Geodesic stability for memoryless binary long-lived consensus

Geo desic stabilit y for memoryless binar y long-live d consensus Cristina G. F ernandes ∗ and Ma y a Stein † Septem ber 24, 2018 Abstract The determination of the stabilit y of th e long-lived consensus prob- lem is a fundamental ope n problem in distributed systems. W e con- centrate o n the memoryless binar y case w ith g eodesic paths. W e offer a conjecture o n the sta bilit y in this case, exhibit tw o classes of colour- ings which attain this conjecture d b ound, a nd improv e the known low er bo unds for all co lourings. W e also intro duce a related para meter, which measures the stability only for certain g eodesics, and for which we a lso prov e low er bo unds. 1 In tro d uction The c onsensus pr oblem in distributed systems consists of the follo wing: giv en a set of v alues, eac h coming from a pro ce ssor or sensor, decide on a repr e- sen tativ e v alue, meaning the consensus of the give n v alues. The long-live d consensus pr oblem consists of rep eatedly solving related instances of the consensus p roblem. In [5], D olev and Ra jsbaum int ro duce the concept of stability of long-liv ed consensu s, wh ere one wish es the r ep resen tativ e v alues, pro duced b y an algorithm for a sequence of input instances, to change as few times as possible (there migh t b e s ome cost asso ciated to a c hange). So the question is how to choose the outputs in a wa y that they are stable in time. In the case with memory , the algo rithm ma y use the v alue pro duced f or the p revious instances in the sequence to decide on the v alue of the curr en t instance. That is n ot allo wed in th e so called memoryless case. See also [1]. ∗ Instituto de Matem´ atica e Estat ´ ıstica, Universidade de S˜ ao Paulo, Brazil, cris@ime.u sp.br . Pa rtial supp ort by CNPq 309657/200 9-1 and 475064/2010-0. † Cen tro d e Modelamiento Ma tem´ atico, Universidad de Chile, Santiago , Chile, mstein@dim .uchile.c l . Supp ort by F ondecyt 11090141 and F ap esp 05/54051-9. 1 W e w ill consid er binary-v alued consens u s, w it h the inpu t sequences b e- ing a geo desic path. Th e case with memory is completely solv ed in [5] and also, for the memoryless case, some b ounds for the minimum num b er of c hanges are shown, whic h w e will imp ro v e here. Da vidovitc h , Dolev, and Ra jsbaum [3 ] consider multi- v alued consensus . Bec k er et al. [2] study a ver- age in sta bilit y for binary consensus using r a ndom walks instead of geo desic paths. W e need a few definitions in order to prop erly state th e problem. The n -hyp er cub e is H n := { 0 , 1 } n . W rite 0 n for (0 , 0 , . . . , 0), and similar. The b al l B t (0 n ) of r adius t around 0 n consists of all element s of H n with at most t en tries iden tical to 1. In the same wa y , we define B t (1 n ). A c olouring of H n is a fun ct ion f : H n → { 0 , 1 } . W e say that a colour- ing f r esp e cts B t (0 n ) and B t (1 n ) if f ( x ) = 0 for eac h x in B t (0 n ) and f ( x ) = 1 for eac h x in B t (1 n ). O bserv e that if n < 2 t + 1, the t wo balls B t (0 n ) and B t (1 n ) intersect, and n o colo uring can r espect B t (0 n ) and B t (1 n ). As we are not intereste d in this case, w e sa y t is valid (for n ) if n ≥ 2 t + 1. A ge o desic P (in H n ) is a sequence ( x 0 , x 1 , . . . , x n ) with x i ∈ H n for i = 0 , 1 , . . . , n , so that there is a p ermutati on ( p 0 , p 1 , . . . p n ) of (0 , 1 , . . . n ) suc h that th e ℓ th entry of x j differs from the ℓ th entry of x j − 1 if and only if j = p ℓ . W e then sa y that P fix e d the ℓ th entry at time j . W e d enote b y inst ( f , P ), for instability , the num b er of colour-jumps of P in the co louring f , that is, the num b er of indices i where f ( x i ) 6 = f ( x i − 1 ). An y suc h ind ex i sh all b e called a jump of P (in f ). Let i nst ( f ) b e the maxim u m v alue of inst ( f , P ) o ver all geo d esics P . The connection of these concepts and the memoryless consens us problem in distributed systems is as follo w s . Eac h p oin t of H n represent s a set of n input v alues (one from ea c h sensor). A colouring of H n corresp onds to an assignmen t of a r epresen tativ e v alue for eac h p ossible set of inpu t v alues. W e pr efer colourings that resp ect the b al ls of a certain radiu s as the output v alue should in some w a y b e representat iv e. A geo desic stands f o r a s lowly c hanging system of in puts (one sensor at a time), and its instabilit y is the n um b er of c hanges of the represent ativ e v alue. W e r emark that, if one considers arb itrary paths instead of geo desics, there is no b ound on the instabilit y as the path might go bac k and f orth b et w een t wo p oints with a differen t output v alue (see [5]). No w, a colouring th at resp ects B t (0 n ) and B t (1 n ) and has low ins ta bilit y is a go od candidate f or a consens u s algorithm. One is therefore interested in the lo w est p ossible instabilit y . 2 Problem 1.1 (Dolev & Ra js baum [5]) . Given n ∈ N , and t valid for n , find the minimum value inst ( n, t ) for inst ( f ) over al l c olourings f of H n that r esp e ct B t (0 n ) and B t (1 n ) . Dolev and R a jsbaum [5] prov e several sp ecial cases: inst ( n, t ) ≥ 1 for n > 4 t , inst ( n, 0) = 1, inst ( n, 1) = 3, and inst (2 t + 1 , t ) = 2 t + 1. In Section 4 , we establish a lo wer b ound of ⌈ t − 1 n − 2 t ⌉ + ⌊ t − 1 n − 2 t ⌋ +3 on inst ( n, t ) (Theorem 4.1) that holds f or all v alues of n and t . A similar lo w er b ound also holds for the r el ated parameter winst ( n, t ), whic h measures the maximum instabilit y of a colouring considering on ly a sp ecial class of geo desics. This parameter is introd uced in Section 3. In Section 4.2, we improv e our b ound to inst (2 t + 2 , t ) ≥ t + 2 + ( t + 1) mo d 2 for t ≥ 1 for the sp ecial case of n = 2 t + 2 (Theorem 4.2). The basic tool for this result is Lemma 4.3, whic h serv es to extend b o unds for smaller v alues of t to larger v alues of t . This to ol is extended in Section 4.3 to arbitrary v alues of n . In [5], it is also s ho wn th at inst ( n, t ) ≤ 2 t + 1. W e conjecture th a t this b ound is in deed the correct v alue. Conjecture 1.2 (Ma in conjecture) . L et n ∈ N , and t b e v a lid for n . Then inst ( n, t ) = 2 t + 1 . If one can solve Problem 1.1, it w ould b e interesting to find all optimal colourings, i.e., colourings for which the b ound inst ( n, t ) is attained. In Section 2 we exhibit t wo new classes, maj t ( k ) and b k t , of colourings th at ha ve instabilit y exactly 2 t + 1. Previously on ly one such colouring (namely maj t (2 t + 1) in our language) wa s kn o w n [5]. 2 Candidates for op timal colourings W e present t w o classes of colourings that resp ect the balls B t (0 n ) and B t (1 n ) and ha v e instabilit y 2 t + 1. 2.1 The ma jority colourings F or a p ositiv e o dd v alue k , d efine maj t ( k ) to b e the colouring th at assigns to eac h p oin t x ∈ H n \ ( B t (0 n ) ∪ B t (1 n )) the colour that app ears on the ma jorit y of the first k entries of x . The balls B t (0 n ) and B t (1 n ) are coloured canonically with 0 and 1, resp ectiv ely . F or a p ositiv e ev en v alue k , define the auxiliary class maj ′ t ( k ) as the class of colourings f th at assign to eac h p oin t x outsid e B t (0 n ) and B t (1 n ) 3 (whic h are coloured canonically) the colour that app ears on the ma jorit y of the fi rst k en tries of x , and an y colour if b oth colours app ear equally often in the first k entrie s of x . Let maj t ( k ) ⊆ maj ′ t ( k ) b e the class of those colourings in maj ′ t ( k ) for w hic h f ( x ) 6 = f ( y ) whenev er x and y restricted to their fi r st k entrie s are the complemen t of eac h other. In what follo ws, w e often abuse notation and, for a p ositiv e ev en v alue k , write maj t ( k ) for an arbitrary elemen t of maj t ( k ). Prop osition 2.1. L et k , t , n ∈ N , with 0 < k ≤ 2 t + 1 ≤ n . Then inst ( maj t ( k )) = 2 t + 1 . The pr oof of Prop osition 2.1 splits into t wo parts: in Lemma 2.2 we sho w that no geo desic jum ps more than 2 t + 1 in maj t ( k ), and in Lemma 2.3 we present a geo d esic that jumps that muc h. Before we turn to these lemmas, let us remark that, when k > 2 t + 1 and o dd, it is easy to find a geo d esic th at jump s k times in maj t ( k ). I n deed, w e ma y start at the p oin t (01) ⌊ n/ 2 ⌋ 0 and then at eac h step switc h an en try , from the fi rst to the last. Eac h of the k fir st steps is a jump . T his sho ws that maj t ( k ), for k large and o dd, has ins tability larger than 2 t + 1. The remainder of this section is dev oted to th e p r oof of Prop osition 2.1, i.e., to Lemma 2.2 and Lemma 2.3. F or a geod esic P = ( x 0 , x 1 , . . . , x n − 1 , x n ), the path Q = ( x n , x n − 1 , . . . , x 1 , x 0 ) is also a geo desic, and is called the r everse of P . Clearly , inst ( f , P ) = inst ( f , Q ) for any colouring f . Lemma 2.2. If 0 < k ≤ 2 t + 1 ≤ n , then inst ( maj t ( k )) ≤ 2 t + 1 . Pr o of. Su pp ose otherwise. Then there is a geo desic P = ( x 0 , x 1 , x 2 , . . . , x n ) in H n with inst ( maj t ( k ) , P ) ≥ 2 t + 2. Let m b e s o that the m th jump of P is the first jump that fixes one of the last n − k en tr ies (as k < 2 t + 2 th er e is suc h an m , 1 ≤ m ≤ n ). Sup p ose P is c hosen such that m = m ( P ) is as large as p ossible. Ou r plan is to m odify P to a geo desic P ′ with m ( P ′ ) > m ( P ), th u s obtaining a contradictio n. Let i + 1 b e th e fi r st jump of P , and let ℓ b e the (2 t + 2)nd jump of P . W e assum e th a t maj t ( k )( x ℓ ) = 1, and thus maj t ( k )( x i ) = 1. The other case is analogous. As ℓ is the (2 t + 2)nd jump of P , there are t + 1 jumps j with j < ℓ and maj t ( k )( x j ) = 0. Hence, P fi xed ( t + 1) 0’s b efore time ℓ , and therefore x ℓ / ∈ B t (1 n ). Thus, since maj t ( k )( x ℓ ) = 1, the ma j o rit y of the fi rst k entrie s of x ℓ is not 0: it is 1 or k is even an d x ℓ has as many 0’s as 1’s in its first k en tries. W e can u se the same argum en t on the reverse of P to obtain that 4 x i / ∈ B t (1 n ), and th us the ma jority of th e first k en tries of x i is 1 or k is ev en and x i has as man y 0’s as 1’s in its first k en tries. Thus w e sho wed that the first k entries of x i c ontain at le ast as many 1 ’s as 0 ’s , (1) and the same holds for x ℓ . Because maj t ( k )( x i ) = maj t ( k )( x ℓ ) = 1, the first k entries of x i and of x ℓ are n ot th e complement of ea c h other. S o, by (1), at least one en try within the k fir st, say th e first en try , is 1 in b ot h x i and x ℓ . This implies that all x j with i ≤ j ≤ ℓ start with a 1. Let S b e the set of those of the fi rst k en tries of x i that do not c h ange in P b et w een x i and x ℓ . W e hav e ju st seen that s := | S | ≥ 1. Let z 1 b e obtained from x i b y c hanging the first ent ry to 0, and for 1 < j ≤ s let z j b e obtained from z j − 1 b y changing another of the entries in S . Then the first k entries of z s ar e the c omplement of the first k entries of x ℓ . (2) Let h b e the (2 t + 1)st jum p of P . Then maj t ( k )( x h − 1 ) = 1 and maj t ( k )( x h ) = 0. There are t jum ps j ≤ h − 1 w it h maj t ( k )( x j ) = 1, eac h fi xing a 1 distinct from the fir st en try . Th us in total x h and x ℓ − 1 ha ve at least ( t +1) 1’s, and cannot b e in B t (0 n ). In the same wa y , w e see th at x i +1 / ∈ B t (0 n ). Consider P ′ = ( z s , z s − 1 , . . . , z 1 , x i , x i +1 , . . . , x ℓ , y 0 , y 1 , . . . , y n − s − ℓ + i − 1 ), where the y j ’s are arbitrarily c h ose n to complete P ′ to a geodesic. W e claim that P ′ has a j u mp in its first s + 1 steps. (3) Then we are done b ecause, by the choic e of P , all the fi rst m + 1 jumps of P ′ fix one of the fi rst k entries, cont radicting our c h o ice of P . It remains to prov e (3). As x i +1 / ∈ B t (0 n ) and maj t ( k )( x i +1 ) = 0, there are at least as many 0’s as 1’s among the first k en tries of x i +1 . So, since the fi rst en tr y of x i is 1, bu t the firs t ent ry of z 1 is 0, there are also at least as many 0’s as 1’s among the fi rst k entries of z 1 . No w, as x i / ∈ B t (1 n ), also z 1 / ∈ B t (1 n ). Hence, if the first k en- tries of z 1 con tain more 0’s than 1’s, it follo ws th a t maj t ( k )( z 1 ) = 0. As maj t ( k )( x i )=1, the geo desic P ′ has the ju mp x i , which is as desired for (3). So w e may assume that the first k en tries of z 1 con tain exactly as many 0’s as 1’s. Th us by (1) an d (2), and b y the definition of z s , it follo ws that z s has at least as m an y 0’s as 1’s in its first k entries, and s o at least as m any 5 0’s as z 1 has. Therefore, z 1 / ∈ B t (1 n ) implies th at z s / ∈ B t (1 n ) and hence, maj t ( k )( z s ) = 0. This finish es the pro of of (3), and th u s the pro of of the lemma. Lemma 2.3. If 0 < k ≤ 2 t + 1 ≤ n , then inst ( maj t ( k )) ≥ 2 t + 1 . Pr o of. W e will pr o ve the follo wing stronger assertion. Ther e exists a ( t + 1) -ge o desic P such that inst ( maj t ( k ) , P ) ≥ 2 t + 1 and the last p oint of P is c olour e d 0 . (4) W e shall pro ce ed b y induction on k . O b serv e that (4 ) h olds for k = 1 and for k = 2. Indeed, for k = 1, consid er the follo w ing ( t + 1)-geo d esic. P = ( 1 t +1 0 n − t − 1 , [1] 1 t 00 n − t − 1 , [0] 1 t 010 n − t − 2 , [1] 1 t − 1 0 2 10 n − t − 2 , [0] 1 t − 1 0 2 1 2 0 n − t − 3 , [1] 1 t − 2 0 3 1 2 0 n − t − 3 , [0] 1 t − 2 0 3 1 3 0 n − t − 4 , [1] . . . 110 t − 1 1 t − 2 0 n − 2 t +1 , [0] 110 t − 1 1 t − 1 0 n − 2 t , [1] 10 t 1 t − 1 0 n − 2 t , [0] 10 t 1 t 0 n − 2 t − 1 , [1] 0 t +1 1 t 0 n − 2 t − 1 , [0] 0 t +1 1 t +1 0 n − 2 t − 2 , [0] 0 t +1 1 t +2 0 n − 2 t − 3 , [0] 0 t +1 1 t +3 0 n − 2 t − 4 , [0] . . . 0 t +1 1 n − t − 1 ) [0] . Note that P jump s 2 t + 1 times. F or k = 2, consider either P , or the ( t + 1)-geodesic P ′ obtained from P b y changing the t wo p oin ts 6 10 t 1 t − 1 0 n − 2 t and 10 t 1 t 0 n − 2 t − 1 to 010 t − 1 1 t − 1 0 n − 2 t and 010 t − 1 1 t 0 n − 2 t − 1 . If maj t ( k )(1 0 t 1 t − 1 0 n − 2 t ) = 0, w e choose P , otherwise we choose P ′ . So supp ose we are given a k ≥ 3. Then t ≥ 1 and n ≥ 3. Consider maj t ( k ) on ˜ H n := { x ∈ H n : x (1) = 0 and x (2) = 1 } , and ob s erv e that this is equiv alen t to considering maj t − 1 ( k − 2) on H n − 2 . Hence, by induction, w e kn o w there exists a t -geo desic ˜ P in H n − 2 that is as in (4) for t − 1. In particular, ˜ P jumps at least 2( t − 1) + 1 = 2 t − 1 times. Abusing notation s lightly , w e shall consid er ˜ P as a path in ˜ H n . No w we extend ˜ P t o a geo desic in H n adding tw o more jum p s. By (4), w e know that ˜ P ends in a p oin t y with maj t ( k )( y ) = 0, and with exactly t + 1 entries equ al to 0 (among these the first en try). Supp ose the fi rst p oin t of ˜ P , let u s call this p oin t a , is coloured 1 in maj t ( k ). Then we ad d the p oin ts y ′ := (1 , 1 , y (3) , y (4) , . . . ) and y ′′ := (1 , 0 , y (3) , y (4) , . . . ) to the end of P ′ and obtain a geo desic P as desired. Indeed, y ′ ∈ B t (1 n ) as y ′ has exact ly t 0’s, hence maj t ( k )( y ′ ) = 1, and so w e h av e ou r fir st extra jump . Note that y ′′ has exactly as m any 1’s as y (in particular, y ′′ / ∈ B t (1 n )), and moreo v er, y ′′ has exact ly as man y 1’s in the first k entries as y . Thus, maj t ( k )( y ′′ ) 6 = maj t ( k )( y ) on ly if k is ev en and y and y ′′ ha ve as many 0’s as 1’s in their first k en tries. But in this ca se, the definition of maj t ( k ) implies that 0 = maj t ( k )( y ′′ ) 6 = maj t ( k )( a ) = 1. Therefore, maj t ( k )( y ′′ ) = 0, and we ha ve the second extra ju mp, implying that P is as in (4). It r ema ins to analyse the case wh ere maj t ( k )( a ) = 0. In this case, n ot e that as ˜ P starts and fin ishes with colour 0, it jumps an ev en n um b er of times, that is, at least 2( t − 1) + 2 = 2 t times. Th u s w e need to add only one more ju mp. If we bu ild P in the same w ay as ab o v e, P jump s at least 2 t + 1 times, bu t for the same reasons as ab o v e, it en d s in a p oint coloured 1. So, instead, let P b e obtained from ˜ P by adding at its b eginning the t wo p oint s a ′′ := (1 , 0 , a (3) , a (4) , . . . ) an d a ′ := (1 , 1 , a (3) , a (4) , . . . ). O bserv e that since a ′′ is the complemen t of y , it has the opp osite colour, i.e., maj t ( k )( a ′′ ) = 1. Hence b et ween a ′′ and a we hav e at least one ju mp. So P is a well-e nding ( t + 1)-geo desic with at least 2 t + 1 jum p s, as desired. 2.2 The partition colourings W e p r esen t a second class of colourings, the colourings b k t , wh ic h resp ect the balls B t (0 n ) and B t (1 n ) and h a ve instabilit y 2 t + 1. Before that, we defin e the auxiliary colouring a Q j that will b e used in the d efinition of b k t . 7 Let m , s an d t b e suc h that m ≥ ( s + 1)( t + 1). Let Q b e a partition of [ m ] in to s + 1 sets of size at least t + 1 eac h. F or j = 0, 1, let a Q j b e the follo wing colouring of H n . Let a Q j ( x ) = j if and only if, in at least one of the sets in Q , all en tries are j . It is not d ifficult to see that a Q j resp ects b oth B s ( j m ) and B t ((1 − j ) m ). Consider a geo desic P = ( x 0 , x 1 , . . . , x m ) in H m . Note that, if i is a jump of P in a Q j , then for some set Q in Q we ha ve that x ℓ ( q ) = j f o r all q ∈ Q either for ℓ = j − 1 or for ℓ = j , but not for b oth. W e sa y that the jump i is asso cia ted to th is set Q . T hus there are at most t w o ju mps in P associated to th e same set Q in Q . This implies that a Q j jumps at most 2 |Q| = 2( s + 1) times. No w, let k , s , t , and n b e suc h that k is o dd, s ≥ − 1, t = s + ( k + 1) / 2, and n ≥ ( s + 1)( t + 1) + k . Note that k ≤ 2 t + 1 b ecause s ≥ − 1. Let Q b e a partition of [ n − k ] into s + 1 sets of size at least t + 1 eac h. (If s = − 1, then n = k and Q = ∅ .) W e shall define the colouring b k t = b k t ( Q ) using a Q 0 and a Q 1 . W e abu se notation and assu me that a Q j ( y ) = 1 − j if Q or y is emp t y . F or ea c h p oin t x , if th e ma j orit y of the firs t k entries of x is 1, then let b k t ( x ) = a Q 0 ( x ′ ), where x ′ is x withou t the first k en tries. If the ma jority of the first k entries of x is 0, then let b k t ( x ) = a Q 1 ( x ′ ). In b oth cases, we sometimes abuse n o tation an d write that b k t = a Q j in x . It is not d iffi cu lt to see that b k t resp ects the balls B t (0 n ) and B t (1 n ). Indeed, let us supp ose th e ma jorit y of the first k en tries of some p oin t x is 1, and h ence b k t = a Q 0 (the other case is symm etric). If x has at m o st t en tries equal to 0, clearly n o set in Q can only consist of 0’s, and so b k t ( x ) = 1. On th e other hand, if x h as at most t 1’s, th en x ′ has at most t − ( k + 1) / 2 = s 1’s and ther efore, as |Q| = s + 1, there is a set in Q that only consists of 0’s. Th us b k t ( x ) = 0 in this case. Hence, in either case, b k t ( x ) is as desired. Observe that, for t = 0 and k = 1, we h av e s = − 1, and hence n = 1. In this case, b 1 0 = maj 0 (1). Prop osition 2.4. L et k , t , n ∈ N b e suc h tha t k is o dd, k ≤ 2 t + 1 and n ≥ ( t + 1 − k +1 2 )( t + 1) + k = ( t +1)(2 t +1) − k ( t − 1) 2 . Then inst ( b k t ) = 2 t + 1 . Pr o of. Let P b e a geod esic in H n . T o pro ve that P ju mps at most 2 t + 1 times in b t k , firs t note that at most k jump s of P are asso ciated to its fir s t k en tries. Second, note that P h as at most tw o j u mps associated to eac h set Q in Q . Indeed, if P has one jump asso ciat ed to Q while b k t = a Q j , th en P has at most one more jum p asso ciated to Q while b k t = a Q 1 − j . Sim ilarly , if P has 8 t wo jumps asso ciated to Q wh ile b k t = a Q j , then P has no ju mps asso ciat ed to Q wh ile b k t = a Q 1 − j . Also, it is n ot hard to find a geo desic in H n that ju mps 2 t + 1 times in b k t . Consider a p oin t x 0 with ( k + 1) / 2 1’s in the first k en tries, and exactly one 1 in eac h of the sets in Q . Then x 0 has exactly t + 1 entries equal to 1. T ake a geo d esic that starts in x 0 , and jumps k times by c hanging alternativ ely 1’s to 0’s and 0’s to 1’s w it hin the first k en tries. After that, w e ha v e that b k t = a Q 1 . So we can ju m p twice p er set Q in Q by changing all entries in Q to 1 firs t, and then c h anging the unique entry in Q that started with a 1 to a 0. 3 W ell-ending geo desics and k - d efined colourings A geo desic in H n is called an m -ge o desic if it starts in a p oin t of H n whic h has exactly m entries equal to 1. (It then ends in a p oin t which has exactly m en tr ies that equal 0.) Let f b e a colouring of H n and t f b e the maxim u m t such that f resp ects B t (0 n ) and B t (1 n ). (If f (0 n ) = 1 or f (1 n ) = 0, th en set t f = − 1.) If P is a geod esic whose first p oin t is coloured 1 in f , or whose last p oin t is coloured 0 in f , we say P ends wel l (in f ). Let winst ( f ) d enot e the maxim u m v alue of inst ( f , P ), tak en ov er all wel l-ending ( t f + 1)-geodesics P . In analogy to Pr o blem 1.1, we ask the follo wing. Problem 3.1. Given t valid for n , which is the smal lest value winst ( n, t ) such that winst ( n , t ) = winst ( f ) for some c olouring f with t f = t ? Observe that winst ( f ) ≤ inst ( f ) for ev er y colo uring f . Moreo ver, min t ≤ s ≤ ( n − 1) / 2 { winst ( n , s ) } ≤ inst ( n, t ) f or all t v alid for n . Call a colouring f k -define d 1 if there are k indices such th at f ( x ) = f ( y ) for an y tw o p oin ts x , y ∈ H n \ ( B t (0 n ) ∪ B t (1 n )) that coincide in all en tries giv en by these k ind ic es. A k -defined colouring that is not ( k − 1)-defined is called strictly k -define d . F or instance, maj t ( k ) is strictly k -defined and a Q 0 is strictly n -d efi ned. Let t b e v alid for n . F or the next lemma, let F n ( t ) denote the set of all strictly n -defined colourings f of H n with t f = t , and let F t + 2 s ≥ t 0’s and at least t 1’s, so, this part of P en ters th e balls exactly when P ′ do es. Thus the part of P that goes through the p oints p i 0 n − k − t + s 1 t − s for i = 0 , . . . , k jumps exactly when P ′ do es, that is, at least g ( s ) times. So, by our assumption on g , it follo ws that P jump s at least 2( t − s ) + g ( s ) ≥ g ( t ) times, co mpleting the pro of of the lemma. 12 4 Lo w er b ounds on ins t ( n, t ) and w inst ( n, t ) 4.1 The zig-zag b ound In this section w e p ro v e lo wer b ounds f or inst ( n, t ) and winst ( n, t ). Recall that any low er b ound on winst ( f ) also serv es as a lo w er b ound for inst ( f ). W e start with a b ound for all v alues of n and v alid t , whic h we obtain from a fairly basic zig-zag argumen t. Later, in Theorem 4.2 and Prop ositio n 4.5, the b ounds f rom Theorem 4.1 will b e impro v ed for the sp ecial cases n = 2 t + 2 an d n = 2 t + 3. Theorem 4.1 (The zig-zag b ound) . L et n ∈ N and let t ≥ 0 b e valid for n . Then (a) winst ( n, t ) ≥ ⌊ t n − 2 t ⌋ + ⌈ t n − 2 t ⌉ + 1 , (b) inst ( n, t ) ≥ ⌊ t − 1 n − 2 t ⌋ + ⌈ t − 1 n − 2 t ⌉ + 3 , i f t ≥ 1 . W e remark that Th eo rem 4.1 (a) pr o ves Conjecture 1.2 for t = 0 and Theorem 4.1 (b) prov es Conjecture 1.2 for t = 1. This has b een sho w n earlier in [5]. W e dedicate th e r est of th is su bsecti on to the pr oof of Theorem 4.1. Pr o of of The or em 4.1. Let f b e a colouring with t f ≥ 0 and let t = t f . F or (a), our aim is to find a w ell-ending ( t + 1)-geodesic P that jumps at least ⌊ t n − 2 t ⌋ + ⌈ t n − 2 t ⌉ + 1 times in f . As t = t f , there is a p oin t x ∈ H n that h a s exactly ( t + 1) 1’s or ( t + 1) 0’s, an d that is coloured 1 or 0, resp ectiv ely . Sa y the f orm er holds for x (the other case is sy m metric) . W e let P start in x , th en ent er B t (0 n ), then go to B t (1 n ), come b ac k to B t (0 n ), go to B t (1 n ) again, etc ., u n til P h as used up all of its en tries. F or example, if x = 1 t +1 0 n − t − 1 , we let P pass n ext through 1 t 0 n − t and then through 1 t 0 t 1 n − 2 t , throu gh 1 t − ( n − 2 t ) 0 n − t 1 n − 2 t , throu gh 1 t − ( n − 2 t ) 0 t 1 2 n − 4 t , and so on. W e can do this un til one of th e follo wing tw o things happ ens. Firstly , coming from B t (0 n ), w e might en d in the complemen t of x with ( t + 1) 0’s (just b efore r ea c hing B t (1 n )). T his will happ en exactly w hen n = ℓ ( n − 2 t ) for some o dd ℓ , whic h is the case if and only if n − 2 t divides t . Th en w e will ha v e jump ed at least ℓ times and ℓ = n n − 2 t = 2 t n − 2 t + 1 =  t n − 2 t  +  t n − 2 t  + 1 . 13 Secondly , on our w a y fr om B t (1 n ) to B t (0 n ), w e might reac h a p oin t of H n \ ( B t (0 n ) ∪ B t (1 n )) wh ic h has no more u n used 1’s. Th is happ ens if and only if n − 2 t d oes n ot divide t . Then we ha ve to return in the d irect ion of B t (1 n ) to end in the complement of x (if w e are not already there). In this case, we hav e jump ed at least 1 + 2 ·  n − ( n − 2 t ) 2( n − 2 t )  + 1 = 2 ·  t n − 2 t  + 2 =  t n − 2 t  +  t n − 2 t  + 1 times, b ecause at least one j ump is ac h ie v ed durin g the first n − 2 t steps, then w e get at least 2 ju mps for eve ry 2( n − 2 t ) steps, and fi nally w e jump at least once more in the last part of P wh en going through B t (1 n ). Note that, b y the construction of P , we ha ve to end up in one of the tw o situations just describ ed. Th is completes the p roof of (a). F or (b), the pro of is similar, the difference b eing that w e let P start inside B t (0 n ), h a ve x as its second p oin t, then re-en ter B t (0 n ), and then go on in a zig-zag fashion as b efore. W e will obtain 2 jumps in the b eginning, at least on e jump dur in g the next n − 2 t steps of P , and then 2 ju mps eve ry 2( n − 2 t ) s te ps. Finally w e migh t ensu re another jum p dep ending on whether n − 2 = ℓ ( n − 2 t ) for some o dd ℓ or not. More precisely , if n − 2 = ℓ ( n − 2 t ) for some o dd ℓ , that is, if n − 2 t d ivides t − 1, then we get ℓ + 2 = n − 2 n − 2 t + 2 =  t − 1 n − 2 t  +  t − 1 n − 2 t  + 3 jumps, and otherwise, we also get 2 + 1 + 2 · ⌊ n − 2 − ( n − 2 t ) 2( n − 2 t ) ⌋ + 1 =  t − 1 n − 2 t  +  t − 1 n − 2 t  + 3 jumps, whic h is as d esired. Clearly , we n eed here that t ≥ 1, b ecause otherwise we could not enter B t (0 n ) t w ice in the b eginning. 4.2 Better b ounds for one strip In this su bsectio n w e will concen trate on th e case wh en H n con tains, b esides the b al ls, only one ‘strip ’ of p oin ts w hic h all ha ve the same num b er of en tries equal to 0 and equ a l to 1. That is, we treat the case n = 2 t + 2. F r om Th eo rem 4.1 , we ha ve that winst (2 t + 2 , t ) ≥ t + 1 and inst (2 t + 2 , t ) ≥ t + 2 for t ≥ 1. T he follo win g result impr o ves th is b ound. Theorem 4.2. i nst (2 t + 2 , t ) ≥ winst (2 t + 2 , t ) ≥ t + 3 for al l t ≥ 2 . 14 W e will pro v e Th eo rem 4.2 by com bining the next t wo lemmas. T he first of these is a to ol for extending b ound s for small v alues of t to larger v alues of t . Lemma 4.3. L et y 0 , t 0 and t ∈ N with t ≥ t 0 . If winst (2 t 0 + 2 , t 0 ) ≥ y 0 for some t 0 ≥ 0 then winst (2 t + 2 , t ) ≥ y 0 + t − t 0 . Pr o of. W e pro ceed b y ind uction on t . The base, for t = t 0 , follo ws directly from the h yp othesis of the lemma. F or t > t 0 , consider a colouring f with t f = t of the hyper cu be H n of d imension n = 2 t + 2. Define a colouring g of the h y p ercub e H n − 2 b y assig ning to eac h x ′ in H n − 2 the v alue g ( x ′ ) = f (01 x ′ ). Then g is suc h that t g = t − 1. Indeed, an y p oin t of H n − 2 \ ( B t − 1 (0 n − 2 ) ∪ B t − 1 (1 n − 2 )) is a w itness to this. W e may th u s apply the induction h yp othesis to obtain a w ell-ending t -geo desic ˜ P in H n − 2 that jump s at least y 0 + t − 1 − t 0 times in g . E xte nding eac h p oin t x ′ of ˜ P to the p oint 01 x ′ of H n , w e obtain a path P ′ in H n that j umps at least y 0 + t − 1 − t 0 times in f . Let 01 a and 01 z b e the first and last p oin t of P ′ resp ectiv ely . Note that since ˜ P is well -ending, either g (01 a ) = 1 or g (01 z ) = 0 (or b oth). W e extend P ′ to P by add ing to its b eginning the p oin ts 00 a and 10 a , if g (01 a ) = 1, and the p oint s 11 z and 10 z to its end otherwise. As we th u s pass once more through either B t (0 n ) or B t (1 n ), our extension P of P ′ jumps at least once more than P ′ , that is, y 0 + t − t 0 times in total. Clearly , P is a ( t + 1)- geod esic , and so is its reverse, b ecause n = 2 t + 2. No w at least one of the t wo, P or its rev erse, has to b e well-e nding, wh ic h completes th e pr oof of the lemma. The next lemma take s care of the base case t = t 0 for Lemma 4.3. It also confirms Conjecture 1.2 for n = 2 t + 2 and small v alues of t . Lemma 4.4. winst (2 t + 2 , t ) ≥ 2 t + 1 for t = 0 , 1 , 2 . Pr o of. Th e case t = 0 is trivial. F or t = 1, let f b e a colouring of H 4 suc h that t f = 1. Note that there are tw o p oin ts x and y in H 4 with exactly t + 1 = 2 en tries equal to 1, differin g in exactly t w o entrie s (that is, su c h that || x − y || 2 = 2), and s uc h that f ( x ) = f ( y ). F or example, t wo of the three p oints 1100, 1010, 1001 must ha ve the same colour in f . No w it is easy to constru ct a w ell-ending 2-geo desic that starts in x and jump s at least three times. F or t = 2, let f b e a colouring of H 6 suc h that t f = 2. Observe that w e only n ee d to fi nd thr ee p oint s x , y , z , all with exactly t + 1 = 3 entries equal to 1, suc h that || x − y || 2 = || y − z || 2 = 2, || x − z || 2 = 4, and f ( x ) = f ( y ) = 15 f ( z ). In deed, if we ha v e suc h p oin ts, it is easy to construct a well-ending 3-geodesic that starts in x and jumps at least fi v e times. The pro of of the existence of x , y and z is a case analysis. By rearranging the order of the entries, w e may assume the p oin ts x = 111000 and y = 11010 0 hav e the s a me colour j in f . If on e among x ′ = 10011 0, y ′ = 10010 1 and z ′ = 010101 has colour j , then we may tak e it as our third p oint z . I f not, then x ′ , y ′ and z ′ all ha v e colour 1 − j and form a triple of p oin ts as desired. Pr o of of The or em 4.2. Th e statemen t is an imm ediat e consequen ce of Lemma 4.3 and Lemma 4.4 for t = 2. 4.3 The extension metho d for more str ips W e no w extend the results from the previous subsection to the general case, when we ha v e more ‘strips ’. The main resu lt of this subsection, Prop osi- tion 4.5, is an extension of Lemma 4.3. Prop osition 4.5. L et n , y 0 , t 0 ∈ N and let t ≥ t 0 b e v alid for n and such that n − 2 t divides t − t 0 . If winst ( n, t 0 ) ≥ y 0 , then winst ( n, t ) ≥ y 0 + 2 t − t 0 n − 2 t . Clearly , Pr op osition 4.5 ca n b e u s ed in the same wa y as Lemma 4. 3 to impro v e Theorem 4.1. T he next lemma tak es care of the b ase case t = t 0 for Prop osition 4.5, for the case n = 2 t + 3. It also confi r ms Conjecture 1.2 for n = 5 and t = 1. Lemma 4.6. winst (5 , 1) ≥ 3 , winst (7 , 2) ≥ 4 , and winst (9 , 3) ≥ 4 . Pr o of. W e start proving that winst (5 , 1) ≥ 3. Let f b e a colouring of H 5 with t f = 1. W e sa y a p oin t x in H 5 is go o d (in f ) if th ere is a j ∈ { 0 , 1 } so that x has exactly t w o en tr ie s equal to j and f ( x ) = j . Also, we sa y th at t wo p oin ts x and y in H 5 are neig h b ours if || x − y || 2 = 2 and they h a ve the same num b er of en tries equal to 1. First of all, w e observe that, if there are t w o go o d p oin ts x and y that are neighbour s, then it is easy to construct a wel l-ending 2-geo desic that jumps the required n um b er of times (in th e same wa y as in Lemma 4.4). So w e ma y assume that if x and y are go o d in f , th en they are not neigh b ours. (6) Second, we may assum e that if x is go od in f then its complement is not go od in f . (7) 16 Indeed, if a p oint x and its complemen t are go od in f , then we ma y obtain a w ell-ending 2-geod esic as desired b y starting out at x , going to B 1 ( j 5 ), then going to B 1 ((1 − j ) 5 ), and then end ing at the complemen t of x . As t f = 1, there is a p oin t w that is go od in f . By symmetry , w e can assume that w = 00011. No w, b ecause of (6), at most one of the p oin ts 11000 , 10100 and 01100 is goo d. So, at least one of them, sa y 11000 , has colour 0 in f . Consid er the 2-geodesic (0001 1[1] , 00001[0] , 01001[0] , 11001[?] , 11000[0] , 11100[1]) . Its third p oin t has colour 0 b ecause of (6) and its last p oint has colour 1 b ecause of (7). So this well- ending 2-geodesic only ju m ps less than three times if f (1100 1) = 0. Bu t in this case, w e u se (6) to see that f (1101 0) = 1, and consider the well-e nding 2-geo desic (0001 1[1] , 00010[0] , 10010[0] , 11010[1] , 11000[0] , 11100[1]) , that ju mps 4 > 3 times. (Again, f (10010 ) = 0 b ecause of (6).) T h is concludes the pro of that winst (5 , 1) ≥ 3. The idea for the other cases is similar to the one u sed in the pro of of Lemma 4.3. W e red uce the problem to 5 en tries, obtaining as ab o v e a ‘partial’ geod esic that jumps at least thr ee times, and extend it so that it jumps least four times, as needed. F or winst (7 , 2), let f b e a colouring of H 7 with t f = 2. Let w be a p oin t in H 7 with exactly 3 entries equal to j and such that f ( w ) = j . By symmetry , w e ma y assume that the firs t t wo en tries of w are 01. Define a colouring g of the h yp ercub e H 5 b y assigning to eac h x ′ in H 5 the v alue g ( x ′ ) = f (01 x ′ ). Then g is su c h that t g = 1. Indeed, the p oin t w ′ in H 5 suc h that w = 01 w ′ serv es as a witness to this. As winst (5 , 1) ≥ 3, there is a we ll-ending 2-geo desic ˜ P in H 5 that jump s at least three times in g . Extend ing eac h p oin t x ′ of ˜ P t o the p oint 01 x ′ of H 7 , w e obtain a path P ′ in H 7 that j umps at least three times in f . If ˜ P jumps exactly three times, then it ends w ell in b oth of its end s. Thus we can extend P ′ in on e of its ends, passing by the neigh b ouring ball, so that it jumps once more, and the result will b e a well-ending 3-geo d esic as desired. If, on the other h and, ˜ P jumps at least four times, then we ju st extend it in an y wa y so that the resulting geo desic is s till wel l-ending. This completes the p roof that winst (7 , 2) ≥ 4. T he pro of that winst (9 , 3) ≥ 4 is similar, so w e omit it. 17 Corollary 4.7. L et t ≥ 1 . Then winst (2 t + 3 , t ) ≥ 2 + 2 t + ( t mo d 3) 3 . Pr o of. W e obtain the b ound by app lying Prop osition 4.5 to n = 2 t + 3 and the base cases obtained f rom Lemma 4.6: t 0 = 1 with y 0 = 3, t 0 = 2 with y 0 = 4, and t 0 = 3 with y 0 = 4. This b ound improv es b y one the b ound fr om Theorem 4.1 (a) for n = 2 t + 3 and t mo d 3 = 0 or 1, and by tw o for t mo d 3 = 2. The rest of this section is dedicated to th e p roof of Pr oposition 4.5. Pr o of of Pr op osition 4.5. W e p roceed by induction on i = i ( n, t ) := t − t 0 n − 2 t . The base, for i = 0 (i.e., t = t 0 ), follo ws directly fr om the hyp othesis of the lemma. F or i > 0, consid er a colouring f of the hyp ercub e H n with t f = t . By the defin iti on of t f , there is an x in H n with exactly t + 1 entries equal to f ( x ). As t is v alid f o r n , we kno w that x h as at least t en tries equal to 1 − f ( x ). So, as n − 2 t ≤ t − t 0 ≤ t , we ma y assu me that x = 0 n − 2 t 1 n − 2 t x ′ , where x ′ ∈ H n ′ for n ′ := n − 2( n − 2 t ). Define a colouring g of the hyp ercub e H n ′ b y assigning to eac h x ′′ in H n ′ the v alue g ( x ′′ ) = f (0 n − 2 t 1 n − 2 t x ′′ ). Th en g is suc h that t g = t ′ := t − ( n − 2 t ). Indeed, g resp ects the balls B t ′ (0 n ′ ) and B t ′ (1 n ′ ) b ecause f resp ects the balls B t (0 n ) and B t (1 n ), and the p oint x ′ has exactly t − ( n − 2 t ) + 1 = t ′ + 1 en tries equal to g ( x ′ ) = f ( x ). Note that t ′ is v alid for n ′ and that n ′ − 2 t ′ = n − 2 t divides t ′ − t 0 . Moreo v er, i ( n ′ , t ′ ) = t − t 0 − ( n − 2 t ) n − 2 t = i ( n, t ) − 1 . So, w e ma y apply the induction hypothesis to H n ′ and g to obtai n a w ell- ending ( t ′ + 1)-geo desic ˜ P in H n ′ that jumps at least y 0 + 2 t − t 0 n − 2 t − 2 times in g . W e su pp ose that the fi rst p oin t ˜ a of ˜ P is suc h that g (˜ a ) = 1. In other w ord s, w e sup pose that ˜ P ends w ell in its firs t p oin t. The other case is analogous. Extending eac h p oin t x ′′ of ˜ P to th e point 0 n − 2 t 1 n − 2 t x ′′ of H n , we obtain a path P ′ in H n that jumps at least y 0 + 2 t − t 0 n − 2 t − 2 times in f . Let z = 0 n − 2 t 1 n − 2 t z ′ b e the last p oin t of P ′ . If f ( z ) = 0, then we extend P ′ to P b y adding to its end the p oin ts 0 n − 2 t − 1 1 n − 2 t +1 z ′ , 0 n − 2 t − 1 101 n − 2 t − 1 z ′ , 18 0 n − 2 t − 1 10 2 1 n − 2 t − 2 z ′ , . . . 0 n − 2 t − 1 10 n − 2 t z ′ , 10 n − 2 t − 2 10 n − 2 t z ′ , 1 2 0 n − 2 t − 3 10 n − 2 t z ′ , . . . 1 n − 2 t 0 n − 2 t z ′ . As we th us p ass once through B t (1 n ), and then through B t (0 n ), our geod esic P jumps at least t wo times more than P ′ . On the other hand , if f ( z ) = 1, then we extend P ′ to P b y add ing to its end the p oint s 0 n − 2 t +1 1 n − 2 t − 1 z ′ , 0 n − 2 t +2 1 n − 2 t − 2 z ′ , 0 n − 2 t +3 1 n − 2 t − 3 z ′ , . . . 0 2 n − 4 t − 1 1 z ′ , 10 2 n − 4 t − 2 1 z ′ , 1 2 0 2 n − 4 t − 3 1 z ′ , . . . 1 n − 2 t 0 n − 2 t − 1 1 z ′ , 1 n − 2 t 0 n − 2 t z ′ . W e passed once through B t (0 n ), and then thr ough B t (1 n ), thus again our geo desic P has at least t wo more jumps than P ′ . So, in either case, P ju mps at least y 0 + 2 t − t 0 n − 2 t times in tota l. By con- struction, P is a well-e nding ( t + 1)-geo desic, as d esired. References [1] Aspn es, J., Busc h, C., Dole v, S., F atourou, P ., Ge orgiou, C., Shv arts- man, A., Spirakis, P ., and W attenhofer, R., Eight op en pr oblems in distribute d c omputing , Bullet in of th e Europ ean Asso ciation for Th eo - retical Comp uter Science, Distributed Computing Column, 90:109–12 6, Octob er 2006. 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