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CS1104: Computer Organisation http://www.comp.nus.edu.sg/~cs1104

CS1104: Computer Organisation http://www.comp.nus.edu.sg/~cs1104. School of Computing National University of Singapore. Lecture 5: Karnaugh Maps. Function Simplification Algebraic Simplification Half Adder Introduction to K-maps Venn Diagrams 2-variable K-maps 3-variable K-maps

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CS1104: Computer Organisation http://www.comp.nus.edu.sg/~cs1104

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  1. CS1104: Computer Organisation http://www.comp.nus.edu.sg/~cs1104 School of Computing National University of Singapore

  2. Lecture 5: Karnaugh Maps • Function Simplification • Algebraic Simplification • Half Adder • Introduction to K-maps • Venn Diagrams • 2-variable K-maps • 3-variable K-maps • 4-variable K-maps • 5-variable and larger K-maps Lecture 5: Karnaugh Maps

  3. Lecture 5: Karnaugh Maps • Simplification using K-maps • Converting to Minterms Form • Simplest SOP Expressions • Getting POS Expressions • Don’t-care Conditions • Review • Examples Lecture 5: Karnaugh Maps

  4. Function Simplification • Why simplify? • Simpler expression uses less logic gates. • Thus: cheaper, less power, faster (sometimes). • Simplification techniques: • Algebraic Simplification. • simplify symbolically using theorems/postulates. • requires skill but extremely open-ended. • Karnaugh Maps. • diagrammatic technique using ‘Venn-like diagram’. • easy for humans (pattern-matching skills). • simplified standard forms. • limited to not more than 6 variables. Function Simplification

  5. Function Simplification • Simplification techniques: • Quine-McCluskey tabulation technique. • tabulation technique based on certain ‘cancellation theorems’. • simplified standard forms. • tedious, repetitive step-by-step technique. • boring to humans BUT suitable for computers. • larger variables possible, but computationally intensive. Function Simplification

  6. Algebraic Simplification • Algebraic simplification aims to minimise (i) number of literals, and (ii) number of terms • But sometimes conflicting. • Let’s aim at reducing the number of literals. Algebraic Simplification

  7. Algebraic Simplification • Example: (x+y).(x+y').(x'+z) (6 literals) = (x.x+x.y'+x.y+y.y').(x'+z) (assoc.) = (x+x.(y'+y)+0).(x'+z) (idemp.,assoc., compl.) = (x+x.(1)+0).(x'+z) (complement) = (x+x+0).(x'+z) (identity 1) = (x).(x'+z) (idemp, identity 0) = (x.x'+x.z) (assoc.) = (0+x.z) (complement) = x.z (identity 0) Number of literals reduced from 6 to 2. Algebraic Simplification

  8. Algebraic Simplification • Find minimal SOP and POS expressions of f(x,y,z) = x'.y.(z + y'.x) + y'.z x'.y.(z+y'.x) + y'.z = x'.y.z + x'.y.y'.x + y'.z (distributivity) = x'.y.z + 0 + y'.z (complement, null element 0) = x'.y.z + y'.z (identity 0) = x'.z + y’.z (absorption) = (x' + y').z (distributivity) Minimal SOP of f = x'.z + y'.z (2 2-input AND gates and 1 2-input OR gate) Minimal POS of f = (x' + y').z (1 2-input OR gate and 1 2-input AND gate) Algebraic Simplification

  9. Algebraic Simplification • Find minimal SOP expression of f(a,b,c,d) = a.b.c + a.b.d + a'.b.c' + c.d + b.d' a.b.c + a.b.d + a'.b.c' + c.d + b.d' = a.b.c + a.b + a'.b.c' + c.d + b.d' (absorption) = a.b.c + a.b + b.c' + c.d + b.d' (absorption) = a.b + b.c' + c.d + b.d' (absorption) = a.b + c.d + b.(c' + d') (distributivity) = a.b + c.d + b.(c.d)' (DeMorgan) = a.b + c.d + b (absorption) = b + c.d (absorption) Number of literals reduced form 13 to 3. Algebraic Simplification

  10. Algebraic Simplification • Difficulty – needs good algebraic manipulation skills. • Advantage – very open-ended (to your desired form!) Algebraic Simplification

  11. S X Half adder (X+Y) C Y Half Adder • Half-Adder is a circuit which adds two single bits (called X,Y) together, to produce a result of two bits (called C, S). • A black-box representation of this circuit is: Truth table representation is: Half Adder

  12. X Y S C Half Adder • In sum-of-minterms forms: C = X.Y S = X'.Y + X.Y' • Algebraic simplification could simplify S to: S = X'.Y + X.Y' = XY • Giving: Half Adder

  13. Introduction to K-maps • Systematic method to obtain simplified sum-of-products (SOPs) Boolean expressions. • Objective: Fewest possible terms/literals. • Diagrammatic technique based on a special form of Venn diagram. • Advantage: Easy with visual aid. • Disadvantage: Limited to 5 or 6 variables. Introduction to K-maps

  14. a'b' ab' ab a'b a b Venn Diagrams • Venn diagram to represent the space of minterms. • Example of 2 variables (4 minterms): Venn Diagrams

  15. a'b' ab' ab a'b a b Venn Diagrams • Each set of minterms represents a Boolean function. Examples: { a.b, a.b' }  a.b + a.b' = a.(b+b') = a { a‘.b, a.b }  a‘.b + a.b = (a'+a).b = b { a.b }  a.b { a.b, a.b', a‘.b }  a.b + a.b' + a‘.b = a + b { }  0 { a‘.b',a.b,a.b',a‘.b }  1 Venn Diagrams

  16. 2-variable K-maps • Karnaugh-map (K-map) is an abstract form of Venn diagram, organised as a matrix of squares, where • each square represents a minterm • adjacent squares always differ by just one literal(so that the unifying theorem may apply: a + a' = 1) • For 2-variable case (e.g.: variables a,b), the map can be drawn as: 2-variable K-maps

  17. Alternative 1: Alternative 2: OR OR a a Alternative 3: a'b' ab' m0 m2 OR b a a b b ab m3 a'b m1 m0 m1 b ab a'b m3 m1 b a m3 m2 a'b' m0 ab' m2 b a'b' a'b a ab ab' 2-variable K-maps • Alternative layouts of a 2-variable (a, b) K-map and others… 2-variable K-maps

  18. equivalent to: a a 1 0 b equivalent to: 0 1 b b 0 1 b a 0 1 a 2-variable K-maps • Equivalent labeling: 2-variable K-maps

  19. C = ab S = ab' + a'b b b 0 0 0 1 a a 1 0 0 1 2-variable K-maps • The K-map for a function is specified by putting • a ‘1’ in the square corresponding to a minterm • a ‘0’ otherwise • For example: Carry and Sum of a half adder. 2-variable K-maps

  20. b bc a 00 01 11 10 b 0 1 a'b'c' a'b'c a'bc a'bc' bc 00 01 11 10 a a ab'c' ab'c abc abc' 0 1 m0 m1 m3 m2 OR c a m4 m5 m7 m6 c Note Gray code sequence 3-variable K-maps • There are 8 minterms for 3 variables (a, b, c). Therefore, there are 8 cells in a 3-variable K-map. Above arrangement ensures that minterms of adjacent cells differ by only ONE literal. (Other arrangements which satisfy this criterion may also be used.) 3-variable K-maps

  21. bc a 00 01 11 10 0 1 m0 m1 m3 m2 m4 m5 m7 m6 3-variable K-maps • There is wrap-around in the K-map: • a'.b'.c' (m0) is adjacent to a'.b.c' (m2) • a.b'.c' (m4) is adjacent to a.b.c' (m6) Each cell in a 3-variable K-map has 3 adjacent neighbours. In general, each cell in an n-variable K-map has n adjacent neighbours. For example, m0 has 3 adjacent neighbours: m1, m2 and m4. 3-variable K-maps

  22. b bc a 00 01 11 10 0 1 1 0 0 1 a 0 1 0 0 c Quick Review Questions (1) Textbook page 104. 5-1. The K-map of a 3-variable function F is shown below. What is the sum-of-minterms expression of F? 5-2. Draw the K-map for this function A: A(x, y, z) = x.y + y.z’ + x’.y’.z Quick Review Questions (1)

  23. y yz 00 01 11 10 wx 00 01 11 10 m0 m1 m3 m2 m4 m5 m7 m6 x m12 m13 m15 m14 w m8 m9 m11 m10 z 4-variable K-maps • There are 16 cells in a 4-variable (w, x, y, z) K-map. 4-variable K-maps

  24. y yz wx m0 m1 m3 m2 m4 m5 m7 m6 x m12 m13 m15 m14 w m8 m9 m11 m10 z 4-variable K-maps • There are 2 wrap-arounds: a horizontal wrap-around and a vertical wrap-around. • Every cell thus has 4 neighbours. For example, the cell corresponding to minterm m0 has neighbours m1, m2, m4 and m8. 4-variable K-maps

  25. 5-variable K-maps • Maps of more than 4 variables are more difficult to use because the geometry (hyper-cube configurations) for combining adjacent squares becomes more involved. • For 5 variables, e.g. vwxyz, need 25 = 32 squares. 5-variable K-maps

  26. v ' v y y yz yz 00 01 11 10 00 01 11 10 wx wx 00 01 11 10 m0 m1 m3 m2 00 01 11 10 m16 m17 m19 m18 m4 m5 m7 m6 m20 m21 m23 m22 x x m12 m13 m15 m14 m28 m29 m31 m30 w w m8 m9 m11 m10 m24 m25 m27 m26 z z 5-variable K-maps • Organised as two 4-variable K-maps: Corresponding squares of each map are adjacent. Can visualise this as being one 4-variable map on TOP of the other 4-variable map. 5-variable K-maps

  27. Larger K-maps • 6-variable K-map is pushing the limit of human “pattern-recognition” capability. • K-maps larger than 6 variables are practically unheard of! • Normally, a 6-variable K-map is organised as four 4-variable K-maps, which are mirrored along two axes. Larger K-maps

  28. w b a‘.b' a‘.b ef 10 11 01 00 cd 00 01 11 10 m18 m19 m17 m16 ef 00 01 11 10 cd m22 m23 m21 m20 00 01 11 10 m0 m1 m3 m2 m30 m31 m29 m28 m4 m5 m7 m6 m26 m27 m25 m24 m12 m13 m15 m14 10 11 01 00 10 11 01 00 m58 m59 m57 m56 m8 m9 m11 m10 m40 m41 m43 m42 m62 m63 m61 m60 m44 m45 m47 m46 a m54 m55 m53 m52 m36 m37 m39 m38 m50 m51 m49 m48 m32 m33 m35 m34 cd 10 11 01 00 cd 00 01 11 10 ef a.b' a.b ef Larger K-maps Try stretch your recognition capability by finding simplest sum-of-products expression for S m(6,8,14,18,23,25,27,29,41,45,57,61). Larger K-maps

  29. Simplification Using K-maps • Based on the Unifying Theorem: A + A' = 1 • In a K-map, each cell containing a ‘1’ corresponds to a minterm of a given function F. • Each group of adjacent cells containing ‘1’ (group must have size in powersof twos: 1, 2, 4, 8, …) then corresponds to a simpler product term of F. • Grouping 2 adjacent squares eliminates 1 variable, grouping 4 squares eliminates 2 variables, grouping 8 squares eliminates 3 variables, and so on. In general, grouping 2n squares eliminates n variables. Simplification Using K-maps

  30. Simplification Using K-maps • Group as many squares as possible. • The larger the group is, the fewer the number of literals in the resulting product term. • Select as few groups as possible to cover all the squares (minterms) of the function. • The fewer the groups, the fewer the number of product terms in the minimized function. Simplification Using K-maps

  31. y yz 00 01 11 10 wx 00 01 11 10 1 1 x 1 1 w 1 1 Simplification Using K-maps • Example: F (w,x,y,z) = w’.x.y'.z' + w'.x.y'.z + w.x'.y.z' + w.x'.y.z + w.x.y.z' + w.x.y.z = m(4, 5, 10, 11, 14, 15) (cells with ‘0’ are not shown for clarity) z Simplification Using K-maps

  32. y yz 00 01 11 10 wx 00 01 11 10 A 1 1 x 1 1 w B 1 1 z Simplification Using K-maps • Each group of adjacent minterms (group size in powers of twos) corresponds to a possible product term of the given function. Simplification Using K-maps

  33. y yz 00 01 11 10 wx 00 01 11 10 A 1 1 x 1 1 w B 1 1 z Simplification Using K-maps • There are 2 groups of minterms: A and B, where: A = w'.x.y'.z' + w‘.x.y'.z = w'.x.y'.(z' + z) = w'.x.y' B = w.x'.y.z' + w.x'.y.z + w.x.y.z' + w.x.y.z = w.x'.y.(z' + z) + w.x.y.(z' + z) = w.x'.y + w.x.y = w.(x'+x).y = w.y Simplification Using K-maps

  34. Simplification Using K-maps • Each product term of a group, w'.x.y' and w.y, represents the sum of minterms in that group. • Boolean function is therefore the sum of product terms (SOP) which represent all groups of the minterms of the function. F(w,x,y,z) = A + B = w'.x.y' + w.y Simplification Using K-maps

  35. Simplification Using K-maps • Larger groups correspond to product terms of fewer literals. In the case of a 4-variable K-map: 1 cell = 4 literals, e.g.: w.x.y.z, w'.x.y'.z 2 cells = 3 literals, e.g.: w.x.y, w.y'.z' 4 cells = 2 literals, e.g.: w.x, x'.y 8 cells = 1 literal, e.g.: w, y', z 16 cells = no literal, e.g.: 1 Simplification Using K-maps

  36. 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 P P P Simplification Using K-maps • Other possible valid groupings of a 4-variable K-map include: Simplification Using K-maps

  37. 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 O O Simplification Using K-maps • Groups of minterms must be (1) rectangular, and (2) have size in powers of 2’s. Otherwise they are invalid groups. Some examples of invalid groups: Simplification Using K-maps

  38. Converting to Minterms Form • The K-map of a function is easily drawn when the function is given in canonical sum-of-products, or sum-of-minterms form. • What if the function is not in sum-of-minterms? • Convert it to sum-of-products (SOP) form. • Expand the SOP expression into sum-of-minterms expression, or fill in the K-map directly based on the SOP expression. Converting to Minterms Form

  39. A AB 00 01 11 10 CD 00 01 11 10 1 1 D 1 1 C B Converting to Minterms Form • Example: f(A,B,C,D) = A(C+D)'(B'+D') + C(B+C'+A'D) = A(C'D')(B'+D') + BC + CC' + A'CD = AB'C'D' + AC'D' + BC + A'CD AB'C'D' + AC'D' + BC + A'CD = AB'C'D' + AC'D'(B+B') + BC + A'CD = AB'C'D' + ABC'D' + AB'C'D' + BC(A+A') + A'CD = AB'C'D' + ABC'D' + ABC + A'BC + A'CD = AB'C'D' + ABC'D' + ABC(D+D') + A'BC(D+D') + A'CD(B+B') = AB'C'D' + ABC'D' + ABCD + ABCD' + A'BCD + A'BCD' + A'B'CD 1 1 1 Converting to Minterms Form

  40. Simplest SOP Expressions • To find the simplest possible sum of products (SOP) expression from a K-map, you need to obtain: • minimum number of literals per product term; and • minimum number of product terms • This is achieved in K-map using • bigger groupings of minterms (prime implicants) where possible; and • no redundant groupings (look foressential prime implicants) Implicant: a product term that could be used to cover minterms of the function. Simplest SOP Expressions

  41. 1 1 1 1 1 1 P 1 O 1 1 1 1 1 Simplest SOP Expressions • A prime implicant is a product term obtained by combining the maximum possible number of minterms from adjacent squares in the map. • Use bigger groupings (prime implicants) where possible. Simplest SOP Expressions

  42. 1 1 1 1 P O 1 1 1 1 1 1 1 1 1 1 1 1 Essential prime implicants Simplest SOP Expressions • No redundant groups: • An essential prime implicant is a prime implicant that includes at least one minterm that is not covered by any other prime implicant. Simplest SOP Expressions

  43. b A bc AB a 00 01 11 10 00 01 11 10 CD 0 1 1 1 0 1 1 1 1 00 01 11 10 1 1 a 0 1 0 0 D 1 1 1 c C 1 1 1 B Quick Review Questions (2) Textbook page 104. 5-3. Identify the prime implicants and the essential prime implicants of the two K-maps below. Quick Review Questions (2)

  44. Simplest SOP Expressions • Algorithm 1 (non optimal): 1. Count the number of adjacencies for each minterm on the K-map. 2. Select an uncovered minterm with the fewest number of adjacencies. Make an arbitrary choice if more than one choice is possible. 3. Generate a prime implicant for this minterm and put it in the cover. If this minterm is covered by more than one prime implicant, select the one that covers the most uncovered minterms. 4. Repeat steps 2 and 3 until all the minterms have been covered. Simplest SOP Expressions

  45. Simplest SOP Expressions • Algorithm 2 (non optimal): 1. Circle all prime implicants on the K-map. 2. Identify and select all essential prime implicants for the cover. 3. Select a minimum subset of the remaining prime implicants to complete the cover, that is, to cover those minterms not covered by the essential prime implicants. Simplest SOP Expressions

  46. All prime implicants A AB 00 01 11 10 CD 1 1 00 01 11 10 1 1 D 1 1 1 C 1 1 B Simplest SOP Expressions • Example: f(A,B,C,D) =  m(2,3,4,5,7,8,10,13,15) Simplest SOP Expressions

  47. A AB 00 01 1110 CD 00 01 11 10 1 1 Essential prime implicants 1 1 D 1 1 1 C 1 1 B A A Minimum cover AB AB 00 01 11 10 00 01 11 10 CD CD 1 1 1 1 00 01 11 10 00 01 11 10 1 1 1 1 D D 1 1 1 1 1 1 C C 1 1 1 1 B B Simplest SOP Expressions Simplest SOP Expressions

  48. A'BC' AB'D' BD A AB A'B'C 00 01 11 10 CD 1 1 00 01 11 10 1 1 D 1 1 1 C 1 1 B Simplest SOP Expressions f(A,B,C,D) = B.D + A'.B'.C + A.B'.D' + A'.B.C' Simplest SOP Expressions

  49. A AB 00 01 11 10 CD 1 00 01 11 10 1 1 1 D 1 1 1 C 1 B Quick Review Questions (3) Textbook page 104. 5-4. Find the simplified expression for G(A,B,C,D). 5-5 to 5-7. Quick Review Questions (3)

  50. A AB 00 01 11 10 CD 1 00 01 11 10 1 0 0 1 1 1 0 D 1 1 0 1 C 0 0 1 1 B Getting POS Expressions • Simplified POS expression can be obtained by grouping the maxterms (i.e. 0s) of given function. • Example: Given F=m(0,1,2,3,5,7,8,9,10,11), we first draw the K-map, then group the maxterms together: Getting POS Expressions

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