values of the parameters. The key results in this chapter are:
• The family of problems with constraints that L = 2 and K = 3 are solved for both paths and cycles. We show that when L = 2 and K = 3, an MCI exists on a path if and only if the number of vertices in the path is no greater than (N−1)[(m−1)N−1]+2, and an MCI exists on a cycle if and only if the number of vertices in the cycle is no greater than (N −1)[(m−1)N −1]. Structural properties of MCIs in this case are analyzed, and algorithms are presented which output MCIs on paths or cycles as long as the MCIs exist.
• The family of problems with the constraint thatK =L+1 are studied for paths.
A scheme using a ‘hierarchical-chain’ structure is presented for constructing MCI on paths. It is shown that the scheme solves the MCI problem for paths that are asymptotically as long as the longest path on which MCIs exist, and clearly, for shorter paths as well.
The rest of the chapter is organized as follows. In Section 4.2, we firstly derive an upper bound for the lengths of paths and cycles on which MCIs exist. We then prove a tighter upper bound for paths for the case of L= 2 and K = 3. In Section 4.3, we present an optimal construction for MCI on paths for the case of L= 2 and K = 3, which meets the upper bound presented in Section 4.2. In Section 4.4, we study the MCI problem for paths when K =L+ 1. In Section 4.5 we generalize our results on MCI from paths to cycles.
Proof: Let G = (V, E) be a path (or cycle) of n vertices with an MCI on it. We can find at most bnLc non-overlapping clusters in G. Let S ⊆ {1,2,· · · , N} be an arbitrary set of L distinct integers. Then since the interleaving on G is an MCI, among those bnLc non-overlapping clusters, at most m−1 of them are assigned only (all or part of the) integers in S and no integers in {1,2,· · · , N} −S. S can be one of ¡N
L
¢ possible sets. SobnLc ≤(m−1)¡N
L
¢. So we get n≤(m−1)L¡N
L
¢+ (L−1).
2
We comment that for the same set of parametersN,K,mandL, if an MCI exists on a path of n=n0 vertices, then an MCI also exists on any path ofn < n0 vertices.
That is because given an MCI on a path, the interleaving on the path’s connected subgraph (which is a shorter path) is also an MCI for the same set of parameters N, K, m and L. However, such an argument does not necessarily hold for cycles.
The upper bound of Proposition 4.1 is in fact loose. In the remainder of this section, we shall prove a tighter upper bound for MCI on paths for the case of L= 2 and K = 3. We present it as the following theorem. Later study will show that this upper bound is exact.
Theorem 4.1 WhenL= 2andK = 3, if there exists an MCI on a path ofnvertices, then n≤(N −1)[(m−1)N −1] + 2.
Before proving the above theorem, we firstly define some notations that will be used throughout this chapter as follows. LetG= (V, E) be a path. We denote the n vertices in the path Gbyv1, v2, · · ·, vn. For 2≤i≤n−1, the two vertices adjacent tovi arevi−1 andvi+1. A connected subgraph ofGinduced by verticesvi, vi+1,· · · , vj (j ≥i) is denoted by (vi, vi+1,· · · , vj). If a set of integers are interleaved on G, then c(vi) denotes the integer assigned to vertex vi. The integers assigned to a connected subgraph of G, (vi, vi+1,· · · , vj), are denoted by [c(vi)−c(vi+1)− · · · −c(vj)].
The following lemma reveals a certain relationship between the structure of MCIs and the lengths of paths.
Lemma 4.1 Let the values of N, K, m and Lbe fixed, where N ≥4, K = 3, m≥2 and L = 2. Let nmax denote the maximum value of n such that an MCI exists on
a path of n vertices. Then in any MCI on a path of nmax vertices, no two adjacent vertices are assigned the same integer.
Proof: Let G= (V, E) be a path of nmax vertices. Assume that there is an MCI on G. We’ll prove this lemma by showing that in this MCI, if two adjacent vertices ofG are assigned the same integer, then there exists a longer path that also has an MCI, which would clearly contradict the definition of nmax.
Assume that in the MCI on G, there are two adjacent vertices that are assigned the same integer. Then without loss of generality (WLOG), one of the following four cases must be true (because we can always get one of the four cases by permuting the names of the integers and by reversing the indices of the vertices):
Case 1: There exist 4 consecutive vertices in G— vi, vi+1, vi+2,vi+3 — such that c(vi) = 1, c(vi+1) = c(vi+2) = 2, and c(vi+3) = 1 or 3.
Case 2: There existx+2≥5 consecutive vertices inG—vi,vi+1,· · ·,vi+x,vi+x+1
— such that c(vi) = 1, c(vi+1) = c(vi+2) =· · ·=c(vi+x) = 2, and c(vi+x+1) = 1 or 3.
Case 3: c(v1) =c(v2) = 1, and c(v3) = 2.
Case 4: c(v1) =c(v2) = · · ·=c(vx) = 1 and c(vx+1) = 2, where x≥3.
We analyze the four cases one by one.
Case 1: In this case, we insert a vertex v0 between vi+1 and vi+2, and get a new path of nmax + 1 vertices. Call this new path H, and assign the integer ‘4’ to v0. Consider any m non-overlapping clusters in H. If none of those m clusters contains v0, then clearly they are also m non-overlapping clusters in the pathG, and therefore have been assigned at least K = 3 distinct integers. If the m clusters contain all the three vertices vi+1, v0 and vi+2, then either vi or vi+3 is also contained in the m clusters because each cluster contains L = 2 vertices, and therefore the m clusters have been assigned at least K = 3 distinct integers: ‘1,2,4’ or ‘2,3,4’. WLOG, the only remaining possibility is that one of the m clusters contains vi+1 and v0 while none of them contains vi+2. Note that among the m clusters, the m −1 of them which don’t contain v0 are also m−1 clusters in the pathG, and they together with (vi+1, vi+2) are m non-overlapping clusters in G and therefore are assigned at least
K = 3 distinct integers. Since c(vi+1) = c(vi+2), the original m clusters including (vi+1, v0) must also have been assigned at leastK = 3 distinct integers. Now we can conclude that the interleaving on H is also an MCI. But H’s length is greater than nmax, which contradicts the definition of nmax.
Case 2: In this case, we insert a vertex v0 between vi+1 and vi+2, and insert a vertex v00 betweenvi+x−1 andvi+x, and get a new path ofnmax+ 2 vertices. Call this new pathH, assign the integer ‘4’ tov0, and assign the integer ‘3’ tov00. Consider any m non-overlapping clusters in H. If neither v0 nor v00 is contained in the m clusters, then clearly those m clusters are also m non-overlapping clusters in the path G, and therefore are assigned at leastK = 3 distinct integers. If bothv0 andv00are contained in the m clusters, then at least one vertex in the set {vi+1, vi+2,· · ·, vi+x−1, vi+x} is also in the m clusters, and therefore the m clusters have at least these 3 integers:
‘2’, ‘3’ and ‘4’. WLOG, the only remaining possibility is that the m clusters con- tain v0 but not v00. (Note that the cluster containing v0 is assigned integers ‘2’ and
‘4’.) When that possibility is true, if the m clusters contain vi+x+1, then they are assigned at least 3 distinct integers — ‘1,2,4’ or ‘2,3,4’; if themclusters don’t contain vi+x+1, then they don’t containvi+x either — then we divide the m clusters into two groups A and B, where A is the set of clusters none of which contains any vertex in {v0, vi+2, vi+3,· · · , vi+x−1}, and B is the set of clusters none of which is in A. Say there areyclusters in B. Then, if the cluster containingv0 also contains vi+1 (respec- tively,vi+2), there exist a set C of yclusters in the path G that only contain vertices in{vi+1, vi+2,· · · , vi+x−1, vi+x} (respectively,{vi+2, vi+3,· · · , vi+x−1, vi+x}), such that (in both cases) the mclusters inA∪C are non-overlapping inG. Thosem clusters in A∪C are assigned at least K = 3 distinct integers since the interleaving on G is an MCI; and they are assigned no more distinct integers than the original m clusters in A∪B are, becausec(vi+1) =c(vi+2) =· · ·=c(vi+x) and either vi+1 or vi+2 is in the same clusters as v0. So the m clusters inA∪B are assigned at least K = 3 distinct integers. Now we can conclude that the interleaving on H is also an MCI. And that again contradicts the definition ofnmax.
Case 3: In this case, we insert a vertexv0 betweenv1 andv2, and assign the integer
‘3’ to v0. The rest of the analysis is very similar to that for Case 1.
Case 4: In this case, we insert a vertex v0 between v1 and v2, and insert a vertex v00 between vx−1 and vx, assign the integer ‘3’ to v0, and assign the integer ‘2’ to v00. The rest of the analysis is very similar to that for Case 2.
So a contradiction exists in all four cases. Therefore, this lemma is proved.
2
The next two lemmas derive upper bounds on the lengths of paths, respectively for the case of N ≥4 and N = 3.
Lemma 4.2 Let the values of N, K, m and Lbe fixed, where N ≥4, K = 3, m≥2 and L= 2. Let nmax denote the maximum value of n such that an MCI exists on a path of n vertices. Then nmax ≤(N −1)[(m−1)N −1] + 2.
Proof: Let G= (V, E) be a path ofnmax vertices. Assume there is an MCI on G. By Lemma 4.1, no two adjacent vertices inGare assigned the same integer. We color the vertices in G with three colors — ‘red’, ‘yellow’ and ‘green’ — through the following three steps:
Step 1, for 2≤ i ≤ nmax −1, if c(vi−1) = c(vi+1), then color vi with the ‘red’
color;
Step 2, for 2 ≤ i ≤ nmax, color vi with the ‘yellow’ color if vi is not colored
‘red’ and there exists j such that these four conditions are satisfied: (1) 1 ≤ j < i, (2) vj is not colored ‘red’, (3) c(vj) = c(vi), (4) the vertices betweenvj andvi — that is, vj+1, vj+2, · · ·, vi−1 — are all colored ‘red’;
Step 3, for 1≤i≤nmax, if vi is neither colored ‘red’ nor colored ‘yellow’, then color vi with the ‘green’ color.
Clearly, each vertex of G is assigned exactly one of the three colors.
If we arbitrarily pick two different integers — say ‘i’ and ‘j’ — from the set {1,2,· · · , N}, then we get a pair [i, j] (or [j, i], equivalently). There are totally ¡N
2
¢
such un-ordered pairs. We partition those ¡N
2
¢ pairs into four groups ‘A’, ‘B’, ‘C’
and ‘D’ in the following way:
(1) A pair [i, j] belongs to group A if and only if the following two conditions are satisfied: (i) at least one ‘green’ vertex is assigned the integer ‘i’ and at least one ‘green’ vertex is assigned the integer ‘j’, (ii) for any two ‘green’ vertices that are assigned integers ‘i’ and ‘j’ respectively, there is at least one ‘green’ vertex between them.
(2) A pair [i, j] belongs to group B if and only if the following two conditions are satisfied: (i) at least one ‘green’ vertex is assigned the integer ‘i’ and at least one
‘green’ vertex is assigned the integer ‘j’, (ii) there exist two ‘green’ vertices that are assigned integers ‘i’ and ‘j’, respectively, such that there is no ‘green’ vertex between them.
(3) Apair [i, j] belongs to groupCif and only if one of the following two conditions is satisfied: (i) at least one ‘green’ vertex is assigned the integer ‘i’ and no ‘green’
vertex is assigned the integer ‘j’, (ii) at least one ‘green’ vertex is assigned the integer
‘j’ and no ‘green’ vertex is assigned the integer ‘i’.
(4) A pair [i, j] belongs to groupD if and only if no ‘green’ vertex is assigned the integer ‘i’ or ‘j’.
E is the set of edges inG= (V, E). For any 1≤i6=j ≤N, letE(i, j)⊆E denote such a subset of edges of G: an edge of G is in E(i, j) if and only if one endpoint of the edge is assigned the integer ‘i’ and the other endpoint of the edge is assigned the integer ‘j’. Letz(i, j) denote the number of edges in E(i, j). Upper bounds forz(i, j) are derived below.
For any pair [i, j] in group A or group C, z(i, j) ≤ 2m −2. That’s because otherwise there would exist m non-overlapping clusters in G — note that a cluster contains exactly one edge — each of which is assigned only integers ‘i’ and ‘j’, which would contradict the assumption that the interleaving on Gis an MCI.
Now consider a pair [i, j] in group B. z(i, j)≤ 2m−2 for the same reason as in the previous case. Assume z(i, j) = 2m−2. Then in order to avoid the existence of m non-overlapping clusters in G each of which is assigned only integers ‘i’ and ‘j’, thez(i, j) = 2m−2 edges inE(i, j) must be consecutive in the pathG, which means, WLOG, that there are 2m−1 consecutive vertices vy+1, vy+2, · · ·, vy+2m−1 (y ≥ 0)
whose assigned integers are in the form of [c(vy+1)−c(vy+2)− · · · −c(vy+2m−1)] = [i−j−i−j− · · · −i−j −i]. According to the definition of ‘group B’, there exist a ‘green’ vertex vk1 and a ‘green’ vertex vk2, such that vk1 is assigned the integer ‘i’, vk2 is assigned the integer ‘j’, and there is no ‘green’ vertex between them. Therefore every vertex between vk1 and vk2 is either ‘red’ or ‘yellow’. By the way the vertices are colored, it’s not difficult to see that either ‘k1 < k2 and vk2−1 is assigned the integer c(vk1) = i’ (let’s call this ‘case 1’), or ‘k2 < k1 and vk1−1 is assigned the integer c(vk2) = j’ (let’s call this ‘case 2’). If ‘case 1’ is true, then since there is an edge between vk2−1 (which is assigned the integer ‘i’) and vk2 (which is assigned the integer ‘j’), vk2 is in the set {vy+1, vy+2,· · · , vy+2m−1}. However, it’s simple to see that every vertex in the set {vy+1, vy+2,· · · , vy+2m−1} that is assigned the integer
‘j’ must have the color ‘red’ — so vk2 must be ‘red’ instead of ‘green’ — therefore a contradiction exists. Now if ‘case 2’ is true, since there is an edge between vk1−1 (which is assigned the integer ‘j’) and vk1 (which is assigned the integer ‘i’), both vk1−1 and vk1 are in the set {vy+2, vy+3,· · · , vy+2m−1}. Then again since every vertex in the set {vy+1, vy+2,· · · , vy+2m−1} that is assigned the integer ‘j’ must have the color ‘red’, and since the color ofvk1 is ‘green’, it is simple to see that all the vertices in the set {vy+1, vy+2,· · · , vk1−1} that is assigned the integer ‘i’ must have the color
‘red’. Then since the color of vy+1 is ‘red’, the vertex vy exists and it must have been assigned the integer ‘j’ — and that contradicts the statement that all the edges in E(i, j) are in the subgraph (vy+1, vy+2,· · · , vy+2m−1). Therefore a contradiction always exists when z(i, j) = 2m−2. So z(i, j) 6= 2m−2. So for any pair [i, j] in group B, z(i, j)≤2m−3.
Now consider a pair [i, j] in group D. By the definition of ‘group D’, no ‘green’
vertex is assigned the integer ‘i’ or ‘j’. Let{vk1, vk2,· · · , vkt}denote the set of vertices that are assigned the integer ‘i’, where k1 < k2 <· · · < kt. If {vk1, vk2,· · · , vkt} 6=∅, by the way vertices are colored, it’s simple to see that vk1 cannot be ‘yellow’ — so vk1 must be ‘red’. Then similarly, vk2, vk3, · · ·, vkt must be ‘red’, too. Therefore all the vertices that are assigned the integer ‘i’ are of the color ‘red’. Similarly, all the vertices that are assigned the integer ‘j’ are of the color ‘red’. Assume there is an
edge whose two endpoints are assigned the integer ‘i’ and the integer ‘j’, respectively.
Then since the two vertices adjacent to a ‘red’ vertex must be assigned the same integer, there exists an infinitely long segment (subgraph) of the path G to which the assigned integers are in the form of ‘· · · − i−j −i−j −i−j − · · ·’, which is certainly impossible. Therefore a contradiction exists. So for any pair [i, j] in group D, z(i, j) = 0.
Let ‘x’ denote the number of distinct integers assigned to ‘green’ vertices, and let
‘X’ denote the set of those x distinct integers. It’s simple to see that exactly ¡x
2
¢
pairs [i, j] are in group A or group B, where i∈ X and j ∈ X — and among them at least x−1 pairs are in group B. It’s also simple to see that exactly x(N −x) pairs are in group C and exactly ¡N−x
2
¢ pairs are in group D. By using the upper bounds we’ve derived for z(i, j), we see that the number of edges in G is at most [¡x
2
¢−(x−1)] ·(2m−2) + (x−1)·(2m−3) +x(N −x)·(2m−2) +¡N−x
2
¢·0 = (1−m)x2+ (2mN −2N −m)x+ 1, whose maximum value (at integer solutions) is achieved when x=N−1 — and that maximum value is (N−1)[(m−1)N −1] + 1.
Sonmax, the number of vertices in G, is at most (N −1)[(m−1)N −1] + 2.
2
Lemma 4.3 Let the values of N, K, m and L be fixed, whereN = 3, K = 3, m≥2 and L= 2. Let nmax denote the maximum value of n such that an MCI exists on a path of n vertices. Then nmax ≤(N −1)[(m−1)N −1] + 2.
Proof: Let G = (V, E) be a path of n vertices that has an MCI on it. We need to show that n≤(N −1)[(m−1)N −1] + 2.
If in the MCI on G, no two adjacent vertices are assigned the same integer, then with the same argument as in the proof of Lemma 4.2, it can be shown that n ≤ (N −1)[(m−1)N −1] + 2.
Now assume there are two adjacent vertices in G that are assigned the same integer. Clearly we can find t non-overlapping clusters in G, such that n ≤ 2t+ 2 and at least one of the t clusters contains two vertices that are assigned the same integer. Among those t non-overlapping clusters, let x, y, z, a, b and c respectively
denote the number of clusters that are assigned only the integer ‘1’, only the integer
‘2’, only the integer ‘3’, both the integers ‘1’ and ‘2’, both the integers ‘2’ and ‘3’, and both the integers ‘1’ and ‘3’. Since the interleaving is an MCI, anym non-overlapping clusters are assigned at least K = 3 distinct integers. Therefore x+y+a ≤ m−1, y+z+b ≤ m−1, z+x+c ≤ m−1. So 2x+ 2y+ 2z+a+b+c ≤ 3m−3. So x+y+z+a+b+c≤3m−3−(x+y+z). Sincex+y+z ≥1,t=x+y+z+a+b+c, and n≤2t+ 2, we getn≤2(x+y+z+a+b+c+ 1)≤2[3m−3−(x+y+z) + 1]≤ 6m−6 = (N −1)[(m−1)N −1] + 2.
Therefore this lemma is proved.
2
So with Lemma 4.2 and Lemma 4.3 proved, we see that Theorem 4.1 becomes a natural conclusion.