The equation to find the quantity of combinations of CSEGs that do NOT include 'redundant CSEGs is represented by the expression: y(n) = n!, whereas n represents the cardinal number.
To determine the quantity of all combinations of CSEGs including 'redundant' CSEGs, use the cardinal number as the base and exponent of a simple exponential equation: y(n) = n^n
The symbol, ‡, shows that the set can be 'reduced' and/or 'normalized' to another CSEG set.
By 'reduce' and 'normalize,' I mean that the segments can be represented as a smaller and simpler set. For example, the set <5432> can be reduced to <3210> by means of subtracting each digit by '2' because '2' is the distance between '0' and the lowest digit in the set, '2.' Likewise, I use the term 'normalize' to signify the process of truncating any empty intervals. For example, the set <4301> cannot be 'reduced' because it contains a zero, and if the four digits were deduced by any constant, the set would display a negative number. Yet, the number '2' is not represented in this set, thus any digit greater than '2' can be reduced by '1' in order to use that position in the contour, rendering the set <3201>. (Since '4' and '3' are greater than '2,' then '1' should be deducted from '4' and '3,' but NOT '0' and '1.')
...
While this footnote is already tangental, I can't seem to find a point where I would like to stop. The math is becoming increasingly interesting as I continue to look further into contour set-theory:
As previously discussed, the contours list derived from the n^n equation rendered some sets that simply fail to qualify as contour sets. These sets were marked by the symbol, ‡ and I will call these 'empty sets' for the sake of this explanation. This led me to conclude that there should exist a different equation that generates all possibilities of CSEG sets, as does the n^n equation, but eliminates these pesky empty sets. When I eliminate these sets to the ones listed, I found that cardinal-1 CSEGS have only 1 set, cardinal-2 CSEGS have only 3 sets, and cardinal-3 CSEGS have 13 sets. I could have gone on, but I would rather look for an equation that does it so I don't have to list out thousands of sets and meticulously pick out the ones that don't work… and not make any errors. So, I knew that I was looking for an equation that would produce the following series: 1, 3, 13… when I plugged in the cardinal numbers, 1, 2, 3… After playing with a few equations, I realized that I was not going to find this on my own. I executed a few google searches with terms like "1 2 13 series" and found a few math-related sites that steered me in the right direction. I eventually found that the series continues on as follows: 1, 3, 13, 75, 541, 4683, 47293, 545835… and that they are called 'ordered Bell numbers.' At first, I wasn't sure if I should trust that my series 1, 3, 13… was enough information to assume that it was related to the first series I found on the internet that contains those numbers. But as it turns out, it is the correct series and here is why:
A few of the websites that I ran across containing information on this series kept bringing up the following 'horse race' thought experiment: In how many ways, counting ties, can horses cross a finish line? ^a ^b ^c I considered comparing my mathematical contour-set problem with the horse race puzzle. Is it the same problem? YES! Imagine that we have two horses named Horse A and Horse B that set out on a race. There are three possible outcomes: They tie for first place <00>, Horse A wins while Horse B comes in second place <01>, and the third outcome, Horse B wins while Horse A comes in second place <10>. We wouldn't say Horse A and Horse B tied for second place <11>. Rather, we would 'reduce' the sets by deducting '1' as I have demonstrated with the contour sets so that '0' is the lowest number. Likewise, if we had three horses—Horse A, Horse B, and Horse C—there would be thirteen possible outcomes. Just as in contour theory, situations such as set <202> would never occur in the horse race outcome because that would be like saying "Horse B won first place, while Horse A and Horse C tied for third place." In the same way that I 'normalized' the contours that didn't use up all the values between 0 and the highest value in the set, horse race judges would 'truncate' the middle value and give both Horse A and Horse C the benefit of the doubt by awarding them both second place rather than the third place medal.
By 'reduce' and 'normalize,' I mean that the segments can be represented as a smaller and simpler set. For example, the set <5432> can be reduced to <3210> by means of subtracting each digit by '2' because '2' is the distance between '0' and the lowest digit in the set, '2.' Likewise, I use the term 'normalize' to signify the process of truncating any empty intervals. For example, the set <4301> cannot be 'reduced' because it contains a zero, and if the four digits were deduced by any constant, the set would display a negative number. Yet, the number '2' is not represented in this set, thus any digit greater than '2' can be reduced by '1' in order to use that position in the contour, rendering the set <3201>. (Since '4' and '3' are greater than '2,' then '1' should be deducted from '4' and '3,' but NOT '0' and '1.')
...
While this footnote is already tangental, I can't seem to find a point where I would like to stop. The math is becoming increasingly interesting as I continue to look further into contour set-theory:
As previously discussed, the contours list derived from the n^n equation rendered some sets that simply fail to qualify as contour sets. These sets were marked by the symbol, ‡ and I will call these 'empty sets' for the sake of this explanation. This led me to conclude that there should exist a different equation that generates all possibilities of CSEG sets, as does the n^n equation, but eliminates these pesky empty sets. When I eliminate these sets to the ones listed, I found that cardinal-1 CSEGS have only 1 set, cardinal-2 CSEGS have only 3 sets, and cardinal-3 CSEGS have 13 sets. I could have gone on, but I would rather look for an equation that does it so I don't have to list out thousands of sets and meticulously pick out the ones that don't work… and not make any errors. So, I knew that I was looking for an equation that would produce the following series: 1, 3, 13… when I plugged in the cardinal numbers, 1, 2, 3… After playing with a few equations, I realized that I was not going to find this on my own. I executed a few google searches with terms like "1 2 13 series" and found a few math-related sites that steered me in the right direction. I eventually found that the series continues on as follows: 1, 3, 13, 75, 541, 4683, 47293, 545835… and that they are called 'ordered Bell numbers.' At first, I wasn't sure if I should trust that my series 1, 3, 13… was enough information to assume that it was related to the first series I found on the internet that contains those numbers. But as it turns out, it is the correct series and here is why:
A few of the websites that I ran across containing information on this series kept bringing up the following 'horse race' thought experiment: In how many ways, counting ties, can horses cross a finish line? ^a ^b ^c I considered comparing my mathematical contour-set problem with the horse race puzzle. Is it the same problem? YES! Imagine that we have two horses named Horse A and Horse B that set out on a race. There are three possible outcomes: They tie for first place <00>, Horse A wins while Horse B comes in second place <01>, and the third outcome, Horse B wins while Horse A comes in second place <10>. We wouldn't say Horse A and Horse B tied for second place <11>. Rather, we would 'reduce' the sets by deducting '1' as I have demonstrated with the contour sets so that '0' is the lowest number. Likewise, if we had three horses—Horse A, Horse B, and Horse C—there would be thirteen possible outcomes. Just as in contour theory, situations such as set <202> would never occur in the horse race outcome because that would be like saying "Horse B won first place, while Horse A and Horse C tied for third place." In the same way that I 'normalized' the contours that didn't use up all the values between 0 and the highest value in the set, horse race judges would 'truncate' the middle value and give both Horse A and Horse C the benefit of the doubt by awarding them both second place rather than the third place medal.
The equation:
The equation to find the quantity of sets given the cardinal number turns out to be much more complicated than I expected! I must take this opportunity to give Travis Richardson, my roommate, credit for working this out. The first process is to create a triangular array using the equation called the "sterling numbers of the second kind."
Stirling Numbers of the second kind equation: ^d
This equation creates the following triangular array if we plug in each combination of numbers for n and k whereas n = 1-5 and k = 1-5:
Then, we look for "Ordered Bell numbers" using this equation: ^e
The 'S' in the equation signifies that we need to look up those coordinates on the table. When those numbers are plugged in, it produces the series we're looking for.
Let n = 3:
After crunching through all of these numbers, I discovered Robert Morris’s article, “New Directions in the Theory and Analysis of Musical Contour.”^9 I realized that Dr. Morris had already worked out these equations including the Stirling Numbers of the Second Kind and the Ordered Bell Numbers, and then some. Thus, this entire footnote could actually be replaced by a single citation of Robert Morris's article; however, I am choosing to show my work anyway. I think that my path to finding this equation is just about as exciting as the statistical contours of Berg's Clarinet piece. Although it is a little disappointing to go through all this work only to find that it was completed by a brilliant music theorist nineteen years ago, I must admit that I am not ashamed of having independently stumbled upon the same conclusions as Robert Morris.
