1 / 61

Disciplined Software Engineering Lecture #9

Disciplined Software Engineering Lecture #9. Watts S. Humphrey Software Engineering Institute Carnegie Mellon University Pittsburgh, PA 15213 Sponsored by the U.S. Department of Defense. Design Notation - Overview. The importance of notation Boolean algebra Karnaugh maps useful symbols

stacia
Télécharger la présentation

Disciplined Software Engineering Lecture #9

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Disciplined Software Engineering Lecture #9 • Watts S. Humphrey • Software Engineering Institute • Carnegie Mellon University • Pittsburgh, PA 15213 • Sponsored by the U.S. Department of Defense

  2. Design Notation - Overview • The importance of notation • Boolean algebra • Karnaugh maps • useful symbols • using Boolean functions • Simplification • Formal methods

  3. The Importance of Notation • Many software development defects are caused by misunderstandings and confusion. • The notation used to represent program requirements and designs should be • precise and concise • convenient to use • commonly understood • suitable for representing a broad range of software functions • Formal notations meet these criteria.

  4. Class Properties- Mathematical Notation English Symbol Example Union, sum A or B, A+B Intersection, product A and B, A*B, AB Complement, not ' A', not A All, universal 1 None, empty set 0

  5. Boolean Algebra • The algebra of classes • defines and manipulates class relationships • consistency - whether statements are self- consistent • validity - whether statements represent reality • Class representations • Venn diagrams • basic relationships • the universal and empty sets

  6. Venn Diagrams Humans Humans Non-adult Women Women Non- adult Men Men Adult Men Adult Women Universal Class A’B’C’ A’BC’ AB’C’ ABC’ ABC AB’C A’BC A’B’C

  7. Properties of Classes - 1 • Reflexive: for every x, x<=x • Idempotent: xx=x, x+x=x • Antisymmetric: if x<=y and y<=x then x=y • Transitive: if x<=y and y<=z then x<=z • Commutative: xy=yx, x+y=y+x • Associative: x(yz)=(xy)z, x+(y+z)=(x+y)+z • Distributive: x(y+z)=xy+xz, x+yz=(x+y)(x+z)

  8. Properties of Classes - 2 • Empty and universal classes: • xx’=0, x+x’=1, 0<=x<=1 • x0=0, x+1=1, x+0=x, x1=x • Consistency: x<=y, xy=x, x+y=y • are equivalent • Absorption: x(x+y)=x+xy=x • DeMorgan’s theorem: • (x’)’=x, (xy)’=x’+y’, (x+y)’=x’y’

  9. An Example • Suppose you had a switch that you wanted to be on under the following conditions • A or B or C are on • not when A and B are on • not when A and C are on • You could represent this as follows • On = (A+B+C)*[(A*B)+(A*C)]’

  10. Simplifying Boolean Expressions • Consider the following guidelines when you simplify Boolean expressions • if there are any primed parenthetical expressions, use DeMorgan’s theorem to remove them • look for common terms that can be factored by the distributive law • make substitutions to reduce complexity • attempt to reduce expressions to a union of intersections

  11. Simplification Example - 1 • Use the light bulb example • On = (A+B+C)*[(A*B)+(A*C)]’ • Apply DeMorgan’s theorem • On = (A+B+C)*[(A’+B’)*(A’+C’)] • Now, A’ is common to the last terms so • On = (A+B+C)*[A’+(B’*C’)] • Now, substitute x for B+C, and x’ for B’*C’ • On = (A+x)*(A’+x’)

  12. Simplification Example - 2 • While this expression is fully simplified, it should be converted to a union of intersections • On = (A+x)*(A’+x’) • On = [(A+x)*A’]+[(A+x)*x’] • On = [(A*A’)+(x*A’)]+[(A*x’)+(x*x’)] • On = (x*A’)+(A*x’) • Now, replacing x with B+C • On = [(B+C)*A’]+[A*(B’*C’)] • On = (A*B’*C’)+[A’*(B+C)]

  13. Simplification Exercises • Simplify the following expressions • 1. F = X+(X’*Y) • 2. F = (X+Y)+[(X*Z)+Y] • 3. F = (X’*Y’*Z’)+(X*Y*Z’)+(X*Y’*Z’) • 4. F = [X’*(Y+Z’)]’*(X+Y’+Z)*(X’*Y’*Z’)’

  14. Exercise 1 Answer • Starting with F = X+(X’*Y) • Use the distributive law • F = (X+X’)*(X+Y) • Now, with the property of the universal class • X+X’ = 1 AND 1*(X+Y) = X+Y • So the answer is • F = X+Y

  15. Exercise 2 Answer • Starting with F = (X+Y)+[(X*Z)+Y], • Apply the distributive law to give • F = (X+Y)+(X+Y)*(Z+Y) • Now, factor out the common X+Y term • F = (X+Y)*[1+(Z+Y)] • By the property of the universal class, 1+Z+Y=1 • so the answer is • F = X+Y

  16. Exercise 3 Answer • Starting with F = (X’*Y’*Z’)+(X*Y*Z’)+(X*Y’*Z’) • Factor out the common Z’ term • F = Z’*(X’Y’+XY+XY’) • Factoring out the common X term • F = Z’*[X’Y’+X(Y+Y’)] = Z’*(X’Y’+X) • Now, with the distributive law • F = Z’*(X’+X)*(Y’+X) = Z’*(Y’+X)

  17. Exercise 4 Answer • Starting with F = [X’*(Y+Z’)]’*(X+Y’+Z)*(X’*Y’*Z’)’ • First apply DeMorgan’s theorem, giving • F = (X+Y’Z)*(X+Y’+Z)*(X+Y+Z) • Factoring out the common X term gives • F = X+Y’Z*(Y’+Z)*(Y+Z) • Factoring out the common Z term • F = X+Y’Z*(Z+Y’Y) = X+Y’Z

  18. Karnaugh Maps - 1 • For expressions of 6 variables or less, Karnaugh maps can aid the simplification process. • Karnaugh maps are a structured way to represent all possible combinations of the variables in an expression. • Factoring is then done by associating related terms to achieve simplification.

  19. Karnaugh Maps - 2 • Boolean expressions can be represented in minterm form. • The minterms are the products of the variables in all possible combinations. • The minterms for X and Y are: • X’*Y’, X*Y’, X’*Y, XY • There are 2**n minterms for n variables.

  20. Karnaugh Maps - 3 XY’ X’Z YZ YZ X X 00 01 11 10 00 01 11 10 0 0 X X 1 1 X X YZ XY’+X’Z X 00 01 11 10 0 X X 1 X X

  21. A 4-Variable Karnaugh Map YZ WX 00 01 11 10 00 X 01 X X X 11 X X 10

  22. A 5 Variable Karnaugh Map XYZ VW 110 111 101 100 000 001 011 010 00 01 11 10

  23. Map Factoring - 1 • When two adjacent cells both have 1s, then those cells can be factored, eliminating the variable that is different for the two cells. • For example, the 0100 and 0101 cells can be factored, eliminating the variable z. • This is shown on the following map.

  24. Map Factoring - 2 YZ WX 00 01 11 10 00 X 01 X X X 11 X X 10

  25. Map Factoring - 3 • Similarly, when four adjacent cells have 1s, then those cells can be factored, eliminating the two variables that are different for the four cells. • For example, the 0101, 0111, 1101, and 1111 cells can be factored, eliminating the variables w and y. • This is shown on the following map.

  26. Map Factoring - 4 YZ WX 00 01 11 10 00 X 01 X X X 11 X X 10

  27. Map Factoring - 5 • When all three of the possible factorings are made, the resulting expression is • F = w’yz + wxy’ + xz • This is shown on the following map.

  28. Map Factoring - 6 YZ WX 00 01 11 10 00 X 01 X X X 11 X X 10 W’YZ+W’XY’+XZ

  29. Simplification Exercises • Using Karnaugh Maps, simplify the following expressions • 1. F = X+(X’*Y) • 2. F = (X+Y)+[(X*Z)+Y] • 3. F = (X’*Y’*Z’)+(X*Y*Z’)+(X*Y’*Z’) • 4. F = [X’*(Y+Z’)]’*(X+Y’+Z)*(X’*Y’*Z’)’

  30. Exercise 1 Answer YZ The function is F = X+(X’*Y) X 00 01 11 10 0 0 0 1 1 1 1 1 1 1 YZ Factoring this gives F = X+Y X 00 01 11 10 0 0 0 1 1 1 1 1 1 1

  31. Exercise 2 Answer YZ The function is F = (X+Y)+[(X*Z)+Y] = X+Y+XZ+Y = X+Y+XZ X 00 01 11 10 0 0 0 1 1 1 1 1 1 1 YZ Factoring this gives F = X+Y X 00 01 11 10 0 0 0 1 1 1 1 1 1 1

  32. Exercise 3 Answer YZ The function is F = (X’*Y’*Z’)+(X*Y*Z’)+(X*Y’*Z’) = X’Y’Z’+XYZ’+XY’Z’ X 00 01 11 10 0 1 0 0 0 1 1 0 0 1 YZ Factoring this gives F = Y’Z’+XZ’ = Z’(Y’+X) X 00 01 11 10 0 1 0 0 0 1 1 0 0 1

  33. Exercise 4 Answer - 1 X+Y’Z The function is F = [X’*(Y+Z’)]’*(X+Y’+Z)*(X’*Y’*Z’)’ = (X+Y’Z)*(X+Y’+Z)*(X+Y+Z) This can be covered in 3 maps. YZ X 00 01 11 10 0 0 1 0 0 1 1 1 1 1 X+Y’+Z X+Y+Z YZ YZ X 00 01 11 10 X 00 01 11 10 0 0 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1

  34. Exercise 4 Answer - 2 The total function is the product of these maps, or a map with 1s where all 3 maps have 1s and 0s elsewhere. This is YZ X 00 01 11 10 0 0 1 0 0 1 1 1 1 1 Factoring this map gives the answer F = X+Y’Z YZ X 00 01 11 10 0 0 0 1 0 1 1 1 1 1

  35. Factoring Zeros • When factoring 1s, the 1s on a map are covered. This gives the minterm form of the expression. • The same map can be factored by covering the 0s and then applying DeMorgan’s Theorem. • This gives the maxterm form of the expression, or the intersection of unions.

  36. Example of Zeros Factoring The following map could be factored by covering 1s as follows, giving F = X+Y+Z YZ X 00 01 11 10 0 0 1 1 1 1 1 1 1 1 Covering the 0s, however, gives F’ = X’Y’Z’ Applying DeMorgan’s Theorem gives the answer F = X+Y+Z YZ X 00 01 11 10 0 0 1 1 1 1 1 1 1 1

  37. Another Factoring Example The following map could be factored by covering 1s as follows, giving the minterm result F = XY’+X’Y+Z YZ X 00 01 11 10 0 0 1 1 1 1 1 1 1 0 Covering the 0s, however, gives F’ = X’Y’Z’+XYZ’ Applying DeMorgan’s Theorem gives the answer in maxterm form F = (X+Y+Z)*(X’+Y’+Z) = (X+Y)*(X’+Y’)+Z YZ X 00 01 11 10 0 0 1 1 1 1 1 1 1 0

  38. Factoring Out Zeros • Sometimes, a 1s factoring will be complicated by one or two badly placed 0s. This can be handled as follows • Factor the map as if the inconvenient 0s were 1s. • Then, treating each of these factors as a small map, factor out the 0s.

  39. Factoring Out 0s Example - 1 Except for two inconvenient 0s, this map has a very simple factoring F = Y’+W’Y = Y’+W’ YZ WX 00 01 11 10 00 1 0 1 1 Because of the 0 at W’X’Y’Z’ and the 0 at W’XYZ, however, we must factor out 0s. 01 1 1 1 0 11 0 1 1 0 10 1 0 0 1

  40. Factoring Out 0s Example - 1 The first factor, Y’, creates the mini map in the variables W, X, and Z. On this map, W’X’Z’ is factored out: F1 = Y’*(W’X’Z’)’ = Y’*(W+X+Z) This factoring is essentially Y’ and not W’X’Z’. YZ WX 00 01 11 10 00 1 0 1 1 01 1 1 1 0 11 0 1 1 0 The second factoring is F2 = W’Y*(XZ)’ = W’Y*(X’+Z’) So the function is F = Y’*(W+X+Z)+W’Y*(X‘+Z’) 10 1 0 0 1

  41. Factoring With More Variables • Karnaugh maps are possible with up to 6 variables. Beyond that, special tricks are needed. • If the expression involves 6 or fewer variables in most of the terms with an occasional additional variable, an approach much like factoring out 0s can be used. • This method is called exception factoring and can be used with any number of variables.

  42. Exception Factoring - 1 • To factor the following function on a 4-variable Karnaugh map • F = AX’Y’+WXY’+W’Y’Z+XY’Z+W’XY’Z’+WX’Y’ • 1. Construct the function, using the variables • W, X, Y, and Z. • 2. For the term with the A, instead of 1, enter an • A in that square. • 3. In factoring out zeros, treat A as A+0.

  43. Exception Factoring - 2 First, construct the map, using only the terms containing W, X, Y, and Z. Next, enter the A term. Put an A wherever there are 0s and nothing where there is a 1. F1 = WXY’+W’Y’Z+XY’Z+W’XY’Z’+WX’Y’ F2 = AX’Y’ YZ YZ WX WX 00 01 11 10 00 01 11 10 00 00 0 0 1 0 0 A A 0 01 01 1 0 1 0 0 0 0 0 11 11 0 1 1 0 0 0 0 0 10 10 1 0 0 1 A 0 0 A

  44. Exception Factoring - 3 First, factor out the 1s, assuming the A was a 1 F1 = Y’ Next, factor out the A term, treating it as A+0 F = Y’*[A+(W’X’Z’)’] = Y’*(A+W+X+Z) YZ WX 00 01 11 10 00 0 A 1 0 01 1 0 1 0 11 0 1 1 0 10 1 0 0 1

  45. Other Useful Symbols • The following basic mathematical symbols are also useful in defining programming functions. • - means “is a member of the set” • - means “is not a member of the set” • - means “for all members of the set” • - means “there exists a member of the • set”

  46. Notation Examples • To update a file when dataset D is a member of the file • (D File) :: Update • To add a number x(i), when every number i is an even integer • i when (i even) :: add x(i) • To clear a register when any number n is negative • i when (i < 0) :: Clear

  47. Function Completeness • A set of functions is complete if it covers all possible conditions. • For a complete set of functions, the Karnaugh map for the sum (or union) of the functions covers all map positions. • There may be multiply-covered map positions.

  48. Function Orthogonality • A set of functions is orthogonal if no two functions cover the same conditions. • This means that the product (intersection) of any function with any other function is zero. • i, j where (i j) => F(i)*F(j) = 0 • The map may not be completely covered.

  49. Complete and Orthogonal • A set of functions is complete and orthogonal if • the functions cover all elements • no two functions cover the same elements

  50. Complete and Orthogonal Functions f, g, h are complete f, g, h are orthogonal 00 01 11 10 00 01 11 10 0 0 f g gh fh f g 1 1 g fh f fg f h g f, g, h are complete and orthogonal 00 01 11 10 0 f f g h 1 g h h f

More Related