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Thue - Morse sequence and free square string

By Glen Davis,2015-11-06 08:24
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Thue - Morse sequence and free square string

Thue - Morse sequence and free square string

    String of hello appeared two consecutive l.A string of prototype appeared two consecutive ot.A string of nonsense appeared two consecutive nse.If a string of continuous appeared two segments of the same, in other words the string contained tangible such as XX (where X represents a substring), we said that the string contains a "square" (square).If a string is not square, we say this string is "no square" (square - free).

    If only use two kinds of characters, for example character 0 and 1 character, we can construct some very limited free square string length.In fact, we can construct the following six square string: 0, 1, 01, 10, 010, 101.If, however, allows the use of three kinds of characters, for example the 0, 1, 2, we can not only construct any length from square string, also can construct the infinite long square string.Before you continue reading, you might as well give it a try first by yourself.

    Let's look at a discussion and just did not seem related problem.You can find the rule of the following sequence?(considering string is essentially a character sequence, so here we often mix "string" and "sequence" of these two concepts.)

    0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1,...

    Answer: they respectively 0, 1, 2, 3, 4,...How many binary representation of the number 1, including 1 0 (have an even number of digital, 1 has an odd number number 1.If I'm not clear, see the table below you should see. Dec0 1 2 3 4 5 6 7 8 9 10 11

    ima

    l n

    umb

    er

    A b0 1 10 11 100 101 110 111 100100101101

    ina0 1 0 1

    ry

    num

    ber

    1 tAn An An An An An An An An An An An

he eveoddoddeveoddeveeveoddoddeveeveodd

    numn n nu nun n nun nn n nu nun nn n nu

    berumbmbembeumbmbeumbumbmbembeumbumbmbe

     of er r r er r er er r r er er r

    The0 1 1 0 1 0 0 1 1 0 0 1

     se

    que

    nce

    We might as well put this sequence is represented by t.Sequence t still has a lot of the definition of equivalence, for example, we can recursively defined, when n is even, t (n) = t (n / 2), when n is odd, t (n) = 1, t (n - 1);Finally, a (0) = 0, the whole sequence is uniquely identified.You will find that define the way although it is new, but still did not change the essence behind.If n is even, then the last of the binary expression of n is the number 0, divided by 2 is actually pretty and remove the number 0, the parity of the number of Numbers 1 did not change obviously.If n is odd, then the last of the binary expression of n is the number 1, and the last of the n - 1 binary expression is 0, the two binary number only in the last one is different, so the parity of the number of the number 1 is certainly the opposite.Therefore, such recursive down continuously, the resulting sequence and the sequence of t the same just now.Due to all the even number n, t = t (n) (n / 2) set up all the time, so the sequence, and a very flattering properties: the sequence of t (1), t (3), t (5),...Are removed, retain only the t (0), t (2), t (4),..., thus obtained a new infinite sequence, it is identical to that of the original sequence!In addition, because for all the odd number n, t (n) = 1 - (n - 1) t = 1, t (n - 1)/(2) set up all the time, you can also choose to remove the t (0), t (2), t (4),..., t (1), t (3), t (5),..., giving a new infinite sequence of it and the original sequence of each item is just the opposite.

    Introduced in sequence t, a lot of places would adopt another way of the definition of equivalent:starting from 0, continuously perform "invert and rear" operation, the resulting sequence is a sequence of t.The so-called "invert", is to turn all of 0 to 1, all the all become 0 1;The so-called "rear", is take the string in the back of the current string.Starting from 0, 1, after the not add it gets behind 0 01.01 invert after 10, add it gets behind 01, 0110;After 0110 the not receive 1001, add it at the back of the 0110 get 01101001...Constant along the way, we will get the sequence t.

    0-01-0110-0110-0110100110010110 -...

    Why is that?Because the essence of this sequence generation method is still in the statistics of the number of binary Numbers 1 number.We continually according to the law of the binary number, the use of t (0) to the value of t

    (2, 1), introduced t (2) the value of t (2, 1).Sequence of t, for example, the first 4 digits represent the 00, 01, 10 and 11 of these four number 1 in the binary number number parity, then the next sequence t number should represent the 4, 100, 101, 110 and 111 the number of the four number 1 in the binary number parity.Four binary number before and after the four binary number difference only lies in the front of the Numbers 1, they contain the parity of the number of Numbers 1 should be just the opposite.Therefore, if the sequence is t the first 4 digits are 0, 1, 1, 0, then the sequence t number should be completely, in turn, the next four, respectively is 1, 0, 0, 1.

    From 1906 to 1906, the Norwegian mathematician Axel Thue published a series of papers, carried on the detailed research to the sequence for the first time, became the combinatorics on words staple of this new branch of mathematics.In 1921, American mathematician Marston Morse sequence in the proposed by the Thue differential topology, and the sequence was eventually named to Thue - Morse sequence.

    Thue - there are a lot of beautiful nature Morse sequence.If a string of continuous appeared two segments of the same, but they have a character of the cross, in other words the string of the form such as aXaXa model, where X represents a substring, a representative one character at a time, so we say aXa "overlap" happened in this string (overlap).For example, the word came ana overlap among banana, word Mississippi has also appeared in the issi overlap.If there is no overlap in a string, we said is the string of "no overlapping" (overlap - free).Let's to prove Thue - one of the most important properties of Morse sequence: it is to avoid overlapping.

    We use the reduction to absurdity.If Thue - Morse sequence has overlapping substring, so in all of the overlapping substring there must be a minimum of overlapping substring.This means that Thue - Morse sequence will contain aXaXa mode, where X said a substring, a for a character, and the length of X has minimum.First, we prove that the length of X is unlikely to be an odd number.Otherwise, the location of the three as the parity of the number of the same, so if westart from the first a, interval to read each character, will be like a result, aX, aX 'X' which is drawn from the X string length is the half of X (to remove the whole).However, we have already mentioned above Thue - Morse sequence: the properties of interval to extract characters, get a new string or Thue - Morse sequence, or after the invert is Thue - Morse sequence.This shows that aX, aX 'a, or take it after the results, in fact is the substring Thue - Morse sequence.Therefore, there exist more shorter than aXaXa Thue - Morse sequence of overlapping phenomenon, this has to do with the length of X of the pram contradictions when.

    Next, we prove that the length of X and can not be even.First noticed that 4 n and 4 n + 1, 4 n + 2, 4 n + 3 binary expression, only the last two digits

    are different, which in turn is 00, 01, 10 and 11.Therefore, t (4 n), t (4 n + 1), t (4 n + 2) and t (4 n + 3) values or in turn is 0, 1, 1, 0, or, in turn, is 1, 0, 0, 1.So, if we put the number four number four Thue - Morse sequence as a group, you will find Thue - Morse sequence one by one (0, 1, 1, 0) and (1, 0, 0, 1).

    If length is greater than or equal to 4 X even, then whatever aXaXa in Thue - Morse sequence appeared in what place, before a aXa will inevitably contains a quad in the middle of the two, might as well assume that this is the first of aXa paragraphs I and I + 1.In addition, don't forget the length of X is an even number, therefore before a aXa need to move to an odd number of unit and to the right after a aXa overlap.This contradiction: move to the right after an odd number of units, aXa's first paragraphs I and I + 1 will correspond to the other two in front of a quad or two behind, so the aXa before the first paragraphs I and I + 1 is two the same number, after an item of the ith aXa and the I + 1 item are two different Numbers, this is clearly absurd.

    If X is equal to the length of the 2, aXa length is very short, so much so that it will happen, no one in the middle of the quad two fall in the scope of the aXa before, at this time in front of the reasoning is failed.But it doesn't matter, if really happened this kind of situation, aXaXa position can only be like the image below, before a aXa paragraphs 1 and 2 correspond to a quad after two, but after a aXa item 1 and 2 will correspond to the next item in the middle of the quad, two contradictions still exist.

    Finally, if the length of X is 0?It's less likely.In Thue - Morse sequence, any three consecutive characters will cover to a quad behind the front two or two, which contains two different Numbers.Therefore, never visible in Thue - Morse sequence such as aaa substring occurs.

    To sum up, Thue - Morse sequence could not contain like aXaXa substring, namely Thue - Morse sequence is no overlap.

    With this conclusion, we can solve the problem of this article was originally mentioned.Using Thue - Morse sequence, we can get a free square infinite long string, which contains only 0, 1, 2, 3 kinds of characters.Method is very simple:

    just listed in Thue - Morse sequence of adjacent two how many between 0 to 1.In Thue - Morse sequence, the first number between 0 and the second number 0 with two Numbers 1, 2 a number between 0 and 0 3 with a Numbers 1, 3 a number between 0 and 4 0 0 with Numbers 1, 4 a number between 0 and 0 5 with two Numbers 1...Then, we get the one with 2, 1, 0, 2 at the beginning of infinite string.Note that due to Thue - Morse sequence is not possible in the three or more than three consecutive Numbers 1, so will not appear in the string is greater than or equal to 3 number, only the number 0, 1 and 2.

    Thue - Morse sequence: 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0,... New sequence: 2, 1, 0, 2, 0, 1, 2,...

    Why is the sequence of the resulting from square?Very simple.If there is a new sequence inside square, such as a 1 a 2 a 3...A n a 1 a 2 a 3...A n, this means that has arisen in Thue - Morse sequence 1 0 0 1 0 0 1...1 0 0 1 0 0 0 1...1 0 0 (1 1 continuous a number 1, and so on), then formed the overlapping substring, and Thue - Morse sequence to avoid overlaps.

    Starting from Thue - Morse sequence to avoid overlaps, we can draw a lot of interesting implications.Thue - for example, Morse sequence must be free of cubic, namely Thue - Morse sequence does not exist in the form such as XXX substring.Might as well the reason is simple: if X = aX ', so XXX is actually aX, aX, aX, aX of "aX" a formed overlapping substring, and contradiction with Thue - Morse sequence to avoid overlaps.This can be further introduced, Thue never happen - Morse sequence cycle.The reason is simple: if Thue - Morse sequence began circulating from somewhere, it's directly with Thue - Morse sequence from the cubic conflicts.

    At the same time, Thue - Morse sequence is a "repetition sequences" (recurrent sequence), meaning that each substring in Thue - Morse sequence will appear infinitely many times.The reasons behind this fact is also very simple.For example, we out of t in Thue - Morse sequence (6) to t (10), so they are 0, 1, 1, 0, 0, said a binary number 0110, 0111, 1000, 1001, 1010 of the parity of the number of the number 1.So, 0, 1, 1, 0, 0 will appear countless times in the future.In number to a binary number 11, 0110, 11, 0111, 11, 1000, 11, 1001, 11, 1010, we will once again be 0, 1, 1, 0, 0;In number to a binary number, 101, 0110, 101, 0111, 101, 1000, 101, 1001, 101, 1010, we will once again be 0, 1, 1, 0, 0.Obviously there are infinitely many such opportunity, for example, in a few to binary number 110100100, 0110, 110100100, 0111, 110100100, 1000, 110100100, 1001, 110100100, 1010, we will once again be 0, 1, 1, 0, 0.

    Construct a sequence is simple repetition, any a cycle sequence which meet the requirements, such as 0, 1, 0, 1, 0, 1,....While Thue - Morse sequence tells us: there is not a repetition of sequence cycle sequences.

    Is presented in the end, we don't prove two Thue - Morse sequence related conclusion.For which A positive integer k 2 or more, there are two equal to the size of the integer set A = {1 A, 2 A, 3 A,..., n} and B = {B 1, 2 B, 3 B,..., b, n},

    a 1 + a 2 + a 3 + + a n = b 1 + b 2 + b 3 + + b n

    a 1 + a 2 + a 3 + + a n = b 1 + b 2 + b 3 + + b n

    ……

    a 1 + a 2 + a 3 + + a n = b 1 + b 2 + b 3 + + b n

    All the number in the set A and set all the number from 1 to the power, and in the B to k equal to the power and all?Of course, set A and set B must be two different collections.The answer is, for all the positive integer k 2 or more, meet the requirements of the solution are present.Using Thue - Morse sequence, we can draw a n = 2, the structure of the solution method is as follows: remove Thue - Morse sequence, the former two, namely t (0), t (1),..., t (2, 1), t (I) = 0, if put into A collection, I if t (I) = 1, will put the I in the collection of B.

    i 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

    t0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0

    (i)

    When k = 3, as shown in the table above, according to the rules, we should put 0, 3, 5, 6, 9, 10, 12, 15 is divided into a group, 1, 2, 4, 7, 8, 11, 13, 14 is divided into another group.Amazing thing appeared: the following three equation is true really!

    0 + 3 + 5 + 6 + 12 + 9 + 10 + 15 = 1 + 2 + 4 + 7 + 8 + 11 + 13 + 14 0 + 3 + 5 + 6 + 12 + 9 + 10 + 15 = 1 + 2 + 4 + 7 + 8 + 11 + 13 + 14 0 + 3 + 5 + 6 + 12 + 9 + 10 + 15 = 1 + 2 + 4 + 7 + 8 + 11 + 13 + 14

    Thue - Morse sequence can also help construct magic square.1977, Adler Allan and Shuo - Yen Robert Li, an algorithm is given, can use Thue - Morse sequence structure of 2 x 2 magic square (where n 2 or higher).First of all, from left to right from up to down to 1 to 2 in the number of fill in the 2 x 2 squares.Then, if the number of the ith Thue - Morse sequence is 0 (that is, t (I, 1) = 0), I out of the box.Finally, bring out all the number of reverse order back into the grid, we had a magic square.Shown below are examples of n = 2.Due to Thue - Morse sequence of 1, 4, 6, 7, 10, 11, 13, 16 number is zero, so we put these number to withdraw from a phalanx of 4 * 4;Put them in reverse order back after,

    you can verify and phalanx of each row, each column and the sum of the Numbers on the two diagonals are 34.

    i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

    t0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0

    (i

1)

    Online while wandering chanced upon Thue - Morse sequence, then intoWikipediaRelated entry cannot extricate oneself, then to the references mentioned behind these entries, good had a combinatorics on words.Actually, Thue - Morse sequence as well as a lot of things worth talking with, but the energy is limited, can only write here.The interested reader might as well take a look at the Combinatorics on Words: Christoffel Words and Repetitions in Words and Automatic Sequences:, Applications, these two books Generalizations.

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