ONLINE AND RANDOM DOMINATION OF GRAPHS
arXiv:1509.08876v1 [math.CO] 29 Sep 2015
CHRISTOPHER COSCIA, JONATHAN DEWITT, FAN YANG, AND YIGUANG ZHANG Abstract. In this paper, we study a random algorithm for graph domination. We consider the expected size of the dominating set for the algorithm in several families of graphs. We then provide a much refined analysis of the worst case of this algorithm and enumerate the cases in which the algorithm has worst-case performance on the n-path Pn . The case of dominating the path graph has connections to previous work of Bouwer and Star, and of Gessel on greedily coloring the path. We also enumerate the cases in which the algorithm has best-case performance on Pn .
1. Introduction In this paper, we consider an online algorithm for graph domination, which was introduced in [4]. The algorithm is as follows: Let G be a graph on n vertices. Randomly label the vertices with distinctSlabels 1 through n, and let vi be the vertex labeled with the number i. Then let Vi equal ij=1 {vj }. After vi is revealed we see the entirety of the vertex induced subgraph induced by Vi , G[Vi ]. Then if vi is not dominated in this subgraph, we add it to our dominating set D. We repeat this procedure until the entire graph has been revealed. In [4], it is shown that this algorithm is near optimal for dense graphs. Using a similar procedure, it is shown in [5] that the domination number of a random graph is concentrated on two numbers. In this paper we study the complementary problem of how well this algorithm does for specific families of non-dense graphs. We begin by studying the performance of this algorithm on the path on n vertices, Pn . We are then able to leverage this information to learn about many other families of graphs. The result of this investigation demonstrates that the algorithm does not perform as well on these sparse graphs as it does on the dense graphs where it has previously been applied. Throughout this paper we will use G to denote a graph. We then define γo (G) to be the expected online domination number of the graph, i.e., the expected size of the dominating set created when our algorithm is run on the graph. After determining, in Section 2, the expected number of vertices in the dominated path, we then consider in Section 3 the orders, realized by permutations, in which the nodes may be revealed which gives the worst case behavior. We enumerate these worst-case permutations. This enumerative work leads to connections between this problem and the work of Bouwer and Star [1] and of Gessel [2] on greedy colorings of the path. In their work they studied greedy colorings which color the paths with two colors as opposed to three. In particular they enumerate these best case permutations. We also show that when the length of the path is odd, that the best case behavior for the path coloring problem coincides the worst case behavior for the domination problem, and when the length of the path is even, that these two problems differ. We enumerate the number of permutations in the worst case. We 2010 Mathematics Subject Classification. 05A05, 05C85, 05C38, 05C69, 05A17. Key words and phrases. permutations, graph domination, random algorithms, path graphs, asymptotics. 1
then turn to the problem of enumerating the permutations in which the algorithm gives best case performance. For an example of an application, note that this random algorithm can be run in a distributed fashion. For instance, imagine a sequence of lights along a strip. The lights are turning on in a random order and we need the entire strip to be illuminated. When a light turns on it notices that its position is already illuminated or not. If it is not illuminated then the light begins to shine its light. Otherwise it stays dark. The expected number of lights left on once all of the lights awaken is equal to the expectation that we compute in this paper. 2. A Study of Pn In particular, we are interested in computing the asymptotic behavior of γo (Pn ) where Pn is the path on n vertices. For simplicity, we define γo (Pn ) = 0 for n ≤ 0. For ease of notation, we define Pn = ([n], {(i, i + 1) | i ∈ [n − 1]}), so that our vertices come pre-labeled. Theorem 2.1. lim γo (Pn )/n =
n→∞
e2 − 1 2e2
Proof. We first claim that γo (Pn ) = 1 +
2 n
n−2 X
! γo (Pi )
i=1
To see this, suppose that the first vertex revealed in the online procedure is vertex i. Then i must enter the dominating set. As i is in the dominating set, we see that the vertices i − 1 and i + 1 are already covered, whereas i − 2 and i + 2 are not. Thus we see that in order to finish dominating Pn we must separately dominate the two subgraphs two away from i which are isomorphic to Pi−2 and Pn−i−1 , respectively. Next, note that when we consider each possible i ∈ [n], that we encounter two of Pj for j ∈ [n − 2]. Then as each i is equally likely and expectations are linear, we have the above formula. P Now, let F (x) be the generating function for γo (Pn ), i.e. F (x) = i≥0 γo (Pn )xn . We now multiply the nth version of the recurrence by xn and sum over n ≥ 0 to find xF 0 (x) =
x x2 + 2F (x) . (x − 1)2 1−x
Cancelling the x, we have the equation: F 0 (x) =
1 x + 2F (x) . 2 (x − 1) 1−x
By solving the differential equation, we see that F (x) = Ce−2(x+log(x−1)) +
1 . 2(x − 1)2
which then allows us to evaluate the function and determine that the constant is −1/2. We now wish to determine the coefficients of F , which after a small amount of manipulation we 2
see is equivalent to determining the coefficients of 1 −1 −2x 1 e F (x) = + . 2 2 (x − 1) 2(x − 1)2 1 We already know that n + 1 is the coefficient of xn in 2(x−1) 2 , so it will suffice to determine 1 n −2x gn , the coefficient of x in e (x−1)2 . To this end, note that ! n ! n i X X 1 −2 xi (j + 1)xj . e−2x = (x − 1)2 i! i=0 i=0
Then collecting the xn terms in the product we see that we have: n X −2j gn = (n + 1 − j) j! j=0 So, we see that γo (Pn ) = γo (Pn ) =
−1 2
−1 g 2 n
+
n+1 , 2
and thus ! n n X n+1 n − 1 X (−2)j n + 1 (−2)j + = + . (n + 1 − j) j! 2 2 j! 2 j=0 j=0
We are interested in the asymptotic behavior of gn , so we consider limn→∞ gn /n and see that gn looks like ! n n + 3 −1 X −2j n+1 lim + n→∞ n 2 j! 2n j=0 We can then evaluate the limit and recognize the power series as e−2 to find that: 1 1 lim gn /n = − 2 ≈ .4323... n→∞ 2 2e Now that we know this we can deduce γo for a variety of other families of graphs. Note that if we consider γo (Cn ), then after we remove the first vertex, that we are left with a Pn−3 and thus γo (Cn ) = 1 + γo (Pn−3 ). Now consider ?n , the star graph with n leaves. Note that on the first round of domination, either one of the leaves is revealed or the center is revealed. If a leaf is revealed then every other leaf must be in the dominating set, otherwise, every vertex is already dominated, in which case we see that n 1 n2 + 1 γo (?n ) = n+ = = O(n). n+1 n+1 n+1 For the wheel graph with n spokes, Wn , things are similar. If we first pick the center we are done, otherwise, we will never pick the center when it is revealed, in which case we are simply dealing with a path. So, we find 1 n γo (Wn ) = + γo (Pn−3 ) n+1 n+1 For the multipartite graph with partitions of size pi , note that once one vertex is placed, all vertices outside that partition are dominated and thus will never be in the dominating 3
set. Thus the remaining vertices in the partition must be added to a dominating set. So, to find γo (Kpi ), we simply weight by the size of each partition, and thus we find P 2 p γo (Kpi ) = P i . pi 3. Worst Case Permutations In this section we consider the orders in which vertices are revealed which maximizes the number of nodes included in the dominating set of the graph. Let π be a permutation of [n]. Then let γ(π) denote the number of nodes in the dominating set of Pn when the nodes of Pn are revealed in the order π1 , π2 , . . . , πn . In addition let Γ(π) be the dominating set obtained. When n is even we see that at most n/2 vertices may be included in the dominating set as no two vertices in the dominating set may be adjacent. Further the permutation 135...246... achieves this number. Morever we see that as any dominating set created by our algorithm is an independent dominating set, and each such dominating set can be achieved, by simply listing those dominated vertices first in our permutation, we see that the number of distinct worst case dominating sets is equal to the number of maximal independent dominating sets on the graph. Note that in the case of the path this is equal to the number of maximal independent sets in the graph. Proposition 3.1. The number of maximal independent-dominating sets in Pn is equal to n/2 + 1 when n is even and is equal to 1 when n is odd. Moreover, when n is odd the unique such set consists of the odd numbered nodes. When n is even the maximal independentdominating set consists either of all the even nodes, all of the odd nodes, or it is of the form {1, 3, 5, . . . , 2j − 1, 2j + 2, 2j + 4, ..., n} for some j ∈ N. Proof. First suppose that the length of the path is 2k + 1, i.e odd. The node 1 must always be in the independent dominating set. If it were not, then 2 must be in the independent set in order for it to be maximal. In this case we see that 3 cannot be in the set and so finding the largest maximal independent set on the remaining nodes reduces to considering the path on the nodes {4, 5, ..., 2k + 1}. This new path is of length 2k − 2. However, because this is a worst case permutation there must be k + 1 nodes in the independent set and so there remain k nodes that must be added to the independent set. However, at most k − 1 nodes can be placed on the remaining path. Thus 1 must be in the set and inductively every odd node in the path must be in the independent set. We conclude that there is a unique maximal independent set of P2k+1 . We now turn to the even case, in which the path has 2k nodes. Note that if there is a sequence of two consecutive vertices i and i + 1 that are not in the dominating set, then the dominating set will consist of vertices of a single parity. Otherwise, there is at least one pair of vertices i and i+1 which are both not in the dominating set. First suppose there is exactly one consecutive non-dominated consecutive pair, which occurs between i and i + 3. We claim that i must be odd. If i were even, the remaining entries 1, 2, ..., i − 2 would be a string of even length, whereas i + 5, i + 6, ..., 2k would be a string of odd length. In particular, we see that each of these strings can accomodate at most i/2 − 1 and k − i/2 − 2 nodes. This sums to k − 3 nodes. However, we as we have added two vertices to the dominating set already, this shows that we will only manage choose k − 1 nodes. However, k nodes are required for the set to be maximal, so this is impossible. One may then check that dominating every 4
other node and then leaving a gap between i and i + 3 gives a maximal set provided that i even. We now claim that there can be at most one such pair of consecutive non-dominated vertices. If there were two gaps then we can take the right most node of the left most gap, and the leftmost node of the rightmost gap, and translate those two nodes and all those nodes between them one step to their left. Then there is a vacant spot next to where the right most node that was moved originated, which we can dominate. So we see that the independent set was not maximal. Now that we have determined the possible worst case configurations, we study which permutations actually correspond to this configuration. We begin here with the study of the single configuration when n is odd. This is the dominating set {1, 3, 5, ..., n}. Let Fn be the set of permutations which achieve the worst case bound. The following table summarizes the values of |Fn | for small values of n and was generated by explicitly testing each permutation in Sn . In the case n = 3 the four permutations in Fn are 123, 132, 312, 321, i.e. those that do not start with 2. n |Fn | 1 1 2 2 3 4 4 24 5 56 6 640 7 1632 8 30464 9 81664 10 2251008 11 6241280 Proposition 3.2. Let f (n) = |Fn |. For n odd, f (n) satisfies the recurrence relation
f (n) = 2(n − 1)f (n − 2) + (n − 1)(n − 2)
n−2 X i=3, i odd
n−3 f (i − 2)f (n − i − 1) i−2
or equivalently, k−1 X 2k − 2 f (2j − 1)f (2k − 2j − 1). f (2k + 1) = 4kf (2k − 1) + (4k − 2k) 2j − 1 j=1 2
In particular, it is possible to enumerate |Fn | for n odd without knowing any values for n even. P Proof. Let fi (n) = {π ∈ Sn : π ∈ Fn and π(1) = i}. Then f (n) = ni=1 fi (n) and fj (n) = 0 when j is even. When i is 1 or i = n = 2k + 1, the first vertex dominated is an endpoint of the path. It then remains to dominate 2k − 1 more vertices, and the label corresponding to neighbor of the endpoint of the path may appear anywhere in the permutation (as it will never be placed into the dominating set). The number of permutations that give rise 5
to a maximal dominating set on the 2k − 1 vertices is f (2k − 1), and there are 2k possible positions for the neighboring vertex, so f (1) = f (2k + 1) = 2kf (2k − 1). For i odd and not 1 or 2k + 1, choosing the ith vertex as the first to dominate splits the path into two parts of odd length. The first i − 2 vertices must be dominated, and the final n − i − 1 vertuces must be dominated, while the (i − 1)st and (i + 1)st vertices may appear anywhere in the permutation as they will not be in the dominating set. Thus we must place an ordering on these two sets of vertices and interleave them in any way we wish; there are n−3 f (i − 2)f (n − i − 1) ways to do this. We must then choose positions i−2 in the permutation for the two neighbors of the first vertex chosen, so in this case fi (n) = (n − 1)(n − 2) n−3 f (i − 2)f (n − i − 1). i−2 It turns out that the subsequence {f (2k +1)}k∈N has been encountered in conjunction with weakly alternating permutations. We say that a permutation π ∈ Sn is weakly alternating if for every even index i we have either πi−1 < πi or πi+1 < πi , i.e. there is a weak peak at i. Let Wn be the set of weakly alternating permutations of order n. Proposition 3.3. We have |Fn | = |Wn | for odd n. In particular, the map that sends π 7→ π −1 is a bijection from Fn into Wn . Proof. Suppose that π ∈ Fn for odd n. The dominating set for Γ(π) consists of the oddlabeled vertices along the path. For each even i ∈ [n], one of i − 1 and i + 1 must be in the dominating set by the time that i is revealed as otherwise i would be added to the dominating set. Moreover, requiring that one of i − 1 and i + 1 appear before i in the permutation is clearly sufficient to ensure that i is not dominated and that every odd node is dominated. Now consider π −1 . We claim that π −1 is weakly alternating. For some even i ∈ [n] appearing at index k in π we have either i − 1 or i + 1 appearing at some index j < k. Without loss of generality suppose that π(j) = i − 1. We see that π −1 (i − 1) = j and π −1 (i) = k. Thus π −1 does have a weak peak at every even i. Now suppose that σ is a weakly alternating permutation. Then we have without loss of generality that σ(i) = k and σ(i − 1) = j < k, and so σ −1 (j) = i − 1 and it appears before σ −1 (k) = i. Thus the map that sends π → π −1 is a bijection between Fn and Wn for odd n. We now turn to the problem of enumerating Wn in the case when n is odd. Recall that the complement of a permutation π ∈ Sn is defined to be the permutation c(π) ∈ Sn where c(π)i = n + 1 − πi [3]. Note that the complement of a weakly alternating permutation with weak peaks at even indices is a permutation with no local maxima in even positions. Permutations of odd length with no local maxima in even positions are enumerated in OEIS sequence A113583. 3.1. Connection with Path Coloring. Permutations with no local maxima in even positions have been studied previously in relationship to greedy graph colorings of the path. Recall the algorithm for greedily coloring a path is to simply label each node with the lowest number so that no two adjacent vertices receive the same label. In particular, suppose that we color the nodes in the order that they appear in π. Then the number of permutations in which the odd vertices are colored 1 and the even vertices are colored 2 is equal to the number of permutations of odd length with no even local maxima [1]. However, the form of the extremal permutations differs from the graph coloring problem when the length of the path being colored is even. In particular, letting D(n) be the number of permutation in Sn 6
which lead to all of the odd vertices of Pn being dominated, Gessel discovered in [2] that letting G(x) = D(1)x/1! + D(3)x3 /3! + · · · , sinh x . cosh x − x sinh x Also letting H(x) = D(0)/0! + D(2)x2 /2! + D(4)x4 /4! + · · · he finds 1 (3.2) H(x) = . cosh x − x sinh x These formulas were studied prior to the work of Gessel in a different form in [1]. G(x) =
(3.1)
3.2. Enumerating F2n . We now turn to the remaining problem of enumerating the worstcase permutations of even length. In particular we establish the following. Proposition 3.4. Let f (n) = |Fn | and define f (0) = 1. With values of f (n) for n odd as in Proposition 3.2, when n is even, f (n) = f (2k) satisfies the recurrence relation f (n) = 2(n − 1)f (n − 2) + (n − 1)(n − 2)
n−1 X n−3 i−2
i=2
f (i − 2)f (n − i − 1)
or equivalently, 2
f (2k) = (4k − 2)f (2k − 2) + (4k − 6k + 2)
2k−1 X i=2
2k − 3 f (i − 2)f (2k − i − 1) i−2
The proof is similar to the case that n is odd, except that choosing a vertex other than the first or last splits the path into an odd-length path and an even-length path, which must be dominated separately, and we cannot throw out the cases in which i is even (in fact, by symmetry, the procedure is the same as when i is odd as when i is odd). Theorem 3.5. Let Fn be the set permutations which achieve the upper bound for the number of vertices in the dominating set when the random algorithm is run on the path. Then letting F (x) be the exponential generating function for |Fn | sinh x 1 (3.3) F (x) = + . cosh x − x sinh x (cosh x − x sinh x)2 One immediate advantage of this view is that the coefficients of the exponential generating function are exactly the probabilities that a randomly chosen permutation in Sn gives rise to a maximal-size dominating set by this online algorithm. Proof. We have already established this for even powers of x in our remarks in Proposition 3.3. So, the even case is all that remains. To finish the proof we show combinatorially that n X 2n |F2n | = O(2n − 2i)O(2i) . 2i i=0 Suppose that π is a permutation in F2n . If all of the nodes in Γ(π) have the same parity, then we have that either π is a permutation counted by O(2n) or the reverse of π is a permutation counted by O(2n). This deals with the cases in the above equation where i is 0 7
or n. Otherwise, from Proposition 3.1 we see that Γ(π) has the form {1, 3, 5, ..., 2j − 1, 2j + 2, 2j + 4, ..., 2n}. Now let π 0 be the permutation π restricted to the set [2j] and let π 00 be the permutation restricted to the remaining elements. By the form of Γ(π) we have that π 0 is counted by O(2n). Similarly we see that subtracting 2j from the elements of π 00 and reversing their order that there is a similar inclusion in O(2n). Moreover, considering taking two permutations σ counted by O(2n − 2j) and τ counted by O(2j) for 0 < j < n. Reverse the entries of τ , add 2n − 2j to each entry and now place the entries of this new permutation among the entries of σ to create a permutation π. This map is clearly an injection and moreover there are 2n ways of merging the two permutations as each way of merging them 2j is selected by the choice of 2i entries which the image of τ occupies (for example, if we take σ = 534216 and τ = 135642 then τ 7→ 246531 7→ 8(10)(12)(11)97. We can then join these permutations, for instance, as π = 58(10)342(12)(11)1697 and Γ(π) = {1, 3, 5, 8, 10, 12}. That the nodes 2j and 2j + 1 are never dominated by this procedure follows from noticing that these nodes are only dominated if they appear before 2j − 1 or 2j + 2 respectively in the permutation π and then observing that when we are composing two permutations we are effectively just relabeling the nodes that the algorithm runs on. Then as neither of 2j and 2j + 1 are dominated we see that the nodes less than 2j are effectively dominated independently from the nodes greater than 2j + 1. It should be noted here that as the proofs of the formulas for G(x) and H(x) given in [2] are purely combinatorial. So, as the above proof is also combinatorial, we see that Theorem 3.5 can be obtained by purely combinatorial means. 4. Best Case Permutations In this section we enumerate the permutations for which the algorithm gives the optimal result. Given a permutation π ∈ Sn with corresponding dominating set Γ(π) we see that each vertex of Γ(π) dominates at most three elements of Pn . Thus we see that the minimum size of a dominating set is dn/3e. Dominating sets achieving this bound are easy to construct: pick all of the nodes congruent to 2 modulo 3 as well as the final node. We enumerate these permutations in three different cases. In particular, the resulting formula depends strongly on the congruence class of n modulo 3. Let Bn be the set of permutations of length n which achieve the lower bound on the size of the dominating set. Proposition 4.1. For n ≡ 0 mod 3, we have that n! (4.1) |Bn | = n/3 . 3 Proof. Note that for any dominating set Γ(π) for π ∈ Sn we have that Γ(π) = {2, 5, 8, ..., n − 1} as if there is a node which is dominated by multiple nodes then not every node is dominated. In particular in order to have the node i be in the dominating set because neither i−2, i−1, i+1 nor i+2 appears in the dominating set we must have that i appears before i−1 and i + 1 in the resulting permutation. Moreover, this is sufficient for a permutation to be a member of Bn . Now partition the n indices into groups of three and for each i ≡ 2 mod 3 between 1 and n, assign one such i to each block of the partition. Then to the three entries of the block assigned i place the entries i, i − 1, i + 1 in this order or the order i, i + 1, i − 1. 8
Note that this process gives each of the permutations in Bn and moreover that counting along the way gives the formula above. We next turn to permutations where n ≡ 2 mod 3. Proposition 4.2. For n ≡ 2 mod 3, we have that for n > 2 n (n − 5)! n − 2 n (n − 2)! |Bn | = 24 + 2 . 5 3(n−5)/3 3 2 3(n−2)/3 Proof. We consider two cases separately. Either Γ(π) contains one of 1 or n or it does not. In the case that Γ(π) contains 1 then we see that π restricted to {3, ..., n} gives a minimal dominating set. In particular because this set contains a number of nodes congruent to 3 modulo 4 so we see that the permutation π when restricted to {3, ..., n} is of the type in Bn−2 . In particular we see that we may add the entries 1 and 2 in exactly one way to such a permutation thus there are n (n − 2)! 2 3(n−2)/3 such permutations. The analysis when n ∈ Γ(π) is similar. Next suppose that neither 1 nor n is in the dominating set. Then we claim that there exists a vertex which is dominated by two nodes. Each vertex dominates three nodes so as there are dn/3e members of the dominating set and thus number is greater than n/3 by exactly 1/3 we see that there is a unique node that is dominated by both of its neighbors. We now count permutations π ∈ Bn which have a unique pair of indices i, i + 2 in their dominating set. For such a permutation, note that if we restrict π to {1, ..., i−2} and {i+4, ..., n} that each of these sets must contain a number of nodes congruent to 0 modulo 3 as otherwise the remaining n − 2 nodes we can use to dominate cannot possible dominate the rest of the graph. We can count the number of ways to dominate the nodes {1, ..., i − 2} and {i + 4, ..., n} analogously with the previous proposition by breaking these nodes into blocks of three. Counting by hand gives that there are 24 possible ways to order the nodes i − 1, i, i + 1, i + 2, i + 3 which will give i and i + 2 in n the dominating set. We then see that there are 5 ways of combining the permutation on {i − 1, i, i + 1, i + 2, i + 3} and the permutation defined on the remaining nodes. Then notes that there are ((n − 5)/3 + 1) distinct choices for the pair of nodes i, i + 2 in the dominating set. Multiplying all of these choices together then gives the result. We now turn to the most complicated case, n ≡ 1 mod 3. This case is more complicated because fewer nodes need to dominate 3 other nodes, which in the previous cases has given the permutations a large amount of structure. Proposition 4.3. For n ≡ 1 mod 3, we have that for n > 7 n n − 4 (n − 7)! n n − 2 (n − 7)! n − 4 n (n − 4)! |Bn | = 720 · + 2 24 +9 7 3 3(n−7)/3 2 5 3(n−7)/3 3 4 3(n−4)/3 10 n (n − 4)/3 (n − 10)! n (n − 4)! + 242 . +6 (n−4)/3 5 10 2 3(n−10)/3 4 3 Proof. We enumerate different subsets of Bn separately with each subset being defined basically in terms of how it dominates the end nodes. We begin by considering those π ∈ Bn such that neither 1 nor n is contained within Γ(π). As each node of the dominating set 9
dominates three other nodes and there are dn/3e nodes in the dominating set, we see that there exist exactly two nodes i and j, which are dominated by two other nodes. Further we see that i and j may have a common dominator or not. Suppose that i and j have a common dominator, then we see that our dominating set contains three nodes i, i + 2, i + 4. Moreover, note that if we restrict π to {1, ..., i − 2} and {i + 6, ..., n} that we have n − 7 nodes left to dominate with our remaining dn/3e − 3 nodes. In particular, this shows that we must have a multiple of three nodes in each of these blocks and moreover, because we have so few remaining nodes we see that the dominating set is determined. Moreover, computing by brute force gives that there are 720 ways to dominate the nodes i − 1, ..., i + 5. Then there are ((n − 7)/3 + 1) possible values that i may take, so we have obtained the first term in the formula. Next, suppose that i and j do not have a common dominator. Then we see that there are exactly two pairs i, i + 2 and j, j + 2 in Γ(π). There are 24 ways to dominate each block i − 1, ..., i + 3, individually. Moreover, by counting the number of nodes that need to be dominated outside of the blocks we see that each of the groups {1, ..., i − 1}, {i + 3, ..., j − 1} and {j + 3, ..., n} must be a multiple of three. So, we see that there are (n − 10)/3 + 2 choose 2 ways to pick i and j. Then multiplying all of this information together as before gives the final term in the expression. Now suppose that both 1 and n are contained within the dominating set. Then we see that π restricted to {3, ..., n − 3} must dominate n − 4 nodes with dn/3e − 2 nodes, so we see that we may treat these remaining nodes as though they belong to the case where n ≡ 0 mod 3. Note that there are 4 ways to dominate when we restrict to 1, 2, n − 1, n so joining these with a permutation that dominates {3, ..., n − 3} gives the second to last term in the equation. Next suppose that 1 is in the dominating set but that n is not. Note that the case where n is in the dominating set, but 1 is not is analogous, so we multiply the relevant terms by 2. We now consider the case in which both 3 and 1 are dominated. In this case we see that restricting to the nodes {5, ..., n} that we have n − 4 nodes to dominate and dn/3e − 2 = (n − 4)/3 nodes we may use in our dominating set. Thus we may consider π restricted to 5, ..., n as being a permutation in Bn−4 . There are 9 ways to dominate 1 and 3 but not 2 and 4. Thus combining as before we obtain the second term within the parentheses. Now suppose that 1 is in the dominating set but neither are n nor 3. In this case we consider the restriction of π to {3, ..., n} and identify this with a permutation in Bn−2 . So, we see that each such permutation arises by combining a permutation in Bn−2 which does not include 1 nor n − 4 in its dominating set with the permutation 1 2. From our analysis in Proposition 4.2, we see that such permutations have a pair of entries in their dominating set i, i + 2. So selecting that pair of entries and enumerating as before gives us the first term within the parentheses. As these different cases are exhaustive, we have finished the proof. For completeness, we also include the following table, which gives the values of |Bn | for small n.
10
n |Bn | 1 1 2 2 3 2 4 24 5 64 6 80 7 3408 8 9856 9 13440 10 1377792 11 4139520
5. Further Questions There are a number of questions which arise from study of the random online algorithm. One obvious question concerns the enumeration of other types of permutations with respect to the number of vertices they will cause to lie in the dominating set. The best case permutations, which minimize the size of the dominating set, seem like a natural place to start. There is another different question which is more closely related to the probabilistic section of the paper. For instance, given the distribution of Sn under the map γ on the line, one can appropriately normalize and then ask whether this distribution is converging to some limit distribution. We have done some tests of this and the results seem encouraging, cf. Figure 5. Note that because γ(π) for π ∈ Sn is between about n/2 and n/3 that normalizing by n and translating will result in a limiting distribution with compact support, which will not be normal. Figure 1. γ(π) plotted for 40000 randomly chosen permutations π in S2000 .
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6. Acknowledgments This research was conducted during the Summer 2014 REU program at East Tennessee State University. CC and JD were were supported by NSF Grant 1263009. YZ was supported by Grant No. 14-12 from the Acheson J. Duncan Fund for the Advancement of Research in Statistics at The Johns Hopkins University. We would like Anant Godbole for his suggestion of the problem. We also thank Curtis Greene for suggesting that we consider best and worst case permutations. References [1] [2] [3] [4]
I. Bouwer and Z. Star. A question of protocol. American Mathematical Monthly, 95(2):118–121, 1988. I. M. Gessel. A coloring problem. American Mathematical Monthly, 98(6):530–533, 1991. S. Kitaev. Patterns in permutations and words. Springer Science & Business Media, 2011. S. E. Nikoletseas and P. G. Spirakis. Near-optimal dominating sets in dense random graphs in polynomial expected time. In J. van Leeuwen, editor, Graph-Theoretic Concepts in Computer Science: 19th International Workshop, WG ’93, Utrecht, The Netherlands, June 16-18, 1993 Proceedings, pages 1–10. Springer, 1994. [5] B. Wieland and A. P. Godbole. On the domination number of a random graph. The Electronic Journal of Combinatorics, 8(1):R37, 2001.
Department of Mathematics, 6188 Kemeny Hall, Dartmouth College, Hanover, NH, 037553551. E-mail address:
[email protected] Department of Mathematics and Statistics, Haverford College, 370 Lancaster Avenue, Haverford, PA, 19041, USA. E-mail address:
[email protected] Department of Mathematical Sciences, Carnegie Mellon University, 5000 Forbes Avenue, Pittsburgh, PA,15213, USA E-mail address:
[email protected] Deparment of Applied Mathematics & Statistics, the Johns Hopkins University, 3400 North Charles Street, Baltimore, MD, 21218, USA. E-mail address:
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