Quivering Recreations


Last time, we used a set of arrows and vertices to represent a sequence of notes:

such as the 'Cantus 1' line found in the first section of C T Padbrué's Synphonia in nuptias (for the marriage betwixt Steyn & van Napels)

If the player doesn't work, there's some ogg vorbis over here.
Fragment Padbrué Synphonia
as a graph:reducible (eliminating uni-cycles and bi-cycles) to:
Representation as graph of pitch transitions Distillation of pitch transitions
whereupon we recovered the following three 'distillations' of the three lines
Three Distillations

We concocted a couple of examples using the reduced lines (as before, if the player doesn't work, there's some mp3 over here).

These directed graphs are known as quivers, consisting as they do of bunches of arrows, often many together (all facing the same direction) between vertex pairs since we've eliminated arrows which cancel each other out (i.e. all bi-cycles).

We wondered how many paths (if any) - consuming all arrows, exactly once - could be used to traverse each of these three graphs. It turns out (via nothing more sophisticated than brute force exhaustive searching) that for these three distillations there are (3, 130, 55) such paths. We show the three for cantus 1 and three typical ones for the other lines:

Cantus 1
(8 arrows)
Cantus 2
(13 arrows)
(11 arrows)
C F E D C B♭ A D F
C B♭ A D C F E D F
C B♭ A D F E D C F
C D E C D A F E A B♭ C A B♭ C
D E C D A F E A B♭ C A B♭ C D
A F E A B♭ C A B♭ C D E C D A
G D C F D A B♭ C G A B♭ G
B♭ G D C G A B♭ C F D A B♭
D A B♭ C F D C G A B♭ G D

Mutation Primer

The following describes an operation which - as someone said back in the 1990s when it was first being investigated - is elementary but complex.

We consider the idea of a quiver mutation as applied to these reduced melodic lines. It's necessary to perform these reductions before we mutate since the rules of mutation require it. The mutation is performed - reversibly - on a vertex of the quiver. In our case the vertex represents a musical pitch, but we don't change that pitch. Instead we alter its connections to its preceding and succeeding pitches by swapping their places and completing a circuit. Basically, the rules are - at the selected vertex:

  1. Reverse all arrows entering and leaving the mutating vertex, M.
  2. For every pair of vertices (P, S) now entering and exiting M (a 2-path from P(redecessor) to S(uccessor) via M), 'complete the circuit' PMS by adding a new arrow from S to P.
  3. Maintain the 'no bi-cycle' constraint on the quiver by removing any newly introduced (by circuit completion) oppositely-directed arrow pairs.

We'll try to maintain a certain consistency of terminology here in distinguishing

  • an arrow - a single directed edge (with one 'arrowhead') from one vertex to another, different one (no uni-cycles)
  • a path - a trip from one vertex to another via arrows
Thus all arrows are paths (perforce of length 1) but not all paths are arrows. A path may start and finish at the same vertex - this is a cycle, or circuit. Thus all circuits are paths but not all paths are circuits. Note that quiver rules eliminate (and mutation rules maintain) all circuits which are also paths of length 2 (no bi-cycles).

Note that in this specific discussion we're referring to a path of length 2 passing through a mutating vertex as a 2-path. The two (non-mutating) vertices must be different otherwise we'd have a bi-cycle between it and the mutating vertex.

We illustrate a sequence of mutations - successively on vertices 1, 2, 3, 2 - in the following figure:

Mutation chain on vertices 1, 2, 3, 2

The first mutation (μ₁) is on vertex labeled ① and reverses arrows ③→① and ①→②. As a consequence there's a 2-path from ②, via ①, to ③ and we 'complete the tri-cycle' by introducing a new arrow ③→②, thus producing the second quiver in the sequence. Since there were no opposing arrows from ② to ③, it remains.

The second mutation (μ₂) at vertex ② reverses arrows ③→② and ②→①. Consequently there's a 2-path from ① to ③ so we complete ②'s tri-cycle by adding a new arrow from ③ to ①. This time there's already an opposing arrow, ①→③. This would introduce a (forbidden) bi-cycle, so we annihilate the pair.

Note that at this stage (deftly avoiding the word 'point') a reapplication of μ₂ would exactly reverse what we have just done. All quiver mutations performed twice in succession at the same vertex are 'do nothings'.

The remaining mutations on vertices ③ and ② proceed as shown. The final mutation on ② turns that vertex from being a sink (no arrows leave it) into being a source (no arrows enter it).

Note that, in general, mutations on distinct nodes ⓜ and ⓝ do not commute, i.e. a quiver resulting from mutation by μₘ then μₙ is unlikely to be the same as the one resulting from μₙ then μₘ.

For this particular set of three vertices, 14 distinct configurations are possible with various applications of the three mutations:

14 quiver configurations on three vertices

In the above, we've labeled the configurations with a τ to suggest a triangle, πₙ to suggest a path through vertex n, and σₙ to suggest a source at vertex n. The inverses, e.g. τ⁻¹ and σₙ⁻¹ (a sink instead of a source) are essentially the same but with all arrows reversed. The following graph (edges bi-directional) shows all 21 ways of mutating between the 14 configurations.

all single-arrow mutations on three vertices

The operations μ₁, μ₂ and μ₃ permute τ, π₁, π₂, π₃, σ₁, σ₂, σ₃, τ⁻¹, π₁⁻¹, π₂⁻¹, π₃⁻¹, σ₁⁻¹, σ₂⁻¹, σ₃⁻¹ without affecting the 'shape' of that graph (the vertex labels just - in this case - swap places seven pairs at a time). Thus we have a group which acts upon, and thereby represents the symmetries of, the structure held by these vertices. For example the first generator, μ₁, effects the seven simultaneous transpositions: (π₁,τ)(π₂,σ₂⁻¹)(π₃,σ₃)(σ₁,σ₁⁻¹)(σ₂,π₂⁻¹)(σ₃⁻¹,π₃⁻¹)(τ⁻¹,π₁⁻¹). And mutations μ₂ and μ₃ act - mutatis mutandis - similarly. The group generated is (C₂×C₂×C₂×C₂×C₂×C₂)⋉S₇, the semidirect product of the group (of order 64) formed from the direct product of 6 cyclic groups of order 2 with the group (of order 5040) of permutations of 7 symbols. This quiver group is of order 322560 = 7!(₈C₇)² = 7! × 64 = 2¹⁰×3²×5×7. The order of this group is 'coincidentally' the same as the number of ways 7 rooks may be placed on a standard 8×8 chessboard such that none may attack another.

Musical Mutations

Needless to say, mutations on quivers with more than three vertices can get considerably more complicated. In general, there will be no limit to the number of configurations. If we consider the six possible mutations possible on the vertices of the above Cantus 1 distillation we can see, for example, that there are four 2-paths through the D vertex, i.e. ADC, ADF, EDC, EDF. All four of these 2-paths will be inverted by a mutation at D and will thereby engender four new arrows AC, AF, EC and EF; the last of these cancelling the existing arrow FE. Thus a quiver with - originally - 8 arrows becomes a quiver with 11. Mutations will always preserve the vertex set but, in general, the number of arrows will typically increase at each application as arrow generation will typically exceed any consequent cancellations.

Although it's possible to have finite quiver groups on six vertices, our trio of Padbrué 'compressings' - all, as it happens, with six vertices - are of infinite order. The six possible 'first mutations', for each quiver, are shown in the following diagram:

Three Reductions of Padbrué's Pavane, with their Six Mutations

The first column (with the uncoloured background) carries the reduction into quivers of each of the three lines in the Pavanne. The remaining six colums show, for each melody, the quivers resulting from mutations on each of the original quiver's six pitches in the order C, D, E, F, A and B♭ for canti 1 and 2 and in the order C, D, F, G, A and B♭ for the bass line. The green backgrounds denote quivers which may be fully traversed, in that it's possible to walk a path from some start vertex to some end vertex consuming each arrow exactly once. The pink backgrounds indicate quivers where this is impossible. This answers - in the negative - the question of its always being possible 'to find at least one path using all arrows in the graph' asked in the previous post.

Clearly a fully traversed path of n arrows will have n + 1 pitches in it from start to finish. Note that such paths are not cycles (Hamiltonian or otherwise) since no 'loop' is completed by terminating at the start node (one may certainly end at the same pitch one began on, but as each pitch may (in general) be visited many times in a traversal, such visits count as separate 'versions').

The following table shows the number of arrows in each of these (first) mutations and also - where possible - the number of full path traversals.

Cantus Iμ(C)μ(D)μ(E)μ(F)μ(A)μ(B♭)
# traversals:3614-3-5
Cantus IIμ(C)μ(D)μ(E)μ(F)μ(A)μ(B♭)
# traversals:130-31513010-315
# traversals:5544445121234234

We've already shown above (and in the previous post) some melodies arising from full traversals of the unmutated quivers. We see that it's always possible to find a full traversal after one mutation at D for all lines (e.g. CDACB♭AFDECF for cantus 1, B♭CEDCADCAFEAB♭CAB♭ for cantus 2, and GADFAB♭GAB♭CDG for the bass line. Mutations at F for both canti provide melodies FCEFDCB♭AD and ECDEFAB♭CDAB♭CA. There's no such animal available for the bass - not because there are no possible traversals but because there's no F to mutate on. Instead we can mutate on G and find full traversals, one of which is FDGB♭CAB♭DAGCF. All three lines can provide full traversals after a mutation on B♭, e.g. ADCAB♭CFEDF, DECDACB♭ACB♭AFEACD and B♭ACFDCGDACB♭AGB♭.

Abstraction to Application

We present a short study using the mutations mentioned. As always, if the player doesn't work, there's some ogg vorbis over here.

Mutations on a Padbrué Pavane

Naturally one need not stop at one mutation, nor need one mutate on the same pitch, simultaneously, for each line. As long as one does not mutate a quiver on the same pitch twice (which simply takes you back to where you were) the mutations are - in general - endless (but not for the simple 14 state explanatory model presented earlier). In this particular example, we're always going to be in F major (or D minor, or E Locrian, G Dorian etc), but clearly 'seed' material from more chromatic pieces will tend to stay chromatic.

Note that there's no guarantee that, after each mutation, a full 'all-arrow-consuming' traversal is possible across a quiver one ends up with. But if they do exist then the more often one mutates, the more such possible traversals will exist and the longer will they become. In any case, there's no law forcing you to use only mutations which result in fully traversible quivers.


Wedding Weeding

On a bright cold day in February of MDCXLII (though we doubt the clocks were striking XIII), Matthias Steyn and Marie van Napels got married in Haarlem.

Here's the first section of Cornelis Thymanszoon Padbrué's Synphonia in nuptias written for the occasion.

Padbrué's Pavane

Each of the three lines is a sequence of single notes and can be written as a path without regard to their length or octave. The first line is C, D, E, F, E, D, C♯, D, E, F, G, F, E, etc and it can be represented as a directed graph:

Pavane Cantus 1

where the arrows represent the transitions between consecutive notes in the melodic line, and the numbers on the arrows indicate how many times such a transition happens within it.

We're now ready to play a game with this graph.

La Règle du jeu

  • Remove all unicycles and bicycles
  • Remove any resulting isolated notes

All of which sounds distressingly bike-unfriendly, notwithstanding an evident tricycle tolerance, but these are technical terms.

A unicycle is an arrow leaving a note and going right back to it without any intervening notes. This particular line has none, so there's nothing to do here. But Cantus II has a unicycle at bar 13 where a C is 'restarted' in bar 14. The Bass line is, like the first, devoid of unicycles.

A bicycle is an arrow from one note to another where there's also an arrow from that second note back to the first. There's an obvious one on the left of the above figure between G and A and also on the right between D and C♯.

Where there are multiple arrows in both directions between a pair of notes, we simply cancel opposite pairs and any that remain - perforce in only one direction - are retained. Thus the single arrow from A to B♭ will cancel one of the two from B♭ to A, leaving only one arrow from B♭ to A. Thus we end up with:

Reduced Pavane Cantus 1

Notice that we've completely lost two notes, C♯ and G. This is because both these notes were attached to each of their immediate neighbours by completely cancelling bicycles.

Note also that we've lost the numbers on the arrows. This is because bicycle cancellations - mostly - leave only one arrow which we shan't bother labelling with a 1. If two (or more) arrows remain, perforce in the same direction from one note to another, we shall draw them in explicitly - similarly unlabelled.

In fact this occurs in both of the other lines which we shall now process. First, Cantus II:

Reduction of Cantus 2

where, for example, the unicycle at C has been removed and one of the three transitions from B♭ to C has been cancelled by the lone transition from C to B♭, leaving the two arrows shown. Notice again that G has been banished since it was connected exclusively by cancelling bicycles.

Finally, we'll go straight to the de-cycled version of the bass line:

Reduced Base

New Paths for Old

Obviously it's possible to find a path through each of the original graphs since that's how they were constructed in the first place.

The question is,

  • Is it possible to find a path through each reduced graph, each arrow being used exactly once?
  • You're allowed to start wherever you like. Let's reproduce the three graphs together here:

    Three reduced lines

    Inspection shows a path C, B♭, A, D, C, F, E, D, F across the first line. It's 9 notes long and consumes all 8 arrows.

    Musical considerations (the original is in either F major or D minor [OK so it might be B♭ Lydian, etc, too]) indicate that since we're beginning the first line with a C that we might consider the piece as being in F major and that we start with an A in line 2 and F in line 3 (or vice versa).

    For line 2 we can find a path (of all 13 arrows) A, F, E, C, A, B♭, C, D, E, A, B♭, C, D, A.

    And for line 3 we can find F, D, C, G, A, B♭, G, D, A, B♭, C, F, a path with 12 notes consuming all 11 arrows.

    This is convenient - it also ends in F major, albeit sans its 5th. It's beginning to look like we can get a half-decent 8 bar sequence out of this. For example:

    A Padbrué Reduction

    And here's another one:

    Another Padbrué Reduction

    If your browser supports it, these two chunks of pressed Padbrué may be heard (twice each) with the player below.

    These are two (re-)compositions from the same three paths, and it's clear that - musically - there's no real limit to expressions of such 'distillations' although they'll probably all sound a bit 'samey'.

    But is there a limit to the number of paths through any of these note-gardens? Are these the only paths? Can you find any others?

    One is tempted to believe that - given their construction from a guaranteed initial path - there will always be at least one path in the reduction. But the pruning is pretty brutal - we can, after all, lose notes - so have we just been lucky with these three examples?


    By the Way, thanks to
    • musescore for music composition, pics and midis
    • mftext for converting midi to text
    • awk for turning text into graphviz dot files
    • graphviz for the graph-drawing

    All free!


    Amateur League

    The results are in

    As promised in the last post, we have now calculated the lengths of aluminium tubing (of a certain grade) required to produce a set of tubular chimes - and even that is much too grandiose a term for the rather toy-like set of factory-cut hollow rods purchased, at the very reasonable price of about thirty quid (including delivery), for the experiment.

    The material arrived the next day, and I measured the lengths principally because I needed accurate data for my calculations below - and only secondarily to check that they were as ordered (which they pretty much were, and certainly within the vendor's advertised tolerance). I also weighed them, principally to check that the density matched the theoretical density used in the frequency calculations - which of course it did, to within about a half of a percent. As I have no machinery to check its Young's modulus of elasticity, I continue to rely on its wikipedially reported value.

    Measuring their frequencies was the most difficult task, and involved suspending the tubes with a hot-glued thread connecting the top of each tube to a shelf, and recording them - with my trusty H2N Zoom recorder - being lightly struck by a similar aluminium tube. Here's a composite of the noises made by this 'instrument'.

    Thirteen Tings

    Do not be fooled by the levels. Although the samples in this recording are untreated (apart from being chopped up into segments), those small cylinders cannot shift very much air and the sounds are quite quiet. It's the H2N being held (by hand) up close.

    The samples were brought into Cockos Reaper whereupon the use of its built-in band-pass filter, ReaEQ, (to cut out the highly audible harmonics) and frequency counter (ReaTune) plugins proved adequate to the task. Here are all the measurements, plus a picture of the set of tubes.

    lengthweightfrequency$\kappa$13 Pipelings
    25110.6351432.4 Thirteen Chimes

    As expected, the frequencies were not even close to the design targets (respectively 165, 175, 185, 196, 208, 220, 233, 247, 262, 277, 294, 311, 330) and were - on average - 3.14 times higher (to the three significant figures employed throughout this exercise). The final column in the above table is $\kappa$, the diffusivity mentioned in the previous post, and is calculated as $L^2 \times f$ for each row ($L$ being the metre value). The average value for $\kappa$ is 32.2, the same multiple (3.14, weirdly close to $\pi$) of the value of 10.2 expected.

    But - and also as expected - the instrument does indeed play a very serviceable chromatic scale. It may not start on the 165 Hertz concert pitch E (as requested) but instead a rather higher 514 Hertz non-concert pitch flat C, not quite a quarter tone below the 'proper' one. That's quite a way out, but our measured value of $\kappa$ could now be used to calculate a new series of $L$ values from $L=\sqrt{\kappa/f}$ - provided of course we were to use exactly the same material. All lengths would, accordingly, be multiplied by a factor of 1.77 (the square root of 3.14).

    For this particular kind of aluminium alloy the previous post shows that $\kappa = 1420 \sqrt{D_o^2 + D_i^2}$, where $D_o = 0.006 m$ (i.e. 6mm) and $D_i = 0.004 m$ (i.e. 4mm). It's perfectly reasonable to incorporate the 'blame' for the (fudge) factor of 3.14 into the 'material constant' of 1420 and claim that in practice this constant should be 4450. This leaves us free to use larger inner and outer diameters as long as it's the same alloy to allow us to shift more air and make louder instruments. For example, if we have our heart set on a more robust 165 Hertz concert pitch E, we choose a larger diameter tubing (say half an inch with a 6.2mm bore). This would give us a new value for $\kappa$ - based on $D_o = 0.0127\,m$ and $D_i=0.0062\,m$ - of $4450 \times 0.01413 = 63$. From this we'd calculate a length of 618 mm.

    And so on


    So why was the calculation incorrect? Why would we need to use 4450 rather than 1420? As the 1420 was itself calculated - as $(22.4/8\pi) \sqrt{E/\rho}$ - from easily verifiable physical properties of a particular alloy of aluminium, the only places where the discrepancy could occur are in the value of 22.4 and from the way that $I$ was defined (in terms of $D_o$ and $D_i$). Perhaps an engineer can let me know where I went wrong?

    Finally, here's a plot of frequency versus length of the tubes received in the mail. The least squares fit (from LibreCalc) is $f = 33.6 \times L^{-1.97}$, satisfyingly close to the model of $L^2 f = \kappa$ (with my measured average of $\kappa=32.2$).

    length frequency plot
    Thirteen frequencies of chromatic tubical pieces