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Can anyone please help me try to understand what will the state diagram look like for this question:

Let $L = \{x_0y_0z_0x_1y_1z_1\ldots x_{n-1}y_{n-1}z_{n-1} \mid \exists n,x,y,z \, x+y=z \land x=\sum_{j=0}^{n-1} x_j2^j \land y=\sum_{j=0}^{n-1} y_j2^j \land z=\sum_{j=0}^{n-1} z_j2^j \land \forall i \, x_i,y_i,z_i \in \{0,1\}\}.$ Show that $L$ is regular. It is sufficient to draw a state diagram.

Try to design your state diagram in a systematic way. Arrange your states in 3 columns (with 1 exception), such that before reading the $j$th bit you are in column $j$. If at this time, the DFA has to remember 2 bits, you could draw the states in lexicographical order of these 2 bits.

Any help is much appreciated. I've spent more than an hour just trying to understand the question and still could not figure it out.

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1 Answers1

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The alphabet for this language is $\Sigma=\{0,1\}$. We also have that $x_i,y_i,z_i\in\Sigma$. So, notice that $$\sum_{j=0}^{n-1}x_j2^j$$ is simply the decimal equivalent if we assume that the symbols of $x$ are expressed as a number in binary. Similar can be said for $y$ and $z$.

Using this, one can see that the language is effectively the those strings such that if you take every third symbol and form $x,y,z$, the numerical equivalent adds up.

For this to be true, for every $x_i,y_i$ and carry from the past additions, the corresponding $z_i$ should be the correct symbol. If it were not so, then irrespective of the future symbols, we cannot have $x+y=z$, and the string can never be accepted.

So, at each iteration, the DFA needs to remember the current values of $x_i,y_i$ and the current carry. If the corresponding $z_i$ is correct and the carry is $0$, it can return to the start state and continue, as this is not going to affect future considerations in any way. This is a final state as at any point when the DFA is in this state, the current sets of $x_i$s, $y_i$s, and $z_i$s are such that the string would be accepted.

If the value of $z_i$ is correct but the carry is not $0$, then the DFA goes to a different state corresponding to carry being $1$. This is not a final state as if the carry was $1$ at the end, it would imply that $x+y\ne z$ as $x,y,z$ must have the same number of symbols.

If the value of $z_i$ is incorrect, then the symbol at that "bit" can never be corrected in the future, and the DFA can be sent to a sink state where the string is rejected irrespective of the remaining symbols.

Using the above, the DFA I have drawn (using JFLAP) is as follows:

DFA

The first column is when $x_i,y_i,z_i$ have been received. $q_0$ corresponds to a carry of $0$ and is that start and final state. $q_7$ corresponds to a carry of $1$. The second column is the set of states when $x_i$ has been received, and the third is when $y_i$ has been received. Then, when $z_i$ is received, if it matches $z_i=x_i+y_i+\text{carry}$, the DFA goes to state $q_0$ or $q_7$ depending on the carry. Else, it goes to $q_{14}$, which is a sink.

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