# What does it mean to prove that a set of binary integers is regular?

I'm not exactly sure what this question is asking me to do:

Show that the set of binary integers (given as strings over $\{0, 1\}$) that are divisible by $3$ is regular, by giving a DFA that recognizes it. Leading 0s are allowed. The empty string should be accepted. Briefly explain your answer.

Is this asking me to construct a DFA where the set of binary integers add up to be divisible or the number of elements is divisible?

I understand that by constructing the DFA I prove that it is regular.

• The question asks you to construct a DFA that accepts a binary string iff this string represents a binary number that is divisible by 3. For example, 0100 should not be accepted, as it represents the number 4, which is not divisible by 3. The string 1001 should be accepted, as it represents 9. Sep 29, 2014 at 15:21

The question asks you to construct a DFA for the language

$$L = \{x \in \{0,1\}^* : x \text{ is the binary representation of an integer that is divisible by three}\}.$$

In other words, the DFA should accept a string if and only if this string is the binary representation of a number that is divisible by three.

• Writing "$x$ is divisible by 3" is unclear when $x$ is a string. It would be more accurate to define $val(x)=\sum_{i=1}^n 2^{n-i}x_i$ where $x_i$ is the i-th bit of $x$. Sep 29, 2014 at 16:33

Hint: Make three states, $q_0, q_1, q_2$, representing the fact that in state $q_i$ you have seen a string of 0s and 1s that, when interpreted as a binary number, represents a number with remainder of $k$ when divided by 3. For example, having read $\mathtt{00101}$, the FA would be in state $q_2$, since the binary string $\mathtt{00101}$ corresponds to the decimal value 5, which has a remainder of 2 when divided by 3.

Now fill in the transitions: If, say we're in state $q_2$ having read a string that corresponds to a number $n$, what state would the FA be on reading a $\mathtt{0}$? What happens when you add a zero to the right of the representation of $n$, going, say from $\mathtt{00101}$ to $\mathtt{001010}$? Use this idea to fill in all the transitions and then decide what the final state(s) should be.

• Ah! My personal favourite automaton. Sep 29, 2014 at 20:57
• @HendrikJan. Oops. I didn't think of looking for an old answer first. Sorry for stepping on your toes Sep 29, 2014 at 21:04
• No, my toes were certainly not hurt. Sorry if I made that impression. I hesitated to give the link, as you are kindly guiding towards a solution, where I blurt it out. Couldn't resist though. Sep 29, 2014 at 21:30

The question want us to design a Deterministic Finite Automaton (DFA) that accepts Binary Representation of Integers that is divisible by 3.

$$\mathcal{L}=\{x\in\{0,1\}^∗:x$$ is the binary representation of an integer that is divisible by three$$\}$$

A word $$w=a_{1}a_{2}...a_{n}\in \Sigma ^{*}$$ is an accepting word for the automaton if $$\overline {\delta }(q_{0},w)\in F$$, that is, if after consuming the whole string $$w$$, the machine is in an accept state.

Approach : We need to divide the Binary Representation of Integer by 3, and note the remainder. If after consuming/scanning [From Left to Right] the entire string, remainder is Zero, then we should end up in one of the Final States, else we should be in Non-Final States.

Now, DFA is defined by Quintuple$$/5$$-Tuple $$(Q,F,q_0,\Sigma,\delta)$$. We will obtain these five components step-by-step.

$$Q$$ : Finite Set of States
We need to note remainder. On dividing any integer by $$3$$, we can get remainder as $$0,1$$ or $$2$$. Hence, we will have Three States $$Z, V$$ and $$T$$ respectively. $$Q=\{Z,V,T\}$$ If after scanning certain part of Binary String, we are in state $$T$$, this means that integer defined from Left to this part will give remainder $$T$$wo when divided by $$3$$. Similarly, $$V$$ for remainder $$1$$, and $$Z$$ for remainder $$0$$.
$$\hspace{6cm}$$ Now, we can write these three states (by Euclidean Division Algorithm) as

$$Z=3m\qquad V=3m+1\qquad T=3m+2\qquad$$ where $$m \in \mathbb{Z}$$

$$F :$$ a set of accept states, $$F\subseteq Q$$
Now we want strings which are divisible by $$3$$, or which gives remainder $$0$$ when divided by $$3$$, or which after complete scanning should end up in state $$Z$$, and gets accepted. Hence, $$F=\{Z\}$$
$$\hspace{8cm}$$ $$q_0$$ : an initial/start state $$q_0\in Q$$
An Empty String $$(\varepsilon)$$ directly gets into $$q_0$$. And $$\varepsilon$$ should get accepted. Thus $$q_0$$ be one of the Final State. And since we have only one Final State. Hence,

$$q_0=Z$$ $$\hspace{6cm}$$ $$\Sigma :$$ Alphabet (a finite set of input symbols)
Since we are scanning/reading a Binary String. Hence,

$$\Sigma=\{0,1\}$$

$$\delta :$$ Transition Function $$(δ : Q × Σ → Q)$$
Now this $$\delta$$ tells us that if we are in state $$a\in Q$$ and next input to be scanned is $$b\in \Sigma$$, then at which state $$c\in Q$$ should we go.

In context of this problem, if the string upto this point gives remainder $$1/V$$ when divided by $$3$$, and if we append $$1$$ to string, then what remainder will resultant string give.

Now, this can be mathematically analyzed by observing how magnitude of a binary string changes on appending 0 and 1.

In Decimal (Base-$$10$$), if we add/append $$0$$, then magnitude gets multiplied by

$$10$$
Example : $$64$$, on appending $$0$$ it becomes $$640$$

Also, if we append $$6$$ to decimal, then

Magnitude gets multiplied by $$10$$, and then we add $$6$$ to multiplied magnitude.

In Binary (Base-$$2$$), if we add/append $$0$$, then magnitude gets multiplied by

$$2$$ (The Positional Weight of each Bit get multiplied by $$2$$)
Example : $$(1010)_2$$ [which is $$(10)_{10}$$], on appending $$0$$ it becomes $$(10100)_2$$ [which is $$(20)_{10}$$]

Similarly, In Binary, if we append $$1$$, then

Magnitude gets multiplied by $$2$$, and then we add $$1$$.
Example : $$(10)_2$$ [which is $$(2)_{10}$$], on appending $$1$$ it becomes $$(101)_2$$ [which is $$(5)_{10}$$]

Thus, we can say that for Binary String $$x$$,

• $$x0=2|x|$$
• $$x1=2|x|+1$$

We will use these relation to analyze three States

• Any string in $$Z$$ can be written as $$3m$$

• On $$0$$, it becomes $$2(3m)$$, which is $$3(2m)$$, nothing but state $$Z$$.
• On $$1$$, it becomes $$2(3m)+1$$, which is $$3(2m)+1$$, that is $$V$$.
[This can be read as if a Binary String is presently divisible by $$3$$, and we append $$1$$, then resultant string will give remainder as $$1$$]
$$\hspace{3cm}$$ • Any string in $$V$$ can be written as $$3m+1$$

• On $$0$$, it becomes $$2(3m+1)=6m+2$$, which is $$3(2m)+2$$, state $$T$$.
• On $$1$$, it becomes $$2(3m+1)+1=6m+3$$, which is $$3(2m+1)$$, state $$Z$$.
[If $$m\in \mathbb{Z}$$, then $$(2m+1)\in \mathbb{Z}$$]
$$\hspace{3cm}$$ • Any string in $$T$$ can be written as $$3m+2$$

• On $$0$$, it becomes $$2(3m+2)=6m+4$$, which is $$3(2m+1)+1$$, state $$V$$.
• On $$1$$, it becomes $$2(3m+2)+1=6m+5$$, which is $$3(2m+1)+2$$, state $$T$$.
$$\hspace{3cm}$$ Hence, the final DFA combining all Five Components is Since, we have constructed DFA. Hence, the language $$\mathcal{L}$$ is regular.