Avoiding Three Consecutive Blocks of the Same Length and Sum Julien Cassaigne 1 , James D. Currie 2 , Luke Schaeffer 3 , Jeffrey Shallit 3 ematiques de Luminy 1 Institut de Math´ University of Winnipeg 2 University of Waterloo 3 April 25th, 2013 Luke Schaeffer (Waterloo) Additive Cubes April 25th, 2013 1 / 28
Introduction Pattern avoidance Problem Find an infinite word w over a finite alphabet Σ such that no factor matches a given pattern. A kth power is a word of the form x k for some x ∈ Σ ∗ . murmur is a square . An abelian kth power is a word of the form x 1 · · · x k , where each x i is a permutation of x 1 . reappear is an abelian square. Let Σ ⊆ Z . An additive kth power is a word of the form x 1 · · · x k such that | x 1 | = · · · = | x k | � � x 1 = · · · = x k . facade is an additive square if we let a = 1, b = 2, . . . Luke Schaeffer (Waterloo) Additive Cubes April 25th, 2013 2 / 28
Introduction Additive Powers Pirillo and Varricchio (1994) discuss avoiding additive k th powers. Independently, Halbeisen and Hungerb¨ uhler (2000) considered additive squares. Theorem (Dekking, 1979) Abelian 4 th powers are avoidable over a binary alphabet. Corollary Additive 4 th powers are avoidable over a binary alphabet. Questions Are additive squares/cubes avoidable? How many symbols are required? Luke Schaeffer (Waterloo) Additive Cubes April 25th, 2013 3 / 28
Introduction Our Result Theorem Suppose Σ = { 0 , 1 , 3 , 4 } and ϕ : Σ ∗ → Σ ∗ is the morphism ϕ (0) = 03 ϕ (1) = 43 ϕ (3) = 1 ϕ (4) = 01 . Then the fixed point w := ϕ ω (0) = 031430110343430 · · · avoids additive cubes. The morphism was found by brute force search (Shallit). Luke Schaeffer (Waterloo) Additive Cubes April 25th, 2013 4 / 28
Introduction Proof Outline Start with an infinite tree T representing all prefixes of w . Based on recursive structure of w . Construct a tree T 4 representing all triples of consecutive blocks. Store information (state) at each node such that we can compute the state of a child from its parent and the edge label, and determine whether the node represents an additive cube given the state. We use two vectors in N 4 . Use linear algebra to show that, along a (hypothetical) path to an additive cube, the vectors are bounded. Exhaustively check the remaining (finite) search space for additive cubes. Luke Schaeffer (Waterloo) Additive Cubes April 25th, 2013 5 / 28
Search Tree Recursive Structure . . . w = 0 3 1 4 3 0 1 1 0 3 4 3 4 3 0 3 . . . w = 0 3 1 4 3 0 1 1 0 3 4 3 4 3 0 3 Luke Schaeffer (Waterloo) Additive Cubes April 25th, 2013 6 / 28
Search Tree Quotients and Remainders x div ϕ . . . w = 0 3 1 4 3 0 1 1 0 3 4 3 4 3 0 3 x mod ϕ . . . w = 0 3 1 4 3 0 1 1 0 3 4 3 4 3 0 3 x Luke Schaeffer (Waterloo) Additive Cubes April 25th, 2013 7 / 28
Search Tree Quotients and Remainders x div ϕ . . . w = 0 3 1 4 3 0 1 1 0 3 4 3 4 3 0 3 x mod ϕ . . . w = 0 3 1 4 3 0 1 1 0 3 4 3 4 3 0 3 x Idea Build a tree with a node for each prefix. For each x , draw an edge from x div ϕ to x labelled x mod ϕ . Luke Schaeffer (Waterloo) Additive Cubes April 25th, 2013 7 / 28
Search Tree T 0 ε 0 1 ε 2 ε 4 3 4 ε ε 0 5 6 7 ε 4 ε 4 ε 0 Luke Schaeffer (Waterloo) Additive Cubes April 25th, 2013 8 / 28
Search Tree T 4 Three consecutive blocks are delimited by four positions: the start of each block, and the end of the last block. Definition Suppose V is the set of nodes in T . We define a tree T 4 on nodes V 4 such that there is an edge from ( x 1 , x 2 , x 3 , x 4 ) ∈ V 4 to ( y 1 , y 2 , y 3 , y 4 ) ∈ V 4 labelled ( a 1 , a 2 , a 3 , a 4 ) ∈ { ε, 0 , 4 } ∗ if and only if there is an edge from x i to y i labelled a i for i = 1 , 2 , 3 , 4. Any triple of blocks b 1 b 2 b 3 corresponds to a node in T 4 . Next Step Annotate each node with information to identify additive cubes. Luke Schaeffer (Waterloo) Additive Cubes April 25th, 2013 9 / 28
Search Tree Add some “state” to each node such that we can compute the state of the child given the state of the parent and edge label, and additive cubes can be identified. Example Associate a word with each node in T . Let ε be the word for the root node, and compute the word for a child as follows: y x → ϕ ( x ) y . − Then (by induction) node i is associated with w [0 .. i − 1]. We can recursively compute w [0 .. i 1 − 1] , w [0 .. i 2 − 1] , w [0 .. i 3 − 1] , w [0 .. i 4 − 1] for a node ( i 1 , i 2 , i 3 , i 4 ) in T 4 . Given w [0 .. i 1 − 1] , w [0 .. i 2 − 1] , w [0 .. i 3 − 1] , w [0 .. i 4 − 1], we can check if w [ i 1 .. i 4 − 1] is an additive cube. Luke Schaeffer (Waterloo) Additive Cubes April 25th, 2013 10 / 28
Linear Algebra Parikh vectors Definition The Parikh map, ψ : Σ ∗ → N Σ , maps a word x to a vector ψ ( x ) that counts the number of occurrences of each symbol a ∈ Σ in x . For example, ψ (034343) = (1 , 0 , 3 , 2). Idea Store ψ ( x ) instead of x . Luke Schaeffer (Waterloo) Additive Cubes April 25th, 2013 11 / 28
Linear Algebra Parikh Vector Operations We can compute ψ ( ϕ ( x ) y ) given ψ ( x ) and y . ψ ( ϕ ( x )) = M ψ ( x ) where M is the incidence matrix of ϕ : 1 0 0 1 0 0 1 1 M := . 1 1 0 0 0 1 0 0 ψ ( xy ) = ψ ( x ) + ψ ( y ) y ψ ( x ) → ψ ( ϕ ( x ) y ) = M ψ ( x ) + ψ ( y ) − We can detect additive cubes. | x | = ψ ( x ) · (1 , 1 , 1 , 1) � x = ψ ( x ) · (0 , 1 , 3 , 4) Luke Schaeffer (Waterloo) Additive Cubes April 25th, 2013 12 / 28
Linear Algebra Block Differences Let b 1 = w [ i 1 .. i 2 − 1], b 2 = w [ i 2 .. i 3 − 1], b 3 = w [ i 3 .. i 4 − 1] be three consecutive blocks. Given t 1 = ψ ( w [0 .. i 1 − 1]), t 2 = ψ ( w [0 .. i 2 − 1]), t 3 = ψ ( w [0 .. i 3 − 1]) and t 4 = ψ ( w [0 .. i 4 − 1]), we can tell if b 1 b 2 b 3 is an additive cube. It suffices to have the Parikh vector for each block: ψ ( b 1 ) = t 2 − t 1 ψ ( b 2 ) = t 3 − t 2 ψ ( b 3 ) = t 4 − t 3 or even just the block differences : u := ψ ( b 2 ) − ψ ( b 1 ) = t 3 − 2 t 2 + t 1 v := ψ ( b 3 ) − ψ ( b 2 ) = t 4 − 2 t 3 + t 2 to detect additive cubes. Luke Schaeffer (Waterloo) Additive Cubes April 25th, 2013 13 / 28
Linear Algebra Proposal Keep two block difference vectors, u = ψ ( x 2 ) − ψ ( x 1 ) v = ψ ( x 3 ) − ψ ( x 2 ) . On transition ( a 1 , a 2 , a 3 , a 4 ), we compute u ′ , v ′ where u ′ = Mu − f ( a 1 , a 2 , a 3 ) v ′ = Mv − f ( a 2 , a 3 , a 4 ) with f ( a , b , c ) = ψ ( a ) − 2 ψ ( b ) + ψ ( c ). Luke Schaeffer (Waterloo) Additive Cubes April 25th, 2013 14 / 28
Linear Algebra Eigenbasis Idea Change basis so the matrix is in Jordan canonical form. Entries are complex numbers, not integers. Eigencoordinates are decoupled for individual analysis. Suppose M = P − 1 DP , where D is a diagonal matrix with diagonal elements λ 1 , λ 2 , λ 3 , λ 4 , the eigenvalues of M . λ 1 . = 1 . 69028 λ 2 = − 1 . 50507 λ 3 . = 0 . 40739 + 0 . 47657 i λ 4 = 0 . 40739 − 0 . 47657 i Note that | λ 3 | = | λ 4 | . = 0 . 62696. Luke Schaeffer (Waterloo) Additive Cubes April 25th, 2013 15 / 28
Linear Algebra Coordinates along a path Recall the equation u ′ = Mu − f ( a 1 , a 2 , a 3 ) For each coordinate i = 1 , 2 , 3 , 4 in the eigenbasis, we have u ′ i = λ i u i − f i ( a 1 , a 2 , a 3 ) . Note that u ′ i − λ i u i = f i ( a 1 , a 2 , a 3 ) is bounded. Question Suppose λ ∈ C and ( z j ) ∞ j =0 is a sequence of complex numbers with z 0 = 0 and | z j +1 − λ z j | ≤ B for all j . What can we say about the asymptotic behaviour of such sequences? Luke Schaeffer (Waterloo) Additive Cubes April 25th, 2013 16 / 28
Linear Algebra Inside the Unit Circle ( | λ i | < 1) Theorem Let λ ∈ C be a complex number such that | λ | < 1 . Suppose ( z j ) ∞ j =0 is a complex sequence such that z 0 = 0 and | z j +1 − λ z j | ≤ B B for all j. Then | z j | ≤ 1 − λ for all j. Luke Schaeffer (Waterloo) Additive Cubes April 25th, 2013 17 / 28
Linear Algebra Inside the Unit Circle ( | λ i | < 1) Theorem Let λ ∈ C be a complex number such that | λ | < 1 . Suppose ( z j ) ∞ j =0 is a complex sequence such that z 0 = 0 and | z j +1 − λ z j | ≤ B B for all j. Then | z j | ≤ 1 − λ for all j. Since λ 3 . = 0 . 40739 + 0 . 47657 i and λ 4 = 0 . 40739 − 0 . 47657 i are inside the unit circle, Corollary For any node in the tree, the third and fourth eigencoordinates of u and v are bounded. Luke Schaeffer (Waterloo) Additive Cubes April 25th, 2013 17 / 28
Linear Algebra Inside the Unit Circle - Corollaries For three consecutive blocks x 1 x 2 x 3 , the block difference vectors u = ψ ( b 2 ) − ψ ( b 1 ) v = ψ ( b 3 ) − ψ ( b 2 ) are close to a plane (2-dimensional subspace). If b 1 b 2 b 3 is an additive cube then we have two linear equations per vector: � � � � ψ ( b 2 ) − ψ ( b 1 ) · (1 , 1 , 1 , 1) = 0 ψ ( b 3 ) − ψ ( b 2 ) · (1 , 1 , 1 , 1) = 0 � � � � ψ ( b 2 ) − ψ ( b 1 ) · (0 , 1 , 3 , 4) = 0 ψ ( b 3 ) − ψ ( b 2 ) · (0 , 1 , 3 , 4) = 0 So u = ψ ( b 2 ) − ψ ( b 1 ) and v = ψ ( b 3 ) − ψ ( b 2 ) are bounded. Luke Schaeffer (Waterloo) Additive Cubes April 25th, 2013 18 / 28
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