# Self Doubt

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"Every cyclic K-map given cyclic function and every cyclic function can be the self-dual function."

What is the meaning of cyclic K-map?

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this might be https://gateoverflow.in/173362/self-doubt

is it important?

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this might be https://gateoverflow.in/173362/self-doubt

This is not sufficient.

doubt came from the comment of the BEST answer

https://gateoverflow.in/29414/identifying-self-dual-function

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ok

but not getting much information about it , in internet
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every prime implicant which is covered by at least two prime implicant and there is no essential prime implicant present in k-map called as cyclic k-map. 0

1) then there are more than 1 cyclic configuration of K-map

https://gateoverflow.in/173362/self-doubt    should have more than 2 possibilities

2) Is every self-dual function leads to cyclic k-map?

otherwise it just " every cyclic k-map lead to self-dual "

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https://gateoverflow.in/173362/self-doubt    should have more than 2 possibilities

How? if you will solve above k-map with prime implicant table then you will get only two possibilities which is mentioned in above link.

Is every self-dual function leads to cyclic k-map?

otherwise it just " every cyclic k-map lead to self-dual "

I don't know, how this can be true? Because in self dual function number of minterms should be equal to no. of maxterms. But this is not the case in cyclic function

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How? if you will solve above k-map with prime implicant table then you will get only two possibilities which is mentioned in above link. I don't know, how this can be true? Because in self dual function number of minterms should be equal to no. of maxterms. But this is not the case in cyclic function

then the given statement (in the question) should be false, right?

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your k-map(1,2,3,4,5,6) and mine(0,1,2,5,6,7(don't consider digit which is mentioned in k-map)) are totally different.

And yes this statement should be wrong from my point of view.
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in that question they didn't mention anything about minterms..

So, if you fix the minterms, then you have 2 possibilities.

But fixing the minterms have more than 1 choice.
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yes @shaik I haven't checked that question previously but according to that question yes there are more than 2 possibilities.

## Related questions

1
338 views
Find least literal count of POS for the following k-map. X 1 0 1 0 1 X 0 1 X X 0 X 0 0 X Ans: Minterms = 2 quads + 2 pairs = 2*2 + 3*2 = 10 literals OR, = 1 quad + 2 pairs = 1*2 + 3*2 = 8 literals Which is the correct approach? Should I favour quad over pair or should it be according to requirement? Since here minimum no.of literals are asked to find so is 2nd approach correct?