Thank you so much for the Tutorial. I have a doubt. what if I give 'aaa' or 'bbb' as input? The Moore Machine will output 11 while there is only one 'aa' or 'bb'. I am a bit confused..
If we give input as "abba" as specified in the question that this machine must give output '1' when ever the sequence 'aa' or 'bb' is encountered. Then for input : "abba" it must give output '1'. But the given figure . does not give output as '1'. I think for this given ans question must be , print '1' whenever the string end with either 'aa' or 'bb' .
when we write aaa or a more then 2 times or b more then 2 times it give wrong answer may be moore machine over lap but mealy machine give answer correctly so how conversen is correct .please help
N doesnt describe the states, it is the output you get back, in combination with the input. So when the Input is in Mealy M: aabb, the output will be (0101). When we give the same Input now in the Moore M. we get the output (0 + "0101"). So Moore M. got n+1.
test it for input abaaaa it gets trapped in B1 state and prints 3 times one, it should print 2 times one, NOT 3. Thus, it has a wrong logic! Either the question should change or both diagrams need readjustment for B1, C1 states->they should not loop to themselves.Instead they should point to B0, C0 respectively in order to fully fit to the question's requirements.
The input "abaaaa" does in fact have the sequence "aa" in it 3 times. You're making the counting error; not the machine. ab *aa* aa, aba *aa* a, and abaa *aa*
Yes because there are two consecutive possibilities of "bb". If the input string is "bbbb" then we get the output as 00111 as there are three consecutive possibilities of "bb".
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hello how are you doing now after 5 year which company are you in and your package
Can't thank you enough! Everything in this Course is just on point, love it!
Thanks for making me understand the topic very easily.
Excellent explanation, thanks sir
I would love to see Data Structures & Algorithm, that's something only a person like you could explain in plain words.
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What happens if input is an empty string? Output is zero with the initial state. Note the input is from sigma star which includes empty string.
Thank you so much for this knowledge from these videos, it has helped me a lot.
thanks for teaching us, sir until when the course of toc will be completed by you
We should get the output 010 for aaa, but in your Mealy Machine, we will get 011.
Isn't that wrong?
we get output 011 for aaa, in both mealy and moore machine.
@@imposter1721 so wouldn't that make both of them incorrect?
@@rahqanshaikh 0011 is right for Moore machine
Thank you so much for the Tutorial.
I have a doubt. what if I give 'aaa' or 'bbb' as input? The Moore Machine will output 11 while there is only one 'aa' or 'bb'.
I am a bit confused..
The question itself states that either of the is sufficient
thanks for these easy explainatoins........
thanx for helping us m facing so difficulties in this suubject
If we give input as "abba" as specified in the question that this machine must give output '1' when ever the sequence 'aa' or 'bb' is encountered. Then for input : "abba" it must give output '1'. But the given figure . does not give output as '1'.
I think for this given ans question must be , print '1' whenever the string end with either 'aa' or 'bb' .
when we write aaa or a more then 2 times or b more then 2 times it give wrong answer may be moore machine over lap but mealy machine give answer correctly so how conversen is correct .please help
yeah sir can you please take a look at this (Neso)
No it's correct, the thing is when we give aaa we get input 1 twice because the aa sequence is repeated twice
thank you
Sir please create table after conversation !!!
it also prints 1 after scanning 'aa' for example 'aaa' (or 'bb')
a a a
0 0 1 1
right ?
Right. At state B1, it loops back to B1 with input a.
Yes there is problem in aaa or bbb
no offence can you do the conversion in terms of transition tables i mean using transition table. Thanks
This example exactly same as the previous lecture except that {a,b} is interchanged with {0,1}. smh.
mealy machine have n states while moore machine should have n+1 states but in the example, 5 states are drawn. How it is not wrong?
N doesnt describe the states, it is the output you get back, in combination with the input. So when the Input is in Mealy M: aabb, the output will be (0101). When we give the same Input now in the Moore M. we get the output (0 + "0101"). So Moore M. got n+1.
what will be the output of state A?
In this case, it is 0, because the condition says only from 'aa' or 'bb' we get '1'.
test it for input abaaaa it gets trapped in B1 state and prints 3 times one, it should print 2 times one, NOT 3. Thus, it has a wrong logic! Either the question should change or both diagrams need readjustment for B1, C1 states->they should not loop to themselves.Instead they should point to B0, C0 respectively in order to fully fit to the question's requirements.
check this input for mealy machine. output are same in both mealey and moore machine. So, conversion is correct. Purpose was to show how to convert.
The input "abaaaa" does in fact have the sequence "aa" in it 3 times. You're making the counting error; not the machine. ab *aa* aa, aba *aa* a, and abaa *aa*
we are getting 3 three times output 1 because it is detecting overlapping sequence..
If input is aaaa then what's the output ??
The output is "00111" as there are three consecutive possibilities of "aa".
what if we print 'bbb'? we will get output as 0011
Yes because there are two consecutive possibilities of "bb".
If the input string is "bbbb" then we get the output as 00111 as there are three consecutive possibilities of "bb".
@@rohitvadhya7153 is this sure one?
@@shanmugapriya7554Yes 👍