Given a list, e.g. (f: f FALSE (g: g FALSE (h: h TRUE FALSE))), write an operator that removes all leading FALSEs and returns only the tail that starts with TRUE. For this example the operator should return just (h: h TRUE FALSE).
This is an exercise, in fact a level, in this game called "functional" that I've become obsessed with. In the previous level we were required to generalized \Omega into the y-combinator so I imagine that this level requires the y-combinator in order to handle a FALSE prefix of arbitrary length.
I'm able to handle a single FALSE prefix with (b: c: IF b (f: f b c) c). Imagining that operator as f I'm guessing the answer should look something like (b: c: IF b (f: f b c) (Y c)). The game is rejecting that answer complaining about "no reduction (grew too big)".
I'm clearly baffled by the y-combinator. Can someone show me how to use it correctly?
Also, what is this crazy syntax the game is using? I don't see it used anywhere else.
As requested, a link to functional's page on Steam is here. I also recently uncovered a link the projects page on github here.
Try this: Y(f: x: (FST x) x (f(SND x))).
By using Y-combinator, it returns the whole list (x) if the head (FST x) is TRUE and recursively calls itself taking the tail (SND x) as an argument otherwise.
Related
I am trying to study SML (for full transparency this is in preparation for an exam (exam has not started)) and one area that I have been struggling with is higher level functions such as map and foldl/r. I understand that they are used in situations where you would use a for loop in oop languages (I think). What I am struggling with though is what each part in a fold or map function is doing. Here are some examples that if someone could break them down I would be very appreciative
fun cubiclist L = map (fn x=> x*x*x) L;
fun min (x::xs) = foldr (fn (a,b) => if (a < b) then a else b) x xs;
So if I could break down the parts I see and high light the parts I'm struggling with I believe that would be helpful.
Obviously right off the bat you have the name of the functions and the parameters that are being passed in but one question I have on that part is why are we just passing in a variable to cubiclist but for min we pass in (x::xs)? Is it because the map function is automatically applying the function to each part in the map? Also along with that will the fold functions typically take the x::xs parameters while map will just take a variable?
Then we have the higher order function along with the anonymous functions with the logic/operations that we want to apply to each element in the list. But the parameters being passed in for the foldr anonymous function I'm not quite sure about. I understand we are trying to capture the lowest element in the list and the then a else b is returning either a or b to be compared with the other elements in the list. I'm pretty sure that they are rutnred and treated as a in future comparisons but where do we get the following b's from? Where do we say b is the next element in the list?
Then the part that I really don't understand and have no clue is the L; and x xs; at the end of the respective functions. Why are they there? What are they doing? what is their purpose? is it just syntax or is there actually a purpose for them being there, not saying that syntax isn't a purpose or a valid reason, but does they actually do something? Are those variables that can be changed out with something else that would provide a different answer?
Any help/explanation is much appreciated.
In addition to what #molbdnilo has already stated, it can be helpful to a newcomer to functional programming to think about what we're actually doing when we crate a loop: we're specifying a piece of code to run repeatedly. We need an initial state, a condition for the loop to terminate, and an update between each iteration.
Let's look at simple implementation of map.
fun map f [] = []
| map f (x :: xs) = f x :: map f xs
The initial state of the contents of the list.
The termination condition is the list is empty.
The update is that we tack f x onto the front of the result of mapping f to the rest of the list.
The usefulness of map is that we abstract away f. It can be anything, and we don't have to worry about writing the loop boilerplate.
Fold functions are both more complex and more instructive when comparing to loops in procedural languages.
A simple implementation of fold.
fun foldl f init [] = init
| foldl f init (x :: xs) = foldl f (f init x) xs
We explicitly provide an initial value, and a list to operate on.
The termination condition is the list being empty. If it is, we return the initial value provided.
The update is to call the function again. This time the initial value is updated, and the list is the tail of the original.
Consider summing a list of integers.
foldl op+ 0 [1,2,3,4]
foldl op+ 1 [2,3,4]
foldl op+ 3 [3,4]
foldl op+ 6 [4]
foldl op+ 10 []
10
Folds are important to understand because so many fundamental functions can be implemented in terms of foldl or foldr. Think of folding as a means of reducing (many programming languages refer to these functions as "reduce") a list to another value of some type.
map takes a function and a list and produces a new list.
In map (fn x=> x*x*x) L, the function is fn x=> x*x*x, and L is the list.
This list is the same list as cubiclist's parameter.
foldr takes a function, an initial value, and a list and produces some kind of value.
In foldr (fn (a,b) => if (a < b) then a else b) x xs, the function is fn (a,b) => if (a < b) then a else b, the initial value is x, and the list is xs.
x and xs are given to the function by pattern-matching; x is the argument's head and xs is its tail.
(It follows from this that min will fail if it is given an empty list.)
So I'm trying to define a function apply_C :: "('a multiset ⇒ 'a option) ⇒ 'a multiset ⇒ 'a multiset"
It takes in a function C that may convert an 'a multiset into a single element of type 'a. Here we assume that each element in the domain of C is pairwise mutually exclusive and not the empty multiset (I already have another function that checks these things). apply will also take another multiset inp. What I'd like the function to do is check if there is at least one element in the domain of C that is completely contained in inp. If this is the case, then perform a set difference inp - s where s is the element in the domain of C and add the element the (C s) into this resulting multiset. Afterwards, keep running the function until there are no more elements in the domain of C that are completely contained in the given inp multiset.
What I tried was the following:
fun apply_C :: "('a multiset ⇒ 'a option) ⇒ 'a multiset ⇒ 'a multiset" where
"apply_C C inp = (if ∃s ∈ (domain C). s ⊆# inp then apply_C C (add_mset (the (C s)) (inp - s)) else inp)"
However, I get this error:
Variable "s" occurs on right hand side only:
⋀C inp s.
apply_C C inp =
(if ∃s∈domain C. s ⊆# inp
then apply_C C
(add_mset (the (C s)) (inp - s))
else inp)
I have been thinking about this problem for days now, and I haven't been able to find a way to implement this functionality in Isabelle. Could I please have some help?
After thinking more about it, I don't believe there is a simple solutions for that Isabelle.
Do you need that?
I have not said why you want that. Maybe you can reduce your assumptions? Do you really need a function to calculate the result?
How to express the definition?
I would use an inductive predicate that express one step of rewriting and prove that the solution is unique. Something along:
context
fixes C :: ‹'a multiset ⇒ 'a option›
begin
inductive apply_CI where
‹apply_CI (M + M') (add_mset (the (C M)) M')›
if ‹M ∈ dom C›
context
assumes
distinct: ‹⋀a b. a ∈ dom C ⟹ b ∈ dom C ⟹ a ≠ b ⟹ a ∩# b = {#}› and
strictly_smaller: ‹⋀a b. a ∈ dom C ⟹ size a > 1›
begin
lemma apply_CI_determ:
assumes
‹apply_CI⇧*⇧* M M⇩1› and
‹apply_CI⇧*⇧* M M⇩2› and
‹⋀M⇩3. ¬apply_CI M⇩1 M⇩3›
‹⋀M⇩3. ¬apply_CI M⇩2 M⇩3›
shows ‹M⇩1 = M⇩2›
sorry
lemma apply_CI_smaller:
‹apply_CI M M' ⟹ size M' ≤ size M›
apply (induction rule: apply_CI.induct)
subgoal for M M'
using strictly_smaller[of M]
by auto
done
lemma wf_apply_CI:
‹wf {(x, y). apply_CI y x}›
(*trivial but very annoying because not enough useful lemmas on wf*)
sorry
end
end
I have no clue how to prove apply_CI_determ (no idea if the conditions I wrote down are sufficient or not), but I did spend much thinking about it.
After that you can define your definitions with:
definition apply_C where
‹apply_C M = (SOME M'. apply_CI⇧*⇧* M M' ∧ (∀M⇩3. ¬apply_CI M' M⇩3))›
and prove the property in your definition.
How to execute it
I don't see how to write an executable function on multisets directly. The problem you face is that one step of apply_C is nondeterministic.
If you can use lists instead of multisets, you get an order on the elements for free and you can use subseqs that gives you all possible subsets. Rewrite using the first element in subseqs that is in the domain of C. Iterate as long as there is any possible rewriting.
Link that to the inductive predicate to prove termination and that it calculates the right thing.
Remark that in general you cannot extract a list out of a multiset, but it is possible to do so in some cases (e.g., if you have a linorder over 'a).
I'm new to functional programming and I'm trying to implement a basic algorithm using OCAML for course that I'm following currently.
I'm trying to implement the following algorithm :
Entries :
- E : a non-empty set of integers
- s : an integer
- d : a positive float different of 0
Output :
- T : a set of integers included into E
m <- min(E)
T <- {m}
FOR EACH e ∈ sort_ascending(E \ {m}) DO
IF e > (1+d)m AND e <= s THEN
T <- T U {e}
m <- e
RETURN T
let f = fun (l: int list) (s: int) (d: float) ->
List.fold_left (fun acc x -> if ... then (list_union acc [x]) else acc)
[(list_min l)] (list_sort_ascending l) ;;
So far, this is what I have, but I don't know how to handle the modification of the "m" variable mentioned in the algorithm... So I need help to understand what is the best way to implement the algorithm, maybe I'm not gone in the right direction.
Thanks by advance to anyone who will take time to help me !
The basic trick of functional programming is that although you can't modify the values of any variables, you can call a function with different arguments. In the initial stages of switching away from imperative ways of thinking, you can imagine making every variable you want to modify into the parameters of your function. To modify the variables, you call the function recursively with the desired new values.
This technique will work for "modifying" the variable m. Think of m as a function parameter instead.
You are already using this technique with acc. Each call inside the fold gets the old value of acc and returns the new value, which is then passed to the function again. You might imagine having both acc and m as parameters of this inner function.
Assuming list_min is defined you should think the problem methodically. Let's say you represent a set with a list. Your function takes this set and some arguments and returns a subset of the original set, given the elements meet certain conditions.
Now, when I read this for the first time, List.filter automatically came to my mind.
List.filter : ('a -> bool) -> 'a list -> 'a list
But you wanted to modify the m so this wouldn't be useful. It's important to know when you can use library functions and when you really need to create your own functions from scratch. You could clearly use filter while handling m as a reference but it wouldn't be the functional way.
First let's focus on your predicate:
fun s d m e -> (float e) > (1. +. d)*.(float m) && (e <= s)
Note that +. and *. are the plus and product functions for floats, and float is a function that casts an int to float.
Let's say the function predicate is that predicate I just mentioned.
Now, this is also a matter of opinion. In my experience I wouldn't use fold_left just because it's just complicated and not necessary.
So let's begin with my idea of the code:
let m = list_min l;;
So this is the initial m
Then I will define an auxiliary function that reads the m as an argument, with l as your original set, and s, d and m the variables you used in your original imperative code.
let rec f' l s d m =
match l with
| [] -> []
| x :: xs -> if (predicate s d m x) then begin
x :: (f' xs s d x)
end
else
f' xs s d m in
f' l s d m
Then for each element of your set, you check if it satisfies the predicate, and if it does, you call the function again but you replace the value of m with x.
Finally you could just call f' from a function f:
let f (l: int list) (s: int) (d: float) =
let m = list_min l in
f' l s d m
Be careful when creating a function like your list_min, what would happen if the list was empty? Normally you would use the Option type to handle those cases but you assumed you're dealing with a non-empty set so that's great.
When doing functional programming it's important to think functional. Pattern matching is super recommended, while pointers/references should be minimal. I hope this is useful. Contact me if you any other doubt or recommendation.
I need to find a function P such that (using Beta - reduction)
P(g, h, i) ->* (h, i, i+1).
I am allowed to use the successor function succ. From wikipedia I got
succ = λn.λf.λx.f(n f x)
My answer is P = λx.λy.λz.yz(λz.λf.λu.f(z f u))z
but I'm not quite sure about it. My logic was the λx would effectively get rid of the g term, then the λy.λz would bring in the h and i via the yz. Then the succ function would bring in i+1 last. I just don't know if my function actually replicates this.
Any help given is appreciated
#melpomene points out that this question is unanswerable without a specific implementation in mind (e.g. for tuples). I am going to presume that your tuple is implemented as:
T = λabcf.f a b c
Or, if you prefer the non-shorthand:
T = (λa.(λb.(λc.(λf.f a b c))))
That is, a function which closes over a, b, and c, and waits for a function f to pass those variables.
If that is the implementation in mind, and assuming normal Church numerals, then the function you spec:
P(g, h, i) ->* (h, i, i+1)
Needs to:
take in a triple (with a, b, and c already applied)
construct a new triple, with
the second value of the old triple
the third value of the old triple
the succ of the third value of the old triple
Here is such a function P:
P = λt.t (λghi.T h i (succ i))
Or again, if you prefer non-shorthand:
P = (λt.t(λg.(λh.(λi.T h i (succ i)))))
This can be partially cleaned up with some helper functions:
SND = λt.t (λabc.b)
TRD = λt.t (λabc.c)
In which case we can write P as:
P = λt.T (SND t) (TRD t) (succ (TRD t))
Here is what I have and the error that I am getting sadly is
Error: This function has type 'a * 'a list -> 'a list
It is applied to too many arguments; maybe you forgot a `;'.
Why is that the case? I plan on passing two lists to the deleteDuplicates function, a sorted list, and an empty list, and expect the duplicates to be removed in the list r, which will be returned once the original list reaches [] condition.
will be back with updated code
let myfunc_caml_way arg0 arg1 = ...
rather than
let myfunc_java_way(arg0, arg1) = ...
Then you can call your function in this way:
myfunc_caml_way "10" 123
rather than
myfunc_java_way("10, 123)
I don't know how useful this might be, but here is some code that does what you want, written in a fairly standard OCaml style. Spend some time making sure you understand how and why it works. Maybe you should start with something simpler (eg how would you sum the elements of a list of integers ?). Actually, you should probably start with an OCaml tutorial, reading carefully and making sure you aunderstand the code examples.
let deleteDuplicates u =
(*
u : the sorted list
v : the result so far
last : the last element we read from u
*)
let rec aux u v last =
match u with
[] -> v
| x::xs when x = last -> aux xs v last
| x::xs -> aux u (x::v) x
in
(* the first element is a special case *)
match u with
[] -> []
| x::xs -> List.rev (aux xs [x] x)
This is not a direct answer to your question.
The standard way of defining an "n-ary" function is
let myfunc_caml_way arg0 arg1 = ...
rather than
let myfunc_java_way(arg0, arg1) = ...
Then you can call your function in this way:
myfunc_caml_way "10" 123
rather than
myfunc_java_way("10, 123)
See examples here:
https://github.com/ocaml/ocaml/blob/trunk/stdlib/complex.ml
By switching from myfunc_java_way to myfunc_caml_way, you will be benefited from what's called "Currying"
What is 'Currying'?
However please note that you sometimes need to enclose the whole invocation by parenthesis
myfunc_caml_way (otherfunc_caml_way "foo" "bar") 123
in order to tell the compiler not to interpret your code as
((myfunc_caml_way otherfunc_caml_way "foo") "bar" 123)
You seem to be thinking that OCaml uses tuples (a, b) to indicate arguments of function calls. This isn't the case. Whenever some expressions stand next to each other, that's a function call. The first expression is the function, and the rest of the expressions are the arguments to the function.
So, these two lines:
append(first,r)
deleteDuplicates(remaining, r)
Represent a function call with three arguments. The function is append. The first argument is (first ,r). The second argument is deleteDuplicates. The third argument is (remaining, r).
Since append has just one argument (a tuple), you're passing it too many arguments. This is what the compiler is telling you.
You also seem to be thinking that append(first, r) will change the value of r. This is not the case. Variables in OCaml are immutable. You can't do anything that will change the value of r.
Update
I think you have too many questions for SO to help you effectively at this point. You might try reading some OCaml tutorials. It will be much faster than asking a question here for every error you see :-)
Nonetheless, here's what "match failure" means. It means that somewhere you have a match that you're applying to an expression, but none of the patterns of the match matches the expression. Your deleteDuplicates code clearly has a pattern coverage error; i.e., it has a pattern that doesn't cover all cases. Your first match only works for empty lists or for lists of 2 or more elements. It doesn't work for lists of 1 element.