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I've been doing past paper questions and keep coming up against these questions that deal with 3 valued logic. My notes mention it but don't give examples that relate to those asked in exams.

I understand the basis that True = 1, False = 0 & Unknown = 1/2 as well as And = Min, Or = Max and Not(x) = 1-x. However I do not know how to apply it to questions such as those below:

In SQL, discuss the possible truth values of the following expression:
R.a > R.b OR R.a <= 0 OR R.b >= 0
Justify your answer.

And:

The phone and age fields of the Owner table might have null values in them. Considering all possible combinations, show which of the three truth values might be returned by the expression:
phone = ’141-3304913’ OR age <50 OR age >= 50

Any help in clarifying these for me would be really appreciated :)

Luca Geretti
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Sophie Pocket
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    Unknown = 1/2?? I've never seen that before. Do you have a reference? – Mark Byers Apr 25 '12 at 12:21
  • the unknown is a half, It was in our lecture notes: Our notes say: "• To understand how AND, OR, and NOT work in 3-valued logic, think of TRUE = 1, FALSE = 0, and UNKNOWN = ½. • AND = MIN; OR = MAX, NOT(x) = 1-x. • Example: TRUE AND (FALSE OR NOT (UNKNOWN)) = MIN(1, MAX(0, (1 - ½ ))) = MIN(1, MAX(0, ½ ) = MIN(1, ½ ) = ½. " – Sophie Pocket Apr 25 '12 at 12:26
  • This confused me, somewhat... – Sophie Pocket Apr 25 '12 at 12:26
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    Wiki says 1/2's exist: http://en.wikipedia.org/wiki/Three-valued_logic#Representation_of_values - news to me! – Morphed Apr 25 '12 at 12:33
  • If I ever had heard of this before, I would be hard pressed to not push it directly out of my mind. The logical conniptions used to keep UNKNOWN + UNKNOWN != TRUE seem a little odd. – Andrew Wolfe Apr 25 '12 at 12:39
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    I'm just going by what our lecturer has told us, and he'll be marking our exam too. – Sophie Pocket Apr 25 '12 at 12:40
  • @SophiePocket: It seems that those rules actually work! Interesting... I've never seen it presented in this way before. `TRUE AND (FALSE OR NOT (UNKNOWN)) = UNKNOWN` can be thought of as `MIN(1, MAX(0, (1 - ½))) = ½`. – Mark Byers Apr 25 '12 at 12:57
  • I assume so but I don't know how to do them or apply it. Ah yes, the theory I understand :) But when presented with questions such as the ones in my question I get confused as to how he wants it presented. I know you won't know what format he wants, but the understanding of how to do questions like these would make me happy :) – Sophie Pocket Apr 25 '12 at 12:59

1 Answers1

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I will focus on the concrete example, which is more proper for clarifying things. Put simply, your logical expression is made of a conjunction of three clauses

C1: phone = '141-3304913'
C2: age < 50
C3: age >= 50

for which tri-boolean logic states that the result is 

True, if any clause is true
False, if all clauses are false
Unknown, in all the other cases

Consequently, if the value associated with True is the largest, with False is the smallest, and with Unknown is any intermediate value, then taking the MAX for a conjunction proves correct. Similarly, a disjunction works with the MIN function. Negation works as long as we interpret any value between 0 and 1 (excluded) as Unknown; clearly, if we take 1/2 then the negation function is "stable", but that does not really matter in mathematical terms.

More operatively, the clauses clearly react to the following values (instances) of your phone variable P and your age variable A:

P1 such that P1 = '141-3304913'
P2 such that P2 <> '141-3304913'
P3 such that P3 = NULL
A1 such that A1 < 50
A2 such that A2 >= 50
A3 such that A3 = NULL

In terms of satisfaction of the clauses, we have

P1 -> C1 = 1
P2 -> C1 = 0
P3 -> C1 = 1/2
A1 -> C2 = 1, C3 = 0
A2 -> C2 = 0, C3 = 1
A3 -> C2 = C3 = 1/2

In general there exist 3*3 possible combinations, since each of your two variables takes three possible values:

P1 A1: C1 = 1, C2 = 1, C3 = 0 -> MAX(1,1,0) = 1 -> true
P1 A2: C1 = 1, C2 = 0, C3 = 1 -> MAX(1,0,1) = 1 -> true
P1 A3: C1 = 1, C2 = 1/2, C3 = 1/2 -> MAX(1,1/2,1/2) = 1 -> true
P2 A1: C1 = 0, C2 = 1, C3 = 0 -> MAX(0,1,0) = 1 -> true
P2 A2: C1 = 0, C2 = 0, C3 = 1 -> MAX(0,0,1) = 1 -> true
P2 A3: C1 = 0, C2 = 1/2, C3 = 1/2 -> MAX(0,1/2,1/2) = 1/2 -> unknown
P3 A1: C1 = 1/2, C2 = 1, C3 = 0 -> MAX(1/2,1,0) = 1 -> true
P3 A2: C1 = 1/2, C2 = 0, C3 = 1 -> MAX(1/2,0,1) = 1 -> true
P3 A3: C1 = 1/2, C2 = 1/2, C3 = 1/2 -> MAX(1/2,1/2,1/2) = 1/2 -> unknown

In particular, since C2 and C3 are mutually exclusive, you never get False as a result of the conjunction.

The expression R.a > R.b OR R.a <= 0 OR R.b >= 0 instead presents these cases:

R.a <= 0, R.a > 0, R.a = unknown
R.b >= 0, R.b < 0, R.b = unknown
R.a - R.b > 0, R.a - R.b <= 0, R.a - R.b = unknown

Apparently we have three variables and 27 possible cases, but several related to R.a - R.b can be trivially ruled out.

Luca Geretti
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  • Thanks, but how do you incorporate the OR values etc into it? No answers are given in past papers so I have no clue what the correct answer should look like. – Sophie Pocket Apr 25 '12 at 12:54
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    I edited to point out the mathematics behind, I hope it suffices. – Luca Geretti Apr 25 '12 at 13:04
  • I now understand how the 2nd example (R.a <...) works :) how does your C1, C2, C3 work? I'm not sure how C1 can be 1, 0 and 1/2. Other than that I understand how you have worked out the logic :D – Sophie Pocket Apr 25 '12 at 13:15
  • I'm not sure I understand: is my current answer sufficiently detailed, or do you need me to solve the R.a,R.b example too? – Luca Geretti Apr 25 '12 at 13:18
  • oh no I understand the R.a, R.b one. I just don't understand how you got to: C1 = 1; C1 = 0; C1 = 1/2 C2 = 1, C3 = 0; C2 = 0, C3 = 1; C2 = C3 = 1/2 – Sophie Pocket Apr 25 '12 at 13:20
  • Like what C2 and C3 stands for as it only appears in the Age clauses. – Sophie Pocket Apr 25 '12 at 13:29
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    That should be clear now: the problem probably lied in the distinction between variable name and variable instance. – Luca Geretti Apr 25 '12 at 13:33
  • Thanks :D I now understand the answer :) thank you for all your help and for putting up with all my questions and struggles :) – Sophie Pocket Apr 25 '12 at 13:36