On Balanced Separators, Treewidth, and Cycle Rank

0.32 MB·English

by  Hermann Gruber

On Balanced Separators, Treewidth, and Cycle Rank
Hermann Gruber
knowledgepark AG, Leonrodstr. 68, D-80636 Mu¨nchen, Germany
Abstract
We investigate relations between different width parameters of graphs, in particular balanced sepa-
0 rator number, treewidth, and cycle rank. Our main result states that a graph with balanced separator
1 number k has treewidth at least k but cycle rank at most k·(cid:0)1+logn(cid:1), thus refining the previously
0 k
known bounds, as stated by Robertson and Seymour (1986) and by Bodlaender et al. (1995). Further-
2
more, weshowthattheimprovedboundsarebestpossible. Amongotherresults, wealsodeterminethe

c cyclerankofpowersofpathgraphs,thusansweringaquestionrecentlyraisedbyNovotnyetal.(2009).
e
D Keywords: vertexseparator,treewidth,pathwidth,bandwidth,cyclerank,orderedcoloring,vertexranking,
hypercube
6
MSC: 05C40, 05C35

]
M
1 Preliminaries
D
. Throughoutthispaper, logxdenotesthebinarylogarithmofx, andlnxdenotesthenaturallogarithmofx.
s
c
[
1.1 Vertex separators

1
v We assume the reader is familiar with basic notions in graph theory, as contained in [7]. In a connected
4 graph G, a subset X of the vertex set V(G) is a vertex separator for G, if there exists a pair x,y of vertices
4 lying in distinct components of V(G)−X, or if V(G)−X contains less than two vertices. Beside vertex
3
separators, also so-called edge separators are studied in the literature. As we shall deal only with the former
1
kind of separators, we will mostly speak just of separators when referring to vertex separators.
.
2 If G has several connected components, we say X is a separator for G if it is a separator for some
1
component of G. A separator X is (inclusion) minimal if no other separator is properly contained in it. A
0 (cid:108) (cid:109)
1 separatorX isbalanced, ifeverycomponentofG\X containsatmost |V(G)|−|X| vertices.; anditiscalled
2
:
v strictly balanced, if every component of G\X contains at most |V(G)|−|X| vertices.
i 2
X
Lemma 1. Let G be a graph with a balanced separator of size at most k. Then G has a balanced separator
r
a of size exactly k.
Proof. AssumeX isabalancedseparatorofsizex<|V(G)|. ThenwecanextendS toabalancedseparator
ofsizex+1asfollows: LetC ,C ,...bethecomponentsofG−X, orderedbydecreasingcardinality. Letv
1 2
be any vertex in the largest component C . We claim that X ∪{v} is again a balanced separator for G.
1
Namely, we have
(cid:24) (cid:25) (cid:24) (cid:25)
|V(G)|−|X| |V(G)|−|X∪{v}|
|C \{v}|≤ −1≤ .
1 2 2
The components C in G−X with i ≥ 2 are also components of G−(X ∪{v}), so it remains to bound
i
(cid:106) (cid:107)
the cardinality of C . The latter is the second largest component of G−X, thus |C | ≤ |V(G)|−|X| , and
2 2 2
consequently
(cid:22) (cid:23) (cid:24) (cid:25)
|V(G)|−|X∪{v}|+1 |V(G)|−|X∪{v}|
|C |≤ ≤ .
2 2 2
For the components C with i>2, we have of course |C |≤|C |, and the proof is completed.
i i 2
1
Arguablytheveryfirstresultongraphseparators,provedbyJordan[13]inthe19thcentury,isperhaps
most naturally phrased in terms of non-strictly balanced separators: namely, every tree admits a balanced
separator consisting of a single vertex. However, the usage of strictly balanced separators seems to predom-
inate in the more recent literature (see e.g. [3, 15, 20]), perhaps for the reason that the use of the ceiling
operator (cid:100)x(cid:101) seems unappealing. But the monotonicity property stated in Lemma 1 ceases to be true for
strictly balanced separators. This is already witnessed by very simple examples, such as P , the path
2k+1
graph of odd order 2k+1. In any case, we can always turn a balanced separator into a strictly balanced
separator by adding at most one vertex.
The(strict) balanced separator numberofagraphG,denotedbys(G)(resp. ˜s(G)forthestrictversion),
is defined as the smallest integer k such that for every Q ⊆ V(G), the induced subgraph G[Q] admits a
(strictly)balancedseparatorofsizeatmostk. ObservethatforanygraphG,wehave˜s(G)≤s(G)≤˜s(G)+1.
1.2 Width parameters
ThecyclerankofagraphG,denotedbyr(G),isastructuralcomplexitymeasureongraphs,andisinductively
definedasfollows: IfGhasnoedges,thenr(G)=1; ifGhasseveralcomponents,thenr(G)isthemaximum
cyclerankamongtheconnectedcomponentsofG;otherwise,r(G)=1+min r(G−v). Iteasilyfollows
v∈V(G)
from the definition of cycle rank that for every vertex v ∈ V(G) holds r(G−v) ≤ 1+r(G), and similarly
r(G−X)≤|X|+r(G−X) for every vertex subset X ⊆V(G).
The notion of cycle rank was originally devised by Eggan and Bu¨chi [8] as a notion on digraphs, and
appears in the literature under many different names, such as ordered chromatic number [14], vertex ranking
number[2],tree-depth[18],orminimum elimination tree height[3,16]. Althoughallthesenotionsultimately
refertothesameconcept,somesourcesuseadifferentnormalization. Forinstance,theminimumelimination
tree height of a graph is equal to the cycle rank minus 1.
Other structural complexity measures studied in this paper include the treewidth and pathwidth of
graphs; more background information on these two measures can be found in [1, 15, 21]. For a graph G,
let V = {U ,U ,...,U } be a collection of subsets of V(G). A tree T = (V,E) with vertex set V is called
1 2 r
a tree decomposition, if all of the following hold: (i) The collection V covers the vertex set of the graph G,
in the sense that V = (cid:83) U; (ii) For every edge (u,v) ∈ E, there is a tree node U ∈ V such that both u
U∈V
and v are in U; and (iii) If two tree nodes U and U are connected in the tree by a path, then U ∩U
1 2 1 2
is a subset of each tree node visited along this path. The width of a tree decomposition T = (V,E) is
defined as max{|U|−1 | U ∈ V}, and the treewidth of G is defined as the minimum width among all tree
decompositionsforG. ApathdecompositionisatreedecompositionwhereT isrequiredtobeapathgraph;
and the pathwidth of G is the minimum width among all path decompositions for G.
2 Main Results
We recall the following well-known result [3, Thm. 11], which relates the strict balanced separator num-
ber˜s(G), treewidth tw(G), pathwidth pw(G), and the cycle rank r(G) of a graph G:
Theorem 2. Let G be a graph of order n≥2, and let ˜s(G) denote its strict balanced separator number, let
tw(G) and pw(G) denote its treewidth and pathwidth, respectively, and let r(G) denote its cycle rank. Then
˜s(G)−1≤tw(G)≤pw(G)≤r(G)≤1+˜s(G)·logn.
As Bodlaender et al. [3] note, the bounds stated in this theorem are spread across the literature, and
(variations of) some of these bounds were discovered independently by several groups of authors.
Knownexamplesofgraphshavingalogarithmicgapbetweentreewidthandpathwidtharethecomplete
binarytreesoforder2d−1,whichhavetreewidth1butpathwidth(cid:100)d/2(cid:101),see[21,22]. Fortherelationbetween
pathwidth and cycle rank, a similar role is played by the path graphs of order n, which have pathwidth 1
but cycle rank 1+(cid:98)logn(cid:99), see [17]. Indeed, the cycle rank of trees of order n can be no larger than this [14].
Butobservethatthestrictbalancedseparatornumberofapathofordernequals2forn≥4, soTheorem2
givesonlyanupperboundof2·(1+logn)foralltreesofdiameteratleast3. Comparingthiswiththeupper
bound on the cycle rank of trees [14] noted above, we see that the former bound is a forteriori unsharp by
a factor 2. The situation is even worse for graphs whose balanced separator number is linear in n, such as
2
complete graphs or expanders: their cycle rank can trivially be at most n. so the estimate is off by a factor
of Ω(logn) in this case. Therefore, our first aim will be to refine the rightmost inequality of Theorem 2.
The following recurrence will play a crucial role in our investigation.
Definition 3. For integers k,n≥1, let R (n) be given by the recurrence
k
(cid:18)(cid:24) (cid:25)(cid:19)
n−k
R (n)=k+R ,
k k 2
with R (r )=r for r ≤k.
k 0 0 0
As it turns out, the function R (n) can serve as upper bound on the cycle rank of a graph in terms of
k
its order and its balanced separator number:
Lemma 4. Let G be a graph of order n whose balanced separator number is at most k. Then for the cycle
rank r(G) holds
r(G)≤R (n).
k
Proof. The overall structure of the argument is the same as e.g. in [3, 18], but here we derive a somewhat
stronger statement. We prove the statement by induction on the order n of G. The base cases n≤k of the
induction are easily seen to hold, since the cycle rank of a graph is always bounded above by its order.
For the induction step, assume n > k. Let X be a balanced separator for G of size exactly k. Using
Lemma 1, we know that such a separator exists. Denote the connected components of G−X by C ,...,C .
1 p
Then r(G) ≤ k + r(G − X), and by definition of cycle rank, r(G − X) ≤ maxp r(G[C ]). As X is a
i=1 i
balanced separator, we have |C |≤(cid:100)n−k(cid:101) for 1≤i≤p, so we can apply the induction hypothesis to obtain
i 2
maxp r(G[C ])≤R ((cid:100)n−k(cid:101)). Putting these pieces together, we have r(G)≤k+R ((cid:100)n−k(cid:101)), as desired.
i=1 i k 2 k 2
In fact, we shall see next that this bound is best possible for all k and each n≥k+1. As in [5], let Pk
n
denotethethekthpowerofapathgraphofordern,inwhichtwodistinctverticesu,v inV(P )=1,2,...,n
n
areadjacentiff|v−u|≤k. DeterminingthecyclerankofthisgraphisaproblemrecentlyraisedbyNovotny
et al. [19]. We show that the cycle rank of Pk is exactly R (n). To this end, we will need the following fact
n k
from [6, 16].
Lemma 5. Let G be an undirected graph. Then
r(G)=min{|S|+r(G−S)}, (1)
S
where S runs over all inclusion minimal separators of G.
The theorem reads as follows:
Theorem 6. Let k,n∈N, and let Pk denote the kth power of a path of order n. Then
n
r(cid:0)Pk(cid:1)=R (n).
n k
Proof. In the case n = k−1, the graph Pk is a clique of order n, and it is known that these graphs have
n
cycle rank n. So we assume n≥k−2 in the following.
We will use Lemma 5 to compute the cycle rank of Pk recursively in terms of its inclusion minimal
n
separators. Fortunately, can completely characterize the inclusion minimal separators for this graph:
Claim 6.1. For k ≤ n−2, the set of inclusion minimal separators of Pk consists precisely of the integer
n
intervals [2;k+1], [3;k+2], ..., [n−k;n−1].
Proof. We show that if S ⊆V(Pk) is a separator in Pk, then S contains an interval of size at least k. Since
n n
every such interval is a separator in Pk, as long as its endpoints exclude both 1 and n, this clearly implies
n
the claimed statement.
Let S be a separator in Pk, and let (a,b) be a pair of vertices separated by S such that the difference
n
|a−b| is minimal among all pairs separated by S. Then by definition of Pk, the difference |a−b| must be
n
strictly greater than k, since otherwise a and b would be adjacent. Also, all vertices ranging from a+1 to
b−1 are necessarily in S, since each of them is adjacent to both a and b, again by definition of Pk. Thus S
n
contains an interval containing at least k numbers. This completes the proof of the claim.
3
Returning to our main objective, observe that by removing an interval [r;r+k−1] of size k from Pk,
n
the resulting graph Pk −[r;r+k−1] has two components—provided 2 ≤ r ≤ n−k. It follows from the
n
definition of Pk that each of these components is isomorphic to the kth power of some path graph of order
n
smaller than n. Plugging Claim 1 into the recurrence from Lemma 5, we thus obtain:
r(Pk)= min (cid:8)k+r(Pk−[r;r+k−1])(cid:9)
n n
2≤r≤n−k
=k+ min max(cid:8)r(Pk),r(Pk )(cid:9)
r n−k−r
2≤r≤n−k
=k+ min max(cid:8)r(Pk),r(Pk )(cid:9).
r n−k−r
2≤r≤(cid:100)n−k(cid:101)
2
Next, note that for r ≤n−k−r, the graph Pk is an induced subgraph of Pk , so
r n−k−r
max(cid:8)r(Pk),r(Pk )(cid:9)=r(Pk ).
r n−k−r n−k−r
In the same fashion, Pk is an induced subgraph of Pk , and hence
n−k−(r+1) n−k−r
(cid:16) (cid:17)
min r(Pk )=r Pk .
2≤r≤(cid:100)n−k(cid:101) n−k−r (cid:100)n−2k(cid:101)
2
Putting these observations together yields the recurrence
(cid:16) (cid:17)
r(Pk)=k+r Pk ,
n (cid:100)n−k(cid:101)
2
with base cases r(cid:0)Pk(cid:1)=n for n≤k.
n
Since this recurrence is virtually identical to the one used for defining R (n), the proof is completed.
k
Given the importance of the function R (n) from Definition 3 in our context, we will analyze this
k
recurrence in more detail. In particular, we derive an upper bound in closed form. For each fixed k, that
upper bound is sharp infinitely often, so our estimate is essentially best possible.
Theorem 7. Let k,n≥1 be integers. Then
(cid:16) n(cid:17)
R (n)≤k· 1+log ,
k k
with equality iff n=k·(2j −1) for some j ∈N\{0}.
Proof. Instead of analyzing the recurrence R (·) directly, we first look at the problem from a different
k
perspective. To this end, for a given positive integer r, let N (r) denote the smallest integer n such that
k
R (n)≥r. Then by R (2r+k−1)=R (2r+k)=k+R (r), we have N (r+k)=2·N (r)+k−1, and
k k k k k k
the recurrence terminates with N (r )=r for r ≤k.
k 0 0 0
To get a closed form for N (r), we make some use of finite calculus (see [10]): The backward difference
k
of N , denoted by ∇N , is defined as ∇N (i)=N (i)−N (i−1). For convenience, let us define N (0)=0,
k k k k k k
such that ∇N (1) is well-defined.
k
Claim 7.1. For all integers i,j with i≥1 and (i−1)·k <j ≤i·k holds
∇N (j)=2i−1.
k
Proof. We prove this by induction on i. The base cases where i = 1 and j ≤ k are easily verified, so it
remains to perform the induction step. For (i−1)·k <j ≤i·k, we have
∇N (j)=2·N (j−k)+k−1−2·N (j−1−k)−k+1
k k k
=2·∇N (j−k)
k
=2·2i−2,
wherethelaststepaboveholdsbyinductionhypothesis, since(i−2)·k <j−k ≤(i−1)·k. Thiscompletes
the proof of the claim.
4
Now that we have a closed form for the backward differences, it is not difficult to derive a closed form
for N (r):
k
Claim 7.2. For all integers k,r ≥1 holds
N (r)=(k+r modk)·2(r−rmodk)/k−k.
k
Proof. TelescopingsumsyieldsN (r)=N (r)−N (0)=(cid:80)r ∇N (i). Writerasr =k·s+twiths=rdivk
k k k i=1 k
and t=r modk. We arrive at a simplified form for N (r) by grouping the summands appropriatey, and by
k
rewriting those using Claim 7.1 afterwards:
s r
(cid:88) (cid:88) (cid:88)
N (r)= ∇N (j)+ ∇N (j)
k k k
i=1j>(i−1)·k j=s·k+1
j≤i·k
s
=(cid:88)(cid:0)k·2i−1(cid:1)+t·2s =k·(2s−1)+t·2s.
i=1
=(k+r modk)·2(r−rmodk)/k−k.
In the last line, we made use of the facts t=r modk and s= r−rmodk.
k
Unfortunately, this closed form seems impossible to resolve after r. We thus resort to giving an upper
bound that is tight infinitely often; for the ease of reading, the proof of the following claim is found in the
appendix.
Claim 7.3. For all integers k,r with k ≥2 and r ≥1 holds
N (r)≥k·(2r/k−1),
k
with equality iff r modk =0.
We are finally in position to analyze the recurrence R (n). The special case k = 1 is well known and
k
not difficult to prove by induction (see e.g. [17]), so in the following we assume k ≥2. For a given integer n,
let r =R (n). Using the definition of the function N (r) and Claim 7.3, we have
k k
n≥N (r)≥k·(2r/k−1), (2)
k
with both inequalities being sharp iff n = N (r) and r modk = 0. From Claim 7.2, we can derive that the
k
latter two conditions are in turn equivalent to the requirement that n=k·(2j −1) for some j ∈N.
With r = R (n), we can solve Ineq. (2) after r to get R (n) ≤ k · (cid:0)1+logn(cid:1), with equality iff
k k k
n=k·(2j −1) for some j ∈N\{0}. This completes the proof of Theorem 7.
The above result settles the relation between balanced separator number and cycle rank. We turn our
attentiontotheotherendofthechainofinequalitiesinTheorem2,namelytotherelationbetweenbalanced
separator number and treewidth. Here the optimal inequality will follow by refining a known result, so the
remainder of this section is necessarily less self-contained than the rest of this paper.
Recall that a graph is called chordal iff every cycle C of order greater than 3 in G has a chord, i.e., an
edge connecting two vertices not adjacent in C. In other words, a chordal graph has no induced cycle of
order greater than 3.
Inthefollowing,wederiveastrengthenedversionofatheoremoriginallyduetoGilbertetal.[9]—these
authors employed a more permissive notion of balanced separator.
Theorem 8. Let G be a chordal graph whose largest clique has order p. Then G contains a clique of order
at most p−1 that is a balanced separator for G.
5
Proof. The proof follows the one given by Gilbert et al. [9] for a similar statement. If G itself is a clique,
then removing a single vertex yields a balanced separator of order |V(G)|−1=p−1. So we assume in the
following that G is not a clique.
ChooseC tobethecliqueinGwhichminimizesthemaximumorderamongtheconnectedcomponents
of G−C; in case there are several such cliques, take C itself to be of minimum order among these cliques.
Our aim is to show that then C is a balanced separator for G. Let A be a component of G\C that is of
maximum order. The capstone of the original proof consists of deriving the following:
Fact [9]. Component A contains a vertex v adjacent to all vertices in C.
For a proof of this, the reader is referred to [9, Fact 3]. Notice that this fact implies that |C| ≤ p−1,
(cid:108) (cid:109)
and thus it only remains to show that |A|≤ |V(G)|−|C| . For the sake of contradiction, assume this is not
2
the case. Then, letting B = V(G)\(C ∪A), we must have |A| > |B|, and thus |B| ≤ |A|−1. Take a
vertex v ∈ A that is adjacent to all vertices in C. Then G−(C ∪{v}) falls apart into the disconnected
subgraphs G[A]−v and G[B], and the maximum order among the connected components of G−(C∪{v})
is at most |A|−1. Thus the clique C∪{v} would have been preferable to C, a contradiction.
It is well known that if a graph has treewidth at most k, then G is subgraph of some chordal graph
whose largest clique is of order at most k+1, compare [1, 21]. The following corollary is now immediate:
Corollary 9. For any graph G, we have s(G)≤tw(G).
Again, for every possible value of s(G), there are infinitely many graphs for which this bound is tight,
as witnessed by the kth power of a path Pk. Also, the inequality is tight in the case of trees. In this way,
n
Corollary 9 gives a proper generalization of Jordan’s classical result on balanced separators in trees [13].
Since ˜s(G) ≤ s(G)+1, Corollary 9 also implies the inequality ˜s(G) ≤ tw(G)+1 given by Robertson and
Seymour [20, Proposition 2.5]. The latter inequality is off by 1 for all trees containing a path of order 4.
Taking the statements of Lemma 4, of Theorem 7, and of Corollary 9 together, we have the following
improvement over Theorem 2:
Theorem 10. Let G be a graph of order n≥2, and let s(G) denote its balanced separator number, let tw(G)
and pw(G) denote its treewidth and pathwidth, respectively, and let r(G) denote its cycle rank. Then
(cid:18) (cid:19)
n
s(G)≤tw(G)≤pw(G)≤r(G)≤s(G)· 1+log .
s(G)
3 Applications
An immediate consequence of Theorem 10 is a tight upper bound for the cycle rank of a graph in terms of
its treewidth:
Corollary 11. Let G be a graph of order n ≥ 2, and let tw(G) and r(G) denote its treewidth and its cycle
rank, respectively. Then
(cid:18) (cid:19)
n
r(G)≤tw(G)· 1+log .
tw(G)
Quite obviously, an analogous statement holds for cycle rank versus pathwidth. Our next application
concernstherelationbetweenthecyclerankandanotherwidthparameterforgraphs,namelythebandwidth.
The latter is defined in the following.
A linear layout of a graph G of order n is a bijection (cid:96) of V(G) into the integer interval [1;n]. The
bandwidth of a linear layout (cid:96) is defined as max |(cid:96)(v)−(cid:96)(u)|; and the bandwidth of G, denoted by
{u,v}∈E(G)
bw(G), is defined as the minimum bandwidth among all linear layouts for G.
Thebandwidthofagraphisknowntobeboundedbelowbyitspathwidth, seee.g.[1, Thm.44], sothe
latterisacommonlowerboundforbothcyclerankandbandwidth. Buthowarethesetworelated? Itisnot
difficult to prove that the bandwidth of the star graph K on n+1 vertices equals (cid:98)n+1(cid:99), cf. [21]. On the
1,n 2
6
other hand, it is clear that r(K ) = 2, so there seem to be no interesting upper bound on the bandwidth
1,n
of a graph in terms of its cycle rank.
For the converse direction, we obtain the following result as a corollary to Theorem 6:
Corollary 12. Let G be a graph of bandwidth at most k. Then
r(G)≤R (n),
k
and this bound is tight if G is the kth power of the path graph of order n.
Proof. Abasicfactaboutthebandwidthbw(G)ofagraphGofordernisthatbw(G)≤kiffGisisomorphic
to a subgraph of Pk [5]. Now we have r(Pk) = R (n), and the cycle rank can never increase when taking
n n k
subgraphs.
As a final application, we determine a sublinear upper bound on the cycle rank of the d-dimensional
hypercube. Recallthatthevertexsetofthed-dimensionalhypercubeis{0,1}d,andtwoverticesareadjacent,
if their vector representations differ in exactly one coordinate. Bounding the cycle rank of this graph from
below has turned out to be useful in formal language theory, namely for comparing the relative succinctness
of different variants of the regular expression formalism, see [11].
Theorem 13. Let H denote the d-dimensional hypercube of order n=2d. Then
d
(cid:18) (cid:19)
nloglogn
r(H )=O √ .
d
logn
Proof. Harper [12] proved that for the bandwidth of the d-dimensional hypercube holds
d (cid:18) (cid:19)
(cid:88) d
bw(H )= .
d (cid:98)d/2(cid:99)
i=0
Using Stirling’s approximation, one can show that this sum is in Θ(d−1/22d). To estimate the cycle rank of
this graph, we use Corollary 12 and Theorem 7 to obtain:
(cid:18) 2d (cid:19)
r(H )≤bw(H )·log
d d bw(H )
d
(cid:32) (cid:33)
2d
=O d−1/2·2d·log
(cid:0) (cid:1)
Θ d−1/22d
(cid:18) (cid:19)
nloglogn
=O √ ,
logn
as desired.
As observed in [11], the hypercube of order n has cycle rank at least Ω(√n ), so this upper bound is
logn
tightuptoafactorofO(loglogn). Incontrast,utilizingtheknownfactthatthepathwidthofthehypercube
is equal to its bandwidth [4] together with the previously known upper bound from Theorem 2 would result
in an upper bound that even exceeds the trivial upper bound of n.
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8
Appendix
Recall that for integer k ≥1, the function N :N→N is inductively defined by N (r )=r for r ≤k, and
k k 0 0 0
N (r+k) = 2·N (r)+k−1. In Claim 7.3, we stated that for k ≥ 2 holds N (r) ≥ k·(2r/k −1), with
k k k
equality iff r modk =0.
Proof of Claim 7.3. To prove our claim, we consider the univariate real-valued function f given by
k,r
f (x)=(k+x)·2(r−x)/k−k, (3)
k,r
with k and r positive real numbers greater than 1. Differentiating after x gives
(cid:16) (cid:16) x(cid:17) (cid:17)
f(cid:48) (x)=2(r−x)/k· 1− 1+ ·ln2 . (4)
k,r k
An easy computation yields f(cid:48) (x) ≥ 0 if and only if x ≤ k ·(1/ln2−1), so the function f (x)
k,r k,r
has a unique maximum at x = k·(1/ln2−1), and it is monotonically increasing (resp. decreasing) for
0
x < x (resp. x > x ). Now let us restrict the domain of f to the closed interval I = [0;k−1]. Since
0 0 k,r
0<1/ln2−1<1/2, the number x is contained within I for k ≥2.
0
Elementary calculus shows that this restricted function will attain its absolute minimum at the left or
right boundary of I. Let us calculate which one is the case here:
(cid:18)(cid:18) (cid:19) (cid:19)
1
f (k−1)−f (0)=k·2r/k· 2− ·2(−k+1)/k−1 . (5)
k,r k,r k
Byevaluatingtheright-handsideat k =1andbydifferentiatingafter k, wecan deducethat this expression
is strictly greater than 0 for all real-valued k > 1. Thus, f (k−1) > f (0), and we conclude that the
k,r k,r
function f :I →R attains its absolute minimum at x =0.
k,r 1
WereturntothefunctionN (r),whosedomainarethepositiveintegers. RecallthatClaim(7.2)states
k
that
N (r)=(k+r modk)·2(r−rmodk)/k−k.
k
Combining this with (3), we get
N (r)=(k+r modk)·2(r−rmodk)/k−k =f (r modk). (6)
k k,r
Wesawabovethatthefunctionf (·),whenrestrictedtotheinterval[0;k−1],attainsitsabsoluteminimum
k,r
for x =0. So we can deduce that
1
N (r)≥f (0)=k·(2r/k−1)
k k,r
holds for all r ≥1 and k ≥2, and that equality holds if and only if r modk =0. This completes the proof
of the claim.
9