Karnaugh mapsReview: Standard forms of expressionsTerminology: MintermsTerminology: Sum of minterms formRe-arranging the truth tableKarnaugh map simplificationsMore two-variable examplesA three-variable Karnaugh mapWhy the funny ordering?Example K-map simplificationUnsimplifying expressionsMaking the example K-mapK-maps from truth tablesGrouping the minterms togetherReading the MSP from the K-mapPractice K-map 1Solutions for practice K-map 1Four-variable K-mapsExample: Simplify m0+m2+m5+m8+m10+m13K-maps can be tricky!February 2, 2004 CS 231 1Karnaugh 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 we’ll 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 minimaltwo-level implementation.February 2, 2004 CS 231 2Review: Standard forms of expressions• We can write expressions in many ways, but some ways are more useful than others• A sum of products (SOP) expression contains:– Only OR (sum) operations at the “outermost” level– Each term that is summed must be a product of literals• The advantage is that any sum of products expression can be implemented using a two-level circuit– literals and their complements at the “0th” level– AND gates at the first level– a single OR gate at the second level• This diagram uses some shorthands…– NOT gates are implicit– literals are reused– this is notokay in LogicWorks!f(x,y,z) = y’ + x’yz’ + xzFebruary 2, 2004 CS 231 3Terminology: Minterms• A minterm is a special product of literals, in which each input variable appears exactly once.• A function with n variables has 2nminterms (since each variable can appear complemented or not)• A three-variable function, such as f(x,y,z), has 23 = 8 minterms:• Each minterm is true for exactly one combination of inputs:x’y’z’ x’y’z x’yz’ x’yzxy’z’ xy’z xyz’ xyzMinterm Is true when… Shorthandx’y’z’ x=0, y=0, z=0 m0x’y’z x=0, y=0, z=1 m1x’yz’ x=0, y=1, z=0 m2x’yz x=0, y=1, z=1 m3xy’z’ x=1, y=0, z=0 m4xy’z x=1, y=0, z=1 m5xyz’ x=1, y=1, z=0 m6xyz x=1, y=1, z=1 m7February 2, 2004 CS 231 4Terminology: Sum of minterms form• Every function can be written as a sum of minterms, which is a special kind of sum of products form• The sum of minterms form for any function is unique• If you have a truth table for a function, you can write a sum ofminterms expression just by picking out the rows of the table where the function output is 1.x y z f(x,y,z) f’(x,y,z)000 1 0001 1 0010 1 0011 1 0100 0 1101 0 1110 1 0111 0 1f = x’y’z’ + x’y’z + x’yz’ + x’yz + xyz’= m0+ m1+ m2+ m3+ m6= Σm(0,1,2,3,6)f’ = xy’z’ + xy’z + xyz= m4+ m5+ m7= Σm(4,5,7)f’ contains all the minterms not in fFebruary 2, 2004 CS 231 5Re-arranging the truth table• A two-variable function has four possible minterms. We can re-arrange these minterms into a Karnaugh map.• Now we can easily see which minterms contain common literals.– Minterms on the left and right sides contain y’ and y respectively.– Minterms in the top and bottom rows contain x’ and x respectively.x y minterm00 x’y’01 x’y10 xy’11 xyY010 x’y’ x’yX1 xy’ xyY0 10x’y’ x’yX1xy’ xyY’ YX’ x’y’ x’yXxy’ xyFebruary 2, 2004 CS 231 6Karnaugh map simplifications• Imagine a two-variable sum of minterms:x’y’ + x’y• Both of these minterms appear in the top row of a Karnaugh map, which means that they both contain the literal x’.• What happens if you simplify this expression using Boolean algebra?x’y’ + x’y = x’(y’ + y) [ Distributive ]= x’ • 1 [ y + y’ = 1 ]= x’ [ x • 1 = x ]Yx’y’ x’yXxy’ xyFebruary 2, 2004 CS 231 7More two-variable examples• Another example expression is x’y + xy.– Both minterms appear in the right side, where y is uncomplemented.– Thus, we can reduce x’y + xy to just y.• How about x’y’ + x’y + xy?– We have x’y’ + x’y in the top row, corresponding to x’.– There’s also x’y + xy in the right side, corresponding to y.– This whole expression can be reduced to x’ + y.Yx’y’ x’yXxy’ xyYx’y’ x’yXxy’ xyFebruary 2, 2004 CS 231 8A three-variable Karnaugh map• For a three-variable expression with inputs x, y, z, the arrangement of minterms is more tricky:• Another way to label the K-map (use whichever you like):Yx’y’z’ x’y’z x’yz x’yz’X xy’z’ xy’z xyz xyz’ZYm0m1m3m2Xm4m5m7m6ZYZ00 01 11 100 x’y’z’ x’y’z x’yz x’yz’X1 xy’z’ xy’z xyz xyz’YZ00 01 11 100m0m1m3m2X1m4m5m7m6February 2, 2004 CS 231 9Why 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.• “Adjacency” includes wrapping around the left and right sides:• We’ll use this property of adjacent squares to do our simplifications.x’y’z + x’yz= x’z(y’ + y)=x’z• 1=x’zx’y’z’ + xy’z’ + x’yz’ + xyz’= z’(x’y’ + xy’ + x’y + xy)= z’(y’(x’ + x) + y(x’ + x))= z’(y’+y)=z’Yx’y’z’ x’y’z x’yz x’yz’X xy’z’ xy’z xyz xyz’ZYx’y’z’ x’y’z x’yz x’yz’X xy’z’ xy’z xyz xyz’ZFebruary 2, 2004 CS 231 10Example K-map simplification • Let’s consider simplifying f(x,y,z) = xy + y’z + xz.• First, you should convert the expression into a sum of minterms form, if it’s 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)000 0001 1010 0011 0100 0101 1110 1111 1f(x,y,z) = x’y’z + xy’z + xyz’ + xyz= m1+ m5+ m6+ m7February 2, 2004 CS 231 11Unsimplifying 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 it’s
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