**2 Answers**

written 3.1 years ago by |

*Answer:*

*General steps to solve any function using Quine McCluskey technique.*

*The function should be in "minterm" form. If it is in Maxterm form then first convert it into minterm form by just writing the missing numbers.*

For example: f=M( 2,3,4,5) for 3 bits then the function in Minterm are f=m(0,1,6,7).*Write the binary designations (binary equivalents) for all given minterms.**Arrange the minterms in groups according to the number of 1s present**.*( for example: 0 will have zero 1s, 1 has one 1s, 7 has three 1s.).The boolean algebraic theorem$A+\overline A = 1$ is applied to pair of minterms in which one variable is different and all other variables are same. (applicable to minterms of adjacent groups only).*CONCEPT:*The matched pair of minterms are checked and a new table is created.*All the minterms in the adjacent groups are compared to see if a group of 'four' can be made by matching.**Repeat the process of grouping till no further combination of minterms is possible.*

**SOLUTION:**

*Step 1 :*

GROUP (Number of 1s) | MINTERM | BINARY DESIGNATION |
---|---|---|

0 | 0 | 0000 |

1 | 1 |

2 8 | 0001 0010 1000 | | 2 | 3 5 9 | 0011 0101 1001 | | 3 | 7 11 14 | 0111 1011 1110 |

*Step 2: Pair minterms with single bit difference.*

MINTERM | BINARY DESIGNATION |
---|---|

0-1 |

0-2 0-8 | 000_ 00_0 _000 | | 1-3 1-5 1-9 2-3 8-9 | 00_1 0_01 _001 001_ 100_ | | 3-7 3-11 5-7 9-11 | 0_11 _011 01_1 10_1 | | 14 | 1110 |

*Step 3:*

MINTERMS | BINARY DESIGNATION | P.I. (PRIME IMPLICANT) |
---|---|---|

0-1-2-3 |

0-1-8-9 0-2-1-3 0-8-1-9 | 00_ _ _ 00 _ 00_ _ _ 00 _ | $\overline A.\overline B$ $\overline B.\overline C$ $\overline A.\overline B$ $\overline B.\overline C$ | | 1-3-5-7 1-3-9-11 1-5-3-7 1-9-3-11 | 0 _ _ 1 _ 0 _ 1 0 _ _ 1 _ 0 _ 1 | $\overline A.D$ $\overline B.D$ $\overline A.D$ $\overline B.D$ | | 14 | 1110 | $A.B.C.\overline D$ |

*Step 4: Prime implicants table.*

PRIME IMPLICANTS | 0 | 1 | 2 | 3 | 5 | 7 | 8 | 9 | 11 | 14 |
---|---|---|---|---|---|---|---|---|---|---|

$\overline A.\overline B$ - 0,1,2,3 | X | X | X * | X | ||||||

$\overline B.\overline C$ - 0,1,8,9 | X | X | X * | X | ||||||

$\overline A.D$ - 1,3,5,7 | X | X | X * | X * | ||||||

$\overline B.D$ - 1,3,9,11 | X | X | X | X * | ||||||

$A.B.C.\overline D$ - 14 | X * |

[ In prime implicants table first select the minterms which are repeated only once in a column amd write their equivalent expression in final function. After completion of all such PIs write expressions for the prime implicants which are not represented by any of the expressions taken in the final function ].

** ANSWER:** $f= \overline A.\overline B + \overline B.\overline C + \overline A.D + \overline B.D + A.B.C.\overline D$

*TIP: Verify the final function by reducing the given using Kmaps for cross-checking the answer. Only in this question has the final function include all the expressions of the prime implicants, it is not possible in all problems.*