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.
Lets say I have array,and number of rows as follows and I want to decide in which order they come.
set of int: people = 1..10;
int row= 3
include "alldifferent.mzn";
constraint alldifferent(position);
array[people] of var people: position;
I have to people in the front row to come before people in behind rows; regardless of a column. Two people are in the same row if person1 mod row=person2 mod row, e.g. persons 2,5,8 are in the same row. I want to force this order in decision variable positions.
I found a naive way of doing so involving forall and exists which I believe is not efficient:
constraint forall(person in position where person>row)(exists(smallerpos in 1..person-1)(smallerpos div row = person div row));
I'm not sure I understand the problem correctly, but the model is quite faster if you add all_different(position), e.g.
include "globals.mzn"; % adding this
set of int: people = 1..10;
int row= 3
array[people] of var people: position;
constraint
forall(pos1,pos2 in people where pos1 < pos2 /\ (position[pos1] mod row)=(position[pos2] mod row))
(
position[pos1] < position[pos2]
)
/\ all_different(position) % adding this
;
Without all_different it takes 0.9s to get all 4200 solutions, with all_different it takes 0.25s.
Note: I first thought of the following solution, but since it don't give the same number of solution it's probably not correct. It only give 1201 solutions, but it's faster: 0.05s. As mentioned above, I probably don't understand your problem correctly.
constraint
forall(r in 0..row-1) (
increasing([position[p] | p in people where p mod row == r])
)
/\ all_different(position)
;
This works but is less efficient.
constraint forall(pos1,pos2 in people where pos1 < pos2 /\ (position[pos1] mod rows)=(position[pos2] mod rows))(
position[pos1] < position[pos2] );
Another proposition is, this one is not any more efficeint either
constraint forall(pos1,pos2 in people where (position[pos1] mod rows)=(position[pos2] mod rows))(
pos1 < pos2 <-> position[pos1] < position[pos2] );
I am having trivial problems converting integer division to a floating point solution in Emacs Lisp 24.5.1.
(message "divide: %2.1f" (float (/ 1 2)))
"divide: 0.0"
I believe this expression is first calculating 1/2, finds it is 0 after truncating, then assigning 0.0 to the float. Obviously, I'm hoping for 0.5.
What am I not seeing here? Thanks
The / function performs a floating-point division if at least one of its argument is a float, and an integer quotient operation (rounded towards 0) if all of its arguments are integers. If you want to perform a floating-point division, make sure that at least one of the arguments is a float.
(message "divide: %2.1f" (/ (float 1) 2))
(or of course if they're constants you can just write (/ 1.0 2) or (/ 1 2.0))
Many programming languages work this way.
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 am studying for my algorithms class. I have a question in context to the Masters theorem:
How is n.log2(n) polynomially larger than n^(log4(3))
(log2(x) = log to the base 2 of x
log4(x) = log to the base 4 of x)
(Note: This is a solved problem on page.95 of 'Introduction to Algorithms' by Cormen et.al.)
log4(3) is less than 1, and so n^(log4(3)) is less than n^1 = n, which is less than n*log2(n).