Applied Mathematics for Database Professionals phần 10 doc

In 3VL, predicates can change their truth values in time, but only from UNKNOWN to TRUE orfrom UNKNOWN to FALSE.. Truth Tables of Three-Valued Logic Tables D-1 through D-3 show the three

■ Note ukasiewicz also introduced the Polish notation, which allows expressions to be written

unambigu-ously without the use of brackets This is the basis of the Reverse Polish Notation (RPN), which many pocket

calculators and expression compilers use

In 3VL, predicates can change their truth values in time, but only from UNKNOWN to TRUE orfrom UNKNOWN to FALSE This is rather slippery, because it means that in 3VL the value UNKNOWN

seems to be somewhat special This description is too vague anyway; we need formal rules,

definitions, and truth tables We’ll look at those details in a minute

3VL is counterintuitive, as opposed to the classical 2-valued logic (2VL) The main reason

is that you lose the tertium non datur (the principle of the excluded middle) That is, if P is

unknown, then ¬P is unknown as well, which in turn means that P ∨ ¬P is unknown

Con-sider the following famous Aristotelian example “Tomorrow a sea battle will take place” is

neither TRUE nor FALSE—it is unknown—yet the sentence “Either tomorrow there will be a sea

battle, or tomorrow there won’t be a sea battle” is certainly true (in reality) However, it is

unknown in 3VL, hence the counterintuitive nature

Another problem arises from the fact that if two predicates are “possible” (that is, theirtruth value could possibly be TRUE), then their conjunction is also “possible.” However, this is

obviously wrong if one of them is a negation of the second Ludwik Stefan Borkowski

(1914–1993) proposed to “fix” this problem by introducing a 4-valued logic (4VL); this indeed

makes the preceding problem disappear, but the solution is unsatisfactory and ad hoc

Furthermore, 3VL does not solve the problem of handling missing information; it does

not allow you to represent the reason why information is missing At the end of this section,

you’ll see that Ted Codd proposed a 4VL in 1990, to distinguish between applicable and

inap-plicable values However, a 4VL doesn’t tell you what to do if you don’t know whether a certain

attribute value is applicable or not; for example, if commission is applicable for sales reps

only, but you don’t know the job of a certain employee, what should you do with the

commis-sion attribute of that employee? One NULL implies 3VL, two NULLs (applicable and inapplicable)

imply 4VL, n NULLs imply (n+2)-valued logic Using the technique of full induction, you can

prove that you’ll end up with an infinite number of (meanings of ) NULLs, and a corresponding

infinite number of truth values To repeat part of this paragraph’s opening sentence: 3VL does

not solve the problem of handling missing information

Even worse, 3VL is not only counterintuitive in itself; it is ill-implemented in the SQLstandard Probably one of the biggest blunders in the SQL standard in this respect is that in

an attribute of the BOOLEAN data type, a NULL represents the logical value UNKNOWN Wasn’t a NULL

supposed to represent the fact that you don’t know the value? There are more examples of

such mistakes in the SQL standard

You’ll have to revisit all tautologies (and therefore all rewrite rules) from 2VL to checkwhether they’re still valid in 3VL; you’ll find out that several tautologies don’t hold anymore

Actually, you could even argue that we should use different names and symbols for theconnectives in 3VL; after all, they are different from the “corresponding” connectives in 2VL

Truth Tables of Three-Valued Logic

Tables D-1 through D-3 show the three standard truth tables for the negation, conjunction,

and disjunction connectives in 3VL

Table D-1.Three-Valued Truth Table for NOT (Negation)

As you can see, P ∧ Q is only TRUE if both P and Q are TRUE The result is FALSE if at least one

of the operands is FALSE, and the three remaining combinations result in UNKNOWN

Missing Operators

You almost never see a truth table for the implication and the equivalence in 3VL The most

likely reason for the absence of the implication is that it is tough to come up with a good

definition It turns out that in 3VL, the implication connective is not definable in terms of

conjunction and negation, or disjunction and negation Remember the following implication

rewrite rule from 2VL:

( P ⇒ Q ) ⇔ (¬P ∨ Q )

If you’d define the implication in 3VL according to this rewrite rule, you’d end up in thesituation that P ⇒ P would evaluate to UNKNOWN if P is UNKNOWN; obviously, in any logic you

would like this to be TRUE The last row in Table D-4 shows the problem

and Equivalence Connectives

ukasiewicz proposed a truth table for the implication where the two U? values in the last row

are both a T instead—so he “fixed” the problem by sacrificing a tautology (the implication

rewrite rule)

Three-Valued Logic, Tautologies, and Rewrite Rules

If you want to use any of the rewrite rules we identified in 2VL, you’ll have to check them first

against the truth tables of 3VL; as you’ll find out, several tautologies and rewrite rules from

2VL are not necessarily also valid in 3VL Listing D-1 shows two examples of such tautologies

P ∨ ¬P

P ⇔ P

Handling Three-Valued Logic

Consider the following two rather obvious statements (in an SQL context):

• A row shows up in a table (or in the result of a query) or it doesn’t

• A constraint gets violated or it doesn’t

There are no other possibilities; therefore, these are two-valued issues However, in 3VL,predicates have three possible outcomes: TRUE, FALSE, or UNKNOWN Unfortunately, there is nointuitive interpretation for this third possible outcome To give you an idea of the inconsisten-cies, look at Table D-5 where you see the difference between evaluating the predicate, say P,

of a WHERE clause of an SQL query compared with checking the predicate of an SQL CHECKconstraint

P(row) WHERE P(row) CHECK(P(row))

T Row accepted Row satisfies constraint

F Row rejected Row violates constraint

U Row rejected Row satisfies constraint

As you can see, FALSE and UNKNOWN lead to the same result for WHERE clauses in queries,whereas TRUE and UNKNOWN lead to the same effect in constraints The third line in the precedingtable shows the discrepancy; a row will violate the predicate (and therefore not be retrieved)when the predicate is part of a WHERE clause However, that row will satisfy the predicate whenevaluated by a CHECK constraint that is defined on a table that receives this row

If you want to get control over 3VL in ISO-standard SQL, it is a good idea to think in terms

of the three SQL functions introduced in Table D-6 They allow you to create a 2VL layer on top

of 3VL, thus providing more explicit control over the three Boolean values TRUE, FALSE, andUNKNOWN

P IS TRUE (P) IS FALSE (P) IS UNKNOWN (P)

Listing D-2 shows some rewrite rules based on the operators defined in Table D-6

IS TRUE ( P ) ⇔ ( P ∧ ¬ ( IS UNKNOWN (P) ) )

IS FALSE ( P ) ⇔ (¬P ∧ ¬ ( IS UNKNOWN (P) ) )

IS TRUE ( ¬P ) ⇔ ( IS FALSE (P) )

IS FALSE ( ¬P ) ⇔ ( IS TRUE (P) )

IS TRUE ( P ∧ Q ) ⇔ ( IS TRUE (P) ∧ IS TRUE (Q) )

IS TRUE ( P ∨ Q ) ⇔ ( IS TRUE (P) ∨ IS TRUE (Q) )

IS FALSE ( P ∧ Q ) ⇔ ( IS FALSE (P) ∨ IS FALSE (Q) )

IS FALSE ( P ∨ Q ) ⇔ ( IS FALSE (P) ∧ IS FALSE (Q) )

IS TRUE ( ∃x∈S: P ) ⇔ ( ∃x∈S: (IS TRUE (P) ) )

IS FALSE ( ∃x∈S: P ) ⇔ ( ∀x∈S: (IS FALSE (P) ) )

IS FALSE ( ∀x∈S: P ) ⇔ ( ∃x∈S: (IS FALSE (P) ) )

IS UNKNOWN ( ∃x∈S: P ) ⇔ ( ¬∃x∈S: ( IS TRUE (P) ) ∧ ∃y∈S: (IS UNKNOWN (P) ) )

IS UNKNOWN ( ∀x∈S: P ) ⇔ ( ¬∃x∈S: ( IS FALSE (P) ) ∧ ∃y∈S: (IS UNKNOWN (P) ) )

You can prove all these tautologies by using three-valued truth tables, or by using a bination of 3VL rewrite rules you proved before

com-Four-Valued Logic

In E F Codd’s The Relational Model for Database Management: Version 2 (Addison-Wesley,

1990), he proposes a revision of the first version of the relational model, RM/V1 Earlier, in

1979, he presented a paper in Tasmania with the title “Extending the Database Relational

Model to Capture More Meaning,” naming the extended version RM/T (T for Tasmania) The

features of RM/T were supposed to be gradually incorporated into the sequence of versions

RM/V2, RM/V3, and so on

The most debatable sections of the RM/V2 book are Chapter 8 (“Missing Information”)and Section 12.4 (“Manipulation of Missing Information”) In these sections, Codd proposes a

4VL in an attempt to make a distinction between the two most common reasons why

informa-tion is missing: applicable and inapplicable, represented with the two discernible values A and

I, respectively The truth tables he provides appear in Table D-7

Note that the last two truth tables use a slightly different format from all other truth tablesyou saw so far in this book They are formatted as a matrix where the four rows represent thefour possible truth values for P, and the four columns represent the four possible values for Q,respectively This can be done because the disjunction and conjunction connectives are com-mutative; in the regular truth table format, those two truth tables would have needed sixteenrows each.

Note that introducing the two discernible values A and I also implies that you have torevisit the outcome of arithmetic operators and string concatenation; Ted Codd proposed thebehavior shown in Listing D-3, where x denotes a regular numeric attribute value and sdenotes a regular character string value

Answers to Selected Exercises

Chapter 1 Answers

Exercise 1

a. This is a FALSE proposition

b. Predicate It is equivalent to the predicate x > 0

c. This is a FALSE proposition

d. This is a TRUE proposition

e. This is a FALSE proposition

If you compare the truth tables for A ∨ B and A | B, you’ll again notice a pattern:

Truth Table for (P ∨Q) ⇔(Q ∨P)

The following table proves the second De Morgan law ¬(P ∧ Q) ⇔ (¬P ∨ ¬Q) by usingavailable rewrite rules.

¬(P ∧ Q) ⇔ Double negation (twice)

¬(¬¬P ∧ ¬¬Q) ⇔ First law of De Morgan (right to left)

Exercise 5

You can use truth tables or existing rewrite rules Alternatively, you can bring up a valuation

for the involved variables for which the predicate evaluates to FALSE (that is, you give a counterexample)

Chapter 2 Answers

Exercise 1

a TRUE; 3 is an element of set A.

b. Meaningless; the left operand should be a set

c TRUE; the empty set is a subset of every set.

d FALSE; there are only five elements in A, none of which is the empty set.

a TRUE; all elements in set A are greater than 1.

b TRUE; 5 is an element that exists in B for which, when chosen for variable x, mod(x,5) = 0.

c FALSE; if you choose element 2 in A for x and choose 1 in B for y, then x + y = 3, which

is not greater than or equal to 4 Because there exists such a combination for x and y,the given predicate does not hold for all values x and y, and is therefore FALSE Thepredicate would be TRUE if you redefine set B to {3,5,7,9}

d TRUE; for each element that you can choose from A, there is always an element in y

that can be picked such that x + y = 11 For x = 2, 4, 6, 8, pick y = 9, 7, 5, 3,respectively

e FALSE; there is no value in set B such that if you add that value to every available value

in set A, the result of every such addition would always be 11 The predicate would be

TRUE if you redefined set A such that it only holds one element, which effectively

down-grades the inner universal quantification to an existential quantification For A = {2}you can choose y = 9 For A = {4} you can choose y = 7 For A = {6} you can choose

y = 5 For A = {8} you can choose y = 3 Another option would be to redefine set A tothe empty set The inner universal quantification will then be TRUE, irrespective of theexpression (x + y = 11) As long as B has at least one element, the full predicate willthen always be TRUE

f TRUE; choose 2 in A for both x and y You might think that it is not allowed to choose the

same value twice, but it is in this case Variables x and y are bound independently to set

A The binding of y to (the inner) A is not influenced by the value you choose for x that

is bound to the outer A

¬( ∃x∈S: ∀y∈T: P(x,y) )

Note that you can view this expression as follows: ¬(∃x∈S: R(x) ), where R(x) = ( ∀y∈T: P(x,y) ) Now you can apply the third rewrite rule of Listing 2-15, which has thefollowing result:

( ∀x∈S: ¬R(x) )

Substituting the expression for R(x) back into this gives the following result:

a. The expression evaluates to { (X;1), (Y;3), (Z;2), (R;1) } This is a function

b. The expression evaluates to { (X;1) } This is a function

c. The expression evaluates to { (X;1), (Y;3), (X;2) } This is not a function; the firstcoordinate X appears twice

Exercise 7

a { { (a;1), (b;1) }, { (a;1), (b;2) } }

Exercise 8

a { (ean;9786012), (price;24.99) }

c { (descr;'A very nice book') }

Exercise 11

a. The left side evaluates to { {(empno;104)}, {(empno;106)}, {(empno;102)} }: a set

of three functions The right side evaluates to { {(empno;103)}, {(empno;104)},{(empno;105)}, {(empno;106)} }; this is a set of four functions The proposition isFALSE

b. This expression evaluates to { {(deptno;10)}, {(deptno;20)} }⊂{ {(deptno;10)},{(deptno;20)} , {(deptno;30)} } This proposition is TRUE; the set of two functions atthe left is indeed a proper subset of the set of three functions at the right

Chapter 5 Answers

Exercise 1

a. This is not a table; not all tuples have the same domain

b. This is a table; in fact it is the empty table

c. This is a table over {partno}

d. This is a table over {suppno,sname,location}

e. This is a table over {partno,pname,price}

Exercise 2

{ p | p∈∏(chr_PART) ∧ p(name)='drill' ⇒ p(price)≥400 }

Exercise 4

E6, the join of S and SP, is a table over {partno,suppno,available,reserved,sname,location}

It holds the six tuples from SP that have been extended with the supplier name and location

E8 represents the Cartesian join of S and P It is a table over {partno,pname,price,suppno,sname,location}, and holds eight tuples

Exercise 5

P1 states that for all pairs of parts that can be chosen from table P, the two parts have different

part numbers This is a FALSE proposition because the formal specification allows a pair to

hold the same part twice

P2 states that for all pairs of different parts (chosen from table P), the two parts have

differ-ent part numbers This is a TRUE proposition

c. If you rewrite the implication into a disjunction, you’ll see that this proposition statesthat there are six parts in PAR1 for which we either have 10 or less items in stock, or thatcost 10 or less This is a TRUE proposition; all six parts in PAR1 satisfy the implication

Exercise 4

This involves specifying three subset requirements and one additional constraint to state that

every tuple in EMP1 has a corresponding tuple in one of the specializations

TRN1⇓{empno} ⊆ EMP1⇓{empno} ∧

MAN1⇓{empno} ⊆ EMP1⇓{empno} ∧

CLK1⇓{empno} ⊆ EMP1⇓{empno} ∧

(∀ e∈EMP1: e(job)='TRAINER'⇒(∃ t∈TRN1: t(empno)=e(empno)) ∧

e(job)='MANAGER'⇒(∃ m∈MAN1: m(empno)=e(empno)) ∧

e(job)='CLERK' ⇒(∃ c∈CLK1: c(empno)=e(empno)))

Exercise 5

This involves specifying three subset requirements and a few additional constraints statingthat all tuples in EMP1 are covered by exactly one tuple in one of the specializations

TRN1⇓{empno} ⊆ EMP1⇓{empno} ∧

MAN1⇓{empno} ⊆ EMP1⇓{empno} ∧

CLK1⇓{empno} ⊆ EMP1⇓{empno} ∧

TRN1⇓{empno} ∩ MAN1⇓{empno} = ∅∧

Chapter 7 Answers

Exercise 2

Predicate o(STATUS)='CONF' ⇒ o(TRAINER)≠-1 is equivalent to predicate o(TRAINER)=-1 ⇒

o(STATUS)∈{'CANC','SCHD'} This is a manifestation of the following rewrite rule:

When designing this constraint, you might want to check with the users whether the TERMtable structure should play a role in this constraint