The Mixing Time of Glauber Dynamics for Colouring Regular Trees

The Mixing Time of Glaub er Dynamics for Colouring Regular T rees ∗ Leslie Ann Goldb erg † , Mark Jerrum ‡ and Marek Karp i nski § No v em b er 27, 202 1 Abstract W e consider Metrop olis Glaub er dynamics for sampling prop er q - colourings of the n -ve rtex complete b -ary tree w hen 3 ≤ q ≤ b/ 2 ln( b ). W e giv e b oth upp er and lo w er b ound s on the mixing time. F or fixed q and b , our u pp er b ound is n O ( b/ log b ) and our lo wer b ound is n Ω( b/q log( b )) , wh ere the constan ts imp licit in the O () and Ω() notation do n ot dep end u p on n , q or b . 1 In tro du c tion This pap er pro ve s b oth upp er and lo w er b ounds on the mixing time of Glaub er dynamics for colourings of regular t r ees. It answ ers in particular the question of Hay es, V era and Vigo da [10], asking whether the mixing time of Glaub er dynamics is sup er-p olynomial for the complete b -a r y tree with q = 3 and b = O (1 ). W e show that the mixing time is not sup er-p olynomial — it is n Θ( b/ log( b )) . More generally , w e consider Me trop olis G laub er dynamics for s ampling prop er q -colourings of the n -ve rtex complete b -ar y tree when 3 ≤ q ≤ b/ 2 ln( b ). W e giv e b ot h upp er and low er b ounds on the mixing time, pinning do wn the dep endance o f the mixing time on n , b and q . F or fixed q and b , our upp er b ound is n O ( b/ log b ) and our lo w er b ound is n Ω( b/q log( b )) , where the constan ts implicit in the O ( ) and Ω() notation do no t dep end up on n , q or b . ∗ Partly fund ed by the EPSRC grant s EP/E06 2 482/ 1 and EP/E06 4906/ 1 † Department o f Computer Science, Universit y o f Liverp o ol, Liverpo o l L69 3BX, UK ‡ School of Mathematical Sciences, Queen Mary , University of London, Mile End Road, London E1 4NS, UK § Department of Computer Science, Universit y of Bonn, R¨ omerstr. 164, 5 3117 Bo nn, Germany 1 2 Previous w ork There has b een quite a bit of w ork on Marko v c hains for sampling the prop er q -colo urings of an input graph. Muc h of this work fo cusses on Glaub er dy- namics , which is a general term fo r a Mark ov c hain whic h up da t es the colour of one v ertex a t a time. Proper colourings corresp ond to configurations in the zer o-temp er atur e Potts mo del from statistical ph ysics, a nd there is a close connection b etw een the mixing time of Glaub er dynamics and the qualita tiv e prop erties of the mo del. In particular, rapid mixing, sp ecifically O ( n log( n )) mixing f o r an n - v ertex sub-gra ph of an infinite graph, often coincides with the case in whic h the mo del has a unique infinite-volume Gibbs measure o n the infinite graph. See W eitz’s PhD t hesis [19] and Martinelli’s lecture notes [12] f o r an exp osition of this material. Martinell, Sinclair and W eitz [13] consider G laub er dynamics on the com- plete n -v ertex tree with branc hing factor b . They sho w that for q ≥ b + 3, Glaub er dynamics f o r sampling prop er q - colourings mixes in O ( n lo g( n )) time for a rbitrary b oundary conditions. This result is optimal in the sense that for q ≤ b + 2 t here ar e b oundary conditions for whic h Glaub er dynamics is not eve n ergo dic. It is also in teresting to determine whether Glaub er dynamics is rapidly mixing for smaller q in the absen ce of boundary conditions. Ha yes , V era, and Vigo da [10 ] show ed that there is a C > 0 suc h t hat for all q > C ( b + 1) / log ( b + 1), Glaub er dynamics mix es in p olynomial time. In f act, their result applies to a ll planar graphs with maxim um degree b + 1. They ask in Section 6 whether the mixing time is sup er-p olynomial for the complete b -ary tree with q = 3 and b = O (1). As noted ab ov e, we show giv e upp er and low er b ounds sho wing tha t the mixing time is p olynomial in this case. As noted in [10], the only previous rapid mixing r esults for q < b + 1 were for 3-colourings of finite subregions of the 2-dimensional in teger lattice [9, 11] and random graphs [6]. [3] considers r e c onstruction on the complete tree with branc hing factor b . They show that for C = 2 and q > C ( b + 1) / ln( b + 1) non-reconstruction holds, meaning that, o v er random colourings of the lea v es, t he exp ected influence on the ro ot is v anishing. It is kno wn [14] that the exp ected influence is non-v a nishing f or a sufficien tly la r g e q satisfying q ≤ (1 − ε )( b + 1) / ln( b + 1) for some ε > 0. This non-v anishing influence implies [2, 3] that the mixing time o f Glaub er dynamics exceeds O ( n log( n )). In The orems 1 and 2, w e giv e upper and low er bo unds for the mixing time for fixed q and b when 3 ≤ q ≤ b/ 2 ln( b ). Our upp er b ound is n O ( b/ log b ) and our lo w er b ound is n Ω( b/q log( b )) . 2 3 Pro o f tec hn iques The upp er b ound ar gumen t is based o n canonical paths. The low er b ound argumen t is based on conductance. Essen tially , the argumen t is that it tak es a while to mov e from a colouring in whic h the colour of the ro ot is for ced to b e one colour by the induced colouring on the leav es to a colouring in whic h the colour of the ro ot is forced to b e a nother colour. This is similar to the r e cursive majority idea [2] used to pro ve a lo w er b o und for the Ising mo del. 4 The problem Fix b ≥ 2 and fix q ≥ 3. Let [ q ] = { 0 , . . . , q − 1 } . Supp ose T is a complete b -ary tr ee of heigh t H — meaning that there are H edges on a path fro m the ro ot r ( T ) to any leaf. L et V b e the set of v ertices of T and n = | V | . Let L b e the set of lea v es of T . Note that n = b H +1 − 1 b − 1 so H = log(( b − 1) n + 1) log( b ) − 1 . (1) The hei g ht h ( v ) of a vertex v ∈ V is the n um b er of edges on a path fro m v do wn to a leaf. So a v ertex v ∈ L has h ( v ) = 0 and h ( r ( T )) = H . F or an y ve rtex v of T , T v denotes the subtree of T ro o ted a t v . F or an y subtree T v , let V ( T v ) b e the set of v ertices of T v and let L ( T v ) b e t he set of lea ves . A pr op er q -c ol o uring of T v is a lab elling of the v ertices with elemen ts of [ q ] suc h that neigh b ouring vertice s receiv e differen t colours. Let Ω( T v ) b e the set of prop er q -colour ing s of T v and Ω = Ω( T r ( T ) ) b e the set of prop er q -colo urings of T . F or a colo uring x ∈ Ω, let x ( T v ) denote the restriction of x to the v ertices in the subtree T v . Similarly , for a set U ⊆ V ( T v ) and a colouring x ∈ Ω( T v ), x ( U ) denotes the restriction of x to U . Let M be the Me trop olis Glaub er dynamics for sampling from Ω. T o mo ve from one colouring to another, t his c hain selects a v ertex v and a colour c uniformly at ra ndo m. The v ertex v is re-coloured with c if and only if this results in a prop er colouring. If q ≥ 3 then the set of prop er colouring s is connected and M conv erges to the uniform distribution o n Ω, whic h w e call π . The g oal is to study the mixing time of M as a function of n , b and q . Let P b e the transition matrix of M . The variation d i s tanc e b etw een 3 distributions θ 1 and θ 2 on Ω is || θ 1 − θ 2 || = 1 2 X i | θ 1 ( i ) − θ 2 ( i ) | = max A ⊆ Ω | θ 1 ( A ) − θ 2 ( A ) | . F or a state x ∈ Ω, the mixing time of M from start ing state x is τ x ( M , δ ) = min  t > 0 : || P t ′ ( x, · ) − π ( · ) | | ≤ δ for a ll t ′ ≥ t  . The mixing time of M is give n b y τ ( M , δ ) = max x τ x ( M , δ ) . Our results are as follows, where lg denotes t he ba se-2 logarithm and ln denotes the natural log arithm. Theorem 1. Supp os e q ≥ 3 . L et M b e the Metr op olis Glaub er dynamics fo r sampling pr op e r q -c olourings of the n -vertex c omplete b -ary tr e e. Then for fixe d q an d b the m ixing time τ ( M , 1 / ( 2 e )) is n O ( b/ log( b )) wher e the c onstant implicit in the O () notation do es not dep en d up on n , q or b . In p articular, τ ( M , 1 / (2 e )) ≤ 3 bq 2 (1 + lg ( n )) n 3+ 3 b ln( b ) . Theorem 2. Supp ose 3 ≤ q ≤ b/ 2 ln( b ) . L et M b e the Metr op olis Glaub er dynamics for sampling pr op er q -c olourings of the n -vertex c omplete b -ary tr e e. Then for fixe d q a n d b the mixing time τ ( M , 1 / (2 e )) is n Ω( b/q log( b )) wher e the c onstant i m plicit in t he Ω() notation do es not dep e n d up on n , q o r b . In p articular, τ ( M , 1 / (2 e )) ≥  1 2 − 1 2 e  2 9 n b − 2 6( q − 1) l n( b ) . 5 Bounds o n H The calculations a r ising in the deriv ation of Theorems 1 and 2 in volv e H . It is clear fr om Equation (1) that H = Θ(log ( n ) / log( b )). Since w e giv e explicit b ounds in t he statement of the theorems, w e also require upper a nd low er b ounds on H . W e record these here. Note that the b o unds can b e impro v ed, but w e prefer to av oid the complication. Lemma 3. H + 1 ≤ lg ( n ) + 1 and H ≤ ln( n ) / ln( b ) . If n ≥ b 3 then H − 1 ≥ ln( n ) / 3 ln( b ) . 4 Pr o of. F or the first upp er b ound, use Equation (1) to see that H + 1 = lo g b (( b − 1) n + 1) ≤ lo g b ( bn ) = 1 + log b ( n ) ≤ 1 + lg ( n ) , since n ≥ 1 and b ≥ 2. F o r the second upp er b ound, note tha t H = ln  n  b − 1 + 1 n  ln( b ) − 1 = ln( n ) ln( b ) − ln( b ) − ln( b − 1 + 1 / n ) ln( b ) ≤ ln( n ) ln( b ) . Finally , fo r the low er b ound, note that H − 1 = ln  n  b − 1 + 1 n  ln( b ) − 2 = ln( n ) ln( b ) + ln  b − 1 + 1 n  ln( b ) − 2 . Dropping the non-negative middle term, this is at least ln( n ) ln( b ) − 2 , whic h give s the result since ln( n ) / 3 ln( b ) ≥ 1. 6 The upp er b oun d In this section w e pro ve Theorem 1. W e will use the canonical paths metho d of Jerrum a nd Sinclair [18]. Let M ′ b e the trivial Marko v chain on Ω tha t mo ves from a state x t o a new state y by selecting y u.a.r. fro m Ω. Let P ′ b e the tra nsition matrix of M ′ . Clearly , for an y δ ′ > 0, τ ( M ′ , δ ′ ) = 1. w e will definine canonical paths b et wee n pairs of colouring s in Ω. Thes e canon- ical paths will constitute what is called a n ( M , M ′ )-flo w. Then Theorem 1 follo ws from the fo llowing prop osition (which is Observ ation 13 in the exp os- itory pap er [8]) taking A ( f ) to b e t he c ongestion of the flow and c to b e 1 /q . The pro o f o f Prop osition 4 com bines Diaconis a nd Saloff Coste’s comparison metho d [4] with upp er and lo w er b ounds o n mixing time [1, 5, 17] a long lines first prop o sed by Ra ndall and T etali [16]. See [8] for details. Prop osition 4. Supp ose that M is a r eversible er go dic Markov c h a in with tr ansition m a trix P and stationary distribution π and that M ′ is a n other r e- versible er go dic Markov c h ain with the sam e s tationa ry distribution. Supp ose that f is a ( M , M ′ ) -flow. L et c = min x P ( x, x ) , and assume c > 0 . Then, for any 0 < δ ′ < 1 / 2 , τ x ( M , δ ) ≤ max  A ( f )  τ ( M ′ , δ ′ ) ln(1 / 2 δ ′ ) + 1  , 1 2 c  ln 1 δ π ( x ) . 5 F or eac h pair of distinct colourings x, y ∈ Ω w e will construct a path γ x,y from x to y using tra nsitions of M . This giv es an ( M , M ′ )-flo w f with congestion A ( f ) = max z ,w 1 π ( z ) P ( z , w ) X x,y :( z ,w ) ∈ γ x,y | γ x,y | π ( x ) P ′ ( x, y ) = nq | Ω | max z ,w X x,y :( z ,w ) ∈ γ x,y | γ x,y | , (2) where the maximum is ov er pairs of distinct states z and w in Ω with P ( z , w ) > 0 (hence , P ( z , w ) = 1 /nq ) and | γ x,y | denotes the le ngth of γ x,y , whic h is the n um b er of transitions on the path. W e will prov e the follo wing lemma b elo w. Lemma 5. T he c an onic al p aths c orr esp on d to an ( M , M ′ ) -flow f with A ( f ) ≤ bq ( H + 1) n 2 9 bH . Theorem 1 follows . Combining Prop osition 4 with δ ′ = 1 / 2 e 2 and Lemma 5, w e get τ x ( M , δ ) ≤ bq ( H + 1) n 2 9 bH  1 2 + 1  ln( | Ω | /δ ) . Since | Ω | ≤ q n , τ x ( M , 1 / ( 2 e )) ≤ bq ( H + 1) n 2 9 bH  1 2 + 1  ln(2 eq n ) ≤ ( H + 1) bq n 2 3 2 (2 + n ln( q ))9 bH ≤ ( H + 1) bq 2 n 3 3 e 3 bH . Theorem 1 then follows by applying the t wo upp er b ounds in Lemma 3. Pro of of L emma 5 Defining the canonical paths: a sp ecial case W e start by defining paths b etw een colouring s x and y for the sp ecial case in whic h, for all v ∈ V , y ( v ) = x ( v ) + 1 (mo d q ). The sequence of colourings on the path is defined to b e the sequence of colourings visited b y pro cedure Cycle + b elo w when it is called with the input T , which is initia lly coloured x . Here is the description of pro cedure Cycle + ( b T ), where ˆ x is a global v ari- able, represen ting the curren t colouring of t r ee T , and the input parameter b T may b e any o f the subtrees T v . 6 1. Let b T 1 , . . . , b T b b e the subtrees root ed at the c hildren of r ( b T ) and let S =  i : x ( r ( b T i )) + 1 6 = x ( r ( b T )) (mo d q )  , 2. F or eac h i ∈ S do Cycle + ( b T i ). 3. Recolour the ro o t r ( b T ) so that ˆ x ( r ( b T )) = x ( r ( b T )) + 1 (mo d q ). 4. F or eac h i / ∈ S do Cycle + ( b T i ). Since q ≥ 3 , w e are guaran t eed that x ( r ( b T )) + 1 6 = ˆ x ( r ( b T i )) (mo d q ), for all i , after line 2; this ensures t hat the ro ot can b e recoloured in line 3. Analysis of the sp ecial case Supp ose w e observ e a tr ansition at some p o in t during the execution o f the pro cedure Cycle + ( b T ), in which the colouring ˆ x is tr a nsformed by adding 1 to the colour of some v ertex v ( mo dulo q ). Ho w many initial colouring s x ( b T ) (and hence ho w man y final colourings y ( b T )) are consis ten t with this observ ed transition? W e will let s ( h ) denote the maxim um n umber of consisten t initial colour- ings x ( b T ), maximised o ver all trees b T of heigh t h and o ver a ll p ossible tran- sitions. W e will compute an upp er b ound on s ( h ). Case 1: Supp ose that v = r ( b T ). The subtrees b T i with i ∈ S hav e already b een pro cessed by the time that the transition ta k es place, so ˆ x ( T i ) = y ( T i ) for these trees. The subtrees with i / ∈ S are y et to b e pr o cessed, so for these trees w e ha v e ˆ x ( T i ) = x ( T i ). How ev er, we do not kno w the set S from observing the transition from ˆ x . Thu s, as man y as 2 b initial colourings x ( b T ) ma y b e consisten t with the observ ed t r ansition from ˆ x . Case 2: Otherwise, v is in one of the subtrees b T k ro oted at one of the c hildren of r ( b T ). Then, b y the argumen t of Case 1, there are tw o choices for the initial colouring x ( T i ) of ev ery subtree with i 6 = k ; also there ar e t wo p o ssibilities for x ( r ( b T )), since w e don’t know whether line (3) has b een executed at the point o f the transition. Then s ( h ) s atisfies the recurrence s ( h ) ≤ max { 2 b , 2 b s ( h − 1) } with initial condition s ( 0) = 1. Solving the recurrence, we disco ve r that at most s ( h ) ≤ 2 bh (3) initial colourings x ( b T ) are consisten t with the observ ed tra nsition, so there are at most s ( H ) ≤ 2 bH initial colourings x of T consisten t with an observ ed transition of the pro cedure Cycle + ( T ) 7 Defining the canonical paths: the general case Let Cycle − b e defined analog ously to Cycle + but implemen ting the p erm u- tation of colours that subtracts 1 (mo dulo q ) from ev ery colour; that is, y ( v ) = x ( v ) − 1 (mo d q ) for all v ∈ V . Let F ⊂ [ q ] b e a set of “ forbidden colours” of size a t most tw o. Giv en Cycle + and Cycle − it is easy to implemen t a pro cedure Cycle ( b T , F ) that systematically recolours the tr ee b T so that the new colour assigned to r ( b T ) a voids the for bidden colo urs F : simply a pply Cycle + or Cycle − or neither in order to bring a colour not in F to the ro ot of b T . If w e observ e a transition during the execution of Cycle ( b T , F ) w e can tell whether it comes from Cycle + or fr o m Cycle − . The recursiv e pro cedure R e c olour , to b e describ ed presen tly , pro vides a systematic approac h to transforming an arbitr ary initial colouring x to an arbitrary final colouring y using single-ve rtex up dates. In doing so, it defines canonical paths b et w een arbitra r y pairs o f prop er colourings x and y o f T . The sequen ce of colouring s o n the path γ x,y is defined to b e the sequence of colourings visited b y pro cedure R e c olour when it is called with the input T (whic h is initially coloured x ) and with colouring y . Lik e Cycle + , the pro cedure R e c olour takes a n argumen t b T , whic h is the tree whic h will b e recoloured from x ( b T ) to y ( b T ). It also takes the argumen t y . As b efore, ˆ x is a glo bal v ariable represen ting the current colouring of the tree T , whic h is initially coloured x . Here is t he des cription of pro cedure R e c olour ( b T , y ) . 1. Let b T 1 , . . . , b T b b e the subtrees ro o t ed at the children of r ( b T ). 2. F or eac h i , 1 ≤ i ≤ b , do Cycle ( b T i , { x ( r ( b T )) , y ( r ( b T )) } ). (This step p erm utes the colours in a subtree, to allow the ro ot to b e recoloured in the following step.) 3. Assign the ro ot r ( b T ) its final colour y ( r ( b T )). 4. F or eac h i , 1 ≤ i ≤ b , do R e c olour ( b T i , y ). Analysis of the canonical paths Supp ose w e observ e a tra nsition at some p oint during the execution of a pro cedure call R e c olour ( b T , y ) when b T has heigh t h . Let P ( h ) b e an upp er b ound on the num b er of pairs ( x ( b T ) , y ( b T )) consisten t with this transition, maximised ov er all trees b T of height h and o ver a ll p ossible transitions. Let C ( h ) = q ( q − 1) ( b h +1 − 1) / ( b − 1) − 1 8 b e the num b er o f prop er colourings of a b -ary tree of height h . Note that P ( H ) is an upp er b ound on the n umber o f canonical paths γ x,y using a given transition. In order to compute the congestion A ( f ) using Equation (2), w e need to compute an upper b o und on P ( H ). W e will compute an upp er b ound on P ( h ) b y induction on h . The base case is P (0 ) = 1. No w supp o se h > 0. Supp ose that the transition starts at a colouring ˆ x and changes the colour of v ertex v from ˆ x ( v ) to a new colour. Case 1: First, supp ose v = r ( b T ). W e star t b y b ounding the num b er of colourings x ( b T ) that are consisten t with the transition. F rom the transition, w e kno w the initial colour of t he ro ot, x ( r ( b T )). F or eac h subtree b T i , we kno w that the initial colouring x ( b T i ) can b e obta ined b y p erm uting the colours in ˆ x ( b T i ). There a r e three p ossible p erm utat io ns (corresp onding to adding − 1 , 0 or 1 mo dulo q ). So the n umber of possibilities for x ( b T ) is at most 3 b . Next w e b ound the n umber of consisten t colourings y ( b T ). The colour y ( r ( b T )) is fixed b y the transition, but we know not hing ab out the colourings of t he subtrees b T i b ey ond the fact t ha t they must b e consisten t with the ro ot b eing coloured y ( r ( b T )). Th us there are at most (( q − 1) C ( h − 1) /q ) b p ossibilities fo r y ( b T ). Ov erall, w e hav e the upp er b ound P ( h ) ≤ (3( q − 1) C ( h − 1) /q ) b (4) in the case v = r ( T ). Case 2: No w supp ose v is con tained in one of the subtrees b T k . It could b e tha t the tra nsition under consideration is emplo y ed during Step 2 of R e c olour (T yp e A), or in Step 4 (T yp e B). Case 2A: Consider first pairs of T yp e A. Ho w man y pairs ( x ( b T ) , y ( b T )) of initial and final colourings may use the transition? W e’ll b ound this num- b er b y considering separately the pair s ( x ( r ( b T )) , y ( r ( b T ))) and ( x ( b T i ) , y ( b T i )) and m ultiplying the results. F or the ro ot, x ( r ( b T )) = ˆ x ( r ( b T )), while there are q p ossibilities fo r y ( r ( b T )). F or i < k , there are at most t hree p ossibilities for the colouring x ( b T i ), and at most C ( h − 1) for y ( b T i ). F or i > k , x ( b T i ) is fixed by the tr ansition, while there are at most C ( h − 1) p o ssibilities for y ( b T i ). No w consider the possibilities for x ( b T k ) and y ( b T k ), starting with x ( b T k ). Give n the transition from ˆ x ( v ) to its new colo ur w e can tell whether t he instance of Cycle ( b T k , { x ( r ( b T )) , y ( r ( b T )) } ) is a pplying Cycle + to b T k or Cycle − to b T k . In either case, (3 ) guarantees that the n umber of initial colourings x ( b T k ) that are consisten t with the transition is at most 2 b ( h − 1) . Since the n um- b er of p ossibilities for y ( b T k ) is a t mo st C ( h − 1), the n um b er for the pair ( x ( b T k ) , y ( b T k )) is b ounded b y 2 b ( h − 1) C ( h − 1 ) . This giv es an upp er b ound of 3 b q (2 h − 1 C ( h − 1 )) b on the total n um b er of pairs ( x ( b T ) , y ( b T )) suc h that the 9 giv en transition is a T yp e A transition. Case 2B: Finally , consider pairs of T yp e B. F or the ro ot, x ( r ( b T )) is arbitrary , while y ( r ( b T )) = ˆ x ( r ( b T )), so there are q p ossibilities in all. F or i < k , there are at most C ( h − 1) p ossibilities for t he colouring x ( b T i ), while y ( b T i ) is fixed. F or i > k , there are t hree p ossibilities for x ( b T i ), while there are at most C ( h − 1) p ossibilities for y ( b T i ). Inductiv ely , the n umber of p ossibilities f o r the pair ( x ( b T k ) , y ( b T k )) is P ( h − 1). This g ives an upp er b ound of 3 b q C ( h − 1) b − 1 P ( h − 1) on the total num b er of pairs ( x ( b T ) , y ( b T )) suc h that the giv en transition is a Ty p e B transition. Completing Case 2: Summing the b ounds on the n umber of pairs ( x ( b T ) , y ( b T )) suc h that the g iv en transition is a T yp e A or T yp e B transition w e obtain an upp er b ound of P ( h ) ≤ 3 b q C ( h − 1) b − 1  2 ( h − 1) b C ( h − 1 ) + P ( h − 1)  (5) on the to t a l num b er of canonical pat hs using a give n transition in the case v 6 = r ( b T ). Notice that (5) a lw ays dominates (4) since h ≥ 1. Now let χ ( h ) = P ( h ) /C ( h ). Since q b − 1 C ( h ) = ( q − 1) b C ( h − 1 ) b , w e hav e the recurrence: χ ( h ) ≤  3 q q − 1  b  2 ( h − 1) b + χ ( h − 1)  , (6) with initia l condition χ (0) = q − 1 . Now note that the recurrence (6) satisfies χ ( h ) ≤ 9 bh . Completing the Analysis: Let λ ( h ) b e an upp er b ound on the n umber of up dates p erfor med b y R e c olor ( b T , y ) when b T has heigh t h . Th us, λ ( H ) is an upp er b ound o n the length of a canonical path γ x,y . No w, b y Equation (2), A ( f ) = nq | Ω | max z ,w X x,y :( z ,w ) ∈ γ x,y | γ x,y | ≤ λ ( H ) nq | Ω | P ( H ) = λ ( H ) n q χ ( H ) , so to prov e Lemma 5 w e need an upp er b ound o n χ ( h ). The subroutine Cycle creates paths of length ( b h +1 − 1) / ( b − 1) . The recurrence gov erning λ ( h ) is th us λ ( h ) = ( b h +1 − 1) / ( b − 1 ) + bλ ( h − 1), with initial condition λ (0) = 1. Note that λ ( h ) ≤ ( h + 1) b h +1 . This can b e v erified b y induction o n h . F o r the inductiv e step, λ ( h ) = h X j =0 b j + bλ ( h − 1) ≤ h X j =0 b j + bhb h , whic h is at most ( h + 1) b h +1 since P h j =0 b j ≤ b h +1 for b ≥ 2. Thus λ ( H ) ≤ ( H + 1) b H +1 ≤ b ( H + 1) n . P utting it all together, the congestion A ( f ) is b ounded ab o v e b y q nχ ( H ) λ ( H ) whic h pro ve s Lemma 5. 10 7 The lo w er b ou n d Supp ose 2 q ≤ b/ ln( b ) . (7) The lo w er b ound pro of will use the following fact. Lemma 6. If q ≥ 3 and 2 q ≤ b/ ln( b ) then b − 2 ≥ 2( q − 1) ln( q − 1) . Pr o of. By (7), q − 1 ≤ q ≤ b/ 2 ln( b ) so 2( q − 1 ) ln( q − 1 ) ≤ b ln( b ) ln  b 2 ln( b )  = b ln( b ) (ln( b ) − ln(2 ln( b ))) = b − b ln(2 ln( b )) ln( b ) ≤ b − 2 , where the final inequalit y holds since q ≥ 3 so b ≥ 6 so b ≥ 2 ln( b ) / ln(2 ln( b )). Giv en a colouring x ∈ Ω, define F ( x ) = { w ∈ V | ∀ y ∈ Ω( T w ) with y ( L ( T w )) = x ( L ( T w )) w e hav e y ( w ) = x ( w ) } . Informally , F ( v ) is the set of v ertices w of T whose colour is for c e d by x ( L ( T w )). Our low er b o und will b e based o n a conductance argumen t whic h sho ws that it tak es a while to mov e from a colouring x in which r ( T ) is f o rced to b e one colour to a colouring y in whic h r ( T ) is forced t o b e ano ther colour. It is useful to note that F ( x ) can b e defined recursiv ely using the structure o f T . If w is a c hild of v w e sa y that w is c -p ermitting fo r v in x if either x ( w ) 6 = c or w 6∈ F ( x ) (or b oth). Observ ation 7. If h ( v ) = 0 then v ∈ F ( x ) . If h ( v ) > 0 then v ∈ F ( x ) if and only if, for every c olour c 6 = x ( v ) , ther e is a ch ild w of v which is no t c -p ermitting for v in x . The re cursiv e definition of F ( x ) illus trates the connection betw een our conductance argumen t a nd low er-b ound argumen ts based on r e cursive ma- jority functions [2, 15 ]. Consider a colouring x c ho sen uniformly at random from Ω. Supp ose v is a v ertex at height h , and let u h = P r( v 6∈ F ( x )). Note tha t the ev ents v 6∈ F ( x ), with v ra nging ov er all ve rtices at heigh t h , are indep enden t and o ccur with same probability , namely u h . 11 Lemma 8. u h ≤ 1 /b . Pr o of. The pro of is b y induction on h . Note that u 0 = 0 . F or the inductiv e step, let v b e a v ertex at height h > 0. Consider a colouring x c hosen uni- formly at ra ndom from Ω. F ix a colour c 6 = x ( v ) and a c hild w of v . The probabilit y that x ( w ) = c is 1 / ( q − 1). T o see this, think ab out construct- ing the colouring down w ards fro m the ro ot: Eac h vertex chooses a colour uniformly at random fr om the colours not used b y its paren t . Also, the probabilit y that w ∈ F ( x ) is 1 − u h − 1 and this is indep enden t of the pro ba- bilit y that x ( w ) = c . (The recursiv e definition of F ( x ) makes it easy to see that these ev ents are indep enden t.) So the proba bility that w is c -p ermitting for v in x is 1 − (1 − u h − 1 ) / ( q − 1). These ev ents are indep endent for different c hildren w of v so the probabilit y that eve ry child w is c -p ermitting for v in x is  1 − 1 − u h − 1 q − 1  b . By Observ ation 7, the ev ent v 6∈ F ( v ) o ccurs when there exists a colour c 6 = x ( v ) suc h that ev ery c hild w if c -p ermitting for v in x , so b y the union b ound: u h = Pr( v 6∈ F ( x )) ≤ ( q − 1)  1 − 1 − u h − 1 q − 1  b ≤ ( q − 1 ) exp  − b (1 − u h − 1 ) q − 1  ≤ ( q − 1 ) exp  − b − 1 q − 1  (8) ≤ ( q − 1 ) b − 2 (9) ≤ b − 1 , where (8) applies the induction hy p othesis a nd (9) uses assumption ( 7 ). Consider a v ertex v of T with h ( v ) ≥ 1 and a leaf ℓ that is a descendan t of v . Consider x ∈ Ω. Sa y that v is ℓ - lo ose in x if there is a c 6 = x ( v ) suc h that ev ery child w of v , except p ossibly the one on the path to ℓ , is c -p ermitting for v in x . Let Ψ v,ℓ b e the probability that v is ℓ -lo ose in x when x is c hosen u.a.r. from Ω. Let ε = ( q − 1) exp  − b − 2 q − 1  . Lemma 9. Consid e r a vertex v of T with h ( v ) ≥ 1 and a le af ℓ that is a desc endant of v . Ψ v,ℓ ≤ ε . 12 Pr o of. The calculation v ery similar to the calculation in the pro of of Lemma 8, with b − 1 r eplacing b . Let h = h ( v ). Then Ψ v,ℓ ≤ ( q − 1)  1 − 1 − u h − 1 q − 1  b − 1 ≤ ( q − 1) exp  − b − 2 q − 1  , where w e ha v e used the fact u h − 1 ≤ b − 1 . W e are no w ready to give the lo w er b ound argumen t. The c o n ductanc e of a set S ⊆ Ω is giv en by Φ S ( M ) = P x ∈ S P y ∈ S π ( x ) P ( x, y ) + P x ∈ S P y ∈ S π ( x ) P ( x, y ) 2 π ( S ) π ( S ) . The conductance of M is Φ( M ) = min S Φ S ( M ), where t he min is o v er all S ⊂ Ω with 0 < π ( S ) < 1. The inv erse of t he conductance of M giv es a lo wer b ound on the mixing time of M . In particular, τ ( M , 1 / (2 e )) ≥ (1 / 2 − 1 / (2 e )) / Φ( M ) . (10) Equation (10) is due to Dye r, F rieze a nd Jerrum [7]. The form ulation used here is Theorem 17 of the exp ository pap er [8]. F or c ∈ [ q ], let S c = { x ∈ Ω | ( r ( T ) ∈ F ( x )) ∧ ( x ( r ( T )) = c ) } . Let S q = { x ∈ Ω | r ( T ) 6∈ F ( x ) } . Clearly , S 0 , . . . , S q form a part ition of Ω. Let S = S 0 ∪ · · · ∪ S ⌊ q / 2 ⌋− 1 . Then Φ( M ) ≤ Φ S ( M ). No w by Lemma 8 w e hav e 0 ≤ π ( S q ) ≤ 1 /b . Also, by symmetry , π ( S c ) = π ( S c ′ ) for c, c ′ ∈ [ q ]. So  1 − 1 b  ⌊ q / 2 ⌋ q − 1 ≤ π ( S ) ≤ ⌊ q / 2 ⌋ q − 1 . Since b ≥ 6 and q ≥ 3 this giv es 5 6 · 1 2 ≤ π ( S ) ≤ 2 3 , so π ( S ) π ( S ) ≥ 1 3 · 2 3 = 2 9 Th us Φ S ( M ) ≤ 9 4   X x ∈ S X y ∈ S π ( x ) P ( x, y ) + X x ∈ S X y ∈ S π ( x ) P ( x, y )   , and by rev ersibilit y Φ( M ) ≤ 9 2 X x ∈ S X y ∈ S π ( x ) P ( x, y ) ≤ 9 2 X x,y π ( x ) P ( x, y ) , (11) 13 where the summation is ov er x and y for whic h r ( T ) ∈ F ( x ) and either r ( T ) 6∈ F ( y ) or x ( r ( T ) ) 6 = y ( r ( T )). No t e t hat if x and y c on t r ibute to the summation in (11) then since P ( x, y ) > 0, they differ on a single v ertex. Since r ( T ) ∈ F ( x ) we cannot mo v e from x to a prop er colouring y b y c ha nging the colour of r ( T ). Th us the o nly p ossibilit y is that r ( T ) 6∈ F ( y ) and x and y differ on a leaf. Also, giv en the dynamics, w e ha v e P ( x, y ) = 1 / ( nq ). Lemma 10. Φ( M ) ≤ 9 2 ε H − 1 . Pr o of. F rom Equation (11) and the discussion ab o v e w e ha ve Φ( M ) ≤ 9 2 X x,y π ( x ) P ( x, y ) where the sum is o ve r all colourings x a nd y for whic h r ( T ) ∈ F ( x ) a nd r ( T ) 6∈ F ( y ) and x and y diffe r on exactly one leaf , ℓ . Letting c = y ( ℓ ) , w e can write Φ( M ) ≤ 9 2 X x ∈ Ω X ℓ ∈ L X c ∈ [ q ] 1 x,ℓ,c π ( x ) 1 nq , where 1 x,ℓ,c is the indicator for the ev en t that r ( T ) 6∈ F ( y ) when y denotes the colouring formed fro m x b y recolouring leaf ℓ with colour c . Multiplying b y the q p ossibilities for c a nd noting that π ( X ) = 1 / | Ω | , we get Φ( M ) ≤ 9 2 1 | Ω | 1 nq q X x ∈ Ω ,ℓ ∈ L 1 x,ℓ , where 1 x,ℓ is the indicator v ariable for the ev ent that there is a colour c suc h that, when y is obtained from x by c hanging the colour of leaf ℓ to c , w e ha ve r ( T ) 6∈ F ( y ). This ev ent implies that eve ry ve rtex v on the path from ℓ to r ( T ) is ℓ -lo o se in x . When x is c hosen uniformly a r andom these eve n ts are indep enden t and b y Lemma 9 they all ha v e probabilit y a t most ε . So Φ( M ) ≤ 9 2 1 | Ω | 1 n b H | Ω | ε H − 1 , where b H is the n umber of ℓ in the summation and | Ω | is the n um b er of x . Theorem 2 follows f rom Lemma 10 since, b y Equation ( 10), the lemma implies τ ( M , 1 / (2 e )) ≥ (1 / 2 − 1 / (2 e )) 2 9 ε − ( H − 1) . Also 14 ε − ( H − 1) =  1 ( q − 1) exp( − ( b − 2) / ( q − 1))  H − 1 = e ( H − 1) ( b − 2 q − 1 − ln( q − 1) ) . 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