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I'm working on coming up with prime forms. I've been using this calculator online to check my work.

My problem is that the prime form I compute for a certain PCS doesn't agree with the calculator, but I can't find my error. I need help figuring out where I'm going off the tracks, or if the online calculator is wrong (which seems unlikely).

I'm starting with the PCS: [8, 11, 10, 1, 0]

Showing my work, I enumerate these rotations:

[0, 1, 8, 10, 11]
[1, 8, 10, 11, 12]
[8, 10, 11, 12, 13]
[10, 11, 12, 13, 20]
[11, 12, 13, 20, 22]

I then choose as the normal form [8, 10, 11, 12, 13], because the distance between the first and last items (13 - 8 = 5) is the least of all of the other rotations.

I then transpose the normal form to [0, 2, 3, 4, 5] (through the interval of 4)

So I get to a prime form of [0, 2, 3, 4, 5].

The online calculator (when I click on the buttons for [0 1 8 10 11]) yields a different prime form of (0,1,2,3,5).

So to recap, my prime form is [0, 2, 3, 4, 5] and the calculator says the correct prime form is [0, 1, 2, 3, 5].

What's going wrong here?

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You might find this more detailed pc-set calculator useful to know about. –  LiberalArtist Sep 1 at 4:14

2 Answers 2

up vote 3 down vote accepted

As the PC Set Calculator you link to shows, [0, 1, 2, 3, 5] is the Prime Inversion of [0, 2, 3, 4, 5]. However, as [0, 1, 2, 3, 5] has a lower interval at the bottom, it is considered the correct prime form of these two equivalent PC Sets.

Another page on the website you link to gives a rigorous method for determining the Prime Form of any PC Set. The last of 8 steps asks you to find:

Which form, the original or the inverted, is most packed to the left (has the smallest numbers)? That will be the Prime Form.

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This seems to be the missing part - the prime form is the most packed to the left, whether that's the way I did it, or in inverted form. In several other online sources of information, it steps you through the algorithm I did, and doesn't explain the inversion part of this. Thanks!. –  FrobberOfBits Aug 28 at 17:27

I think you forgot to test inversions -- you should measure the distance between the first two and last two classes, and find the form that packs the smaller intervals to the left.

If you invert [0 2 3 4 5], you get [0 1 2 3 5], which succeeds in packing the shorter distance to the left of the form (starts with a distance of 1 instead of 2).

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