type Set = Int => Boolean
/**
* Returns whether all bounded integers within `s` satisfy `p`.
*/
def forall(s: Set, p: Int => Boolean): Boolean = {
def iter(a: Int): Boolean = {
if (a > bound) true
else if (contains(s, a) && !p(a)) false
else iter(a + 1)
}
iter(-bound)
}
/**
* Returns whether there exists a bounded integer within `s`
* that satisfies `p`.
*/
def exists(s: Set, p: Int => Boolean): Boolean = !forall(s, (x => !p(x)))
/**
* Returns a set transformed by applying `f` to each element of `s`.
*/
def map(s: Set, f: Int => Int): Set = (x => exists(s, (y: Int) => f(y) ==
x))
so for this piece of code. I don't understand function map.
I see its input are 2 arguments, which are set and method f. But the "body" part, I tried so hard but still don't get it. And what is that "y", and why using f(y) == x makes it apply method f to set elemtns?
need some explanation for me.
thank you!
To be succinct:
If you say: val set2 = map(set1, f),
then set2(x) will returns true if and only if there exits y in set1 such as f(y) == x
That's exactly what exists(set1, y => f(y) == x) is checking.
To put it an other way, an integer is in set2 only if you can obtain it by applying f to an element of set1.
We can try to understand this piece of code by applying it backwards.
The map method here would return true for every given x and function f, if x is a result of the function f applied to the elements of the original set.
It is done by checking that if we go over the original map and apply f to every element of this map, at least one of them will be equal to x (that is what the part (x => exists(s, (y: Int) => f(y) == x)) does).
Regarding exists itself, it is a statement that if we go over all elements of the set (using forall method) with a given predicate p, for at least one of the elements this predicate will not be false (this is the part !forall(s, (x => !p(x)))).
And what is that "y"
exists takes a function Int => Boolean as its second argument. This function is (y: Int) => f(y) == x, and y is simply the name we give to its argument, same as x in x => !p(x).
and why using f(y) == x makes it apply method f to set elemtns?
This definition says "x is a member of map(s, f) when exists(s, y => f(y) == x). Now consider a simple case: s is the set {1, 2} represented by a function x => (x == 1) || (x == 2), and f = z => z + 1 over it. Then we have
s2 = map(s, z => z + 1) = x => exists(s, y => y + 1 == x)
by inlining f into definition of map. You can check that:
2 is a member of s2, i.e. s2(2) is exists(s, y => y + 1 == 2) is exists(s, y => y == 1) is true.
0 is not a member of s2, i.e. s2(2) is exists(s, y => y + 1 == 0) is exists(s, y => y == -1) is false.
Thinking a bit more, you should be able to list all members of s2, and then to generalize to any s and f.
Related
I am having a Matrix of size D by D (implemented as List[List[Int]]) and a Vector of size D (implemented as List[Int]). Assuming value of D = 3, I can create matrix and vector in following way.
val Vector = List(1,2,3)
val Matrix = List(List(4,5,6) , List(7,8,9) , List(10,11,12))
I can multiply both these as
(Matrix,Vector).zipped.map((x,y) => (x,Vector).zipped.map(_*_).sum )
This code multiplies matrix with vector and returns me vector as needed. I want to ask is there any faster or optimal way to get the same result using Scala functional style? As in my scenario I have much bigger value of D.
What about something like this?
def vectorDotProduct[N : Numeric](v1: List[N], v2: List[N]): N = {
import Numeric.Implicits._
// You may replace this with a while loop over two iterators if you require even more speed.
#annotation.tailrec
def loop(remaining1: List[N], remaining2: List[N], acc: N): N =
(remaining1, remaining2) match {
case (x :: tail1, y :: tail2) =>
loop(
remaining1 = tail1,
remaining2 = tail2,
acc + (x * y)
)
case (Nil, _) | (_, Nil) =>
acc
}
loop(
remaining1 = v1,
remaining2 = v2,
acc = Numeric[N].zero
)
}
def matrixVectorProduct[N : Numeric](matrix: List[List[N]], vector: List[N]): List[N] =
matrix.map(row => vectorDotProduct(vector, row))
Could anybody kindly explain to me why the following
/**
* Returns a set transformed by applying `f` to each element of `s`.
*/
def map(s: Set, f: Int => Int): Set = x => exists(s, y => f(y) == x)
is not equivalent to
def map(s: Set, f: Int => Int): Set = x => exists(s, f(x))
where "exists" is a function that returns whether there exists a bounded integer within s(the first argument) that satisfies p(the second argument).
Why do you need to specify "y => f(y) == x"? Thanks a million!
exists's second argument has the type Int => Boolean (right?), in other words, it expects a function from Int to Boolean. Now, f(x) doesn't conform to that type - it has the type Int. So - y => f(y) == x creates a function with the correct type, that returns true if its input equals x.
If the excess characters bug you - you can also shorten it a bit using the anonymous argument '_':
def map(s: Set, f: Int => Int): Set = x => exists(s, f(_) == x)
This is a implementation of the Y-combinator in Scala:
scala> def Y[T](func: (T => T) => (T => T)): (T => T) = func(Y(func))(_:T)
Y: [T](func: (T => T) => (T => T))T => T
scala> def fact = Y {
| f: (Int => Int) =>
| n: Int =>
| if(n <= 0) 1
| else n * f(n - 1)}
fact: Int => Int
scala> println(fact(5))
120
Q1: How does the result 120 come out, step by step? Because the Y(func) is defined as func(Y(func)), the Y should become more and more,Where is the Y gone lost and how is the 120 come out in the peform process?
Q2: What is the difference between
def Y[T](func: (T => T) => (T => T)): (T => T) = func(Y(func))(_:T)
and
def Y[T](func: (T => T) => (T => T)): (T => T) = func(Y(func))
They are the same type in the scala REPL, but the second one can not print the result 120?
scala> def Y[T](func: (T => T) => (T => T)): (T => T) = func(Y(func))
Y: [T](func: (T => T) => (T => T))T => T
scala> def fact = Y {
| f: (Int => Int) =>
| n: Int =>
| if(n <= 0) 1
| else n * f(n - 1)}
fact: Int => Int
scala> println(fact(5))
java.lang.StackOverflowError
at .Y(<console>:11)
at .Y(<console>:11)
at .Y(<console>:11)
at .Y(<console>:11)
at .Y(<console>:11)
First of all, note that this is not a Y-combinator, since the lambda version of the function uses the free variable Y. It is the correct expression for Y though, just not a combinator.
So, let’s first put the part which computes the factorial into a separate function. We can call it comp:
def comp(f: Int => Int) =
(n: Int) => {
if (n <= 0) 1
else n * f(n - 1)
}
The factorial function can now be constructed like this:
def fact = Y(comp)
Q1:
Y is defined as func(Y(func)). We invoke fact(5) which is actually Y(comp)(5), and Y(comp) evaluates to comp(Y(comp)). This is the key point: we stop here because comp takes a function and it doesn’t evaluate it until needed. So, the runtime sees comp(Y(comp)) as comp(???) because the Y(comp) part is a function and will be evaluated only when (if) needed.
Do you know about call-by-value and call-by-name parameters in Scala? If you declare your parameter as someFunction(x: Int), it will be evaluated as soon as someFunction is invoked. But if you declare it as someFunction(x: => Int), then x will not be evaluated right away, but at the point where it is used. Second call is “call by name” and it is basically defining your x as a “function that takes nothing and returns an Int”. So if you pass in 5, you are actually passing in a function that returns 5. This way we achieve lazy evaluation of function parameters, because functions are evaluated at the point they are used.
So, parameter f in comp is a function, hence it is only evaluated when needed, which is in the else branch. That’s why the whole thing works - Y can create an infinite chain of func(func(func(func(…)))) but the chain is lazy. Each new link is computed only if needed.
So when you invoke fact(5), it will run through the body into the else branch and only at that point f will be evaluated. Not before. Since your Y passed in comp() as parameter f, we will dive into comp() again. In the recursive call of comp() we will be calculating the factorial of 4. We will then again go into the else branch of the comp function, thus effectively diving into another level of recursion (calculating factorial of 3). Note that in each function call your Y provided a comp as an argument to comp, but it is only evaluated in the else branch. Once we get to the level which calculates factorial of 0, the if branch will be triggered and we will stop diving further down.
Q2:
This
func(Y(func))(_:T)
is syntax sugar for this
x => func(Y(func))(x)
which means we wrapped the whole thing into a function. We didn’t lose anything by doing this, only gained.
What did we gain? Well, it’s the same trick as in the answer to a previous question; this way we achieve that func(Y(func)) will be evaluated only if needed since it’s wrapped in a function. This way we will avoid an infinite loop. Expanding a (single-paramter) function f into a function x => f(x) is called eta-expansion (you can read more about it here).
Here’s another simple example of eta-expansion: let’s say we have a method getSquare() which returns a simple square() function (that is, a function that calculates the square of a number). Instead of returning square(x) directly, we can return a function that takes x and returns square(x):
def square(x: Int) = x * x
val getSquare: Int => Int = square
val getSquare2: Int => Int = (x: Int) => square(x)
println(square(5)) // 25
println(getSquare(5)) // 25
println(getSquare2(5)) // 25
Hope this helps.
Complementing the accepted answer,
First of all, note that this is not a Y-combinator, since the lambda version of the function uses the free variable Y. It is the correct expression for Y though, just not a combinator.
A combinator isn't allowed to be explicitly recursive; it has to be a lambda expression with no free variables, which means that it can't refer to its own name in its definition. In the lambda calculus it is not possible to refer to the definition of a function in a function body. Recursion may only be achieved by passing in a function as a parameter.
Given this, I've copied the following implementation from rosetta code that uses some type trickery to implement Y combinator without explicit recursion. See here
def Y[A, B](f: (A => B) => (A => B)): A => B = {
case class W(wf: W => A => B) {
def get: A => B =
wf(this)
}
val g: W => A => B = w => a => f(w.get)(a)
g(W(g))
}
Hope this helps with the understanding.
I don't know the answer, but will try to guess. Since you have def Y[T](f: ...) = f(...) compiler can try to substitute Y(f) with simply f. This will create an infinite sequence of f(f(f(...))). Partially applying f you create a new object, and such substitution becomes impossible.
I was reading this blog post and i was not able to understand a part of the code.
object O {
def maximum(x: List[Int]): Int = x match {
case Nil => error("maximum undefined for empty list")
case x :: y :: ys => maximum((if(x > y) x else y) :: ys)
case x :: _ => x
}
}
Please explain the code maximum((if(x > y) x else y) :: ys)
How the if condition can be a part of the method maximum ?
I understand that if condition is not exactly a parameter.
In Scala, if is an expression, not a statement.
Try this in the REPL:
scala> val x=1; val y=0
x: Int = 1
y: Int = 0
scala> val test=if(x > y) x else y
test: Int = 1
if evaluates to 1 and 1 is assigned to test. In Java if could be expressed with the conditional operator (x > y) ? x : y
Now, you have a function called maximum that takes a List[Int] as a parameter.
maximum((if(x > y) x else y) :: ys) calls maximum (recursively) with a list obtained prepending one between x and y (depending on what the if evaluates to) to ys.
I need to write some function NTimesComposition(f:(int * int -> int), n:int) which receives some function f and integer n and after doing composition of f, n times, like this f(x,(f(x,f(x,y)))) <- (here for example n = 3) I began to write it on smlnj, but it seems more complicated than I thought thanks in advance for any idea:
NTimesComposition(f:(int * int -> int), n:int)
if n = 1 then fn(x,y) => f(x, y ) else NTimesComposition...//here I'm stuck, must be recurstion
You already got it for n = 1 and you most likely just forgot to pass the (x, y) in the recursive call for n > 1. Obviously here it needs to be something of the form fn (x,y) => f (x, ...) where the ... part is where your recursive calls is going to be.
If you had forgot the (x,y) in the recursive part making it fn (x,y) => NTimesComposition (f, n-1) then you would end up building a chain of anonymous functions as "long" as your argument n describes. That would result in a different type of your NTimesComposition function depending on what n you supply which is not valid due to the way SML's type system works (Hindley-Milner).
The following two functions will do the job for you
fun foo (f, 1) = (fn xy => f xy)
| foo (f, n) = (fn (x,y) => f(x, foo (f, n-1) (x,y)))
and
fun baz (f, 1) xy = f xy
| baz (f, n) (x,y) = f(x, foo (f, n-1) (x,y))
where the first resembles your code the most using the anonymous function.