Balanced Word to Balanced (Sturmian?) Sequence
$begingroup$
Let $E in {0,1}^{n}, nin mathbb{N}$, be a balanced finite word: for every two subwords $U,V$ of the same length, the number of $1$'s in $U$ differs from the number of $1$'s in $V$ by at most one.
- Can $E$ be continued to an infinite balanced sequence?
- Can $E$ be continued to an infinite Sturmian sequence? (A subcase of the previous question.)
Furthermore, in both cases, what is the cardinality of the set of possible extensions of $E$?
This came up when I tried to solve an exercise, where I have to show that Sturmian sequences are dense in the space of balanced sequences (with regard to the usual topology of symbolic dynamics) - if the the answer to the second question is "yes", than we can take any prefix of length $n$ of a balanced sequence, continue it to a Sturmian sequence, and get by the definition of the metric that the distance between the two sequences is $leq frac{1}{2^n}$.
general-topology dynamical-systems combinatorics-on-words
$endgroup$
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$begingroup$
Let $E in {0,1}^{n}, nin mathbb{N}$, be a balanced finite word: for every two subwords $U,V$ of the same length, the number of $1$'s in $U$ differs from the number of $1$'s in $V$ by at most one.
- Can $E$ be continued to an infinite balanced sequence?
- Can $E$ be continued to an infinite Sturmian sequence? (A subcase of the previous question.)
Furthermore, in both cases, what is the cardinality of the set of possible extensions of $E$?
This came up when I tried to solve an exercise, where I have to show that Sturmian sequences are dense in the space of balanced sequences (with regard to the usual topology of symbolic dynamics) - if the the answer to the second question is "yes", than we can take any prefix of length $n$ of a balanced sequence, continue it to a Sturmian sequence, and get by the definition of the metric that the distance between the two sequences is $leq frac{1}{2^n}$.
general-topology dynamical-systems combinatorics-on-words
$endgroup$
add a comment |
$begingroup$
Let $E in {0,1}^{n}, nin mathbb{N}$, be a balanced finite word: for every two subwords $U,V$ of the same length, the number of $1$'s in $U$ differs from the number of $1$'s in $V$ by at most one.
- Can $E$ be continued to an infinite balanced sequence?
- Can $E$ be continued to an infinite Sturmian sequence? (A subcase of the previous question.)
Furthermore, in both cases, what is the cardinality of the set of possible extensions of $E$?
This came up when I tried to solve an exercise, where I have to show that Sturmian sequences are dense in the space of balanced sequences (with regard to the usual topology of symbolic dynamics) - if the the answer to the second question is "yes", than we can take any prefix of length $n$ of a balanced sequence, continue it to a Sturmian sequence, and get by the definition of the metric that the distance between the two sequences is $leq frac{1}{2^n}$.
general-topology dynamical-systems combinatorics-on-words
$endgroup$
Let $E in {0,1}^{n}, nin mathbb{N}$, be a balanced finite word: for every two subwords $U,V$ of the same length, the number of $1$'s in $U$ differs from the number of $1$'s in $V$ by at most one.
- Can $E$ be continued to an infinite balanced sequence?
- Can $E$ be continued to an infinite Sturmian sequence? (A subcase of the previous question.)
Furthermore, in both cases, what is the cardinality of the set of possible extensions of $E$?
This came up when I tried to solve an exercise, where I have to show that Sturmian sequences are dense in the space of balanced sequences (with regard to the usual topology of symbolic dynamics) - if the the answer to the second question is "yes", than we can take any prefix of length $n$ of a balanced sequence, continue it to a Sturmian sequence, and get by the definition of the metric that the distance between the two sequences is $leq frac{1}{2^n}$.
general-topology dynamical-systems combinatorics-on-words
general-topology dynamical-systems combinatorics-on-words
edited Dec 8 '18 at 9:45
co.sine
asked Dec 6 '18 at 8:30
co.sineco.sine
362112
362112
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