I have two fields with range type.
Are these two where expressions will give same results?
where range1 && range2
where not isempty( range1 * range2 )
They will, indeed.
Both overlap operator (&&) and intersection (*) are evaluated correctly, including cases where the 2 ranges have a common bound and their intersection:
only contains 1 element (= the bound is included in both)
contains no elements (= the bound is not included in at least 1 of the 2 ranges)
The following query tests pretty much all the cases (= intersections with more than 1 point, exactly 1 point, 0 points but "close", "truly" 0 points, and all the combinations of lower/upper bounds inclusions):
WITH r(range) AS (
VALUES (numrange(0,1,'[]')), (numrange(1,2,'[]')), (numrange(0,2,'[]')), (numrange(5,6, '[]')),
(numrange(0,1,'[)')), (numrange(1,2,'[)')), (numrange(0,2,'[)')), (numrange(5,6, '[)')),
(numrange(0,1,'(]')), (numrange(1,2,'(]')), (numrange(0,2,'(]')), (numrange(5,6, '(]')),
(numrange(0,1,'()')), (numrange(1,2,'()')), (numrange(0,2,'()')), (numrange(5,6, '()'))
)
SELECT * FROM (
SELECT r1.range, r2.range, r1.range && r2.range AS UsingOverlap, NOT isempty(r1.range * r2.range) AS UsingIntersect
FROM r r1, r r2
) T
Feel free to add WHERE UsingOverlap <> UsingIntersect.
Related
I have a 16-bit WORD and I want to read the status of a specific bit or several bits.
I've tried a method that divides the word by the bit that I want, converts the result to two values - an integer and to a real, and compares the two. if they are not equal, then it it equates to false. This appears to only work if i am looking for a bit that the last 'TRUE' bit in the word. If there are any successive TRUE bits, it fails. Perhaps I just haven't done it right. I don't have the ability to use code, just basic math, boolean operations, and type conversion. Any ideas? I hope this isn't a dumb question but i have a feeling it is.
eg:
WORD 0010000100100100 = 9348
I want to know the value of bit 2. how can i determine it from 9348?
There are many ways, depending on what operations you can use. It appears you don't have much to choose from. But this should work, using just integer division and multiplication, and a test for equality.
(psuedocode):
x = 9348 (binary 0010000100100100, bit 0 = 0, bit 1 = 0, bit 2 = 1, ...)
x = x / 4 (now x is 1000010010010000
y = (x / 2) * 2 (y is 0000010010010000)
if (x == y) {
(bit 2 must have been 0)
} else {
(bit 2 must have been 1)
}
Every time you divide by 2, you move the bits to the left one position (in your big endian representation). Every time you multiply by 2, you move the bits to the right one position. Odd numbers will have 1 in the least significant position. Even numbers will have 0 in the least significant position. If you divide an odd number by 2 in integer math, and then multiply by 2, you loose the odd bit if there was one. So the idea above is to first move the bit you want to know about into the least significant position. Then, divide by 2 and then multiply by two. If the result is the same as what you had before, then there must have been a 0 in the bit you care about. If the result is not the same as what you had before, then there must have been a 1 in the bit you care about.
Having explained the idea, we can simplify to
((x / 8) * 2) <> (x / 4)
which will resolve to true if the bit was set, and false if the bit was not set.
AND the word with a mask [1].
In your example, you're interested in the second bit, so the mask (in binary) is
00000010. (Which is 2 in decimal.)
In binary, your word 9348 is 0010010010000100 [2]
0010010010000100 (your word)
AND 0000000000000010 (mask)
----------------
0000000000000000 (result of ANDing your word and the mask)
Because the value is equal to zero, the bit is not set. If it were different to zero, the bit was set.
This technique works for extracting one bit at a time. You can however use it repeatedly with different masks if you're interested in extracting multiple bits.
[1] For more information on masking techniques see http://en.wikipedia.org/wiki/Mask_(computing)
[2] See http://www.binaryhexconverter.com/decimal-to-binary-converter
The nth bit is equal to the word divided by 2^n mod 2
I think you'll have to test each bit, 0 through 15 inclusive.
You could try 9348 AND 4 (equivalent of 1<<2 - index of the bit you wanted)
9348 AND 4
should give 4 if bit is set, 0 if not.
So here is what I have come up with: 3 solutions. One is Hatchet's as proposed above, and his answer helped me immensely with actually understanding HOW this works, which is of utmost importance to me! The proposed AND masking solutions could have worked if my system supports bitwise operators, but it apparently does not.
Original technique:
( ( ( INT ( TAG / BIT ) ) / 2 ) - ( INT ( ( INT ( TAG / BIT ) ) / 2 ) ) <> 0 )
Explanation:
in the first part of the equation, integer division is performed on TAG/BIT, then REAL division by 2. In the second part, integer division is performed TAG/BIT, then integer division again by 2. The difference between these two results is compared to 0. If the difference is not 0, then the formula resolves to TRUE, which means the specified bit is also TRUE.
eg: 9348/4 = 2337 w/ integer division. Then 2337/2 = 1168.5 w/ REAL division but 1168 w/ integer division. 1168.5-1168 <> 0, so the result is TRUE.
My modified technique:
( INT ( TAG / BIT ) / 2 ) <> ( INT ( INT ( TAG / BIT ) / 2 ) )
Explanation:
effectively the same as above, but instead of subtracting the two results and comparing them to 0, I am just comparing the two results themselves. If they are not equal, the formula resolves to TRUE, which means the specified bit is also TRUE.
eg: 9348/4 = 2337 w/ integer division. Then 2337/2 = 1168.5 w/ REAL division but 1168 w/ integer division. 1168.5 <> 1168, so the result is TRUE.
Hatchet's technique as it applies to my system:
( INT ( TAG / BIT )) <> ( INT ( INT ( TAG / BIT ) / 2 ) * 2 )
Explanation:
in the first part of the equation, integer division is performed on TAG/BIT. In the second part, integer division is performed TAG/BIT, then integer division again by 2, then multiplication by 2. The two results are compared. If they are not equal, the formula resolves to TRUE, which means the specified bit is also TRUE.
eg: 9348/4 = 2337. Then 2337/2 = 1168 w/ integer division. Then 1168x2=2336. 2337 <> 2336 so the result is TRUE. As Hatchet stated, this method 'drops the odd bit'.
Note - 9348/4 = 2337 w/ both REAL and integer division, but it is important that these parts of the formula use integer division and not REAL division (12164/32 = 380 w/ integer division and 380.125 w/ REAL division)
I feel it important to note for any future readers that the BIT value in the equations above is not the bit number, but the actual value of the resulting decimal if the bit in the desired position was the only TRUE bit in the binary string (bit 2 = 4 (2^2), bit 6 = 64 (2^6))
This explanation may be a bit too verbatim for some, but may be perfect for others :)
Please feel free to comment/critique/correct me if necessary!
I just needed to resolve an integer status code to a bit state in order to interface with some hardware. Here's a method that works for me:
private bool resolveBitState(int value, int bitNumber)
{
return (value & (1 << (bitNumber - 1))) != 0;
}
I like it, because it's non-iterative, requires no cast operations and essentially translates directly to machine code operations like Shift, And and Comparison, which probably means it's really optimal.
To explain in a little more detail, I'm comparing the bitwise value to a mask for the bit I am interested in (value & mask) using an AND operation. If the bitwise AND operation result is zero, then the bit is not set (return false). If the AND operation result is not zero, then the bit is set (return true). The result of the AND operation is either zero or the value of the bit (1, 2, 4, 8, 16, 32...). Hence the boolean evaluation comparing the AND operation result and 0. The mask is created by taking the number 1 and shifting it left (bit wise), by the appropriate number of binary places (1 << n). The number of places is the number of the bit targeted minus 1. If it's bit #1, I want to shift the 1 left by 0 and if it's #2, I want to shift it left 1 place, etc.
I'm surprised no one rates my solution. It think it's most logical and succinct... and works.
To form a matrix consisting of identical rows, one could use
x:1 2 3
2 3#x,x
which produces (1 2 3i;1 2 3i) as expected. However, attempting to generalise this thus:
2 (count x)#x,x
produces a type error although the types are equal:
(type 3) ~ type count x
returns 1b. Why doesn't this work?
The following should work.
q)(2;count x)#x,x
1 2 3
1 2 3
If you look at the parse tree of both your statements you can see that the second is evaluated differently. In the second only the result of count is passed as an argument to #.
q)parse"2 3#x,x"
#
2 3
(,;`x;`x)
q)parse"2 (count x)#x,x"
2
(#;(#:;`x);(,;`x;`x))
If you're looking to build matrices with identical rows you might be better off using
rownum#enlist x
q)x:100000?100
q)\ts do[100;v1:5 100000#x,x]
157 5767696j
q)\ts do[100;v2:5#enlist x]
0 992j
q)v1~v2
1b
I for one find this more natural (and its faster!)
So I've spent hours trying to work out exactly how this code produces prime numbers.
lazy val ps: Stream[Int] = 2 #:: Stream.from(3).filter(i =>
ps.takeWhile{j => j * j <= i}.forall{ k => i % k > 0});
I've used a number of printlns etc, but nothings making it clearer.
This is what I think the code does:
/**
* [2,3]
*
* takeWhile 2*2 <= 3
* takeWhile 2*2 <= 4 found match
* (4 % [2,3] > 1) return false.
* takeWhile 2*2 <= 5 found match
* (5 % [2,3] > 1) return true
* Add 5 to the list
* takeWhile 2*2 <= 6 found match
* (6 % [2,3,5] > 1) return false
* takeWhile 2*2 <= 7
* (7 % [2,3,5] > 1) return true
* Add 7 to the list
*/
But If I change j*j in the list to be 2*2 which I assumed would work exactly the same, it causes a stackoverflow error.
I'm obviously missing something fundamental here, and could really use someone explaining this to me like I was a five year old.
Any help would be greatly appreciated.
I'm not sure that seeking a procedural/imperative explanation is the best way to gain understanding here. Streams come from functional programming and they're best understood from that perspective. The key aspects of the definition you've given are:
It's lazy. Other than the first element in the stream, nothing is computed until you ask for it. If you never ask for the 5th prime, it will never be computed.
It's recursive. The list of prime numbers is defined in terms of itself.
It's infinite. Streams have the interesting property (because they're lazy) that they can represent a sequence with an infinite number of elements. Stream.from(3) is an example of this: it represents the list [3, 4, 5, ...].
Let's see if we can understand why your definition computes the sequence of prime numbers.
The definition starts out with 2 #:: .... This just says that the first number in the sequence is 2 - simple enough so far.
The next part defines the rest of the prime numbers. We can start with all the counting numbers starting at 3 (Stream.from(3)), but we obviously need to filter a bunch of these numbers out (i.e., all the composites). So let's consider each number i. If i is not a multiple of a lesser prime number, then i is prime. That is, i is prime if, for all primes k less than i, i % k > 0. In Scala, we could express this as
nums.filter(i => ps.takeWhile(k => k < i).forall(k => i % k > 0))
However, it isn't actually necessary to check all lesser prime numbers -- we really only need to check the prime numbers whose square is less than or equal to i (this is a fact from number theory*). So we could instead write
nums.filter(i => ps.takeWhile(k => k * k <= i).forall(k => i % k > 0))
So we've derived your definition.
Now, if you happened to try the first definition (with k < i), you would have found that it didn't work. Why not? It has to do with the fact that this is a recursive definition.
Suppose we're trying to decide what comes after 2 in the sequence. The definition tells us to first determine whether 3 belongs. To do so, we consider the list of primes up to the first one greater than or equal to 3 (takeWhile(k => k < i)). The first prime is 2, which is less than 3 -- so far so good. But we don't yet know the second prime, so we need to compute it. Fine, so we need to first see whether 3 belongs ... BOOM!
* It's pretty easy to see that if a number n is composite then the square of one of its factors must be less than or equal to n. If n is composite, then by definition n == a * b, where 1 < a <= b < n (we can guarantee a <= b just by labeling the two factors appropriately). From a <= b it follows that a^2 <= a * b, so it follows that a^2 <= n.
Your explanations are mostly correct, you made only two mistakes:
takeWhile doesn't include the last checked element:
scala> List(1,2,3).takeWhile(_<2)
res1: List[Int] = List(1)
You assume that ps always contains only a two and a three but because Stream is lazy it is possible to add new elements to it. In fact each time a new prime is found it is added to ps and in the next step takeWhile will consider this new added element. Here, it is important to remember that the tail of a Stream is computed only when it is needed, thus takeWhile can't see it before forall is evaluated to true.
Keep these two things in mind and you should came up with this:
ps = [2]
i = 3
takeWhile
2*2 <= 3 -> false
forall on []
-> true
ps = [2,3]
i = 4
takeWhile
2*2 <= 4 -> true
3*3 <= 4 -> false
forall on [2]
4%2 > 0 -> false
ps = [2,3]
i = 5
takeWhile
2*2 <= 5 -> true
3*3 <= 5 -> false
forall on [2]
5%2 > 0 -> true
ps = [2,3,5]
i = 6
...
While these steps describe the behavior of the code, it is not fully correct because not only adding elements to the Stream is lazy but every operation on it. This means that when you call xs.takeWhile(f) not all values until the point when f is false are computed at once - they are computed when forall wants to see them (because it is the only function here that needs to look at all elements before it definitely can result to true, for false it can abort earlier). Here the computation order when laziness is considered everywhere (example only looking at 9):
ps = [2,3,5,7]
i = 9
takeWhile on 2
2*2 <= 9 -> true
forall on 2
9%2 > 0 -> true
takeWhile on 3
3*3 <= 9 -> true
forall on 3
9%3 > 0 -> false
ps = [2,3,5,7]
i = 10
...
Because forall is aborted when it evaluates to false, takeWhile doesn't calculate the remaining possible elements.
That code is easier (for me, at least) to read with some variables renamed suggestively, as
lazy val ps: Stream[Int] = 2 #:: Stream.from(3).filter(i =>
ps.takeWhile{p => p * p <= i}.forall{ p => i % p > 0});
This reads left-to-right quite naturally, as
primes are 2, and those numbers i from 3 up, that all of the primes p whose square does not exceed the i, do not divide i evenly (i.e. without some non-zero remainder).
In a true recursive fashion, to understand this definition as defining the ever increasing stream of primes, we assume that it is so, and from that assumption we see that no contradiction arises, i.e. the truth of the definition holds.
The only potential problem after that, is the timing of accessing the stream ps as it is being defined. As the first step, imagine we just have another stream of primes provided to us from somewhere, magically. Then, after seeing the truth of the definition, check that the timing of the access is okay, i.e. we never try to access the areas of ps before they are defined; that would make the definition stuck, unproductive.
I remember reading somewhere (don't recall where) something like the following -- a conversation between a student and a wizard,
student: which numbers are prime?
wizard: well, do you know what number is the first prime?
s: yes, it's 2.
w: okay (quickly writes down 2 on a piece of paper). And what about the next one?
s: well, next candidate is 3. we need to check whether it is divided by any prime whose square does not exceed it, but I don't yet know what the primes are!
w: don't worry, I'l give them to you. It's a magic I know; I'm a wizard after all.
s: okay, so what is the first prime number?
w: (glances over the piece of paper) 2.
s: great, so its square is already greater than 3... HEY, you've cheated! .....
Here's a pseudocode1 translation of your code, read partially right-to-left, with some variables again renamed for clarity (using p for "prime"):
ps = 2 : filter (\i-> all (\p->rem i p > 0) (takeWhile (\p->p^2 <= i) ps)) [3..]
which is also
ps = 2 : [i | i <- [3..], and [rem i p > 0 | p <- takeWhile (\p->p^2 <= i) ps]]
which is a bit more visually apparent, using list comprehensions. and checks that all entries in a list of Booleans are True (read | as "for", <- as "drawn from", , as "such that" and (\p-> ...) as "lambda of p").
So you see, ps is a lazy list of 2, and then of numbers i drawn from a stream [3,4,5,...] such that for all p drawn from ps such that p^2 <= i, it is true that i % p > 0. Which is actually an optimal trial division algorithm. :)
There's a subtlety here of course: the list ps is open-ended. We use it as it is being "fleshed-out" (that of course, because it is lazy). When ps are taken from ps, it could potentially be a case that we run past its end, in which case we'd have a non-terminating calculation on our hands (a "black hole"). It just so happens :) (and needs to ⁄ can be proved mathematically) that this is impossible with the above definition. So 2 is put into ps unconditionally, so there's something in it to begin with.
But if we try to "simplify",
bad = 2 : [i | i <- [3..], and [rem i p > 0 | p <- takeWhile (\p->p < i) bad]]
it stops working after producing just one number, 2: when considering 3 as the candidate, takeWhile (\p->p < 3) bad demands the next number in bad after 2, but there aren't yet any more numbers there. It "jumps ahead of itself".
This is "fixed" with
bad = 2 : [i | i <- [3..], and [rem i p > 0 | p <- [2..(i-1)] ]]
but that is a much much slower trial division algorithm, very far from the optimal one.
--
1 (Haskell actually, it's just easier for me that way :) )
Do they have a reason for doing so? I mean, in the sum of minterms, you look for the terms with the output 1; I don't get why they call it "minterms." Why not maxterms because 1 is well bigger than 0?
Is there a reason behind this that I don't know? Or should I just accept it without asking why?
The convention for calling these terms "minterms" and "maxterms" does not correspond to 1 being greater than 0. I think the best way to answer is with an example:
Say that you have a circuit and it is described by X̄YZ̄ + XȲZ.
"This form is composed of two groups of three. Each group of three is a 'minterm'. What the expression minterm is intended to imply it that each of the groups of three in the expression takes on a value of 1 only for one of the eight possible combinations of X, Y and Z and their inverses." http://www.facstaff.bucknell.edu/mastascu/elessonshtml/Logic/Logic2.html
So what the "min" refers to is the fact that these terms are the "minimal" terms you need in order to build a certain function. If you would like more information, the example above is explained in more context in the link provided.
Edit: The "reason they used MIN for ANDs, and MAX for ORs" is that:
In Sum of Products (what you call ANDs) only one of the minterms must be true for the expression to be true.
In Product of Sums (what you call ORs) all the maxterms must be true for the expression to be true.
min(0,0) = 0
min(0,1) = 0
min(1,0) = 0
min(1,1) = 1
So minimum is pretty much like logical AND.
max(0,0) = 0
max(0,1) = 1
max(1,0) = 1
max(1,1) = 1
So maximum is pretty much like logical OR.
In Sum Of Products (SOP), each term of the SOP expression is called a "minterm" because,
say, an SOP expression is given as:
F(X,Y,Z) = X'.Y'.Z + X.Y'.Z' + X.Y'.Z + X.Y.Z
for this SOP expression to be "1" or true (being a positive logic),
ANY of the term of the expression should be 1.
thus the word "minterm".
i.e, any of the term (X'Y'Z) , (XY'Z') , (XY'Z) or (XYZ) being 1, results in F(X,Y,Z) to be 1!!
Thus they are called "minterms".
On the other hand,
In Product Of Sum (POS), each term of the POS expression is called a "maxterm" because,
say an POS expression is given as: F(X,Y,Z) = (X+Y+Z).(X+Y'+Z).(X+Y'+Z').(X'+Y'+Z)
for this POS expression to be "0" (because POS is considered as a negative logic and we consider 0 terms), ALL of the terms of the expression should be 0. thus the word "max term"!!
i.e for F(X,Y,Z) to be 0,
each of the terms (X+Y+Z), (X+Y'+Z), (X+Y'+Z') and (X'+Y'+Z) should be equal to "0", otherwise F won't be zero!!
Thus each of the terms in POS expression is called a MAXTERM (maximum all the terms!) because all terms should be zero for F to
be zero, whereas any of the terms in POS being one results in F to be
one. Thus it is known as MINTERM (minimum one term!)
I believe that AB is called a minterm is because it occupies the minimum area on a Venn diagram; while A+B is called a MAXTERM because it occupies a maximum area in a Venn diagram. Draw the two diagrams and the meanings will become obvious
Ed Brumgnach
Here is another way to think about it.
A product is called a minterm because it has minimum-satisfiability where as a sum is called a maxterm because it has maximum-satisfiability among all practically interesting boolean functions.
They are called terms because they are used as the building-blocks of various canonical representations of arbitrary boolean functions.
Details:
Note that '0' and '1' are the trivial boolean functions.
Assume a set of boolean variables x1,x2,...,xk and a non-trivial boolean function f(x1,x2,...,xk).
Conventionally, an input is said to satisfy the boolean function f, whenever f holds a value of 1 for that input.
Note that there are exactly 2^k inputs possible, and any non-trivial boolean-function can satisfy a minimum of 1 input to a maximum of 2^k -1 inputs.
Now consider the two simple boolean functions of interest: sum of all variables S, and product of all variables P (variables may/may-not appear as complements). S is one boolean function that has maximum-satisfiability hence called as maxterm, where as P is the one having minimum-satisfiability hence called a minterm.
I want to return a data.frame from a function if TRUE, else return NA using return(ifelse(condition, mydf, NA))
However, ifelse strips the column names from the data.frame.
Why are these results different?
> data.frame(1)
X1
1 1
> ifelse(TRUE, data.frame(1), NA)
[[1]]
[1] 1
Some additional insight from dput():
> dput(ifelse(TRUE, data.frame(1), 0))
list(1)
> dput(data.frame(1))
structure(list(X1 = 1), .Names = "X1", row.names = c(NA, -1L),
class = "data.frame")
ifelse is generally intended for vectorized comparisons, and has side-effects such as these: as it says in ?ifelse,
‘ifelse’ returns a value with the same shape as ‘test’ ...
so in this case (test is a vector of length 1) it tries to convert the data frame to a 'vector' (list in this case) of length 1 ...
return(if (condition) mydf else NA)
As a general design point I try to return objects of the same structure no matter what, so I might prefer
if (!condition) mydf[] <- NA
return(mydf)
As a general rule, I find that R users (especially coming from other programming languages) start by using if exclusively, take a while to discover ifelse, then overuse it for a while, discovering later that you really want to use if in logical contexts. A similar thing happens with & and &&.
See also:
section 3.2 of Patrick Burns's R Inferno ...
Why can't R's ifelse statements return vectors?