R ICH WORDS ◮ A palindrome is a finite word that reads the same forwards and backwards. ◮ Examples: 1001 , 01010 , kayak , racecar Theorem (Droubay, Justin, Pirillo 2001): Every word of length n contains at most n distinct nonempty palindromes as factors. ◮ A finite word of length n is called rich if it contains n distinct nonempty palindromes. ◮ The word 01101 contains the palindromes
R ICH WORDS ◮ A palindrome is a finite word that reads the same forwards and backwards. ◮ Examples: 1001 , 01010 , kayak , racecar Theorem (Droubay, Justin, Pirillo 2001): Every word of length n contains at most n distinct nonempty palindromes as factors. ◮ A finite word of length n is called rich if it contains n distinct nonempty palindromes. ◮ The word 01101 contains the palindromes 0 ,
R ICH WORDS ◮ A palindrome is a finite word that reads the same forwards and backwards. ◮ Examples: 1001 , 01010 , kayak , racecar Theorem (Droubay, Justin, Pirillo 2001): Every word of length n contains at most n distinct nonempty palindromes as factors. ◮ A finite word of length n is called rich if it contains n distinct nonempty palindromes. ◮ The word 01101 contains the palindromes 0 , 1 ,
R ICH WORDS ◮ A palindrome is a finite word that reads the same forwards and backwards. ◮ Examples: 1001 , 01010 , kayak , racecar Theorem (Droubay, Justin, Pirillo 2001): Every word of length n contains at most n distinct nonempty palindromes as factors. ◮ A finite word of length n is called rich if it contains n distinct nonempty palindromes. ◮ The word 01101 contains the palindromes 0 , 1 , 11 ,
R ICH WORDS ◮ A palindrome is a finite word that reads the same forwards and backwards. ◮ Examples: 1001 , 01010 , kayak , racecar Theorem (Droubay, Justin, Pirillo 2001): Every word of length n contains at most n distinct nonempty palindromes as factors. ◮ A finite word of length n is called rich if it contains n distinct nonempty palindromes. ◮ The word 01101 contains the palindromes 0 , 1 , 11 , 0110 ,
R ICH WORDS ◮ A palindrome is a finite word that reads the same forwards and backwards. ◮ Examples: 1001 , 01010 , kayak , racecar Theorem (Droubay, Justin, Pirillo 2001): Every word of length n contains at most n distinct nonempty palindromes as factors. ◮ A finite word of length n is called rich if it contains n distinct nonempty palindromes. ◮ The word 01101 contains the palindromes 0 , 1 , 11 , 0110 , and 101 ,
R ICH WORDS ◮ A palindrome is a finite word that reads the same forwards and backwards. ◮ Examples: 1001 , 01010 , kayak , racecar Theorem (Droubay, Justin, Pirillo 2001): Every word of length n contains at most n distinct nonempty palindromes as factors. ◮ A finite word of length n is called rich if it contains n distinct nonempty palindromes. ◮ The word 01101 contains the palindromes 0 , 1 , 11 , 0110 , and 101 , so it is rich.
R ICH WORDS ◮ A palindrome is a finite word that reads the same forwards and backwards. ◮ Examples: 1001 , 01010 , kayak , racecar Theorem (Droubay, Justin, Pirillo 2001): Every word of length n contains at most n distinct nonempty palindromes as factors. ◮ A finite word of length n is called rich if it contains n distinct nonempty palindromes. ◮ The word 01101 contains the palindromes 0 , 1 , 11 , 0110 , and 101 , so it is rich. ◮ The word 0120 contains only the palindromes 0 , 1 , and 2 , so it is not rich.
R ICH WORDS ◮ A palindrome is a finite word that reads the same forwards and backwards. ◮ Examples: 1001 , 01010 , kayak , racecar Theorem (Droubay, Justin, Pirillo 2001): Every word of length n contains at most n distinct nonempty palindromes as factors. ◮ A finite word of length n is called rich if it contains n distinct nonempty palindromes. ◮ The word 01101 contains the palindromes 0 , 1 , 11 , 0110 , and 101 , so it is rich. ◮ The word 0120 contains only the palindromes 0 , 1 , and 2 , so it is not rich. ◮ An infinite word is called rich if all of its finite factors are rich.
R EPETITIONS IN RICH WORDS Theorem (Pelantov´ a and Starosta, 2013): Every infinite rich word contains a square.
R EPETITIONS IN RICH WORDS Theorem (Pelantov´ a and Starosta, 2013): Every infinite rich word contains a square. ◮ This result holds over any finite alphabet.
R EPETITIONS IN RICH WORDS Theorem (Pelantov´ a and Starosta, 2013): Every infinite rich word contains a square. ◮ This result holds over any finite alphabet. ◮ So, what types of powers can be avoided by infinite rich words on k letters?
R EPETITIONS IN RICH WORDS Theorem (Pelantov´ a and Starosta, 2013): Every infinite rich word contains a square. ◮ This result holds over any finite alphabet. ◮ So, what types of powers can be avoided by infinite rich words on k letters? ◮ Cubes?
R EPETITIONS IN RICH WORDS Theorem (Pelantov´ a and Starosta, 2013): Every infinite rich word contains a square. ◮ This result holds over any finite alphabet. ◮ So, what types of powers can be avoided by infinite rich words on k letters? ◮ Cubes? ◮ If so, what about fractional powers between 2 and 3?
R EPETITIONS IN RICH WORDS Theorem (Pelantov´ a and Starosta, 2013): Every infinite rich word contains a square. ◮ This result holds over any finite alphabet. ◮ So, what types of powers can be avoided by infinite rich words on k letters? ◮ Cubes? ◮ If so, what about fractional powers between 2 and 3? ◮ We are asking for the repetition threshold for rich words on k letters, denoted RRT ( k ) .
R EPETITIONS IN RICH WORDS Theorem (Pelantov´ a and Starosta, 2013): Every infinite rich word contains a square. ◮ This result holds over any finite alphabet. ◮ So, what types of powers can be avoided by infinite rich words on k letters? ◮ Cubes? ◮ If so, what about fractional powers between 2 and 3? ◮ We are asking for the repetition threshold for rich words on k letters, denoted RRT ( k ) . ◮ We will determine RRT ( 2 ) .
R EPETITIONS IN RICH WORDS Theorem (Baranwal and Shallit, 2019): There is an infinite √ binary rich word with critical exponent 2 + 2 / 2.
R EPETITIONS IN RICH WORDS Theorem (Baranwal and Shallit, 2019): There is an infinite √ binary rich word with critical exponent 2 + 2 / 2. √ ◮ Note: 2 + 2 / 2 ≈ 2 . 707.
R EPETITIONS IN RICH WORDS Theorem (Baranwal and Shallit, 2019): There is an infinite √ binary rich word with critical exponent 2 + 2 / 2. √ ◮ Note: 2 + 2 / 2 ≈ 2 . 707. ◮ They conjectured that this is the smallest possible critical exponent among infinite binary rich words, i.e., that √ RRT ( 2 ) = 2 + 2 / 2.
R EPETITIONS IN RICH WORDS Theorem (Baranwal and Shallit, 2019): There is an infinite √ binary rich word with critical exponent 2 + 2 / 2. √ ◮ Note: 2 + 2 / 2 ≈ 2 . 707. ◮ They conjectured that this is the smallest possible critical exponent among infinite binary rich words, i.e., that √ RRT ( 2 ) = 2 + 2 / 2. √ ◮ The irrationality of 2 + 2 / 2 makes this hard to prove!
R EPETITIONS IN RICH WORDS Theorem (Baranwal and Shallit, 2019): There is an infinite √ binary rich word with critical exponent 2 + 2 / 2. √ ◮ Note: 2 + 2 / 2 ≈ 2 . 707. ◮ They conjectured that this is the smallest possible critical exponent among infinite binary rich words, i.e., that √ RRT ( 2 ) = 2 + 2 / 2. √ ◮ The irrationality of 2 + 2 / 2 makes this hard to prove! ◮ Baranwal and Shallit: RRT ( 2 ) ≥ 2 . 7
B ARANWAL AND S HALLIT ’ S CONSTRUCTION Define morphisms f and h by f ( 0 ) = 0 f ( 1 ) = 01 f ( 2 ) = 011 h ( 0 ) = 01 h ( 1 ) = 02 h ( 2 ) = 022 .
B ARANWAL AND S HALLIT ’ S CONSTRUCTION Define morphisms f and h by f ( 0 ) = 0 f ( 1 ) = 01 f ( 2 ) = 011 h ( 0 ) = 01 h ( 1 ) = 02 h ( 2 ) = 022 . The infinite word f ( h ω ( 0 )) is rich and has critical exponent √ 2 + 2 / 2.
B ARANWAL AND S HALLIT ’ S CONSTRUCTION Define morphisms f and h by f ( 0 ) = 0 f ( 1 ) = 01 f ( 2 ) = 011 h ( 0 ) = 01 h ( 1 ) = 02 h ( 2 ) = 022 . The infinite word f ( h ω ( 0 )) is rich and has critical exponent √ 2 + 2 / 2. ◮ The proof was completed using the automatic theorem proving software Walnut .
A N IRRATIONAL REPETITION THRESHOLD ?
A N IRRATIONAL REPETITION THRESHOLD ? √ ◮ One way to show that RRT ( 2 ) = 2 + 2 / 2 would be to give a structure theorem for infinite binary rich words with critical exponent less than some number close to (but √ larger than) 2 + 2 / 2.
A N IRRATIONAL REPETITION THRESHOLD ? √ ◮ One way to show that RRT ( 2 ) = 2 + 2 / 2 would be to give a structure theorem for infinite binary rich words with critical exponent less than some number close to (but √ larger than) 2 + 2 / 2. ◮ One would hope that every infinite binary rich word with critical exponent less than 14 / 5 looks like f ( h ω ( 0 )) .
A N IRRATIONAL REPETITION THRESHOLD ? √ ◮ One way to show that RRT ( 2 ) = 2 + 2 / 2 would be to give a structure theorem for infinite binary rich words with critical exponent less than some number close to (but √ larger than) 2 + 2 / 2. ◮ One would hope that every infinite binary rich word with critical exponent less than 14 / 5 looks like f ( h ω ( 0 )) . ◮ Unfortunately, this is not the case!
A N IRRATIONAL REPETITION THRESHOLD ? √ ◮ One way to show that RRT ( 2 ) = 2 + 2 / 2 would be to give a structure theorem for infinite binary rich words with critical exponent less than some number close to (but √ larger than) 2 + 2 / 2. ◮ One would hope that every infinite binary rich word with critical exponent less than 14 / 5 looks like f ( h ω ( 0 )) . ◮ Unfortunately, this is not the case! ◮ Fortunately, it is not much worse than this.
A NOTHER STRUCTURE THEOREM Every infinite binary rich word with critical exponent less than 14 / 5 looks like either u = f ( h ω ( 0 )) or v = f ( g ( h ω ( 0 ))) . f ( 0 ) = 0 g ( 0 ) = 011 h ( 0 ) = 01 f ( 1 ) = 01 g ( 1 ) = 0121 h ( 1 ) = 02 f ( 2 ) = 011 g ( 2 ) = 012121 h ( 2 ) = 022
A NOTHER STRUCTURE THEOREM Every infinite binary rich word with critical exponent less than 14 / 5 looks like either u = f ( h ω ( 0 )) or v = f ( g ( h ω ( 0 ))) . f ( 0 ) = 0 g ( 0 ) = 011 h ( 0 ) = 01 f ( 1 ) = 01 g ( 1 ) = 0121 h ( 1 ) = 02 f ( 2 ) = 011 g ( 2 ) = 012121 h ( 2 ) = 022 Theorem (Currie, Mol, and Rampersad, 2020+): Let w be an infinite rich word over the binary alphabet { 0 , 1 } with critical exponent less than 14 / 5. For every n ≥ 1, a suffix of w has the form f ( h n ( w n )) or f ( g ( h n ( w n ))) for some infinite word w n over { 0 , 1 , 2 } .
A N IRRATIONAL REPETITION THRESHOLD ! Theorem (Currie, Mol, and Rampersad, 2019+): The repetition √ threshold for binary rich words is 2 + 2 / 2. Proof:
A N IRRATIONAL REPETITION THRESHOLD ! Theorem (Currie, Mol, and Rampersad, 2019+): The repetition √ threshold for binary rich words is 2 + 2 / 2. Proof: ◮ If an infinite binary rich word has critical exponent less than 14 / 5, then it contains all factors of u = f ( h ω ( 0 )) or all factors of v = f ( g ( h ω ( 0 ))) .
A N IRRATIONAL REPETITION THRESHOLD ! Theorem (Currie, Mol, and Rampersad, 2019+): The repetition √ threshold for binary rich words is 2 + 2 / 2. Proof: ◮ If an infinite binary rich word has critical exponent less than 14 / 5, then it contains all factors of u = f ( h ω ( 0 )) or all factors of v = f ( g ( h ω ( 0 ))) . ◮ Baranwal and Shallit showed that the critical exponent of u √ is 2 + 2 / 2.
A N IRRATIONAL REPETITION THRESHOLD ! Theorem (Currie, Mol, and Rampersad, 2019+): The repetition √ threshold for binary rich words is 2 + 2 / 2. Proof: ◮ If an infinite binary rich word has critical exponent less than 14 / 5, then it contains all factors of u = f ( h ω ( 0 )) or all factors of v = f ( g ( h ω ( 0 ))) . ◮ Baranwal and Shallit showed that the critical exponent of u √ is 2 + 2 / 2. ◮ So it suffices to show that v has critical exponent at least √ 2 + 2 / 2.
A N IRRATIONAL REPETITION THRESHOLD ! Theorem (Currie, Mol, and Rampersad, 2019+): The repetition √ threshold for binary rich words is 2 + 2 / 2. Proof: ◮ If an infinite binary rich word has critical exponent less than 14 / 5, then it contains all factors of u = f ( h ω ( 0 )) or all factors of v = f ( g ( h ω ( 0 ))) . ◮ Baranwal and Shallit showed that the critical exponent of u √ is 2 + 2 / 2. ◮ So it suffices to show that v has critical exponent at least √ 2 + 2 / 2. ◮ In fact, we show that v is rich, and has critical exponent √ exactly 2 + 2 / 2.
A N IRRATIONAL REPETITION THRESHOLD ! Theorem (Currie, Mol, and Rampersad, 2019+): The repetition √ threshold for binary rich words is 2 + 2 / 2. Proof: ◮ If an infinite binary rich word has critical exponent less than 14 / 5, then it contains all factors of u = f ( h ω ( 0 )) or all factors of v = f ( g ( h ω ( 0 ))) . ◮ Baranwal and Shallit showed that the critical exponent of u √ is 2 + 2 / 2. ◮ So it suffices to show that v has critical exponent at least √ 2 + 2 / 2. ◮ In fact, we show that v is rich, and has critical exponent √ exactly 2 + 2 / 2. ◮ Our proof technique can also be applied to u , providing an alternate proof of Baranwal and Shallit’s result.
E STABLISHING RICHNESS
E STABLISHING RICHNESS ◮ For a binary word w , let ∆( w ) denote the sequence of first differences of w modulo 2.
E STABLISHING RICHNESS ◮ For a binary word w , let ∆( w ) denote the sequence of first differences of w modulo 2. ◮ e.g., ∆( 0111001 ) =
E STABLISHING RICHNESS ◮ For a binary word w , let ∆( w ) denote the sequence of first differences of w modulo 2. ◮ e.g., ∆( 0111001 ) =
E STABLISHING RICHNESS ◮ For a binary word w , let ∆( w ) denote the sequence of first differences of w modulo 2. ◮ e.g., ∆( 0111001 ) = 1
E STABLISHING RICHNESS ◮ For a binary word w , let ∆( w ) denote the sequence of first differences of w modulo 2. ◮ e.g., ∆( 0111001 ) = 1
E STABLISHING RICHNESS ◮ For a binary word w , let ∆( w ) denote the sequence of first differences of w modulo 2. ◮ e.g., ∆( 0111001 ) = 10
E STABLISHING RICHNESS ◮ For a binary word w , let ∆( w ) denote the sequence of first differences of w modulo 2. ◮ e.g., ∆( 0111001 ) = 10
E STABLISHING RICHNESS ◮ For a binary word w , let ∆( w ) denote the sequence of first differences of w modulo 2. ◮ e.g., ∆( 0111001 ) = 100
E STABLISHING RICHNESS ◮ For a binary word w , let ∆( w ) denote the sequence of first differences of w modulo 2. ◮ e.g., ∆( 0111001 ) = 100
E STABLISHING RICHNESS ◮ For a binary word w , let ∆( w ) denote the sequence of first differences of w modulo 2. ◮ e.g., ∆( 0111001 ) = 1001
E STABLISHING RICHNESS ◮ For a binary word w , let ∆( w ) denote the sequence of first differences of w modulo 2. ◮ e.g., ∆( 0111001 ) = 1001
E STABLISHING RICHNESS ◮ For a binary word w , let ∆( w ) denote the sequence of first differences of w modulo 2. ◮ e.g., ∆( 0111001 ) = 10010
E STABLISHING RICHNESS ◮ For a binary word w , let ∆( w ) denote the sequence of first differences of w modulo 2. ◮ e.g., ∆( 0111001 ) = 10010
E STABLISHING RICHNESS ◮ For a binary word w , let ∆( w ) denote the sequence of first differences of w modulo 2. ◮ e.g., ∆( 0111001 ) = 100101
E STABLISHING RICHNESS ◮ For a binary word w , let ∆( w ) denote the sequence of first differences of w modulo 2. ◮ e.g., ∆( 0111001 ) = 100101 ◮ Fact: ∆( u ) and ∆( v ) are Sturmian words .
E STABLISHING RICHNESS ◮ For a binary word w , let ∆( w ) denote the sequence of first differences of w modulo 2. ◮ e.g., ∆( 0111001 ) = 100101 ◮ Fact: ∆( u ) and ∆( v ) are Sturmian words . ◮ Thank you, Edita Pelantov´ a!
E STABLISHING RICHNESS ◮ For a binary word w , let ∆( w ) denote the sequence of first differences of w modulo 2. ◮ e.g., ∆( 0111001 ) = 100101 ◮ Fact: ∆( u ) and ∆( v ) are Sturmian words . ◮ Thank you, Edita Pelantov´ a! ◮ By a theorem of Rote (1994), this means that u and v are complementary symmetric Rote words .
E STABLISHING RICHNESS ◮ For a binary word w , let ∆( w ) denote the sequence of first differences of w modulo 2. ◮ e.g., ∆( 0111001 ) = 100101 ◮ Fact: ∆( u ) and ∆( v ) are Sturmian words . ◮ Thank you, Edita Pelantov´ a! ◮ By a theorem of Rote (1994), this means that u and v are complementary symmetric Rote words . ◮ By a theorem of Blondin-Mass´ e et al. (2011), every complementary symmetric Rote word is rich.
E STABLISHING RICHNESS ◮ For a binary word w , let ∆( w ) denote the sequence of first differences of w modulo 2. ◮ e.g., ∆( 0111001 ) = 100101 ◮ Fact: ∆( u ) and ∆( v ) are Sturmian words . ◮ Thank you, Edita Pelantov´ a! ◮ By a theorem of Rote (1994), this means that u and v are complementary symmetric Rote words . ◮ By a theorem of Blondin-Mass´ e et al. (2011), every complementary symmetric Rote word is rich. ◮ Therefore, both u and v are rich!
E STABLISHING THE CRITICAL EXPONENT ◮ We still want to determine the critical exponent of v .
E STABLISHING THE CRITICAL EXPONENT ◮ We still want to determine the critical exponent of v . ◮ To do this, we relate the repetitions in v to the repetitions in ∆( v ) .
E STABLISHING THE CRITICAL EXPONENT ◮ We still want to determine the critical exponent of v . ◮ To do this, we relate the repetitions in v to the repetitions in ∆( v ) . v = 001010010110100101001011 · · · ∆( v ) = 01111011101110111101110 · · ·
E STABLISHING THE CRITICAL EXPONENT ◮ We still want to determine the critical exponent of v . ◮ To do this, we relate the repetitions in v to the repetitions in ∆( v ) . v = 00101001011 01001 01001 01 1 · · · ∆( v ) = 01111011101110111101110 · · ·
E STABLISHING THE CRITICAL EXPONENT ◮ We still want to determine the critical exponent of v . ◮ To do this, we relate the repetitions in v to the repetitions in ∆( v ) . v = 00101001011 01001 01001 01 1 · · · ∆( v ) = 01111011101 11011 11011 1 0 · · ·
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