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How many illegitimate states does synchronous mod-6 counter have?
1) 3

2) 2

3) 1

4) 0
in Digital Logic by Active (5k points)
edited by | 172 views
how we defined the illegitimate states ??

I heard this term ist time

can anyone explain me properly :3
What I felt after reading that word was the states which doesn't participate in counting loop are considered as illegitimate states.

Its my assumption


yes ...but answer given is 3 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 ???



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 ??

But how to know which counter to consider?Its just said synchronous counter..
Hmm point !

but if it's given that it's a synchronous counter (ring counter) ..and mod n (n>=3) then it always gives 3 illegitimate states


For n=3 we only get 2. Plz check once

hmmm 2 :/
Multiple Sources are saying are different answers.

When I googled this question, answer was mentioned as 3 in 1 website, 1 in another website, 0 in another website. Now here in this website, you guys are saying 2 :\

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