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by super_normal
546 days ago
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id prefer it if you would write this in haskell so i can understand what is going on there better. i am saying of the first 1,000 digits after the ones place, each digit of the sequence appears an even number of times, every consecutive subsequence of two appears an even number of times, and every consecutive subsequence of three happens exactly once. please, if you could do some more intelligable work and show your result as standard deviations away from the the exact de bruijn counts i would really appreciate it. etcetcera. (i dont know where you got the 10 symbols from. im talking about over the entire 1,000 length decimal sequence. etcetera.) |
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I use the Python decimal package to calculate 17^(1/7) to 4000 decimal places. I show the first 12 characters of the ASCII representation ("1.4989198720"), then take the first 1000 characters after the decimal place. I show the first and last 10 characters of this string ("4989198720...6163659068") so that you can verify it against the string you are testing.
Then I use a list comprehension to check each 3-digit string (e.g., 001, 002, 003, ..., 999) to find out whether it is a substring of the concatenation of `s` with itself, and list the ones that don't appear. I manually abbreviated the list, but I'm saying that in my string, 000, 002, 004, 005, 009, [and some other numbers], 993, 994, 995, 998, and 999 did not appear. (The concatenation `s+s` is checked so that the subsequences that occur at the wraparound---here,684 and 849---are found)
Because some of these length-3 sequences are missing, to my understanding this does not constitute a de Bruijn sequence of order 3 on a size-10 alphabet. However, I'm also not entirely sure whether this is what you were claiming.