DETERMINISTIC FINITE AUTOMATA (DFA) EXAMPLE - 1 (STRINGS STARTS WITH) IN AUTOMATA THEORY || TOC
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- Опубликовано: 19 сен 2024
- DETERMINISTIC FINITE AUTOMATA (DFA) EXAMPLE - 1
Design DFA which accepts all strings over given alphabet which starts with given substring.
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Instagram : / sundeepsaradhikanthety
Incredible explanation, very minimalistic yet very robust method and very clean and clear drawing and writing! Definitely the best teacher of comp theory in the world! Thanks Sundeep!
Good explanation sir.....Very helpful for Students who r taking Semester Exams Now.......Very Simple and Easy Explanation.......ధన్యవాదాలు సార్
its amazing sir i enjoying the automata. great work sir ,i have started watching video today and lets see when i fill left comment in the last video
Man, you’re the goat
Fantastic explanation ❤
Very clearly explained sir!
Excellent Explanation.🙏
Sir as we Known DFA has only 1 transaction state, but u make 2 transactions from ( q1, 0 ) -> q1 & dead state. I think that's wrong. @14:00
how?
Sir erase karna bhul gaye
Great Explanation sir , Every topic became easy after watching each video..
its interesting tutorial thanks sir
Awesome explain sir ❤ love you from Bihar ❤
Nice video I learned more knowledge 😊
Thank you sir
but the diagram was a NFA it has multiple path for other state . can you please explain
Thank you
nice explaination sir😊
Good explanation sir
Sir why you gave t
Dead state in 2nd example
Can't we give transition to q0 or q2???
Sir.... shouldn't we use 1 on q1 instead of zero and dead state we will get 01 onle there @14:21
god gifted
sir is it possible for dfa that finall state may have 2 two inputs for same state ?
one doubt sir, shouldn't the number of states = number of input symbols + one which is no of states should be 3 ??? but ur solving with 2 states
???
amazing
Thank you it helps a lot
Sir can u explain the problem The set of all strings such that each block of five consecutive symbols contains atleast two 0's where alphabet is 0,1
Thank you sir🙌
Is there any other answer for the string starting with ab??
🎯 Key Takeaways for quick navigation:
00:32 🚀 *Constructing DFA for "Starts With 0" Problem*
- The problem is to construct a DFA that accepts strings starting with the digit '0'.
- The language L is defined as strings over the alphabet {0, 1} that start with '0'.
- The minimum string length is 1, requiring two states for the DFA, with transitions for both '0' and '1'. A dead state is added to handle strings not starting with '0'.
11:08 🧩 *Constructing DFA for "Starts With 01" Problem*
- The problem is to construct a DFA that accepts strings starting with the substring '01'.
- The language L is defined as strings over the alphabet {0, 1} that start with '01'.
- The minimum string length is 2, requiring three states for the DFA. A dead state is introduced to handle strings not starting with '01', ensuring all transitions are defined for every state and input symbol.
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Sir why fa has only one initial state????
Sir in the 2nd example q1 state has 2 0's as input then this is not dfa right?!
I think sir forgot to erase the self loop of 0 on q1
Very difficult to understand 🙂
Same bro
Is this DFA applicable in the industry??? Why we are learning this it's so hard to study no doubt your explain is so good but is it applicable anywhere !!! NO !!!
Tomorrow Internal still now I can't understand 🥴🙄
14:16 not self loop in q1 bcz string start with 01. -q0--0-->q1--1-->q2
@14:21 what if we pass q1 to q0 when we get input 0 in q1?
It accepts words starting 00 also, so to avoid that we move it to the dead state if we come across those words.
@@krishnachaitanya8353 But then it wouldn't be a DFA right ? I am new to this concepts but from q1 -> 0 -> {Dead State, q1} occurs ?
Why can't we pass q1 to q0 with input 0?
In NFA
If it goes from q1 to q0 with the input of 0,it can accept 00 too. But it is wrong. Bcoz there is a string which start with 01 only.
Good explanation sir