tabular representation

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Tabular Representation

Based on:

Ryszard Janicki, David L. Parnas, and Jeffery Zucker.Tabular Representations in Relational Documents. Relational Methods in Computer Science, Springer-Verlag, 1996.

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222

Outline

• Motivation• Simple tabular notation (raw table)• Full tabular notation (well-done table)

– Cell connection graph– Table predicate rule– Table relation rule

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333

Motivation---Modeling Requirements

• Monitored vs. controlled environmental quantities– Abstracted to mathematical variables whose values changed

over time, i.e., time function.– Monitored: to be measured by the system (mi)– Controlled: to be controlled by the system (ci)

• Requirements documented with two relations– NAT: describes the environment

• dom (NAT): set of vectors of time-function m’s• ran (NAT): set of vectors of time-function c’s• (m, c) NAT iff the environment allows

– REQ: describes the effect of the system• dom (REQ): set of vectors of time-function m’s• ran (REQ): set of vectors of time-function c’s considered permissible• (m, c) REQ iff the system should permit

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444

Motivation---Modeling Requirements

• Monitored vs. controlled environmental quantities– Abstracted to mathematical variables whose values changed

over time, i.e., time function.– Monitored: to be measured by the system (mi)– Controlled: to be controlled by the system (ci)

• Requirements documented with two relations– NAT: describes the environment

• dom (NAT): set of vectors of time-function m’s• ran (NAT): set of vectors of time-function c’s• (m, c) NAT iff the environment allows

– REQ: describes the effect of the system• dom (NAT): set of vectors of time-function m’s• ran (NAT): set of vectors of time-function c’s considered permissible• (m, c) NAT iff the system should permit

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Requirements can be documented as mathematical relations or functions!

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Motivation

• Describe the following function in Z and OCL

f(x,y) = 0 if x 0 y = 10

x if x < 0 y = 10

y2 if x 0 y > 10

-y2 if x 0 y < 10

x + y if x < 0 y > 10

x – y if x < 0 y < 10

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Motivation

• Describe the following function in Z and OCL

f(x,y) = 0 if x 0 y = 10

x if x < 0 y = 10

y2 if x 0 y > 10

-y2 if x 0 y < 10

x + y if x < 0 y > 10

x – y if x < 0 y < 10

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In Z

f: Z Z Z

f(x,y) = 0 if x 0 y = 10

x if x < 0 y = 10

y2 if x 0 y > 10

-y2 if x 0 y < 10

x + y if x < 0 y > 10

x – y if x < 0 y < 10

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In Z

f: Z Z Z

x: Z, y: Z

(x 0 y = 10 f(x,y) = 0) (x < 0 y = 10 f(x,y) = x) (x 0 y > 10 f(x,y) = y2 ) (x 0 y < 10 f(x,y) = -y2) (x < 0 y > 10 f(x,y) = x + y) (x < 0 y < 10 f(x,y) = x – y)

f(x,y) = 0 if x 0 y = 10

x if x < 0 y = 10

y2 if x 0 y > 10

-y2 if x 0 y < 10

x + y if x < 0 y > 10

x – y if x < 0 y < 10

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In OCL

context Math::f(x: Integer, y: Integer): Integer

pre: true

post: result = if x >= 0 and y = 10 then 0 else

if x < 0 and y = 10 then x else

if x >= 0 and y > 10 then y*y else

if x >= 0 and y < 10 then –y*y else

if x < 0 and y > 10 then x + y else

if x < 0 and y < 10 then x – y else 0

endif endif endif endif endif endif

f(x,y) = 0 if x 0 y = 10

x if x < 0 y = 10

y2 if x 0 y > 10

-y2 if x 0 y < 10

x + y if x < 0 y > 10

x – y if x < 0 y < 10

1010

In OCL

context Math::f(x: Integer, y: Integer): Integer

pre: true

post: result = if x >= 0 and y = 10 then 0 else

if x < 0 and y = 10 then x else

if x >= 0 and y > 10 then y*y else

if x >= 0 and y < 10 then –y*y else

if x < 0 and y > 10 then x + y else

if x < 0 and y < 10 then x – y else 0

endif endif endif endif endif endif

f(x,y) = 0 if x 0 y = 10

x if x < 0 y = 10

y2 if x 0 y > 10

-y2 if x 0 y < 10

x + y if x < 0 y > 10

x – y if x < 0 y < 10

Not very readable or checkable!

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Use a Table?

y = 10 y > 10 y < 10

x 0

x < 0

f(x,y) = 0 if x 0 y = 10

x if x < 0 y = 10

y2 if x 0 y > 10

-y2 if x 0 y < 10

x + y if x < 0 y > 10

x – y if x < 0 y < 10

x y2 x + y

x -y2 x - y

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Tabular RepresentationElements• Header: an indexed set of cells, H = {hi | i I}, where I = {1,2, …, k}• Grid indexed by headers H1, …, Hn, with Hj = {hi

j | i Ij}, j = 1,.., n: an indexed set of cells G = {g | g I}, where I = I1 … In

Raw table skeleton• A collection of headers plus a grid indexed by this collection

h11 h2

1 h31

h12

h22

g11 g21 g31

g12 g22 g32

H1 = {hi1 | i = 1, 2, 3}

H2 = {hi2 | i = 1, 2}

G = {gij | i = 1, 2, 3 and j = 1, 2}

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Why Tabular Representations of Relations?

• Conventional math descriptions– Too complex to parse to be really useful– Lengthy and hard to read and understand

• Digital system– Not continuous <-> continuous function of

analog– Domain and range: tuple of distinct types

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ExerciseUsing the tabular notation, specify a program that reserves a golf tee time. • The standard green fee is $65 on weekdays (Monday-Friday) and $80 on

weekend (Saturday and Sunday). • However, an El Paso resident pays a reduced green fee of $45 and $60 on

weekdays and weekend, respectively. • A senior (of age 60+) pays only $40 and $50 on weekdays and weekend,

respectively.• A junior (of age <17) pays only $20 and $30 on weekdays and weekend,

respectively.

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Sample Solution

weekdays weekend

age < 17 $20 $30

17 age < 60

and resident

$45 $60

17 age < 60

and non-resident

$65 $80

age 60 $40 $50

Q: Nested tables for resident/non-resident (thanks to Elsa)?

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Exercise

• Specify the following function.

g(x,y) = x + y if (x < 0 y 0) (x < y y < 0)

x - y if (0 x < y y 0) (y x < 0 y < 0)

y - x if (x y y 0) (x 0 y < 0)

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Inverted Table

• A header specifies the output of the function.

x + y x - y y - x

y 0

y < 0

x < 0 0 x < y x y

x < y y x < 0 x 0

H1

H2 G

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Cell Connection Graph (CCG)

• Characterizes information flow, i.e., where do I start reading the table and where do I get the result?

• A relation interpreted as an acyclic directed graph– Each arch must either start from or end at the grid G.

H1

H2 H3G

H1

H2 H3G

H1

H2 H3G

H1

H2 H3G

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Cell Connection Graph (CCG)

• Characterizes information flow, i.e., where do I start reading the table and where do I get the result?

• A relation interpreted as an acyclic directed graph– Each arch must either start from or end at the grid G.

H1

H2 H3G

H1

H2 H3G

H1

H2 H3G

H1

H2 H3G

But, how the domain and values of the relation specified are determined? E.g., how to combine the cells?

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Well-Done Table Skeleton

• Table skeleton with– Table predicate rule, PT specifying the domain

– Table relation rule, RT specifying the relation

H1

H2G

H1 H2

G

PT(H1,H2) = H1 H2

RT(G) = G

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Exercise

• Write the PT and RT of the following table and explain how to interpret the table.

H1

H2G

G

H1 H2

H1

H2G

H1

H2G

H1

H2

2222

Examplef(x,y) = 0 if x 0 y = 10

x if x < 0 y = 10

y2 if x 0 y > 10

-y2 if x 0 y < 10

x + y if x < 0 y > 10

x – y if x < 0 y < 10

y = 10 y > 10 y < 10

x 0

x < 0

x y2 x + y

x -y2 x - y

H1 H2

G

G

H1

H2

2323

ExerciseUsing the full tabular notation, specify a program that reserves a golf tee time. • The standard green fee is $65 on weekdays (Monday-Friday) and $80 on

weekend (Saturday and Sunday). • However, an El Paso resident pays a reduced green fee of $45 and $60 on

weekdays and weekend, respectively. • A senior (of age 60+) pays only $40 and $50 on weekdays and weekend,

respectively.• A junior (of age <17) pays only $20 and $30 on weekdays and weekend,

respectively.

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Exercise1. Using Z and the tabular notation, specify a program that hires an

employee for a company. The program takes an employee’s name, gender, SSN, job position, and salary, and adds the employee to the company’s employee database. It should detect when some of the employee’s information is missing or invalid:– when the gender is not specified– when the job position is not specified– when the salary is less than 0– when there already exists an employee with the same SSN

2. Compare the Z specification and the tabular specification. Is there any significant difference, and if so, which is better and why?

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Solution

emps’ = emps {e} r = ok

emps’ = emps r {s_err, g_err}

emps’ = emps r {n_err, p_err}

emps’ = emps r ALL_ERR

e.gener e.gender =

e.ssn ok

e.ssn not ok

Assume a state variable emps, an input e and an output r. ALL_ERR denotes {n_err, s_err, g_err, p_err}.

e.pos

e.pos =

e.salary > 0 e.salary 0H1 H2 H3 H4

G

H1

H4

H3

H2 G

Incomplete

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Solution

ssn ok pos ok

ssn ok (pos ok)

(ssn ok) pos ok

(ssn ok) (pos ok)

salary > 0 gender ok

salary < 0 (gender ok)

salary 0 gender ok

salary 0 (gender ok)

H1 H2

G

H1

H2 G

es’ = es {e} r = ok

es’ = es r {g_err}

es’ = es r {s_err}

es’ = es r {g_err, s_err}

es’ = es r {p_err}

es’ = es r {p_err, g_err}

es’ = es r {p_err, s_err}

es’ = es r {p_err, g_err,

s_err}

es’ = es r {n_err}

es’ = es r {n_err, g_err}

es’ = es r {n_err, s_err}

es’ = es r {n_err, g_err,

s_err}

es’ = es r {n_err, p_err}

es’ = es r {n_err, p_err,

g_err}

es’ = es r {n_err, p_err,

s_err}

es’ = es r {n_err, p_err,

g_err, s_err}