Write $\bar n$ for the decimal expansion of $n$ (with no leading 0). Let $a$ and $b$ be integers, with $a > 0$. Consider the language of the decimal expansions of the multiples of $a$ plus a constant:

$$M = \{ \overline{a\,x+b} \mid x\in\mathbb{N} \}$$

Is $M$ regular? context-free?

(Contrast with Language of the graph of an affine function)

I think this would make a good homework question, so answers that start with a hint or two and explain not just how to solve the question but also how to decide what techniques to use would be appreciated.


5 Answers 5


Very simple: Suppose numbers are written in decimal (other bases are handled by a trivial modification). Build a DFA, with $a$ states 0, 1, ..., $a - 1$. Start state is 0, and from state $q$ on input the digit $d$ go to state $(10 q + d) \bmod a$. Accepting state is $b \bmod a$ (might need a tweak if $b > a$).

  • 1
    $\begingroup$ Very nice -- much better than mine! $\endgroup$ Jan 24, 2013 at 3:16

It is regular. Let's first work in binary, which will generalize to any base > 1. Let $M_{a,b}$ be the language in question. For a = 1, b = 0 we get

$M_{1,0} = \{1, 10, 11, 100, 101, ...\}$

which is all strings over $\{0,1\}$ without leading zeroes, which is regular (construct a regular expression for it).

Now for any $a$, with b still 0 we get $M_{a,0}$ from $M_{1,0}$ by multiplying numerically by a, that is, performing the transformation $\bar x \rightarrow \overline {ax}$ on each string of $M_{a,0}$. That can be done bitwise by a sequence of shifts and additions of $x$ that depend on the bits of fixed string $\bar a$. The two transformations we need are:

$\bar x \rightarrow \overline {2x}$ which is $\bar x \rightarrow \bar x0$


$\bar x \rightarrow \overline {2x+x}$

Concatenating a 0 on the right clearly preserves regularity. We thus have only to prove that the second operation preserves regularity. The way to do that is with a finite-state transducer working on $\bar x$ from right to left. That's the hardest step. I suggest doing it with a pseudo-code program and some finite auxilliary memory (like some bit variables) rather than using states. As long as the memory is of a fixed size over all inputs and you scan strictly right to left, it's a finite state transduction and preserves regularity.

Finally, to get $M_{a,b}$ from $M_{a,0}$ we need to numerically add $b$ to each string, but that's done with a similar transducer $T_b$ which depends on the fixed number b.

To do the same in any base, show additionally how to perform multiplication by a single digit $d$ in that base, using a transducer $S_d$ that depends on d. With that, we can multiply by multi-digit numbers and still remain in the regular languages. Or, we can finesse this by saying that multiplication by $d$ is just repeated addition.

You wanted only hints. Each of those steps depends on a fairly complex theorem/technique, so proving those to get a complete proof will be the extra work.

  • $\begingroup$ Your FA does not get $\bar{x}$ as an input, so I don't see how what you write shows that the language at hand is regular. Note that not every program that uses finite memory corresponds to FA: it is important that can go only once and from left to right over the input, considering every input symbol exactly once. $\endgroup$
    – Raphael
    Mar 22, 2012 at 7:28
  • $\begingroup$ @Raphael You can go from right to left if you like, what matters is being consistent. I can't find a flaw in David's proof sketch; invoking transducers is a bit less elementary than I envisioned, but it looks valid. $\endgroup$ Mar 22, 2012 at 9:36
  • $\begingroup$ @Gilles: First of all, he does not explain how to interleave multiplication by $a$ and adding of $b$ to the result in one pass; he even reduces multiplication by $a$ to "a sequence of shifts and additions of $x$". Every single shift and addition is fine, but how to do the sequence? Secondly, and more importantly, he shows how to build a transducer that computes $\bar{x} \mapsto \overline{ax + b}$; that does not immediately give you a FA that accepts $M$. For instance, multiplying numbers is easy but factoring is (allegedly) not. So you need at least an additional argument. $\endgroup$
    – Raphael
    Mar 22, 2012 at 10:57
  • $\begingroup$ I'm not constructing an FSA. I'm starting with a set easily shown to be regular ($M_{1,0}$) and then transforming all strings in it with a finite series of operations, each of which preserves regularity. This requires a number of passes (transducers). But that's OK, as the sequence of transducers and structure of each is fixed in advance based only on $a$ and $b$. Each pass (transducer) preserves regularity, so there's no need to interleave them in one pass. Yes, not "elementary". But constructing an FSA in one go, with one pass, would be awfully complex. $\endgroup$ Mar 22, 2012 at 12:35
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    $\begingroup$ @Raphael -- yes, that's very powerful. In fact, many non-regular families are also closed under finite-state transducers. And, you can use transducers as reduction mechanisms, getting a whole theory of "structural" complexity of non-regular languages. $\endgroup$ Mar 22, 2012 at 23:17

Hint #1: first solve the more popular problem "write a automaton that recognizes the decimal/binary representations of numbers divisible by 3" when the least signigicant bit appears first.

Intermediate question: prove that $\{ \overline{a\,x+b}\mid a\,x+b≥0 \quad x\in\mathbb{Z}\}$ is regular.

Hint #2: The graph of the function $(n \mapsto 10n+d)$ "modulo $a$" is finite. You can compute it for each $d$ in $\{0, \dots, 9\}$ which is both the set of digits and the language of your automaton.

Hint #3: now, replace $x∈\mathbb Z$ with $x∈\mathbb N$. What does this change? Try to use the fact that regular languages are stable by intersection instead of building an ad-hoc automaton.

Hint #4: now assume that $a$ is a prime number (so that $\mathbb Z/a\mathbb Z$ is a field) and that we are still in the case where $x∈\mathbb Z$. We now use a representation where the first bit is the most significant bit. Can you build the automaton the same way?

Hint #5: you don't always have to build an automaton to prove a language is regular. How can you use previous results to prove that $M$ is regular? (with the most significant bit first)

  • $\begingroup$ Feel free to comment if you feel that this is not appropriate. $\endgroup$
    – jmad
    Mar 22, 2012 at 2:35
  • $\begingroup$ Hint #1 is a large step. In hint #4, it's important to realize that $a\in\{2,5\}$ and $a\bot10$ are different. Going via $\mathbb{Z}$ feels like a detour, you have to manage the sign character: why not stay in $\mathbb{N}$? $\endgroup$ Mar 22, 2012 at 9:41
  • $\begingroup$ @Gilles: I wanted to say $\overline{ax+b}$ when $ax+b≥0$ and $x∈\mathbb Z$, because $3x+1234$ is tedious to recognize. $\endgroup$
    – jmad
    Mar 22, 2012 at 10:20
  • $\begingroup$ @Gilles: I think Hint #1 is too cool to be spoiled. You are probably right about Hint #4. $\endgroup$
    – jmad
    Mar 22, 2012 at 10:26

I am following the idea of @vonbrand:

Hint 1:

Solve first for $b=0$. You might use the Myhill-Nerode Theorem.

We define the following relation $\bar x\approx \bar y :\iff x\equiv y \mod a$. This is obviously an equivalence relation. Furthermore it is right-congruent, since if we append a digit $d$ we get $\bar x \approx \bar y \iff 10x+d \equiv 10y+d \mod a \iff \bar x d \approx \bar y d$. Finally it saturates $L$, since $L$ is the equivalence class $[0]$. Since $\approx$ has a finite number of classes we have by the Myhill-Nerode Theorem that it is regular. The associated FSA is minimal and has $a$ states.

Hint 2:

Solve the general case, reuse the automaton induced by the $b=0$ case.

We know that the language is regular for $b=0$. Hence simply take the $a$ state FSA $M$ for the $b=0$ language. Now we construct a FSA for $L$. Assume that $b$ has $k$ digits. Then the FSA branches like a 10-ary tree of depth $k$ and contains all paths to numbers with $k$ digits. All states that can be reached with a number not in the form $ax+b$ are rejecting otherwise accepting. Finally we connect the tree-part of the FSA with the automaton $M$, according to the remainder by the division by $a$.

  • $\begingroup$ I understand the technique, but not the details. Isn't Hint 1 also adressing the $a=1$ case? Moreover, for mod 10 I would expect 10 states (not $a$)? $\endgroup$ Jan 20, 2013 at 13:54

The language is regular.

Hint: cast out nines

Proof idea

For $a=9$ and $b < 9$,

build an automaton with $9$ states labeled $0$ through $8$. $0$ is the initial state, and the one final state is $b$. From state $s$, on digit $d$, transition to state $(s + d) \;\mathrm{mod}\; 9$.

To handle other values of $a$ that are coprime with $10$,

group digits in packets to find some $k$ such that $a$ divides $10^k-1$ (e.g. take $k=3$ if $a=37$ because $999 = 27 \times 37$).

To handle values of $a$ whose only prime factors are $2$ and $5$,

note that it's all about a finite number of digits at the end.

To generalize to all values of $a$ and $b$,

use the fact that union and intersection of regular languages are regular, that finite languages are regular, and that the multiples of $a_1 \cdot a_2$ are exactly the multiples of both when $a_1$ and $a_2$ are coprime.

Note that we use whichever technique is convenient; the three main elementary techniques (regular expressions, finite automata, set-theoretic properties) are all represented in this proof.

Detailed proof

Let $a = 2^p 5^q a'$ with $a'$ coprime with $10$. Let $M' = \{\overline{a'\,x+b} \mid x\in\mathbb{Z} \wedge a'\,x+b \ge 0\}$ and $M'' = \{\overline{2^p 5^q\,x+b} \mid x\in\mathbb{Z} \wedge 2^p 5^q\,x+b \ge 0\}$. By elementary arithmetic, the numbers equal to $b$ modulo $a$ are exactly the numbers equal to $b$ modulo $a'$ and to $b$ modulo $2^p5^q$, so $M \cap \{\overline{x} \mid x \ge b\} = M' \cap M'' \cap \{\overline{x} \mid x \ge b\}$. Since the intersection of regular languages is regular, and $\{\overline{x} \mid x \ge b\}$ is regular because it is the complement of a finite (hence regular) language, if $M'$ and $M''$ are also regular, then $M \cap \{\overline{x} \mid x \ge b\}$ is regular; and $M$ is therefore regular since it is the union of that language with a finite set. So to conclude the proof it suffices to prove that $M'$ and $M''$ are regular.

Let us start with $M''$, i.e. numbers modulo $2^p 5^q$. The integers whose decimal expansion is in $M''$ are characterized by their last $\mathrm{max}(p,q)$ digits, since changing digits further left means adding a multiple of $10^{\mathrm{max}(p,q)}$ which is a multiple of $2^p 5^q$. Hence $0^* M'' = \aleph^* F$ where $\aleph$ is the alphabet of all digits and $F$ is a finite set of words of length $\mathrm{max}(p,q)$, and $M'' = (\aleph^* F) \cap ((\aleph \setminus \{0\}) \aleph^*)$ is a regular language.

We now turn to $M'$, i.e. numbers modulo $a'$ where $a'$ is coprime with $10$. If $a' = 1$ then $M'$ is the set of decimal expansions of all naturals, i.e. $M' = \{0\} \cup ((\aleph \setminus \{0\}) \aleph^*)$, which is a regular language. We now assume $a' > 1$. Let $k = a'-1$. By Fermat's little theorem, $10^{a'-1} \equiv 1 \mod a'$, which is to say that $a'$ divides $10^k-1$. We build a deterministic finite automaton that will recognize $0^* M'$ as follows:

  • The states are $[0,k-1] \times [0,10^k-2]$. The first part represents a digit position and the second part represents a number modulo $10^k-1$.
  • The initial state is $(0,0)$.
  • There is a transition labeled $d$ from $(i,u)$ to $(j,v)$ iff $v \equiv d 10^i + u \mod 10^k-1$ and $j \equiv i + 1 \mod k$.
  • A state $(i,u)$ is final iff $u \equiv b \mod a'$ (note that $a'$ divides $10^k-1$).

The state $(i,u)$ reached from a word $\overline{x}$ satisfies $i \equiv |\overline{x}| \mod k$ and $u \equiv x \mod 10^k-1$. This can be proved by induction over the word, following the transitions on the automaton; the transitions are calculated for this, using the fact that $10^k \equiv 1 \mod 10^k-1$. Thus the automaton recognizes the decimal expansions (allowing initial zeroes) of the numbers of the form $u + y 10^k$ with $u \equiv b \mod a'$; since $10^k \equiv 1 \mod a'$, the automaton recognizes the decimal expansions of the numbers equal to $b$ modulo $a'$ allowing initial zeroes, which is $0^* M'$. This language is thus proved regular. Finally, $M' = (0^* M') \cap ((\aleph \setminus \{0\}) \aleph^*)$ is a regular language.

To generalize to bases other than $10$, replace $2$ and $5$ above by all the prime factors of the base.

Formal proof

Left as an exercise for the reader, in your favorite theorem prover.


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