I am playing with the CVC5 example at https://github.com/cvc5/cvc5/blob/main/examples/api/python/pythonic/linear_arith.py.
from cvc5.pythonic import *
slv = SolverFor('QF_LIRA')
x = Int('x')
y = Real('y')
slv += And(x >= 3 * y, x <= y, -2 < x)
slv.push()
print(slv.check(y-x <= 2/3))
slv.pop()
slv.push()
slv += y-x == 2/3
print(slv.check())
slv.pop()
It works as it is supposed to work.
However, whenever I try to print the content of the formula (i.e., print(slv)), it raises the following error: Cannot print: Kind.CONST_INTEGER
The same happens with literals that compound the formula: i.e., print(x >= 3); but not with variables: print(x) returns x.
I would like to have this printing capability, since Z3 allows it and I am trying my (originally-in-Z3-made) implementation with different SMT sovlers. Any idea?
Note that print(slv) does return info ([]), when it is empty. I tried using str(), but the error persists and indeed I guess print() uses str() before printing.
PS: I am using CVC5, should I use CVC4 or is CVC5 mature enough?
I think this is a bug. You should report it to the CVC5 folks at https://github.com/cvc5/cvc5/issues. (i.e., they should be able to handle this case just fine.)
In the interim, you can use the following workaround:
print(slv.sexpr())
which prints:
(and (let ((_let_1 (to_real x))) (and (>= _let_1 (* 3.0 y)) (<= _let_1 y) (> x (- 2)))))
which takes a bit of squinting to see that this is what you asserted, but it should do the trick.
Can anybody please help me how this code works,
I am not getting it by myself, some help would be greatly appreciated.
Prime number in scala using recusion:
def isPrime(n: Int): Boolean = {
def isPrimeUntil(t: Int): Boolean =
if (t<=1) true
else n%t != 0 && isPrimeUntil(t-1)
isPrimeUntil(n/2)
}
The number n is prime if and only if there's no number t such that t != 1, t!= n, n % t = 0.
So, if you find some number from 2 to n-1 such that n % t = 0, n is composite, otherwise it is prime.
One more thing, you may see that there's no need to search for divisors among the numbers greater than n/2.
So, all the algorithm does is checks n % d for each t from n/2 to 2. As soon as it is found, the algorithms stops ans says it's composite (returns False). Otherwise it gets to t = 1 and assures the number is prime (returns True).
Just to mention, it's enough to consider the numbers from ceil(sqrt(n)) to 2, which results in better time complexity (O(sqrt(n)) vs O(n)).
isPrime(7) --> isPrimeUntil(3) --> (3 <= 1)? no
(7%3 != 0)? yes
isPrimeUntil(2) --> (2 <= 1)? no
(7%2 != 0)? yes
isPrimeUntil(1) --> (1 <= 1)? yes
isPrime(7) is true. No divisor was found between 1 and 7/2.
isPrime(9) --> isPrimeUntil(4) --> (4 <= 1)? no
(9%4 != 0)? yes
isPrimeUntil(3) --> (3 <= 1)? no
(9%3 != 0)? no
isPrime(9) is false. Found that 9 is divisible by 3.
If you have a local Scala REPL, you should paste this function in there and play around with it. If not, there's always Scastie. I made a Scastie snippet, in which I changed the formatting to my liking, added comments and a demo range.
There are examples of Scala that make it look almost like Malbolge. This one is not that bad.
Let's follow it through with a composite number like 102. Calling isPrime(102) causes isPrimeUntil(51) to be invoked (as 51 is half 102). Since 51 is greater than 1, the nested function calculated 102 % 51, which is 0, so, by "short-circuit evaluation" of logical AND, the nested function should return false.
Now let's try it with 103. Calling isPrime(103) causes isPrimeUntil(51) to be invoked (as 51 is half 103 and the remainder of 1 is simply discarded). Since 51 is greater than 1, the nested function calculated 103 % 51, which is 1, so the nested function calls itself as primeUntil(50). Since 50 is greater than 1, the... so on and so forth until calling itself as primeUntil(1). Since t == 1, primeUntil returns true and the recursion stops.
This gives the wrong answer for negative composite numbers. Plus, as others have mentioned, it is inefficient to start the recursion at n/2. This would be an improvement:
def isPrime(n: Int): Boolean = {
def isPrimeUntil(t: Int): Boolean = {
if (t <= 1) true else n % t != 0 && isPrimeUntil(t - 1)
}
isPrimeUntil(Math.floor(Math.sqrt(Math.abs(n))).toInt)
}
Hmm... it's still giving the wrong answer for −1, 0, 1. But hopefully now you understand this function.
For an example function like:
singleDigits = (list) ->
return false for i in list when i > 9
true
I'm wondering whether or not that would be possible without the lone trueat the end
e.g.
singleDigits = (list) -> return true unless false for i in list when i > 9
(I know that is not working, just to illustrate what I'm asking)
I think you could use the some method in Array.
someDigits = (list) -> list.some (digit) -> digit > 9
The advantage over reduce is that it will stop in the moment the predicate becomes true, whereas reduce will still finish going over the entire array.
See Array.prototype.some reference.
How about:
singleDigits = (list) ->
list.reduce (previous = true, next) -> previous and next < 10
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 :) )
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?