|
Home
| |
Computer Music
Related Reading: Chapters 20
and 21 in SOE.
Due:
Anytime during Reading Week, but no later that midnight May 4.
Preliminaries:
- Read Chapters 20 and 21 in SOE.
- Download and unzip the
SOE source code.
- Copy (or move) all the files in SOE/haskore/src to SOE/src.
- To test the system, fire up GHCi in the directory SOE/src,
and load the module MDL. Then try running:
 | "testNT cMajArp" on Windows XP, or |
| "testWin95 cMajArp" on older versions of Windows, or |
| "testLinux cMajArp" on Linux. |
If none of these work, then type "test cMajArp", which
will write a MIDI file called "test.mid" in SOE/src,
which you can then double-click to play using whatever your computer's default
MIDI player is.
Assignment:
- Write a simple monophonic (i.e. one-voice) melody in
MDL. This could be your own invention, or a simple melody that you know.
If you don't know music, remember that a melody is just a sequence of notes --
so write anything that you want!
- If the above melody is called
m, then compose a song in which you are not allowed to introduce new
notes, but you can use m as many times as you
like, and you must use each of the following operators at least once:
| sequential composition (i.e.
:+:) |
| parallel composition (i.e.
:=:) |
| delay |
| revM |
| Tempo |
| Trans |
| Instr
|
Write a function called invert
that takes a Music value and "inverts" each
note in it with respect to a given pitch. For example, if the given
pitch is (C,5), then the "inversion" of (A,4) is (Ef,5), and so on. The
type of invert should be
Pitch -> Music -> Music. (Hint: use the
functions absPitch and
pitch to make your life easier.) Test
your solution.
Extra credit: Prove that
invert p . invert p = id, where
p is any pitch and id is the identity function.
Solutions
Here are solutions to problems 3 and 4. However, I have not tested
this code, so take it with a grain of salt!
Problem 3
> invert :: Pitch -> Music -> Music
> invert p (Note p0 d) =
> let p' = pitch (2 * absPitch p - absPitch p0)
> in Note p' d
> invert p (Rest d) = Rest d
> invert p (m1 :+: m2) = invert p m1 :+: invert p m2
> invert p (m1 :=: m2) = invert p m1 :=: invert p m2
> invert p (Tempo r m) = Tempo r (invert p m)
> invert p (Trans n m) = Trans (-n) (invert p m)
> invert p (Instr i m) = Instr i (invert p m)
Problem 4
Prove that invert p . invert p = id.
Unfortunately, this theorem is in fact false! The reason is that its
proof relies on two lemmas:
Lemma:
pitch .
absPitch = id
absPitch .
pitch = id
But the first one is actually
false because, for example:
pitch (absPitch
(Df,0)) = pitch 1 = (Cs,0)
In other words, a D-flat gets converted into a C-sharp.
The right theorem involves the notion of "interpretation" discussed in
Chapter 21, since a D-flat indeed "sounds" the same as a C-sharp. But
instead of doing that, I will just assume that the two lemmas above
are correct, and prove a false theorem :-).
Proof by induction.
Base cases:
invert p (invert p (Note p0 d))
= invert p (Note (pitch (2 * absPitch p - abspitch p0)) d)
= Note (pitch (2 * absPitch p - abspitch (pitch (2 * absPitch p - absPitch
p0)))) d
= Note (pitch (2 * absPitch p - (2 * absPitch p - absPitch p0))) d
= Note (pitch (absPitch p0)) d
= Note p0 d
invert p (invert (Rest d))
= invert p (Rest d)
= Rest d
Induction steps:
invert p (invert p (m1 :+: m2))
= invert p (invert p m1 :+: invert p m2)
= invert p (invert p m1) :+: invert p (invert p m2)
= m1 :+: m2
Similarly for :=:, the other binary operator.
invert p (invert p (Tempo r m))
= invert p (Tempo r (invert p m))
= Tempo r (invert p (invert p m))
= Tempo r m
Similarly for Instr.
Last but not least:
invert p (invert p (Trans n m))
= invert p (Trans (-n) (invert p m))
= Trans n (invert p (invert p m))
= Trans n m
|
