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Prime Implicant chart for minimizing Cyclic Boolean functions

  • Difficulty Level : Hard
  • Last Updated : 10 Jun, 2021

Prerequisite – K-Map (Karnaugh Map), Implicants in K-Map 
A functions is said to be a Cyclic Boolean Function if there is no Essential prime Implicant in its respective K-Map. 

Properties of Cyclic functions: 
 

  • Every prime Implicant is of same size.
  • Every minterm is covered by at least two prime Implicants (which means no essential prime Implicants).
  • Having no essential prime Implicants means that there exist more than one minimized Solution / Expression for such functions, which will further be realized using digital circuits.
  • For a cyclic function we can have two minimal forms with no overlapping of prime Implicants.

Example: 
Find the minimal expression for the following function. 
 

f(w, x, y, z) = \Sigma(0, 2, 4, 5, 10, 11, 13, 15) 

 



As we can see in the above K-Map that there exist no essential prime Implicants. Here we can use prime Implicant chart to solve it easily. 

Steps to solve above function using prime Implicant chart: 

 

  • Step-1: 
    Draw prime Implicant chart as below.The horizontal entries denote the given minterms which are mapped against all prime Implicants (vertically).The square boxes are crossed (‘x’) whenever a prime Implicant covers a particular minterm in K-Map. 

    For example ‘WXZ’ prime Implicant is covering 13 and 15 therefore the corresponding squares are crossed(denoted by ‘x’). 

    Note that, A, B, C, D .., are variables used to denote all the prime Implicants. 

     

 

  • Step-2: 
    Arbitrarily choose any prime Implicant; check (\checkmark   ) the prime Implicant and the corresponding covered minterms as well. Now delete the row of the prime Implicant and corresponding columns of its minterms. 

    In our example prime Implicant A is chosen which is covering minterms 2 and 10. Therefore. delete the row of A and the columns of 2 and 10. The arbitrarily chosen Prime Implicant (in our example A) must be present in the final minimal expression. 



     

 

 

  • Step-3: 
    Find all such prime Implicant which are being covered by other prime Implicant completely and remove their corresponding rows (since these are non essential prime Implicants.). 

    In our example H is covering {0, 4} and B is covering {0} it means H is covering all the minterms that are covered by B, so delete B. Similarly D is covering C completely, so delete C. 

     

 

 

  • Step-4: 
    Now follow the standard procedure of prime Implicant chart mentioned in the sub steps given below: 
    1. Find the minterm which is covered by only one prime Implicant. 

       

    2. Check (\checkmark   ) that minterm, its corresponding prime Implicant and all other minterms which are covered by that corresponding prime Implicant. 

       

    3. If all the minterms are checked (\checkmark   ) stop the procedure, otherwise goto sub-step-1. 
       

Example-2: find the minimal expression for the following cyclic function. 
 



f(x, y, z) = \Sigma(0, 1, 2, 5, 6, 7) 

 

Step-1: 
Draw prime Implicant chart. 

 

Step-2: 
A is chosen arbitrarily.Now the row of A and the columns of the corresponding minterms (0 and 2) are deleted. 

 

 

Step-3: 
Since B and F are completely covered by C and E respectively hence deleting prime Implicants B and F . 

 

Step-4: 
Now follow the standard procedure of prime Implicant chart mentioned in Example-1. 

Minterm 1 is covered by prime Implicant C only therefore check (\checkmark   ) C along with all the minterms which are covered by it (1 and 5). 

Minterm 6 is covered by prime Implicant E only therefore check (\checkmark   ) E along with all the minterms which are covered by it (6 and 7). 

 

Now since all the min terms (1, 5, 6, 7) are checked(\checkmark   ) therefore stop the procedure. 

 

Final Minimal Expression: A + C + E  

 

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