Information Processing Letters 86 (2003) 191–196
www.elsevier.com/locate/ipl
Minimum feedback vertex sets in shuffle-based
interconnection networks ✩
Rastislav Královič a,∗ , Peter Ružička b
a Department of Computer Science, Comenius University, Bratislava, Slovak Republic
b Institute of Informatics, Comenius University, Bratislava, Slovak Republic
Received 4 June 2002
Communicated by F. Dehne
Abstract
Given a graph G, the problem is to construct a smallest subset S of vertices whose deletion results in an acyclic subgraph.
The set S is called a minimum feedback vertex set for G.
Tight upper and lower bounds on the cardinality of minimum feedback vertex sets have been previously obtained for some
hypercube-like networks, such as meshes, tori, butterflies, cube-connected cycles and hypercubes. In this paper we construct
minimum feedback vertex sets and determine their cardinalities in certain shuffle-based interconnection networks, such as
shuffle-exchange, de Bruijn and Kautz networks.
 2002 Elsevier Science B.V. All rights reserved.
Keywords: Combinatorial problems; Feedback vertex set; Shuffle exchange; de Bruijn; Kautz
1. Introduction
Consider minimum feedback vertex set problem:
Given a graph G = (V , E), the problem is to find
a smallest subset S of V whose removal induces
an acyclic subgraph G = (V − S, E ). The set S is
called a minimum feedback vertex set for G. In graphtheoretic community the set S is called the decycling
set and its cardinality is denoted as the decycling
number of the graph G.
✩
Supported in part by grant from VEGA 1/7155/20.
* Corresponding author.
E-mail addresses: [email protected] (R. Královič),
[email protected] (P. Ružička).
The corresponding problem for the removal of the
minimum number of edges in order to eliminate all of
the cycles in the graph is known as the cycle rank of
the graph and equals to |E| − |V | − k, where k is the
number of components.
The minimum feedback vertex set problem has
been widely studied [1–7,9–12,14]. The problem is
known to be NP-hard for general graphs [8] and the
best known approximation algorithm is of approximation ratio 2 [1]. As already mentioned in [5], there
are also polynomial time algorithms for a number of
topologies, as permutation graphs, interval graphs, reducible flow graphs, cocomparability graphs, convex
bipartite graphs and cyclically reducible graphs.
The problem was originally formulated in the area
of combinatorial circuit design. Other applications of
0020-0190/02/$ – see front matter  2002 Elsevier Science B.V. All rights reserved.
doi:10.1016/S0020-0190(02)00504-5
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R. Královič, P. Ružička / Information Processing Letters 86 (2003) 191–196
the problem (see [5,6]) are in operating systems to resource allocation mechanisms that prevent deadlocks,
in artificial intelligence to the constraint satisfaction
problem and Bayesian inference, in synchronous distributed systems to the study of monopolies and in optical networks to converters placement problem.
The size of minimum feedback vertex sets has been
studied for some hypercubic graphs, as meshes, tori,
butterflies, cube-connected cycles and hypercubes.
The achieved results were tight but not exact. The best
known lower and upper bounds on the size of feedback
vertex sets for d-dimensional hypercubes [13] are
2d−1 −
2d−1 − (d − 1)
d −1
and 2d−1 −
2d−1
,
d
respectively. In [2] exact bounds for hypercubes have
been presented for d 8 and tight bounds for 9 d 13. The best known lower and upper bounds for d-dimensional meshes of size nd [5] are
(d − 1)nd − dnd−1 + 1
2d − 1
and
(d − 1)nd
,
2d − 1
respectively. And for d-dimensional butterflies the
lower and upper bounds [5] are
(d − 1)2d + 1
3
and
(d + 1/2)2d
,
3
respectively.
In this paper we give exact cardinalities of minimum feedback vertex sets for certain shuffle-based
interconnection networks. For d-dimensional shuffleexchange graphs over binary alphabets, the minimum feedback vertex set is of size 2d−2 for odd d
and 2d−2 − 1 for even d. For d-dimensional binary
de Bruijn graphs the size of minimum feedback vertex set is (2d − 2)/3 and for d-dimensional Kautz
graphs over ternary alphabets the size is 2d−1 .
The paper is structured as follows. Preliminary
definitions are presented in Section 2. In Section 3
we apply a simple but general lower bound from [3]
to shuffle-based interconnection networks of interest.
In Section 4 we construct a minimum feedback vertex
set in shuffle-exchange, binary de Bruijn and ternary
Kautz graphs.
2. Preliminaries
The d-dimensional shuffle-exchange graph has 2d
vertices. Each vertex corresponds to a unique d-bit
binary number, and two vertices u and v are connected
by an edge if either u and v differ precisely in the last
bit or u is a left or right cyclic shift of v. If u and v
differ in the last bit, the edge is called an exchange
edge. Otherwise, the edge is called a (left or right)
shuffle edge.
The d-dimensional binary de Bruijn digraph consists of 2d vertices and 2d+1 arcs. Each vertex corresponds to a unique d-bit binary string, and there is an
arc from a vertex u1 u2 . . . ud to vertices u2 . . . ud 0 (left
arc) and u2 . . . ud 1 (right arc). In addition to having
outdegree 2, every vertex of the binary de Bruijn digraph also has indegree 2. The underlying undirected
version of d-dimensional de Bruijn digraph without
multiple edges is called d-dimensional de Bruijn graph
DBd .
In the d-dimensional ternary Kautz digraph, each
vertex corresponds to a unique string u1 u2 . . . ud ,
where each ui ∈ {0, 1, 2}, and ui = ui+1 for i = 1,
2, . . . , d − 1. Two vertices u and v are joined by an
(left) arc if and only if the first d − 1 letters of v are
equal to the last d − 1 letters of u. The underlying
undirected version of d-dimensional Kautz digraph
without multiple edges is called d-dimensional Kautz
graph Kd .
3. Lower bounds on size
The following theorem is useful for some nearregular graphs.
Theorem 3.1 [3]. For feedback vertex set S in a graph
G = (V , E) with maximum degree ∆ it holds that
|E| − |V | + 1
|S| .
∆−1
We apply this theorem to shuffle-exchange, binary
de Bruijn, and ternary Kautz graphs.
Lemma 3.2. The number of edges in d-dimensional
shuffle-exchange graph SEd , d 3, is 3 · 2d−1 − 3 for
d even and 3 · 2d−1 − 2 for d odd.
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Corollary 3.3. Any feedback vertex set in SEd , d 3,
is of size at least 2d−2 − 1 for d even and at least 2d−2
for d odd.
Lemma 3.4. The number of edges in d-dimensional
binary de Bruijn graph DBd is 2d+1 − 3 for d 2.
Corollary 3.5. Any feedback vertex set in DBd , d 2,
is of size at least (2d − 2)/3.
Lemma 3.6. The Kautz graph Kd has 3 · 2d−1 vertices
and 3(2d − 1) edges.
Corollary 3.7. Any feedback vertex set in Kautz graph
Kd has at least 2d−1 vertices.
4. Upper bounds on size
We present upper bounds on the size of minimum feedback vertex sets in shuffle-exchange, binary
de Bruijn, and ternary Kautz graphs.
Theorem 4.1. There is a feedback vertex set in SEd of
size 2d−2 − 1 for even d and 2d−2 for odd d.
Proof. First we construct a feedback set S of size
2d−2 . Consider a set of vertices
S = 0β1 | β ∈ {0, 1}d−2 .
We show that every cycle in SEd contains at least one
vertex from S. First note that vertices containing all
zeroes or all ones have degree one so they cannot be
contained in any cycle. For the rest of the proof it
is therefore sufficient to deal only with vertices that
contain both zeroes and ones.
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Let us start by considering cycles containing exclusively shuffle edges. For an arbitrary vertex α we find a
value k such that a k-times right shift of α is in S. Note
that k may be larger than the length of the cycle without causing any troubles. As α contains at least one 1
we may assume without loss of generality that α ends
with 1. Then α = β0a 1b for some 0 < a, b d, where
β is either empty or ends with 1. Then k = a + b right
shifts yield a string 0a 1b β from S.
Now assume that there is a cycle C with at least one
exchange edge. Without loss of generality this edge
changes last bit from 1 to 0 in a vertex α. If α starts
with 0, then it holds α ∈ S. So consider α = 1k β1r
for some 0 < k, r d, where β starts and ends with
0. After the exchange operation we move to a vertex
α = 1k β1r−10. We continue the proof by induction
on k. As a basis, let k = 1 and distinguish two cases.
If r > 1, then α = 10γ 10 for some γ . The cycle must
continue by a shuffle edge but both shifts yield a vertex
in S. If r = 1, then α = 10γ 00 for some γ . In this case
left shift leads to a vertex in S, so suppose the cycle
C continues by a series of right shifts followed by an
exchange. First of the shifts yields a vertex 0γ 0 for
some γ . Look at the first place in C where the vertex
is not of the form 0γ 0: clearly, it must be of the form
0γ 1 ∈ S.
In the induction step k > 1 we again distinguish two
cases. If r > 1, then α = 11γ 10 for some γ . Right
shift leads to a vertex 011γ 1 ∈ S. So suppose that C
continues with a number of left shifts followed by an
exchange. First of these shifts leads to 1γ 101. Find the
first vertex in C which is not of the form 1γ 1 for some
γ . It is either 0γ 1 ∈ S or 1k β 1r 0 for 0 < k < k and
induction hypothesis applies.
The last case we consider is k > 1, r = 1, i.e.,
α = 11γ 00. C continues by a sequence of shifts to one
direction (left or right) followed by an exchange. Right
Fig. 1. Minimum feedback set in SE 4 . Black vertices form the feedback set of size 2d−2 . Arrows denote the transformation to reduce the size
by 1.
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shifts yield a situation similar to the case k = r = 1 and
left shifts a situation similar to k > 1, r = 1.
Now suppose d = 2k. We show how to reduce the
size of S by 1. Let S be a set of vertices obtained
from S by replacing every vertex of the form 0a (01)b ,
0 a d, 0 < b k by a vertex 0a (01)b−1 00 and
by removing vertex 0 . . . 0. We show that S is a
feedback set. First note that every cycle containing
(01)k contains also (01)k−1 00 as (01)k has degree
2. Consider an arbitrary cycle C. Because S is a
feedback set, C contains a vertex α from S. Suppose
α∈
/ S , i.e., α = 0a (01)b , a 2. As α has degree
3, either 0a (01)b−1 00 or 0a−1 (01)b 0 is in C. The
vertex 0a (01)b−1 00 is in S , so suppose that only
β = 0a−1 (01)b 0 is in C. Again, β has degree 3,
so either β1 = 0a−1 (01)b 1 or β2 = 0a−2(01)b 00 is
in C. However, β1 is in S because it is in S and
ends with 11, and β2 is in S as a replacement for
0a−2 (01)b+1 . ✷
Theorem 4.2. There is a feedback vertex set in DBd of
size 13 (2d − 2).
Proof. We shall treat the odd and even dimensions
separately and construct feedback sets of size 13 (2d −
1) for even d and 13 (2d − 2) for odd d.
Let d = 2k and let S be the set of vertices starting
with 0. For an arbitrary cycle C we show that it
contains at least one vertex from S. Suppose for the
sake of contradiction that every vertex in C starts with
1 and distinguish two cases. If C contains a left edge
(i.e., left shift with an arbitrary bit written in the last
position), then C cannot contain left edges only. To see
why, look at a vertex α ∈ C. Either α contains a zero
bit or α = 11 . . . 1 and the next edge leads to 1 . . . 10.
In any case, there is a vertex that contains a zero bit
in C and a number of left edges would bring this zero
bit to the beginning. Hence, C contains a “turn”—a left
edge followed by a right edge. The left edge goes from
vertex 1β to vertex βa for a string β and some a. The
following right edge cannot lead to the vertex 1β again
so it must lead to 0β.
If C contains only right edges, it must contain a
vertex α of the form α = 1r 0β for some r > 0 and
some string β. Look at the number of zeroes in the first
r + 1 positions of the vertices in C. In α this number is
one, and in the following vertex it is zero. Because C
is a cycle there must be a moment when the number of
zeroes in the first r + 1 positions increases. But since C
contains only right edges it must be the case that there
is a zero in the beginning.
So we have proved that S is a feedback vertex set.
S forms a binary tree: vertices of the form 01β are
leaves and a vertex 00β has children 0β0 and 0β1.
The root is the vertex 0 . . . 0 which has only one child
and the rest forms a complete binary tree with 2k − 2
levels. Let S contain every second level of this tree.
We prove that S is a feedback vertex set. Indeed,
every cycle contains a vertex v from S. If v ∈
/ S, v
has only one neighbor outside S (leaves of the tree are
Fig. 2. Minimum feedback set in a de Bruijn graph of even dimension (DB4 ). The labels of vertices are decimal representations of their binary
strings.
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195
Fig. 3. Minimum feedback set in a de Bruijn graph of odd dimension (DB5 ). The labels of vertices are decimal representations of their binary
strings.
in S , inner vertices have three neighbors in the tree
and degree four and vertex 0 . . . 0 has one neighbor in
the tree and degree two). It follows that at least one
of v’s neighbors in the cycle is in S . The number of
vertices in S is 1 + 22 + 24 + · · · + 22k−2 = 13 (2d − 1)
as d = 2k.
Now, let d = 2k + 1. Let S = S1 ∪ S2 , where
S1 = 00β | β ∈ {0, 1}d−2 and
S2 = 11β | β ∈ {0, 1}d−2 .
In a way similar to the previous case we start by
showing that S is a feedback vertex set. Consider a
cycle C such that each vertex in C has different bits
in first two positions. First note that C must contain
both right and left edges. Indeed, a sequence of right
edges produces a prefix of alternating zeroes and ones.
Similarly, a sequence of left edges requires the starting
vertex to have a prefix of alternating zeroes and ones.
It follows that if C contains left or right edges only, it is
a degenerated cycle of length two with vertices (01)k 0
and (10)k 1.
Hence there is a “turn” in C where a left edge is
followed by a right edge. Without loss of generality,
let the vertex just before the turn be 01β. The left edge
leads to 1βa for some a and the following right edge
must lead to 11β.
Each of the sets S1 , S2 forms a tree in a way
analogous to the case of even dimension (S2 is a
symmetric image of S1 obtained by negating all bits).
Again, we take S to be every second level of both
trees. By the same argument as above we get that S is a feedback vertex set. The number of vertices in
S is 2(1 + 22 + 24 + · · · + 22k−2 ) = 13 (2d − 2) as
d = 2k + 1. ✷
Theorem 4.3. There is a feedback vertex set in Kautz
graph Kd of size 2d−1 .
Proof. Let S consists of all vertices starting with 0.
For the sake of contradiction, let C be a cycle that does
not contain a vertex from S. Note that C must contain
a vertex that has 0 in some position—otherwise every
vertex would be either 121212 . . . or 212121 . . . and C
would be a degenerated cycle of length two. It follows
that C cannot contain only left edges, as they would
bring the 0 to the beginning. On the other hand, C
cannot contain right edges exclusively: a sequence of
right edges would produce a prefix of alternating 1’s
and 2’s leading again to a degenerated cycle. So C
contains a “turn”: a left edge followed by a right edge.
Let the vertex just before the turn be without loss of
generality of the form 1aβ where a ∈ {0, 2}. The left
edge leads to aβb, hence a = 2. The next right edge
cannot lead back to 12β so it must lead to 02β.
Hence we have a feedback vertex set containing one
third of all vertices. ✷
Note that the argument used in the above proof is
valid also for symmetrical minimum vertex sets S =
{101, 102, 120, 121} and S = {212, 210, 201, 202} in
Fig. 4.
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Fig. 4. Minimum feedback set in Kautz graph K3 .
5. Conclusions
We have solved Minimum Feedback Vertex Set
Problem for shuffle-based interconnection networks
(shuffle-exchange, de Bruijn, Kautz). The question is
how to extend these results to shuffle-based interconnection topologies over larger alphabets.
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