ANY intuitively means: anything that matches in the set.
However, I found out it, it doesn't behave as expected in two cases:
x > ANY(...) The value must be greater than the smallest value in
the list to evaluate to TRUE.
x < ANY(...) The value must be
smaller than the biggest value in the list to evaluate to TRUE
(Source)
In other words, the entire result of sub-query introduced by ANY has to be compared to x, which runs counter to meaning of ANY.
Why is it so, or I am just misunderstanding the semantics of <ANY & >ANY operators ?
x > ANY(...) The value must be greater than the smallest value in the list to evaluate to TRUE.
x < ANY (...) The value must be smaller than the biggest value in the list to evaluate to TRUE
Both of the above statements assert the Algorithm that oracle might using for comparison. for example.
select * from X where id > Any ( 1 , 2 , 3 , 4 , 5 )
in above code can be drill down as
id > 1 OR id > 2 OR id > 3 and so on..
so only id > 1 will assert it true. (Algo Optimization) Same way they did it for x < ANY (...)
x > ANY(...) The value must be greater than the smallest value in the list to evaluate to TRUE
This is stating a consequence. It may take some thinking about. x may be greater than every value in the list, and it will still return TRUE. The opposite may be easier to understand:
x > ANY(...) will be FALSE if the value is less than or equal to the smallest value in the list, because it must also be less than or equal to all other members of the list
Or yet another way:
x > ANY(...) will be true if there is at least one value in the list, y say, that x is greater than. If there are a set of such y values, the smallest value in the list must logically be one member of that set.
Related
I am working with a legacy code base written in VB and have run into a conditional operator that I don't understand and cannot figure out what to search for to resolve it.
What I am dealing with is the following code and the variables that result as true. The specific parts that I do not understand are (1) the relationship between the first X and the first parens (-2 and (2) the role of X < 2
If X is a value below -2 it evaluates as false.
If X is a value above 2 it evaluates as true.
If Y is below 5 it evaluates to true as expected.
X = 3
Y = 10
If X > (-2 And X < 2) Or Y < 5 Then
'True
Else
'False
End If
I'm gonna leave off the Or Y < 5 part of the expression for this post as uninteresting, and limit myself to the X > (-2 And X < 2) side of the expression.
I haven't done much VB over the last several years, so I started out with some digging into Operator Precedence rules in VB, to be sure I have things right. I found definitive info on VBA and VB.Net, and some MSDN stuff that might have been VB6 but could also have been the 2013 version of VB.Net. All of them, though, gave the < and > comparison operators higher precedence over the And operator, regardless of whether you see And as logical or bitwise.
With that info, and also knowing that we must look inside parentheses first, I'm now confident the very first part of the expression to be evaluated is X < 2 (rather than -2 And X). Further, we know this will produce a Boolean result, and this Boolean result must be then converted to an Integer to do a bitwise (not logical) And with -2. That result (I'll call it n), which is still an Integer, can at last be compared to see if X > n, which will yield the final result of the expression as a Boolean.
I did some more digging and found this Stack Overflow answer about converting VB Booleans to Integers. While not definitive documentation, I was once privileged to meet the author (Hi #JaredPar) and know he worked on the VB compiler team at Microsoft, so he should know what he's talking about. It indicates that VB Boolean True has the surprising value of -1 as an integer! False becomes the more-normal 0.
At this point we need to talk about the binary representation of negative numbers. Using this reference as a guide (I do vaguely remember learning about this in college, but it's not the kind of thing I need every day), I'm going to provide a conversion table for integers from -3 to +3 in an imaginary integer size of only 4 bits (short version: invert the bit pattern and add one to get the negative representation):
-3 1101
-2 1110
-1 1111 --this helps explain **why** -1 was used for True
0 0000
1 0001
2 0010
3 0010
Stepping back, let's now consider the original -2 And X < 2 parenthetical and look at the results from the True (-1) and False (0) possible outcomes for X < 2 after a bitwise And with -2:
-2 (1110) And True (1111) = 1110 = -2
-2 (1110) And False (0000) = 0000 = 0
Really the whole point here from using the -1 bit pattern is anything And True produces that same thing you started with, whereas anything And False produces all zeros.
So if X < 2 you get True, which results in -2; otherwise you end up with 0. It's interesting to note here that if our language used positive one for True, you'd end up with the same 0000 value doing a bitwise And with -2 that you get from False (1110 And 0001).
Now we know enough to look at some values for X and determine the result of the entire original expression. Any positive integer is greater than both -2 and 0, so the expression should result in True. Zero and -1 are similar: they are less than two, and so will be compared again only as greater than -2 and thus always result in True. Negative two, though, and anything below, should be False.
Unfortunately, this means you could simplify the entire expression down to X > -2. Either I'm wrong about operator precedence, my reference for negative integer bit patterns is wrong, you're using a version of VB that converts True to something other than -1, or this code is just way over-complicated from the get-go.
I believe the sequence is as follows:
r = -2 and X 'outputs X with the least significant bit set to 0 (meaning the first lesser even number)
r = (r < 2) 'outputs -1 or 0
r = (X > r) 'outputs -1 or 0
r = r Or (Y < 5)
Use AndAlso and OrElse.
Was fooling around with trying to reduce the length of the code so that it gives off fewer headaches to look at and debug, and I came across this curious little fact:
Debug.Print 5<9<8 'returns "True"
At first I thought this was because it just checked the first set, but then I found that
Debug.Print 5<4<8 '*also* returns "True"
Does VBA interpret this kind of triple inequality as an Or statement? I can't imagine why someone would choose to make that the interpretation VBA makes because it's almost certainly the less used option, but I struggle to think of another explanation.
Also, what is a quick and pretty way of writing If 5 < X < 8 Then (to use sample numbers), without having to resort to endless And statements, ie If 5 < x And X < 8 Then? It's okay for one statement, but the doubling of length adds up quick, especially since variables aren't typically named X.
Edit: okay, it's certainly not an Or because VBA also says that Debug.Print 8<6<2 is True. What on earth is it thinking?
I have no clue but my educated guess would be that it first evaluates the left side of the equation (5<9) which gives TRUE. Then, it proceeds to evaluate the rest (TRUE<8) and implicitly converts TRUE to its integer value (I believe this to be -1 in VB).
-1<8 -> TRUE
Works with the second case as well since FALSE will convert to 0 and 0<8.
Basically it would have everything to do with implicit conversion of boolean to integer and their respective value in VBA.
It's to do with the way VBA evaluates expressions and implicit conversion. The first part of the equation is evaluated and the result stored as a numeric value (the boolean is implicitly converted to an integer)
(well.... technically a boolean is just an integer, but we'll just go along like so...)
'// True = -1
'// False = 0
Debug.Print 5 < 9 < 8
Debug.Print CInt(5 < 9) '// Prints -1
Debug.Print -1 < 8 '// = True
Which is why the following gives "False" instead:
Debug.Print 5 < 9 < -1
Because
Debug.Print Cint(5 < 9) '// True = -1
Debug.Print -1 < -1 '// False
If you want to find out if something is in the middle of two other numbers then you have to use the And operator to force a separate evaluation (either side of the operator is then evaluated and compared logically)
Debug.Print (3 < 5 And 5 < 4) '// False
Looking at it from a parse tree perspective might shed more light about why it works that way.
Excluding whatever instruction comes after the THEN token, the parse tree for If 5 < X < 8 Then might look something like this (quite simplified):
The comparison operators being a binary operator, there's an expression on either side of it, and in order to resolve the Boolean expression for the IfBlockStatement, VBA needs to visit the tree nodes in a specific order: because VBA parses expressions left to right, the 5 < X part stands on its own as an expression, and then the result of that expression is used to resolve the {expression} < 8 part of the expression.
So when VBA resolves 5 < X, because that's a ComparisonExpression the result is a Boolean value; when that Boolean value then needs to be compared to the 8 integer literal, VBA performs an implicit type conversion and actually compares CInt({Boolean}) < 8, which will evaluate to True regardless of the result of the first expression, since False converts to 0 and True converts to -1 when expressed as an integer literal, and both are < 8.
These mechanics are built into how the runtime works, so in order to evaluate if X is between 5 and 8, you need to build your expression so that it's parsed as such:
If X > 5 And X < 8 Then
That gives you two distinct expression trees joined by a LogicalAndOperator, which then works off a valid Boolean expression on either sides.
5<9<8 = True<8 = True
5<4<8 = False<8 = True
The other answers covered up nicely the first part of your question, but didn't satisfactorily cover up the second part of it, i.e. What is a quick and pretty way of writing If 5 < X < 8 Then (to use sample numbers), without having to resort to endless And statements, i.e. If 5 < x And X < 8 Then?
There are two ways. The first:
Select Case X
Case 5 To 8
...
End Select
Here, the value before the To keyword must be the smaller value of the two. Also note that while this will work for integers, I have no idea if it works for types like Double and such (I suspect it won't though).
The second way, which works irrespective of whether the interval bounds are integers or not, is not necessarily shorter, but it evaluates things in a single comparison:
If Sgn(x - 5) + Sgn(x - 8) = 0 Then ...
This is an interesting way of evaluating whether a value is between some bounds, because it can also provide information on whether the value is equal to one of those bounds or is "outside" them (and on which "side" it is). For example, on a -∞..0..+∞ axis:
if x = 4, the expression above is -2, thus x is to the left of the (5..8) interval
if x = 5, the expression above is -1, thus x is the left bound of the (5..8) interval
if x = 6, the expression above is 0, thus x is inside the (5..8) interval, i.e. between its bounds
if x = 8, the expression above is 1, thus x is the right bound of the (5..8) interval
if x = 9, the expression above is 2, thus x is to the right of the (5..8) interval
Of course, if you want to include the bounds in the interval, say, test If 5 <= x And X <= 8 Then, the comparison above becomes If Abs(Sgn(x - 5) + Sgn(x - 8)) < 2 Then ..., which is another shortcut to check if the expression is -1, 0 or 1.
In the end, none of the ways above are as short as a Between(x, 5, 8) hypothetical function, but at least they are alternatives to the "classical" method.
I'm sorry the title is so confusingly worded, but it's hard to condense this problem down to a few words.
I'm trying to find the minimum value of a specific equation. At first I'm looping through the equation, which for our purposes here can be something like y = .245x^3-.67x^2+5x+12. I want to design a loop where the "steps" through the loop get smaller and smaller.
For example, the first time it loops through, it uses a step of 1. I will get about 30 values. What I need help on is how do I Use the three smallest values I receive from this first loop?
Here's an example of the values I might get from the first loop: (I should note this isn't supposed to be actual code at all. It's just a brief description of what's happening)
loop from x = 1 to 8 with step 1
results:
x = 1 -> y = 30
x = 2 -> y = 28
x = 3 -> y = 25
x = 4 -> y = 21
x = 5 -> y = 18
x = 6 -> y = 22
x = 7 -> y = 27
x = 8 -> y = 33
I want something that can detect the lowest three values and create a loop. From theses results, the values of x that get the smallest three results for y are x = 4, 5, and 6.
So my "guess" at this point would be x = 5. To get a better "guess" I'd like a loop that now does:
loop from x = 4 to x = 6 with step .5
I could keep this pattern going until I get an absurdly accurate guess for the minimum value of x.
Does anybody know of a way I can do this? I know the values I'm going to get are going to be able to be modeled by a parabola opening up, so this format will definitely work. I was thinking that the values could be put into a column. It wouldn't be hard to make something that returns the smallest value for y in that column, and the corresponding x-value.
If I'm being too vague, just let me know, and I can answer any questions you might have.
nice question. Here's at least a start for what I think you should do for this:
Sub findMin()
Dim lowest As Integer
Dim middle As Integer
Dim highest As Integer
lowest = 999
middle = 999
hightest = 999
Dim i As Integer
i = 1
Do While i < 9
If (retVal(i) < retVal(lowest)) Then
highest = middle
middle = lowest
lowest = i
Else
If (retVal(i) < retVal(middle)) Then
highest = middle
middle = i
Else
If (retVal(i) < retVal(highest)) Then
highest = i
End If
End If
End If
i = i + 1
Loop
End Sub
Function retVal(num As Integer) As Double
retVal = 0.245 * Math.Sqr(num) * num - 0.67 * Math.Sqr(num) + 5 * num + 12
End Function
What I've done here is set three Integers as your three Min values: lowest, middle, and highest. You loop through the values you're plugging into the formula (here, the retVal function) and comparing the return value of retVal (hence the name) to the values of retVal(lowest), retVal(middle), and retVal(highest), replacing them as necessary. I'm just beginning with VBA so what I've done likely isn't very elegant, but it does at least identify the Integers that result in the lowest values of the function. You may have to play around with the values of lowest, middle, and highest a bit to make it work. I know this isn't EXACTLY what you're looking for, but it's something along the lines of what I think you should do.
There is no trivial way to approach this unless the problem domain is narrowed.
The example polynomial given in fact has no minimum, which is readily determined by observing y'>0 (hence, y is always increasing WRT x).
Given the wide interpretation of
[an] equation, which for our purposes here can be something like y =
.245x^3-.67x^2+5x+12
many conditions need to be checked, even assuming the domain is limited to polynomials.
The polynomial order is significant, and the order determines what conditions are necessary to check for how many solutions are possible, or whether any solution is possible at all.
Without taking this complexity into account, an iterative approach could yield an incorrect solution due to underflow error, or an unfortunate choice of iteration steps or bounds.
I'm not trying to be hard here, I think your idea is neat. In practice it is more complicated than you think.
I am trying to verify that a variable is NOT equal to either this or that. I tried using the following codes, but neither works:
if x ~=(0 or 1) then
print( "X must be equal to 1 or 0" )
return
end
if x ~= 0 or 1 then
print( "X must be equal to 1 or 0" )
return
end
Is there a way to do this?
Your problem stems from a misunderstanding of the or operator that is common to people learning programming languages like this. Yes, your immediate problem can be solved by writing x ~= 0 and x ~= 1, but I'll go into a little more detail about why your attempted solution doesn't work.
When you read x ~=(0 or 1) or x ~= 0 or 1 it's natural to parse this as you would the sentence "x is not equal to zero or one". In the ordinary understanding of that statement, "x" is the subject, "is not equal to" is the predicate or verb phrase, and "zero or one" is the object, a set of possibilities joined by a conjunction. You apply the subject with the verb to each item in the set.
However, Lua does not parse this based on the rules of English grammar, it parses it in binary comparisons of two elements based on its order of operations. Each operator has a precedence which determines the order in which it will be evaluated. or has a lower precedence than ~=, just as addition in mathematics has a lower precedence than multiplication. Everything has a lower precedence than parentheses.
As a result, when evaluating x ~=(0 or 1), the interpreter will first compute 0 or 1 (because of the parentheses) and then x ~= the result of the first computation, and in the second example, it will compute x ~= 0 and then apply the result of that computation to or 1.
The logical operator or "returns its first argument if this value is different from nil and false; otherwise, or returns its second argument". The relational operator ~= is the inverse of the equality operator ==; it returns true if its arguments are different types (x is a number, right?), and otherwise compares its arguments normally.
Using these rules, x ~=(0 or 1) will decompose to x ~= 0 (after applying the or operator) and this will return 'true' if x is anything other than 0, including 1, which is undesirable. The other form, x ~= 0 or 1 will first evaluate x ~= 0 (which may return true or false, depending on the value of x). Then, it will decompose to one of false or 1 or true or 1. In the first case, the statement will return 1, and in the second case, the statement will return true. Because control structures in Lua only consider nil and false to be false, and anything else to be true, this will always enter the if statement, which is not what you want either.
There is no way that you can use binary operators like those provided in programming languages to compare a single variable to a list of values. Instead, you need to compare the variable to each value one by one. There are a few ways to do this. The simplest way is to use De Morgan's laws to express the statement 'not one or zero' (which can't be evaluated with binary operators) as 'not one and not zero', which can trivially be written with binary operators:
if x ~= 1 and x ~= 0 then
print( "X must be equal to 1 or 0" )
return
end
Alternatively, you can use a loop to check these values:
local x_is_ok = false
for i = 0,1 do
if x == i then
x_is_ok = true
end
end
if not x_is_ok then
print( "X must be equal to 1 or 0" )
return
end
Finally, you could use relational operators to check a range and then test that x was an integer in the range (you don't want 0.5, right?)
if not (x >= 0 and x <= 1 and math.floor(x) == x) then
print( "X must be equal to 1 or 0" )
return
end
Note that I wrote x >= 0 and x <= 1. If you understood the above explanation, you should now be able to explain why I didn't write 0 <= x <= 1, and what this erroneous expression would return!
For testing only two values, I'd personally do this:
if x ~= 0 and x ~= 1 then
print( "X must be equal to 1 or 0" )
return
end
If you need to test against more than two values, I'd stuff your choices in a table acting like a set, like so:
choices = {[0]=true, [1]=true, [3]=true, [5]=true, [7]=true, [11]=true}
if not choices[x] then
print("x must be in the first six prime numbers")
return
end
x ~= 0 or 1 is the same as ((x ~= 0) or 1)
x ~=(0 or 1) is the same as (x ~= 0).
try something like this instead.
function isNot0Or1(x)
return (x ~= 0 and x ~= 1)
end
print( isNot0Or1(-1) == true )
print( isNot0Or1(0) == false )
print( isNot0Or1(1) == false )
I have a question about the SQL standard which I'm hoping a SQL language lawyer can help with.
Certain expressions just don't work. 62 / 0, for example. The SQL standard specifies quite a few ways in which expressions can go wrong in similar ways. Lots of languages deal with these expressions using special exceptional flow control, or bottom psuedo-values.
I have a table, t, with (only) two columns, x and y each of type int. I suspect it isn't relevant, but for definiteness let's say that (x,y) is the primary key of t. This table contains (only) the following values:
x y
7 2
3 0
4 1
26 5
31 0
9 3
What behavior is required by the SQL standard for SELECT expressions operating on this table which may involve division(s) by zero? Alternatively, if no one behavior is required, what behaviors are permitted?
For example, what behavior is required for the following select statements?
The easy one:
SELECT x, y, x / y AS quot
FROM t
A harder one:
SELECT x, y, x / y AS quot
FROM t
WHERE y != 0
An even harder one:
SELECT x, y, x / y AS quot
FROM t
WHERE x % 2 = 0
Would an implementation (say, one that failed to realize on a more complex version of this query that the restriction could be moved inside the extension) be permitted to produce a division by zero error in response to this query, because, say it attempted to divide 3 by 0 as part of the extension before performing the restriction and realizing that 3 % 2 = 1? This could become important if, for example, the extension was over a small table but the result--when joined with a large table and restricted on the basis of data in the large table--ended up restricting away all of the rows which would have required division by zero.
If t had millions of rows, and this last query were performed by a table scan, would an implementation be permitted to return the first several million results before discovering a division by zero near the end when encountering one even value of x with a zero value of y? Would it be required to buffer?
There are even worse cases, ponder this one, which depending on the semantics can ruin boolean short-circuiting or require four-valued boolean logic in restrictions:
SELECT x, y
FROM t
WHERE ((x / y) >= 2) AND ((x % 2) = 0)
If the table is large, this short-circuiting problem can get really crazy. Imagine the table had a million rows, one of which had a 0 divisor. What would the standard say is the semantics of:
SELECT CASE
WHEN EXISTS
(
SELECT x, y, x / y AS quot
FROM t
)
THEN 1
ELSE 0
END AS what_is_my_value
It seems like this value should probably be an error since it depends on the emptiness or non-emptiness of a result which is an error, but adopting those semantics would seem to prohibit the optimizer for short-circuiting the table scan here. Does this existence query require proving the existence of one non-bottoming row, or also the non-existence of a bottoming row?
I'd appreciate guidance here, because I can't seem to find the relevant part(s) of the specification.
All implementations of SQL that I've worked with treat a division by 0 as an immediate NaN or #INF. The division is supposed to be handled by the front end, not by the implementation itself. The query should not bottom out, but the result set needs to return NaN in this case. Therefore, it's returned at the same time as the result set, and no special warning or message is brought up to the user.
At any rate, to properly deal with this, use the following query:
select
x, y,
case y
when 0 then null
else x / y
end as quot
from
t
To answer your last question, this statement:
SELECT x, y, x / y AS quot
FROM t
Would return this:
x y quot
7 2 3.5
3 0 NaN
4 1 4
26 5 5.2
31 0 NaN
9 3 3
So, your exists would find all the rows in t, regardless of what their quotient was.
Additionally, I was reading over your question again and realized I hadn't discussed where clauses (for shame!). The where clause, or predicate, should always be applied before the columns are calculated.
Think about this query:
select x, y, x/y as quot from t where x%2 = 0
If we had a record (3,0), it applies the where condition, and checks if 3 % 2 = 0. It does not, so it doesn't include that record in the column calculations, and leaves it right where it is.