On finding a minimum spanning tree in a network with random weights

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1 On finding a minimum spanning tree in a network with random weights Colin McDiarmid Department of Statistics University of Oxford Harold S. Stone IBM T.J. Watson Center 30 August 1996 Theodore Johnson University of Florida Abstract We investigate Prim s standard tree-growing method for finding a minimum spanning tree, when applied to a network in which all degrees are about d and the edges e have independent identically distributed random weights w(e). We find that when the kth edge e k is added to the current tree, where k = o( d), the probability that this edge e k is incident to the node that was most recently added to the tree equals k + o(1) as d. We also find for example that, if the edge weights are uniformly distributed on (0, 1), then the expected value of w(e k ) is asymptotic to ( k )/d. 1 Introduction and statement of main results We investigate how Prim s tree-growing method for finding a minimum spanning tree operates on a network with random edge weights. Our main results are Theorems 1, 2 and 3 below. In order to prove these theorems we develop some basic results in combinatorial probability which are perhaps of independent interest. These are stated in section 2 and proved in section 3. The proofs of the main theorems are then completed in the next sections. 1

2 The tree-growing method for finding a minimum spanning tree in a network is sometimes called Prim s method, and was described in Jarńik [15], Prim [21] and Dijkstra [10]. It is one of the standard ways of implementing the greedy approach, and works as follows, see for example [1, 8, 20]. It starts say with vertex 1. At each stage k, if the current tree with k 1 edges is not spanning, it finds a minimum weight edge e k between the current tree T k and the rest of the graph, and adds this edge to the tree. We shall refer to this method as the tree-growing method. We investigate how the tree-growing method works on a suitable random network. For each d = 1, 2,..., let G d be a connected graph such that all vertex degrees are in d + O( d). In particular, we may be interested in the case when G d is the complete graph K d+1. Suppose that in G d the edge weights are nonnegative independent random variables, with a common continuous distribution function F. We find that we can analyse the first few steps of Prim s tree-growing method quite precisely. We first consider which vertex in the current tree is incident to the next edge added, then consider the weight of that new edge, and finally consider how the growing tree fits inside the (unique) optimal spanning tree. Theorem 1 If k = o( d), then the probability that e k, the kth edge added to the tree, is incident with the node that was added most recently to the tree equals o(1) as d. 2 2k Indeed, let us define a random variable Z k taking values in {1,..., k}, by setting Z k = s when the edge e k is incident with the node in the tree which was added s steps before. Then the total variation distance between the random variables Z k and V k is O(k/ d), where the distribution of V k is given in Theorem 6 below. If k = o( d) and k, then Z k converges in distribution to the random variable V described in Corollary 7 below, that is P (Z k = j) t t 2 e t /t! for j = 1, 2,.... t=j Recall that, given random variables X and Y taking values in some set S, the total variation distance d(x, Y ) between X and Y (or between their distributions) is the supremum over all measurable subsets A of S of the quantity P (X A) P (Y A). Note that we use V k and V to denote random variables not sets of vertices. 2

3 The above result shows that, at least in the early stages, the random tree constructed by Prim s minimum spanning tree algorithm grows in a manner very different from that in other models of random trees (see for example the books [17, 5] and recent papers [3, 7, 9, 16, 19]). Theorem 2 Let W k denote the random weight w(e k ) of the kth edge added to the tree, and let the random variable Ŵk be obtained by picking uniformly at random one of the k smallest values out of kd independent random variables, each with distribution function F. Then the total variation distance between W k and Ŵk is O(k/ d). Suppose that the common distribution of the edge-weights is uniform on (0, 1). If k = o( d/ log d), then E(W k ) = ( o(1))/d as d ; and 2 2k if k = o( d) and k, then d W k is asymptotically uniformly distributed on (0, 1). Indeed, suppose more generally that the (positive) edge-weights have finite mean and that their common distribution function has a right derivative α > 0 at 0. If k = o( d/ log d) then E(W k ) = ( o(1))/(αd) as d ; and 2 2k if k = o( d) and k, then αd W k is asymptotically uniformly distributed on (0, 1). Let Td denote the (unique) random minimum spanning tree in a random network as above. Our interest here is in Prim s tree growing method for finding Td rather than in Td itself. However, there has been considerable interest in Td. Consider the case when the underlying graph is the complete graph K d. The total weight w(td ) was studied by Frieze [11], who showed that, if the edge weights are uniformly distributed on (0, 1), then E(w(Td )) ζ(3) = 1/i as d. i=1 The weight w(t d ) was subsequently investigated further by Timofeev [22], Frieze and McDiarmid [12], Avram and Bertsimas [6], Aldous [2, 4] and Janson [14]. The focus of the paper [2] is the local structure of the tree T d. In particular, it is shown that the degree of vertex 1 has a certain limit distribution, which is a mixture of Poissons. How do the growing trees T k constructed by the algorithm sit inside the (unique) optimal spanning tree T d? 3

4 Theorem 3 Suppose that the underlying graph in the random network is the complete graph K d, and the edge weights are uniformly distributed on (0, 1). Let k as d. If k = o(d 1 2 ), then the expected degree of vertex 1 in the tree T k π 2 /6 1.65, and thus the expected number of edges in the optimal tree Td between vertex 1 and the vertices not in T k 2 π 2 / If k = o(d 1 2 / log d), then the expected total weight of the edges incident with vertex 1 in T k is (1 + o(1))π 2 /6d 1.65/d, and thus the expected weight of the edges in the optimal tree Td between vertex 1 and the vertices not in T k is (1 + o(1))(2ζ(3) π 2 /6)/d 0.75/d. 2 Probability results In this section we present some basic results in combinatorial probability that will be used in the proofs of the theorems above, but which are perhaps of independent interest. Proofs of these results are presented in the next section. Let k be a positive integer, and let [k] denote the set {1, 2,..., k}. Given a vector x = (x 1,..., x k ) in [k] k, define t j = t j (x) for j = 1, 2,..., k in turn by setting t j to be the least index i distinct from t 1,..., t j 1 such that x i [j] (where we let t j = if there is no such i). Note that t j is the number of indices one has to transverse in x in order to find one more element of x that is in [j]. Thus t 1 is the index of the first 1 in x, t 2 is the index of the first 2 or of the second 1 in x, and so on. We are interested in particular in the index t k observe that always t k [k] and in the value v(x) = k + 1 x tk (x). For example, if k = 4 and x = (3, 1, 4, 2) then (t 1,..., t 4 ) = (2, 4, 1, 3) and v(x) = 5 x 3 = 1. Observe also that v(x) = k if and only if t k (x) = k and x k = 1. We shall consider a random vector X = (X 1,..., X k ) taking values in [k] k, where the X i are uniformly distributed on [k] or have a similar distribution, and investigate the distributions of t k (X) and of v(x). However, before we do that let us note a useful alternative noniterative characterisation of t k, in terms of congestion at the large values in {1,..., k}. Lemma 1 Let x [k] k and let j [k]. Then t j (x) t if and only if, for some r 1, there are at least r indices i t with x i {j r + 1,..., j}. 4

5 It is convenient to move our point of reference to the lower end of {1,..., k}. Given a vector x = (x 1, x 2,...) of positive integers, define u(x) to be the least t such that for some j we have {i t : x i [j]} j (and let u(x) = if there is no such t). In other words, u(x) is the first index i for which the number of elements of x up to that point that fall into some set [j] is at least as large as j (so that we have congestion amongst the small values). Further, let w(x) = x u(x). Thus w(x) is equal to the first element of x that is the jth element of x in the set [j] for some j (and so causes congestion). From Lemma 1 we immediately obtain: Lemma 2 For any x [k] k, t k (x) = u(k+1 x), and so v(x) = w(k+1 x), where k+1 x denotes the vector with ith co-ordinate k+1 x i. In the above example, with k = 4 and x = (3, 1, 4, 2), we have u(5 x) = u(2, 4, 1, 3) = 3 = t 4 (x) and w(2, 4, 1, 3) = 1 = v(x). The random vectors X = (X 1,..., X k ) taking values in [k] k which we consider will always have the symmetry property that P (X = x) = P (X = k+1 x). Thus t k (X) and u(x) will have the same distribution, as will v(x) and w(x). The results below are phrased in terms of t k (X) and v(x), as that is what we need for our applications to tree-growing in random networks, though the proofs are in terms of the corresponding quantities u(x) and w(x). Theorem 4 Suppose that the random vector X = (X 1,..., X k ) takes values in [k] k, and is obtained by sampling uniformly, with or without replacement, from a universe of nk objects, where n have the value i for each i = 1,..., k. Then t k (X) is uniformly distributed on [k]. Now let us restrict our attention to the case of the uniform distribution: Lemma 6 below will allow us to apply the results to other cases. Let V k denote the random variable v(x), when the co-ordinates X i are independent and uniform on [k]. Corollary 5 For any k 1, P (V k = 1) = k and P (V k = k) = 1 k 2. The above result follows easily from the preceding theorem. With a little more work, we obtain the complete distribution of V k, though not in a particularly attractive form. 5

6 Theorem 6 For each j [k], P (V k = j) = k 1 t t=0 r=j 1 ( ) t (r + 1) r 1 k (r+1) (1 r + 1 r k )t r (1 t r k r 1 ), where the summand for r = t = k 1 is interpreted to mean k 2. It is straightforward to deduce from Theorem 6 (or directly from Theorem 4) that, if k 2, then P (V k = 2) = k (k 1 k )k 1. Thus, as k, P (V k = 1) 1 and P (V 2 k = 2) 1 1. The following 2 e corollary to Theorem 6 extends these limiting results. Corollary 7 As k, V k converges in distribution to a random variable V, where P (V = j) = t t 2 e t /t! for j = 1, 2,.... Note that P (V = 1) = 1 2 infinite mean). t=j and that V is finite almost surely (though it has 3 Proofs of probability results We first prove the combinatorial result Lemma 1, which provides a noniterative definition of the t j s, and which immediately yields Lemma 2. It is possible to give a proof using Hall s theorem but quicker to work from first principles. Proof of Lemma 1 Let 1 r j, let I = {i t : x i {j r+1,..., j}}, and suppose that I r. Note that I is the set of indices in [t] such that the corresponding elements of x are among the r largest elements of the set [j], and therefore are not in any of the sets [1], [2],..,[j r]. Thus none of t 1,..., t j r are in I, and so each of the r quantities t j r+1,..., t j will be picked from [t]. Thus t j (x) t. For the converse result, assume that for each r 1 there are at most r 1 indices i t with x i {j r + 1,..., j}. We must show that t j (x) > t. This 6

7 is clearly true if j = 1, since by our assumption x i > 1 for each i = 1,..., t. Suppose that it is false, and that j > 1 is the smallest index for which it is false (for any x). It is easy to see that t j (x) > t if and only if t j ((x 1,..., x t )) > t, if and only if t j (y) > t, where y is obtained by rearranging the co-ordinates x 1,..., x t in nondecreasing order (consider the effect of a single transposition). We shall show that t j (y) > t, which gives a contradiction and so completes the proof. If y 1 > j then clearly t j (y) > t and we are done, so suppose that y 1 j. Also, by our assumption x i j for each i = 1,..., t. Let s be the maximum i t such that y i j. Then s 1 and y s j 1. Now let r 1 and let I = {i s 1 : y i {j r,..., j 1}}. If I then y s {j r,..., j 1}, and so I = {i s : y i {j r,..., j 1}} 1 = {i t : x i {j r,..., j}} 1 r 1. Hence, by the minimality of j as a counterexample, we must have t j 1 ((y 1,..., y s 1 )) > s 1, that is t j 1 ((y 1,..., y s )) s. But now, since y 1 y 2, we have t j (y) = t j ((y 1,..., y s )) > t j 1 ((y 1,..., y s )) s. Hence t j (y) > s, and so t j (y) > t, and we are done. Next we prove Theorem 4. Define an equivalence relation on [k] k by setting x y if for some z we have x i y i + z (mod k) for each i = 1,..., k. Note that each equivalence class has exactly k members. Let us say that the random vector X = (X 1,..., X k ) taking values in [k] k is locally uniform if P (X = x) = P (X = y) whenever x y. The following lemma is a more general version of Theorem 4. Lemma 3 In each equivalence class, for each j = 1,..., k there is exactly one member x with u(x) = j. Hence, if the random vector X taking values in [k] k is locally uniform, then u(x) is uniformly distributed on [k]. Proof Let x y with x y. It suffices to show that u(x) u(y). Suppose for a contradiction that u(x) = u(y) = t. We may assume that 1 y t < x t k. Let z = x t y t, so that 1 z < k. Let r 1 be such that there are r indices i t with x i [r], and suppose that r is as small as possible. Note that x t r (or else u(x) < t), and so z < r. Hence, by the 7

8 minimality of r, there are at most z 1 indices i t with x i [z]. Hence there are at least r z + 1 indices i t with x i {z + 1,..., r}. But now there are at least r z + 1 indices i t with y i [r z], which forces u(y) < t. Proof of Corollary 5 Let X be uniformly distributed on [k] k. Recall that V k and w(x) have the same distribution. By Lemma 3, Thus we have Similarly, P (u(x 1,..., X i 1 ) > i 1) = (1 (i 1)/k). P (w(x) = 1) = P (X u(x) = 1) = P (u(x 1,..., X i 1 ) > i 1 and X i = 1) = i=1 i=1 (1 (i 1)/k)(1/k) = (k + 1)/2k. P (w(x) = k) = P (X u(x) = k) = P (u(x 1,..., X k 1 ) > k 1 and X k = k) = 1/k 2. Next we shall prove Theorem 6. Given a vector x = (x 1,..., x t ) of positive integers, define ρ(x) as follows: if u(x) t then ρ(x) =, and if not then ρ(x) is the maximum r 0 such that {i t : x i [r+1]} r. For example, if k = 4 and x = (3, 2, 4) then ρ(x) = 3. Given t 0 and a vector x of length at least t, let ρ t (x) denote ρ((x 1,..., x t )) if t 1, and let ρ 0 (x) = 0. Thus 0 ρ t (x) t or ρ t (x) =. Note that, if ρ t 1 (x) = r and x t [r + 1], then u(x) = t. Lemma 4 Let X be uniformly distributed on [k] k, and let 1 t k 1 and 0 r t. Then ( ) t P (ρ t (X) = r) = (r + 1) r 1 k r (1 r + 1 r k )t r (1 t r k r 1 ), 8

9 where for r = t = k 1 the formula is interpreted to mean 1/k. Proof Call x = (x 1,..., x r ) a trigger if ρ(x) = r. Thus x is a trigger if and only if each x i [r + 1] and u(x) > r. Observe that, if ρ(x) is finite, then it is the maximum length of a subsequence of x that is a trigger. (The null sequence is also a trigger.) Consider first the case r = t. Now ρ t (x) = t if and only if (x 1,..., x t ) is a trigger. Hence, by Lemma 3, P (ρ t (X) = t) = ( t + 1 k )t 1 t + 1, which is as required. Now let 0 r < t. By the above, ρ t (x) = r if and only if, setting S = {i t : x i r + 1}, we have S = r, (x i : i S) is a trigger and u(x i r 1 : i [t] \ S) > t r. Now, as above, the probability p 1 that (X 1,..., X r ) is a trigger is given by p 1 = ( r + 1 k )r 1 r + 1, and the probability p 2 that each of X r+1,..., X t is at least r + 2 and u(x r+1 r 1,..., X t r 1) > t r is given by But p 2 = (1 r + 1 k )t r (1 t r k r 1 ). P (ρ t (X) = r) = and so we obtain the formula as required. ( ) t p 1 p 2, r Proof of Theorem 6 Let X be uniformly distributed on [k] k. Note that u(x) = t and w(x) = j j 1 ρ t 1 (x) < and x t = j. Hence, for j [k], P (w(x) = j) = = P (u(x) = t, X t = j) t=1 t 1 P (ρ t 1 (X) = r) 1 t=1 r=j 1 k. 9

10 Theorem 6 now follows from Lemma 4. Proof of Corollary 7 Let X be uniformly distributed on [k] k. Consider a fixed j 1, and let k. Then by Lemma 4 and its proof, Hence, as k, and it follows that P (w(x) = j) P (w(x) = j + 1) = 1 P (ρ t 1 (X) = j 1) k t=1 = 1 ( ) t 1 j j 2 k (j 1) (1 j k t=j j 1 k )t j (1 t j k j ) (j j 2 /(j 1)!) 1 ( t k t=1 k )(j 1) exp( jt k )(1 t k ) 1 (j j 1 /j!) x j 1 (1 x)e jx dx 0 = j j 2 e j /j!. P (w(x) = j) P (w(x) = j + 1) j j 2 e j /j!, P (w(x) = j) p j = t t 2 e t /t!. t=j It remains to check that S = j 1 p j satisfies S = 1. Observe first that S = t t t 2 e t /t! = t t 1 e t /t!. t 1 t 1 Now consider the following special case f(z) = k 0(k + 1) k 1 z k /k! of the exponential generating series, which is known to satisfy the identity f(z) = exp(zf(z)) (see for example (5.66) and (5.67) in [13]). Now zf(z) = t t 1 z t /t!, t 1 which equals S when z = 1/e. But by the identity above, x = f(1/e) satisfies x = exp(x/e), and so x = e. Hence, S = (1/e)f(1/e) = 1 as required. 10

11 4 The procedures SIM n In this section we introduce and investigate procedures SIM n, as a step towards completing the proofs of the theorems on tree-growing. It is convenient to work initially with more general procedures SIM A. We find that we can analyse the procedures SIM n quite precisely, using our earlier results. In the next section we shall introduce a procedure SIM which simulates the tree-growing method. This procedure is rather complicated to analyse, but we can handle it by relating it to the procedures SIM n. We shall prove the theorems on the tree-growing method by trapping SIM between SIM n and SIM, where is the maximum degree of the graph, δ is the minimum degree and n = δ k + 1. Let k and be positive integers. Let A = (A 1,..., A k ), where each A l [ ], and A l k. The procedure SIM A operates on a vector y = (y i,j : i [k], j [ ]) of distinct real numbers, as follows. It maintains sets A A l A l for each l [k], initially set equal to A l. At each step l [k], let the minimum value in (y i,j : i [l], j A A i ) occur at (i A l, jl A ) (so i A 1 = 1); let wl A be the minimum value y i A l,jl A ; and then delete jl A from A A. Also, let i A l zl A = l + 1 i A l. Lemma 5 Let A = (A 1,..., A k ) and B = (B 1,..., B k ), where each A l B l [ ], and A l k. Consider the operation of the two procedures SIM A and SIM B on some vector y. (a) For each l [k], we always have A A l Bl B, and so wl A wl B. (b) If jl B A A for each l [k], then SIM A and SIM B behave identically on i B l y, so that in particular i A l = i B l and wl A = wl B for each l [k]. (c) Now suppose further that k l=1 B l \ A l = h. Then {(i A l, j A l ) : l [k]} {(i B l, j B l ) : l [k]} 2h. Proof Parts (a) and (b) are straightforward, so consider part (c). It is sufficient to consider the case h = 1. (Note that even in this case, the ordered k-tuples ((i A l, jl A ) : l [k]) and ((i B l, jl B ) : l [k]) may be quite different.) Let l 0 [k] and suppose that B l0 \ A l0 = {j 0 }. Then, for each m = 1,..., k, {(i A l, jl A ) : l [m]} = {(i B l, jl B ) : l [m]} 11

12 if (l 0, j 0 ) {(i B l, j B l ) : l [m]}, and {(i A l, j A l ) : l [m]} {(i B l, j B l ) : l [m]} = 2 if (l 0, j 0 ) {(i B l, j B l ) : l [m]}. For, consider some step m < k where the latter relation holds: either (i A m+1, j A m+1) = (i B m+1, j B m+1), or (i B m+1, j B m+1) {(i A l, j A l ) : l [m]}. Consider SIM A, where each A l = [n] (and k n ). In this case we write SIM n, A n l, i n l, jl n, wl n for SIM A, A A l, i A l, jl A, wl A. Now let Y = (Y i,j : i [k], j [ ]) be a family of independent random variables, each with the continuous distribution function F. When SIM n operates on a random point Y n = (Y i,j : i [k], j [n]), we write Il n, Jl n, Zl n, Wl n for i n l, jl n, zl n, wl n. We need to relate the distributions obtained by sampling with and without replacement as in Theorem 4. Lemma 6 Let k and n be positive integers; let X be uniformly distributed on [k] k ; and let Y be obtained by sampling k times uniformly without replacement from a universe of nk objects, where n have the value i for each i = 1,..., k, as in Theorem 4. Then d(x, Y) (k 1)/2n. Proof Let Z 1, Z 2,..., Z k be independent, with each uniformly distributed on [k] [n]. Let R be the event that Z i = Z j for some 1 i < j k. Let f((i, j)) = i and if z = (z 1,..., z k ) then let f(z) = (f(z 1 ), f(z 2 ),..., f(z k )). Let A [k] [n]. Then P (X A) = P (f(z) A), and It follows that 0 P (Y A) P ((f(z) A) (Z R)) P (Z R). P (X A) P (Y A) P (Z R) ( ) k /(kn). 2 We shall need the following notation. Given 1 j k and 1 n, list the jn random variables (Y i,l : i [j], l [n]) in increasing order, as S j,n = (S j,n 1, S j,n 2,..., S j,n jn ). In the next lemma we establish the connection between the earlier probability results and the analysis of SIM n. 12

13 Lemma 7 (a) The total variation distance between Zk n d(zk n, V k ) (k 1)/2n. Further and V k satisfies P (Z n k = k) 1 k 2 1 n + 1 k 2 n. (b) The distribution of Wk n is exactly that of a value picked uniformly at random from S k,n 1, S k,n 2,..., S k,n k, the k smallest values of the kn random variables in Y n. Proof Define the random vector X = (X 1, X 2,..., X k ), by letting X l = i if S k,n l is Y i,j for some j. Note that the distribution of X is that of sampling without replacement as in Theorem 4. Then Ik n = X tk (X) and so Z n k = k + 1 I n k = k + 1 X tk (X) = v(x). Hence by Lemma 6 d(zk n, V k ) (k 1)/2n, as required. Now consider P (Zk n = k) = P (Ik n = 1). Let A be the family (A 1,..., A k ) where A 1 = [n+1] and A 2 =... = A k = [n]. The procedure SIM A generates the random variable Ik A. Then, arguing as in the proof of Corollary 5, we find P (Ik A = 1) = 1 n + 1 k kn + 1, and so Also, 1 k 2 P (IA k = 1) 1 k k 2 n. P (Ik n = 1) P (Ik A = 1) k kn + 1 < 1 n. Thus P (Ik n = 1) 1 k < 1 2 n + 1 k 2 n. This completes the proof of part (a). To prove (b), recall that Wk n = S k,n t k (X). By Theorem 4, t k(x) is uniformly distributed on [k]. Also, note that X is independent of S k,n and hence so is t k (X). This completes the proof. 13

14 5 The procedure SIM In this section we introduce the simulation SIM of the tree growing method, and relate it to the procedures SIM n discussed in the last section. We first recall the tree-growing method for finding a minimum spanning tree, and introduce some notation. Let G = (V, E) be a connected graph with edge-weights w(e) that are all distinct. If T is a subtree in G containing the vertex v, we let E(v, T ) denote the set of edges between v and the vertices not in T. The tree-growing method operates as follows. At step 0, it chooses a vertex v 1 (without looking at the edge costs), and forms the initial tree T 1 consisting only of v 1. At step 1, let the minimum value of w(e) for e E(v 1, T 1 ) occur at e 1. Let I 1 = 1. Let v 2 be the end node of e 1 not in the current tree T 1. Add the edge e 1 and vertex v 2 to the tree T 1 to form the tree T 2. Now let 1 < l < n, and consider step l. We start with a tree T l in G on vertices v 1,..., v l with edges e 1,..., e l 1. Let the minimum value of w(e) for e i [l] E(v i, T l ) occur at e l E(v Il, T l ). Let v l+1 be the end node of e l not in the current tree T l. Add the edge e l and vertex v l+1 to the tree T l to form the tree T l+1. Suppose that the connected graph G has n vertices, maximum degree and minimum degree δ. Let k [n]. When the tree-growing method is applied for k steps to the graph G, where the edges have independent nonnegative costs each with the continuous distribution function F, it yields random variables I l, w(e l ) for l [k]. Let Z k = k + 1 I k. Note that Z k = s if and only if the edge e k added to the tree T k is incident with the vertex v k that was added s steps earlier to T k. Recall that Y = (Y i,l : i [k], l [ ]) is a family of independent random variables, each with distribution function F. Also, given 1 j k and 1 n, the jn random variables (Y i,l : i [j], l [n]) listed in increasing order form S j,n = (S j,n 1, S j,n 2,..., S j,n jn ). Let us denote the random variable S j,n j by the shorter form Sj n. Lemma 8 There is a randomised procedure SIM, which operates on G together with the family Y, such that the following properties hold. Let n = δ k

15 (a) SIM yields random variables I l, W l = w(e l ) for l [k], which have exactly the same joint distribution as with the tree-growing method. SIM yields also random variables J 1,..., J k described in the proof. (b) Wl W l Wl n Sl n almost surely for each l [k]. (c) Let the random variable D be the number of indices l [k] such that J l [n]. Then (i) On the event D = 0, I l = Il n (ii) Always and W l = W n l for each l [k]. {(I l, J l ) : l [k]} {(I n l, J n l ) : l [k]} 2D. (iii) P (D > 0) E(D) ɛ, where ɛ = k( δ + 2k 2)/. Note that we are using the same notation for the random variables generated by the tree-growing method and for those generated by the simulating procedure SIM: it should always be clear to which we are referring. Proof We first describe the procedure SIM. It is very similar to the treegrowing method, except that initially the random weights are not assigned to specific edges, and we delay making assignments as long as possible. At step 0, it starts with the initial tree T 1 consisting only of the vertex v 1, as with the tree-growing method. At step 1, let A 1 = [ E(v 1, T 1 ) ] (so that A 1 equals the degree of v 1 in G). Let the minimum value in {Y 1,j : j A 1 } occur at (I 1, J 1 ) (so that I 1 = 1). Let e 1 be an edge picked uniformly at random from the set E(v 1, T 1 ), let W 1 = w(e 1 ) be the minimum value Y I1,J 1, and let v 2 be the end node of e 1 not in the current tree T 1. Add the edge e 1 and vertex v 2 to the tree T 1 to form the tree T 2. Delete J 1 from A 1. Now let 1 < l k and consider step l. It starts with a tree T l in G on vertices v 1,..., v l with edges e 1,..., e l 1. For each i [l 1], we have a set A i [ ] with A i = E(v i, T l ). Further, if some Y i,j has been picked as a minimum value, then j A i. Let A l = [ E(v l, T l ) ]. Let the minimum value in i [l]{y i,j : j A i } occur at (I l, J l ). Let e l be an edge picked uniformly at random from the set E(v Il, T l ), let W l = w(e l ) be the minimum value Y Il,J l, and let v l+1 be the end node of e l not in the current tree T l. Add the edge e l and vertex v l+1 to the tree T l to form the tree T l+1. Delete J l from A Il. Also, for each i [l] \ {I l }, if there is an edge in G between v i and v l+1 then 15

16 delete the last member of A i. [These last deleted indices correspond to the nontree edges between T l and the vertex v l+1.] This completes the description of the procedure SIM. Next we show that this procedure does indeed simulate precisely the action of the tree-growing method, in that the joint distribution of the e l, W l = w(e l ), I l, v l, T l is exactly the same as with that method. To see this, let l [k], and suppose that the joint distributions are identical up to the start of step l. Assume that each process has generated the same values v 1,..., v l, e 1,..., e l 1, w(e 1 ),..., w(e l 1 ), T 1,..., T l. It follows easily that A i = E(v i, T l ) for each i [l]. With the tree-growing process, we are now to find the minimum value of w(e) for e i [l] E(v i, T l ). For each i [l] and e E(v i, T l ), the information that we have learned about w(e) is precisely that w(e) > b i, where the bound b i = max{w(e i ),..., w(e l 1 )} (and b l = 0). Hence, all the random variables involved are still (conditionally) independent, and for each i [l] the random variables (w(e) : e E(v i, T l )) are identically distributed, each having the distribution of a random edge-weight conditioned on being greater than b i. It follows that, for each fixed i [l], each edge e E(v i, T l ) is equally likely to be the next minimum e l. But this corresponds exactly to what happens with SIM, where for each i [l] the family of random variables (w(e) : e E(v i, T l )) is replaced by the family (Y i,j : j A i ) which has exactly the same distribution (that is, conditional on the past history, the random variables (Y i,j : i [l], j A i ) are independent, and the distribution of Y i,j is that of a random edge-weight conditioned on being at least b i ). Hence, the two conditional distributions on the behaviour of step l are identical. This completes the proof of part (a) of the lemma, that SIM does indeed simulate the tree-growing process. Now we relate the procedure SIM to the procedures SIM n and SIM. In the first k 1 steps, from each set A l at most k 1 elements can be deleted. It follows that always A n l A l A l for each l [k 1], and part (b) of the lemma follows as in Lemma 5 (a). Now consider part (c). The subparts (i) and (ii) follow as in Lemma 5. It remains only to consider subpart (iii). In the first k 1 steps, we always have A l \ [n] A l n (k 1) ɛ/k. 16

17 Hence P (J l [n]) is always at most ɛ/k, and so we obtain the desired upper bound for E(D). 6 Completing the proofs Proof of Theorem 1 Let k [δ], let ɛ = k( δ + 2k 2)/ and let n = δ k + 1. Then, by Lemmas 7 and 8, the total variation distance between the random variables Z k and V k satisfies d(z k, V k ) P (D > 0) + d(z n k, V k ) ɛ + (k 1)/2n. In order to prove Theorems 2 and 3 we need two more lemmas. The first lemma gives some results which we shall need concerning the random variables Sk n ( = S k,n k ) and Wk n under the conditions on the distribution function F stated in Theorem 2. Lemma 9 Suppose that the common distribution function F of the (positive) random variables Y i,j has right derivative α > 0 at 0. (a) If the mean is finite, and if n, then αne(s n k ) 1 and αne(wk n ) ( k ) 0 uniformly over k [n]. (b) If both k and n, then the distribution of αnwk n uniform distribution on (0, 1). converges to the Proof Part (a) follows from Lemma 7(b) in a straightforward way. To prove part (b), let 0 < t < 1 be fixed. Given n > t, let N be the number of indices (i, j) such that Y i,j < F 1 (t/n), so that N has the binomial distribution B(kn, t/n). Also let M = min(n/k, 1). Then P (W n k F 1 (t/n)) = kn i=0 P (N = i) min(i/k, 1) = E(M). 17

18 Clearly E(M) E(N)/k = t; and it may be shown that in fact E(M) t as k, by using standard bounds on the tail of the binomial distribution. Thus P (Wk n F 1 (t/n)) t as k. But by our assumptions on F, αnf 1 (t/n) t as n. Hence P (αnwk n t) t as k, n, as required. Lemma 10 Let k [n] with k = O(n 1 2 ), and let Ŝn k = max j [k] S n j. Then E(S n k 1 (D>0) ) E(Ŝn k D) (4/α)(log n/n) E(D) + o(1/n). Proof Let j [k], let 0 t n, and assume that F (t/n) = p 1 2. Then j 1 ( ) P (Sj n jn > t/n) = p i (1 p) jn i i=0 i ( ) j jn (1 p) jn i=0 i ( ) jn + j (1 p) jn j = ρ j, (e(n + 1)) j exp( jnp) where ρ = exp ( (nf (t/n) 1 log (n + 1))). Hence P (Ŝn k > t/n) ρ j ρ/(1 ρ). Now set t = (4/α) log n. Then ρ = n 3+o(1), and so P (Ŝn k > t/n) n 3+o(1). Now Ŝn k Y 1, Y k,1, and E(Y j,1 Ŝn k > t/n) E(Y j,1 Y j,1 > t/n) 2E(Y 1,1 ). Hence and so E(Ŝn k 1 ( Ŝ n k >t/n)) 2kE(Y 1,1)P (Ŝn k > t/n) kn 3+o(1) ; E(Ŝn k D) (t/n)e(d) + ke(ŝn k 1 ( Ŝk n >t/n)) (t/n)e(d) + o(1/n), 18

19 as required. Proof of Theorem 2 By Lemma 8, the vectors (w(e 1 ),..., w(e k )) and (W 1,..., W k ) have the same distribution, where the W i refer to the random variables generated by SIM. By the same lemma, the total variation distance between (W 1,..., W k ) and (W1 n,..., Wk n ) is at most P (D > 0) < ɛ. Thus in particular, d(w(e k ), Wk n ) < ɛ. But Ŵk and Wk n have the same distribution, by Lemma 7 (b), and so we have d(w(e k ), Ŵk) < ɛ. Further, always W k Wk n Sk n, and W k = Wk n on D = 0. Hence W k Wk n Sk n 1 (D>0), and so E(W k ) E(W n k ) E(S n k 1 (D>0) ) (4/α)(log n/n) ɛ + o(1/n), by Lemmas 8 and 10. Lemma 9 now completes the proof. Proof of Theorem 3 Let n = δ k +1. The expected degree of vertex 1 in the tree T k is E( 1 (Zj =j)) = E( 1 (Z n j =j)) + E (1 (Zj =j) 1 (Z n j =j)). The first term on the right hand side here equals k P (Zj n = j), and by Lemma 7 this differs from k 1/j 2 by at most 2k/n. We must show that the second term the error term is small. But by Lemma 5(c), on the event D i, (1 (Zj =j) 1 (Z n j =j)) i. Hence the absolute value of the error term is at most E(D), and E(D) < ɛ by Lemma 8. Thus the expected degree of vertex 1 in the tree T k differs from k 1/j 2 by at most 2k/n + ɛ. Now we consider weights. Note that Z j = j if and only if I j = 1, and similarly Z n j = j if and only if I n j = 1. Thus the expected weight of the edges between vertex 1 and the vertices in the tree T k equals E(W j 1 (Ij =1)) = E(Wj n 1 (I n j =1)) + E (W j 1 (Ij =1) Wj n 1 (I n j =1)). 19

20 Again, we have a main first term and an error term. To consider the first term, note that on the event I n j = 1, we have W n j = S n j, and note also that S n j and I n j are independent. Thus by Lemma 7. Hence, by Lemma 9, E(W n j 1 (I n j =1)) = E(S n j 1 (I n j =1)) = E(S n j )P (I n j = 1) = (1 + o(1))e(s n j )/j 2, E(Wj n 1 (I n j =1)) = (1 + o(1)) E(S n j )/j 2 = (1 + o(1)) 1 αd 1 j 2. It remains to show that the error term above is o(1/d). But 0 W j, Wj n Sj n Ŝn k. Hence, by Lemma 8, E (W j 1 (Ij =1) Wj n 1 (I n j =1)) E(Ŝn k D) = o(1/d), by Lemma Concluding remarks We have been able to understand quite well the early stages of the operation of Prim s tree-growing method on a random network. Perhaps we were fortunate that there were such elegant underlying probability results. It would be interesting to find out what happens at later stages in the process, and indeed to investigate similarly the behaviour of other algorithms on this or other problems on random networks. What we learn may be of value in designing detailed implementations of the algorithms. We are grateful to the referees for a careful reading of the manuscript, and we would like to acknowledge their helpful comments. 20

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