How Many Different DFA's Can Be Designed Part-2
Duration: 5 min
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The video lecture, presented by Sanchit Jain from Knowledge Gate, addresses a fundamental problem in automata theory: calculating the total number of possible Deterministic Finite Automata (DFAs) that can be designed under specific constraints. The instructor starts by presenting a problem statement on the screen: "Q how many DFA can be designed, with a fixed initial state, over an alphabet $\Sigma = \{a, b\}$, and the number of states be two." However, he immediately modifies the problem, crossing out "two" and writing a superscript "3", indicating that the example will use three states ($n=3$) for clarity. He begins the derivation by drawing a transition table. The rows represent the states ($q_0, q_1, q_2$) and the columns represent the input symbols ($a, b$). He explains that for every state and every input symbol, there must be a defined transition to a next state. Since there are 3 states available, each cell in the transition table has 3 possible choices. With 3 states and 2 input symbols, there are a total of $3 imes 2 = 6$ transitions. Therefore, the total number of possible transition functions is $3^6$. Next, the instructor considers the final states. He explains that for a set of 3 states, any subset of these states can be designated as final states. The number of subsets of a set with 3 elements is $2^3$, which equals 8. He writes $8 imes 3^6$ on the board, representing the total number of DFAs for this specific case. This combines the choices for transitions and the choices for final states. The problem specifies a "fixed initial state," which simplifies the calculation by removing the need to choose an initial state from the $n$ available states. In the second half of the video, the instructor generalizes this specific example to derive a universal formula. He introduces variables $n$ for the number of states ($|Q| = n$) and $m$ for the size of the input alphabet ($|\Sigma| = m$). He reiterates that the number of transitions is $n imes m$, and since each transition can go to any of the $n$ states, the number of transition functions is $n^{n imes m}$. Similarly, the number of ways to choose final states remains $2^n$ because each of the $n$ states has 2 possibilities (final or non-final). The video concludes with a slide summarizing the result: "If no of states is $|Q| = n$, and no of input alphabet $|\Sigma| = m$, Then no of states will be $n^{n imes m} * 2^n$". This formula provides a direct method to calculate the number of possible DFAs for any given number of states and alphabet size, assuming a fixed initial state. The instructor emphasizes that this formula is a standard result used in competitive exams and theoretical computer science to count the complexity of DFA design. The slide also contains a typo where it says "no of states will be" instead of "no of DFAs will be", but the mathematical expression is correct.
Chapters
0:00 – 2:00 00:00-02:00
The instructor introduces the problem statement on the screen: "Q how many DFA can be designed, with a fixed initial state, over an alphabet $\Sigma = \{a, b\}$, and the number of states be two." He immediately corrects the number of states from "two" to "3" by writing a superscript. He then draws three columns of circles labeled 0, 1, and 2 to represent the states. He proceeds to draw a transition table with rows labeled $q_0, q_1, q_2$ and columns labeled $a, b$. He explains that for each of the 6 cells in the table (representing transitions), there are 3 possible next states. He writes the number 3 in each cell to illustrate this. He calculates the total number of transition functions as $3^6$. He then explains that for 3 states, there are $2^3 = 8$ ways to choose the final states. He writes the final calculation $8 imes 3^6$ on the board to represent the total number of DFAs for this specific case.
2:00 – 4:40 02:00-04:40
The instructor generalizes the specific example to derive a universal formula. He introduces variables $n$ for the number of states ($|Q| = n$) and $m$ for the size of the input alphabet ($|\Sigma| = m$). He explains that the number of transitions is $n imes m$, and since each transition can go to any of the $n$ states, the number of transition functions is $n^{n imes m}$. Similarly, the number of ways to choose final states remains $2^n$ because each of the $n$ states has 2 possibilities (final or non-final). He writes the final formula $n^{n imes m} * 2^n$ on the board. The video concludes with a slide summarizing the result with the text: "If no of states is $|Q| = n$, and no of input alphabet $|\Sigma| = m$, Then no of states will be $n^{n imes m} * 2^n$". The instructor notes that this formula is a standard result used in competitive exams and theoretical computer science to count the complexity of DFA design. The slide also contains a typo where it says "no of states will be" instead of "no of DFAs will be", but the mathematical expression is correct.
The video provides a step-by-step derivation of the formula for the number of possible DFAs. It starts with a concrete example using 3 states and an alphabet of size 2, calculating the number of transition functions ($3^6$) and final state combinations ($2^3$). It then generalizes this to $n$ states and an alphabet of size $m$, resulting in the formula $n^{n imes m} * 2^n$. The lecture emphasizes the combinatorial logic behind choosing transitions and final states, assuming a fixed initial state.