would only be true for the forward mapping.

Interruptus: Oh, OK......but in the backward mapping the pitch class O is always assigned to face number 11, right?

John: Yes...

Interruptus: ...That doesn't seem like it would be too interesting.

John: ...

Methodus: Yeah. For all streams you would have pitch class O assigned to either face 1 or face 11. And as I understand it, throughout the piece the instruments assigned to faces 1 and 11 will be bass clarinet 1 and oboe 1, right?

John: Yes, that's right.

Methodus: Suppose a stream were to go around one of these faces. The pitches of the notes produced by the edges around these faces would always be calculated relative to pitch class O.

John: No, that's not right......oh, I think I forgot to mention that these pitch classes that are assigned to faces will be transposed. Did I say that?

Methodus: No, I don't remember you saying anything about transposing these pitch classes. How does that work?

John: Well...as I said, pitch classes are assigned to faces. For a given stream, the particular pitch-class mapping that is used will depend upon the stream type. But the pitch-class mapping is not used as is. The pitch class that is assigned to the current face of the first edge of a stream will be transposed so that it matches the pitch class that was assigned to this face in the parent stream.

Methodus: So, in effect, the same pitch class is used for this particular face in both the child and parent?

John: Yes.

Methodus: What about the other faces of the child stream?

John: Those would be transposed by the same amount.

Methodus: So for a given stream, you use the stream type to select a particular pitch-class mapping. With this mapping, you assign a pitch class to each face. Then you transpose all of these pitch classes by the same amount in order to make the pitch class that you use for the current face of the first edge of the stream be equal to the pitch class that was used for this face in the parent stream. Did I get that right?

John: Yes, that is correct.

Methodus: What about the first...

Interruptus: Earlier you said that the probability of generating a stream type of a given length is not a decreasing function of length. Do you know which length would be most likely to occur when you generate a stream type randomly?

John: Yes, if you generated a random sequence of stream types, you could expect that the majority of the stream types would be of length 12.

Methodus: What would be the length with the second highest frequency?

John: That would be 9.

Methodus: So...this probability function has more than one peak?

John: Yes, there is a global maximum at 12 and a local maximum at 9. Let's say P(k) (p-of-k) is the probability of generating a stream type of length k. The complete list of probabilities would be:

P(6) = 0.06250 (p-of-six equals point-oh-six-two-five-oh)
P(7) = 0.06250
P(8) = 0.06250
P(9) = 0.11719
P(10) = 0.08203
P(11) = 0.10938
P(12) = 0.12109
P(13) = 0.10742
P(14) = 0.09497
P(15) = 0.07471
P(16) = 0.05011
P(17) = 0.03012
P(18) = 0.01588
P(19) = 0.00700
P(20) = 0.00223
P(21) = 0.00037

Methodus: These are all rounded to the fifth decimal place, right?

John: Yes. But, P(6), P(7) and P(8) are all exactly 0.06250.

Methodus: Why would these be equal?

John: The number of different stream types of length 6, 7 and 8 is 2, 4 and 8, respectively. But the probability of generating a given stream type of length 7 is half that of length 6. The same may be said of lengths 8 and 7.

Methodus: So for these cases, as n grows there are more possibilities. But the chance of getting any one of these possibilities is reduced.

John: Yes. For lengths 6, 7 and 8, these two factors are in balance. So the overall probability remains constant.

Methodus: But then it jumps up at length 9.

John: Yes, the probability of getting a stream type of length 9 is substantially higher because the number of stream types of length 9 is more than twice that of length 8. And 6 of the stream types of length 9 have a forced edge. A given stream type of length 9 that has one forced edge is just as likely to occur as a given stream type of length 8 that has no forced edges.

Methodus: So, how many stream types are there of length 9?

John: There are 18.

Methodus: What about for the other lengths?

John: Here is the complete list. Let n(k) (n-of-k) be the number of stream types of length k.

n(6) = 2 = 2 + 0 (n-of-six equals two, which equals two plus zero)
n(7) = 4 = 4 + 0
n(8) = 8 = 8 + 0
n(9) = 24 = 18 + 6
n(10) = 42 = 42 + 0
n(11) = 96 = 80 + 16
n(12) = 204 (two-oh-four) = 160 (one-sixty) + 44
n(13) = 360 (three-sixty) = 280 (two-eighty) + 80
n(14) = 586 = 394 + 192
n(15) = 872 = 520 + 352
n(16) = 1130 (eleven-hundred-thirty) = 618 (six-hundred-eighteen) + 512 (five-hundred-twelve)
n(17) = 1284 (twelve-eighty-four) = 594 (five-ninety-four) + 690 (six-ninety)
n(18) = 1290 (twelve-ninety) = 498 + 792
n(19) = 1074 (one-thousand-seventy-four) = 314 (three-fourteen) + 760
n(20) = 628 = 88 + 540
n(21) = 196 = 0 + 196

Methodus: ...So the length with the largest number of possibilities is 18......for these sums at the end...would that be the number of stream types with no forced edges plus the number of stream types with one forced edge?

John: Yes. I broke it out this way because you would need both of these numbers to calculate the probabilities.

Methodus: Oh, OK. So...for example, let's take...umm...length 14. To calculate the probability of getting a stream type of length 14, I would multiply 394 (three-ninety-four) with one-half to the power 13. To that I would add 192 (one-ninety-two) times one-half to the power 12. I would get...uh...0.0949707... (point-oh-nine-four nine-seven-oh-seven...)

Interruptus: I can understand why these probabilities P(6) through P(21) would have to add up to 1. After all they are probabilities. Still, the form of this decomposition of 1 has a particular structure to it. I wonder if these numbers n(6) through n(21) are the only nontrivial solution to this form of decomposition.

John: Well...let's see, the sum of n(6) through n(21) can be written as a weighted sum of the powers of one-half from 5 through 19. The coefficients in this sum would be 2, 4, 14, 18, 58, 124 (one-twenty-four), 240 (two-forty), 472, 746, 1032 (one-thousand-thirty-two), 1308 (thirteen-oh-eight), 1386 (thirteen-eighty-six), 1258 (twelve-fifty-eight), 854 and 284.

Methodus: OK. I can see that. You just added the first term in the sum that forms n(k) with the second term of the sum that forms n(k+1) (n-of-k-plus-one). I mean 2, 4, 14, 18, 58 are (2+0) (two plus zero), (4+0), (8+6), (18+0), (42+16).

John: Right. Now, you could break up these coefficients in other ways. For example, you could have (2+0), (2+2), (8+6), (18+0), (42+16), and so on. That would give you a different sequence of numbers m(6) through m(21). They would have the same probabilities. For that you would have:

m(6) = 2 = 2 + 0
m(7) = 2 = 2 + 0
m(8) = 10 = 8 + 2
m(9) = 24 = 18 + 6
m(10) = 42 = 42 + 0
m(11) = 96 = 80 + 16
and so on.

Methodus: OK. But what about these coefficients? Are they unique? I mean suppose I asked you to solve the following problem. Find a sequence of 15 positive integers a5 (a five), a6, and so on up through a19, such that the following sum equals 1: a5 times the 5th power of one-half, plus a6 times the 6th power of one-half, and so on up through a19 times the 19th power of one-half. Is there another solution to this problem other than these coefficients 2, 4, 14, 18, 58, and so on?

John: I'm not sure. I haven't given that any thought......I don't think that would be right because you can write one of the powers of one-half in terms of other powers of one-half.

Methodus: Oh...right......I think I have another solution. If you change the first number from 2 to 3, that will increase the sum by one-half to the power of 5. You can make it up by reducing the coefficient of the 10th power of one-half. You would reduce 124 (one-twenty-four) by 32 to get 92. So you would have the coefficients 3, 4, 14, 18, 58, 92, 240 (two-forty), and so on.

John: Yeah, that would be another solution.

Methodus: I wonder if the original solution that we had is the only one for which the coefficients add up to 7800.

John: That might be worth...

Interruptus: For Dodecahedron, how did you assign duration mappings to particular stream types?

John: Well...as I said earlier, I constructed 30 different duration mappings. Each of these was assigned to a particular class of stream types. The same duration mapping was used for all stream types that end on a particular edge, relative to the first edge. There was one exception. The stream types that end on the edge / were subdivided. If the direction of the last edge was forward, one duration mapping was used. If it was backward, a different duration mapping was used.

Interruptus: Yes, I know. But, what was your rationale for assigning a particular duration mapping to a particular last edge?

John: Oh. For that, I paired duration mappings with characteristics. For example, I assigned the duration mapping 111111 (one one one, one one one) to edge w because the characteristic for that edge is accelerando.

Methodus: By 111111, do mean the duration mapping that was created using the partition 1+1+1+1+1+1?

John: Yes, that's right. I'll refer to these duration mappings using a 6-digit name, where the i-th digit is the number of duration units that is assigned to the face number i.

Methodus: Why did you assign this duration mapping to an edge for which the characteristic is accelerando?

John: I felt that it would be easier for the listener to perceive an accelerando if the duration of each note was the same. Similarly, I assigned the duration mapping 222221 to edge q, for which the characteristic is decelerando.

Methodus: What about 222222?

John: I assigned that to edge o, which is staccato. I did that so that the durations would not be all the same...Maybe I should give you the complete list of assignments.

Methodus: Yes that would be helpful.

John: OK. Here's the complete table.

221111 f crescendo
211113 g crescendo
111411 b decrescendo
112222 e decrescendo
121114 m sharp-or-flat-to-natural
121161 n sharp-or-flat-to-natural
111131 c natural-to-sharp-or-flat
135111 d natural-to-sharp-or-flat
222222 o staccato
111223 l staccato
123141 h tenuto
231122 j tenuto
161111 v note-level-crescendo
112511 i note-level-decrescendo
111111 w accelerando
311312 u accelerando
222221 q decelerando
211212 t decelerando
411311 - sharp
111112 + sharp
115111 r flat
111133 s flat
322122 * exaggerate-frequency-range
142212 z exaggerate-frequency-range
114221 x senza-vibrato
711111 y senza-vibrato
212331 p pitch-bend-up-or-down
333111 k pitch-bend-up-or-down
121215 / forward shift-attack-time
111441 / backward shift-attack-time

Methodus: How did you assign the other duration mappings?

John: Well...I assigned 711111 to edge y because I wanted to have some long notes for which the absence of vibrato would be perceptible. Similarly, I assigned 161111 to edge v because I wanted notes that would be sufficiently long for a note-level crescendo to develop. The same may be said of my assignment of 112511 to edge i, except for that the characteristic is note-level-decrescendo. To some extent, that's also why I assigned 212331 and 333111 to edges for which the characteristic is pitch-bend-up-or-down.

Methodus: What about the others?

John: For the most part, with the other assignments I tried to ensure that there would be a relatively balanced distribution of durations over the various characteristics.

Methodus: What do you mean by balanced?

John: For any given one of these 30 duration mappings, the total number of duration units over all the faces ranges from 12 through 24.

Methodus: Right...because all of these duration mappings are based on partitions of numbers in the range 6 through 12...and each of the components of a partition gets assigned to two faces.

John: I assigned 111112 and 411311 to edges for which the characteristic is sharp. In one case the total number of duration units is 14. For the other it's 22.

Methodus: Oh...I see. You mean balanced in terms of the total number of duration units.
 

Previous Page      Next Page

Lines in the Air: A One-Act Play

Other Work by John Greschak

Public Domain