Continued Fractions and Structural Complexity

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Continued Fractions and Structural Complexity

Karl Palmen

Dear Calendar People

 

I think that the structural complexity  of calendar cycle of leap years and common years with the leap years spread as smoothly as possible is correlated to the number of continued fraction iterations of the mean year.

 

I show this with two examples:

 

33-year cycle with 8 leap years:  leap years (4, 8, 12, 16, 20, 24, 28, 32 ) pattern cccLcccLcccLcccLcccLcccLcccLcccLc

30-year cycle with 4 leap years: leap years (2, 5, 7, 10, 13, 15, 18, 21, 24, 26, 29)  pattern cLccLcLccLccLcLccLccLccLcLccLc

 

Both cycles have the leap years spread as smoothly as possible, but the 33-year cycle is clearly simpler in structure than the 30-year cycle. The 33-year cycle has its leap years equally spaced within the cycle while the 30-year cycle has a complicated sequence of intervals of 3 and 2.

 

Now I show the continued fractions for these two cycles.  I start with the proportion of leap years, invert it, take the fractional part of that, invert it, and on till one gets a whole number.

 

33-year cycle with 8 leap years

8/33,

33/8 = 4+1/8,

 8.

 

30-year cycle with 11 leap years

11/30

30/11 = 2+8/11

8/11 = 1+3/8

3/8 = 2+2/3

3/2 = 1+1/2

2.

 

Now I show a pattern of the cycle (with a leap year at the end) for each step of the continued fractions:

 

33-year cycle with 8 leap years

8/33: ccccLcccLcccLcccLcccLcccLcccLcccL

33/8 = 4+1/8:  cccccccL

8.

 

30-year cycle with 11 leap years

11/30: ccLccLcLccLccLcLccLccLccLcLccL

30/11 = 2+8/11: cLLcLLLcLLL

11/8 = 1+3/8: ccLccLcL

8/3 = 2+2/3: cLL

3/2 = 1+1/2: cL

2.

 

I thought of an alternative continued fraction method which, can lead to faster convergence. It is done by taking the signed fractional part from the nearest integer. For the 33-year cycle, this makes no difference, but for the 30-year cycle it is shortened to just one step more than the 33-year cycle and the accompanying cycles have leap years in the minority:

 

30-year cycle with 11 leap years

+11/30: ccLccLcLccLccLcLccLccLccLcLccL

+30/11 = 3-3/11 cccLcccLccL

-11/3 = -4-1/3 ccL

3.

 

Now one can see that the cycle gets progressively simpler in structure in each step.

The last step has the simplest cycle with just 1 leap year, which would thus occur in equal intervals.

The penultimate step has two intervals between the leap years and one interval (shown in red) occurs just once in the cycle.

The antepenultimate step has two intervals between the leap years and both intervals occur more than once in the cycle.

 

Karl

 

16(02(04

 

 

 

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