Scholarly article on topic 'Rank numbers for bent ladders'

Rank numbers for bent ladders Academic research paper on "Mathematics"

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Academic research paper on topic "Rank numbers for bent ladders"

Discussiones Mathematicae Graph Theory 34 (2014) 309-329 doi: 10.7151 /dmgt. 1739

RANK NUMBERS FOR BENT LADDERS

Peter Riohter1, Emily Leven2 Anh Tran3, Bryan Ek4 , Jobby Jacob4 and

Darren A. Narayan4

1 Department of Mathematics, University of Rochester

Rochester, NY 14642-0002, USA

2 Department of Mathematics, University of California

San Diego, La Jolla, California, 92093-0112, USA

3 Department of Mathematics, Temple University Philadelphia PA 19122, USA

4 School of Mathematical Sciences

Rochester Institute of Technology Rochester, NY 14623, USA

e-mail: prichter@u.rochester.edu esergel07@gmail. com anh.van.tran@temple.edu bte1759@rit.edu jxjsma@rit.edu dansma@rit.edu

Abstract

A ranking on a graph is an assignment of positive integers to its vertices such that any path between two vertices with the same label contains a vertex with a larger label. The rank number of a graph is the fewest number of labels that can be used in a ranking. The rank number of a graph is known for many families, including the ladder graph P2 x Pn. We consider how "bending" a ladder affects the rank number. We prove that in certain cases the rank number does not change, and in others the rank number differs by only 1. We investigate the rank number of a ladder with an arbitrary number of bends.

Keywords: graph colorings, rankings of graphs, rank number, Cartesian

product of graphs, ladder graph, bent ladder graph.

2010 Mathematics Subject Classification: 05C78, 05C15, 05C76.

1. Introduction

A coloring f : V(G) ^ {1,2,... , k} is a k-ranking of G if f (u) = f (v) implies every u — v path contains a vertex w such that f (w) > f (u). The rank number of a graph, xr (G), is the minimum k such that G has a k-ranking. A k-ranking that uses xr (G) labels will be referred to as a xr-ranking. When the value of k is clear we will refer to a k-ranking simply as a ranking.

Research on rank numbers was sparked by its applications to the scheduling of manufacturing systems, Cholesky factorizations of matrices and VLSI layout [11, 14]. The optimal tree node ranking problem is identical to the problem of generating a minimum height node separator tree for a tree graph. Node separator trees are extensively used in VLSI layout [11]. These models are suitable for communication networks design where information flow between nodes needs to be monitored. Similar models are applicable in the design of management organizational structures. A matrix application was observed by Kloks, Müller, and Wong [10].

It was shown by Bodlaender et al. [2] that for a given bipartite graph G and a positive integer t, deciding if xr (G) < t is NP-Complete. However rank numbers have been determined for several families of graphs including: paths, cycles, split graphs, complete multipartite graphs, Mübius ladder graphs, caterpillars, powers of paths and cycles, and some grid graphs [1, 2, 3, 4, 6, 7, 12], and [13].

In 2009, Novotny, Ortiz, and Narayan [12] determined the rank number of the ladder graph Ln = P2 x Pn and showed xr (P2 x Pn) = Llog2 (n + 1)J + [log2 (n +1 — (2^2 nJ-l))j + 1 = [log2 (n + 1)J + |_log2 (^jj + 1.

This result was also shown by Chang, Kuo, and Lin [4]. We consider how the rank number behaves if the ladder has one or more 'bends'. It turns out that in many cases the rank number does not change, and in others it differs by only 1. In this paper we determine rank numbers for the two extreme cases of bent ladders: the first where there is a single bend (L-shaped) and in the other the number of bends is maximized (similar to a staircase).

2. Preliminaries

We begin by recalling a definition of Ghoshal, Laskar, and Pillone [6].

Definition 1. A k-ranking is minimal if decreasing any label violates the ranking property.

The operation of a reduction was introduced by Ghoshal, Laskar, and Pillone [6].

Definition 2. Given a graph G and a set S C V(G) the reduction of G is a graph GS such that V(GS) = V(G) - S and for vertices u and v, {u, v} € E(GS)

if and only if there exists a u — v path in G with all internal vertices belonging to S.

We present a generalization of Lemma 5 in [12] that will be used for bent ladders and staircase ladders. A 1-bridge is a set of two adjacent vertices x and y along with four edges that connect two graphs together as shown in Figure 1. Recall that a vertex separating set of a connected graph G is a set of vertices whose removal disconnects G. A graph is k-connected if any vertex separating set contains at least k vertices.

Lemma 3. Let G be the union of two 2-connected graphs H\ and H2 that are connected by a 1-bridge, where xr (Hi) = xr (H2). Then xr (G) > xr (Hi) + 2.

Figure 1. A 1-bridge.

Proof. Assume that xr (Hi) = xr (H2). Let the two added vertices be labeled x and y. We consider cases for different minimal rankings of G. We will show in each case there is a vertex with a label greater than or equal to xr (Hi) + 2.

Case (i). There exists a vertex in each copy of Ls labeled xr (Hi). Since the highest two labels are unique in the ranking, we have xr (G) > xr (Hi) + 2.

Case (ii). There exists a vertex in each copy of Ls labeled xr (Hi) + 1. Since the vertex with the highest label must be unique it follows that xr (G) > xr (Hi) + 2.

Case (iii). There exists a vertex u in one copy of Hi labeled xr (Hi) and one vertex v in the other copy of Hi labeled xr (Hi) + 1. Without loss of generality assume v is in the copy of Hi on the right side. Since the ranking of G is minimal the vertices in the copy of Hi on the left side include labels 1,2,..., xr (Hi) and vertices in the copy of Hi on the right side include labels 1,2,... ,xr (Hi) — 1, xr (Hi) + 1. Let w and z be the two vertices in G labeled xr (Hi) — 1. Note that there are two edge disjoint paths from w to x and two edge disjoint paths from z to x. Hence there must be a path from w to z that avoids both u and v. Hence either x or y must be labeled at least xr (Hi) + 2. ■

We define Ln to be critical if Xr (Ln) = Xr (Ln-1) + 1 for n > 2. It was shown by Novotny, Ortiz, and Narayan [12] that a ladder Ln is critical if and only if n = 2k - 1, or 2k + 2k-1 - 1 for any k > 1.

Lemma 4. Let k > 2 and n = 2k - 2 or 2k+2k-1 - 2. Then xr (Ln) -Xr (Ln) = 1.

Proof. Recall that Xr (P2 x Pn) = Llog2(n + 1)J + [log2 (M^)

Case (i). n = 2k -2. Xr (Ln) - Xr (Ln) = ([log2(2k - 2 + 1)J + |_log2 (2(2k-2+1))J)

- ([log2(2k-1 -j 1 + 1)J + [log^2(2k-1-^)J)

= (log2(2k - 1)J - [log2(2k-1)J

log2 log2

2(2k-2+1) 3

2k+1 —2 3

2(2k-1 —1+1)\ 3

Since there is only one power of 2 between 2fc+g1 2 and 23-, log2 2

log2 (2k)

Case (ii). n = 2k + 2k-1 - 2. Xr(Ln) - xr (Lf) = ([log2(2k + 2k—1 - 2 + 1)J + [log^2(2fe+2fe31—2+1))j) - ([log2(2k—1 + 2k—2 - 1 + 1)j + [log^2(2fc-1+23fc-2—1+ ~ log2(2k + 2k—1 - 2 + l)j -l (log2(2k—1 + 2k—2 - 1 + 1)J

2(2k +2k-1 —2+1) 3

■>fc-1

2(2fc-1+2fc-2-1+1)\

log2(2k + 2fc"1 - 2 + 1)( - [log2(2fc—1 + 2k-2 - 1 + 1)J

3-2k-2)

= 1 + (k - 1) - (k - 1) = 1.

3. Bent Ladders

We define a bent ladder BLn(a, b) to be the union of La and Lb that are joined at a right angle with a single L2, so that n = a + b + 2. We note that b is implicitly determined by n. An example of a bent ladder is shown in Figure 2.

Theorem 5. Let BLn(a, b) be the bent ladder composed of La, Lb, and L2 where n = a + b + 2.

Xr (Ln) — 1 if n = 2k — 1 for some k € Z+ Then Xr (BLn(a, b)) = ^ and a = 2 or 3 (mod 4),

Xr (Ln) other-wise.

Figure 2. The bent ladder BLn(a). We use a series of lemmas to prove the result. Lemma 6. xr (BLn) < xr (Ln) for all n.

Proof. It was shown in [12] that there exists a xr-ranking of Ln where the label 1 is placed on alternating vertices. We consider such a labeling here. Label the vertices in La as they are in the first a rungs of Ln. Label the vertices of Lb as they are in the last b rungs of Ln. The remaining four vertices are labeled as shown in Figure 2. The remaining two rungs at the bend will have two vertices labeled 1. Let d, e be the other two labels. Without loss of generality assume d < e and e is on the rung adjacent to a vertex in La. Let c be the label on the rung of Lb adjacent to the vertex on the joining L2 that is not labeled 1.

Now we show that the labeling f of BLn is a ranking. It may be helpful to refer to Figure 3. We consider two vertices x and y where f (x) = f (y). If x and y are both in La or both in Lb then the ranking condition must be met. Finally consider the case where x € V(La) and y € V(Lb). There are two x,y paths in Ln one passing through d and another passing through e. Hence e > d > f (x). Since d or e will be on the path from x to y in BLn the ranking property is met. Hence the labeling f of BLn is a ranking.

Figure 3. Transforming a ladder into a single bent ladder

We give the following definition that will be used in the next lemma. Given a vertex x we say that x has path access to i if there exists a path from x to a vertex labeled i that avoids any vertex with a label larger than i.

Lemma 7. Let k > 2 and n = 2k — 2 or 2k + 2k-1 — 2. Then in any xr -ranking of Ln, the highest two labels occur diagonally opposite in the central two rungs, and there is a vertex labeled 1 on each end of the ladder that has path access to each of the labels i = 2,..., xr(Ln).

Proof. The lemma is true for all xr-rankings of L2 and L4. Suppose the lemma holds for Ln. Consider L2n+2. Not placing the labels xr(Ln) + 1 and xr(Ln)+2 on the center two rungs will leave the ladder L n to be labeled with only xr (Ln) — 2 labels which is impossible by Lemma 4.

We define a sequence {gn} that will be used in the following lemma. Let hi = a + 1 where 2a is the highest power of 2 that divides i. Then replace each t > 2 in {hn} with the terms 2t — 2 and 2t — 1 in either order. Finally add 1 to each of the terms to get the sequence {gn}. ■

Lemma 8. Let f be minimal xr-ranking of L2k+2k-i-2 where vitj is the vertex in the i-th row and j-th column. Then f (vi)j-) = 1 if i + j is even and f (vi)j-) = gj if i + j is odd.

Proof. This lemma is true for all xr-rankings of L4. Suppose the lemma holds for Ln. Consider L2n+2. Note that if n = 2k + 2k-1 — 2 then 2n + 2 = 2fc+1 + 2k — 2. By Lemma 8 the highest two labels must lie on opposite corners of the center two rungs. The remaining structure follows by induction. ■

We illustrate an example of a labeling in Figure 4.

Figure 4. Note that labels within an oval may be interchanged.

We next define a sequence {wn} that will be used in the upcoming Lemma. Let zi = a + 1 where 2a is the highest power of 2 that divides i. Then replace each t > 1 in {zn} with the terms 2t and 2t + 1 in either order to obtain {wn}.

Lemma 9. Let h be a xr-ranking of L2k-2. Let vitj be the vertex in the i-th row and j-th column. Then if i + j is odd then h(vi)j-) = Wj. If j = 1 (mod 4) and i = 1 or j = 2 (mod 4) and i = 2 then h(vi)j-) = 1. If j = 3 (mod 4) and i = 1 or j = 0 (mod 4) and i = 2 then h(vi)j-) = 1 or 2.

Proof. Observe that the lemma holds for L2. Then note that if n = 2k — 2 then 2n + 2 = 2fc+1 — 2. Suppose the lemma holds for Ln. Consider L2n+2• By Lemma 8 the highest two labels must lie on opposite corners of the center two rungs. The remaining structure follows by induction. ■

We illustrate an example of a labeling in Figure 5.

Figure 5. An example of a labeling. Lemma 10. xr (Ln) — 1 < xr (BL„) < xr (Ln) for all n.

Proof. Contract the three vertices and z at the bend in BLn into a single vertex labeled with m = max{x,y, z}, and note that this gives a valid ranking of Ln-1. See Figure 6.

Hence xr (BL„) > Xr (Ln-i). By Lemma 6, Xr (Ln-i) < Xr (BLn) < Xr (Ln). Noting that xr (Ln) = xr (Ln-1) or xr (Ln-1) + 1 gives the desired result. ■

ha h

(a) ib)

Figure 6. A valid ranking of Ln_i.

The combination of Lemmas 6 and 10 gives the rank numbers for all BLn where Ln is not a critical ladder. We consider the following case involving the non-critical ladder L10.

Example 11. Let n = 10. Recall that xr (Lg) = xr (L10) = 5. By Lemma 6 we have xr (BL10) < xr (L10) = 5. Lemma 10 implies that the labels in

any k-ranking of BLio can be used to form a k-ranking of Lg. Then we have

5 = xr (Lg) < xr (BLio) < Xr (Lio) = 5. Hence Xr (BLio) = 5.

However we see in this next example that this approach cannot be extended to critical ladders.

Example 12. Let n = 11. Recall that xr (Lio) = 5 and xr (Ln) = 6. Lemmas

6 and 10 give that 5 = xr (Lio) < Xr (BLio) < xr (Lii) = 6. Hence 5 < Xr (BLio) < 6.

As a result we must consider cases of xr (BLn) where Ln is a critical ladder separately. We address these cases in the next three lemmas.

Lemma 13. For k > 2, xr (BL2fc+2fc-i-^ = xr (L2k+2fc-i-^.

Proof. It was shown in [12] that xr (L2j +2j-i-i) = 2j + 1. We proceed by induction on k. For the base case k = 2, it is easy to verify that xr (BL5) = xr (L5) = 5. Assume that xr (BL2j+2J-i-i) = xr (L2j+2j-i-i) for some j. Consider BL2j+i+2j-i as one copy of BL2j +2J-i-i and one copy of L2j+2J-i-i joined by a 1-bridge. By induction we have that xr (BL2j+2J-i-i) = xr (L2j+2j-i-i) = 2j + 1. Application of the bridge lemma gives that xr (BL2j+i+2J-i) = xr (L2j+i+2j-i) = 2(j + 1) + 1. ■

In our next two lemmas we investigate xr (BL2k_l). Let BL2k-i be composed of ladders La and Lb joined by a L2. We make the following observations which will be helpful in the proofs of Lemmas 14 and 15. We have that a + b = 2k - 3 = 1 (mod 4). In Lemma 14 we consider the case where a = 0 (mod 4) (which implies that b = 1 (mod 4)). In Lemma 15 we consider the case where a = 2 (mod 4) (which implies that b = 3 (mod 4)).

Lemma 14. Let k > 3. Consider BL2k-i as two ladders, La and Lb, joined by an L2. If a = 0 (mod 4) or a = 1 (mod 4), then xr (BL2k_^ = xr (L2k_l) .

Proof. Recall that xr (L2j_l) = 2j [12]. We proceed by induction on k. For the base case k = 3, there is only one bent ladder where a = 0 (mod 4) or a = 1 (mod 4). This is precisely the case where a = 1 and b = 4. Since this graph is composed of two copies of L3 joined by a 1-bridge, we have that xr (BL7) = xr (L7) = 6. Assume that xr (BL2j_l) = xr (L2j_l) for some j. Consider BL2j+i_i as one copy of BL2j_l and one copy of L2j_l joined by an Ll. By induction we have that xr (BL2j_l) = xr (L2j_l) = 2j. Application of the Lemma 4 gives that xr (BL2j+i_i) = xr (L2j+i_i) = 2(j + 1). ■

Lemma 15. Consider BL2k_i as two ladders, La and Lb, joined by an L2. If a = 2 (mod 4) or a = 3 (mod 4), then xr (BL2k_l) = xr (L2k— 1.

Proof. Without loss of generality suppose that a = 2 (mod 4). By Lemma 10, Xr (BL2fc_i) = xr (L2fc_0 or xr (L2k— 1. We exhibit an explicit ranking using xr (L2k— 1 = xr (L2fc_2) labels. Rank L2k_2 with xr labels choosing all vertices marked with a star to be 1. Then at the a-th rung of L2k_2, relabel the vertex labeled 1 with 2. If the vertices of the (a — 1) rung are 1 and 3 this gives a ranking. Otherwise if the vertices of the (a — 1) rung are 1 and 2 then the the vertices of the (a — 2) rung are 1 and 3; exchanging the labels 2 and 3 on these 2 rungs gives a ranking. Expand the vertex labeled 2 at the a-th rung into three vertices x,y, and z as follows, where x = 1, y = 2, z = 1. Note that this is a ranking of BL2k_1 using xr (L2k— 1 labels. ■

Proof of Theorem 5. Note that the theorem holds for n = 2,3, and 4. We proceed by induction on j for all values of n in the interval 2j + 2j_1 — 1 < n < 2J+1 + 2j — 1. Suppose n = 2j + 2j_1 — 1. By Lemma 13, xr (BLn) = xr (Ln). Suppose 2j + 2j_1 < n < 2j+1 — 2. Since xr (L2fc+2fc-i_^ = xr (L2j+i_2) by Lemma 6 we have xr (BLn) < xr (Ln) = xr Lj+2j-i_1). Hence xr (BLn) = xr (L2j+2j-i_1) = xr (Ln). If n = 2j+1 — 1 then by Lemmas 14 and 15 the claim holds. If 2j+1 < n < 2j+1+ 2j — 2 then contracting the three vertices x,y, and z at the bend of BLn gives a ranking of Ln_1. Note that in this case, xr (Ln) = xr (Ln_1). Hence xr (BLn) > xr (Ln_1) = xr (Ln). Finally by Lemma 6, xr (BLn) = xr (Ln). This completes the inductive step. ■

Corollary 16. For all n = 2k — 1, xr (BLn) = xr (Ln) regardless of where the ladder is bent.

4. Staircase Ladders

In this section we investigate ladders with a maximum number of bends. We call these graphs staircase ladders. We define a staircase ladder SLn to be a graph with n — 1 subgraphs S1, S2,..., Sn_1 each of which are isomorphic to C4. The staircase ladder is placed on a grid with the vertices of the subgraphs as follows: v(S1) = {(0, 0), (0,1), (1,1), (1, 0)}, v(S2) = {(1, 0), (1,1), (2,1), (2, 0)}, v(Ss) = {(1,1), (1, 2), (2, 2), (2,1)}, v(s4) = {(2, 2), (2, 3), (3, 3), (3, 2)}. For 0 < j < P^l, v(s2j+1) = {(j,j), (j,j + 1), (j + 1,j + 1), (j + 1,j)}. For 0 < j < , v(s2j) = {(j + 1, j), (j + 1, j + 1), (j + 2, j + 1), (j + 2, j)}.

The graph of SLg is shown in Figure 7. The staircase ladders SLn has n — 1 induced subgraphs isomorphic to C4 (squares).

Theorem 17. We have

xr(Ln+1) if n = 2k + 2k_1 — 2 for some k > 3,

xr (SLn) | xr (Ln) otherwise. We use a series of lemmas to establish the result.

Figure 7. The graph SL8. Lemma 18. For all n > 1, xr(SL2n+2) > Xr(SLn) + 2.

Proof. This follows from Lemma 4. ■

Lemma 19. For all j > 2, Xr (SL2j-2) > XrLj-2).

Proof. It is clear that the result holds for j = 2. Suppose the statement holds for j - 1. By Lemma 18, Xr (SL2J-2) > Xr(SL23-1-2) +2 > Xr(L23-1-2) + 2 = Xr (L2j-2). Hence the lemma holds for all j. ■

Lemma 20. For all n > 1, Xr(SLn) < Xr (P^O + 1 = Xr (Ln+i)-

Proof. Consider the following labeling of SLn. Label all vertices of degree 2 with 1, except for the bottom left and top right corners. The reduction of this graph is P2+i. Labeling the remaining vertices using the labels {2,3,..., Xr (Pi+O + 1 = Xr (Ln+i)} gives the desired result. ■

We recall the labeling h of Ln defined in Lemma 9. For a staircase graph SLn let vi,j be the j-th vertex of the path along the top of the staircase and let v2,j be the j-th vertex along the bottom of the staircase graph. We then label the vertices of the staircase using a labeling a where a(vi,j) = h(vi,j) and a(v2,j) = h(v2,j) and a(v2)2i+i) = 1 for all 1 < i < [^J. An example of a staircase labeled with a is given in Figure 8.

Lemma 21. For all j > 2, Xr (SL23-2) = Xr(L;y-2), Xr (SL23-i) = Xr (SL23-2) +1 = Xr(L2j-i). Furthermore in a Xr-ranking of Xr (SL2j-2) the corner vertices with label 1 have path access to all labels in the set {1,2,... ,Xr (SL2j-2) + 1}, and every Xr-ranking of Xr (SL2j-2) has the recursive structure defined by a.

Proof. Note that the lemma is true for all Xr-rankings of Xr (SL2) and Xr (SL6). Suppose the lemma holds for SL2j-2, for all k > 3. Consider SL2j+i-2 as the union of two copies of SL2j-2 connected by four central vertices, label the vertex of degree 4 as Xr (SL2j-2) + 1, and label the other two vertices 1 if they are adjacent to a vertex labeled 2, or 1 or 2 otherwise. Note that this is a ranking

Figure 8. Note that labels within an oval may be interchanged.

of SL2j+i_2 using xr (SL2j-2) + 2 labels having the recursive structure described above. Note that the vertices labeled 1 on the ends have path access to all labels {1,2,..., Xr (^-2) + 2}. Hence Xr (^^2^+1-2) < Xr -2) + 2 = Xr (¿2i-2) = Xr (L2i+i_2)- By Lemma 19, Xr (SL2j+i-2) = Xr CLy^^).

Let r = Xr (SL2j-2). We next prove that there does not exist a Xr-ranking of SL2j+i-2 with a different structure that the one given above. Consider SL2j+i-2 as the union of two copies of SL2j -1 sharing a single vertex a, plus an extra vertex d. See Figure 9.

If either copy of SL2k-2 uses r labels then a and d must have labels greater than r. Hence we will only consider the case where one of the top two labels is in each copy. If a is neither r + 1 nor r + 2, and these labels occur once in the left and right copies of SL2j-1. Note that the label r must be unique. Note that a = r, otherwise there will exist a ranking of SL2j-1 that uses r labels.

First we consider the case where one of the vertices b or c is labeled using r +1 or r + 2. Then SL2j-2 can be ranked with r labels, which is a contradiction.

Next we consider the case where neither b or c is labeled r + 1. Then since the removal of r, r + 1, and r + 2 do not disconnect the graph, there is a path between vertices labeled r — 1 from each copy of SL2j -1. To show Xr (SL2j+i-i) = Xr (SL2j+i-2) + 1, suppose Xr (SL2j+i-i) = Xr (SL2j+i-2). Consider SL2j+i-1 as a copy of SL2j+i-2 connected to an extra rung, L1. The extra

Figure 9. Joining of two staircase ladder graphs.

rung is connected to a corner vertex of SL2j+i_2 labeled 1 with path access to {1,2,...,xr (SL2j+i_2)} which is a contradiction. Hence Xr (SL2j+i_1) > Xr (SL2j+i_2) + 1 = Xr (L2j+i_2) + 1 = Xr (L2j+i_i) = Xr (L2i+i)• By Lemma 20, Xr (SL2j+i_1 ) = Xr (L2j+i) = Xr (SL2j+i_1). This completes the inductive step.

Lemma 22. For j > 3, Xr (SLj+2j-i_2) = Xr (Lj+2j-i_i)•

Proof. We first show that Xr (SL10) > 7. Let SL^ be the graph consisting of SL4 along with a pendant edge as shown in Figure 10.

Figure 10. SL4 with a pendant edge.

Figure 11. SL4, SL5, and SL6-

By inspection we can see that xr (SL4) = 5. Since SL10 is a 2-connected graph and is the union of two copies of SL4 plus two additional vertices we have that xr (SL10) > 7 by Lemma 3. Let j > 2 and suppose that the claim holds for j. Then xr (SL2J+i+2j-2) > xr (SL2j +2j-i-2) + 2 by Lemma 18. So Xr (SL2j +2J-1-2) > xr (SL2j+2j-1—2) + 2 = xr (L2j+2j-i-i). By Lemma 20, xr (SL2j+i+2j-2) = xr (L2j +2j-i—1) and the claim holds. The result clearly holds for 1 < n < 3 since SLn = Ln. We next consider the cases 4 < n < 6. It is known that xr(L4) = 4 and xr(L5) = 5 [12]. Since L4 is a subgraph of SL4 we have that xr(SL4) > 4. The labeling in Figure 11 (a) gives the reverse inequality. Since SL5 contains the subgraph SL4 (as shown in Figures 11 (a) and (b)) it follows that xr(SL5) > 5. The labeling shown in Figure 11 (c) shows that xr) < 5. Hence 5 = xr (SL5) = xr (L5).

Let n > 7. The proof proceeds by induction on j for values of n in the interval [2j - 1,2j+1 - 1) for j > 3.

2k - 1 < n < 2k + 2fc+1 - 3. Note that Xr (Ln) = Xr (Ln+i). By Lemma 20, we have Xr (SL2j-1) < Xr (SLn) < Xr (Ln+i). By Lemma 21, Xr (SL2j-i) = Xr (L2j-i) = Xr (Ln+1). Hence Xr (SLn) = Xr (Ln). If n = 2k + 2k-1 - 2 then Xr (SLn+1) by Lemma 22. If 2k+2k+1-1 < n < 2k+1-2, then Xr (SL2j+2j-i-2) = Xr (L2j+2j-i-1) < Xr (SLn) < Xr (L2J+1-2) = Xr (SL2j+i-2) by Lemmas 21 and 22. But since Xr (L2j +2^-1-1) = Xr ^¿+1-2), Xr (L2j +2J-1-2) = Xr (L2J+1 -2). Hence Xr (SLn) = Xr (Ln) as desired. This completes the inductive step. ■

5. Ladders with Multiple Bends

With ladders with a single bend, the direction of the bend is not important, as they will result in isomorphic graphs. However for ladders with multiple bends both the directions and the locations of the bends can have an impact on the rank number. We will use the notation BL^ to denote a bent ladder of length n with m bends.

It was shown by Novotny, Ortiz, and Narayan [12] that a ladder can be optimally ranked so that there is an pattern of alternating ones (see Figure 11(a)). In some cases the labelling pattern from a ladder graph can be adapted to fit a bent ladder, as is the case in Figure 11 (b). However if the bends are in a different direction and in different places, the rank number can increase. We will define a 'bad bend' when the alternating labeling is forced to label a vertex of degree 4 with a 1. A bend is defined to be a 'good bend' otherwise. We describe this in the next example.

Example 23. We start with the ladder P2 x P10. It was shown by Novotny, Ortiz, and Narayan [12] that Xr (P2 x P10) = 6.

The labeling shows that the rank number of the graphs in Figures 12 (a) and (b) is less than or equal to 6. However the graph shown in Figure 12 (c) has a bad bend and a rank number of at least 7. To see this note that a 6-ranking would force the two circled vertices to be labeled 5 and 6 and labeling the remaining vertices with 1, 2, 3, and 4 does not permit a ranking.

The following two lemmas can serve as the base cases for generalizing the upper bound for the rank number of a ladder with an arbitrary number of bends. In Lemma 24 we consider a ladder with only good bends, and in Lemma 25 we consider a ladder with one bad bend.

Lemma 24. Let BLn be a ladder with two good bends. Then Xr (BL;;) < Xr (Ln) for all n.

Figure 12. One ladder with two bent ladders.

Proof. Consider a xr-ranking of Ln+1 labeled with alternating ones. We label the vertices in La as they are in the first a rungs of Ln+1. Label the vertices of Lb as they are in the rungs from a + 3 < m < a + 2 + b and label the vertices in Lc as they are in the last c rungs of Ln+1. The remaining vertices can be labeled as shown in Figure 13. The first graph shows the case when the length of the middle ladder is odd and the second graph shows where the length of the middle ladder is even. Let c and f be the vertices on the ends of Lb that are not labeled 1.

xidiihiv

Lb Lc

La Lb Lc

Figure 13. Ladder graphs with the central ladder having an odd or even length.

The two configurations of a ladder with two bends are shown below in Figure 14. We consider two vertices x and y where f (x) = f (y). If x and y are both in La, Lb, and Lc then the ranking condition must be met. If x € V(La) and y € V(Lb), then the paths between them must go through either d or e. If x € V(La) and y € V(Lc) then the path must go through d or e and g or h. Finally, if x € V(Lb)

and y € V(Lc) then the path between them must go through g or h. ■

La l 1 M i i a

m 1 C 1 u

La 1 M | f

1 c n i Lc

Figure 14. A double bent ladder with the central ladder having odd or even length. Here M = max(d, e) and m = min(d, e).

Lemma 25. Let BLn be a ladder with one good bend and one bad bend. Then Xr (BL^) < xr (Ln+1) for all n.

Proof. Consider a xr-ranking of Ln+1 labeled with alternating ones. We label the vertices in La as they are in the first a rungs of Ln+1. Label the vertices of Lb as they are in the rungs from a + 3 < m < a + 2 + b and label the vertices in Lc as they are in the last c rungs of Ln+1. The remaining vertices can be labeled as shown in Figure 15. The first graph shows the case when the length of the middle ladder is odd and the second graph shows where the length of the middle ladder is even. Let c and f be the vertices on the ends of Lb that are not labeled 1. Without loss of generality, we assume the bend between Lb and Lc is the bad bend.

.v i d i i h i k i

La Lb Lc

l 1 c f l g l V

X d 1 / 1 S V

La Lb Lc

1 e 1 c 1 h 1 g 1

Figure 15. Ladder graphs with the central ladder having an odd or even length.

The two configurations of a ladder with two bends are shown in Figure 16. We consider two vertices x and y where f (x) = f (y). If x and y are both in La, Lb, and Lc then the ranking condition must be met. if x € V(La) and y € V(Lb), then the paths between them must go through either d or e. If x € V(La) and y € V(Lc) then the path must go through d or e, and g or h and k. Finally, if

x € V(Lb) and y € V(Lc) then the path between them must go through g or h and k. ■

Figure 16. A double bent ladder with the central ladder having odd or even length. Here M = max(d, e) and m = min(d, e).

We next explore the ladders with t > 2 bends. We note that after the first bend there are two choices for each additional bend so number of cases to consider grows exponentially. However present some general results when all of the bends are good bends.

Lemma 26. Let BL^ be a ladder with t good bends. Then xr (BL^) < xr (Ln) for all n.

Proof. We will proceed by induction on t. The base case of t = 1 is trivial. Assume the hypothesis is true t = j. Consider a xr-ranking of BL^. Suppose that xr (BLnQ < xr (Ln). Label the vertices in La as they are in the first

a rungs of BLn and label the vertices of xr (bL^ as they are in the last b rungs of BLn. The remaining vertices are labeled as shown in Figure 17 (a). Consider the labeling of BL^1 shown in Figure 17 (b) where M = max(r, s) and m = min(r, s). To see that this is labeling a ranking consider two vertices x and y where f (x) = f (y). If x and y are both in L„ or BL^ then the ranking condition holds. If x € La and y € BL^ then any path between them must pass through either m or M, and hence is a ranking. Then xr (BL^1) < xr (Ln). The proof then follows by induction.

Lemma 27. Let BL^ be a ladder with t good bends. Then xr (Ln-t) < xr (BL^) < xr (Ln) for all n.

Proof. Consider a xr-ranking of BL^. By Lemma 10, xr (BL^) — xr (Ln-1). Assume the formula holds for xr (BL^). Then we construct BLJ+1 as shown in Figure 18. Let M = max{x, y, z}. Then the labeling is a ranking of BL^—^ Thus

xr (BLn+1) — xr (BLJi-0, and since xr (BLra-l) — xr (L n—1—1), xr (BL+1) — xr (Ln—1—t). The proof then follows by induction on t. ■

Figure 17. Adding an additional good bend to a ladder with multiple bends.

Figure 18. Unbending a ladder.

We can apply the result involving the rank number of a ladder, to obtain new results for some ladders with multiple good bends.

Theorem 28. Let t < 2j-1 - 1. Then

2j when 2j + t - 1 < n < 2j + 2j-1 - 2, 2j + 1 when 2j + 2j-1 +1 - 1 < n < 2j+1 - 2.

Xr (BLn) =

Proof. Let 2j + t - 1 < n < 2j + 2j-1 - 2.

It was shown in [12] we have that xr (Ln) = 2j whenever 2j - 1 < n < 2j+2j-1-2. Then as long as 2j +t-1 < 2j+2j-1 -2 we have xr (Ln-t) = Xr (Ln).

Xr (L2j-1) = Xr (L2j+2j-i-2) = 2j. However we need to stay in this range when we subtract t. Hence we have t < 2j-1 - 1. This inequality insures that our upper bounds are at least as big as our lower bounds. For the sake of completeness, we include the details. Note that t < 2j-1 - 1 ^ 2j-1 +1 < 2j-1 + 2j-1 - 1 = 2 ■ 2j-1 - 1 = 2j - 1 ^ 2j + 2j-1 +1 < 2j+1 - 1 ^ 2j + 2j-1 +1 - 1 < 2j+1 - 2. For the second set of bounds, t < 2j-1 - 1 ^ t + 1 < 2j-1 ^ 2t + 2 < 2j ^ 2j+1 + 2t + 2 < 2j+1 + 2j

^ 2j + t + 1 < 2j + 2j-1 ^ 2j +1 - 1 < 2j + 2j-1 - 2. ■

We next consider the inclusion of bad bends in a ladder.

Lemma 29. Let q be the number of bad bends. Then xr (BL^O < Xr (Ln+q) for all n and m.

Proof. Consider a Xr-ranking of BLm labeled with alternating ones. There are two possible ways to extend BL^ to BLm+1, where the (m + 1)-st bend is either good or bad. If the bend is good then label the vertices in La as they appear in the first a rungs of BL^. Then label the vertices of BL^ as they appear in the last b rungs of BL^. If the bend is bad then label the vertices in La as they appear in the first a rungs of BLm+1. Then label the vertices of BL^ as they appear in the last b rungs of BLm+1. For both cases the remaining vertices can be labeled as shown in Figure 19.

Figure 19. Labelings for a ladder graph with multiple bends.

Figure 20. Adding an additional bend to a ladder with multiple bends, where M = max(d, e) and m = min(d, e). We illustrate the bending of the ladder in Figure 20. We consider two vertices x and y where f (x) = f (y). If x and y are both in V(La) or both in V(BL^) then it is clear that the labeling is a ranking. If x € V(La) and y € V(BL^), then the paths between them must go through r or s as before, meeting the ranking condition. we will have one of two cases (i) xr (BLm+1) < Xr (BL^) < Xr (Lra+q) or (ii) Xr (BLm+1) < Xr (BLm) < Xr (Ln+q+1). In either case we have Xr (BLm+1) < Xr {Ln+q/) where q' is the number of bad bends that are added.

Lemma 30. xr (BL^O — Xr mj) for all m and n.

Proof. The number of bad bends in a ladder with m bends is bounded by |_?fj • Since for every x > y, xr (Lx) > Xr (Ly) we have that xr (BL^O — Xr (Ln+q) — Xr {Ln+ymj j for all n and m. ■

It turns out that bending the ladder has a relatively small impact on the number. We show that the rank number can increase by at most one.

Figure 21.

Theorem 31. For any ladder with multiple bends, the rank number is either Xr (Ln) or Xr (Ln) + 1.

Proof. Let f be a xr-ranking Ln. Let G be the graph obtained by subdividing each horizontal edge of the ladder Ln. Then we construct another ranking f' of G by letting f '(v) = 1 for all of the new vertices v, and let f '(v) = f (v) +1 for all vertices that appear in Ln and G. We can make bends by drawing G on a grid and "making turns" using the new vertices. The edges along the inside corners can be contracted keeping the largest label to obtain a ranking of the desired bent ladder. These steps are illustrated in Figure 21 (a)-(d). ■

Acknowledgements

The authors are grateful to an anonymous referee for several comments and for providing a proof structure for Theorem 31.

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Received 11 August 2011 Revised 4 March 2013 Accepted 28 March 2013

Copyright of Discussiones Mathematicae: Graph Theory is the property of Discussiones Mathematicae Graph Theory and its content may not be copied or emailed to multiple sites or posted to a listserv without the copyright holder's express written permission. However, users may print, download, or email articles for individual use.

Copyright of Discussiones Mathematicae: Graph Theory is the property of Discussiones Mathematicae Graph Theory and its content may not be copied or emailed to multiple sites or posted to a listserv without the copyright holder's express written permission. However, users may print, download, or email articles for individual use.