Conversion of Mealy Machine to Moore Machine (Using Transition Table)

[@cool_guy_Vaibhav]

I just gave my exam, and was able to solve with the help of your video. Thanks

[@shaiksubhani2918]

It's clarity...Great job nd thank you sir...

[@debalinade548]

such an amazing tutorial. Thank you so much sir. I am never going to forget this method in my life.

[@sanchiagarwal8851]

bhai itna bhi kya hi smjh aagaya?

[@akshat680]

chlo btao what is turing machine

[@shyamborse1586]

Khoop chhan 👌👌I understand it very well

[@akshaysubash3176]

Sir, when we draw the state diagram, the state Q2 is isolated. So how will the sequence reach Q2? as there is no continuation from any other states to Q2.

[@SurajMandal-gn2ji]

Final after 5 video i found easy and correct solution

[@swatiyadav4438]

Thanks for explaining in a simple and efficient way 😘😘😘

[@Mahadevlover365]

Apke lecture hai to automata me pass hona mumkin hai😄😄😄💋💋💋💋

[@vits5b4afzal66]

such an amazing tutorial. Thank you so much sir.

[@udiptadeka562]

finally i scored such a good marks all credit goes to you sir thank you

[@maliktarannum5418]

ur explanation is very clear and easy to understand ..its very helpful...thank u

[@sukamaldash3599]

Interesting thing in this question is, there is no way to reach q2 state in mealy machine (or q20 and q21 states in moore machine). Not only we can reach them but also, they are confined within themselves. So, my final question is, should we discard this state in final answer? Or should we keep it as it is in mealy machine?

[@sumanandfriends]

we are supposed to keep them io think

[@gorkijatt7679]

same doubt i m having when i tried to draw the mealey machine using the transition table... q2 state is only called by itself with no link to others states

[@venkatarohitpotnuru38]

@@gorkijatt7679 yeah bro

[@andyanderson222]

I think we should discard it. In one of the previous DFA lesson identical case was discussed and we removed such state. Moore / Mealy machines are something like an extended type of DFA, so same logic should work. Also if we look at this situation with few examples at hand, both machines (with q2 and without it) will give the same result, so q2 have nothing to do here.

[@Sudhanshu-es6lz]

तुम्ही खरंच खूपच छान शिकवता sir

[@jangamneelima9632]

Super sir this lecture video ..I prepared for sem exam easily

[@KuruvaKrishna-uf1bo]

Super explanation sir

[@satadhi]

you guys are doing awesome work !

[@kalaivani4686]

Thanks for reuploading thiz 🤗

[@Ak_Presents_Channel]

But while drawing the transition diagram there is no connection between q2 and all remaining states ..

[@tohidmomin8858]

Great superb sir for clarity

[@kirankaur4009]

One of the best academy

[@aryushgupta8680]

If we construct the mealy machine of the given table then I think there's a problem in state of q2 as it is isolated why??????

[@martincout]

thank you! well explained . the only video that I understood

[@gadhamshettyyamini8314]

Really it's great...sir ..easy to understand

[@kamleshguptaom]

Very neat and clean explaination

[@AyanKhan-fe4qc]

when i making transition diagram q20 and q21 are not linked to any state sois it unreachable state??????

[@Hammadali-ed1un]

Superb Lecture !!!!! Keep it Man !!! Amazing

[@jashansinghsidhusidhu8527]

Thank you sir.This helped me a lot.

[@soumyadipsantra9690]

You saved me Sir..

[@rahulkapoor778]

One thing missed!!!......if ur initial state is returning output 1...that means it can accept a zero length or null length sequence.......which is wrong....so u need to add a new state...say Q..on top of q0 that will have same state transitions as q0 but it's output would be 0.....that means...on top of the table ..u will add Q state with.....row values ...q3 ..and q11...and 0

[@Farahat1234]

Means? I did not understood!

[@vedgupta8444]

Moore machine cannot give output on initial state hence add a new initial state having output 0

[@jayparekh7598]

No bro you are wrong.see the introductory video of moore machine carefully.moore machine can have output without any input and that is not wrong.

[@RahultechIndia-68]

​@@jayparekh7598 no

[@AmanKumar-fh6nu]

exactly ​@@jayparekh7598

[@eswarprasad9773]

When we design the state diagram for this..q20 and q21 are forming a seperate diagram i.e. no connection to rest of the diagram???????

[@abdelrhmanahmed1378]

thats true q2 is unreachable state

[@AyanKhan-fe4qc]

yes ame doubt but now lol

[@crickshorts1888]

Easy because of u❤️❤️

[@ilyasceren9853]

reis allah razı olsun bundan sonra sizin videoları görünce hemen atlamayacağım. "şüphesiz ki, hindular dünyaya ders anlatmak için yollanmışlardır."

[@khushitripura3633]

Thanks u sir !! May God bless you

[@hreg5260]

Wow! easy explanation

[@kashyapchavhan8221]

If output is not given for any state then what will we consider 1 or 0

[@sohininandi5742]

there will always be an output given. that is what a mealy machine is. hope this helps.

[@__-lt4qy]

Easy explanation sir

[@kavyabanka4482]

Sir when state nor present in that transition what is the procedure

[@RobSummers993]

Thank you! Very easy to understand when you expalin it like this.

[@IshwaryaQ]

Tq very much sir I understood very clearly

[@ismail8973]

Thank you wonderful lecture

[@sadiyashaikh8511]

Ur the best.... U helped me alott with all my doubts...

[@user-su1pt5eu5e]

if the starting state split up into two while conversion from mealy machine to moore machine then what will be the starting state for equivalent moore machine??

[@aadityasharma5799]

How you connecting q2 in diagram

[@theforestlover4873]

If i not removed the outputs is it right?

[@gandlakiran3891]

U have to explain the very good sir

[@deepant7036]

thanks man, will help me a lot in tomoros tst

[@mondalshubh]

Thank you soo much sir.

[@srikanyabhattacharya9850]

very good explanation Sir!

[@shraddhamodanwal8293]

pls design it into transition diagram

[@ChristianBurnsShafer]

Transition diagrams have design limitations, whereas transition functions do not. So, it is worthwhile to be able to perform these operations without the use of a diagram.

[@ShahidMd-xz7jy]

for this table neither mealy or morre machine possible because q2 is isolated from rest of the states. please explain if i am wrong

[@mauryaajay]

No bro its true as per my knowledge bcz moore and mealy m/c DFA nature and DFA can exist of isolated states also so its true

[@anitaggarwal3189]

If the starting state is split,then in the final moore machine,which of them will become the initial state.

[@shivammahobiya536]

Well done

[@shivammahobiya536]

I really get all that you explained.

[@kevalramvaishnav5028]

Sir Outpur of Moore Machine should start with 0

[@Have_Fn]

Thank you so much

[@prime_tech7555]

But Sir Moore Machine must have an initial state yeilding 0, and new state must be introduced in order to do so...

[@sgiione]

Thankyou sir

[@Bikilatemesgenyoutube9127]

thank you

[@event9123]

thank you sir 👍👍👍

[@osamanisar8828]

Great .. thank you soo much

[@davidralte4572]

Thank You

[@rakshandachavan7971]

Nice explanation sir 👌👍

[@chocolyte1950]

Sir i have questions how can i remove the transition from the table (a,b) 😂

[@nagasamhithpatibandla4791]

Thanks bro

[@tagegnworku1482]

Thanks my sir so it is best

[@lonely8515]

Thank you so much...

[@anamikasingh9634]

Really very helpful...thanku so much

[@souvikpal8436]

Suppose we need to create 2 states for q00 and q01; now q0 was the initial state now which one should be initial state among q00 and q01

[@ChristianBurnsShafer]

It doesn't matter. You can start at either one and it will result in the same output (except for the first symbol).

[@rlalduhsaka6746]

the only video that make me understand(out of 4)

[@tirtharajghosh2680]

thank u.. Very good explanation

[@ABHISHEKSHARMA-nv1zu]

fantastic .....

[@shivomsaxena1581]

sir is turing machine is the upcomping topic ???

[@btsindianarmygirl995]

Thank u sir

[@hemantagrahari4298]

thank you so much sir...........for this video.........

[@harpreetsandhu7697]

Minimization of finite automata with output

[@amrrasslan4329]

hats off to you sir , you really saved my life xD

[@Jenil911-Pandya]

Q2 is only Reachable by Q2 LOL If we draw it Q2 is separate Machine

[@nathanielkungu6873]

So we'll explained I understood you

[@danilodevedzic5009]

Thank you.

[@mike10829]

faadu hai boss tu

[@amankharemedia]

Love you 😘

[@c.danielpremkumar8495]

Excellent.

[@bhawanishankarsharma6654]

Thank you sir.....

[@abdelrhmanahmed1378]

why tell now on using the fsm with output we dont determine the final state ?! is that because we dont need one to be our final state and the string will follow the state and outputting the result and if we find what we construct our machine for that mean the string is meeting our condition

[@BhagirathSingh]

thanku very much

[@zeynepturhan4847]

q21 = 0 and q20 = 1 İsn't this true ?

[@MrGyan]

I have the same question 🙄

[@rohitbiswas4997]

q2 is split into two, so one giving output 1 is named q21 and one giving output 0 is named q20 just last subscript is written based on output

[@zeynepturhan4847]

I graudated from university but thanks

[@rohitbiswas4997]

Yep afterall three years later haha I was studying for my exam thought of answering XD

[@zeynepturhan4847]

Haha

[@unknown-19_0.]

You are God sir

[@hamzahsohail]

5:27 q2 = cutoo 🤣

[@DeepakKumar-ul1qp]

i definitely get 5 marks .because tomorrow is my xam and i am reading this first today.

[@alemitumulu3969]

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[@ARCHSGAMING]

Best !! Video

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[@shadabsiddiqui28]

Osm

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BKwas

[@ankitjaiswal29]

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Overrated in whole KZbin 🎉 one night batting

[@ashwinthankachan9451]

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[@Karansingh-gh4oy]

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[@shivatheboss4833]

?????????????

[@Farahat1234]

Kch smjh nhi aaya😐😐😢😢