Karnaugh Maps

Karnaugh Maps

Karnaugh maps Last time we saw applications of Boolean logic to circuit design. The basic Boolean operations are AND, OR and NOT. These operations can be combined to form complex expressions, which can also be directly translated into a hardware circuit. Boolean algebra helps us simplify expressions and circuits. Today well look at a graphical technique for simplifying an expression into a minimal sum of products (MSP) form: There are a minimal number of product terms in the expression. Each term has a minimal number of literals. Circuit-wise, this leads to a minimal two-level implementation. June 12, 2002 2000-2002 Howard Huang 1 Re-arranging the truth table A two-variable function has four possible minterms. We can rearrange these minterms into a Karnaugh map. x y minterm 0 0 xy 0 1 xy 1 0 xy

1 1 xy Y X 0 1 0 xy xy 1 xy xy Now we can easily see which minterms contain common literals. Minterms on the left and right sides contain y and y respectively. Y Minterms in the top and bottom rows contain x and x Y Y 1 respectively. 0 X June 12, 2002 0 1 xy xy X X xy xy Karnaugh Maps xy xy xy xy

2 Karnaugh map simplifications Imagine a two-variable sum of minterms: xy + xy Both of these minterms appear in the top row of a Karnaugh map, which means that they both contain the literal x. X xy xy Y xy xy What happens if you simplify this expression using Boolean xy + xy = x(y + y) [ Distributive ] algebra? = x 1 [ y + y = 1 ] = x [x1=x] June 12, 2002 Karnaugh Maps 3 More two-variable examples Another example expression is xy + xy. Both minterms appear in the right side, where y is uncomplemented. Thus, we can reduce xy + xy to just y. X xy xy Y

xy xy How about xy + xy + xy? We have xy + xy in the top row, corresponding to x. Theres also xy + xy in the right side, corresponding to y. Y to x + y. This whole expression can be reduced X June 12, 2002 xy xy xy xy Karnaugh Maps 4 A three-variable Karnaugh map For a three-variable expression with inputs x, y, z, the arrangement of minterms is more tricky: YZ X 0 1 00 xyz xyz 01 xyz xyz 11 xyz xyz 10 xyz xyz X 0

1 00 m0 m4 YZ 01 11 m1 m3 m5 m7 10 m2 m6 Another way to label the K-map (use whichever you like): Y X xyz xyz xyz xyz xyz xyz Y xyz xyz Z June 12, 2002 X m0 m4 m1 m5 m3 m7 m2 m6 Z

Karnaugh Maps 5 Why the funny ordering? With this ordering, any group of 2, 4 or 8 adjacent squares on the map contains common literals that can be factored out. Y X xyz xyz xyz xyz xyz xyz xyz xyz Z xyz + xyz = xz(y + y) = xz 1 = xz Adjacency includes wrapping around the left and right sides: xyz + xyz + xyz + xyz xyz xyz xyz xyz = z(xy + xy + xy + X xyz xyz xyz xyz xy) Z = z(y(x + x) + y(x + x)) = z(y+y) = z to do our simplifications. Well use this property of adjacent squares Y June 12, 2002

Karnaugh Maps 6 Example K-map simplification Lets consider simplifying f(x,y,z) = xy + yz + xz. First, you should convert the expression into a sum of minterms form, if its not already. The easiest way to do this is to make a truth table for the function, and then read off the minterms. You can either write out the literals or use the minterm shorthand. Here is the truth table and sum of minterms for our example: x y z f (x,y,z) 0 0 0 0 0 0 1 1 0 1 0 1 0 1 0 0 1 1 1 1 0 0 1 1

0 1 0 1 0 1 1 1 June 12, 2002 f(x,y,z) = xyz + xyz + xyz + xyz = m1 + m5 + m6 + m7 Karnaugh Maps 7 Unsimplifying expressions You can also convert the expression to a sum of minterms with Boolean algebra. Apply the distributive law in reverse to add in missing variables. Very few people actually do this, but its occasionally useful. xy + yz + xz = (xy 1) + (yz 1) + (xz 1) = (xy (z + z)) + (yz (x + x)) + (xz (y + y)) = (xyz + xyz) + (xyz + xyz) + (xyz + xyz) = xyz + xyz + xyz + xyz In both cases, were actually unsimplifying our example expression. The resulting expression is larger than the original one! But having all the individual minterms makes it easy to combine them together with the K-map. June 12, 2002 Karnaugh Maps 8 Making the example K-map Next up is drawing and filling in the K-map. Put 1s in the map for each minterm, and 0s in the other squares.

You can use either the minterm products or the shorthand to show you where the 1s and 0s belong. In our example, we can write f(x,y,z) in two equivalent ways. f(x,y,z) = xyz + xyz + xyz + xyz f(x,y,z) = m1 + m5 + m6 + m7 Y X xyz xyz xyz xyz xyz xyz Y xyz xyz m0 m4 X m1 m5 Z m3 m7 m2 m6 Z Y In either case, the resulting K-map is shown below. X 0 0 1 1 0 1 0

1 Z June 12, 2002 Karnaugh Maps 9 K-maps from truth tables You can also fill in the K-map directly from a truth table. The output in row i of the table goes into square mi of the Kmap. Remember that the rightmost columns of the K-map are Y switched. x y z f (x,y,z) 0 0 0 0 0 0 1 1 0 1 0 1 0 1 0 0 m0 m4 X m1 m5 m3 m7

m2 m6 Z Y X 0 0 1 1 0 1 0 1 Z 1 1 1 1 0 0 1 1 June 12, 2002 0 1 0 1 0 1 1 1 Karnaugh Maps 10 Grouping the minterms together The most difficult step is grouping together all the 1s in the K-map. Make rectangles around groups of one, two, four or eight 1s. All of the 1s in the map should be included in at least one rectangle.

Do not include any of the 0s. Y X 0 0 1 1 0 1 0 1 Z Each group corresponds to one product term. For the simplest result: Make as few rectangles as possible, to minimize the number of products in the final expression. Make each rectangle as large as possible, to minimize the number of literals in each term. Its all right for rectangles to overlap, if that makes them larger. June 12, 2002 Karnaugh Maps 11 Reading the MSP from the K-map Finally, you can find the MSP. Each rectangle corresponds to one product term. The product is determined by finding the common literals in that rectangle. Y X 0 0 1 1 0 1 0

1 Z Y X xyz xyz xyz xyz xyz xyz xyz xyz Z For our example, we find that xy + yz + xz = yz + xy. (This is one of the additional algebraic laws from last time.) June 12, 2002 Karnaugh Maps 12 Practice K-map 1 Simplify the sum of minterms m1 + m3 + m5 + m6. Y X Z Y X m0 m4 m1 m5 m3 m7 m2 m6

Z June 12, 2002 Karnaugh Maps 13 Solutions for practice K-map 1 Here is the filled in K-map, with all groups shown. The magenta and green groups overlap, which makes each of them as large as possible. Minterm m6 is in a group all by its lonesome. Y X 0 0 1 1 1 0 0 1 Z The final MSP here is xz + yz + xyz. June 12, 2002 Karnaugh Maps 14 Four-variable K-maps We can do four-variable expressions too! The minterms in the third and fourth columns, and in the third and fourth rows, are switched around. Again, this ensures that adjacent squares have common literals. Y Y W

wxyz wxyz wxyz wxyz wxyz wxyz wxyz wxyz wxyz wxyz wxyz wxyz wxyz wxyz X wxyz wxyz Z W m0 m4 m12 m8 m1 m5 m13 m9 m3 m7 m15 m11 m2 m6 X m14 m10 Z Grouping minterms is similar to the three-variable case, but: You can have rectangular groups of 1, 2, 4, 8 or 16 minterms. You can wrap around all four sides. June 12, 2002

Karnaugh Maps 15 Example: Simplify m0+m2+m5+m8+m10+m13 The expression is already a sum of minterms, so heres the K-map: Y W 1 0 0 1 0 1 1 0 0 0 0 0 Y 1 0 X 0 1 W m0 m4 m12 m8 m1 m5 m13 m9 m2 m6 m14 m10 X

Z Z m3 m7 m15 m11 We can make the Y following groups, resulting in the MSP Y xz + xyz. 1 0 0 1 wxyz wxyz wxyz wxyz W 0 0 1 1 1 0 0 0 0 0 X 0 1 W wxyz wxyz wxyz Z June 12, 2002 wxyz wxyz wxyz wxyz wxyz wxyz wxyz X

wxyz wxyz Z Karnaugh Maps 16 K-maps can be tricky! There may not necessarily be a unique MSP. The K-map below yields two valid and equivalent MSPs, because there are two possible ways to include minterm m7. Y 0 0 X 1 1 0 1 1 1 Z Y X 0 0 1 1 0 1 Y 1 1 X 0 0 1 1 0

1 Z Z yz + yz + xy yz + yz + xz 1 1 Remember that overlapping groups is possible, as shown above. June 12, 2002 Karnaugh Maps 17 Prime implicants The challenge in using K-maps is selecting the right groups. If you dont minimize the number of groups and maximize the size of each group: Your resulting expression will still be equivalent to the original one. But it wont be a minimal sum of products. Whats a good approach to finding an actual MSP? First find all of the largest possible groupings of 1s. These are called the prime implicants. The final MSP will contain a subset of these prime implicants. Here is an example Karnaugh map Ywith prime implicants marked: W 1 1 0 0 1 1 1 0 0 0 1 1

0 0 X 0 1 Z June 12, 2002 Karnaugh Maps 18 Essential prime implicants Y W 1 1 0 0 1 1 1 0 0 0 1 1 0 0 X 0 1 Z If any group contains a minterm that is not also covered by another overlapping group, then that is an essential prime implicant. Essential prime implicants must appear in the MSP, since they contain minterms that no other terms include. Our example has just two essential prime implicants: The red group (wy) is essential, because of m0, m1 and m4.

The green group (wxy) is essential, because of m10. June 12, 2002 Karnaugh Maps 19 Covering the other minterms Y W 1 1 0 0 1 1 1 0 0 0 1 1 0 0 X 0 1 Z Finally pick as few other prime implicants as necessary to ensure that all the minterms are covered. After choosing the red and green rectangles in our example, there are just two minterms left to be covered, m13 and m15. These are both included in the blue prime implicant, wxz. The resulting MSP is wy + wxz + wxy. The black and yellow groups are not needed, since all the minterms are covered by the other three groups.

June 12, 2002 Karnaugh Maps 20 Practice K-map 2 Simplify for the following K-map: Y W 0 1 1 0 0 0 1 0 1 1 1 1 0 1 X 1 0 Z June 12, 2002 Karnaugh Maps 21 Solutions for practice K-map 2 Simplify for the following K-map: Y W 0 1 1

0 0 0 1 0 1 1 1 1 0 1 X 1 0 Z All prime implicants are circled. Essential prime implicants are xz, wx and yz. The MSP is xz + wx + yz. (Including the group xy would be redundant.) June 12, 2002 Karnaugh Maps 22 I dont care! You dont always need all 2n input combinations in an n-variable function. If you can guarantee that certain input combinations never occur. If some outputs arent used in the rest of the circuit. We mark dont-care outputs in truth tables and K-maps with Xs. x y z f (x,y,z) 0 0 0 0 0 0 1 1 0

1 0 1 0 1 X 0 1 1 1 1 0 0 1 1 0 1 0 1 0 1 X 1 Within a K-map, each X can be considered as either 0 or 1. You should pick the interpretation that allows for the most simplification. June 12, 2002 Karnaugh Maps 23 Practice K-map 3 Find a MSP for f(w,x,y,z) = m(0,2,4,5,8,14,15), d(w,x,y,z) = m(7,10,13) This notation means that input combinations wxyz = 0111, 1010 and 1101 (corresponding to minterms m7, m10 and m13) are unused. Y W 1 1 0 1

0 1 x 0 0 x 1 0 1 0 X 1 x Z June 12, 2002 Karnaugh Maps 24 Solutions for practice K-map 3 Find a MSP for: f(w,x,y,z) = m(0,2,4,5,8,14,15), d(w,x,y,z) = m(7,10,13) W 1 1 0 1 0 1 x 0 Z Y 0 x 1 0 1 0 X 1 x All prime implicants are circled. We can treat Xs as 1s if we

want, so the red group includes two Xs, and the light blue group includes one X. The only essential prime implicant is xz. The red group is not essential because the minterms in it also appear in other groups. The MSP is xz + wxy + wxy. It turns out the red group is redundant; we can cover all of the minterms in the map without Juneit.12, 2002 Karnaugh Maps 25 Summary K-maps are an alternative to algebra for simplifying expressions. The result is a minimal sum of products, which leads to a minimal two-level circuit. Its easy to handle dont-care conditions. K-maps are really only good for manual simplification of small expressions... but thats good enough for CS231! Things to keep in mind: Remember the correct order of minterms on the K-map. When grouping, you can wrap around all sides of the K-map, and your groups can overlap. Make as few rectangles as possible, but make each of them as large as possible. This leads to fewer, but simpler, product terms. There may be more than one valid solution. June 12, 2002 Karnaugh Maps 26

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