The Main Results

Suppose that D is an r-critical rooted distribution on C_m. Then, by Lemma 6.2.7, we have the following inequalies:

We assume that m = 2k + 1. If D is an r-critical rooted distribution on C2k+1 with the number of pebbles at least f (C ), then

(2^k+ 2^k+1) This is a contradiction. Thus, there is no r-critical distribution D on odd cycle with more than f (C_2k+1) pebbles, and we conclude that c_r(C_2k+1) ≤ f (C_2k+1) − 1.

Moreover, by placing ^2k+1₃⁻¹ − 1 and ^2k+1₃⁻¹ + 1 pebbles on vertices a_k and a_k+1, respectively, we have an r-critical rooted distribution. Thus c_r(C_2k+1) = f (C_2k+1) − 1, if k is odd.

Now we consider the case when k is even.

Since c_r(C_2k+1) ≤ f (C_2k+1) = 2b^2k+1₃ c+1, and the distribution of b^2k+1₃ c and b^2k+1₃ c+1 pebbles on vertices ak and ak+1, respectively, we have an r-critical rooted distribution.

Thus cr(C2k+1) = f (C2k+1), if k is even.

For the case m = 2k, we also let a_i, i = 1, 2, . . . , 2k − 1 be non-rooted vertices which starting from a vertex adjacent to rooted vertex r and continuing around the cycle. By a similar argument as above, we have c_r(C_2k) ≤ f (C_2k) = 2^k and an r-critical rooted distribution which use 2^k pebbles on vertex a_k. Thus, c_r(C_2k) = f (C_2k).

Theorem 6.3.2. Let T be a tree with diameter d. Then c_r(T ) = 2^d. Proof.

Let T be a tree with diameter d. Therefore, there exists two vertices r, r⁰ ∈ V (T ) such that d(r, r⁰) = d. Let r be the rooted vertex of T and the other of vertices are labeled by a_j,i where d(a_j,i, r) = j, i ∈ {1, 2, . . .} and j ∈ {1, 2, . . . , d}.

Since D is an r-critical rooted distribution on T , Xd

D(am,k). For otherwise, if Xd m=1

D(A_m)

2^m > 1, then there will be at least one pebble

which can not be used. This implies that D is not r-critical. So,

Let P be the Petersen graph with rooted vertex r and the other 9 vertices are labeled by a_i where d(a_i, r) = 1, ∀i = {1, 2, 3}, and b_j where d(b_j, r) = 2, ∀j = {1, 2, 3, 4, 5, 6},

If D is an r-critical rooted distribution, then the following properties hold:

1. D(ai) ≤ 1 and D(bj) ≤ 3. For otherwise, D(ai) ≥ 2 or D(bj) ≥ 4, then it can not satisfy the definition of r-critical pebbling number.

2. If D(ai) = 1, then D(ak) = 0 for i 6= k.

3. If D(b_j) ≥ 2, then |b_j| ≤ 3. For otherwise, |b_j| ≥ 4 and D(b_j) ≥ 2, then we will find that there are two vertices bj such that D(ai) ≥ 2 by a pebbling step, this contradicts property 1.

4. If D(b_j) ≥ 2, then D(a_i) = 0 where a_i is adjacent to b_j. For otherwise, D(a_i) = 1 and D(bj) ≥ 2, then we have D(ai) = 2 by a pebbling step, this contradicts property 1.

5. There exists at least one vertex v ∈ V (D) such that D(v) is even . 6. If D(v) = 3, then |v| ≤ 1, and D(v) ≥ 2 where |v| ≤ 3.

Now, we put the pebbles on the vertices. (b1, b3, b5) = (2, 1, 3) or (2, 2, 2). Then will be a pebble on the rooted vertex r by a sequence of pebbling steps. So cr(P ) ≥ 6.

Therefore, we claim that c_r(P ) ≤ 6. Thus, it suffices to show that 7 pebbles on P can not satisfy the above properties.

Moreover, we partition 7 into positive integers not greater than 3 (as the cases).

Case 1 : (3,3,1)

It contradicts property 5.

Case 2 : (3,2,2)

First, without loss of generality, we put 2 pebbles on vertex b₁ and put 3 pebbles on one of the vertices in {b₂,b₃,· · ·,b₆}. Clearly, 3 pebbles are not on b₂, b₃ or b₆, since it can not satisfy the definition of r-critical pebbling number. So, we suppose 3 pebbles on b₄ and b₅ is treated by the same way. Then the last 2 pebbles will be put on b₂, b₃, b₅ or b₆. However, if we put 2 pebbles on either one of them, it will not satisfy the definition of r-critical pebbling number.

Case 3 : (3,2,1,1)

First, without loss of generality, we put 2 pebbles are placed on b₁ and 3 pebbles on b₄. Consider 1 pebble on a₃, b₂, b₃, b₅ or b₆. However, if we put 1 pebble on a₃, b₂ or b₃, it will not satisfy the definition of r-critical pebbling number. So, we can only put 1 pebble on b₅ and b₆, respectively. But, by a sequence of pebbling steps, 1 pebble on b₅ will not be used. It will not satisfy the definition of r-critical pebbling number.

Case 4 : (3,1,1,1,1)

It contradicts property 5.

Case 5 : (2,2,2,1)

First, without loss of generality, we put 2 pebbles on vertex b1. Since b1, b2, . . . , b6

form a cycle. So, there are three cases such that three vertices have 2 pebbles. The

three subcases are (b₁, b₃, b₅), (b₁, b₂, b₃), and (b₁, b₄, b₅). But, these subcases which have 6 pebbles will put a pebble on r by a sequence of pebbling steps. Then the 7th pebble is unnecessary. This means that the rooted distribution D is excessive.

Thus, it will not satisfy the definition of r-critical pebbling number.

Case 6 : (2,2,1,1,1)

First, without loss of generality, we put 2 pebbles on vertex b1. We know that b2, b3

and b₆ can not be put 2 pebbles on either one of them, since it can not satisfy the definition of r-critical pebbling number. So, we suppose 2 pebbles on b₄(respectively on b₅). Then, the other three pebbles will be put on a₃, b₂, b₃, b₅ or b₆ one for each of three vertices. Again, this can not satisfy the definition of r-critical pebbling number.

Case 7 : (2,1,1,1,1,1)

First, without loss of generality, we put 2 pebbles on vertex b₁. Then, a₁ must has no pebble on it, otherwise it can not satisfy the definition of r-critical pebbling number. Consider b₃ and b₆. One of them has 1 pebble and the other has no pebble.

We suppose that b3 has 1 pebble, then a2 must have no pebble on it. So, we just remain only 4 vertices a3, b2, b4, and b5 to put 1 pebble on one of them. But, it will not satisfy the definition of r-critical pebbling number either.

Case 8 : (1,1,1,1,1,1,1)

It contradicts property 5.

This concludes the proof.

Theorem 6.3.4. c_r(Q_n) = 2ⁿ. Proof.

By Lemma 6.1.2, f (Q_n) ≥ c_r(Q_n). Since f (Q_n) = 2ⁿ ( see [4]), c_r(Q_n) ≤ 2ⁿ. Now, by Lemma 6.1.1, since Q_n is a graph with diameter n, so we have c_r(G) ≥ 2ⁿ. Therefore, cr(Qn) = 2ⁿ.

Theorem 6.3.5. cr(C5¤C5) = 25.

Proof.

By Lemma 6.1.2, f (C5¤C5) ≥ cr(C5¤C5). Since f (C5¤C5) = 25 ( see [14]), cr(C5¤C5) ≤ 25. Since we have an r-critical rooted distribution on C₅¤C₅, see Figure 7. Therefore, c_r(C₅¤C₅) = 25.

r

5 2 11 7

Figure 7: An r-critical rooted distribution on C₅¤C₅.

Finally, we consider that the analogous statement about Graham’s Conjecture for r-critical pebbling number, c_r(G¤H) ≤ c_r(G)c_r(H). But c_r(C₃) = 2, and c_r(C₃¤C₃) ≥ 5.

So the inequality can not be satisfied. Therefore, the analog of Graham’s Conjecture on pebbling number does not hold.

7 Conclusion

In this thesis, we mainly study the critical pebbling number of several classes of graphs and we are able to obtain several new results. The results are (1) c_r(C_m) = f (C_m) − 1 if m ≡ 3 (mod 4), and f (C_m) otherwise; (2) c_r(T ) = 2^d, where T is a tree and d is the diameter of T ; (3) c_r(P ) = 6, where P is the Petersen graph; (4) c_r(Q_n) = 2ⁿ, where Q_n is an n-cube; and (5) c_r(C₅¤C₅) = 25. It takes no time to realize that finding the critical pebbling number of a graph is not easy at all. More properties have to be discovered. We also wish that we can do a better job in the near future.

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