On the uniform edge-partition of a tree
Bang Ye Wu
∗Økø
Shu-Te UniversityHung-Lung Wang
ÙûO
National Taiwan University
Shih Ta Kuan
0®
Shu-Te University
Kun-Mao Chao
_
National Taiwan University
Abstract
We study the problem how uniformly one can partition the edge set of a tree with n edges into k connected components, where k ≤ n. The objective is to minimize the ratio of the maximum to the minimum number of edges of the subgraphs in the partition. We show that, for any tree and k ≤
4, there exists a k-split with ratio at most two.
For general k, we propose a simple algorithm that finds a k-split with ratio at most three in O(n log k) time. Experimental results on random trees are also shown.
1
Introduction
In this paper, we study the problem of splitting a tree into k parts with approximately equal num-ber of edges in each part subject to that the edges in each part are connected. How well can one do it?
More formally, we define a k-split of a tree T as follows. Let T be a tree and 1 ≤ k ≤ e(T ). A
k-tuple (T1, T2, . . . , Tk) is a k-split of T if (1) each
Tiis a connected subgraph of T ; and (2) Tiand Tj
are edge disjoint for i 6= j; and (3) the union of all the subgraphs form the whole tree T . To measure how equal a partition is, three natural objectives are usually used.
• To minimize the maximum (min-max). • To maximize the minimum (max-min). • To minimize the ratio of the maximum to the
minimum (min-ratio).
For partitioning into two parts, all the three ob-jectives are equivalent, but they differ when the number of parts is larger than two. For the k-split of a tree, the worst cases for both min-max and max-min objectives are trivial. However, it is much involved for the min-ratio problem. We
∗Corresponding author
show that, in the algorithmic aspect, to find the optimal k-split of a tree with respect to each of the three objectives is NP-hard, even for unweighted tree. We focus on the worst case analysis of the ratio, and prove that, for any tree and k ≤ 4, there exists a k-split with ratio at most two. For gen-eral k, we propose a simple algorithm that finds a k-split with ratio at most three in O(n log k) time. Experimental results on random trees are also shown.
Graph partition is an important problem in computer science. It finds applications in par-allel computing, data storage and segmentation, and operation research. Most of the previous re-search was devoted to the vertex partition, and many variants of the problem have been defined and investigated with different objectives and con-straints. Most of the problems on general graphs are NP-hard [1, 10]. For the vertex partition of a tree, polynomial time algorithms for both the min-max and the min-max-min objectives were developed [5, 7, 15, 17]. Becker and Perl [6] summarized their previous results with some other co-authors and showed that the tree vertex partition problem of several other objective functions can also be solved by using the shifting algorithm. An open problem in that paper is the most uniform vertex
partition-ing problem for trees, in which the objective is to
minimize the difference of the maximum and the minimum weights of the vertex set in the parti-tion. For a special case that the tree is a path, a solution was given in [16]. One can image that the problem is more difficult than the min-max or max-min problem since both the smallest and the largest parts are concerned. In this paper, we shall see that it is similar for the edge partition problem. In addition to the algorithmic aspect, some properties of the tree vertex partition have also been studied [9, 14].
The study on edge partition is helpful for the multiserver routing problem. In such a problem, we are given a network, k identical servers, a set of demand points, and a home location. The servers are required to visit some set of demand points
and return back to the home in a “fair” way such that each demand point is visited by at least one server. In [2], Averbakh and Berman showed that the problem is NP-hard even for an edge weighted tree network and k = 2. They gave approximation algorithms for the problem and a variant that the two servers have different home locations. The au-thors further developed approximation algorithms for the case of k servers and some other variants of the problem [3].
Another application of the tree edge partition is for Multiple Sequence Alignments (MSA), which is important in computational biology. To align-ment a set of n sequences, we insert gaps into the sequences and arrange their characters into columns with n rows, one from each sequence. The MSA problem has typically been formalized as op-timization problems with different objectives. The
sum-of-pairs or SP objective for multiple
align-ment is to minimize the sum, over all pairs of se-quences, of the pairwise distance between them in the alignment. The SP-alignment problem is NP-hard. A progressive approach for MSA incremen-tally builds the alignment by inserting sequences one at a time but the resulting SP-cost heavily de-pends on the inserting order and there is no per-formance guarantee. One way to get an alignment with worst case guarantee is to employ a tree, in which each vertex represents one sequence, and perform the progressive alignments along the tree edges [4, 11, 18, 20]. More details about MSA and algorithms for constructing such a guide tree are referred to [12, 19]. For n sequences, each with length m, the time complexity of such an approach is O(nm2), and it may be very time consuming for
large n and m. Since one tree edge corresponds to performing a pairwise alignment, a k-split of the tree partitions the whole work into k parts and derives a parallel algorithm for the problem. To balance the working load, a k-split with small ratio should be applied.
The remaining sections are organized as follows. In Section 2, we define some notations, explain the computational complexity of the problem, and show some preliminary results. The worst cases of the min-ratio for k = 3 and k = 4 are discussed in Section 3. In Section 4, we show a simple algo-rithm for general k with ratio at most three, and present some experimental results. Finally. con-cluding remarks are given in Section 5.
2
Notations and Preliminaries
Let E(T ) denote the edge set of a tree T and e(T ) denote the number of edges of tree T . Throughout this paper, n = e(T ). An edge with endpoints u and v is denoted by (u, v). Let T be a rooted tree and v be a vertex of T . We use Tv todenote the subtree rooted at v, i.e., the subgraph induced on v and all its descendants. Let u be a child of v. The subgraph Tu ∪ (u, v) is called a
branch of v.
Definition 1: Let T be a tree, 1 ≤ k ≤ e(T ), and p ≥ 1. The ratio of a k-split (T1, T2, . . . , Tk)
of T is defined by maxi{e(Ti)}
mini{e(Ti)}.
By T = A ] B, we denote that T is split into A and B, i.e., the edge sets of the two subgraphs form a partition of E(T ). It is also noted that A and B share a common vertex if T = A ] B. By
T = A ] B ] C, we understand a 3-split (A, B, C)
of T , in which B intersects with both A and C. It includes the case that the three subgraphs share a common vertex.
Problem: Minimum Ratio k-Split Instance: Tree T and an integer 1 <
k < e(T ).
Goal: Find a k-split of T with minimum ratio.
The min-max and the max-min k-split problem are defined similarly except that the objectives are to minimize the maximum subgraph, and to maxi-mize the minimum subgraph respectively. We can easily show that all the three problems are NP-hard by a simple reduction to the following prob-lem.
Problem: 3-Partition
Instance: A bound B ∈ Z+and a set A
of 3m integers ai, 1 ≤ i ≤ 3m, satisfying
B/4 < ai< B/2 and
P
iai= mB.
Question: Can A be partitioned into
m disjoint sets Ai, 1 ≤ i ≤ m, such that
P
a∈Aia = B for 1 ≤ i ≤ m (note that
each Ai must therefore contain exactly
three elements from A)?
Given A and B as an instance of the 3-partition problem, we construct a tree T consisting of a root
r and 3m branches Yi, 1 ≤ i ≤ 3m, incident with
the root, in which Yi is an arbitrary tree of ai
edges. It is easy to see that there exists a m-split of T with ratio one if and only if the answer of the 3-partition problem is “yes”. Since the 3-partition
problem is NP-complete in the strong sense [10], we have the following result.
Theorem 1: The Minimum Ratio k-Split prob-lem is NP-hard.
Obviously, the reduction remains true for the min-max and the min-max-min objective functions. Corollary 2: Both the Min-Max and the Max-Min k-Split problems are NP-hard.
The next lemma appeared in [13]. For our con-venience, we rewrite it and give a proof for the completeness.
Lemma 3 : Let T be a rooted tree. For any 1 ≤ γ ≤ e(T ), we can split T into (T1, T2) at a
vertex v in linear time such that γ ≤ e(T1) ≤ 2γ,
in which v is a vertex satisfying e(Tv) ≥ γ and
e(Tu) < γ for any child u of v.
Proof: In linear time, we can traverse the tree in the post order and compute the number of edges for the subtree rooted at each vertex. Such a ver-tex v can be easily found while traversing the tree. Assume that B1, B2, . . . , Bk are the branches at
v. If e(Tv) = γ, we have done. Otherwise, we
can find j < k such that Pj−1i=1e(Bi) < γ and
Pj
i=1e(Bi) ≥ γ. Since e(Bj) ≤ γ, we have that
Pj
i=1e(Bi) ≤ 2γ. The union
Sj
i=1Bi is the
de-sired subgraph.
Taking γ = n/3 in Lemma 3, we have the fol-lowing result. To show that the bounds are tight, consider a tree consisting of exact three branches, each with n/3 edges, incident with the centroid. A centroid of a tree is a vertex such that no branch contains more than one half of the vertices if we root the tree at its centroid.
Corollary 4: For any tree T , there is a 2-split of T with ratio at most two. The numbers of the two subgraphs are at most 2n/3 and at least n/3. Furthermore, such a 2-split can be found in O(n) time and the bounds are tight.
The following simple result shows an upper and a lower bounds for the sizes of the subgraphs in a
k-split with a limited ratio.
Lemma 5: If (T1, T2, . . . , Tk) is a k-split of T with
ratio r, then, for each subgraph Ti, r(k−1)+1n ≤
e(Ti) ≤ k+r−1rn .
Proof: Let x be the number of edges of the maximum component. Since the number of edges
of the minimum component is no more than the mean of the remainder, i.e., n−x
k−1,
x ≤ r(n − x) k − 1 .
Solving the inequation, we have x ≤ rn
k+r−1.
Simi-larly, let y denote the minimum number of edges. The maximum is no less than the mean of the remainder, n−yk−1, and we have y ≥ r(k−1)(n−y), which implies y ≥ n
r(k−1)+1.
Particularly, for r = 2, the number of edges of each subgraph in a k-split with ratio at most two is between n
2k−1 and k+12n . But the converse of the
lemma is obviously not true. Corollary 4 shows the worst case of k = 2 for the min-max, the max-min, and the min-ratio objectives. For the min-max objective, it is trivial to extend to k-split and show that 2n
k+1 is a tight bound by induction. Given a
tree T , by Lemma 3, we find T = T1]T0such that
n
k + 1 ≤ e(T1) ≤
2n
k + 1.
Suppose by induction hypothesis that T0 can be
split into k − 1 subgraphs, each with at most 2e(T0)/k edges. Since e(T
1) ≥ n/(k + 1), the
num-ber of edges of each subgraph is upper bounded by 2(n − n/(k + 1))
k =
2n
k + 1.
The tightness of the bound can be easily shown by considering an extreme case in which the tree has
k + 1 branches at the root and each has exactly n/(k + 1) edges. Similarly, one can easily show
that n
2k−1 is the tight bound of the k-split with
max-min objective. Also, it can be easily shown that the extreme case for the min-max objective (or the max-min objective) is a witness to that the min-ratio is lower bounded by two. To be a worst case, it need to be shown that any tree can be split with ratio at most two. However, such a simple induction does not work for the min-ratio objective.
In the next lemma, we show that there is a sim-ilar result for the min-ratio but with a stronger condition, and this result will be used as one of the cases for proving the bound of the min-ratio. Lemma 6: Let T = Y ] T0and (T
1, T2, . . . , Tk−1)
be a (k − 1)-split of T0 with ratio at most 2. If n
k+1 ≤ e(Y ) ≤ 2k−12n , then (Y, T1, T2, . . . , Tk−1) is
a k-split of T with ratio at most two.
Proof: Let tmax = maxie(Ti), tmin =
show that tmax
2 ≤ y ≤ 2tmin. By Lemma 5,
tmax ≤ 2(n − y)/k and tmin ≥ (n − y)/(2k − 3).
Since y ≥ n/(k + 1), we have tmax y ≤ 2(n − y) ky = 2n ky − 2 k ≤ 2(k + 1) k − 2 k = 2. Similarly, since y ≤ 2n/(2k − 1), tmin y ≥ (n − y) y(2k − 3) = n y(2k − 3)− 1 2k − 3 ≥ 2k − 1 2(2k − 3)− 1 2k − 3 = 1/2.
3
Worst cases of 3-splits and
4-splits
Now let us consider the 3-split of a tree. By Lemma 6, if a tree T can be split into Y ] T0such
that e(T )k+1 ≤ e(Y ) ≤ 2e(T )2k−1, we can find a 3-split of T with ratio at most two. However, it is not always possible to find such a split. But it does not imply that there is no 3-split of ratio within two. In the following, we show that such a 3-split always exists for any tree. First, we establish a 3-split which will be used as a basis of our discussion for k = 3 and 4. In the remaining paragraphs, we shall use the following notations: Let x = e(X),
y = e(Y ), xi= e(Xi), and yi = e(Yi) for i = 1, 2.
Claim 7: For any k ≥ 3, a tree T can be split into X ] P ] Y such that
n
k+1≤ x, y ≤ k+12n .
Proof: Root T at an arbitrary vertex. By Lemma 3, we split T = X ] T1 at a vertex u such
that n
k+1 ≤ e(X) ≤ k+12n . Then, root T1 at u, and
we can split another subgraph Y , n
k+1 ≤ e(Y ) ≤
2n
k+1, from T1at a vertex v. Note that u and v are
not necessarily distinct.
Claim 8: Let 3 ≤ k ≤ 4 and X be a tree rooted at
u and 2n
2k−1 ≤ x ≤ k+12n . If each branch at u has no
more than n
k+1 edges, X can be split into X1 and
X2at u such that x1≥ x2and 2k−1n ≤ x1≤2k−12n .
Proof: First we show that a subgraph X1 can
be split from X at u such that n
2k−1 ≤ e(X1) ≤ 2n
2k−1. If there exists a branch of more than or
equal to n/(2k−1) edges, the branch is the desired subgraph since n/(k+1) < 2n/(2k−1). Otherwise, the result directly follows Lemma 3.
Second, we show that we can assume that x1≥
x2 without loss of the generality. Suppose that
x1< x2. Since, for k ≤ 5, x2= x − x1< 2n k + 1− n 2k − 1 < 2n 2k − 1, the number of edges of X2 is also in the desired
range, and we may exchange X1 and X2.
Theorem 9: For any tree T , a 3-split of T with ratio at most two can be found in O(n) time. Proof: By Claim 7, we can find T = X ]P0]Y
such that n
4 ≤ y ≤ x ≤
n
2. We consider the following two cases.
• Case 1: y ≤ 2n/5.
In this case T can be split into Y ] T1 such
that n/4 ≤ y ≤ 2n/5. By Corollary 4, there is a 2-split (P1, P2) of T1 with ratio at most
two. By Lemma 6, (Y, P1, P2) is a 3-split of
T with ratio at most two. • Case 2: 2n/5 < y ≤ x ≤ n/2.
As in Claim 8, we split X = X1] X2 such
that n/5 ≤ x1≤ 2n/5 and x1≥ x2, in which
x1 = e(X1) and x2 = e(X2). It should be
noted that X2∪ P0is connected since X1and
X2are split at the vertex shared by X and P0.
If x1≥ n/4, it is similar to Case 1. Otherwise
we have n/5 ≤ x1 < n/4. Since x1 ≥ x/2
and x ≥ y, it follows that x1 ≥ y/2. By
e(X2∪ P0) = n − x1− y, we have
n/4 < e(X2∪ P0) < 2n/5.
Consequently, (X1, X2∪ P0, Y ) is a 3-split of
T with ratio at most two.
We have show that there exists a 3-split with ratio at most 2 in both cases, and the proof is completed since the time complexity is obviouly O(n).
Next, we turn to the 4-splits. We show the following result.
Theorem 10 : For any tree T , there exists a 4-split of T with ratio at most two.
Proof: Similar to the proof of Theorem 9, we start at splitting T into X ] P0] Y as in Claim 7
such that n 5 ≤ y ≤ x ≤ 2n 5 . Case 1: y ≤ 2n/7.
In this case T can be split into Y ] T1 such that
n/5 ≤ y ≤ 2n/7. By Theorem 9, there is a
3-split (P1, P2, P3) of T1with ratio at most two. By
Lemma 6, (Y, P1, P2, P3) is a 4-split of T with ratio
at most two.
Case 2: 2n/7 < y ≤ x ≤ 2n/5.
By Claim 8, we split X = X1] X2such that x1≥
x2 and n/7 ≤ x1 ≤ 2n/7. If x1 ≥ n/5, it is
similar to Case 1, and therefore we assume that
n/7 ≤ x1 < n/5. Similarly we split Y = Y1] Y2
such that y1 ≥ y2 and n/7 ≤ y1 < n/5. Let
P = P0∪ X2, and we have T = X1] P ] Y as in
Figure 1.(a). Remember that 2x1≥ y.
By the property of a centroid, P can be split into three subgraphs P2, P1a and P1b (possibly
null) at its centroid in such a way that each of the subgraphs has no more than de(P )/2e edges. If there are only two branches and e(P ) is an odd number, we add a dummy edge incident with the centroid to simplify the proof. One may check that the correctness is not affected. Therefore we can assume that each of the three subgraphs has no more than e(P )/2 edges. Let P2 be the largest
and P1= P1a∪ P1b. We have
e(P2) ≤ e(P1) ≤ 2e(P2). (1)
Since x1< n/5 and y ≤ 2n/5, we have
e(P1) > n/5 > x1. (2)
Since x1≥ n/7 and y ≥ 2n/7, we have
e(P2) ≤ 1
2(n − x1− y) ≤ 2n
7 ≤ y. (3) By Eqs. (1)–(3) and x1 ≤ y ≤ 2x1, we further
divide this case into the following subcases:
• y/2 ≤ e(P2) ≤ e(P1) ≤ 2x1.
• e(P1) > 2x1.
• e(P2) < y/2.
For each case, we shall show that there exists a desired 4-split.
• Case 2.1: y/2 ≤ e(P2) ≤ e(P1) ≤ 2x1. In this
case (X1, Y, P1, P2) is a desired 4-split.
• Case 2.2: e(P1) > 2x1. we divide into two
subcases.
Case 2.2.1: P1 adjacent to X1. Let Q = P1∪
X1. Split Q into Q1and Q2 such that
e(Q)
3 ≤ e(Q2) ≤ e(Q1) ≤ 2e(Q)
3 . (4) We show that (P2, Y, Q1, Q2) is a desired
4-split. First, since e(P2) ≥ e(P1)/2 > x1 ≥
y/2, we have
e(P2) ≤ y ≤ 2e(P2) (5)
Since e(P1) > 2x1,
e(Q2) ≥1
3(e(P1) + x1) > x1≥ y/2. (6) Since x1< e(P2) and e(P1) ≤ 2e(P2),
e(Q1) ≤ 2
3(e(P1) + x1) < 2e(P2). (7) By Eqs. (4)–(7), the result follows.
Case 2.2.2: P2 adjacent to X1. In this case,
P2 contains X2 (Figure 1.(b)) since e(P2) >
x1 ≥ x2. Let P2a = P2− X2 and e(P1a) ≥
e(P1b). We show that (X, P2a∪ P1b, P1a, Y )
is a desired 4-split.
Since e(P2) ≥ e(P1a), e(P1) > 2x1, and x1≥
x2, we have
e(P2a) + e(P1b) = e(P2) + e(P1b) − x2
≥ e(P1) − x2
> 2x1− x2≥ x
2. (8) Since e(P1a) ≥ e(P1b) and e(P1) > 2x1,
e(P1a) > x1≥x
2. (9)
Since x + y ≥ 4n/7 and x ≥ y, Eqs. (8) and (9) also imply X is the largest subgraph, and the result follows.
• Case 2.3: e(P2) < y/2. We divide into two
subcases.
Case 2.3.1: P2 adjacent to Y (Figure 1.(c)).
We show that (X1, P1, P2∪Y2, Y1) is a desired
4-split. Since y2≤ y1 and e(P2) < y/2 ≤ y1,
we have
(b) X1 X2 X P2 P1b P1a (a) X1 X2 X Y1 Y2 Y P (c) X1 Y1 Y2 P1 P2 (d) X1 Y1 Y2 P2 P 1a P1b
Figure 1: 4-split cases
Since e(P2) ≥ (n − x1− y)/3,
e(P2) + y2 ≥ (n − x1− y1)/3
≥ (n − n/5 − n/5)/3
= n/5 ≥ y1
Combined with Eq. (10), we have
y1≤ e(P2∪ Y2) ≤ 2y1. (11)
By Eq. (2) and e(P1) ≤ 2e(P2) < y ≤ 2x1,
we have
x1≤ e(P1) ≤ 2x1, (12)
and
e(P1) < y ≤ 2y1. (13)
Since e(P2) < y/2 and y2≤ y/2,
e(P2∪ Y2) < y ≤ 2x1. (14)
By Eqs. (11)–(14), the result follows. Case 2.2.2: P1 adjacent to Y (Figure 1.(d)).
Suppose that Y is adjacent to P1a, and here
P1a may be larger or smaller than P1b. In this
case, we show that (X1, P2∪ P1b, P1a∪ Y2, Y1)
is a desired 4-split.
Since y2≤ y1and e(P1a) < e(P2) < y/2 ≤ y1,
we have
e(P1a∪ Y2) ≤ 2y1. (15)
Similarly,
e(P1a∪ Y2) ≤ 2x1. (16)
Since e(P2) < y/2 and e(P1b) ≤ e(P2),
e(P2∪ P1b) < y ≤ 2y1. (17)
Similarly,
e(P2∪ P1b) < 2x1. (18)
Since e(P2∪ P1b) < 2y1, y1 < n/5, and x1 <
n/5, we have
e(P1a∪ Y2) ≥ n − (x1+ y1+ e(P2∪ P1b)) > n/5. (19)
Similarly,
e(P2∪ P1b) > n/5. (20)
By Eqs. (19) and (20), in (X1, P2∪ P1b, P1a∪
Y2, Y1), the subgraph X1or Y1is the smallest
and no less than a half of the maximum (by Eqs. (15) – (18)). Therefore it is a 4-split with ratio at most two.
The next corollary is directly from the above theorem.
Corollary 11: Given a tree T of n edges, a 4-split of T with ratio at most two can be found in
4
On general k
4.1
A simple algorithm
We now propose a simple algorithm which finds a k-split of a tree. Given a tree T and an integer
k, the algorithm starts at the 1-split (T ) and
re-peatedly computes a (i+1)-split from the i-split by 2-splitting the maximum subgraph. We shall show that the algorithm takes only O(n log k) time and always return a k-split with ratio at most three. Algorithm Simple-Split
Input: A tree T and an integer k ≤ e(T ). Output: A k-split of T .
1: Initiate an empty queue Q of trees, and insert T into Q.
2: For i ← 1 to k − 1 do
2.1: Choose a tree Y in Q with maximum number of edges.
2.2: Find a 2-split (Y1, Y2) of Y with ratio at
most two.
2.3: Remove Y from Q. 2.4: Insert Y1 and Y2into Q.
3: Output the k trees in Q as the k-split of T . In the next theorem, we show the performance of the algorithm.
Theorem 12: Given a tree T with n edges and an integer k ≤ n, the algorithm Simple-Split finds a k-split of T with ratio at most 3 in O(n log k) time.
Proof: Let Miand mibe respectively the
max-imum and minmax-imum numbers of edges of trees in the queue Q at i-th iteration. We first claim that the ratio Mi/mi is at most 3 for each i.
Initially Q contains only the input tree T , and
M1/m1 = 1. Suppose that Mi/mi ≤ 3 for some
i. We shall show that Mi+1/mi+1 ≤ 3, and then
the above claim is consequently true by induc-tion. At (i + 1)-th iteration, the maximum tree
Y is chosen and split into Y1 and Y2 with ratio
at most 2. Therefore, Mi+1 ≤ Mi, and mi+1 =
min{mi, e(Y1), e(Y2)}. Since min{e(Y1), e(Y2)} ≥
e(Y )/3 = Mi/3 and Mi/mi≤ 3, we have
Mi+1
mi+1 ≤
Mi
Mi/3≤ 3.
Next, we turn to the time complexity. Let fn(i)
be the total time complexity of executing Step 2.2 in the first i iterations. By Corollary 4, splitting
a tree of Mi edges at i-th iteration takes O(Mi)
time. Since the ratio Mi/mi is at most three, by
Lemma 5, we have
Mi≤
3n
i + 2.
Therefore, for some constant c, fn(1) ≤ cn, and
fn(i) ≤ fn(i − 1) + c 3n
i + 2
for i > 1. Solving the recurrence relation, we have
fn(k) ≤ c k−1 X i=1 3n i + 2 ≤ 3cn k X i=1 1 i = 3cnHk,
in which Hkis the well-known k-th harmonic
num-ber. Since Hk = O(log n), we obtain fn(k) =
O(n log k).
For Step 2.1, 2.3, and 2.4, by simply using a data structure like heap to store the numbers of edges of the trees in the queue, all the operations can be done in totally O(k log k) time. Therefore the total time complexity is O(n log k).
4.2
Experimental results on random
trees
To investigate the practical behavior of the al-gorithm Simple-Split, we implemented the algo-rithm and performed tests on random trees. Be-fore showing the experimental results, we explain how we find the 2-split at Step 2.2. By Corollary 4, we can find a 2-split of ratio at most two by the procedure described in the proof of Lemma 3. However, although it ensures the bound of the worst ratio, the procedure does not try to find the best one. We used the following procedure to find a 2-split. For a given tree Y , we root the tree at its centroid. Initially we regard each branch as a subgraph, and then repeatedly merge the small-est two subgraphs until only two subgraphs are left. To see that the procedure always returns a 2-split (Y1, Y2) of ratio at most two, it is sufficient
to show that the smaller subgraph Y1 contains at
least e(Y )/3 edges. Since the tree is rooted at its centroid, each branch contains no more than
e(Y )/2 edges. If e(Y1) < e(Y )/3, it implies that
e(Y2) < 2e(Y )/3 since Y2 is either a single branch
or obtained by merging two subgraphs smaller than Y1. But e(Y1) + e(Y2) = e(Y ), and it is a
Table 1: The average ratios n 100 500 1000 3000 6000 10000 k = 2 1.21 1.23 1.20 1.17 1.19 1.21 k = 3 1.85 1.84 1.85 1.87 1.89 1.85 k = 4 1.51 1.50 1.46 1.39 1.45 1.48 k = 5 1.97 1.99 2.00 2.06 1.98 1.97 k = 10 2.09 2.11 2.08 2.13 2.09 2.10 k = 20 2.23 2.26 2.21 2.22 2.17 2.17 k = 50 3.00 2.38 2.41 2.41 2.39 2.43 k = 100 1.00 2.43 2.50 2.54 2.51 2.50
Table 2: The distribution of ratios (in percentage)
(n, k) ≤ 1.4 ≤ 1.6 ≤ 1.8 ≤ 2 ≤ 2.2 ≤ 2.4 ≤ 2.6 ≤ 2.8 (100, 4) 45.9 65.3 83.2 93.9 98.4 99.2 99.8 100 (500, 4) 48.0 73.2 88.2 96.7 98.5 99.4 100 100 (100, 10) 0.4 1.7 9.9 48.5 72.0 87.5 97.4 99.8 (500, 10) 0 1.7 9.0 40.2 68.4 86.5 95.7 99.5 (5000, 10) 0.2 1.5 9.3 35.4 68.3 86.4 95.4 99.3 (500, 20) 0 0 2.1 20.8 50.6 76.4 92.3 98.7 (5000, 20) 0 0 0.6 15.1 49.6 76.0 92.1 99.1
For different n (number of edges) and k, we recorded the ratios of the k-splits found by the algorithm. Since the program runs very fast, we do not show the execution time. For each (n, k), hundreds of instances were tested, and the average ratios are shown in Table 1.
Since the worst cases (ratio 3) do exist, show-ing the worst ratios in the test is meanshow-ingless. The more instances we run, the larger the worst ratio is. Instead, we show the distributions for some typical pairs (n, k). In Table 2, we show the per-centage of the ratio in each specified range. For example, the value 65.3 in the cell of the second row and third column means that, for n = 100 and k = 4, there are 65.3% of the instances in the test such that the ratio of the obtained split is less than or equal to 1.6.
By the experimental results, we observed the following.
• For small k, the algorithm performs well, but
the obtained ratios get larger and tend toward the worst case as k increasing. Observing the cases of k = 2, we find that the algorithm splits a tree into two parts quite evenly, and it is also the reason why the performance is good for k = 4 but rather bad for k = 3.
• As long as k is small with respect to n, the
results are almost not affected by n. In Table 1, we can see that the average ratios in each
row are almost the same except for (n, k) = (100, 50) and (100, 100).
• The distributions approximate to the normal
distribution. For each (n, k), the standard de-viation is approximately 0.27. In our test, the obtained ratios of about 70% of the instances are in the range [µ − σ, µ + σ], in which µ is the mean and σ is the standard deviation.
• There are many instances that the algorithm
obtained a ratio larger than two. In this as-pect, it is significant to develop an algorithm always finding a ratio within two for general
k. Even for k = 3 and k = 4, there are
still about respectively 35% and 5% of the instances in our test such that the obtained ratios are larger that two. Therefore, the re-sults for 3-splits and 4-splits in this paper is meaningful.
5
Concluding Remarks
One of the most important open problems in this line of investigation is that whether there ex-ists a k-split with ratio at most two for general k. Our future work includes exact and approxima-tion algorithms for finding the min-ratio k-split for general or fixed k.
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