Path products and Regular expressions, Logic Based Testing

Paths, Path products and Regular expressions: path products & path expression, reduction procedure, applications, regular expressions & flow anomaly detection.
Logic Based Testing: overview, decision tables, path expressions, kv charts, specifications.

1. Software Testing Methodologies
Presented by:
Dr. B.Rajalingam
Associate Professor & HOD
Department of Artificial Intelligence and Data Science (AI&DS)
St. Martin's Engineering College
UNIT 3
Path products and Regular expressions

2. SOFTWARE TESTING METHODOLOGIES
Prerequisites
•1. A course on “Software Engineering”
Course Objectives
To provide knowledge of the concepts in software testing such as testing process, criteria,
strategies, and methodologies.
To develop skills in software test automation and management using latest tools.
Course Outcomes:
•Design and develop the best test strategies in accordance to the development model.
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam 2

3. Syllabus
UNIT - I
Introduction: Purpose of testing, Dichotomies, model for testing, consequences of bugs,
taxonomy of bugs: Flow graphs and Path testing: Basics concepts of path testing, predicates, path
predicates and achievable paths, path sensitizing, path instrumentation, application of path
testing.
UNIT - II
Transaction Flow Testing: transaction flows, transaction flow testing techniques. Dataflow
testing: Basics of dataflow testing, strategies in dataflow testing, application of dataflow testing. :
domains and paths, Nice & ugly domains, domain testing, domains and interfaces testing, domain
and interface testing, domains and testability.
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam 3

4. Syllabus
UNIT - III
Paths, Path products and Regular expressions: path products & path expression, reduction
procedure, applications, regular expressions & flow anomaly detection.
Logic Based Testing: overview, decision tables, path expressions, kv charts, specifications.
UNIT - IV
State, State Graphs and Transition testing: state graphs, good & bad state graphs, state testing,
Testability tips.
UNIT - V
Graph Matrices and Application: Motivational overview, matrix of graph, relations, power of a
matrix, node reduction algorithm, building tools. (Student should be given an exposure to a tool
like JMeter or Win-runner).
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam 4

5. TEXT BOOKS:
1. Software Testing techniques – Baris Beizer, Dreamtech, second edition.
2. Software Testing Tools – Dr. K. V. K. K. Prasad, Dreamtech.
REFERENCE BOOKS:
1. The craft of software testing – Brian Marick, Pearson Education.
2. Software Testing Techniques – SPD(Oreille)
3. Software Testing in the Real World – Edward Kit, Pearson.
4. Effective methods of Software Testing, Perry, John Wiley.
5. Art of Software Testing – Meyers, John Wiley.
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam 5

6. 6
Sub Topic
No’s
Sub Topic name
1 Path Products & expressions: Introduction
2 Path Expression
3 Path Product
4 Reduction Procedure
5 Reduction Procedure: Application
6 Regular Expression
7 Flow anomaly detection
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

7. Paths, Path Products and Regular
Expressions
•Interpret the control flowgraph and identify the path products, path sums and path expressions.
•Identify how the mathematical laws (distributive, associative, commutative etc) holds for the
paths.
•Apply reduction procedure algorithm to a control flowgraph and simplify it into a single path
expression.
•Find the all possible paths (Max. Path Count) of a given flow graph.
•Find the minimum paths required to cover a given flow graph.
•Calculate the probability of paths and understand the need for finding the probabilities.
•Differentiate between Structured and Un-structured flowgraphs.
•Calculate the mean processing time of a routine of a given flowgraph.
•Understand how complimentary operations such as PUSH / POP or GET / RETURN are
interpreted in a flowgraph.
•Identify the limitations of the above approaches.
•Understand the problems due to flow-anomalies and identify whether anomalies exist in the
given path expression.
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam 7

8. Path Products & Expressions
8
PURPOSE: APPLICATIONS
1. 1. How many paths in a flow-graph? Maximum, minimum etc.
2. The probability of getting to a point in a program ?(to a node in a flow graph)
3. The mean processing time of a routine (a flow graph)
4. Effect of Routines involving complementary operations :(Push / Pop & Get /
Return)
5. Check for data flow anomalies.
6. Regular expressions are applied to problems in test design & debugging.
7. Electronics engineers use flow graphs to design & analyze circuits & logic
designers.
8. Software development, testing & debugging tools use flow graph analysis
tools & techniques.
9. These are helpful for test tool builders.
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

9. Path Products & Expressions - MotivationMotivation
9

1.Flow graph is an abstract representation of a program.
2.A question on a program can be mapped on to an equivalent question
on an appropriate flow graph.
3.It will be a foundation for syntax testing & state testing
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

10. Path Products
•Normally flow graphs used to denote only control flow connectivity.
•The simplest weight we can give to a link is a name.
•Using link names as weights, we then convert the graphical flow graph into an
equivalent algebraic like expressions which denotes the set of all possible paths from
entry to exit for the flow graph.
•Every link of a graph can be given a name.
•The link name will be denoted by lower case italic letters.
•In tracing a path or path segment through a flow graph, you traverse a succession of
link names.
•The name of the path or path segment that corresponds to those links is expressed
naturally by concatenating those link names.
•For example, if you traverse links a,b,c and d along some path, the name for that path
segment is abcd. This path name is also called a path product.
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam 10

11. Path Products & Expressions - DefinitionsDefinitions
11
Path Expression:Path Expression:
An algebraic representation of sets of paths in a flow graph.
Regular Expression:
Path expressions converted by using arithmetic laws & weights into an
Algebraic function.
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

12. Path Product
12
•Annotate each link with a name.
•The pathname as you traverse a path (segment) expressed as
concatenation of the link names is the path productpath product.
•Examples of path products between 1 & 4 are:
a b d a b c b d a b c b c b d ….
2 31
b
c
a
4
d
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

13. Path Expression
13
Path ExpressionPath Expression
Simply: Derive using path products.
Example:
{ a b d, a b c b d, a b c b c b d , ….. }
abd + abcbd + abcbcbd + ….
2 31
b
c
a
4
d
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

14. Path Expression
14
Example:
a b
c d
h
fe
g
j
i
{ abcd , abfhebcd , abfigebcd , abfijd }
abcd + abfhebcd + abfigebcd + abfijd
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

15. Path Expression
15
Path name for two successive path segments is the concatenation of
their path products.
X = abcY = def XY = abcdef
a X = aabc X a = abcaXaX = abcaabc
X = ab + cd Y = ef + gh XY = abef + abgh + cdef + cdgh
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

16. Path Segments & Products
16
Loops:
a
1 = a a
2 = aa a
3 = aaa a
n = aaaaa … n
times
X = abc X
1 = abc X
2 = (abc)
2 = abcabc
Identity element
a
0 = 1 X
0 = 1 (path of length 0)
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

17. Path Product
17
Denotes a set of paths in parallel between two nodes.
Path Product
•Not Commutative:
XY ≠ YX in general
•Associative
A ( BC ) = ( AB ) C = ABC : Rule 1
•Commutative
X + Y = Y + X : Rule 2
•Associative
( X + Y ) + Z = X + ( Y + Z ) = X + Y + Z : Rule 3
•Distributive
A ( B + C) = A B + A C : Rule 4
( A + B ) C = A C + B C
•Absorption
X + X = X : Rule 5
X + any subset of X = X
X = a + bc + abcdX + a = X + bc + abcd = X
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

18. Path Products
18
•Loop:
An infinite set of parallel paths.
b
* = b
0 + b
1 + b
2 + b
3 + ……
X
* = X
0 + X
1 + X
2 + X
3 + ……
X
+ = X
1 + X
2 + X
3 + ……
•X X* = X* X = X
+
 a a* = a* a = a
+
X
n
= X
0 + X
1 + X
2 + X
3 + …… + X
n
a c
b
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

19. Path products
19
More Rules…
X
m
+ X
n
= X
n if n ≥ m : Rule 6
= X
m if n < m
X
m
X
n
= X
m+n : Rule 7
X
n
X
* = X
* X
n
= X
* : Rule 8
X
n
X
+ = X
+ X
n
= X
+
: Rule 9
X
* X
+ = X
+ X
* = X
+
: Rule 10
Identity Elements ..
1 : Path of Zero Length
1 + 1 = 1 : Rule 11
1 X

=

X
1 = X : Rule 12
1
n
=
1
n
= 1
* = 1
+
= 1
: Rule 13
1
+
+ 1
=
1
* = 1
: Rule 14
Identity Elements ..
0 : empty set of paths
X + 0 = 0
+
X = X
: Rule 15
X 0 = 0
X = 0
: Rule 16
0
* = 1 + 0 + 0
2 + 0
3 + . . . = 1: Rule 17
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

20. Reduction Procedure
20
 To convert a flow graph into a path expression that
denotes the set of all entry/exit paths.
 Node by Node Reduction Procedure
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

21. Reduction Procedure
21
Initialization Steps:
1.Combine all serial links by multiplying their path expressions.
2.Combine all parallel links by adding their path expressions.
3.Remove all self-loops - replace with links of the form X*
4.Select a non-initial & non-final node. Replace it with a set of equivalent links,
whose path expressions correspond to all the ways you can form a product of the set
of in-links with the set of out-links of that node.
5.Combine any serial links by multiplying their path expressions. ( as in step 1)
6.Combine any parallel links by adding their path expressions.( as in step 2)
7.Remove all the self-loops.( as in step 3)
8.IF there’s just one node between entry & exit nodes, path expression for the flow
graph is the link’s path expression.ELSE, return to step 4.
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam
1 2 3
5
4
a b
f
c
g
ed

22. Reduction Procedure
22
Path Expression for a Flow Graph
•Is not unique
•Depends on the order of node removal.
Cross-Term Step (Step 4 of the algorithm)
•Fundamental step.
•Removes nodes one by one till there’s one entry & one exit node.
•Replace the node by path products of all in-links with all out-links and
interconnecting its immediate neighbors.
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

23. Reduction Procedure Example
23
Processing of Loop Terms:
b
1 2
ca
d
1 2
b*ca
b*d
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

24. Reduction Procedure Example
24
Processing of Loop Terms:Processing of Loop Terms:
1 2 3 45
b c fa
e
d
bd
1 2 45
bc fa
e
1 2 4 5
(bd)* bc fa
e
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

25. Reduction Procedure Example
25
Processing of Loop Terms:Processing of Loop Terms:
1 2 4 5
(bd)* bc fa
e
1 45
a (bd)* bc f
e (bd)* bc
1 4 5
a (bd)* bc ( e (bd)* bc ) * f
a (bd)* bc ( e (bd)* bc )* f
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

26. Reduction Procedure Example
26
Example:Example:
1 3 4 5 6
7 8 9
f
b c da
g
j
m
k
h
2
e
10
i
l
1 3 4 5 6
7 8 9
f
b c da
g
j
im
k
h
2
e
il
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

27. Reduction Procedure Example
27
1 3 4 5 6
bc+gkh
da
gif
imf
2
e
ilh
1 3 4 5 6
8
jf
b c da
g
imf
kh
2
e
ilh
1 3 4 5 6
7 8
f
b c da
g
j
im
kh
2
e
ilh
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

28. Reduction Procedure Example
28
1 3 5 6
b(c+gkh)da
bgif
imf
2
e
ilh
1 3 5 6
(bgif)*b(c+gkh)
da
imf
2
e
ilh
1 3 6
(bgif)*b(c+gkh)da
imf
2
e
ilhd
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

29. Reduction Procedure Example
29
1 3 6
(bgif)*b(c+gkh)da
(ilhd)*imf
2
(ilhd)*e
1 6
a(bgif)*b(c+gkh)d
(ilhd)*imf(bgif)*b(c+gkh)d
2
(ilhd)*e
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

30. Reduction Procedure Example
30
Flow Graph Path Expression :Flow Graph Path Expression :
a(bgif)*b(c+gkh)d {(ilhd)*imf(bgif)*b(c+gkh)d}* (ilhd)*e
1 6
a(bgif)*b(c+gkh)d
2
{(ilhd)*imf(bgif)*b(c+gkh)d}* (ilhd)*e
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

31. Before going into Applications
31
Before that, we learn:
Identities
Structured Flow Graphs (code/routines)
Unstructured Flow graphs (routines)
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

32. Identities / Rules
32
( A + B ) * =( A* + B* ) * : I1
=( A* B* )* : I2
=( A* B )* A* : I3
=( B* A )* B* : I4
=( A* B + A )* : I5
=( B* A + B )* : I6
( A + B + C + . . . ) * =( A* + B* + C* + . . . )*: I7
=( A* B* C* . . . )*: I8
Derived by removing nodes in different orders & applying the series-parallel-loop rules.
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

33. Structured Flow Graphs
33
Reducible to a single link by successive application of the transformations shown
below.
A B A, B
Process
A B
WHILE .. DO ..
A B A
IF THEN .. ELSE ..
B
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam
A B A, B
REPEAT .. UNTIL ..

34. Structured Flow Graphs
34
Properties:Properties:
•No cross-term transformation.
•No GOTOs.
•No entry into or exit from the middle of a loop.
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam
a c
b
d
abcd e
g
h
i
j
e
f Some examples:

35. Unstructured Flow Graphs
35
Some examples – unstructured flow graphs/code:
X
X
Jumping into loops
X
Jumping out of loops
Branching into Decisions
X
Branching out of Decisions
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam

36. Applications
•The purpose of the node removal algorithm is to present one very generalized
concept- the path expression and way of getting it.
•Convert the program or graph into a path expression.
•Identify a property of interest and derive an appropriate set of "arithmetic" rules that
characterizes the property.
•Replace the link names by the link weights for the property of interest.
•The path expression has now been converted to an expression in some algebra, such
as ordinary algebra, regular expressions, or boolean algebra.
•This algebraic expression summarizes the property of interest over the set of all paths.
•Simplify or evaluate the resulting "algebraic" expression to answer the question you
asked.
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam 36

37. How Many Paths in a Flow graph ?
•What is the maximum number of different paths possible?
•What is the fewest number of paths possible?
•How many different paths are there really?
•What is the average number of paths?
•Determining the actual number of different paths is an inherently difficult problem because
there could be unachievable paths resulting from correlated and dependent predicates.
•If we know both of these numbers (maximum and minimum number of possible paths) we have
a good idea of how complete our testing is.
•Asking for "the average number of paths" is meaningless.
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam 37

38. Maximum Path Count Arithmetic
•Label each link with a link weight that corresponds to the number of paths that link represents.
•Also mark each loop with the maximum number of times that loop can be taken.
•If the answer is infinite, you might as well stop the analysis because it is clear that the
maximum number of paths will be infinite.
•There are three cases of interest: parallel links, serial links, and loops
•This arithmetic is an ordinary algebra.
•The weight is the number of paths in each set.
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam 38

39. Lower Path Count Arithmetic
A lower bound on the number of paths in a routine can be
approximated for structured flow graphs.
The arithmetic is as follows:
The values of the weights are the number of members in a
set of paths.
EXAMPLE:
Applying the arithmetic to the earlier example gives us the
identical steps unit l step 3 (C) as below
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam 39

40. Lower Path Count Arithmetic
•From Step 4, the it would be different from the previous
example:
•If you observe the original graph, it takes at least two paths
to cover and that it can be done in two paths.
•If you have fewer paths in your test plan than this
minimum you probably haven't covered. It's another check.
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam 40

41. Calculating the Probability
•Path selection should be biased toward the low - rather than the high-probability paths.
•This raises an interesting question:
•What is the probability of being at a certain point in a routine?
•This question can be answered under suitable assumptions, primarily that all probabilities
involved are independent, which is to say that all decisions are independent and uncorrelated.
•We use the same algorithm as before : node-by-node removal of uninteresting nodes.
•Weights, Notations and Arithmetic:
•Probabilities can come into the act only at decisions (including decisions associated with
loops).
•Annotate each outlink with a weight equal to the probability of going in that direction.
•Evidently, the sum of the outlink probabilities must equal 1
•For a simple loop, if the loop will be taken a mean of N times, the looping probability is N/(N
+ 1) and the probability of not looping is 1/(N + 1).
•A link that is not part of a decision node has a probability of 1.
•The arithmetic rules are those of ordinary arithmetic.
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam 41

42. Mean Processing Time of a Routine
•Given the execution time of all statements or instructions for every link in a flowgraph and
the probability for each direction for all decisions are to find the mean processing time for
the routine as a whole.
•The model has two weights associated with every link: the processing time for that link,
denoted by T, and the probability of that link P.
•The arithmetic rules for calculating the mean time:
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam 42

43. Example
1.Start with the original flow graph annotated with probabilities and processing time.
2.Combine the parallel links of the outer loop.
The result is just the mean of the processing times for the links because there aren't any other links leaving
the first node. Also combine the pair of links at the beginning of the flow graph..
3. Combine as many serial links as you can.
4. Use the cross-term step to eliminate a node and to create the inner self - loop.
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam 43

44. Example
Finally, you can get the mean processing time, by using the arithmetic rules as follows:
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam 44

45. Push/Pop, Get/Return
•This model can be used to answer several different questions that can turn up in debugging.
•It can also help decide which test cases to design.
•Given a pair of complementary operations such as PUSH (the stack) and POP (the stack),
considering the set of all possible paths through the routine, what is the net effect of the
routine? PUSH or POP? How many times? Under what conditions?
•Here are some other examples of complementary operations to which this model applies:
•GET/RETURN a resource block.
•OPEN/CLOSE a file.
•START/STOP a device or process.
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam 45

46. Limitations and Solutions
•The main limitation to these applications is the problem of
unachievable paths.
•The node-by-node reduction procedure, and most graph-theory-based
algorithms work well when all paths are possible, but may provide
misleading results when some paths are unachievable.
•The approach to handling unachievable paths (for any application) is
to partition the graph into sub graphs so that all paths in each of the
sub graphs are achievable.
•The resulting sub graphs may overlap, because one path may be
common to several different sub graphs.
•Each predicate's truth-functional value potentially splits the graph
into two sub graphs. For n predicates, there could be as many as 2n
sub graphs. STM(Unit 3) : Dr. B.Rajalingam 46

47. Regular Expressions and Flow Anomaly
Detection
The Problem:
•The generic flow-anomaly detection problem (note: not just data-flow anomalies, but any flow
anomaly) is that of looking for a specific sequence of options considering all possible paths
through a routine.
•Let the operations be SET and RESET, denoted by s and r respectively, and we want to know if
there is a SET followed immediately a SET or a RESET followed immediately by a RESET
(an
 
ss 
or an 
rr 
sequence).
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam 47

48. Regular Expressions and Flow Anomaly Detection
•A file can be opened (o), closed (c), read (r), or written (w).
•If the file is read or written to after it's been closed, the sequence is nonsensical.
Therefore,
 
cr 
and 
cw 
are anomalous.
•Similarly, if the file is read before it's been written, just after opening, we may have a bug.
Therefore,
 
or 
is also anomalous.
•Furthermore,
 
oo 
and 
cc, though not actual bugs, are a waste of time and therefore should also be
examined.
•A tape transport can do a rewind (d), fast-forward (f), read (r), write (w), stop (p), and skip (k).
•There are rules concerning the use of the transport; for example, you cannot go from rewind to
fast-forward without an intervening stop or from rewind or fast-forward to read or write without
an intervening stop.
•The following sequences are anomalous:
 
df,
 
dr,
 
dw,
 
fd, and
 
fr. Does the flowgraph lead to
anomalous sequences on any path? If so, what sequences and under what circumstances?
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam 48

49. The Method
•Annotate each link in the graph with the appropriate operator or the null operator 1.
•Simplify things to the extent possible, using the fact that a + a = a and 12 = 1.
•You now have a regular expression that denotes all the possible sequences of operators in that
graph. You can now examine that regular expression for the sequences of interest.
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam 49

50. Limitations
•Huang's theorem can be easily generalized to cover sequences of greater length than two
characters.
•Beyond three characters, though, things get complex and this method has probably reached its
utilitarian limit for manual application.
•There are some nice theorems for finding sequences that occur at the beginnings and ends of
strings but no nice algorithms for finding strings buried in an expression.
•Static flow analysis methods can't determine whether a path is or is not achievable. Unless the
flow analysis includes symbolic execution or similar techniques, the impact of unachievable
paths will not be included in the analysis.
•The flow-anomaly application, for example, doesn't tell us that there will be a flow anomaly - it
tells us that if the path is achievable, then there will be a flow anomaly. Such analytical
problems go away, of course, if you take the trouble to design routines for which all paths are
achievable.
October 9, 2025 STM(Unit 3) : Dr. B.Rajalingam 50

51. Sub Topic No’s Sub Topic name
1
Overview
2 Decision Tables
3 Path Expression
4 KV maps
5 Specifications
October 9, 2025 Unit 3: Dr. B.Rajalingam 51

52. Logic based testing: Overview
•Logic is used in a program by programmers.
•Boolean algebra is the way to work with logic – simplification & calculation.
Hardware logic testing –
•hardware logic test design tools and methods use logic & Boolean algebra.
•Hardware design language compilers/translators use logic & Boolean algebra.
•Impact of errors in specifications of a software is high as these are first in and
last out.
•So, higher level language for specs is desired to reduce the number of errors.
•Higher order logic systems are used for formal specifications.
•The tools to simplify, transform and check specs use Boolean algebra.
October 9, 2025 Unit 3: Dr. B.Rajalingam 52

53. Logic based testing: Overview
Knowledge based systems:
•Knowledge based systems and artificial intelligence systems use high
level logic languages which are based on rule bases consisting of rules.
•Rules are predicate expressions containing domain knowledge related
elements combined with logical connectives.
•The answers to queries (problems) are derived based on Boolean
algebraic operations performed on the rule bases.
•Such programs are called inference engines.
October 9, 2025 Unit 3: Dr. B.Rajalingam 53

54. Modeling Logic with Decision Tables
•A matrix representation of the logic of a decision
•Specifies the possible conditions and the resulting actions
•Best used for complicated decision logic.
•Consists of three parts
Condition stubs
Lists condition relevant to decision
Action stubs
Actions that result from a given set of conditions
Rules
Specify which actions are to be followed for a given set of conditions
October 9, 2025 Unit 3: Dr. B.Rajalingam 54

55. Modeling Logic with Decision Tables
•Indifferent Condition
•Condition whose value does not affect which action is taken for two
or more rules
•Standard procedure for creating decision tables
•Name the condition and values each condition can assume
•Name all possible actions that can occur
•List all rules
•Define the actions for each rule
•Simplify the table
October 9, 2025 Unit 3: Dr. B.Rajalingam 55

56. Complete decision table for payroll system
example
9.56
October 9, 2025 Unit 3: Dr. B.Rajalingam 56

57. Constructing a Decision Table
PART 1. Frame The Problem.
•Identify the conditions (decision criteria).
•These are the factors that will influence the decision.
E.g., We want to know the total cost of a student’s tuition.
•What factors are important?
•Identify the range of values for each condition or criteria.
E.g. What are they for each factor identified above?
•Identify all possible actions that can occur.
E.g. What types of calculations would be necessary?
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58. Constructing a Decision Table
PART 2. Create The Table.
•Create a table with 4 quadrants.
•Put the conditions in the upper left quadrant. One row per condition.
•Put the actions in the lower left quadrant. One row per action.
•List all possible rules.
•Alternate values for first condition. Repeat for all values of second condition.
Keep repeating this process for all conditions.
•Put the rules in the upper right quadrant.
•Enter actions for each rule
•In the lower right quadrant, determine what, if any, appropriate actions should
be taken for each rule.
•Reduce table as necessary.
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59. Example
•Calculate the total cost of your tuition this quarter.
•What do you need to know?
•Level. (Undergrad or graduate)
•School. (CTI, Law, etc.)
•Status. (Full or part time)
•Number of hours
•Actions?
•Actions?
•Consider CTI only (to make the problem smaller):
•U/G
•Part Time (1 to 11 hrs.): $335.00/per hour
•Full Time (12 to 18 hrs.): $17,820.00
•* Credit hours over 18 are charged at the part-time rate
•Graduate:
•Part time (1 to 7 hrs.): $520.00/per hour
•Full time (>= 8 hrs.): $520.00/per hour
•Create a decision table for this problem. In my solution I was able to reduce the number of rules from 16 to 4.
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60. 60
Boolean Algebra
A Boolean algebra consists of:
•a set B={0, 1},
•2 binary operations on B (denoted by + & ×),
•a unary operation on B (denoted by '), such
that :
0 + 0 = 0 0 × 0 = 0
1 + 0 = 1 0 × 1 = 0
0 + 1 = 1 1 × 0 = 0
1 + 1 = 1 1 × 1 = 1
0’=1 and 1’=0.
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Rules of a Boolean Algebra
The following axioms (‘rules’) are satisfied for all elements x, y& z of B:
(1) x + y = y + x (commutative axioms)
x× y = y × x
(2) x + (y + z) = (x + y) + z (associative axioms)
x × (y × z) = (x × y) × z
(3) x × (y + z) = (x × y) + (x × z)
x + (y × z) = (x + y) × (x + z) (distributive axioms)
(4) x + 0 = x x × 1 = x (identity axioms)
(5) x + x' = 1 x × x' = 0 (inverse axioms)
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Laws of Boolean Algebra
•In addition to the laws given by the axioms of Boolean Algebra, we can
show the following laws
x'' = x (double complement)
x + x = x x× x = x (idempotent )
(x + y)' = x' × y' (x × y)' = x' + y’ (de Morgan’s laws)
x + 1 = 1 x × 0 = 0 (annihilation)
x + (x × y) = x x× (x + y) = x (absorption)
0' = 1 1' = 0 (complement)
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Exercise
Simplify the Boolean expression
(x' × y) + (x × y)
Solution: (x' × y) + (x × y)
= (y × x') + (y × x) (commutative)
= y × (x' + x) (distributive)
= y × (x + x') (commutative)
= y × 1 (inverse)
= y (identity)
Thus (x' × y) + (x × y) = y
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Boolean Notation
•This means that in effect we’ll be employing Boolean Algebra notation.
•The truth tables can be rewritten as
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Notational Short-cuts
We will employ short-cuts in notation:
(1) In ‘multiplication’ we’ll omit the symbol ×, & write xy for x × y (just as in
ordinary algebra)
(2) The associative law says that
x + (y + z) = (x + y) + z
So we’ll write this as simply x + y + z, because the brackets aren’t necessary.
Similarly, write the product of 3 terms as xyz
(3) In ordinary algebra, the expression x + y × z means x + (y × z), because of the
convention that multiplication takes precedence over addition.
e.g. x + yz means x + (y × z), and not (x + y) × z
Similarly, ab + cd means (a × b) + (c × d)
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Reducing Boolean Expressions
•Is this the smallest possible implementation of this expression? No!
•Use Boolean Algebra rules to reduce complexity while preserving
functionality.
•Step 1: Use idempotent law (a + a = a). So
xyz + xyz’ + x’yz = xyz + xyz + xyz’ + x’yz
•Step 2: Use distributive law
•a(b + c) = ab + ac. So xyz + xyz + xyz’ + x’yz = xy(z + z’) + yz(x + x’)
•Step 3: Use Inverse law (a + a’ = 1). So
xy(z + z’) + yz(x + x’) = xy.1 + yz.1
•Step 4: Use Identity law (a . 1 = a). So
• xy + yz = xy.1 + yz.1 = xyz + xyz’ + x’yz
G = xyz + xyz’ + x’yz
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67. xyF
001
011
100
110
67
Karnaugh maps
•Alternate way of representing Boolean function
•All rows of truth table represented with a square
•Each square represents a min term
01
y
x
0
1
1
00
1
01
y
x
0
1
x’y’
xy’xy
x’y
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Karnaugh maps
•Easy to convert between truth table, K-map, and SOP.
•Unoptimized form: number of 1’s in K-map equals number of
minterms (products) in SOP.
•Optimized form: reduced number of minterms
F(x,y) = x’y + x’y’ = x’
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Karnaugh Maps
•A Karnaugh map is a graphical tool for assisting in the general
simplification procedure.
•Two variable maps.
0
A
10
1
B
01
0
1
F=AB +A’B 0
A
11
1
B
01
0
1
• Three variable maps.
0
A
11
1
0001
0
1
BC
0
11
1
1110
F=AB’C’ +AB C +ABC +ABC  + A’B’C + A’BC’
F=AB +AB +AB 
ABCF
0000
0011
0101
0110
1001
1011
1101
1111
+
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Rules for K-Maps
We can reduce functions by circling 1’s in the K-map.
Each circle represents minterm reduction.
Following circling, we can deduce minimized and-or form.
F(x,y) = x’y + x’y’ = x’

01
y
x
0
1
1
00
1
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Rules for K-Maps
Rules to consider
1.Every cell containing a 1 must be included at least once.
2.The largest possible “power of 2 rectangle” must be enclosed.
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Karnaugh Maps
•A Karnaugh map is a graphical tool for assisting in the
general simplification procedure.
•Two variable maps.
0
A
10
1
B
01
0
1
F=AB +A’B
0
A
11
1
B
01
0
1 F=A+B
•Three variable maps.
F=A+B C +BC 
0
A
11
1
0001
0
1
BC
0
11
1
1110
F=AB +AB +AB 
F=AB’C’ +AB C +ABC +ABC  + A’B’C + A’BC’
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More Karnaugh Map Examples
Examples
g = b'
01
0
1
a
b
c
ab
00011110
0
1
01
0
1
a
b
c
ab
00011110
0
1
01
01
f = a
0010
0111
cout = ab + bc + ac
11
00
0011
0011
f = a
1. Circle the largest groups possible.
2. Group dimensions must be a power of 2.
3. Remember what circling means
October 9, 2025 Unit 3: Dr. B.Rajalingam

74. Specifications
Specification validation procedure / steps
1.Rewrite the specifications using consistent terminology.
2.Identify the predicates on which the cases are based. Name them with suitable letters, such as
A, B, C.
3.Rewrite the specification in English that uses only the logical connectives AND, OR, and NOT,
however stilted it may seem.
4.Convert the rewritten specifications into an equivalent set of Boolean expressions.
5.Identify the default action and cases, if any are specified.
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75. Specifications Continue
6. Enter the Boolean expressions in a KV chart and check for consistency. If the specifications
are consistent, there will be no overlaps, except for the cases that result in multiple actions.
7. Enter the default cases, and check for consistency.
8. If all boxes are covered, the specification is complete.
9. If the specification is incomplete or inconsistent, translate the corresponding boxes of the KV
chart back into English and get a clarification, explanation, or revision.
10. If the default cases were not specified explicitly, translate the default cases back into English
and get a confirmation.
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76. Thank you
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