How to display indexes for UITableView in a complex list?

How do I find (and list) all the possible 7-note combinations in a 12-tone tuning (NOT simply an nCr problem)?

  • Note, We are trying to list each of the scales, not just the total number of scales...and ideally do so in programming logic IE loops We are assuming that each MODE only counts once IE the mode of (indexes) 1 2 3 4 10 11 12 would be the same as 1 2 3 9 10 11 12. By the same token, the scale of 1 3 5 7_8 10 12 _ would be the same as 1 3 5_6 8 10 12_ which would be the same as 1_2 4 6_7 9 11 (we would only display the mode with the lowest total value IE lowness = 2^note1 + 2^note2....) This is not so simple as nCr as a the same note sets in different positions across the octave (IE the above examples) do >not< get counted as original combinations. A simple example (for musicians) the C major and D major are counted as the same thing (due to having the same interval gaps, just starting from a different position/note).

  • Answer:

    This is a very old problem See the wiki article in "source" below

M S at Yahoo! Answers Visit the source

Was this solution helpful to you?

Other answers

this is a rather difficult question. for smaller cases: f(2,n)=floor(n/2) f(3,3)=1 f(3,4)=1 f(3,5)=2 f(3,6)=4 f(3,7)=5 f(3,8)=7 f(3,9)=10 f(n,n)=1 f(n,n+1)=1 f(n,n+2)=floor(n/2)+1

This is a very old problem See the wiki article in "source" below

Paul G

this is a rather difficult question. for smaller cases: f(2,n)=floor(n/2) f(3,3)=1 f(3,4)=1 f(3,5)=2 f(3,6)=4 f(3,7)=5 f(3,8)=7 f(3,9)=10 f(n,n)=1 f(n,n+1)=1 f(n,n+2)=floor(n/2)+1

Joseph

Related Q & A:

Just Added Q & A:

Find solution

For every problem there is a solution! Proved by Solucija.

  • Got an issue and looking for advice?

  • Ask Solucija to search every corner of the Web for help.

  • Get workable solutions and helpful tips in a moment.

Just ask Solucija about an issue you face and immediately get a list of ready solutions, answers and tips from other Internet users. We always provide the most suitable and complete answer to your question at the top, along with a few good alternatives below.