We present and prove a theorem answering the question "how many states does a
minimal deterministic finite automaton (DFA) that recognizes the set of base-b
numbers divisible by k have?"Comment: LaTeX, 7 pages (corrected typo in new version

Alexeev, Boris

English

arXiv

AbstractWe present and prove a theorem answering the question “how many states does a minimal deterministic finite automaton (DFA) recognizing the set of base-b numbers divisible by k have?

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Minimal DFA for testing divisibility 

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Journal of Computer and System Sciences 69 (2004) 235–243
Minimal DFA for testing divisibility
Boris Alexeev
510 Millstone Cir, Athens, GA 30605, USA
Received 28 September 2003; revised 21 January 2004
Abstract
We present and prove a theorem answering the question ‘‘how many states does a minimal deterministic
ﬁnite automaton (DFA) recognizing the set of base-b numbers divisible by k have?’’
r 2004 Elsevier Inc. All rights reserved.
1. Statement of the problem
The following exercise is typical in introductory texts on deterministic ﬁnite automata (DFAs):
‘‘produce an automaton that recognizes the set of binary strings that, when interpreted as binary
numbers, are divisible by k:’’ For example, exercise 1.30 in [2] asks the student to prove that the
language fx j x is a binary number that is a multiple of kg is regular for each kX1; explicitly
presenting an automaton is the easiest solution.
The traditional (and correct) answer constructs a k-state automaton that keeps track not only
of divisibility by k; but also the current residue modulo k: For example, if the input read was 1101,
the machine would remember ‘‘13 mod k’’. The transitions between states are simple: if the
automaton’s current state is ‘‘r mod k’’, and the input symbol read is ‘‘0’’, it moves to state
ð2rÞmod k; if the input symbol read is ‘‘1’’, it moves to state ð2r þ 1Þmod k:
(This example also generalizes to bases other than binary. Furthermore, even if the input string
is encoded in base b; the canonical DFA will still have k states. It will, however, contain b
transitions from each state.)
The traditional answer, unfortunately, in general fails to produce a minimal DFA. This paper
addresses the considerably more difﬁcult question of ‘‘how many states does a minimal DFA that
recognizes the set of base-b numbers divisible by k have?’’ We denote this number by fbðkÞ and derive
a closed-form expression; in the proof, we also describe the states of the minimal DFA in more detail.
The function fbðkÞ may be computed by algorithmic means. The author used two
implementations of the Hopcroft minimization algorithm: an original Perl program and the
ARTICLE IN PRESS
E-mail address: kaif@uga.edu.
0022-0000/$ - see front matter r 2004 Elsevier Inc. All rights reserved.
doi:10.1016/j.jcss.2004.02.001
highly optimized AT&T FSM PackageTM: According to experts in the ﬁeld, no prior work
addresses the general case of this problem except through such computational alleys.
2. Interesting patterns
The function fbðkÞ exhibits very curious behavior. One interesting pattern considers fbðkÞ with b
ﬁxed and k ¼ x  yz for increasing values of z:
Example. Table of fbðkÞ for b ¼ 6 and k ¼ 2z: (That is, x ¼ 1; y ¼ 2; and z ranges from 0 to 10.)
The successive differences of f6ð2zÞ are the powers of 2 and 3; sorted in increasing order!
Example. Table of fbðkÞ for b ¼ 22  5 ¼ 20 and k ¼ 30  5z: (That is, x ¼ 30; y ¼ 5; and z ranges
from 0 to 6.)
Here, the successive differences of f20ð30  5zÞ come in increasing order from two sequences:
f2  4mg ¼ f2; 8; 32; 128;yg and f12  5mg ¼ f12; 60; 300;yg:
We observe that the function fbðkÞ manages to pick terms, in increasing order, from two
unrelated sequences! At ﬁrst, it is hard to imagine a formula that would produce such a function.
Investigating this bizarre behavior was the starting point for this study.
3. Main result
Theorem 1. Let lðx; yÞ ¼ x
gcdðx;yÞ: Then
fbðkÞ ¼ lðk; bNÞ þ
XN
a¼0
minflðba; kÞ; lðk; baÞ 	 lðk; baþ1Þg
¼ min
AX0
lðk; bAÞ þ
XA	1
a¼0
lðba; kÞ
( )
¼ lðk; bA0Þ þ
XA0	1
a¼0
lðba; kÞ;
where A0 is the smallest nonnegative integer a satisfying lðk; baÞ 	 lðk; baþ1Þolðba; kÞ:
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B. Alexeev / Journal of Computer and System Sciences 69 (2004) 235–243236
Remark. The function lðx; yÞ is not symmetric; indeed, lðx; yÞ ¼ lðy; xÞ if and only if x ¼ y:
We use the notation lðk; bNÞ to denote lðk; baÞ for sufﬁciently large a; similarly, the
inﬁnite sum can be truncated when lðk; baÞ 	 lðk; baþ1Þ ¼ 0: This equality certainly holds for
aXlog2 k:
Lemma 6 shows that the three expressions in the theorem are equivalent.
To understand the expressions of fbðkÞ in the theorem, we may draw a table listing a;
lðba; kÞ; lðk; baÞ; and lðk; baÞ 	 lðk; baþ1Þ: The ﬁrst and third expressions may be under-
stood fairly simply as written. However, the second expression is more difﬁcult; it states
that fbðkÞ is the minimal sum one can obtain by summing zero of more elements of the
form lðba; kÞ (as a ranges from 0 to A	 1) and then the following value of lðk; baÞ (that is,
a ¼A).
Example. b ¼ 6; k ¼ 16 ¼ 24: We can calculate fbðkÞ with any of the expressions above (for the
third, use A0 ¼ 2). The minimal terms of the ﬁrst expression appear underlined below;
simultaneously, the minimal ‘‘path’’ 8 ¼ 1þ 3þ 4 (in terms of the second formula above) is
indicated in boldface. Note that other paths such as 15 ¼ 1þ 3þ 9þ 2; 9 ¼ 1þ 8; and 16 ¼ 16
(the trivial path A ¼ 0) yield nonminimal sums.
4. Corollaries to the main result
Corollary 2. The following are upper bounds for fbðkÞ:
fbðkÞpk ¼ lðk; b0Þ
fbðkÞp1þ k
gcdðk; bÞ ¼ lðb
0; kÞ þ lðk; b1Þ
fbðkÞp1þ b
gcdðb; kÞ þ
k
gcdðk; b2Þ ¼ lðb
0; kÞ þ lðb1; kÞ þ lðk; b2Þ
Proof. These follow immediately from the second expression in Theorem 1. &
Corollary 3. The canonical DFA described in Section 1 is minimal if and only if gcdðk; bÞ ¼ 1 or
k ¼ 2:
Proof. The canonical DFA has k states and hence we must determine when fbðkÞ ¼ k:
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B. Alexeev / Journal of Computer and System Sciences 69 (2004) 235–243 237
If gcdðk; bÞ ¼ 1 or k ¼ 2; the ﬁrst expression of Theorem 1 immediately gives fbðkÞ ¼ k:
Otherwise, we have kgcdðk;bÞok 	 1; and by the previous corollary,
fbðkÞp1þ k
gcdðk; bÞok: &
Corollary 4. The successive differences of f6ð2zÞ are powers of 2 and 3, sorted in increasing order.
Proof. Manipulation of the result of the theorem yields
f6ð2zÞ ¼ lð2z; 6NÞ þ
XN
a¼0
minflð6a; 2zÞ; lð2z; 6aÞ 	 lð2z; 6aþ1Þg
¼ 1þ
XN
a¼0
minf3a  J2a	zn;I2z	a	1mg
¼ 1þ
Xz	1
a¼0
minf3a; 2z	a	1g:
It is not difﬁcult to see that as one increments z/z þ 1; a new term of the form minf3a; 2z	a	1g is
added, and the desired property holds. &
Remark. A similar approach may be applied to the general case of fbðx  yzÞ for increasing values
of z: In particular, we can easily prove the pattern we noticed in Section 2 for f20ð30  5zÞ:
Corollary 5. If b ¼ pn (p not necessarily prime, but see the remark) and k ¼ pm  x with gcdðx; pÞ ¼
1; then fbðkÞ ¼ x þ Jmnn:
Proof. We use the ﬁrst expression of the theorem:
fbðkÞ ¼ lðk; bNÞ þ
XN
a¼0
minflðba; kÞ; lðk; baÞ 	 lðk; baþ1Þg
¼ x þ
XN
a¼0
minfJpna	mn;Jpm	nan  x 	 Jpm	ðnþ1Þan  xg:
As long as naom; Jpna	mn ¼ 1 and Jpm	nan  x4Jpm	ðnþ1Þan  x: There are precisely Jm
n
n such a
(since 0paom
n
Þ; so we have
fbðkÞ ¼ x þ
XJmnn	1
a¼0
f1g þ
XN
a¼Jm
n
n
f0g
¼ x þ m
n
l m
;
as desired. &
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Remark. If p is prime, and thus b is a prime power, this corollary completely characterizes fbðkÞ;
as all k can be represented in the form pm  x with gcdðx; pÞ ¼ 1:
5. Proof of the main result
Lemma 6. The three expressions of Theorem 1 are equivalent.
Proof. By looking at the powers of a ﬁxed prime, we see that lðba; kÞ and gcdðk; baÞ are increasing
(not necessarily strictly) functions of a: It is also easy to show that gcdðk; baþ1Þ=gcdðk; baÞ is
decreasing, which immediately implies that lðk; baÞ 	 lðk; baþ1Þ is decreasing. Therefore, in
the sum
XN
a¼0
minflðba; kÞ; lðk; baÞ 	 lðk; baþ1Þg;
one takes A0 elements from the ﬁrst sequence flðba; kÞg and then inﬁnitely many from the
second sequence flðk; baÞ 	 lðk; baþ1Þg: Telescoping the latter, one gets the other two
expressions of the theorem. (The cut-off A0 is the smallest nonnegative integer a satisfying
lðk; baÞ 	 lðk; baþ1Þolðba; kÞ:) &
Proof of Theorem 1. Constructing a DFA directly, as in Section 1, is often difﬁcult because one
must describe the transitions between states in addition to the states themselves. We will use the
Myhill–Nerode Theorem and the accompanying theory of extension invariant equivalence
relations to work with the states of the automaton only.
Deﬁnition. Given a language (set of strings) L over an alphabet S; we deﬁne the extension
invariant equivalence relation BL associated with L as follows: strings x and y in S
 are
equivalent ðxBLyÞ if for any sufﬁx zAS
; xzAL if and only if yzAL: (As is customary, S
 denotes
the set of all ﬁnite strings over S: Later, we use Sþ ¼ S
\feg to denote the set of nonempty strings
over S:)
The Myhill–Nerode Theorem [1, Theorems 3.9–10] establishes that the minimal-state
automaton accepting L has, up to isomorphism, one state corresponding to each equivalence
class ofBL: Therefore, the minimal-state automaton has exactly the number of states as the index
ofBL: (In particular, a language L is regular if and only ifBL has ﬁnite index.) In addition, any
DFA recognizing L can be altered by identifying (‘‘gluing’’) some states together to obtain the
minimal-state automaton.
In this proof, we let S be the set of base-b digits and L the set of base-b numbers divisible by k:
In addition, since we work with only one language at a time, we may write xBy rather than
xBLy:
To begin, we will restate the problem equivalently in a way that will allow us to utilize modular
arithmetic. Because the canonical DFA accepting L has a state for each residue modulo k; the
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B. Alexeev / Journal of Computer and System Sciences 69 (2004) 235–243 239
Myhill–Nerode Theorem implies that the minimal-state DFA will contain states that correspond
to groups of residues modulo k: Therefore, in the pursuing analysis, rather than considering strings
of digits, we discuss residues; in a way, we are projecting S
 onto Zk (in the natural manner). For
example, L now becomes very simple: instead of containing all numbers divisible by k; it contains
the single residue 0 ðmod kÞ: To complete the reduction, we need only bother ourselves with one
further
Deﬁnition. Let rAZk be a residue modulo k and dAS a base-b digit. We deﬁne the concatenation
rd to be the residue b  r þ d ðmod kÞ: Similarly, if d ¼ dn	1?d1d0ASþ is a nonempty string
of digits, let the concatenation rd be what is obtained by successively concatenating individual
digits:
rd  b  ðb  ð?ðb  r þ dn	1Þ?Þ þ d1Þ þ d0Þ  bn  r þ dn	1?d1d0 ðmod kÞ;
where %d denotes d interpreted as an integer. Of course, if d ¼ e; the empty string, rd ¼ re  r:
Finally, extend BL onto Zk: residues x; yAZk are equivalent if for any string zAS
; xz 
0 ðmod kÞ if and only if yz  0 ðmod kÞ:
Now, suppose A is a nonnegative integer. We will describe
lðk; bAÞ þ
XA	1
a¼0
lðba; kÞ ð
Þ
pre-equivalence classes, each a group of residues, which will be a reﬁnement of the equivalence
classes of BL:
The pre-equivalence classes we deﬁne naturally present themselves in packages, a term we
borrow from computer programming to indicate collections of classes. Altogether, there are
Aþ 1 distinct packages, which we number 0;y;A; in addition, we will sometimes refer to
packageA as the distinctive package etcetera. These packages come in the sizes anticipated from
(
): if 0paoA; package a contains lðba; kÞ pre-equivalence classes, while package A contains
lðk; bAÞ pre-equivalence classes.
We now deﬁne the packages. Suppose 0paoA: Package a will consist of those residues r such
that there exists a string d of length a such that rd  0 and no smaller a works; furthermore, these
residues will be grouped according to their corresponding d’s. Mathematically, for each 0pcoba
such that gcdðba; kÞ j c; package a contains the pre-equivalence class fx j ba  x þ c  0g; except
those x that appeared in package a	 1 or earlier. (Note that the equation ba  x þ c  0 has a
solution x iff gcdðba; kÞ j c:) Because there are precisely ba=gcdðba; kÞ ¼ lðba; kÞ such c in the
desired range, these packages have the stated sizes. Before we proceed, note that the union of the
pre-equivalence classes in packages 0 through a consists of all residues x satisfying ba  x þ c  0
with 0pcoba; and no others.
Package etcetera consists of the leftovers; mathematically, it is similar, but there is no restriction
on c: for each 0pcok (only to avoid duplication modulo k), package A contains the pre-
equivalence class fx j bA  x þ c  0g; except those x that have appeared previously. Once again,
we have the necessary number of classes, since k=gcdðk; bAÞ ¼ lðk; bAÞ:
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B. Alexeev / Journal of Computer and System Sciences 69 (2004) 235–243240
Example. b ¼ 6; k ¼ 16 ¼ 24: the pre-equivalence classes for A ¼ 2: This value of A was
chosen so that these groups correspond to the states in the minimal DFA. Strikeouts
indicate that the given value of x satisﬁes ba  x þ c  0 but already appeared in a previous
package.
Recall once more from the statement of the theorem thatA0 is the smallest nonnegative integer
a satisfying lðk; baÞ 	 lðk; baþ1Þolðba; kÞ:
We make three separate claims:
1. for any A; our pre-equivalence classes coincide with the equivalence classes of BL with two
possible exceptions: some pre-equivalence classes may be empty and some pre-equivalence
classes in package etcetera may actually be equivalent (both of these would produce an
overcount);
2. for ApA0; all the pre-equivalence classes are nonempty; and
3. for AXA0; the classes of package etcetera are actually inequivalent.
It follows that for A ¼A0; our pre-equivalence classes are precisely the Myhill–Nerode
equivalence classes of BL:
We begin by afﬁrming (1): if two residues r and s are in the same class of package a; there exists
no string d of length less than a such that rd  0 or sd  0: In addition, r  ba  s  ba; so for any
string d of length at least a; we have rd  sd: Therefore, r and s are equivalent, and the pre-
equivalence classes are a reﬁnement of those of BL:
Moreover, if r and s are in different classes and at least one of r and s is not in package etcetera,
then rfs: Indeed, if r and s are in different packages, the result is obviously true. If r and s are in
different classes of the same package a with aoA; we can also conclude that rfs because r and s
satisfy ba  x þ c  0 for different values of c; therefore, there exists a string d (namely, the d such
that %d ¼ c) of length a such that rd  0 but sdc0:
Before continuing, we note the signiﬁcance of A0: If apA0; then
lðk; ba	1Þ 	 lðk; baÞXlðba	1; kÞ 3 k  gcdðk; b
a	1Þ
gcdðk; baÞ pk 	 b
a	1;
and if a4A0; then
lðk; ba	1Þ 	 lðk; baÞolðba	1; kÞ 3 k  gcdðk; b
a	1Þ
gcdðk; baÞ 4k 	 b
a	1:
Equipped, we proceed in order to (2). Suppose ApA0; then, we claim that for any ﬁxed
0oapA and c such that gcdðk; baÞ j c; there exists an x satisfying
ba  x þ c  0 ðwÞ
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B. Alexeev / Journal of Computer and System Sciences 69 (2004) 235–243 241
which does not satisfy ba	1  x þ c0  0 with 0pc0oba	1: Indeed, consider all x satisfying ðwÞ and
note that these x are spaced apart equally with k
gcdðk;baÞ separation between consecutive solutions.
Multiplying these x by ba	1 yields (possibly duplicate) residues ba	1  x spaced k  gcdðk;ba	1Þgcdðk;baÞ apart.
But, because apApA0;
k  gcdðk; b
a	1Þ
gcdðk; baÞ pk 	 b
a	1;
whence there exists an x satisfying ðwÞ such that ðba	1  xÞmod k is in between 1 and k 	 ba	1; and
such an x cannot satisfy ba	1  x þ c0  0 with 0pc0oba	1: Therefore, all of the classes of
packages 0 through A are nonempty.
We ﬁnish with (3). Suppose AXA0; it sufﬁces to show that if rBs and a is the minimal
a such that ba  r  ba  s; then apA: Assume the contrary: a4A: Then, r and s are both
solutions of ðwÞ for a ﬁxed c: To derive a contradiction, we again focus on the spacing of
solutions of ðwÞ: So, consider all x satisfying ðwÞ; they are spaced k
gcdðk;baÞ apart. As before,
the residues ba	1  x for x satisfying ðwÞ are spaced k  gcdðk;ba	1Þ
gcdðk;baÞ apart. However, because
a4AXA0;
k  gcdðk; b
a	1Þ
gcdðk; baÞ 4k 	 b
a	1
and thus there is not enough room for two distinct ðba	1  xÞmod k in between 1 and k 	 ba	1:
Therefore, either ba	1  r  ba	1  s or one of r and s satisﬁes ba	1  x þ c0  0 with 0pc0oba	1: The
former contradicts the minimality of a; and the second is impossible as well: without loss of
generality, r satisﬁes such an equation. But then, there exists a string d of length a	 1 such that
rd  0: Because rBs; it follows that rd  sd  0 for a string of length a	 1; once again
contradicting the minimality of a! We have reached a contradiction in all cases, therefore our
assumption was false and apA: Therefore, any two residues are ‘‘distinguished’’ at or before
a ¼A; and it follows that any r and s in the package etcetera are equivalent if and only if they are
in the same pre-equivalence class.
At last, we are done. &
Acknowledgments
The author would like to thank Professors Jason Eisner of Johns Hopkins University, Dana
Scott and Klaus Sutner of Carnegie Mellon University, and Michael Sipser of MIT for answering
queries about the problem and related issues and Professor Rodney Canﬁeld of the University of
Georgia for many useful conversations and attention to this work.
For large-scale computations, when speed was crucial, AT&T Research’s FSM Package was
used to compute fbðkÞ; to complement the author’s own programs.
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References
[1] J. Hopcroft, J. Ullman, Introduction to Automata Theory, Languages, and Computation, Addison-Wesley,
Reading, MA, 1979.
[2] M. Sipser, Introduction to the Theory of Computation, PWS Publishing, 1997.
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Minimal DFAs for Testing Divisibility

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