Avoiding Three Consecutive Blocks of the Same Size and Same Sum

Av oiding Three Consecutiv e Blo c ks of the Same Size and S ame Sum Julien Cassaig ne Institut de Math´ emati ques de Luminy Case 907, 163 a v enue de Lum in y 1328 8 Mar seille Cedex 9 F rance cass aigne@im l.univ-mrs.fr James D. Curr ie Department of Math em atics and Stat istics Universit y of Winnip eg 515 P orta ge Aven ue Winnip eg, Manito b a R3B 2E9 Canada j.cu rrie@uwi nnipeg.ca Luk e Schaeffer and Jeffrey Shall it Sc ho ol of Com puter Science Universit y of W at erlo o W aterl oo , ON N2L 3G1 Canada l3sc haef@stu dent.cs.uwaterloo.ca shal lit@cs.u waterloo.ca August 6, 20 1 8 Abstract W e sho w that there exists an infin it e word o ver the alphab et { 0 , 1 , 3 , 4 } contai ning no three consecutiv e blocks of the same size and th e same sum. Th is answ ers an op en problem of P ir il lo and V arricc hio from 1994. 1 1 In tro d u ction Av oidability pro ble ms in words hav e receiv ed m uc h atten tion since the seminal pap ers o f Axel Th ue [16 , 1 7, 4]. Generally speaking, the goal is to construct a n infinite w ord o v er a finite alphab et with no factor (i.e., a contiguous blo c k of sy m b ols) having some prop ert y , or to show that no su c h w ord exis ts. Th ue constructed an infinite word o v er a 2-letter alphab et con ta ining no factor that is an overlap (i.e., a finite word of the form axaxa , where a is a single letter and x is a p ossibly empt y w or d), and he also constructed an infinite word ov er a 3-letter alphab et containing no fa c tor that is a squar e (i.e., a finite no nempty w ord of the form w w ). Th ue used iterated morphisms to construct his w ords. Giv en a finite a lphabet Σ and a morphism h : Σ ∗ → Σ ∗ satisfying h ( a ) = ax for some a ∈ Σ and x ∈ Σ ∗ , we can iterate h to obtain → h ω ( a ) := a x h ( x ) h 2 ( x ) · · · , whic h is infinite iff h i ( x ) 6 = ǫ for all i . It is easy to see that → h ω ( a ) is actually a fixed p oin t of h ; that is, h ( → h ω ( a )) = → h ω ( a ). A sufficien t (but no t necessary) condition f or h i ( x ) 6 = ǫ is that h b e none r asing , that is, h ( a ) 6 = ǫ for all a ∈ Σ. Th ue used the morphism µ (0) = 01 and µ (1) = 10. The fixed p oin t → µ ω (0), know n a s the Thue-Morse word t = 0110100 1 · · · , is o v erlap-free. Such a morphism is called 2-uniform since eac h letter is mapp ed to an image of size 2 . It can also b e show n that the fixed p oin t of t he (non unifo rm) morphism giv en b y 2 → 21 0, 1 → 20 , 0 → 1, is squarefree. Erd˝ os [7] in tro duced the no tion o f ab elian a v oida bility . An ab elian k -th p ower for k ≥ 2 is a finite no nempty w ord of the form x 1 x 2 · · · x k where | x 1 | = · · · = | x k | and eac h x i is a p erm utation of x 1 . Dekking [6] constructed an infinite w ord o v er { 0 , 1 } con ta ining no ab elian 4-th p o w ers, and an infinite w ord o ve r a 3 - letter alphab et con ta ining no ab elian cub es. The former is giv en by the fixed po int o f the morphism a → abb , b → aaab , and the latter is giv en by the fixed p oin t of the morphism a → aabc , b → bbc , and c → acc . Ker¨ anen [1 2] constructed an infinite w o rd o ver a 4-letter alphab et containing no ab elian squares, using an 85-uniform morphism. In a ll three case s the alphab et size is optimal. In what follows we assume our finite a lpha b et Σ is a subset of N . An additive k -th p o w er for k ≥ 2 is a finite nonempt y word of t he form x 1 x 2 · · · x k where | x 1 | = · · · = | x k | and P x 1 = P x 2 = · · · = P x k , where b y P x i w e mean the sum of the eleme n ts app earing in the w ord x i . Since t w o words of the same length o ver { 0 , 1 } ha ve the same s um if and only if t he y are p erm utations of each other, Dekking’s result men tio ne d a b ov e shows that it is p ossible to a v oid additiv e 4th-p o w ers. In a 1994 paper, Pirillo and V arricch io [13] raised the follo wing question: do there exis t infinite w ords a voiding additiv e squares or additive cubes? They raised the ques tion in the con text of se migroup theory , as fo llo ws: Let S b e a semigroup, let k ≥ 1 b e an inte ger, a nd ϕ : Σ + → S b e a morphism. W e sa y that a nonempt y w ord w is a uniform k -p ower, mo d ϕ if it can b e written in the form w = w 1 · · · w k with ϕ ( w 1 ) = · · · = ϕ ( w k ) and | w 1 | = · · · = | w k | . If there exists an in t eger 2 R ( ϕ, k ) suc h that eac h w ord w ∈ Σ + with length ≥ R ( ϕ, k ) contains a fa cto r that is a uniform k -p ow er, mo d ϕ , then we sa y that ϕ is uniformly k -r ep etitive . If for ev ery finite alphab et Σ, ev ery morphism ϕ : Σ + → S is uniformly k -rep etitiv e, then w e sa y that that S is unif o rmly k - r ep etitive . Pirillo and V arricc hio pro v ed Prop osition 1. L et k ≥ 1 b e an inte ger. Then the fol lowing ar e e quivalent: (a) N + is not uniformly k -r ep etitive; (b) every finitely gener ate d and uniform l y k -r ep etitive semigr oup is fini te. In this con text, N + is not unifo r mly k -rep etitiv e if and only there ex ists a n infinite word o ver a finite subset o f N that av oids additive k -th p o wers . Pirillo and V arricc hio observ ed that, b y Dekking’s result men tioned ab o ve, the semigroup N + is no t uniformly 4-rep etitiv e. They remark ed, “W e do not kno w whether N + is uniformly 2- repetitiv e or uniformly 3 - rep etitiv e. This seems to b e a difficult pro blem in com binatorial n umber theory .” This theme w as tak en up again b y Halb eisen and Hungerb ¨ uhler in 200 0 [10], apparen tly not kno wing of the pap er of Pirillo and V arricc hio. They ask ed (in our t e rminology) if it is p ossible to a v oid additiv e squares. Fiv e other recen t pap ers men tioning the problem of av oiding additiv e pow ers are [9, 1 4 , 5, 8, 3]. In this pap er w e show that there exists an infinite w or d ov er the alphab et { 0 , 1 , 3 , 4 } that a voids additiv e cubes. This answ ers one of the op en questions of Pirillo and V arricchio. As a consequence w e get that N + is no t uniformly 3-rep etitiv e. 2 Notation and definitions W e consider the alphabet Σ = { 0 , 1 , 3 , 4 } . Define the morphism ϕ : Σ ∗ → Σ ∗ b y ϕ (0) = 0 3 ϕ (1) = 4 3 ϕ (3) = 1 ϕ (4) = 0 1 Define the infinite w ord w = → ϕ ω (0) = 03 14301103434303 101101103143034343034 3430 3 1 43011031011011031011011 · · · . W e will sho w that w contains no additiv e cub es. By a blo ck we will mean a finite factor of w . The sum of a blo c k is the sum of its sym b ols (in terpreting the sym b ols 0 , 1 , 3 , 4 as in tegers). W e define a double blo ck to b e a pair of consecutiv e blo c ks, and a triple blo ck to b e a triple of consecutiv e blo c ks. 3 2.1 Matrices and eigen v alues Let ψ : Σ ∗ → Z 4 b e the P arikh vec tor map, whic h sends a w ord x ∈ Σ ∗ to a v ector ( | x | 0 , | x | 1 , | x | 3 , | x | 4 ) T ∈ Z 4 , where | x | a is the num b er o f o ccurrences of a in x . W e let M denote t he incidence matrix of ϕ , giv en by M =     1 0 0 1 0 0 1 1 1 1 0 0 0 1 0 0     . Note that ψ ( ϕ ( x )) = M ψ ( x ). The eige n v alues of M are the ro ots of its c har a c teristic p olynomial X 4 − X 3 − 2 X 2 + 2 X − 1 and are, to limited precision, as follo ws: 1 λ 1 . = 1 . 69 0284494616614 λ 2 . = − 1 . 505 0 68413621472 λ 3 . = 0 . 40 7391959502429 + 0 . 47 6 565325929643 i λ 4 . = 0 . 40 7391959502429 − 0 . 4765 6 5325929643 i. Let Λ denote the diagonal matrix of eigen v alues. The eigen v ectors of M are giv en b y columns of the follo wing matrix, Q =     0 . 4723959 4 0 . 1780718 9 0 . 6269630 9 0 . 6269630 9 0 . 5511808 0 0 . 6713843 4 − 0 . 29375 620 − 0 . 0553 4050 i − 0 . 293756 20 + 0 . 055340 50 i 0 . 6055647 7 − 0 . 56439708 0 . 2782428 2 − 0 . 46132816 i 0 . 278242 82 + 0 . 46132816 i 0 . 3260875 9 − 0 . 44608227 − 0 . 37154337 + 0 . 2987 8 887 i − 0 . 3715433 7 − 0 . 29878887 i     . The eigen v ectors a re normalized to ha v e Euclidean norm 1. T ogether these matrices are an eigen v alue decompo s ition of M since M = Q Λ Q − 1 . W e mak e extensiv e use of this decomp osition. In particular, let τ : C 4 → C 4 b e the linear map corresp onding to left- m ultiplication by Q − 1 . Also define linear maps τ j : C 4 → C for j = 1 , 2 , 3 , 4 suc h that τ ( x ) = ( τ 1 ( x ) , τ 2 ( x ) , τ 3 ( x ) , τ 4 ( x )). Then since Q − 1 M = Λ Q − 1 , w e ha ve τ j ( M x ) = λ j τ j ( x ) for all v ectors x . The matrix for τ is just Q − 1 , and w e ha ve Q − 1 . =     0 . 5124 0 . 5979 0 . 3537 0 . 656 9 0 . 1806 0 . 6809 − 0 . 4524 − 0 . 5724 0 . 5788 − 0 . 5749 i − 0 . 3219 + 0 . 218 3 i − 0 . 069 0 + 0 . 616 5 i − 0 . 1662 − 0 . 6810 i 0 . 5788 + 0 . 5749 i − 0 . 3219 − 0 . 218 3 i − 0 . 0690 − 0 . 6165 i − 0 . 166 2 + 0 . 681 0 i     . The ro ws of this matrix giv e us the maps τ 1 , . . . , τ 4 . Th us, f or example, τ 1 ( a, b, c, d ) . = . 5124 + . 5979 b + . 3537 c + . 6569 d . 1 In this pap er, without fur ther co mm ent, w e will fre q uen tly make use of floating p oint a ppro x ima tions to certain algebr aic num b ers. W e leav e it to the r eader to verify that the a ppro x imations we use a re accur ate enough to verify our claims. 4 2.2 Indexing and paren ts Let w [ i ] denote the i th sym b ol of w , with w [0] = 0 being the first sym b ol of w . W e let w [ p , q ) denote the sym b ols fro m p to q excluding the sym b ol at q , as lo ng a s p ≤ q . W e in terpret w [ p, p ) to b e the empt y w ord. Define the function η that maps a p osition p to | ϕ ( w [0 , p )) | . Since w = ϕ ( w ), t he morphism ϕ maps an y pr efix w [0 , p ) of w to some other prefix o f w . Therefore ϕ ( w [0 , p )) is the unique prefix of length η ( p ), that is, w [0 , η ( p )). F or example, w [0 , 3) = 031, and w [0 , 3) = 031 maps to ϕ (031) = 0 3143 = w [0 , 5), and henc e η (3) = 5. Note tha t ϕ is no nerasing, s o it follo ws that | ϕ ( x ) | ≥ | x | . Since ϕ (0) = 03 , it follo ws that | ϕ ( x ) | > | x | for an y nonempt y prefix x of w . Hence η ( p ) ≥ p for all p , and the inequalit y is strict for p > 0. The function η is also clearly a non-decreasing function, so a ≤ b implies η ( a ) ≤ η ( b ). W e also hav e w [0 , η ( p )) w [ η ( p ) , η ( p + 1)) = w [0 , η ( p + 1)) = ϕ ( w [0 , p + 1)) = ϕ ( w [0 , p )) ϕ ( w [ p ]) = w [0 , η ( p )) ϕ ( w [ p ]) , so w [ η ( p ) , η ( p + 1)) = ϕ ( w [ p ]). Th us, the image of w [ p ] is w [ η ( p ) , η ( p + 1)). By definition, some position p maps to ϕ ( w [ p ]) starting at η ( p ), so w e can think of eac h sym b ol in w [ η ( p ) , η ( p + 1)) as arising f rom w [ p ]. W e define a f unc tion to a sso ciate the p ositions in [ η ( p ) , η ( p + 1)) with p , giv en b elo w. Definition 2. F o r a p osition p in w , we let par( p ) denote the p ar ent of p , whic h w e define to b e the unique position t suc h that η ( t ) ≤ p < η ( t + 1). Also, a child of a p osition p is any p osition q suc h that par( q ) = p . P arents hav e t w o elemen tary prop erties, whic h w e presen t without proo f. 1. The inequalit y par( p ) ≤ p holds for all p with strict ine qualit y unless p = 0. 2. If a ≤ b then par( a ) ≤ par( b ). In other w ords, par( x ) is a non- de creasing function. The follo wing table illustrates these c oncepts for t he first few p ositions: p 0 1 2 3 4 5 6 7 8 9 10 11 12 1 3 14 15 w [ p ] 0 3 1 4 3 0 1 1 0 3 4 3 4 3 0 3 η ( p ) 0 2 3 5 7 8 10 12 14 16 17 19 20 2 2 23 25 par( p ) 0 0 1 2 2 3 3 4 5 5 6 6 7 7 8 8 W e no w fo rm an infinite graph T with p ositions a s v ertices and edges from eac h vertex to its c hildren (in the p osition sense). It follow s from these pro perties tha t there is a path in T from 0 to a n y v ertex. Also T is acyclic with the exception of t he lo op at 0. In other 5 w ords, with the exception of a single lo op, T is a n infinite tree with 0 at the ro ot. P art o f this tree is sho wn in Figure 1, where w e see tha t w is obtained b y a lev el-order tra v ersal of T . Indeed, the lev els are equal to a = 0, x = 3, ϕ ( x ) = 1, ϕ 2 ( x ) = 4 3, etc. so that w = a x ϕ ( x ) ϕ ( x ) 2 ϕ ( x ) 3 . . . . Since T is a tree, we are oft e n interes ted in the path from 0 (the r oot) to an arbitrary v ertex. W e define the ancestral sequence of a p osition p to b e the sequence { p i } ∞ i =0 suc h that p 0 = p and p i +1 = par( p i ) for all i ≥ 0. w [0] = 0 w [1] = 3 w [2] = 1 w [3] = 4 w [4] = 3 w [5] = 0 w [6] = 1 w [7] = 1 w [8] = 0 w [9] = 3 w [10] = 4 w [11] = 3 w [12] = 4 w [13] = 3 ε 0 ε ε 4 ε 0 ε ε 0 ε 4 ε 4 Figure 1 : The first 6 lev els of T Supp ose w e are giv en a v ector of p ositions p = ( p 1 , . . . , p k ) such tha t p 1 ≤ p 2 ≤ · · · ≤ p k . Then p i and p i +1 delimit the blo c k b = w [ p i , p i +1 ) f o r eac h 1 ≤ i < k , so we hav e k − 1 consecutiv e blo c ks. W e extend the definition of pa ren ts to v ectors b y the equation par( p 1 , . . . , p k ) := (par( p 1 ) , . . . , par( p k )) . Giv en consecutiv e blo c ks b 1 · · · b k − 1 delimited b y p , define par( b 1 · · · b k − 1 ) to b e the blo c ks delimited by par ( p ). W e a lso extend the definition of ancestral sequence to b e the seq uence of iterated pa ren ts for anything that has parents , e.g., p ositions, ve ctors of p ositions, and consecutiv e blo c ks. 6 2.3 P arikh v ectors of prefixes and blo c ks Define the function σ ( p ) := ψ ( w [0 , p )). That is, σ ( p ) is the P arikh v ector o f the prefix of w up to, but not including, the p osition p . Note that σ ( p ) · (1 , 1 , 1 , 1) = | w [0 , p ) | 0 + | w [0 , p ) | 1 + | w [0 , p ) | 3 + | w [0 , p ) | 4 = | w [0 , p ) | = p. It follo ws that σ is inj e ctiv e. If ( p, q ) is an edge in T then w [0 , q ) con tains ϕ ( w [0 , p )) and perhaps another symbol, so w e exp ec t that σ ( q ) ≈ M σ ( p ). The follo wing lemma mak es this precise. Lemma 3. Given a p osition p , ther e is a bije ction b etwe en childr en of p and pr op er pr efixes of ϕ ( w [ p ]) . (B y a pr op er pr efix of a wor d x , we me an a p ossibly empty pr efix d i ff er ent fr om x .) F urthermor e, if q is a child of p and a is the c orr esp onding pr efix of ϕ ( w [ p ]) then we have σ ( q ) = M σ ( p ) + ψ ( a ) . Pr o of. Recall that ϕ maps t he sym b ol w [ p ] to w [ η ( p ) , η ( p + 1)). By definition, q is a child of p if and only if η ( p ) ≤ q < η ( p + 1). Then σ ( q ) = ψ ( w [0 , q )) = ψ ( w [0 , η ( p )) w [ η ( p ) , q )) = ψ ( w [0 , η ( p )) ) + ψ ( w [ η ( p ) , q )) . Let a = w [ η ( p ) , q ), a prop er prefix of w [ η ( p ) , η ( p + 1)) = ϕ ( w [ p ]). Then σ ( q ) = ψ ( ϕ ( w [0 , p ))) + ψ ( a ) = M ψ ( w [0 , p )) + ψ ( a ) = M σ ( p ) + ψ ( a ) . Th us, every edge ( p, q ) in T has a corresp onding w ord a , as pro per prefix o f ϕ ( w [ p ]). W e can think of the a corresp onding to an edge as an edge lab el. This allo ws us to extend the previous lemm a from a single edge to an y w alk in T . Corollary 4. If p 0 · · · p ℓ is a walk in T with e dges a 1 , . . . , a ℓ then σ ( p ℓ ) = ℓ X i =1 M ℓ − i ψ ( a i ) + M ℓ σ ( p 0 ) . 7 Pr o of. W e use induction on ℓ , the length of the walk . When ℓ = 0 the result is trivial. Otherwise, b y Lemma 3 w e ha v e σ ( p ℓ ) = M σ ( p ℓ − 1 ) + ψ ( a ℓ ) . Then w e a pply the induction hypothesis and simplify to complete the pro of: σ ( p ℓ ) = M ℓ − 1 X i =1 M ℓ − 1 − i ψ ( a i ) + M ℓ − 1 σ ( p 0 ) ! + M 0 ψ ( a ℓ ) = ℓ X i =1 M ℓ − i ψ ( a i ) + M ℓ σ ( p 0 ) . No w supp ose w e apply this corollary to an ancestral seq uence, { p i } ∞ i =0 . Let a i b e the lab el for the edge from p i to p i +1 , for eac h i . Then the c orollary sa ys that σ ( p 0 ) = k − 1 X i =0 M i ψ ( a i ) + M k σ ( p k ) . F or k large enough we get p k = 0 so σ ( p k ) = 0 and th us σ ( p 0 ) = ∞ X i =0 M i ψ ( a i ) . (1) 2.4 A graph homomorphism Define a directed graph Q = (Σ , T ) where v ertices are sym b ols, and with a set of lab elled edges T a s sho wn in Figure 2. Let there b e an edge from c ∈ Σ t o d ∈ Σ lab elled by ℓ ∈ Σ ∗ whenev er ℓ is a prefix of ϕ ( c ) up to, but not including, some sym b ol d in ϕ ( c ). Notice that the map ζ that sends x to w [ x ] maps v ertices in T to v ertices in Q . Then Lemma 3 sa ys that if an edge ( p, q ) is lab elled by a then the w ord aζ ( q ) is a prefix of ϕ ( ζ ( p )), so there is an edge ( ζ ( p ) , ζ ( q )) in Q also lab elled by a . Therefore ζ is a graph homomorphism from T to Q that prese rv es edge lab els . Definition 5. W e sa y a lab elled digraph homomorphism f : A → B is child-bije ctive if f maps children of a t o c hildren of f ( a ) bijectiv ely for all v ertices a ∈ A . W e claim that ζ is child-bijectiv e. Ev ery c hild of ζ ( p ) in Q corresp onds to a prefix of ϕ ( w [ p ]), a nd these prefixes corresp ond to c hildren of p according to Lemma 3, therefore c hildren of ζ ( p ) corresp ond to c hildren of p . F urthermore, if q is a child of p then ζ ( q ) is a c hild o f ζ ( p ), so ζ is indeed c hild-bijectiv e. Child-bijectivit y implies a bijection b et w een w alks starting at p and ζ ( p ) resp ectiv ely . 8 0 4 1 3 ε 0 ε 0 ε 4 ε Figure 2 : The directe d graph Q Prop osition 6. L et f : A → B b e a c h ild-bije ctive lab el le d digr aph homomorp h ism. We define a function ˆ f that sends walk s in A to walks in B so that v 1 · · · v k 7→ f ( v 1 ) · · · f ( v k ) . If we fix some v ∈ A then ˆ f is bije ction b etwe en walks starting at v in A and walks starting at f ( v ) in B . Pr o of. Fix the v ertex v in A . Let X ℓ b e t he set of w alks in A starting at v o f length ℓ and lik ewise let Y ℓ b e the se t of w alks in B starting at f ( v ) of length ℓ . Clearly ˆ f maps w alks of length ℓ to w alks of length ℓ , so it suffices to sho w that ˆ f restricts to a bijection b et w een X ℓ and Y ℓ for e ac h ℓ . Our pro of pr o ceeds b y induction on ℓ . There is only one elemen t in b oth X 0 and Y 0 , so ˆ f restricts to a bijection b et w een X 0 and Y 0 . Let v 0 · · · v ℓ b e a w alk in X ℓ . W e decompo se this walk into a shorter walk v 0 · · · v ℓ − 1 in X ℓ − 1 and the final edge ( v ℓ − 1 , v ℓ ). By induction, ˆ f is a bijection b et w een X ℓ − 1 and Y ℓ − 1 , so the w alk maps to f ( v 0 ) · · · f ( v ℓ − 1 ). Since f is c hild- bijectiv e, f maps neigh b ours of v ℓ − 1 to neighbours of f ( v ℓ − 1 ), so v ℓ go es to f ( v ℓ ). No w w e recomp ose f ( v 0 ) · · · f ( v ℓ − 1 ) and the edge ( f ( v ℓ − 1 ) , f ( v ℓ )), to giv e the w alk f ( v 0 ) · · · f ( v ℓ ) = ˆ f ( v 0 · · · v ℓ ). Since all the maps w ere bijec tiv e, ˆ f is inde ed a bijec tion. By this proposition, ζ giv es us a bijection b et w een walks in T starting at some position p and walks in Q starting at w [ p ]. If w e are only in terested in the edges of a w alk in T , t hen w e can use an e quiv alen t w alk in Q . The following definition illustrates this ide a. Definition 7. Define the set o f v ectors D ℓ ⊆ Z 4 b y D ℓ = ( ℓ − 1 X i =0 M i ψ ( a i ) : a ℓ − 1 · · · a 0 the e dges of a w alk in T ) . 9 By Pro p osition 6, this is eq uiv alen t to the fo llo wing alternate definition of D ℓ : D ℓ = ( ℓ − 1 X i =0 M i ψ ( a i ) : a ℓ − 1 · · · a 0 the e dges of a w alk in Q ) . Later w e will need to find all elemen ts of D 9 , and this second definition is imp ortan t b ecause there are only finitely man y walks of length 9 in Q compared to infinitely man y in T . Th us, it is straightforw ard to en umerate the w alks in Q , and th us ele men ts of D 9 . 3 Comparing bl o c k s equences In this section we are concerned with blo c ks b 0 and c 0 . Beginning in Section 3 .2 , w e take b 0 and c 0 to b e blo c ks with the same length and sum, and let { b i } ∞ i =0 and { c i } ∞ i =0 b e the corresp onding a nc estral se quences. The goal of this section is to sho w that ψ ( b i ) and ψ ( c i ), the P arikh v ectors of corresp onding a ncestors for b 0 and c 0 , are approx imately the same. That is, ψ ( b i ) − ψ ( c i ) is b ounded by a constan t that does not dep end on b 0 and c 0 . There are four main steps to this pro of: 1. W e b ound | τ 3 ( ψ ( b i ) − ψ ( c i )) | and | τ 4 ( ψ ( b i ) − ψ ( c i )) | , proving that ψ ( b i ) − ψ ( c i ) is close to a 2-dimensional subspace . 2. W e sho w that the sum and length conditions force ψ ( b 0 ) − ψ ( c 0 ) to b e in a la t t ic e, L . W e also sho w that t he in t ersection o f t his lattice with the 2- dimensional subspace is trivial, and hence ψ ( b 0 ) − ψ ( c 0 ) belongs to a finite se t of p oin ts. 3. W e bo und | τ 1 ( ψ ( b i ) − ψ ( c i )) | and | τ 2 ( ψ ( b i ) − ψ ( c i )) | . 4. W e sho w that since all the eigenco ordinates are small, ψ ( b i ) − ψ ( c i ) is short , a nd we discuss ho w to en umerate these short vec tors for t he next sec tion. 3.1 Bounding t w o co ordinates W e start by b ounding the third a nd fourth eigenco ordinates. F or this step w e do not require that t he blo c ks ha v e the same length and sum, so we state the theorem for a n y tw o blo c ks. Theorem 8. If b and c ar e blo cks (not n e c ess arily c onse cutive) then | τ 3 ( ψ ( b ) − ψ ( c )) | ≤ C 3 | τ 4 ( ψ ( b ) − ψ ( c )) | ≤ C 3 wher e C 3 . = 2 . 1758 is a c onstant. 10 Pr o of. First, notice that τ 3 is t he complex conjugate of τ 4 . That is, fo r an y x ∈ R 4 w e ha v e τ 3 ( x ) = τ 4 ( x ). Therefore w e only need to pro v e one of the inequalities because | τ 3 ( ψ ( b ) − ψ ( c )) | = | τ 4 ( ψ ( b ) − ψ ( c )) | . W e will pro ve the first inequality . F rom Eq. (1) w e see that σ ( p 0 ) = ∞ X i =0 M i ψ ( a i ) . If our blo c k b is delimited b y p 0 and q 0 , whe re σ ( q 0 ) = ∞ X i =0 M i ψ ( a ′ i ) , then w e let δ i = ψ ( a ′ i ) − ψ ( a i ) and get ψ ( b ) = σ ( q 0 ) − σ ( p 0 ) = ∞ X i =0 M i ψ ( a ′ i ) − ∞ X i =0 M i ψ ( a i ) = ∞ X i =0 M i δ i . No w apply τ 3 to g e t τ 3 ( ψ ( b )) = τ 3 ∞ X i =0 M i δ i ! = ∞ X i =0 λ i 3 τ 3 ( δ i ) . W e consider the magnitude and separate the first 9 terms from the rest: | τ 3 ( ψ ( b )) | =      ∞ X i =0 λ i 3 τ 3 ( δ i )      ≤      8 X i =0 λ i 3 τ 3 ( δ i )      +      ∞ X i =9 λ i 3 τ 3 ( δ i )      . W e b ound the tw o parts separately , using different techniq ues. F or the finite sum we ha ve      τ 3 8 X i =0 M i δ i !      =      τ 3 8 X i =0 M i ψ ( a ′ i ) ! − τ 3 8 X i =0 M i ψ ( a i ) !      = | τ 3 ( α ′ ) − τ 3 ( α ) | , where α = P 8 i =0 M i ψ ( a i ) and α ′ = P 8 i =0 M i ψ ( a ′ i ). Note that α and α ′ are in D 9 , so this is b ounded b y the maxim um o v er all u, v ∈ D 9 of | τ 3 ( u ) − τ 3 ( v ) | . It turns out that there are 11 only 301 v ectors in D 9 , so it is not difficult for a computer program to determine that the maxim um o v er all u, v ∈ D 9 is ac hieve d by u = (24 , 30 , 2 4 , 12) and v = (17 , 25 , 13 , 5 ), and | τ 3 ( u ) − τ 3 ( v ) | . = 1 . 05517. F or the infinite series, w e first compute a n upper b ound for | τ 3 ( δ i ) | . Since δ i = ψ ( a ′ i ) − ψ ( a i ) whe re a i , a ′ i ∈ { ε, 0 , 4 } , it follo ws that | τ 3 ( δ i ) | is le ss than the maxim um o ve r all s, t ∈ { ε, 0 , 4 } of | τ 3 ( ψ ( s ) − ψ ( t )) | . The maxim um turns o ut to b e C = | τ 3 (1 , 0 , 0 , 0) | . = 0 . 8 1582, and is ac hiev ed b y s = 0 , t = ε . Then      ∞ X i =9 λ i 3 τ 3 ( δ i )      ≤ ∞ X i =9 | λ 3 | i | τ 3 ( δ i ) | ≤ C ∞ X i =9 | λ 3 | i = C | λ 3 | 9 1 − | λ 3 | . = 0 . 03 2736 . Com bining these t w o bo unds giv es us | τ 3 ( ψ ( b )) | ≤      8 X i =0 λ i 3 δ i      +      ∞ X i =9 λ i 3 δ i      ≤ 1 . 0 5517 · · · + 0 . 032736 · · · = 1 . 0879 · · · = C 3 / 2 . By t he same reasoning, we get an identic al b ound | τ 3 ( ψ ( c )) | ≤ C 3 / 2, a nd hence | τ 3 ( ψ ( b ) − ψ ( c )) | ≤ | τ 3 ( ψ ( b )) | + | τ 3 ( ψ ( c )) | ≤ C 3 . = 2 . 1758 . 3.2 In tersection with the lattice In this section, w e start to use the fact tha t the blo c ks b 0 and c 0 ha ve the same length and sum. This is true if and only if (1 , 1 , 1 , 1) · ( ψ ( b 0 ) − ψ ( c 0 )) = 0 (0 , 1 , 3 , 4) · ( ψ ( b 0 ) − ψ ( c 0 )) = 0 . Define a lattice of integer p oin ts that mee t these conditions, as fo llo ws: L := { v ∈ Z 4 : (1 , 1 , 1 , 1) · v = 0 a nd (0 , 1 , 3 , 4) · v = 0 } . By construction, ψ ( b 0 ) − ψ ( c 0 ) b elongs to L . If v = ( v 1 , v 2 , v 3 , v 4 ) ∈ L , then (0 , 1 , 3 , 4) · ( v ) = 0 implies v 2 = − 3 v 3 − 4 v 4 . When w e com bine this with the eq uation (1 , 1 , 1 , 1) · v = 0 w e get v 1 = 2 v 3 + 3 v 4 . It follo ws that any v ector in L is of the for m (2 , − 3 , 1 , 0) v 3 + (3 , − 4 , 0 , 1) v 4 where v 3 , v 4 ∈ Z . Since the v ectors ( 1 , − 2 , 2 , − 1) and (1 , − 1 , − 1 , 1) are an orthogo na l basis for this la ttice , ev ery v ector in L can b e written as an integer linear combination of t hese v ectors. Since ψ ( b 0 ) − ψ ( c 0 ) is in L w e can write it in this form. F r o m Theorem 8 w e kno w that | τ 3 ( ψ ( b 0 ) − ψ ( c 0 )) | is bounded, so w e migh t expect ψ ( b 0 ) − ψ ( c 0 ) to be a short v ector in the lattice. But this requires pro of. 12 Prop osition 9. L et v = m (1 , − 2 , 2 , − 1) + n (1 , − 1 , − 1 , 1) b e an arbitr ary ve c tor in L . Then | τ 3 ( v ) | ≥ α | m | (2) | τ 3 ( v ) | ≥ β | n | . (3) wher e α . = 1 . 49 14 and β . = 2 . 16 57 ar e c on s t ants. Pr o of. W e pro v e only Eq. ( 2 ), leaving Eq. (3) to the reader. The equation holds trivially when m = 0, so assume m 6 = 0. W e also use the fact that τ 3 (1 , − 1 , − 1 , 1) . = 0 . 80357 − 2 . 09 0 82 i is no nz ero. Then | τ 3 ( v ) | = | mτ 3 (1 , − 2 , 2 , − 1) + nτ 3 (1 , − 1 , − 1 , 1) | = | m | | τ 3 (1 , − 1 , − 1 , 1) |     n m + τ 3 (1 , − 2 , 2 , − 1) τ 3 (1 , − 1 , − 1 , 1)     ≥ | m | | τ 3 (1 , − 1 , − 1 , 1) |     Im  n m + τ 3 (1 , − 2 , 2 , − 1) τ 3 (1 , − 1 , − 1 , 1)      ≥ | m | | τ 3 (1 , − 1 , − 1 , 1) |     Im  τ 3 (1 , − 2 , 2 , − 1) τ 3 (1 , − 1 , − 1 , 1)      . If w e t a k e α = | τ 3 (1 , − 1 , − 1 , 1) |     Im  τ 3 (1 , − 2 , 2 , − 1) τ 3 (1 , − 1 , − 1 , 1)      . = 1 . 49 14 , then | τ 3 ( v ) | ≥ α | m | , comple ting the proof. Recall from last section tha t | τ 3 ( ψ ( b 0 ) − ψ ( c 0 )) | ≤ 2 . 1758. By Prop osition 9, if τ 3 ( ψ ( b 0 ) − ψ ( c 0 )) = m (1 , − 2 , 2 , − 1) + n (1 , − 1 , − 1 , 1 ) then 1 . 4914 | m | ≤ | τ 3 ( ψ ( b 0 ) − ψ ( c 0 )) | ≤ 2 . 17 5 8 , so | m | ≤ 2 . 1758 1 . 4914 ˙ =1 . 4589 . Therefore m ∈ {− 1 , 0 , 1 } . Similarly , w e deduce that n ∈ {− 1 , 0 , 1 } . T a ble 1 lists all 9 p oss ible v ectors with m, n ∈ {− 1 , 0 , 1 } . W e see that only three of them, (0 , 0 , 0 , 0), (1 , − 2 , 2 , − 1) and ( − 1 , 2 , − 2 , 1), satisfy the constrain t | τ 3 ( ψ ( b 0 ) − ψ ( c 0 )) | ≤ 2 . 1758, so ψ ( b 0 ) − ψ ( c 0 ) m ust be one of these v ectors. 13 v m n | τ 1 ( v ) | | τ 2 ( v ) | | τ 3 ( v ) | (0 , 0 , 0 , 0) 0 0 0 0 0 (1 , − 2 , 2 , − 1) 1 0 0 .63278 1.51365 1.5425 ( − 1 , 2 , − 2 , 1) − 1 0 0.632 78 1.51365 1.5425 (1 , − 1 , − 1 , 1) 0 1 0 .21770 0.62031 2.2399 2 ( − 1 , 1 , 1 , − 1) 0 − 1 0.2 1770 0.62031 2.23992 (2 , − 3 , 1 , 0) 1 1 0.415 08 2.13396 2.37327 ( − 2 , 3 , − 1 , 0) − 1 − 1 0.4150 8 2.13396 2.37327 (0 , 1 , − 3 , 2) − 1 1 0.8504 8 0.89334 3.02667 (0 , − 1 , 3 , − 2) 1 − 1 0.8 5048 0.89334 3.02667 T a ble 1: Short lattice v ectors ordered b y | τ 3 ( v ) | . 3.3 Bounding t w o more co ordinates Theorem 10. L et b and c b e blo cks ( n ot ne c essarily c onse cutive) with the same length an d same sum, and let { b i } ∞ i =0 and { c i } ∞ i =0 b e their anc estr al se quenc es r esp e ctively. Then | τ 1 ( ψ ( b i ) − ψ ( c i )) | ≤ C 1 | τ 2 ( ψ ( b i ) − ψ ( c i )) | ≤ C 2 for al l i , w her e C 1 = 2 | τ 1 (0 , 0 , 0 , 1) | | λ 1 | − 1 . = 1 . 90 32 C 2 = 2 | τ 2 (1 , 0 , 0 , − 1) | | λ 2 | − 1 . = 2 . 9818 . Pr o of. Our proo f is inductive , starting at i = 0. In the last sec tion, w e argued t ha t ψ ( b 0 ) − ψ ( c 0 ) is either (0 , 0 , 0 , 0), (1 , − 2 , 2 , − 1) o r ( − 1 , 2 , − 2 , 1). W e see from table 1 that b oth inequalities are satisfied for these v ectors. Otherwise, | τ 2 ( ψ ( b i +1 ) − ψ ( c i +1 )) | =   τ 2  M − 1 ( ψ ( b i ) − ψ ( c i ) − δ i + δ ′ i )    = | λ 2 | − 1 | τ 2 ( ψ ( b i ) − ψ ( c i ) − δ i + δ ′ i ) | ≤ | λ 2 | − 1 ( | τ 2 ( ψ ( b i ) − ψ ( c i )) | + | τ 2 ( δ i ) | + | τ 2 ( δ ′ i ) | ) . As in the pro of o f Theorem 8 , the quantitie s | τ 2 ( δ i ) | and | τ 2 ( δ ′ i ) | are b ounded b y the maximu m o ver a ll u , v ∈ { ε, 0 , 4 } of | τ 2 ( u − v ) | , whic h turns out to b e | τ 2 (1 , 0 , 0 , − 1) | . = 0 . 7 5301. By induction, w e kno w that | τ 2 ( ψ ( b i ) − ψ ( c i )) | ≤ 2 | τ 2 (1 , 0 , 0 , − 1) | | λ 2 |− 1 . Therefore w e ha v e | τ 2 ( ψ ( b i +1 ) − ψ ( c i +1 )) | ≤ | λ 2 | − 1  2 | τ 2 (1 , 0 , 0 , − 1) | | λ 2 | − 1 + 2 | τ 2 (1 , 0 , 0 , − 1) |  = 2 | τ 2 (1 , 0 , 0 , − 1) | | λ 2 | − 1 . 14 This comple tes the induc tion, and the pro of of the second inequalit y . The pro o f of the first inequalit y is virtually identic al, so it is left to the reader. W e hav e now b ounded all four eigenco ordinates of ψ ( b i ) − ψ ( c i ). In other w ords, τ ( ψ ( b i ) − ψ ( c i )) has b ounded length. Since τ is in v ertible, ψ ( b i ) − ψ ( c i ) will also hav e b ounded length. 3.4 Finding U Define the set U = { x ∈ Z 4 : | τ i ( x ) | ≤ C i for 1 ≤ i ≤ 4 } where C 1 , C 2 , C 3 , C 4 are the constants in Theorems 8 and 10. These theorems pro v e that ψ ( b i ) − ψ ( c i ) is an ele men t o f U . Prop osition 11. I f x ∈ U then | x | ≤ 6 . 28 . Pr o of. Since x is in U , there are b ounds on t he comp onen ts of τ ( x ), and therefore on its length. W e ha v e | τ ( x ) | 2 = 4 X i =1 | τ i ( x ) | 2 ≤ 1 . 9 032 2 + 2 . 98 18 2 + 2 . 17 58 2 + 2 . 17 58 2 . = 21 . 9 8 . Supp ose A is a n arbitra r y matrix a nd ev ery eigenv alue λ of A ∗ A lies b et w een µ min and µ max , where A ∗ denotes the conjugate tra ns p ose of A . Under thes e conditions, a classical theorem in linear algebra (e.g., [2, pp. 415–420]) give s the inequality µ min | x | 2 ≤ | Ax | 2 ≤ µ max | x | 2 . The smallest eigen v alue of τ ∗ τ is µ . = 0 . 55713, so from the low er b ound of the theorem w e get | µ | | x | 2 ≤ | τ ( x ) | 2 and so | x | 2 ≤ | τ ( x ) | 2 | µ | . = 21 . 98 0 . 55713 . = 39 . 455 . Therefore | x | ≤ 6 . 28. Prop osition 11 tells us that w e can en umerate v ectors in U by listing all in teger v ec- tors of length less than 6 . 28 a nd discarding the ones that fail to satisfy the inequali- ties | τ i ( x ) | ≤ C i for eac h i . Our computer progra m for en umerating U lists 503 v ectors. There is not enough space to repro duce the en tire list here, but it can be do wnloaded fro m http://www. student.cs. uwaterloo.ca/~l3schaef/su m cube/ . 15 4 Main graph 4.1 Graph pro ducts Recall the tree T with v ertices represen ting p ositions in w . W e are in terested in finding an analogous graph for tr iple blo c ks. Since eac h triple blo c k is delimited by four p ositions, so it is natura l to consider the graph T × T × T × T = T 4 where the pro duct o f t w o graphs is defined below . Definition 12. Let G 1 and G 2 b e graphs. Define the tensor pro duct G 1 × G 2 where V ( G 1 × G 2 ) = V ( G 1 ) × V ( G 2 ) E ( G 1 × G 2 ) = E ( G 1 ) × E ( G 2 ) F urther, if e 1 and e 2 are lab elled b y ℓ 1 and ℓ 2 resp e ctiv ely then the edge ( e 1 , e 2 ) is lab elled ( ℓ 1 , ℓ 2 ). Note tha t ev ery v ertex p = ( p 1 , p 2 , p 3 , p 4 ) in T 4 has a unique paren t par( p ) b ecause eac h co ordinate p i has a unique paren t par ( p i ) in T . Th us, if a v ertex v in T 4 delimits a triple blo c k b 1 b 2 b 3 then the paren t of v delimits the paren t of b 1 b 2 b 3 . Since eac h no de in T 4 has a unique paren t, w e define the ancestral sequence of a no de v to b e the sequenc e of paren ts starting at the no de. An ancestral sequence in T 4 giv es us four ancestral sequ ences in T for the four co ordinates o f T 4 . Ev ery ancestral sequence in T ev en tually reache s 0, the ro ot, and remains there b ecause 0 is its o wn parent. It follo ws that ev ery ances tral sequence in T 4 ev en tually reac hes (0 , 0 , 0 , 0). Also note that an y cycle in T 4 induces four cycles in T . Since the lo op from 0 to itself is the only cycle in T , it follows tha t the only cycle in T 4 is a lo op from (0 , 0 , 0 , 0) to it s elf. W e conclude tha t T 4 is a tree, except for a lo op at (0 , 0 , 0 , 0), the ro ot of the tree. Earlier we sa w that the g raph homomorphism ζ : T → Q w as c hild-bijectiv e. By Prop osition 6 there is a certain relat io ns hip b et w een walk s in T and w alks in Q . Define a graph homomorphism ξ : T 4 → Q 4 that sends ( p 1 , p 2 , p 3 , p 4 ) to ( ζ ( p 1 ) , ζ ( p 2 ) , ζ ( p 3 ) , ζ ( p 4 )). By t he follow ing propo s ition, ξ is a child-bijection. Prop osition 13. If f 1 : A 1 → B 1 and f 2 : A 2 → B 2 ar e child-bije ctions then f : A 1 × A 2 → B 1 × B 2 sending ( x, y ) to ( f 1 ( x ) , f 2 ( y )) is also a child-b i j e ction . Pr o of. Exercise . Since ξ is a c hild-bijection, Proposition 6 say s that if w e fix some v in T 4 then there is a bijection b e t ween w alks in T 4 starting at v and w alks in Q 4 starting at ξ ( v ). F urthermore, the bijec tion sends a w alk v 0 · · · v ℓ to a w a lk ξ ( v 0 ) · · · ξ ( v ℓ ), and the bijection preserv es edge lab els. 16 4.2 Augmen ting Q with b lo c k v ectors W e kno w that a triple blo c k b 1 b 2 b 3 delimited b y ( p 1 , p 2 , p 3 , p 4 ) will give us a w alk from (0 , 0 , 0 , 0) to ( p 1 , p 2 , p 3 , p 4 ) in T 4 . Then w e kno w the P arikh v ector of eac h blo c k, ψ ( b 1 ) = σ ( p 2 ) − σ ( p 1 ) ψ ( b 2 ) = σ ( p 3 ) − σ ( p 2 ) ψ ( b 3 ) = σ ( p 4 ) − σ ( p 3 ) . If b 1 b 2 b 3 is a n a dditiv e cube, then ψ ( b 2 ) − ψ ( b 1 ) = σ ( p 3 ) − 2 σ ( p 2 ) + σ ( p 1 ) ψ ( b 3 ) − ψ ( b 2 ) = σ ( p 4 ) − 2 σ ( p 3 ) + σ ( p 2 ) m ust b oth b elong t o L and, since the blocks are nonempty , w e ha v e p 1 < p 2 < p 3 < p 4 . The p oin t is that w e can de cide whethe r a vertex in T 4 corresp onds to an additiv e cube. The problem with ξ is that w e cannot tell f rom ξ ( p 1 , p 2 , p 3 , p 4 ) whether ( p 1 , p 2 , p 3 , p 4 ) delimits an additive cub e, because the v ertices in Q 4 do not con t ain any informatio n a b out the Parikh vec tors. O n the other hand, supp ose w e pro ject the w alk from (0 , 0 , 0 , 0) to ( p 1 , p 2 , p 3 , p 4 ) in to Q 4 via ξ . Then since mapping w alks via ξ is bijectiv e, we can reco v er original w alk in T 4 from its image in Q 4 . No w let there b e a walk fro m ( p 1 , . . . , p 4 ) to ( q 1 , . . . , q 4 ) in T 4 , and supp ose w e are giv en t he corresp onding w alk in Q 4 . W e ha ve seen that if w e hav e ( p 1 , . . . p 4 ) then w e can reconstruct t he en tire o r iginal walk T 4 , including ( q 1 , . . . , q 4 ). What do w e need to know ab out ( p 1 , . . . , p 4 ) to reconstruct just σ ( q 3 ) − 2 σ ( q 2 ) + σ ( q 1 )? Let us supp ose for simplicit y that the w alk is length 1, so ( p 1 , . . . , p 4 ) and ( q 1 , . . . , q 4 ) are adjacen t, and let the edge b e lab elled ( a 1 , . . . , a 4 ). Then b y Lemma 3 , σ ( q 3 ) − 2 σ ( q 2 ) + σ ( q 1 ) = ( M σ ( p 3 ) + ψ ( a 3 )) − 2( M σ ( p 2 ) + ψ ( a 2 )) + ( M σ ( p 1 ) + ψ ( a 1 )) = M ( σ ( p 3 ) − 2 σ ( p 2 ) + σ ( p 1 )) + ψ ( a 3 ) − 2 ψ ( a 2 ) + ψ ( a 1 ) , so it suffices to know σ ( p 3 ) − 2 σ ( p 2 ) + σ ( p 1 ). Similarly , w e can compute σ ( q 4 ) − 2 σ ( q 3 ) + σ ( q 2 ) from σ ( p 4 ) − 2 σ ( p 3 ) + σ ( p 2 ). Th us, w e de fine the gra ph G with v ertices 2 in the set V ( G ) = V ( Q 4 ) × Z 4 × Z 4 = Σ 4 × Z 4 × Z 4 . There is a n edge from ( c , u , v ) to ( c ′ , u ′ , v ′ ) if there is an edge ( c , c ′ ) in Q 4 lab elled b y ( a 1 , a 2 , a 3 , a 4 ) and the follo wing consisten cy equations hold: u ′ = M u + ψ ( a 3 ) − 2 ψ ( a 2 ) + ψ ( a 1 ) v ′ = M v + ψ ( a 4 ) − 2 ψ ( a 3 ) + ψ ( a 2 ) . 2 The vertices of G resemble templates as defined in [1 ]. T emplates and G ar e differe nt pe r spectives on the same s tructure. 17 T o see w here these equations come from, define the map g : T 4 → G suc h that g ( p ) = ( ξ ( p ) , σ ( p 3 ) − 2 σ ( p 2 ) + σ ( p 1 ) , σ ( p 4 ) − 2 σ ( p 3 ) + σ ( p 2 )) . Then a tr iple blo c k b 1 b 2 b 3 maps to the fo ur delimiting c haracters w [ p 1 ] , w [ p 2 ] , w [ p 3 ] , w [ p 4 ], with the vec tors ψ ( b 2 ) − ψ ( b 1 ) and ψ ( b 3 ) − ψ ( b 2 ). Thus , w e see t ha t the tw o v ectors are meant to represen t the diff erences betw een pairs of blo c k ve ctors. W e hav e seen that giv en an edge ( ξ ( p ) , ξ ( q )) in Q 4 , w e may compute σ ( q i +2 ) − 2 σ ( q i +1 ) + σ ( q i ) from σ ( p i +2 ) − 2 σ ( p i +1 ) + σ ( p i ) for i = 1 , 2. The consistenc y equations simply enforce this relationship, so that ( g ( p ) , g ( q )) is an edge in G when ( p , q ) is an edge in T 4 . Th us, g is a graph homomorphism from T 4 to G . Prop osition 14. T he gr a p h h o momorphism g is child -bije ctive. Pr o of. Consider an edge ( g ( p ) , ( c , u , v )) in G . By the definition o f edges in G there must b e an edge ( ξ ( p ) , c ) in Q 4 . Since ξ is c hild-bijectiv e, there exists a neighbour q of p in T 4 suc h that ξ ( q ) = c . Then ( g ( p ) , g ( q )) is an edge in G . The consistency equations ha v e a unique solution for g ( q ) giv en g ( p ) and c , so ( c , u , v ) = g ( q ) as required. Since g is c hild-bijective , Prop osition 6 sa ys that a ny walk in G starting at g ( p ) is the image of some w alk from p to q in T 4 , and therefore ends in g ( q ). 4.3 Additiv e cu b es and w alks in G W e sa w in the previous section that w alks in T 4 corresp ond to walks in G . If an ar bitr a ry triple blo c k is delimited b y q then there is a walk in T 4 from (0 , 0 , 0 , 0) to q and therefore a w alk from g (0 , 0 , 0 , 0) to g ( q ). Unfortunately , a walk from g (0 , 0 , 0 , 0) to g ( q ) in G do es not necessarily mean that q delimits a triple blo c k, since w e do not necessarily hav e q 1 < q 2 < q 3 < q 4 . W e c an fix this, but we need a few prop ositions. Note that par( x ) is an increasing function, so if p ≤ q then par( p ) ≤ par( q ). By induction, if p 0 ≤ q 0 then for i th a nc estors p i and q i w e ha v e p i ≤ q i . The con trap ositiv e sa ys that p i < q i implies p 0 < q 0 , whic h has an application in the follo wing lemm a. Lemma 15. Given an a n c estr al se quenc e of blo cks { b i } ∞ i =0 for some nonempty blo ck b 0 , ther e exists k such that b j is nonempty for al l 0 ≤ j ≤ k and b j is empty for al l k < j . F urthermor e, b k is the differ enc e of two p r op er pr efixes of some ϕ ( c ) , so b k is either 0 or 4 . Pr o of. Supp ose b 0 = w [ p 0 , q 0 ) and let { p i } ∞ i =0 and { q i } ∞ i =0 b e the corresponding ancestral sequence s for p 0 and q 0 . R ec all that the a nc estral sequence of a blo c k b 0 is the sequence of paren ts of b 0 , so b i = par( b i − 1 ) = w [par( p i − 1 ) , par( q i − 1 )) = w [ p i , q i ) . An ancestral sequence of p ositions ev entually reac hes 0 so there exists some n 0 ∈ N suc h that p n = q n = 0, and hence b n is empty , for all n > n 0 . Th us, w e can tak e b k to b e the last 18 nonempt y blo c k b ecause there are only finitely man y nonempt y blo c ks. By definition, b j is empt y for all j > k . Since b k is nonempt y we ha v e p k < q k , and th us p j < q k for an y j ≤ k b y the discussion ab o ve. It f o llo ws that b j is nonempt y for any j ≤ k . By the definition of paren ts, η ( q k +1 ) ≤ q k < η ( q k ). It f ollo ws that y = w [ η ( q k +1 ) , q k ) is a prop er prefix of w [ η ( q k +1 ) , η ( q k +1 + 1)) = ϕ ( w [ q k +1 ]). Similarly , x = w [ η ( p k +1 ) , p k ) is a prop er prefix of w [ η ( p k +1 ) , η ( p k +1 + 1)) = ϕ ( w [ p k +1 ]). But p k +1 = q k +1 b ecaus e b k +1 is empt y , so η ( p k +1 ) = η ( q k +1 ) and the prefixes x and y start at the same p osition a nd xb k = y . F or our morphism, we ha v e seen that the prop er prefixes are { ε, 0 , 4 } . The n b k is a suffix of some pro p er prefix, a nd since b k is no ne mpt y w e hav e b k ∈ { 0 , 4 } . Supp ose w e ha v e a triple blo c k delimited by p . Usually we are interes ted in the three blo c ks inside, but w e can also think of it as one big block delimited b y p 1 and p 4 . Then the lemma a pplie s, and mot iv ates the fo llowing definition: X := { ( p 1 , p 2 , p 3 , p 4 ) ∈ N 4 : p 1 ≤ p 2 ≤ p 3 ≤ p 4 , p 1 < p 4 , par( p 1 ) = par( p 2 ) = par( p 3 ) = par( p 4 ) } . W e hav e defined X so t ha t it is precis ely t he set of p ositions that delimit a nonempt y triple blo c k with an empt y paren t. No w let us consider the set g ( X ). Since p is in X , we kno w from the lemma that it delimits a triple blo c k b 1 b 2 b 3 that is a sub w ord of ϕ ( c ) fo r some c . Since ϕ (0) , ϕ (1) , ϕ (3) , ϕ (4) all o ccur in the first sev en characters of w , and g dep en ds on the c on ten t of the triple block and not its absolute p osition, it suffices to compute g ( p ) for all p ∈ X such tha t p 4 < 7. The lemma states tha t b 1 b 2 b 3 is either 0 or 4, so the triple blo c k is one of w [0 , 1) , w [3 , 4) , w [5 , 6). Th us, g ( X ) = { g (0 , 0 , 0 , 1) , g (0 , 0 , 1 , 1) , g (0 , 1 , 1 , 1) , g (3 , 3 , 3 , 4) , g (3 , 3 , 4 , 4) , g (3 , 4 , 4 , 4) , g (5 , 5 , 5 , 6) , g (5 , 5 , 6 , 6) , g (5 , 6 , 6 , 6) } . A w alk in T 4 from ( 0 , 0 , 0 , 0) to q , where q delimits a nonempt y triple blo c k, m ust pass through the s et X . Th us, we might c onsider starting our walk f r om a no de in X instead of from (0 , 0 , 0 , 0). If our walk s tarts at some p ∈ X then w e hav e p 1 < p 4 and hence q 1 < q 4 , whic h helps us to show that q 1 < q 2 < q 3 < q 4 . Hence, additive cubes corresp ond to walks as des crib ed in Theorem 16. Theorem 16. Given an ad ditive cub e delimi te d by q , ther e is a walk in G fr om g ( p ) to g ( q ) , wher e p is in X and g ( q ) is in the se t Z := Σ 4 × L × L . Conversely, if we ar e given a walk in G starting in g ( X ) and endin g a t v ∈ Z , ther e is an additive cub e delimite d by q wher e g ( q ) = v . Pr o of. Supp ose q delimits an additiv e cub e. Then the difference betw een tw o blo c k vec tors is in L , so g ( q ) is in Z . By the previous lemma, there exists some ancestor p ∈ T 4 of q such that p delimits a nonempt y triple blo c k, but par( p ) delimits εεε . By o ur construction of X , p belongs to X , and o ur w a lk from p to q in T 4 maps t o a walk fro m g ( p ) to g ( q ) in G . In the other direction, supp ose we hav e a w alk in G starting at g ( p ) and ending at some v ertex v in Z . The w alk in G corresp onds to a w alk in T 4 from p to s ome q and v = g ( q ) . 19 No w g ( q ) is in Z , so the blo c k v ector differences are in L , and therefore a ll blo c ks ha ve the same length and sum. Since p 1 < p 4 w e know that q 1 < q 4 and thus ev ery blo c k has p ositiv e length giving q 1 < q 2 < q 3 < q 4 . W e conclude that q delimits an additiv e cube. 4.4 Reduction to a finite subgraph The relationship b et w een additiv e cub es a nd w a lks in G is nice, but G is an infinite graph b ecaus e Z 4 is infinite, and it is difficult to apply gra ph algorithms t o an infinite graph. Recall the main result of the previous section, which states that if b 0 and c 0 are blo c ks with the same length a nd sum, and { b i } ∞ i =0 , { c i } ∞ i =0 are their ancestral sequence s resp ectiv ely then ψ ( b i ) − ψ ( c i ) ∈ U , where U is a finite set w e can en umerate. In other w ords, if p delimits an additiv e cub e and q is any no de on the w alk from (0 , 0 , 0 , 0) to p , and q delimits blo c ks d 1 d 2 d 3 then ψ ( d i ) − ψ ( d j ) ∈ U for all i, j . Let g ( q ) = ( c , u , v ) and note tha t b y the definition of g , we ha ve u = ψ ( d 2 ) − ψ ( d 1 ), v = ψ ( d 3 ) − ψ ( d 2 ) and finally u + v = ψ ( d 3 ) − ψ ( d 1 ). It follo ws tha t u , v , and u + v are all in U . Define the set H := { ( c , u , v ) : u , v , u + v ∈ U and c ∈ Σ 4 } . An y no de along the path to an additive cub e mus t b e in H , so we ma y restrict our searc h to the subgraph of G induced b y H , call it G ′ . Notice tha t H is a subset of Σ 4 × U × U , so it con tains at most 4 4 × 503 × 503 . = 64 . 7 × 1 0 6 elemen ts and th us G ′ is finite. W e can up date Theorem 16 so that the w alks m ust lie in G ′ . Corollary 17. Given an additive cub e deli m ite d by q , ther e is a walk in G ′ fr om g ( p ) to g ( q ) , wher e p is in A := g ( X ) ∩ H and g ( q ) is in the set B := Z ∩ H . Conversely, if we ar e giv en a walk in G starting in A and ending at so me β ∈ B , ther e is an additive cub e delimite d by q wher e g ( q ) = β . Since G ′ is a practical size, we can determine whether there is a pat h from A to B within G ′ using a computer, but first we need to b e able to compute A , B and H . Since w e can en umerate the set U , it is straightforw ard to en umerate Σ 4 × U × U and t hen narrow it dow n to H . Also, we can test whether a v ector is in the lattice L using dot pro ducts and therefore w e can determine if an elemen t is in Z = Σ 4 × L × L . Th us, w e can list the elemen ts of B = Z ∩ H b y testing whether eac h elemen t in H is also in Z . Earlier w e sho we d that g ( X ) is a nine-elemen t set, and it is not difficult to chec k that g ( X ) ⊆ H , so A = g ( X ) ∩ H = g ( X ). W e ar e no w r eady to complete the pro o f of o ur main result. Theorem 18. The infinite wor d w c ontains no additive cub es. Pr o of. Our computer searc h of G ′ computes the set of ve rtices R ⊆ H that are reac hable from A . It turns out that | R | = 13557 2, but R ∩ B is empt y , so w e conclude that w con tains no additive cub es. Corollary 19. N + is not uniformly 3 -r ep etitive. 20 5 A t w o-s ided in fi nite w ord a v oidi n g additiv e cub es Ab o v e w e ha v e sho wn that the (one-side d) infinite word → ϕ ω (0) = 03 1430110343430 · · · a voids additiv e cub es. Using the appropriate iterations of ϕ , w e no w pro ve the same res ult for tw o-sided infinite words . (Suc h a word is a map fro m Z to a finite set, in this case, Σ = { 0 , 1 , 3 , 4 } , as opp osed to the o ne -sided infinite w ords — maps from N to Σ — w e ha v e discusse d th us far.) F or notatio n and results inv olving morphisms and t w o- s ided infinite w ords, see [15]. In particular, w e write a tw o-sided infinite word as · · · a − 2 a − 1 a 0 .a 1 a 2 · · · , where the p erio d is written to the left of the character indexed with 1. Also, if h is a morphism satisfying h ( a ) = xa for some w ord x , then w e define → h ω ( a ) to b e the left-infinite w ord · · · h 3 ( x ) h 2 ( x ) h ( x ) xa . Theorem 20. Ther e exists a two-side d infinite wor d over Σ = { 0 , 1 , 3 , 4 } avoiding additive cub es. Pr o of. Note that 3 0 is a factor of ϕ 4 (0). It follows that for all n ≥ 0, the w ord ϕ n (30) con tains no additiv e cube. No w ϕ 2 (3) = 43 , a nd ϕ 2 (0) = 03 1. Letting h = ϕ 2 , w e see that ← h ω (3) . → h ω (0) = · · · 0314303 4 343034343 . 03143 011034343031011011 · · · is a t wo-sided infinite w o r d a v oiding additive cub es. 6 Op en p roblems W e do not kno w if the alphab et size of 4 giv en in this pap er is optimal for av oiding additive cub es . Since an ab elian cub e is necessarily an additive cub e, and w e know it is imp ossible to a void abelian cub e s o ve r an alphabet of size 2, the alphabet size cannot b e 2. But it is still conceiv able that, to av oid a dditive cub es, some alphab et of cardinality 3 migh t suffice. By a depth-first searc h approac h, for example, w e hav e g e nerated a finite w ord of length 1288 o ver the alphab et { 0 , 1 , 2 } a v oiding additive cub es. The mor e difficult question o f whether it is p ossible to a v oid additiv e squares ov er a n alphab et equal to some finite subset of Z remains op en. Since alphab et size 4 is needed to a void ab elian squares, the alphab et must b e at least this large. How ev er, as F reedman has sho wn [8], the longest w ord ov er { a, b, c, d } with a + d = b + c av oiding additiv e squares is of length ≤ 60. Also see [11 ]. 7 Ac kno wledgments W e ar e pleased to thank Thomas Stoll for many discussions on t his problem. 21 References [1] A. Ab erk ane, J. D. Currie, and N. Rampersad. The num b er of ternary w o rds av oiding ab elian cub es grows exp onen tially . J. Inte ger Se quen c es 7 (200 4), 04.2 .7 (electronic). [2] K. E. Atk inson. A n Intr o duction to Numeric al Analysis . John Wiley , 1978. [3] Y.-H. Au, A. 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