An alternative: the decomposition theorem of Restivo-Salemi Restivo and Salemi (1985) discovered an alternative decomposition for finite binary overlap-free words. Theorem. Every finite binary overlap-free word w can be written uniquely in the form x µ ( y ) z , where y is overlap-free, and x , z ∈ { ǫ, 0 , 00 , 1 , 11 } . Furthermore, if | w | ≥ 7, then this decomposition is unique. 10 / 28
An alternative: the decomposition of Restivo-Salemi The Restivo-Salemi decomposition was extended to infinite binary overlap-free words by Allouche, Currie, and JOS (1998). 11 / 28
An alternative: the decomposition of Restivo-Salemi The Restivo-Salemi decomposition was extended to infinite binary overlap-free words by Allouche, Currie, and JOS (1998). Theorem. Every infinite binary overlap-free word w can be written uniquely in the form w = x µ ( y ) where x ∈ { ǫ, 0 , 1 , 00 , 11 } and y is overlap-free. 11 / 28
An alternative: the decomposition of Restivo-Salemi The Restivo-Salemi decomposition was extended to infinite binary overlap-free words by Allouche, Currie, and JOS (1998). Theorem. Every infinite binary overlap-free word w can be written uniquely in the form w = x µ ( y ) where x ∈ { ǫ, 0 , 1 , 00 , 11 } and y is overlap-free. Furthermore, the correct decomposition can be deduced by examining the first 5 symbols of w . 11 / 28
Iterating the Restivo-Salemi decomposition The Restivo-Salemi decomposition can be iterated: w = x 1 µ ( y 1 ) 12 / 28
Iterating the Restivo-Salemi decomposition The Restivo-Salemi decomposition can be iterated: w = x 1 µ ( y 1 ) x 1 µ ( x 2 ) µ 2 ( y 2 ) = 12 / 28
Iterating the Restivo-Salemi decomposition The Restivo-Salemi decomposition can be iterated: w = x 1 µ ( y 1 ) x 1 µ ( x 2 ) µ 2 ( y 2 ) = x 1 µ ( x 2 ) µ 2 ( x 3 ) µ 3 ( y 3 ) = · · · = 12 / 28
Iterating the Restivo-Salemi decomposition The Restivo-Salemi decomposition can be iterated: w = x 1 µ ( y 1 ) x 1 µ ( x 2 ) µ 2 ( y 2 ) = x 1 µ ( x 2 ) µ 2 ( x 3 ) µ 3 ( y 3 ) = · · · = If the sequence of x i contains infinitely many nonempty words, then this gives the decomposition w = x 1 µ ( x 2 ) µ 2 ( x 3 ) · · · . 12 / 28
Iterating the Restivo-Salemi decomposition The Restivo-Salemi decomposition can be iterated: w = x 1 µ ( y 1 ) x 1 µ ( x 2 ) µ 2 ( y 2 ) = x 1 µ ( x 2 ) µ 2 ( x 3 ) µ 3 ( y 3 ) = · · · = If the sequence of x i contains infinitely many nonempty words, then this gives the decomposition w = x 1 µ ( x 2 ) µ 2 ( x 3 ) · · · . Otherwise, we get w = x 1 µ ( x 2 ) µ 2 ( x 3 ) · · · µ i ( x i +1 ) µ ω ( a ) for a ∈ { 0 , 1 } . 12 / 28
Iterating the Restivo-Salemi decomposition The Restivo-Salemi decomposition can be iterated: w = x 1 µ ( y 1 ) x 1 µ ( x 2 ) µ 2 ( y 2 ) = x 1 µ ( x 2 ) µ 2 ( x 3 ) µ 3 ( y 3 ) = · · · = If the sequence of x i contains infinitely many nonempty words, then this gives the decomposition w = x 1 µ ( x 2 ) µ 2 ( x 3 ) · · · . Otherwise, we get w = x 1 µ ( x 2 ) µ 2 ( x 3 ) · · · µ i ( x i +1 ) µ ω ( a ) for a ∈ { 0 , 1 } . Further, this decomposition is unique. 12 / 28
Iterating the Restivo-Salemi decomposition So we can specify an infinite binary overlap-free word by providing (i) the infinite sequence of x i , or 13 / 28
Iterating the Restivo-Salemi decomposition So we can specify an infinite binary overlap-free word by providing (i) the infinite sequence of x i , or (ii) the finite sequence of x i (which is followed by 0 ω ) and a . 13 / 28
Iterating the Restivo-Salemi decomposition So we can specify an infinite binary overlap-free word by providing (i) the infinite sequence of x i , or (ii) the finite sequence of x i (which is followed by 0 ω ) and a . We encode the permissible x i as follows: 13 / 28
Iterating the Restivo-Salemi decomposition So we can specify an infinite binary overlap-free word by providing (i) the infinite sequence of x i , or (ii) the finite sequence of x i (which is followed by 0 ω ) and a . We encode the permissible x i as follows: p 0 = ǫ 13 / 28
Iterating the Restivo-Salemi decomposition So we can specify an infinite binary overlap-free word by providing (i) the infinite sequence of x i , or (ii) the finite sequence of x i (which is followed by 0 ω ) and a . We encode the permissible x i as follows: p 0 = ǫ p 1 = 0 13 / 28
Iterating the Restivo-Salemi decomposition So we can specify an infinite binary overlap-free word by providing (i) the infinite sequence of x i , or (ii) the finite sequence of x i (which is followed by 0 ω ) and a . We encode the permissible x i as follows: p 0 = ǫ p 1 = 0 p 2 = 00 13 / 28
Iterating the Restivo-Salemi decomposition So we can specify an infinite binary overlap-free word by providing (i) the infinite sequence of x i , or (ii) the finite sequence of x i (which is followed by 0 ω ) and a . We encode the permissible x i as follows: p 0 = ǫ p 1 = 0 p 2 = 00 = 1 p 3 13 / 28
Iterating the Restivo-Salemi decomposition So we can specify an infinite binary overlap-free word by providing (i) the infinite sequence of x i , or (ii) the finite sequence of x i (which is followed by 0 ω ) and a . We encode the permissible x i as follows: p 0 = ǫ p 1 = 0 p 2 = 00 = 1 p 3 p 4 = 11 13 / 28
An example of the iterated decomposition Let’s start with h = 001001101001011001101001100101101001011001101001 · · · , the word counting the number of 0’s (mod 2) in the binary expansion of n . Then 14 / 28
An example of the iterated decomposition Let’s start with h = 001001101001011001101001100101101001011001101001 · · · , the word counting the number of 0’s (mod 2) in the binary expansion of n . Then h = 00 µ (101100101101001100101100110100101101001 · · · ) 14 / 28
An example of the iterated decomposition Let’s start with h = 001001101001011001101001100101101001011001101001 · · · , the word counting the number of 0’s (mod 2) in the binary expansion of n . Then h = 00 µ (101100101101001100101100110100101101001 · · · ) = 00 µ (1) µ ( µ (010011010010110011010011001011010 · · · )) 14 / 28
An example of the iterated decomposition Let’s start with h = 001001101001011001101001100101101001011001101001 · · · , the word counting the number of 0’s (mod 2) in the binary expansion of n . Then h = 00 µ (101100101101001100101100110100101101001 · · · ) = 00 µ (1) µ ( µ (010011010010110011010011001011010 · · · )) = 00 µ (1) µ ( µ (0)) µ ( µ ( µ (1011001011010011001011001 · · · ))) 14 / 28
An example of the iterated decomposition Let’s start with h = 001001101001011001101001100101101001011001101001 · · · , the word counting the number of 0’s (mod 2) in the binary expansion of n . Then h = 00 µ (101100101101001100101100110100101101001 · · · ) = 00 µ (1) µ ( µ (010011010010110011010011001011010 · · · )) = 00 µ (1) µ ( µ (0)) µ ( µ ( µ (1011001011010011001011001 · · · ))) 00 µ (1) µ 2 (0) µ 3 (1) µ 4 (0) · · · = 14 / 28
An example of the iterated decomposition Let’s start with h = 001001101001011001101001100101101001011001101001 · · · , the word counting the number of 0’s (mod 2) in the binary expansion of n . Then h = 00 µ (101100101101001100101100110100101101001 · · · ) = 00 µ (1) µ ( µ (010011010010110011010011001011010 · · · )) = 00 µ (1) µ ( µ (0)) µ ( µ ( µ (1011001011010011001011001 · · · ))) 00 µ (1) µ 2 (0) µ 3 (1) µ 4 (0) · · · = p 2 µ ( p 3 ) µ 2 ( p 1 ) µ 3 ( p 3 ) µ 4 ( p 1 ) · · · . = 14 / 28
An example of the iterated decomposition Let’s start with h = 001001101001011001101001100101101001011001101001 · · · , the word counting the number of 0’s (mod 2) in the binary expansion of n . Then h = 00 µ (101100101101001100101100110100101101001 · · · ) = 00 µ (1) µ ( µ (010011010010110011010011001011010 · · · )) = 00 µ (1) µ ( µ (0)) µ ( µ ( µ (1011001011010011001011001 · · · ))) 00 µ (1) µ 2 (0) µ 3 (1) µ 4 (0) · · · = p 2 µ ( p 3 ) µ 2 ( p 1 ) µ 3 ( p 3 ) µ 4 ( p 1 ) · · · . = So h is encoded by the sequence of indices 2313131 · · · = 2(31) ω . 14 / 28
Valid decomposition sequences However, not every sequence of x i gives an infinite overlap-free word. 15 / 28
Valid decomposition sequences However, not every sequence of x i gives an infinite overlap-free word. For example, if x 1 = 00, then x 2 � = 0, for otherwise w begins 00 µ (0) = 0001, which has an overlap. 15 / 28
Valid decomposition sequences However, not every sequence of x i gives an infinite overlap-free word. For example, if x 1 = 00, then x 2 � = 0, for otherwise w begins 00 µ (0) = 0001, which has an overlap. Can we somehow characterize the “legal” sequences of x i that give the overlap-free infinite words? 15 / 28
Valid decomposition sequences However, not every sequence of x i gives an infinite overlap-free word. For example, if x 1 = 00, then x 2 � = 0, for otherwise w begins 00 µ (0) = 0001, which has an overlap. Can we somehow characterize the “legal” sequences of x i that give the overlap-free infinite words? Yes, using a finite automaton. 15 / 28
The automaton Let O denote the set of all infinite overlap-free words. States of the automaton represent subsets of O , as follows: A = O { x ∈ Σ ω : 1 x ∈ O} B = { x ∈ Σ ω : 1 x ∈ O and x begins with 101 } C = { x ∈ Σ ω : 0 x ∈ O} = D { x ∈ Σ ω : 0 x ∈ O and x begins with 010 } E = { x ∈ Σ ω : 0 x ∈ O and x begins with 11 } = F { x ∈ Σ ω : 0 x ∈ O and x begins with 1 } G = { x ∈ Σ ω : 1 x ∈ O and x begins with 1 } = H { x ∈ Σ ω : 1 x ∈ O and x begins with 00 } I = { x ∈ Σ ω : 1 x ∈ O and x begins with 0 } = J { x ∈ Σ ω : 0 x ∈ O and x begins with 0 } K = 16 / 28
We connect states as follows: an arrow from state S to state T is labeled i means w ∈ T ⇐ ⇒ p i µ ( w ) ∈ S . 0 0 H K 0 0 J G 1 1 3 3 1 3 I F 1 0 0 3 E C 0 2 4 A 1 3 1 3 B 0 D 0 1 3 17 / 28
The result for overlaps Theorem. Every infinite binary overlap-free word x is encoded by an infinite path, starting in state A , through the automaton. 18 / 28
The result for overlaps Theorem. Every infinite binary overlap-free word x is encoded by an infinite path, starting in state A , through the automaton. Every infinite path through the automaton not ending in 0 ω codes a unique infinite binary overlap-free word x . If a path i ends in 0 ω and this suffix corresponds to a cycle on state A or a cycle between states B and D, then x is coded by either i ; 0 or i ; 1. If a path i ends in 0 ω and this suffix corresponds to a cycle between states J and K, then x is coded by i ; 0. If a path i ends in 0 ω and this suffix corresponds to a cycle between states G and H, then x is coded by i ; 1. 18 / 28
The special role of 7 3 -powers 7 3 -powers play a special role in the theory of binary words: 19 / 28
The special role of 7 3 -powers 7 3 -powers play a special role in the theory of binary words: Kolpakov & Kucherov (1997) showed that the function measuring the minimum frequency of a letter in α -power-free words is discontinuous at 7 3 . 19 / 28
The special role of 7 3 -powers 7 3 -powers play a special role in the theory of binary words: Kolpakov & Kucherov (1997) showed that the function measuring the minimum frequency of a letter in α -power-free words is discontinuous at 7 3 . Karhum¨ aki and JOS (2004) proved that there are polynomially many α -power-free words for α ≤ 7 3 , but exponentially many such words for α > 7 3 . 19 / 28
The special role of 7 3 -powers 7 3 -powers play a special role in the theory of binary words: Kolpakov & Kucherov (1997) showed that the function measuring the minimum frequency of a letter in α -power-free words is discontinuous at 7 3 . Karhum¨ aki and JOS (2004) proved that there are polynomially many α -power-free words for α ≤ 7 3 , but exponentially many such words for α > 7 3 . Rampersad (2005) showed that the only 7 3 -power-free binary words that are the fixed points of a non-identity morphism are the Thue-Morse word and its complement; furthermore 7 3 is best possible. 19 / 28
The special role of 7 3 -powers 7 3 -powers play a special role in the theory of binary words: Kolpakov & Kucherov (1997) showed that the function measuring the minimum frequency of a letter in α -power-free words is discontinuous at 7 3 . Karhum¨ aki and JOS (2004) proved that there are polynomially many α -power-free words for α ≤ 7 3 , but exponentially many such words for α > 7 3 . Rampersad (2005) showed that the only 7 3 -power-free binary words that are the fixed points of a non-identity morphism are the Thue-Morse word and its complement; furthermore 7 3 is best possible. Currie & Rampersad (2010) showed that 7 3 is the infimum of all exponents α such that there exists an infinite word avoiding α -powers and containing arbitrarily large squares beginning at every position. 19 / 28
Extending Fife to 7 3 -Powers Partial results in the Ph. D. thesis of Narad Rampersad (2007) 20 / 28
Extending Fife to 7 3 -Powers Partial results in the Ph. D. thesis of Narad Rampersad (2007) Done for finite words by Blondel, Cassaigne, and Jungers (2009) 20 / 28
Extending Fife to 7 3 -Powers Partial results in the Ph. D. thesis of Narad Rampersad (2007) Done for finite words by Blondel, Cassaigne, and Jungers (2009) In this talk: a simpler version, but for infinite words. 20 / 28
Extending Fife to 7 3 -Powers Partial results in the Ph. D. thesis of Narad Rampersad (2007) Done for finite words by Blondel, Cassaigne, and Jungers (2009) In this talk: a simpler version, but for infinite words. Relies on a version of the Restivo-Salemi decomposition that works for 7 3 -powers: Theorem. Let 2 < α ≤ 7 3 . Then every infinite binary α -power-free word w can be written uniquely in the form w = x µ ( y ) where x ∈ { ǫ, 0 , 1 , 00 , 11 } and y is overlap-free. 20 / 28
Extending Fife to 7 3 -Powers Partial results in the Ph. D. thesis of Narad Rampersad (2007) Done for finite words by Blondel, Cassaigne, and Jungers (2009) In this talk: a simpler version, but for infinite words. Relies on a version of the Restivo-Salemi decomposition that works for 7 3 -powers: Theorem. Let 2 < α ≤ 7 3 . Then every infinite binary α -power-free word w can be written uniquely in the form w = x µ ( y ) where x ∈ { ǫ, 0 , 1 , 00 , 11 } and y is overlap-free. Furthermore, the correct decomposition can be deduced by examining the first 5 symbols of w . 20 / 28
The Fife-like automaton for 7 3 -powers 4 0 31 0 310 3 3 1 1 0 3 4 4 33 203 1 4 3 0 3 4 40 3 0 3 2 1 4 0 0 ǫ 0 3 2 4 1 1 0 1 2 0 20 1 2 11 3 401 2 2 1 3 0 1 3 130 1 0 13 0 2 21 / 28
Verifying the automaton Each transition in the automaton corresponds to an assertion, such as 22 / 28
Verifying the automaton Each transition in the automaton corresponds to an assertion, such as µ ( x ) is 7 3 -power-free iff x is 7 3 -power-free. 22 / 28
Verifying the automaton Each transition in the automaton corresponds to an assertion, such as µ ( x ) is 7 3 -power-free iff x is 7 3 -power-free. Many of these follow from known results on α -power-free words. 22 / 28
Verifying the automaton Each transition in the automaton corresponds to an assertion, such as µ ( x ) is 7 3 -power-free iff x is 7 3 -power-free. Many of these follow from known results on α -power-free words. Others require some (fairly simple) ad hoc reasoning. 22 / 28
Proof of one assertion F 23 = F 13 : in other words, 00 µ (1 µ ( w )) is 7 3 -power-free iff 0 µ (1 µ ( w )) is 7 3 -power-free. 23 / 28
Proof of one assertion F 23 = F 13 : in other words, 00 µ (1 µ ( w )) is 7 3 -power-free iff 0 µ (1 µ ( w )) is 7 3 -power-free. One direction is obvious. 23 / 28
Proof of one assertion F 23 = F 13 : in other words, 00 µ (1 µ ( w )) is 7 3 -power-free iff 0 µ (1 µ ( w )) is 7 3 -power-free. One direction is obvious. For the other, note that if 0 µ (1 µ ( w )) is 7 3 -power-free, but 00 µ (1 µ ( w )) is not, then the 7 3 -power in it must be a prefix. 23 / 28
Proof of one assertion F 23 = F 13 : in other words, 00 µ (1 µ ( w )) is 7 3 -power-free iff 0 µ (1 µ ( w )) is 7 3 -power-free. One direction is obvious. For the other, note that if 0 µ (1 µ ( w )) is 7 3 -power-free, but 00 µ (1 µ ( w )) is not, then the 7 3 -power in it must be a prefix. However, if 0 µ (1 µ ( w )) is 7 3 -power-free, then w must start with 0 (else 0 µ (1 µ ( w )) would start with 01010). 23 / 28
Proof of one assertion F 23 = F 13 : in other words, 00 µ (1 µ ( w )) is 7 3 -power-free iff 0 µ (1 µ ( w )) is 7 3 -power-free. One direction is obvious. For the other, note that if 0 µ (1 µ ( w )) is 7 3 -power-free, but 00 µ (1 µ ( w )) is not, then the 7 3 -power in it must be a prefix. However, if 0 µ (1 µ ( w )) is 7 3 -power-free, then w must start with 0 (else 0 µ (1 µ ( w )) would start with 01010). So 00 µ (1 µ ( w )) starts with 001001. But this word cannot appear twice, because any letter that precedes it gives a 7 3 -power. 23 / 28
Consequences of the main theorem Theorem. The lexicographically least infinite 7 3 -power-free word is 001001 t . 24 / 28
Consequences of the main theorem Theorem. The lexicographically least infinite 7 3 -power-free word is 001001 t . Proof. Examine the possible paths in the automaton. 24 / 28
More consequences of the main theorem An infinite word ( a n ) n ≥ 0 is k -automatic if there is an automaton with output that, on input n in base k , reaches a state whose associated output is a n . 25 / 28
More consequences of the main theorem An infinite word ( a n ) n ≥ 0 is k -automatic if there is an automaton with output that, on input n in base k , reaches a state whose associated output is a n . Theorem. An infinite 7 3 -power-free word is 2-automatic if and only if (a) it is encoded by the automaton previously shown and (b) the sequence of symbols coding it is ultimately periodic. 25 / 28
Proof of 2-automatic result It suffices to look at the 2-decimation of x 1 µ ( x 2 ) µ 2 ( x 3 ) · · · . 26 / 28
Proof of 2-automatic result It suffices to look at the 2-decimation of x 1 µ ( x 2 ) µ 2 ( x 3 ) · · · . If x 1 empty this is x 2 µ ( x 3 ) µ 2 ( x 4 ) · · · and 26 / 28
Proof of 2-automatic result It suffices to look at the 2-decimation of x 1 µ ( x 2 ) µ 2 ( x 3 ) · · · . If x 1 empty this is x 2 µ ( x 3 ) µ 2 ( x 4 ) · · · and x 2 µ ( x 3 ) µ 2 ( x 4 ) · · · . 26 / 28
Proof of 2-automatic result It suffices to look at the 2-decimation of x 1 µ ( x 2 ) µ 2 ( x 3 ) · · · . If x 1 empty this is x 2 µ ( x 3 ) µ 2 ( x 4 ) · · · and x 2 µ ( x 3 ) µ 2 ( x 4 ) · · · . If | x 1 | = 1 this is x 1 x 2 µ ( x 3 ) µ 2 ( x 4 ) · · · 26 / 28
Proof of 2-automatic result It suffices to look at the 2-decimation of x 1 µ ( x 2 ) µ 2 ( x 3 ) · · · . If x 1 empty this is x 2 µ ( x 3 ) µ 2 ( x 4 ) · · · and x 2 µ ( x 3 ) µ 2 ( x 4 ) · · · . If | x 1 | = 1 this is x 1 x 2 µ ( x 3 ) µ 2 ( x 4 ) · · · and x 2 µ ( x 3 ) µ 2 ( x 4 ) · · · . 26 / 28
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