1. Chapter 3 The Karnaugh Map

2. These K-maps are described by the location. Next, each square will be 1 or 0 depending on the value of the function. ABfm 00?0 01?1 10?2 11?3

3. A three variable function can be configured multiple ways. The K-maps are described by the location. Next, each square will be 1 or 0 depending on the value of the function. ABCfm 000?0 001?1 010?2 011?3 100?4 101?5 110?6 111?7

4. Gray code: only 1 variable changes Consider what DOES NOT change

6. An implicant of a function is a product term that can be used in an SOP. Minterm is a product term that includes each variable of the problem, either uncomplemented or complemented. From the point of view of the map, an implicant is a rectangle of 1, 2, 4, 8,... (any power of 2) 1’s. That rectangle may not include any 0’s. The implicants of F are Minterms – groups of 1Groups of 2Groups of 4 A´B´C´D´A´CDCD A´B´CDBCD A´BCDACD ABC´D´B´CD ABC´DABC´ ABCDABD AB´CD

7. A prime implicant is an implicant that (from the point of view of the map) is not fully contained in any one other implicant. Each set of circles is a prime implicant. An essential prime implicant is a prime implicant that includes at least one 1 that is not included in any other prime implicant. ABD is not essential, all other circles are. f=∑m(0, 3, 7, 11, 12, 13, 15) Consider what DOES NOT change

8. Minimum SOP using K map

9. Map Method 1 1.Find all essential prime implicants. Circle them on the map and mark the minterm(s) that make them essential with an asterisk (*). Do this by examining each 1 on the map that has not already been circled. It is usually quickest to start with the most isolated 1’s, that is, those that have the fewest adjacent squares with 1’s in them. 2.Find enough other prime implicants to cover the function. Do this using two criteria: a. Choose a prime implicant that covers as many new 1’s (that is, those not already covered by a chosen prime implicant). b. Avoid leaving isolated uncovered 1’s.

10. minimumall prime implicants f = y´z´ + wyz + w´xz Essential, can’t cover the 1 some other way. Consider what DOES NOT change

11. x´yz´ + x´yz + xy´z´ + xy´z + xyz x´ y + x y´ + x zx´ y + x y´ + y z Given: function Make: k map Minimize function

12. Some not usedminimum G = A´BC´ + A´CD + ABC + AC´D Given: function Make: k map Minimize function Essential PI *, not covered by any other

13. g = xz + wz + w´yz´ + wx´y g = xz + wz + w´yz´ + x´yz´ g = xz + wz + x´yz´ + w´xy More than one way to skin a cat. All these are minimum. All will give same result

14. F = A´C´D´ + AC´D + A´CD + ACD´ + B´D´ + AB´ - this is the vertical 4 F = A´C´D´ + AC´D + A´CD + ACD´ + B´D´ + B´C - this is the corner 4 F = A´C´D´ + AC´D + A´CD + ACD´ + AB´ + B´C – this is 2 by 2 All variations have this in common – the grayed area F = A´C´D´ + AC´D + A´CD + ACD´ + …

15. Map Method 2 1.Circle all of the prime implicants. 2.Select all essential prime implicants; they are easily identified by finding 1’s that have only been circled once. 3.Then choose enough of the other prime implicants (as in Method 1). Of course, these prime implicants have already been identified in step 1.

16. G= Σm(0,1,3,7,8,11,12,13,15) All PI circledA minimum Another minimum

17. f = a´c´d´ + bc´d + acd + b´cd´f = a´b´d´ + a´bc´ + abd + ab´c No essential PI

18. A prime implicant is a rectangle of 1, 2, 4, 8, … 1’s or X’s not included in any one larger rectangle. Thus, from the point of view of finding prime implicants, X’s (don’t cares) are treated as 1’s. An essential prime implicant is a prime implicant that covers at least one 1 not covered by any other prime implicant (as always). Don’t cares (X’s) do not make a prime implicant essential.

19. minimumother p.i.s F = BD + A´C´D + AB´C Consider what DOES NOT change F=∑m(1,7,10,11,13) + ∑d(5,8,15)

20. g 1 = c´d´ + ab + b´d´ + a´cd g 2 = c´d´ + ab + b´d´ + a´b´c Only essential PI: c’d’ Other PI used in all solutions: ab g3 = c´d´ + ab + ad´ + a´b´c Use b’d’ – the corners Use ad’ – rt top & bot

21. Finding a minimum product of sums expression requires no new theory. The following approach is the simplest: 1.Map the complement of the function. (If there is already a map for the function, replace all 0’s by 1’s, all 1’s by 0’s and leave X’s unchanged.) 2.Find the minimum sum of products expression for the complement of the function (using the techniques of the last two sections). 3.Use DeMorgan’s theorem (P11) to complement that expression, producing a product of sums expression.

22. f = (a´ + c)(a + c´)(a´ + b´ + d´) f = (a´ + c)(a + c´)(b´ + c´ + d´) f =Σm(0,1,4,5,10,11,14)f ’=Σm(2,3,6,7,8,9,12,13,15) f ´ = ac´ + a´c + abd f ´ = ac´ + a´c + bcd f = ∏m(2,3,6,7,8,9,12,13,15) f = a´c´ + ab´c + acd´ Minimum Product of Sum Use DeMorgan

23. Finding a minimum product of sums expression can equally be done using the zero locations : 1.Circle the zeros. 2.Each literal is complemented. 3.Write in POS form. F = (a’ + c)(a + c’)(a’ + b’ + d’) 0 Minimum Product of Sum

24. 5 variable maps Use 2 layers Tedious Still use grey – reflective code Decimal number of grid

25. 5 variable maps Use 2 layers Adjacent

26. 6 variable maps Use 4 layers Very Tedious Forget It!

27. F = A´B´C + A´BE + AB´C´E´ + ABCD´E´ + BDE

28. F = A´C´E´ + ABCD + CD´E + BCE + B´C´DE´ + A´CD´ F = A´C´E´ + ABCD + CD´E + BCE + B´C´DE´ + A´D´E´

29. Multiple functions can be covered simultaneously Can keep separate – that is OK. May be able to save a few gates by combining systems. More important when we get to memory devices.

30. F = A´C´ + ABG = A´C + AB Two functions F & G on separate K-maps Note AB term is common So can use that gate for both.

31. Two functions F & G Note AB term is common So can use that gate for both. Common gate Separate functions, separate gates

32. F = A´B´ + ABC´G = A´B + BC´ F = Σm(0,1,6)G= Σm(2,3,6) Two functions Treat separately Two functions Use common terms F = A´B´ + ABC´G = A´B + ABC´ Which has the least cost? The least number of gates?

33. F = A´C + A´C´D´ + AB´CG = AC´ + A´C´D´ + AB´C F=∑m(0,2,3,4,6,7,10,11) G=∑m(0,4,8,9,10,11,12,13) Circle the essential PI for each function. Then find the common terms. G can be circled many ways.

34. F = A´C + A´C´D´ + AB´CG = AC´ + A´C´D´ + AB´C Build the function with logic symbols. Really see the benefit of sharing functions. Again, that will be more obvious with memory devices.

35. F = B´C + AB´C´ + A´BD + ABD G = C + A´BD H = BC + AB´C + ABD Three functions (3.38) Look for common terms