[math-fun] Re: Path Question
Fred Lunnon writes:
Even if the following were to work, it would hardly qualify as an "explicit" construction: but for what it's worth, suppose we have already to hand a (cyclic) deBruijn sequence S_n in which every binary word of length n occurs as a factor exactly once.
...
Lest this all seem absurdly half-baked, here's an example:
n = 4: 00101101 00001111
n = 5: 00101101 11010100010011 0000011111
A very nice idea! Can it be taken out to infinity? Jim
On 4/17/07, James Propp <propp@math.wisc.edu> wrote:
Fred Lunnon writes: ...
Lest this all seem absurdly half-baked, here's an example:
n = 4: 00101101 00001111
n = 5: 00101101 11010100010011 0000011111
A very nice idea! Can it be taken out to infinity?
Jim
n = 6: 0010110111010100010011 000010101111011001110001101001 000000111111 n = 7: [no such sequence ...] A more thorough investigation would backtrack over all possibilities; but at this juncture the original concept looks fairly moribund! Perhaps it's worth mentioning a bomb-proof but clumsier alternative. deBruijn sequences --- (k!)^k^(n-1)/k^n disitnct ones, modulo cyclic shift --- exist for every span n and base k (with k = 2 here). So at each iteration pick any one with span n' = 2^n, then shift it until the previous one with span n appears at the front. Or we might use binary Galois FSR sequences here, instead of deBruijn. These omit the string 0^n; on the other hand, there are theorems bounding the variation in word frequency (see Lidl & Niederreiter). Fred Lunnon
On 4/17/07, James Propp <propp@math.wisc.edu> wrote:
Fred Lunnon writes:
Even if the following were to work, it would hardly qualify as an "explicit" construction: but for what it's worth, suppose we have already to hand a (cyclic) deBruijn sequence S_n in which every binary word of length n occurs as a factor exactly once.
...
A very nice idea! Can it be taken out to infinity?
A complete search yielded the following maximal sequence, with n = 7 : [01011001001101110001010100101111 00111011010000100011001100001101 01011101001111011111000001011011 00011100101000100100000001111111] For n >= 7, it looks as if no span-n binary deBruijn sequence with factor 0^n1^n can be extended to one of span n+1, even if the "bad" factor is replaced by 0^{n+1}1^{n+1}. WFL
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Fred lunnon -
James Propp