232-01-Quine-Mc Cluskey Method PDF

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3.4 QUINE–MCCLUSKEY METHOD

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¯ þ BCD. FIGURE 3.22 f (A, B, C, D, E) ¼ B¯ CD

¯ CD ¯ þ BCD þ A¯ BD. FIGURE 3.23 f (A, B, C, D, E) ¼ AC¯ E¯ þ B

A ¼ 1 map are – , 1, and 1, respectively, whereas the corresponding entries in the A ¼ 0 map are 0’s. Thus the four-variable Karnaugh map with A in cells BCDE and A in cells BCDE and BC DE shown in Figure 3.21 is equivalent to the maps of Figure 3.20. The map is then reduced in two steps. Step 1. Group all terms employing 1’s and – ’s. The letter variable terms are ignored at this step. Figure 3.22 shown the relevant groupings on the map of Figure 3.21. Step 2. Group the letter variable(s) with the adjacent 1’s and – ’s. The resulting terms are then ORed with the terms derived in step 1 to obtain the minimized function (shown in Figure 3.23).

3.4

QUINE– MCCLUSKEY METHOD

The Karnaugh map approach is not suitable for minimizing Boolean functions having more than six variables. For functions with a large number of variables, a tabular method known as the Quine – McCluskey method is much more effective. The method consists of two steps: 1. Generation of all prime implicants. 2. Selection of a minimum subset of prime implications, which will represent the original function.

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A prime implicant is a product term that cannot be combined with any other product term to generate a term with fewer literals than the original term. As an example, consider a Boolean function f (A, B, C) ¼ ABC þ ABC þ ABC þ ABC þ ABC which after minimization becomes f (A, B, C) ¼ AB þ BC þ AC þ ABC The product terms AB, BC, AC, and ABC are all prime implicants because none of them can be ombined with any other term in the function to yield a term with fewer literals. A prime implicants is called an essential prime implicant if it covers at least one minterm that is not covered by any other prime implicant of the function. Example 3.8

Let us minimize the following Boolean function: f (A, B, C, D) ¼ Sm(1, 4, 5, 10, 12, 13, 14, 15)

The Karnaugh map for the function is shown in Figure 3.24. The prime implicants for the function are BC, AB, ACD, and ACD. The minimized function is f (A, B, C, D) ¼ ACD þ BC þ AB þ AC D The prime implicant AC D is an essential prime implicant because it covers minterm A BCD, which is not covered by any other prime implicant. Similarly, only ACD covers minterm ABC D, B C covers AB C D, and AB covers ABCD; in other words, ACD, B C, and AB are also essential prime implicants. The Quine – McCluskey method for minimization can be formulated as follows: Step 1. Tabulate all the minterms of the function by their binary representations. Step 2. Arrange the minterms into groups according to the number of 1’s in their binary representation. For example, if the first group consists of minterms with n 1’s, the

FIGURE 3.24 Examples of adjacencies in Karnaugh.

3.4 QUINE–MCCLUSKEY METHOD

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second group will consist of minterms with (n þ 1) 1’s and so on. Lines are drawn between different group to simplify identification. Step 3. Compare each minterm in a group with each of the minterms in the group below it. If the compared pair is adjacent (i.e., if they differ by one variable only), they are combined to form a new term. The new term has a dash in the position of the eliminated variable. Both combining terms are checked off in the original list indicating that they are not prime implicatnts. Step 4. Repeat the above step for all groups of minterms in the list. This results in a new list of terms with dashes in place of eliminated variables. Step 5. Compare terms in the new list in search for further combinations. This is done by following step 3. In this case a pair of terms can be combined only if they have dashes in the same positions. As before, a term is checked off if it is combined with another. This step is repeated until no new list can be formed. All terms that remain unchecked are prime implications. Step 6. Select a minimal subset of prime implicants that cover all the terms of the original Boolean function. Example 3.9 procedure:

Let us minimize the following Boolean function using the Quine–McCluskey

f (A, B, C, D, E) ¼ Sm(0, 1, 2, 9, 11, 12, 13, 27, 28, 29) The minterms are first tabulated according to step 1. Minterm

A

B C

D

E

0 1 2 9 11 12 13 27 28 29

0 0 0 0 0 0 0 1 1 1

0 0 0 1 1 1 1 1 1 1

0 0 1 0 1 0 0 1 0 0

0 1 0 1 1 0 1 1 0 1

0 0 0 0 0 1 1 0 1 1

The minterms are then grouped according to the number of 1’s contained in each term, as specified in step 2. This results in list 1 of Figure 3.25. In list 1, terms of group 1 are combined with those of group 2, terms of group 2 are combined with those of group 3, and so on, using step 3. For example, 0(00000) is adjacent to 1(00001). So they are combined to form 0000 – which is the first term in list 2. Both combined terms are checked off in list 1. Since 0(00000) is also adjacent to 2(00010) they are combined to form the term 000– 0, which is also entered in list 2. A line is then drawn under the two terms in list 2 in order to identify them as a distinct group. The next step is to compare the two terms in group 2 of list 1 with the two terms in group 3. Only terms 1(00001) and 9(01001) combine to give 0 – 001; all other terms

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FIGURE 3.25 Determination of prime implicants.

differ in more than one variable and therefore do not combine. As a result, the second group of list 2 contains only one combination. The two terms in group 3 are now compared with the three terms in group 4. Terms 9(01001) and 11(01011) combine to give 010 – 1, terms 9(01001) and 13(01101) combine to give 01– 01, terms 12(01100) and 13(01101) combine to give 0110 – , and terms 12(01100) and 28(11100) combine to give –1100. Thus the third group of list 2 contains four terms. Finally, the three terms in group 4 of list 1 are compared with the two terms in group 5. Terms 13(01101) and 29(11101) combine to give –1101, terms 11(01011) and 27(11011) combine to give –1011, and terms 28(11100) and 29(11101) combine to give 1110– . Therefore the fourth group of list 2 contains three terms. The process of combining terms in adjacent groups is continued for list 2. This results in list 3. It can be seen in Figure 3.25 that certain terms cannot be combined further in list 2. These correspond to the prime implicants of the Boolean function and are labeled PI 1, . . . , PI7. The final step of the Quine – McCluskey proedure is to find a minimal subset of the prime implicants which can be used to realize the original function. The complete set of prime implicants for the given function can be derived from Figure 3.25; are (BC D, ABC D, AB C E, AC DE, AB DE, ABCE, B CDE) In order to select the smallest number of prime implicants that account for all the original minterms, a prime implicant chart is formed as shown in Figure 3.26. A prime implicant

FIGURE 3.26 Prime implicant chart.

3.4 QUINE–MCCLUSKEY METHOD

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FIGURE 3.27 Examples of adjacencies in Karnaugh.

chart has a column for each of the original minterms and a row for each prime implicant. For each prime implicant row, an X is placed in the columns of those minterms that are accounted for by the prime implicant. For example, in Figure 3.26 prime implicant PI1, comprising minterms 12, 13, 28, and 29, has X’s in columns 12, 13, 28, and 29. To choose a minimum subset of prime implicants, it is first necessary to identify the essential prime implicants. A column with a single X indicates that the prime implicant row is the only one covering the minterm corresponding to the column; therefore the prime implicant is essential and must be included in the minimized function. Figure 3.26 has three essential prime implicants, and they are identified by asterisks. The minterms covered by the essential prime implicants are marked with asterisks. The next step is to select additional prime implicants that can cover the remaining column terms. This is usually done by forming a reduced prime implicant chart that contains only the minterms that have not been covered by the essential prime implicants. Figure 3.27 shows the reduced prime implicant chart derived from Figure 3.26. Prime implicant PI4 covers the minterms 1 and 9. Therefore the minimum sumof-products equivalent to the original function is f (A, B, C, D, E) ¼ PI1 þ PI3 þ PI4 þ PI7 ¼ "110 " þ 000 " 0 þ 0 " 001 þ "1011 ¼ BC D þA BC E þ A C DE þ BCDE For some functions, the prime implicant chart may not contain any essential prime implicants. In other words, in every column of a prime implicant chart there are two or more X’s. Such a chart is said to be cyclic. Example 3.10

The following Boolean function has a cyclic prime implicant chart: f (A, B, C) ¼ Sm(1, 2, 3, 4, 5, 6)

The prime implicants of the function can be derived as shown in Figure 3.28. The resulting prime implicant chart as shown in Figure 3.29 is cyclic; all columns have two X’s. As can

FIGURE 3.28 Derivation of prime implicants.

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FIGURE 3.29 Prime implicant chart.

be seen, there is no simple way to select the minimum number of prime implicants from the cyclic chart. We can proceed by selecting prime implicant PI1, which covers minterms 1 and 3. After crossing out row PI1 and columns 1 and 3, we see that PI4 and PI5 cover the remaining columns (Fig. 3.30). Thus the minimum sum-of-products form of the given Boolean function is f (A, B, C) ¼ PI1 þ PI4 þ PI5 ¼ AC þ BC þ A B This is not a unique minimum sum of products for the function. For example, f (A, B, C) ¼ PI6 þ PI2 þ PI3 ¼ A C þ BC þ AB is also a minimal form of the original function. It can be verified from the Karnaugh map of the function (Fig. 3.31) that these are the minimum sum-of-products forms. Note that each minterm in the Karnaugh map can be grouped within two different loops, which ndicates that two different prime implicants can over the same minterm. 3.4.1

Simplification of Boolean Function with Don’t Cares

The Quine – McClusky procedure for minimizing Boolean functions containing don’t care minterms is similar to the conventional procedure in that all the terms, including don’t cares, are used to produce the complete set of prime implicants. However, don’t care terms are not listed as column headings in the prime implicant chart because they need not be included in the final expression.

FIGURE 3.30 Examples of adjacencies in Karnaugh....