Problem
Given two boolean functions f1(a,b) and f2(a,b,c) with a,b and c booleans, I'd like to know if there exists a value of c such that for any combinations of a et b f1(a,b)=f2(a,b,c).
For example if f1(a,b)=a AND b and f2(a,b,c)=a AND b AND c, we can see that f1=f2 when c=1. However, if f1(a,b)=a OR b and f2(a,b,c)=a AND b AND c, no value of c holds for f1=f2.
Failed approach
I tried to use model checking, with a trivial model implemented in nuXmv, to answer this question with a CTL specification such as EF ( AG ( (a & b) = (a & b & c))) but it fails.
Obviously, it works fine with the specification AG ( c=true -> (a & b = a & b & c)) but it requires to have 2^n specifications (with n being the difference between the number of variables between the two functions).
What is the best way/tool/approach in your opinion to tackle the issue.
Thanks for pointing to the right direction.
If I were using nuXmv for this task, I would write the following model
MODULE main
VAR
a : boolean;
b : boolean;
c : boolean;
CTLSPEC ((a & b) = (a & b & c));
and then incrementally remove from the state space all those states for which the property is falsified. For example, given this:
nuXmv > reset; read_model -i test.smv ; go ; check_property
-- specification (a & b) = ((a & b) & c) is false
-- as demonstrated by the following execution sequence
Trace Description: CTL Counterexample
Trace Type: Counterexample
-> State: 1.1 <-
a = TRUE
b = TRUE
c = FALSE
I would then write this (because c is the only watched variable):
MODULE main
VAR
a : boolean;
b : boolean;
c : boolean;
INVAR
c != FALSE;
CTLSPEC ((a & b) = (a & b & c));
which would actually allow for the iterative-refinement search to stop because now the property is true under all possible evaluations of a and b:
nuXmv > reset; read_model -i test.smv ; go ; check_property
-- specification (a & b) = ((a & b) & c) is true
Now, this does not appear to be the ideal solution because it requires parsing the output of nuXmv and correspondingly modify the original file. Compared to your approach, that checks all possible assignments of values, this method enumerates only those assignments which make the formula unsatisfiable. Still, there could be exponentially many counter-examples so it is not that much of an improvement.
One alternative is to use SMT with universal quantification over an EUF formula including the definitions of f1 and f2, where the only free variable is c:
(define-fun f1 ((a Bool) (b Bool)) Bool
(and a b)
)
(define-fun f2 ((a Bool) (b Bool) (c Bool)) Bool
(and a b c)
)
(declare-fun c () Bool)
(assert (forall
((a Bool) (b Bool))
(=
(f1 a b)
(f2 a b c)
)
))
(check-sat)
(get-model)
This gives you the following model with the SMT solver z3:
~$ z3 test3.smt2
sat
(model
(define-fun c () Bool
true)
)
If you would like to know all possible values of c for which f1 = f2, then your best shot is to implement an incremental allsat search on top of z3 (see: plenty of Q/A here on stackoverflow on the topic).
Related
I'm trying to get a grip on how tuples work in Haskell.
I came across this type constructor leftRight :: (Either a b -> c) -> (a -> c, b -> c) and I'm struggling to see what it does.
So we have (Either a b -> c)which means that either a is applied to c or b is applied to c. But the output is a tuple?
Does that mean that it splits the function so that left side of the tuple takes in argument given at a and right side takes argument given at right?
How would this look in code?
"I came across this type constructor..."
It's not a type constructor - it's a function type declaration.
The -> separates out the parameter types. The final one is the return type, and the previous ones are the input types.
Hence leftRight :: (Either a b -> c) -> (a -> c, b -> c) takes one input and returns one output.
Input function: (Either a b -> c)
Output function pair: (a -> c, b -> c)
The parentheses contain the functions.
The first function takes an Either type (left value is the error type, a, and the right value is the OK type, 'b' - it helps me to think of the latin, sinister for left, and dexter for right - your mileage may vary) as the input and returns something of type c.
The second function comes as a tuple of two separate functions, one is a -> c, and one is b -> c.
A concrete version: type a is a String to contain my error message, type bis an Int, and type c is another string.
leftRight :: (Either String Int -> String) -> (String -> String, Int -> String)
So we have Either a b -> c which means that either a is applied to c or b is applied to c
Wrong, or at least badly worded. Nothing is “applied to c” here.
What it actually means is: the function can accept an a-value or a b-value, and in either case produces a c-value.
This is equivalent to having both a function that accepts only a-values and gives c-results, and one that accepts only b-values and gives c-results. The tuple groups both of these functions together.
It might help to look at an example to illustrate:
fryPancake :: Either Butter Margarine -> Pancake
[Assume we've already taken care of the flour, eggs etc. in the batter]
From this you can obtain both
fryPancakeWithButter :: Butter -> Pancake
fryPancakeWithButter b = fryPancake (Left b)
fryPancakeWithMargarine :: Margarine -> Pancake
fryPancakeWithMargarine m = fryPancake (Right m)
Now you just group both of these together:
fryPancake' :: (Butter -> Pancake, Margarine -> Pancake)
fryPancake' = (fryPancakeWithButter, fryPancakeWithMargarine)
...which is the same as
fryPancake' = leftRight fryPancake
The input to leftRight is a function and the output is two functions. The tuple is barely relevant, except that it groups the two functions together into a single output, so leftRight can return both of them at once.
One function's type is a -> c and the other one's type is b -> c. Presumably, the first function wraps the a in Left and then calls the original function, and the second one wraps the b in Right and then calls the original function:
leftRight :: (Either a b -> c) -> (a -> c, b -> c)
leftRight original = (leftFunc, rightFunc)
where
leftFunc aVal = original (Left aVal)
rightFunc bVal = original (Right bVal)
The key is to use function composition. A function of type a -> c can be created from functions of type a -> Either a b and Either a b -> c. You have Left :: a -> Either a b and you have the argument f :: Either a b -> c passed to leftRight.
The same argument lets you construct a function of type b -> c similarly. This gives you
leftRight :: (Either a b -> c) -> (a -> c, b -> c)
leftRight f = let g1 = ...
g2 = ...
in (g1, g2)
I leave the definitions of g1 and g2 as an exercise.
I have following declaration:
data Two a b = Two a b deriving (Eq, Show)
instance (Semigroup a, Semigroup b) => Semigroup (Two a b) where
(Two a b) <> (Two c d) = Two (a <> c) (b <> d)
And tried it in the prelude:
*Main First Lib MonoidLaws Semi> (Two a b) <> (Two c d) = Two (a <> c) (b <> d)
<interactive>:10:3: error:
* Occurs check: cannot construct the infinite type: t1 ~ Two t1 t1
Expected type: t1 -> t -> b
Actual type: Two t1 t1 -> Two t t -> Two b b
* Relevant bindings include
(<>) :: t1 -> t -> b (bound at <interactive>:10:3)
How can I use mappend function from Semigroup for Two datatype in prelude?
Try it with something that's an instance of Semigroup, like List.
For example:
> (Two "1" "2") <> (Two "3" "4")
Two "13" "24"
In your attempt / example, a and b, c and d are not defined, so Haskell sees them as variables. Because you're using = between them, it's assuming you want to do pattern matching, so it's trying to match them to themselves, which is causing an infinite loop (as that is perfectly valid Haskell — to define values in terms of themselves). This is causing an error, though, because it would imply an infinite type, which it definitely seems is not what you want.
It might be worth starting with some simpler things, possibly. A good basic book which I helped author is http://happylearnhaskelltutorial.com but that doesn't deal with instantiating your own typeclasses. Having said that it'll give you a reasonably good understanding of pattern matching, variables, types and values which you'll need before you move on to understanding typeclasses enough to build your own instances.
I have a simulation with lots of calls to functions of the type F = A -> B -> C -> D, where A..D are concrete types.
The objects of type A have a medium lifetime. (It is the genome of codegolf's ratrace.)
The most expensive computation arises from parameter A. I can easily memoize like this:
f1 :: F
f1 a = let expensive = trace "expensive computation!" $ expensiveComputation a
in \b c -> expensive
and hold some pre-processed expensive values via partial application:
preProz :: [B -> C -> D]
preProz = [f1 [], f1 [False], f2 []]
The traces indicate that preProz <*> [[],[[]]] <*> [1,2] does not recompute the values to my delight.
Now I found out that some of my Fs would benefit from pre-processing B, too. This pre-processing is independent from A, and, in fact, memoizing like this has no benefit
f2 a = let expensive = trace "expensive computation!" $ expensiveComputation a
in \b -> let dear = trace "expensive computation!" $ expensiveComputation b
in expensive + dear
because dear is recomputed, even is the bs equal.
What I need is something like:
(B -> e) -> A -> e -> C -> D
where e should be memoized. The type of e is sort-of-existential here.
But this forces me to recompute all values A for every B, which is just as bad, and I cannot save the es, which are private to the function.
How can I memoize along 2 parameters independently?
You need a function that memoizes both a and b together:
f12 a b = ...
in \c -> ...
When you want to memoize a but not b, you use f1 a and when you want to memoize both you use f12 a b.
It would of course be nice to share some implementation between f1 and f12. However, you can do that only by having private functions that take the precomputed results in place of the original values:
f1 a = privateA (precomputeA a)
f12 a b = privateAB (precomputeA a) (precomputeB b)
privateA a' b = privateAB a' (precomputeB b)
private AB a' b' c = ...
If the precomputation of b depends on the precomputation of a, then:
f1 a = privateA (precomputeA a)
f12 a b = let a' = precomputeA a in privateAB a' (precomputeB a' b)
privateA a' b = privateAB a' (precomputeB a' b)
private AB a' b' c = ...
I've purposely not used function composition and eta-reduction, to make things clearer. I've also left out any strictness annotations that you might want to use to control times of evaluation.
Perhaps memoizing isn't quite the right term here. You mean something like "partial application with some precomputation as well."
I'm having some trouble understanding, how this type declaration works.
The type is: (a -> b) -> (b -> c) -> (c -> d) -> a -> d
So, to me I interpret this as a function that takes a function and that function takes another function which outputs a value d.
So, this is how I make my function:
Example :: (a -> b) -> (b -> c) -> (c -> d) -> a -> d
Example f g h x = f ( g ( h (x) )
I'd really appreciate it, if you guys could help me clarify. Thank you!
I think that you already know the theory behind the type you're writing, so I'll try to inject some intuitive way to read it (at least I hope so, your question is not totally clear to me).
When you read something like (a -> b) inside a type, that's a function, as you said. For example (Int -> Bool) is a function.
Let's make an example:
even :: Int -> Bool -- A more generic version of that is in the Prelude
even n = n `rem` 2 == 0
filter :: (Int -> Bool) -> [Int] -> [Int] -- And of that, too
filter _ [] = []
filter f (x:xs)
| f x = x : filter f xs
| otherwise = filter f xs
filteredEven :: [Int]
filteredEven = filter even [1..5] -- it gives [2, 4]
In this example we have a "high order function", a function that get another function and use it in some way.
In a function like the one you're defining you simply use 3 functions (and another parameter). But you can know more.
Each function you declare in the type accept a value returned from the previous one. So a possible solution is the one you have already showed. But the types are generic. There is not a total function that returns a generic value (where total means that it terminate always returning a value different from bottom if all the values are total and different by bottom, so it don't crash or return undefined, for example). So, if you wants a total function you have to have a way to generate the variables requested, from the context of the function (their parameters).
In the example before, using the names used by you, you have to return a value of type d. You only have a way to produce a value of that type, the h function. But to use the h function you have to get a value of type c. You only have the g function for that. But you need a value of type c. Fortunately you have the function f, that in exchange of a value of type a returns the value needed. We have this value (and don't have any other way to obtain a value of that type), so the function can be written. We can't in any way alter the values obtained (call multiple times the functions don't work, for purity and the fact that we have only a way to produce the values), so that's the only way to construct the function, if we wants it to be total:
Example (a -> b) -> (b -> c) -> (c -> d) -> a -> d
Example f g h x = h (g (f x)))
We can write the function in many other ways, but the results they give will be always the same (if Example, f, g and h are total and x is not bottom). So the type can express really well the function, because we can understand how the function works only looking at the type!
In most of programming languages that support mutable variables, one can easily implement something like this Java example:
interface Accepter<T> {
void accept(T t);
}
<T> T getFromDoubleAccepter(Accepter<Accepter<T>> acc){
final List<T> l = new ArrayList<T>();
acc.accept(new Accepter<T>(){
#Override
public void accept(T t) {
l.add(t);
}
});
return l.get(0); //Not being called? Exception!
}
Just for those do not understand Java, the above code receives something can can be provided a function that takes one parameter, and it supposed to grape this parameter as the final result.
This is not like callCC: there is no control flow alternation. Only the inner function's parameter is concerned.
I think the equivalent type signature in Haskell should be
getFromDoubleAccepter :: (forall b. (a -> b) -> b) -> a
So, if someone can gives you a function (a -> b) -> b for a type of your choice, he MUST already have an a in hand. So your job is to give them a "callback", and than keep whatever they sends you in mind, once they returned to you, return that value to your caller.
But I have no idea how to implement this. There are several possible solutions I can think of. Although I don't know how each of them would work, I can rate and order them by prospected difficulties:
Cont or ContT monad. This I consider to be easiest.
RWS monad or similar.
Any other monads. Pure monads like Maybe I consider harder.
Use only standard pure functional features like lazy evaluation, pattern-matching, the fixed point contaminator, etc. This I consider the hardest (or even impossible).
I would like to see answers using any of the above techniques (and prefer harder ways).
Note: There should not be any modification of the type signature, and the solution should do the same thing that the Java code does.
UPDATE
Once I seen somebody commented out getFromDoubleAccepter f = f id I realize that I have made something wrong. Basically I use forall just to make the game easier but it looks like this twist makes it too easy. Actually, the above type signature forces the caller to pass back whatever we gave them, so if we choose a as b then that implementation gives the same expected result, but it is just... not expected.
Actually what came up to my mind is a type signature like:
getFromDoubleAccepter :: ((a -> ()) -> ()) -> a
And this time it is harder.
Another comment writer asks for reasoning. Let's look at a similar function
getFunctionFromAccepter :: (((a -> b) -> b) -> b) -> a -> b
This one have an naive solution:
getFunctionFromAccepter f = \a -> f $ \x -> x a
But in the following test code it fails on the third:
exeMain = do
print $ getFunctionFromAccepter (\f -> f (\x -> 10)) "Example 1" -- 10
print $ getFunctionFromAccepter (\f -> 20) "Example 2" -- 20
print $ getFunctionFromAccepter (\f -> 10 + f (\x -> 30)) "Example 3" --40, should be 30
In the failing case, we pass a function that returns 30, and we expect to get that function back. However the final result is in turn 40, so it fails. Are there any way to implement doing Just that thing I wanted?
If this can be done in Haskell there are a lot of interesting sequences. For example, tuples (or other "algebraic" types) can be defined as functions as well, since we can say something like type (a,b) = (a->b->())->() and implement fst and snd in term of this. And this, is the way I used in a couple of other languages that do not have native "tuple" support but features "closure".
The type of accept is void accept(T) so the equivalent Haskell type is t -> IO () (since every function in Java is essentially IO). Thus getFromDoubleAccepted can be directly translated as
import Data.IORef
type Accepter t = t -> IO ()
getFromDoubleAccepter :: Accepter (Accepter a) -> IO a
getFromDoubleAccepter acc = do
l <- newIORef $ error "Not called"
acc $ writeIORef l
readIORef l
If you want an idiomatic, non-IO solution in Haskell, you need to be more specific about what your actual end goal is besides trying to imitate some Java-pattern.
EDIT: regarding the update
getFromDoubleAccepter :: ((a -> ()) -> ()) -> a
I'm sorry, but this signature is in no way equal to the Java version. What you are saying is that for any a, given a function that takes a function that takes an a but doesn't return anything or do any kind of side effects, you want to somehow conjure up a value of type a. The only implementation that satisfies the given signature is essentially:
getFromDoubleAccepter :: ((a -> ()) -> ()) -> a
getFromDoubleAccepter f = getFromDoubleAccepter f
First, I'll transliterate as much as I can. I'm going to lift these computations to a monad because accept returns void (read () in Haskell-land), which is useless unless there is some effect.
type Accepter m t = t -> m ()
getFromDoubleAccepter :: (MonadSomething m) => Accepter m (Accepter m t) -> m t
getFromDoubleAccepter acc = do
l <- {- new mutable list -}
acc $ \t -> add l t
return (head l)
Of course, we can't make a mutable list like that, so we'll have to use some intuitive sparks here. When an action just adds an element to some accumulator, I think of the Writer monad. So maybe that line should be:
acc $ \t -> tell [t]
Since you are simply returning the head of the list at the end, which doesn't have any effects, I think the signature should become:
getFromDoubleAccepter :: Accepter M (Accepter M t) -> t
where M is an appropriate monad. It needs to be able to write [t]s, so that gives us:
type M t = Writer [t]
getFromDoubleAccepter :: Accepter (M t) (Accepter (M t) t) -> t
And now the type of this function informs us how to write the rest of it:
getFromDoubleAccepter acc =
head . execWriter . acc $ \t -> tell [t]
We can check that it does something...
ghci> getFromDoubleAccepter $ \acc -> acc 42
42
So that seems right, I guess. I'm still a bit unclear on what this code is supposed to mean.
The explicit M t in the type signature is a bit aesthetically bothersome to me. If I knew what problem I was solving I would look at that carefully. If you mean that the argument can be a sequence of commands, but otherwise has no computational features available, then you could specialize the type signature to:
getFromDoubleAccepter :: (forall m. (Monad m) => Accepter m (Accepter m t)) -> t
which still works with our example. Of course, this is all a bit silly. Consider
forall m. (Monad m) => Accepter m (Accepter m t))
= forall m. (Monad m) => (t -> m ()) -> m ()
The only thing a function with this type can do is call its argument with various ts in order and then return (). The information in such a function is completely characterized[1] by those ts, so we could just as easily have used
getFromDoubleAccepter :: [t] -> t
getFromDoubleAccepter = head
[1] As long as I'm going on about nothing, I might as well say that that is not quite accurate in the face of infinity. The computation
crazy :: Integer -> Accepter m (Accepter m Integer)
crazy n acc = crazy (n+1) >> acc n
can be used to form the infinite sequence
... >> acc 3 >> acc 2 >> acc 1 >> acc 0
which has no first element. If we tried to interpret this as a list, we would get an infinite loop when trying to find the first element. However this computation has more information than an infinite loop -- if instead of a list, we used the Last monoid to interpret it, we would be able to extract 0 off the end. So really
forall m. (Monad m) => Accepter m (Accepter m t)
is isomorphic to something slightly more general than a list; specifically a free monoid.
Thanks to the above answers, I finally concluded that in Haskell we can do some different things than other languages.
Actually, the motivation of this post is to translate the famous "single axiom classical logic reduction system". I have implemented this in some other languages. It should be no problem to implement the
Axiom: (a|(b|c)) | ((d|(d|d)) | ((e|b) | ((a|e) | (a|e))))
However, since the reduction rule looks like
Rule: a|(b|c), a |-- c
It is necessary to extract the inner parameter as the final result. In other languages, this is done by using side-effects like mutable slots. However, in Haskell we do not have mutable slots and involving IO will be ugly so I keep looking for solutions.
In the first glance (as show in my question), the getFromDoubleAccepter f = f id seems nonsense, but I realise that it actually work in this case! For example:
rule :: (forall r.a -> (b -> c -> r) -> r) -> a -> c
rule abc a = abc a $ flip const
The trick is still the same: since the existential qualification hides r from the caller, and it is up to the callee to pick up a type for it, we can specify c to be r, so we simply apply the given function to get the result. On the other hand, the given function has to use our input to produce the final answer, so it effectively limiting the implementation to what we exactally want!
Putting them together, let's see what we can do with it:
newtype I r a b = I { runI :: a -> b -> r }
rule :: (forall r. I r a (I r b c)) -> a -> c
rule (I abc) a = abc a (I (\b c -> c))
axiom :: I r0 (I r1 a (I r2 b c))
(I r0 (I r3 d (I r3 d d))
(I r4 (I r2 e b) (I r4 (I r1 a e) (I r1 a e))))
axiom = let
a1 (I eb) e = I $ \b c -> eb e b
a2 = I $ \d (I dd) -> dd d d
a3 (I abc) eb = I $ \a e -> abc a (a1 eb e)
a4 abc = I $ \eb aeae -> runI a2 (a3 abc eb) aeae
in I $ \abc (I dddebaeae) -> dddebaeae a2 (a4 abc)
Here I use a naming convention to trace the type signatures: a variable name is combinded by the "effective" type varialbes (means it is not result type - all r* type variable).
I wouldn't repeat the prove represented in the sited essay, but I want to show something. In the above definition of axiom we use some let bindings variables to construct the result. Not surprisingly, those variables themselves can be extracted by using rule and axiom. let's see how:
--Equal to a4
t4 :: I r0 a (I r1 b c) -> I r2 (I r1 d b) (I r2 (I r0 a d) (I r0 a d))
t4 abc = rule axiom abc
--Equal to a3
t3 :: I r0 a (I r1 b c) -> I r1 d b -> I r0 a d
t3 abc eb = rule (t4 abc) eb
--Equal to a2
t2 :: I r a (I r a a)
t2 = rule (t3 axiom (t3 (t4 axiom) axiom)) axiom
--Equal to a1
t1 :: I r a b -> a -> I r b c
t1 ab a = rule (t3 t2 (t3 (t3 t2 t2) ab)) a
One thing left to be proved is that we can use t1 to t4 only to prove all tautologies. I feel it is the case but have not yet proved it.
Compare to other languages, the Haskell salutation seems more effective and expressive.