The optimal pebbling number of the complete m-ary tree

Discrete Mathematics 222 (2000) 89 00 www.elseier.com/locate/disc The optimal pebbling number of the complete m-ary tree Hung-Lin Fu, Chin-Lin Shiue Department of Applied Mathematics, National Chiao Tung Uniersity, 00 TA Hsueh Road, Hsinchu 30050, Taiwan Receied 4 March 997; reised 3 December 999; accepted 3 December 999 Abstract In this paper, we nd the optimal pebbling number of the complete m-ary tree. c 2000 Elseier Science B.V. All rights resered. MSC: 05C05 Keywords: Optimal pebbling; Complete m-ary tree; Integer linear programming. Introduction Throughout this paper, a conguration of a graph G means a mapping from V (G) into the set of non-negatie integers N {0}. Suppose p pebbles are distributed onto the ertices of G; then we hae the so-called distributing conguration (d.c.) where we let () be the number of pebbles distributed to V (G) and H equals V (H) () for each induced subgraph H of G. Note that now G = p. A pebbling moe consists of remoing two pebbles from one ertex and then placing one pebble at an adjacent ertex. If a d.c. lets us moe at least one pebble to each ertex by applying pebbling moes repeatedly (if necessary), then is called a pebbling of G. The optimal pebbling number of G, f (G), is min{ G is a pebbling of G}, and a d.c. is an optimal pebbling of G if is a pebbling of G such that G = f (G). Note here that the pebbling number f(g) of a graph G is dened as the minimum number of pebbles p such that any distributing conguration with p pebbles is a pebbling of G. The problem of pebbling graphs was rst proposed by Saks and Lagarias [] as a tool for soling a number theoretical problems by Lemke and Kleitman [4]. In terms of pebbling, they expected the pebbling number of an n-cube to be 2 n. Later, Supported in part by the National Science Council of the Republic of China (NSC-85-22-M-009-00). Corresponding author. E-mail address: hlfu@math.nctu.edu.tw (H.-L. Fu). 002-365X/00/$ - see front matter c 2000 Elseier Science B.V. All rights resered. PII: S002-365X(00)00008-X

90 H.-L. Fu, C.-L. Shiue / Discrete Mathematics 222 (2000) 89 00 this problem was soled by Chung []. An alternatie proof of the following theorem in number theory was thus obtained. Theorem. (Clarke et al. [2] and Lemke and Kleitman [4]). For any gien integers a ;a 2 ;:::;a d there exists a nonempty subset X {; 2;:::;d} such that d i X a i and i X gcd(a i;d)6d. In [], Chung also mentioned a conjecture by Graham: gien two graphs G and H, is the pebbling number of the Cartesian product of G and H, f(g H), no bigger than f(g)f(h)? So far, the problem remains unsoled in general. It is worth mentioning that Moews [5] showed that the inequality holds for trees G and H. On the other hand, the study of optimal pebbling number is equally interesting. First, an absolutely nontriial result on paths was obtained by Pachter et al. [7]. Theorem.2 (Pachter et al. [7]). Let P be a path of order 3t + r for 06r62; i.e.; V (P) =3t + r. Then f (P)=2t + r. Recently, Fu and Shiue [3] found f (T ) for T a caterpillar by way of a generalized pebbling on a path. Since the statement of the theorem is too long, we omit it here. Furthermore, in the same paper, they hae proed an analog of the conjecture mentioned aboe. Theorem.3 (Fu and Shiue [3]). For any graph G and H; f (G H)6f (G)f (H). Clearly, Theorem.3 gies an upper bound for the optimal pebbling number of a product graph; for example, an n-cube Q n =Q n K 2. We note here that quite recently an upper bound for f (Q n ) has been obtained. Theorem.4 (Moews [6]). f (Q n )=( 4 3 )n+o(log n). In this paper, we shall focus on the study of the optimal pebbling number of a complete m-ary tree, where a complete m-ary tree with height h, denoted by Th m,is an m-ary tree satisfying that has m children for each ertex not in the hth leel. In Section 2, we rst obtain f (T ) for a complete m-ary tree ith m 3. Then, in Section 3, we show that the optimal pebbling problem of the complete binary tree with height h, Th 2, can be transformed to an instance of an integer linear programming problem(ilp) and we nd an ecient algorithm to nd the optimal solution for the instance of ILP which corresponds to the optimal pebbling number problem for the complete binary tree. 2. m 3 Consider a d.c. of G and let H be a connected induced subgraph of G. Then V (H) ( H for short) is a d.c. of H induced by. InH, the maximum number of

H.-L. Fu, C.-L. Shiue / Discrete Mathematics 222 (2000) 89 00 9 Fig.. pebbles which can be moed to the ertex using pebbling moes in H is denoted by H (). Clearly, we hae H ()6 H for each V (H). Now, we hae the rst result. Theorem 2.. f (T m h )=2h for each m 3. Proof. Let 0 be the ertex in the 0th leel (root) of Th m =T, and be a d.c. such that ( 0 )=2 h and ()=0 for each V (T )\{ 0 }. Then it is obious that is a pebbling of T, and thus f (T )62 h. On the other hand, we will proe f (T ) 2 h by induction on h. This is triial for h=0,soleth. Assume that f (Th m ) 2h and let T 0 contain components T ;T 2 ;:::;T m. Clearly, f (T )=f (T 2 )= = f (T m )=f (Th m ) 2h. Now, let be an arbitrary pebbling of T such that ( 0 )=x 0 and Ti =x i ;i=; 2;:::;m. Then, for i =; 2;:::;m, f (T i )6 2 x 0 + x i + 4 x j: This implies that j {; 2;::: ;i ;i+;:::;m} mf (Th )6 m m 2 x 0 + m+3 4 x + m+3 4 x 2 + + m+3 4 x m6 m 2 T ; ( ) and we hae T 2f (Th m ) 2h. Thus, f (T ) 2 h and this concludes the proof. We note nally that the proof fails when m = 2 because then (m +3)=4 m=2. In fact, f (T 2 3 )67 23 (see Fig. ). 3. m =2 If a d.c. of a complete binary tree ith height h can be obtained by placing x i pebbles on each ertex in the ith leel, i =0; ;:::;h, we denote it by x 0 ;x ;:::;x h T. (We may omit T from this notation when it is clear which tree we are using.) In this section, we mainly proe that the optimal pebbling of a complete binary tree Th 2 can be obtained using a d.c. of this type where x i is een for each i 0; we will call such a d.c. symmetric. See Fig. for an example.

92 H.-L. Fu, C.-L. Shiue / Discrete Mathematics 222 (2000) 89 00 Fig. 2. Since f (T0 2 ) = is easy to see, throughout this section, let T be a complete binary tree with height h. For each ertex V (T )inthekth leel of T, the branch of T including and all its descendants is denoted by T, and the subtree of T obtained by remoing V (T )\{} is denoted by T. If a d.c. has the property that T = x k ;x k+ ;:::;x h T where x i is een for each i k, then we say is symmetric on T, and this implies that Tu (u)= h j=i x j for each u V (T )intheith leel of T, where k6i6h. Lemma 3.. Let be a d.c. of T and be a ertex in the kth leel of hich has two children u and w; and let be symmetric on T u and. If either h k + or either (u) or (w) does not equal ; then the following statements are equialent; in any case; () implies (2); and (2) implies (3). () T ( ) for each V (T ); (2) Tu 3 T u (u)+ 3 T ()+ 6 (w) 3 2h k and Tw 3 (w)+ 3 T ()+ 6 T u (u) 3 2h k ; (3) T ( h ) for each h V (T ) in the hth leel of T. Proof. Let Tu = y k+ ;y k+2 ;:::;y h Tu, u h V (T u ) be an arbitrary ertex in the hth leel of T and P = w; ; u; u k+2 ;:::;u h be the path connecting w and u h. Clearly, u i is in the ith leel of T, i = k + 2;k + 3;:::;h; see Fig. 2. We rst moe pebbles from V (T ) \ V (P) to the ertices of P by applying pebbling moes such that the number of pebbles in P is as large as possible. Then there exists a d.c. of P and m k ;:::;m h which are dened by (w) =m k = Tw (w), () =m k = T (), (u)=m k+ = y k+ + h 2 j=k+2 y j, (u h )=m h = y h and (u i )=m i = y i + h 2 j=i+ y j for i = k +2;k+3;:::;h. Note that T (u h )= P (u h). Now, we are ready to proe our implications. () (2): Clearly, T (u h )=m h + 2 (m h + 2 (m h 2 + 2 (:::(m k + 2 m k ) :::) ) :

H.-L. Fu, C.-L. Shiue / Discrete Mathematics 222 (2000) 89 00 93 By taking away all the oors, we hae ( 2 )h i m i i=k =( 2 )h (k ) Tw (w)+( 2 )h k T ()+ =( 2 )h (k ) Tw (w)+( 2 )h k T ()+ =( 2 )h (k ) Tw (w)+( 2 )h k T ()+ =( 2 )h (k ) Tw (w)+( 2 )h k T ()+ h i=k+ ( 2 )h i y i + 2 ( 2 )h i y i + i=k+ ( 2 )h i + i=k+ h i=k+ h k j=h i+2 j=i+ y j ( 2 )h i+ ( 2 )j + y h j=i+ y i (3 ( 2 )h i+ ( 2 )h k )y i : i=k+ This is equialent to saying that (2 i (k+) 3 )y i + 3 T ()+ 6 (w) i=k+ = Tu 3 T u (u)+ 3 T ()+ 6 (w) 3 2h k : Similarly, we obtain Tw 3 (w)+ 3 T ()+ 6 T u (u) 3 2h k. (2) (3): Without loss of generality, restrict attention to h = u h V (T u ). Let g k = m k and g i = m i + 2 g i, i = k; k +;:::;h. Then T (u h )=m h + 2 (m h + 2 (m h 2 + 2 (:::(m k + 2 m k ) :::) ) = g h : It suces to proe that g h. First, we will proe that g j 2 h j h i=j+ 2i j m i for k 6j6h by induction on j. Note that 2 h j h i=j+ 2i j m i is een for k 6j6h. Since Tu 3 T u (u) + 3 T () + 6 (w) 3 2h k, we hae h i=k ( 2 )h i m i. Thus h i=k 2i (k ) m i 2 h (k ) and g k = m k 2 h (k ) h i=k 2i (k ) m i, so the assertion is true for j = k. Assume that g j 2 h j h i=j+ 2i j m i for some j, k j6h, then g j+ = m j+ + 2 g j m j+ + 2 h j h i=j+ 2i j m i =2 h (j+) h i=(j+)+ 2i (j+) m i, completing the induction. Setting j = h, we then hae g h, so we are done. (3) (): We shall use the bottom-up idea to proe the implication. If h = k +, by our hypothesis, either (u) 2, (w) 2, or T () 2. This implies T () and we are done. Otherwise, let h k +. Assume that we hae shown T ( ) for each V (T u ) in the leels greater than j k + and let j V (T u ) be an arbitrary ertex in the jth leel which has two children u j+ and w j+. By assumption, we hae T (u j+ )= Tuj+ (u j+ )+ 2 ( T j ( j )+ 2 j+ (w j+ ) ) and T (w j+ )= Twj+ (w j+ )+ 2 ( T j ( j )+ 2 T uj+ (u j+ ) ). Note that Tuj+ (u j+ )= Twj+ (w j+ )= h i=j+ y i y j

94 H.-L. Fu, C.-L. Shiue / Discrete Mathematics 222 (2000) 89 00 Fig. 3. is een; this implies that either Tuj+ (u j+ )= Twj+ (w j+ ) 2 ort j ( j ) 2. Hence, T ( j )=T j ( j )+ 2 T uj+ (u j+ )+ 2 j+ (w j+ ) 2, and we obtain T ( ) 2 for each V (T u ) in leels aboe h. Similarly, T ( ) 2 for each V ( ) in leels aboe h. Either of these two facts implies that T (). This concludes the proof of this implication. Lemma 3.2. Let be a symmetric d.c. of T and = x 0 ;x ;:::;x h T. Then is a pebbling of T if and only if h (2i 3 )x i 3 2h+. Proof. Let be the root of hich has two children u and w. Then is symmetric on T u and, respectiely. By Lemma 3., we obtain that is a pebbling of T if and only if and Tu 3 T u (u)+ 3 T ()+ 6 (w) 3 2h Tw 3 (w)+ 3 T ()+ 6 T u (u) 3 2h : Note that Tu = Tw = h i= 2i x i, Tu (u)= Tw (w)= h i= x i and T ()=x 0. Hence, obsering that the two inequalities are identical, we hae the proof. Lemma 3.3. For any pebbling of T; there exists a pebbling of T such that is symmetric on T and T = T. Proof. The proof follows by constructing recursiely starting from the highest leel. To do this, we let be a pebbling of hich is symmetric on T u and where u and w are two children of (see Fig. 3). We will rst construct another pebbling of T from which is symmetric on T u and, by rearranging the pebbles on. Then, by adjusting the number of pebbles on, w, and T u, we will hae a pebbling symmetric on T and with no more pebbles than. The proof ends when reaches the root.

H.-L. Fu, C.-L. Shiue / Discrete Mathematics 222 (2000) 89 00 95 We shall use the same notations as we hae used in the proof of Lemma 3.. Without loss of generality, let Tu Tw. For conenience, we introduce three functions dened on the set of distributing congurations of T. For each d.c. of T,, let u ()= Tu 3 T u (u)+ 3 T ()+ 6 (w); and w ()= Tw 3 (w)+ 3 T ()+ 6 T u (u) ()= u () w (): Now, by obseration, we hae the following facts: () ()= Tu Tw 2 ( T u (u) Tw (w)). (2) By Lemma 3., since is a pebbling of hich is symmetric on T u and, we hae u () 3 2h k and w () 3 2h k. (3) Let Tw = y k+ ;y k+2 ;:::;y h Tw.If() 0, then there exists an index j k+ such that y j 0. (For otherwise, by the denition of, () 0.) Now, dene by Tw = y k+ ;y k+2 ;:::;y j 2 ;y j +4;y j 2;y j+ ;:::;y h Tw, Tu = Tu and T = T. Then is symmetric on T u and, T = T, and (w)= Tw (w) + 2, which implies that ( )=()+, T ()= T ()+ and u ( )= u ()+ 3. Using the aboe facts we can construct a d.c. of T by way of the following procedure: begin Let (0) :=, :=( (0) ) and i:=0; while 2 do if (i) Tw = z (i) k+ ;:::;z(i) h, and j k+ is an index such that z (i) end j 0, let (i+) T := i T, (i+) Tu := i Tu, and let (i+) Tw = z (i) k+ ;z(i) k+2 ;:::;z(i) j 2 ;z(i) j = +; i = i +; +4, z(i) j 2;z (i) j+ ;:::;z(i) h ; Clearly, the d.c. = (i) obtained satises the following properties: (a) is symmetric on T u and ; (b) T = T ; (c) T u = Tu and = Tw ; this implies that T u ; and (d) ( ) 2. Furthermore, by () and (3), if ( ) 0, then =. Most importantly, we hae to proe the following property: (e) is a pebbling of T.

96 H.-L. Fu, C.-L. Shiue / Discrete Mathematics 222 (2000) 89 00 If =, this was assumed. Otherwise, by (3), we hae T () T () which implies that T ( ) T ( ) for each V ( T ). Furthermore, u ( ) u () 3 2h k.by (d), we hae ( )60; this implies that w ( ) u ( ) 3 2h k.if we must hae h k +, so, by Lemma 3., we hae the claim. Now we are ready to dene. Let Tw = z k+ ;z k+2 ;:::;z h Tw and let =0 if z k+ is een and if z k+ is odd. We dene by ( )= ( ) for each ertex V ( T )\{} and T = x k ;x k+ ;:::;x h T where x k = ()+ T u +2, x k+ =z k+ and x j =z j for each j k+. Clearly, T = T, Tu (u)= Tw (w)= (w) is een, and x j is een for each j k. This implies that u ( )= w ( )= Tw 6 Tw (w)+ 3 T ()= 3 l for some integer l, 2 Tw (w) = 2 Tw (w)= 2 ( (w) ), T = T, and is symmetric on T. Again, we hae to proe that is a pebbling of T. Since is a pebbling of T, and (u)= (w) is een and ( )= ( ) for each ertex V ( T ) \{}, it suces to proe that T () T () and u ( )= w ( ) 3 2h k. Since ( ) 2, we hae T u = ( )+ 2 ( T u (u) (w)) 2 ( T u (u) (w) ) = 2 ( T u (u) (w)) : This implies that T () = ()+ T u +2 +2 2 Tw (w) ()+ 2 ( T u (u) (w)) +2 +2 2 ( (w) ) = ()+ 2 T u (u)+ 2 (w) + ()+ 2 T u (u) + 2 (w) = T (): Note that Tw =, Tw (w)= (w) and T ()= T ()+ T u +2 = T ()+( )+ 2 T u (u) 2 (w)+2. Therefore, we hae u ( ) = w ( )= Tw 6 Tw (w)+ 3 T () =( ) 6 ( (w) )+ 3 ( T ()+( ) + 2 T u (u) 2 (w)+2) = 3 (w)+ 3 T ()+ 6 T u (u)+ 3 ( ) 6 w ( )+ 3 ( ) 6 : Now, if ( ) 0, then w ( )+ 3 ( ) 3 2h k. Otherwise we hae w ( )+ 3 ( ) w ( )+( )= u ( ) 3 2h k. This implies that u ( ) = w ( ) 3 2h k 6. But u( )= w ( )= 3 l, l an integer; therefore u( )= w ( ) 3 2h k. Hence by Lemma 3., we conclude that is a pebbling of T.

H.-L. Fu, C.-L. Shiue / Discrete Mathematics 222 (2000) 89 00 97 Theorem 3.4. f (T ) = min{ h 2i x i h (2i 3 )x i 3 2h+ ;x 0 {0; ; 2; 3}; and x i {0; 2}; i=; 2;:::;h}. Proof. By Lemma 3.3, since each pebbling of T,, has a symmetric pebbling such that T = T, f (T ) can be obtained by minimizing h 2i x i where x 0 ;x ;:::;x h is a symmetric pebbling of T. Now it suces to claim that there exists an optimal symmetric pebbling of T, x 0 ;x ;:::;x h, such that x i 4 for each i {0; ;:::;h}. Suppose not. Then in each optimal symmetric pebbling of T, y 0 ;y ;:::;y h, there exists a smallest index 06j6h such that y j 4. Since f (T )62 h, we must hae j h. By Lemma 3.2, it is easy to check that (j) = y 0 ;:::;y j ;y j 4;y j+ +2;y j+2 ;:::;y h is also an optimal symmetric pebbling of T. By applying this operation repeatedly we obtain an optimal symmetric pebbling of T, x0 ;x ;:::;x h, such that x j 4 for j =0; ;:::;h. This is a contradiction. Thus we hae the proof. Clearly, by Theorem 3.4, we can transform the optimal pebbling problem of complete binary tree to the following instance of ILP: min 2 i x i (2 i 3 )x i 3 2h+ ; x 0 {0; ; 2; 3} and x i {0; 2}; i=; 2;:::;h: ( ) Although ILP is NP-complete, we hae an ecient algorithm to sole ( ), and thus we can quickly nd the optimal pebbling number of the complete binary tree. In what follows we present the algorithm with full details. Algorithm OPCBT(h) Input: h (a positie integer). Output: (x 0 ;x ;:::;x h ) (an optimal solution of ( )). begin {Step : initialization of c = h (2i 3 )x i and x i, i =0; ;:::;h:} x 0 :=3; c:=( 3 )x 0; for i = to h do x i :=2; c:=c +(2 i 3 )x i; {Step 2: testing x i = 0 or 2, i =; 2;:::;h:} for i = h to step do if c (2 i 3 )x i 3 2h+ then c:=c (2 i 3 )x i; x i :=0;

98 H.-L. Fu, C.-L. Shiue / Discrete Mathematics 222 (2000) 89 00 {Step 3: testing x 0 =0; ; 2or3:} while c ( 3 ) 3 2h+ and x 0 0 do c:=c ( 3 ); x 0 :=x 0 ; end Theorem 3.5. Algorithm OPCBT produces an optimal solution of ( ); and its time complexity is O(h). Proof. Let (n 0 ;n ;:::;n h ) be the output of Algorithm OPCBT. Clearly, (3; 2;:::;2) is a feasible solution of ( ) in Step ; hence an optimal solution of ( ) exists. Now, we shall claim that (n 0 ;n ;:::;n h ) is the unique optimal solution of ( ). Obsere that for each integer k, 2 2 k 3+ k i= 2 2i. Therefore, if (3; 2;:::;2; 0;n k+ ;:::;n h )is a feasible solution of ( ), then any feasible solution (x 0 ;x ;:::;x k ; 2;n k+ ;:::;n h )is not optimal. On the other hand, if (3; 2;:::;2; 0;n k+ ;:::;n h ) is not a feasible solution of ( ), then for each feasible solution (x 0 ;x ;:::;x k ;x k ;n k+ ;:::;n h ), x k = 2. Hence, if (x0 ;x ;:::;x h ) is an optimal solution, by Step 2, starting from h, we see that x h =n h, and then xh = n h ;:::;x = n. Finally, by Step 3, we hae x0 = n 0. This implies that the output is the unique optimal solution. Since the time complexity of Algorithm OPCBT is easy to see, we conclude the proof. Concluding remark. Let = x 0 ;x ;:::;x h be the optimal pebbling of T obtained from Algorithm OPCBT. Then by Theorem 3.4, for any k h, we hae T = 2 i x i 3 2h+ + = 3 x i k ( 4 )i 2 h+ + 3 ( 4 )k 2 h+ + i= 3 x i =2( 4 2h + 4 2 2 h + + 4 k 2 h )+2 3 4 k 2 h + = 2(2 h 2 +2 h 4 + +2 h 2k )+2 3 2h 2k + 3 x i 3 x i: ( ) Eq. ( ) then suggests that, in order to obtain an optimal pebbling of T, 2 pebbles should be placed on each ertex of the (h 2)th, (h 4)th;:::;and (h 2k)th leels, using the lower leels to ensure that h 2k 2 i x i 2 3 2h 2k + h 3 x i. Therefore, we conclude that when h is suciently large there exists an h such that x h ;x h +;:::;x h = 0; 2;:::;0; 2; 0; 2; 0; 0, and that f (T ) can be approximated by 3 (2h+ + h) with small errors. To be more precise, f (Th 2 )=(2 h+ + h)=3 + O(log h)

H.-L. Fu, C.-L. Shiue / Discrete Mathematics 222 (2000) 89 00 99 and for i O(log h), the alue of x i in an optimal pebbling of T 2 h is 2; if h i is een and positie; and 0; otherwise: To proe this, let each n {0; ;:::;2 h+2 } hae binary expansion n = h+ y i(n)2 i. Write h+ (n)=y 0 (n)+2 y i (n) and (n)=n (n)=3: i= Let = x 0 ;x ;:::;x h be a symmetrical d.c. of Th 2 i =; 2;:::;h. Write n = T. Then with x 0 {0; ; 2; 3}, x i {0; 2} for x 0 = y 0 (n)+2y (n); x =2y 2 (n);:::;x h =2y h+ (n); ( ) so by Lemma 3.2, will be a pebbling of T if and only if (n)=n (n)=3 3 2h+ : ( ) f (Th 2 ) will therefore be the minimal n for which ( ) holds. Since (n+)6 (n)+ for all n, is increasing with n. Let n 0 =2 2 h =3 ; then y i (n 0 ) will be if h i is odd and 0 i h, and 0 otherwise; therefore, (n 0 )=2 h=2, and (n 0 )=n 0 2 3 h=2. Set k = h=2 2 log 2 h, and let h 2, so that k 0. Eidently, h 2k, so y h 2k (n 0 ) = 0. Therefore, for all i 0 no bigger than 2 h 2k, no carry out of y h 2k (n 0 ) will occur when adding i to n 0, and so y h 2k+ (n 0 + i)=y h 2k+ (n 0 );:::;y h+ (n 0 + i)=y h+ (n 0 ): ( ) Therefore, for 06i62 h 2k, and (n 0 + i) 2 h 2 62(h 2k)+67+4 2 log 2 h 67+2log 2 h (n 0 + i) (n 0 + i 2 3 h 2 ) 6 7 3 + 2 3 log 2 h: ( ) Let i 0 = h=3 log 2 h and i = h=3 + log 2 h. We hae 2 h 2k 2 2( 2 log 2 h +) h, sofor h large enough, 06i 0 6i 62 h 2k, and we may apply ( ) toget (n 0 + i 0 )6n 0 + i 0 2 3 h 2 + 7 3 + 2 3 log 2 h 3 2h+ ; for h suciently large; so ( ) does not hold for n = n 0 + i 0, and (n 0 + i ) n 0 + i 2 3 h 2 7 3 2 3 log 2 h 3 2h+ ; for h suciently large; so ( ) does hold for n=n 0 +i. Therefore, if h is large enough, n 0 +i 0 f (T 2 h )6 n 0 + i and f (T 2 h )=2h+ =3+h=3 + O(log h), as desired. The remark about the alue of x i in an optimal pebbling now follows immediatedly from ( ) and ( ).

00 H.-L. Fu, C.-L. Shiue / Discrete Mathematics 222 (2000) 89 00 Acknowledgements The authors would like to express their appreciation to the referees for their eorts in improing the exposition of this paper. A special thanks goes to an anonymous referee who indeed rewrote the concluding remark for this paper. References [] F.R.K. Chung, Pebbling in hypercubes, SIAM J. Discrete Math. 2 (4) (989) 467 472. [2] T.A. Clarke, R.A. Hochberg, G.H. Hurlbert, Pebbling in diameter two graphs and products of paths, J. Graph Theory 25 (997) 9 28. [3] H.L. Fu, C.L. Shiue, Optimal pebbling graph, submitted. [4] P. Lemke, D. Kleitman, An addition theorem on the integers modulo n, J. Number Theory 3 (3) (989) 335 345. [5] D. Moews, Pebbling graphs, J. Combin. Theory Ser. B 55 (992) 244 252. [6] D. Moews, Optimally pebbling hypercubes and powers, Discrete Math. 90 (998) 27 276. [7] L. Pachter, H. Sneily, B. Voxman, On pebbling graphs, Proceedings of the Twenty-sixth Southeastern International Conference on Combinatorics, Graph Theory Comput. Congr. Numer. 07 (995) 65 80.