1. Example where Euphoria defaults on a boolean expression
You get a phone call from your boss where you are ordered to search for =
information about dyslexia .. immediately! You go to the PC and start =
websearching. In the phone conversation you were told which words to =
search for...... it was "dyslexia"(A) and "adult"(B) and something =
else(C) which you don't remember. What to do when you don't know the =
third search criterium?
In this situation you will probably dismiss C from the rule and state:
if A and B then match.....
Is this a valid transformation of boolean logic ... when one variable is =
unknown? What this rule returns is not the same as what your boss asked =
you to do.
Any variable may be nil (=3Duninstanciated). That has a lot of =
implications.
if A and B and C then match....
When A=3DTrue, B=3DTrue and C=3Dnil then this rule CANNOT be solved as =
True or False (i.e result is nil). That is consistent with boolean =
logic (and any other logic too)=20
Logically you could deduct that the paper you return is:
not (what he asked for) and
not (not (what he asked for))
If E =3D what he asked for then
not E and not( not E) =3D> not E and E =3D> E and not E
The paper you returned to your boss was what he asked for and was not he =
asked for ( =3D "maybe" what he asked for)
So how do Euphoria handle this?
atom e,=20
e =3D 2 -- what he asked for
f =3D 3 -- what you gave him
if (f =3D e) and not ( f !=3D e) then print( 1, 1)
else print ( 1, 0)
end if
object line
line =3D gets( 0)
The Euphoria program returns "0" ...... which is wrong!
"not E and E" is unsolvable, neither true nor false.=20
Euphoria cannot handle this (like C and Pascal).
Rom
2. Re: Example where Euphoria defaults on a boolean expression
Rom wrote:
<snip>
> "not E and E" is unsolvable, neither true nor false.
No. "not E and E" is always false, of course.
> Euphoria cannot handle this (like C and Pascal).
No, please try:
integer E
E = 0
? not E and E
E = not E
? not E and E
Regards,
Juergen
3. Re: Example where Euphoria defaults on a boolean expression
From: "Juergen Luethje" <jluethje at gmx.de>
> No, please try:
>=20
> integer E
> E =3D 0
> ? not E and E
> E =3D not E
> ? not E and E
There was a bug in my example:
atom e, f
e =3D 3 -- does not matter what value you state here
f =3D 3 -- does not matter what value you state here
-- (I wrote wrongly "if (f =3D e) and not ( f !=3D e) then ..." )
if (f =3D e) and ( f !=3D e) then print( 1, 1)
else print ( 1, 0)
end if
object line
line =3D gets( 0)
f =3D e and f !=3D e cannot be resolved. The evaluation should return =
nil.
Euphoria will alway return False here. Euphoria will never return =
possible true.=20
Then we are a long way into philosophy.
I tried to exemplify that with the examle. If we have rule (A and B and =
C), and we don't know C, then we cannot know if (A and B and C) is true. =
We cannot conclude from the (A and B) about (A and B and C). That is =
what a program has to be evaluated if we don't know C. That is an =
unresolvable boolean problem (but we all know that if (A and B) is true =
then (A and B and C) may be true. What is unresolvable (=3Dnil) here is =
nothing more mysterious than maybe.=20
But C/Delphi/Euphoria never return unresolvable as a possible result for =
ANY logical expression (implying that all logic statements can be =
solved) . That does not mean that every logical expression can be =
solved, just that these languages cannot return unresolvable as a =
possible result.
Of course an evaluation of an uninstanciated boolean var is =
unresolvable. If a program output True or False here that output is =
rubbish.=20
It is so easy to implement a nil flag (=3D uninstanciated var) to deal =
with uninstanciated vars.
A =3D nil if A -> nil
A =3D True, B =3D nil if A and B -> nil
A =3D False, B =3D nil if A and B -> False
A =3D True, B =3D nil if A or B -> True
A =3D False, B =3D nil if A or B -> nil
3 * nil -> nil
3 / 0 -> nil
Rom
4. Re: Example where Euphoria defaults on a boolean expression
- Posted by dm31 at uow.edu.au
Nov 23, 2002
| atom e, f
|
| e = 3 -- does not matter what value you state here
| f = 3 -- does not matter what value you state here
|
|-- (I wrote wrongly "if (f = e) and not ( f != e) then ..." )
|
| if (f = e) and ( f != e) then print( 1, 1)
| else print ( 1, 0)
| end if
|
| object line
| line = gets( 0)
|
|
|f = e and f != e cannot be resolved. The evaluation should return
nil.
Why can they not be resolved?? Both e and f are true values (!= 0),
and (f = e) is true and (f != e) is not true. true and not true =
true.
This is the simplest example they give in first year logic course at
uni's of a contridiction. ie, when false is always the result. It does
not matter what numbers e & f are, true or false. This should always,
and does always return false.
|Euphoria will alway return False here. Euphoria will never return
possible true.
That is correct. What was the problem??
5. Re: Example where Euphoria defaults on a boolean expression
(myself)
>>|f =3D e and f !=3D e cannot be resolved. The evaluation should =
return nil.
From: <dm31 at uow.edu.au>
> Why can they not be resolved?? Both e and f are true values (!=3D 0),
> and (f =3D e) is true and (f !=3D e) is not true. true and not true =
=3Dtrue.
>>Euphoria will alway return False here. Euphoria will never return=20
>>possible true.=20
From: <dm31 at uow.edu.au>
>That is correct. What was the problem??
I also also feel bad about how I formulated the example.
I withdraw the whole example "if (f =3D e) and (f !=3D e) then ....." =
.... You are right. Big blunder 
(
The problem I tried to exemplify about (f =3D e) and (f !=3D e) cannot =
be translated into an if statement. I cannot state that both (f =3D e) =
is true and (f !=3D e) is true the way I tried to in the example.
To state (f =3De) is true and (f !=3D e) is true cannot be done i =
Euphoria. Euphoria does not support that kind of logic statements. I =
don't have that option with a procedural language.=20
How to evaluate (A and B and C) (...the kind of logical statements =
Euphora supports) when we don't know C is however unresolvable (we can =
neither conclude True nor False, which will leave us with an unresolved =
result). Euphoria forces us to state some value for anything, so you may =
say that C cannot be unknown. That we don't know the value of a var is =
something any program used for programming logic must deal with. I feel =
dealing with that is the first step in any AI-project.
Regards
Rom
6. Re: Example where Euphoria defaults on a boolean expression
On 23 Nov 2002, at 17:31, Rom wrote:
<snip>
> How to evaluate (A and B and C) (...the kind of logical statements Euphora
> supports) when we don't know C is however unresolvable (we can neither
> conclude
> True nor False, which will leave us with an unresolved result). Euphoria
> forces
> us to state some value for anything, so you may say that C cannot be unknown.
> That we don't know the value of a var is something any program used for
> programming logic must deal with. I feel dealing with that is the first step
> in
> any AI-project.
Correct. The first step is to know if you do not know something. This means
knowing if equal(table,"red"), which is bad is table was undefined. Also in Eu,
you can't import a new OOP-ish object of table during runtime. You can add
it to an associated array, but you can't add the methods of how to deal with
tables.
Kat
7. Re: Example where Euphoria defaults on a boolean expression
Rom wrote:
> From: "Juergen Luethje" <jluethje at gmx.de>
>
>> No, please try:
>>
>> integer E
>> E = 0
>> ? not E and E
>> E = not E
>> ? not E and E
>
>
> There was a bug in my example:
>
> atom e, f
>
> e = 3 -- does not matter what value you state here
> f = 3 -- does not matter what value you state here
>
> -- (I wrote wrongly "if (f = e) and not ( f != e) then ..." )
>
> if (f = e) and ( f != e) then print( 1, 1)
> else print ( 1, 0)
> end if
>
> object line
> line = gets( 0)
>
>
> f = e and f != e cannot be resolved. The evaluation should return nil.
>
> Euphoria will alway return False here. Euphoria will never return
> possible true. Then we are a long way into philosophy.
>
> I tried to exemplify that with the examle. If we have rule (A and B
> and C), and we don't know C, then we cannot know if (A and B and C) is
> true. We cannot conclude from the (A and B) about (A and B and C). That
> is what a program has to be evaluated if we don't know C. That is an
> unresolvable boolean problem (but we all know that if (A and B) is true
> then (A and B and C) may be true. What is unresolvable (=nil) here is
> nothing more mysterious than maybe.
Yes, I understand. But "maybe E" is not logically equivalent to
"not E and E". (Well, you talked about that with dm31 at uow.edu.au in the
meantime.)
> But C/Delphi/Euphoria never return unresolvable as a possible result
> for ANY logical expression (implying that all logic statements can be
> solved) . That does not mean that every logical expression can be
> solved, just that these languages cannot return unresolvable as a
> possible result.
>
> Of course an evaluation of an uninstanciated boolean var is
> unresolvable. If a program output True or False here that output is
> rubbish.
>
> It is so easy to implement a nil flag (= uninstanciated var) to deal
> with uninstanciated vars.
>
> A = nil if A -> nil
> A = True, B = nil if A and B -> nil
> A = False, B = nil if A and B -> False
> A = True, B = nil if A or B -> True
> A = False, B = nil if A or B -> nil
> 3 * nil -> nil
> 3 / 0 -> nil
If "nil" means "unknown", then "not nil" means "known" -- but known to
be true, or known to be false? Is there a solution to this logical
pitfall? I would appreciate it very much, if there were a "3-value-logic"
(false/unknown/true) in Euphoria.
For instance I also would prefer
open(<non existing file>, "r") -> nil
to
open(<non existing file>, "r") -> -1
The above mentioned use of "nil" is primarily independent of the problem
of uninstanciated variables, and those things shouldn't be mixed up, I
think. Using nil for such variables looks easy indeed, but I think it's
also easy to genarate a runtime error (as Euphoria currently does), to
deal with them. The question is to me, what brings more benefit?
>From my personal experience (being not an IT professional) it looks to
me, that generating runtime errors for uninstanciated variables is
better for the sake of safety.
Regards,
Juergen
--
Q: How many Microsoft engineers does Bill need to change a light bulb?
A: None. He just declares darkness to be an industry standard.
8. Re: Example where Euphoria defaults on a boolean expression
I use a programming language called 'Progress'. It has an explicit 'unknown'
value that you can assign to any variable. It has been a very useful device.
The symbol for it in the language is '?'.
So ...
if theDate = ? then
theDate = today - 1.
reads...
if the date is not known, set it to yesterday's date.
----------------
cheers,
Derek Parnell
----- Original Message -----
From: "Juergen Luethje" <jluethje at gmx.de>
To: "EUforum" <EUforum at topica.com>
Sent: Sunday, November 24, 2002 5:32 AM
Subject: Re: Example where Euphoria defaults on a boolean expression
Rom wrote:
> From: "Juergen Luethje" <jluethje at gmx.de>
>
>> No, please try:
>>
>> integer E
>> E = 0
>> ? not E and E
>> E = not E
>> ? not E and E
>
>
> There was a bug in my example:
>
> atom e, f
>
> e = 3 -- does not matter what value you state here
> f = 3 -- does not matter what value you state here
>
> -- (I wrote wrongly "if (f = e) and not ( f != e) then ..." )
>
> if (f = e) and ( f != e) then print( 1, 1)
> else print ( 1, 0)
> end if
>
> object line
> line = gets( 0)
>
>
> f = e and f != e cannot be resolved. The evaluation should return nil.
>
> Euphoria will alway return False here. Euphoria will never return
> possible true. Then we are a long way into philosophy.
>
> I tried to exemplify that with the examle. If we have rule (A and B
> and C), and we don't know C, then we cannot know if (A and B and C) is
> true. We cannot conclude from the (A and B) about (A and B and C). That
> is what a program has to be evaluated if we don't know C. That is an
> unresolvable boolean problem (but we all know that if (A and B) is true
> then (A and B and C) may be true. What is unresolvable (=nil) here is
> nothing more mysterious than maybe.
Yes, I understand. But "maybe E" is not logically equivalent to
"not E and E". (Well, you talked about that with dm31 at uow.edu.au in the
meantime.)
> But C/Delphi/Euphoria never return unresolvable as a possible result
> for ANY logical expression (implying that all logic statements can be
> solved) . That does not mean that every logical expression can be
> solved, just that these languages cannot return unresolvable as a
> possible result.
>
> Of course an evaluation of an uninstanciated boolean var is
> unresolvable. If a program output True or False here that output is
> rubbish.
>
> It is so easy to implement a nil flag (= uninstanciated var) to deal
> with uninstanciated vars.
>
> A = nil if A -> nil
> A = True, B = nil if A and B -> nil
> A = False, B = nil if A and B -> False
> A = True, B = nil if A or B -> True
> A = False, B = nil if A or B -> nil
> 3 * nil -> nil
> 3 / 0 -> nil
If "nil" means "unknown", then "not nil" means "known" -- but known to
be true, or known to be false? Is there a solution to this logical
pitfall? I would appreciate it very much, if there were a "3-value-logic"
(false/unknown/true) in Euphoria.
For instance I also would prefer
open(<non existing file>, "r") -> nil
to
open(<non existing file>, "r") -> -1
The above mentioned use of "nil" is primarily independent of the problem
of uninstanciated variables, and those things shouldn't be mixed up, I
think. Using nil for such variables looks easy indeed, but I think it's
also easy to genarate a runtime error (as Euphoria currently does), to
deal with them. The question is to me, what brings more benefit?
>From my personal experience (being not an IT professional) it looks to
me, that generating runtime errors for uninstanciated variables is
better for the sake of safety.
Regards,
Juergen
--
Q: How many Microsoft engineers does Bill need to change a light bulb?
A: None. He just declares darkness to be an industry standard.
==^^===============================================================
This email was sent to: ddparnell at bigpond.com
9. Re: Example where Euphoria defaults on a boolean expression
- Posted by 1evan at sbcglobal.net
Nov 23, 2002
Rom wrote:
>
> atom e,
>
> e = 2 -- what he asked for
> f = 3 -- what you gave him
> if (f = e) and not ( f != e) then print( 1, 1)
> else print ( 1, 0)
> end if
>
> object line
> line = gets( 0)
>
>
>The Euphoria program returns "0" ...... which is wrong!
>
>"not E and E" is unsolvable, neither true nor false.
>Euphoria cannot handle this (like C and Pascal).
>
>Rom
>
In boolean logic, there are two possible states: True and False, 1 and 0.
If E = 1(True) then not E = 0(False)
0 and X (either 1 or 0) is always 0(False)
not E and E = 0
0 and 1 = 0
If E = 0(False) then not E = 1(True)
not E and E = 0
1 and 0 = 0
(3 = 2) is 0(False)
not (3 != 2) is 0(False) -- (3 != 2) is 1(True), not(1) is False(0)
(3 = 2) and not(3 != 2) is 0(False) -- 0 and 0 is 0(False)
10. Re: Example where Euphoria defaults on a boolean expression
- Posted by 1evan at sbcglobal.net
Nov 23, 2002
Rom wrote:
> if (f = e) and ( f != e) then print( 1, 1)
> else print ( 1, 0)
> end if
>
>
>f = e and f != e cannot be resolved. The evaluation should return nil.
>
>
( f = e ) and ( f != e) evaluates to 0 also.
( 3 = 2) and (3 != 2) => 0
0 and 1 => 0
11. Re: Example where Euphoria defaults on a boolean expression
- Posted by 1evan at sbcglobal.net
Nov 23, 2002
Rom wrote:
>How to evaluate (A and B and C) (...the kind of logical statements Euphora
>supports) when we don't know C is however unresolvable (we can neither conclude
>True nor False, which will leave us with an unresolved result).
>Regards
>Rom
>
We can conclude False if either A = 0 or B = 0. However, if A = 1 and B
= 1 then we can not resolve the statement as True or False without
knowing the value of C.
12. Re: Example where Euphoria defaults on a boolean expression
From: <1evan at sbcglobal.net>
> Rom wrote:
>=20
> > if (f =3D e) and ( f !=3D e) then print( 1, 1)
> > else print ( 1, 0)
> > end if
> >
> >
> >f =3D e and f !=3D e cannot be resolved. The evaluation should =
return nil.
> >
> > =20
> ( f =3D e ) and ( f !=3D e) evaluates to 0 also.
> ( 3 =3D 2) and (3 !=3D 2) =3D> 0
> 0 and 1 =3D> 0
The example I formulated was a logical miss from my side. I have =
admitted that. What you are saying is true. Also that (False and True =
and nil) -> False, of course.
Rom
13. Re: Example where Euphoria defaults on a boolean expression
- Posted by 1evan at sbcglobal.net
Nov 23, 2002
Yes, sorry. I read your follow-up post after I sent mine.
Rom wrote:
>
>From: <1evan at sbcglobal.net>
>
>
>>Rom wrote:
>>
>>
>>> if (f = e) and ( f != e) then print( 1, 1)
>>> else print ( 1, 0)
>>> end if
>>>
>>>
>>>f = e and f != e cannot be resolved. The evaluation should return nil.
>>>
>>>
>>( f = e ) and ( f != e) evaluates to 0 also.
>>( 3 = 2) and (3 != 2) => 0
>>0 and 1 => 0
>>
>>
>The example I formulated was a logical miss from my side. I have admitted that.
>What you are saying is true. Also that (False and True and nil) -> False, of
>course.
>
>Rom
>
>==^^===============================================================
>This email was sent to: 1evan at sbcglobal.net
>
>
>
14. Re: Example where Euphoria defaults on a boolean expression
From: "Bernie Ryan" <xotron at bluefrognet.net>
> Rom:=20
> If you are dissatisfied with the way Euphoria handles
> true and false then why don't you write a library/include
> file to place on the contribution page that performs the
> way that you think is correct. This would benefit the other
> users like yourself that need that capability.
I haven't planned that in near future. But I will by a month or two =
finish a small scripting interpreter where nil will be a possible state =
for vars and if ... then ... constructs. Everyone can do a bit toying =
with that when program is finished.=20
Also, defining a new boolean type (with empty/nil as a possible state) =
and defining implications of that type when implemented upon operators =
like > and < and control structures like if ... then ... should be easy =
to post.
Rom
15. Re: Example where Euphoria defaults on a boolean expression
On Saturday 23 November 2002 05:40 pm, you wrote:
>
>
> From: "Bernie Ryan" <xotron at bluefrognet.net>
>
> > Rom:
> > If you are dissatisfied with the way Euphoria handles
> > true and false then why don't you write a library/include
> > file to place on the contribution page that performs the
> > way that you think is correct. This would benefit the other
> > users like yourself that need that capability.
>
> I haven't planned that in near future. But I will by a month or two finish
> a small scripting interpreter where nil will be a possible state for vars
> and if ... then ... constructs. Everyone can do a bit toying with that when
> program is finished.
>
> Also, defining a new boolean type (with empty/nil as a possible state) and
> defining implications of that type when implemented upon operators like >
> and < and control structures like if ... then ... should be easy to post.
Or, you could download Lua, which can assign nil as a value for a variable.
Irv
16. Re: Example where Euphoria defaults on a boolean expression
--- Rom <kjehas at frisurf.no> wrote:
> Of course an evaluation of an uninstanciated boolean var is unresolvable. If
> a program output True or False here that output is rubbish.
I completely disagree. More below...
> It is so easy to implement a nil flag (= uninstanciated var) to deal with
> uninstanciated vars.
I can see value in this, but not in regards to your 'unresolvable' boolean
statements.
> A = nil if A -> nil
> A = True, B = nil if A and B -> nil
> A = False, B = nil if A and B -> False
> A = True, B = nil if A or B -> True
> A = False, B = nil if A or B -> nil
> 3 * nil -> nil
> 3 / 0 -> nil
If something is not true, then it's false. If you really want to implement
this 3-value logic business, then fine, but that's not the world we live in.
Suppose you have no brothers. Please evaluate this statement:
"Your brother is named Joe."
You're telling me this is unresolvable, because 'brother' is uninstanciated.
But the test here is, do you have a brother named Joe? The answer is no,
except the reason is that you don't have a brother, rather than that his name
is Bob.
Please explain more on why this shouldn't work. You've given us plenty of
examples of how your idea would work, but I haven't seen anything to justify
it.
Matt Lewis
__________________________________________________
Do you Yahoo!?
Yahoo! Mail Plus – Powerful. Affordable. Sign up now.
http://mailplus.yahoo.com
