A musical question for math-types.

This is along the same lines as calculating the number of possible positions in chess...

How many distinct 2 bar melodies can be played using a single (standard) octave? Assume 4/4 time w/no time changes, and no notes or rests shorter than 1/16 note.

I think your first criterion is which instrument is being used; that way there's some limitation on which notes can be played.

-mike

A fascinating question. I would guess at billions.
When you say a standard octave, do you include sharps/flats and/or
microtones?

Only the standard sharps and flats - 12 tones in all.

Originally posted by richjohnson
This is along the same lines as calculating the number of possible positions in chess...

How many distinct 2 bar melodies can be played using a single (standard) octave? Assume 4/4 time w/no time changes, and no notes or rests shorter than 1/16 note.

Is there a limit on how long the notes can be?

Originally posted by Cheshire Cat
Is there a limit on how long the notes can be?

Yes - 2 bars (2 linked whole notes)

I must confess that I don't know how to answer this puzzle. If the only type of note possible was an eighth note the answer would be 665416609183179841, but with different lengths of the note I don't know where to go with this number. Alas, if you know the answer and how to get it could you please send it to me so that I can put it out of my mind :-)

ok, now what about this: why not do like so many mathematicians, make the problem easier, solve the easier problem and then try to get back to the original problem.
first take a look at the first 16th. You can choose that one from 12 notes. then put another 16th after it, you can choose this one from 12 notes too, but if you make it the same as the one before you can choose to connect them, so you get an 8th note, or not connect them, which makes 2 16th notes. that makes 13 choices for the second notes. So for music that lasts 2 16th notes you have 12*13=156 choices. now go on like that, put the third 16th in. You also have 13 choices for that one, so for music that lasts 3 16th you have 12*13*13 choices. you can go on like this till you have two bars of music and that makes 12*(13^31) choices, which is about 4.087*10^35

Originally posted by Nuathala
ok, now what about this: why not do like so many mathematicians, make the problem easier, solve the easier problem and then try to get back to the original problem.
first take a look at the first 16th. You can choose that one from 12 notes. then put another 16th after it, you can choose this one from 12 notes too, but if you make it the same as the one bef ...[text shortened]... is till you have two bars of music and that makes 12*(13^31) choices, which is about 4.087*10^35

This sounds okay to me. She must be very smart 😉

Originally posted by Nuathala
ok, now what about this: why not do like so many mathematicians, make the problem easier, solve the easier problem and then try to get back to the original problem.
first take a look at the first 16th. You can choose that one from 12 notes. then put another 16th after it, you can choose this one from 12 notes too, but if you make it the same as the one bef ...[text shortened]... is till you have two bars of music and that makes 12*(13^31) choices, which is about 4.087*10^35

I could be wrong but one reason that this doesn't work is that for the first note there are 13 choices since a rest would be possible. Taking that into consideration, the second sixteenth note would have 13 possibilities as well. This brings us to 169 possibilities for the first two notes. If you keep doing that down the line for each of the 32 sixteenth note possibilities, you have 13*13....32 thirteens or 13 to the thirty second power. This is 4.428 * 10^35. Unfortunately, if you add different lengths of each note, you have vastly more possibilities; since, in a three sixteenth note bar you can have either three sixteenth notes, one eigth note then one sixteenth note, one sixteenth note then one eighth note, or an eighth note and a sixteenth note combined which are four different distinct melodies. Now add 29 more spaces and what do you have? That's where I get stuck.

Originally posted by Cheshire Cat
I could be wrong but one reason that this doesn't work is that for the first note there are 13 choices since a rest would be possible. Taking that into consideration, the second sixteenth note would have 13 possibilities as well. Th ...[text shortened]... d 29 more spaces and what do you have? That's where I get stuck.

Nuathala did not include the rest. But let me explain what she meant.
Every 13 mention in this proof is the number of possible notes and rest.
presume we are at the beginning of the melody, looking at only a three sixteenth bar.
If we place three sixteenth notes we have 13^3 possibilities. If we take an eight note and then a sixteenth note, we have 2*13^2 posibilities. 13 for each position and 2 because we can reverse oreder. If we take a three sixteenth note we have only 13 posibilities. Total possible combinations: 13^3 + 2*13^2 + 13 = 13*(13+1)^2=13*14^2.

But as Nuathala says, one can see an eight note as a combination of an sixteenth note with an enlenghtment. If you write your note bar with only sixteenth notes and $'s if you lengthen the note, (then an eighth note is writen as a sixteenth note and a $ and a three sixteenth note would be writen as an sixteenth note with 2 $s) then one sees that on each position there can be 14 symbols (twelve notes and the $). Therefor there are 14^32 possible combinations.

Unless (I don't know much of this) it is not allowed the lenghten the the last note of a bar into the next bar. Then there are only 13*14^15 possibilities per bar. which makes 13^2*14^30=24,201,432,355,484,595,421,941,037,243,826,300 ~= 24*10^33


Or am I missing something?

you're right, i forgot about the rest, that indeed makes on extra choice for every 16th, so your answer seems right to me. 🙂

Originally posted by Nuathala
you're right, i forgot about the rest, that indeed makes on extra choice for every 16th, so your answer seems right to me. 🙂

What about slurs? On most instruments two notes slurred together are different from two notes played separately, and you can slur across bars (meaning that different notes as well as similar notes can be joined together both within and across bars). You can't slur rests though.

Originally posted by Fiathahel
Nuathala did not include the rest. But let me explain what she meant.
Every 13 mention in this proof is the number of possible notes and rest.
presume we are at the beginning of the melody, looking at only a three sixteenth bar.
If we place three sixteenth notes we have 13^3 possibilities. If we take an eight note and then a sixteenth note, we ha ...[text shortened]... 4,201,432,355,484,595,421,941,037,243,826,300 ~= 24*10^33


Or am I missing something?

Maybe I am wrong on this; but, I think the point you are missing is that even if you connect two space in a three space bar, An eigth not followed by a sixteenth is a different melody than a sixteenth followed by an eigth regardless of what note.

Originally posted by Cheshire Cat
Maybe I am wrong on this; but, I think the point you are missing is that even if you connect two space in a three space bar, An eigth not followed by a sixteenth is a different melody than a sixteenth followed by an eigth regardless of what note.

no, we're not missing that. you can make 8th note-16th note by choosing: any note-same note tied to note before-any note (not tied if same note as note before)
and 16th note-8th note by choosing: any note-any note (not tied if same note as before)-same note tied to note before. so the combination 8th-16th is made starting with an 8th note, and the combination 16th-8th is made starting with 2 16th notes.