Deterministic Finite Automata (Example 1)

TOC: An Example of DFA which accepts all strings that starts with '0'. This lecture shows how to construct a DFA that accepts all binary strings that start with '0' and rejects all that does not start with '0'.
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why we should take only 0 and 1 as inputs?

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What goes to state C, stays in state C.

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1

[@simply_me6308]

Any string starts with '1'

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I had FAFL when I studied my engineering. For some reason, I couldnt master the subject to the extent what i intended. 15 years gone by and I decided to refresh the basic of computer science starting with FAFL and Discrete Math.
Loving watching this series as it is bringing back a lot of memories along with better understanding of basic CS.
Are there any notes that I can download [as in just the images that you have drawn] in a PDF format perhaps ?

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Idu the final state!!

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Thanks a lot from Brazil.
I had difficulties when I studied this subject at college but after watching your classes I am a little more confident.

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Amazing explanation

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Everyone Lecture You Explained Well.......Thanks a Lot Sir

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Thanks!

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Very good explanation ...such an ideal tutorial ...

[@jeremiahmort4294]

Trying to build a DFA was kicking my butt, but after seeing your video I know what I need to do. Thank you so much!

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Wonderful refresher, many thanks

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Thanks for making the video

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Nice explanation with verifying that constructed DFA 👌

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Keep teaching & sharing......

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Great explanation thank u today i have presentation on finite automata and ity types this is helpful

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Thank you so much ❤️

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[@kamaleshsekharbharadwaj226]

Sir why can't we place the self loop in A state for input 1 instead of considering another state C. If input would be 1 then it would remain in A or if 0 then it will go to final state.

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i really want to know why

[@15seconds_craft]

same Question

[@troomanhooman]

Because it will accept inputs that start with 1. Picture this, for example, the input is 100:
Input 1: Stays in state A
Input 0: Goes to state B
Input 0: Stays in state B
The string is then accepted even if it did not start with 0.
But using a separate state C:
Input 1: Goes to state C
Input 0: Stays in state C
Input 0: Stays in state C
The string is then not accepted as it was deadlocked in state C.

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[@ansarisobiya50]

can you explain this sum design a dfa for set of string over (a,b) in which there are atleast two occurance of b between any two occurance of a

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[@hifsaghafoor9455]

sir can we create a self loop for input 1 on state A??

[@oniiichannn5113]

How about ends with 001 for example: 1001, 101001, and 001 are accepter while the strings 1000, 10011, and 0010 are rejected.

[@tejaltatiwar4682]

we dont care about any 1 except at first position

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Ty

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[@vishalprasadgupta714]

Sir is it is possible to have a DFA where there is two transition function {a,b} but there is one state (A) from which if we are taking {a} we reach to some state (say (B)) but on taking {b} we are not reaching any state .

[@prashantkawathe933]

Is it ok to put a self loop on A when the input is 1 ?