Hemanth_13
yes ...but answer given is 3 ..my doubt is how 3 states are illegitimate states here
if we need mod 6 counter
we need 3 flip flips [if we use Johnson counter]
therefore total 8 states we have (0 to 7)
In which states are required to make 6 mod counter
remaining 2 are left
therefore answer should be 2 right
but I search in the Google ....and find correct answer is 3
if you find any solution for this or know how to solve this question
let me know ok ???
Magma
Yes what you said was right in case of Johnson counter if we have n states we get mod 2n counter, so we have two illegitimate states.
But we can use ring counter as well where we get mod n counter. Its huge taking to check with 6 bits 2^6=64 combinations
I tried with 4 bits and i got 3 illegitimate states.
Hemanth_13 thanks !!
you give the example of mod 4 for mod 6 also it gives 3 illegitimate states therefore it means In mod n counter using ring counter where n >= 3 and n is any arbitrary number gives 3 illegitimate states ??
For n=3 we only get 2. Plz check once
