Question

g f g | = Dono mi $ $ $ $ $ Tabulation Method (Quine-McCluskey) Example: f= {(1,2,3,4,7,8, 12, 15) + d 0,5,9,10, 14) Index Impl. Binary Impl. Dec. Index | Impl. Binary Impl. Dec. Index Impl. Binary Impl. Dec. 000. 000000*_d 00- (0,1,2,3) 00-0 0001 0-0 (0, 1,4,5) 0-00 0010 -00 (0,1,8,9) -000 -0-0 0100 (0, 2, 8, 10) 00-1 --00 (0,4,8,12) 1000 0-01 0--1 (1,3,5,7) 0011 -001 1-0 (8, 10, 12, 14) 0101 001- 1001 -010 010- 1010 -100 1100 100- We stop when we can no longer form 0111 10-0 1110 14 d larger ones and name the prime implicants 1-00 1111 0-11 (3.7) Note: We can form larger implicants only by 01-1 combining implicants of adjacent indices 1-10 (10, 14)* 11-0 (12, 14)* Index = # of 1's in the string 7.15) We list all the minterms in binary and (14, 15) decimal form, grouped by their indices. Note: every size-4 implicant will be formed in 2 ways out of size-2 implicants: We mark the d's. These are the size-1 implicants. We continue to size-2, size-4, etc implicants, until no more possible, marking the non-prime implicants with * $ $ $ $ 15 * 1

please expert explain how to get this chart to me

Answer 1

The above method is a tabular method used to minimize the boolean expression by grouping in minterms orderly.
Tabulation Method :- This method is used to reduce the boolean expression of possible. Given f= {(1,2,3,4,7,8, 12, 15) + d (0,5,9,101 14) Initially, we have to write a table in order of minimum to maximum I's. This means numbers (minterms) which contain equal no. of I's are grouped starting from minimum value. No. of Binary No equivalent Decimal No. o d 0000 ooot oolo oloo 1ooo ool olot lool lolo 1100 I-d Joomoood it Ile tooo oll 110 III 4

. In the above table, d denotes don't care. Now, in the next table, we have to compare the binary no. each with other binary nois for the changes in the position of O's & I's. If the changes in the position is only one bit, then group those two minterm no's and leave the changed bit with blank as shown below: representation Grouped Decimal Index Binary no (0. (012) (0,4) (0.8) 10 oo olo o oool Io - 001 ololHHH oool (1,3) (1,5) 0 0 (1,9) (2,3) -o01 ool- -olo 010 - 100 10o- 10_0 100 (2,10) (4,5) (4,12) 18,9) (8.10) (8,12) (To be continued in next page)

(3,7) OL!! 01 - 1 1-10 1-O (5 ) (10,14) (12,14) -tli (7,15) (14,15) H Il- Now, again we have to group the minterms of above table in such a way that the change in the values of bit is only 1 i.e. only 1 position of minterm is different such interms are group and left the one bit changed position with blank (-) as below. Index | Implicant Binary 00-- 0-0- Implicant Decimall (0,1,2,3) (0,1,4,5) (0, 1,8,9) (0,2,8.10) (0,4,8,12) Ic -oo- a com To o II 10- -o-o --00 (1.3.5,1) B (8, 10, 12, 14 A I--0