How can I write a generator for the second argument someBoundedInt which will generate an Int randomly between the values generated for minmaxBound?
val boundedIntProperty = forAll {
(minmaxBound: (Int,Int), someBoundedInt: Int) =>
minmaxBound._1 <= someBoundedInt && someBoundedInt <= minmaxBound._2
}
You can nest calls to forAll like this:
val boundedIntProperty = forAll { (minBound: Int, maxBound: Int) =>
forAll( Gen.choose(minBound, maxBound) ) { someBoundedInt =>
...
}
}
Note that above, minBound can be larger than maxBound sometimes, which will make Gen.choose fail (not produce a value). So you probably want to generate your bounds in a smarter way too.
Related
What is the best way to implement a sorting function that has an internal state received else where?
Something like:
type Sorter[Item] = (Item, Item) => Boolean
type StringSorter = Sorter[String]
def customSorter : StringSorter = (i1,i2) =>
{
val i1Cnt = itemCountMap.get(i1)
val i2Cnt = itemCountMap.get(i2)
if (i1Cnt==None || i2Cnt==None ) {
i1<i2
} else {
i1Cnt.get<i2Cnt.get
}
}
And here is an example:
val l1 = List("a","b","c")
val itemCountMap = Map("a"->2,"b"->3,"c"1)
l1.sortWith(customSorter)
//The returned list will be ["c","a","b"]
I am pretty new to Scala, and in general, lambda functions are not suppose to have states (right?).
Why you ask? 'cause I am using a generic type lists in spark which later, deep in the code of the executors, I want to analyze based on a specific order, and this order may depend on some static list, and I also want to control this order function.
First, this i1Cnt==null will never be true because get on Map never returns null, it returns an Option WHICH IS NOT A NULL, it is very different.
Second, there is no problem with state in a lambda, there is a problem with mutable shared state (and not only in a lambda, but everywhere).
Third, your function would be better if it receives the Map to use instead of relying on a global variable.
Fourth, here is the fixed code.
type Sorter[Item] = (Item, Item) => Boolean
def customSorter(map: Map[String, Int]): Sorter[String] = { (s1, s2) =>
(map.get(s1), map.get(s2)) match {
case (Some(i1), Some(i2)) => i1 < i2
case _ => s1 < s2
}
}
You can see it running here)
I have superficially read a couple of blog articles/Wikipedia about continuation-passing style. My high-level goal is to find a systematic technique to make any recursive function (or, if there are restrictions, being aware of them) tail-recursive. However, I have trouble articulating my thoughts and I'm not sure if what my attempts of it make any sense.
For the purpose of the example, I'll propose a simple problem. The goal is, given a sorted list of unique characters, to output all possible words made out of these characters in alphabetical order. For example, sol("op".toList, 3) should return ooo,oop,opo,opp,poo,pop,ppo,ppp.
My recursive solution is the following:
def sol(chars: List[Char], n: Int) = {
def recSol(n: Int): List[List[Char]] = (chars, n) match {
case (_ , 0) => List(Nil)
case (Nil, _) => Nil
case (_ , _) =>
val tail = recSol(n - 1)
chars.map(ch => tail.map(ch :: _)).fold(Nil)(_ ::: _)
}
recSol(n).map(_.mkString).mkString(",")
}
I did try to rewrite this by adding a function as a parameter but I did not manage to make something I was convinced to be tail-recursive. I prefer not including my attempt(s) in the question as I'm ashamed of their naiveness, so please excuse me for this.
Therefore the question is basically: how would the function above be written in CPS ?
Try that:
import scala.annotation.tailrec
def sol(chars: List[Char], n: Int) = {
#tailrec
def recSol(n: Int)(cont: (List[List[Char]]) => List[List[Char]]): List[List[Char]] = (chars, n) match {
case (_ , 0) => cont(List(Nil))
case (Nil, _) => cont(Nil)
case (_ , _) =>
recSol(n-1){ tail =>
cont(chars.map(ch => tail.map(ch :: _)).fold(Nil)(_ ::: _))
}
}
recSol(n)(identity).map(_.mkString).mkString(",")
}
The first order of business in performing the CPS transform is deciding on a representation for continuations. We can think of continuations as a suspended computation with a "hole". When the hole is filled in with a value, the remainder of the computation can be computed. So functions are a natural choice for representing continuations, at least for toy examples:
type Cont[Hole,Result] = Hole => Result
Here Hole represents the type of the hole that needs to be filled in, and Result represents the type of value the computation ultimately computes.
Now that we have a way to represent continuations, we can worry about the CPS transform itself. Basically, this involves the following steps:
The transformation is applied recursively to an expression, stopping at "trivial" expressions / function calls. In this context, "trivial" includes functions defined by Scala (since they are not CPS-transformed, and thus do not have a continuation parameter).
We need to add a parameter of type Cont[Return,Result] to each function, where Return is the return type of the untransformed function and Result is the type of the ultimate result of the computation as a whole. This new parameter represents the current continuation. The return type for the transformed function is also changed to Result.
Every function call needs to be transformed to accommodate the new continuation parameter. Everything after the call needs to be put into a continuation function, which is then added to the parameter list.
For example, a function:
def f(x : Int) : Int = x + 1
becomes:
def fCps[Result](x : Int)(k : Cont[Int,Result]) : Result = k(x + 1)
and
def g(x : Int) : Int = 2 * f(x)
becomes:
def gCps[Result](x : Int)(k : Cont[Int,Result]) : Result = {
fCps(x)(y => k(2 * y))
}
Now gCps(5) returns (via currying) a function that represents a partial computation. We can extract the value from this partial computation and use it by supplying a continuation function. For example, we can use the identity function to extract the value unchanged:
gCps(5)(x => x)
// 12
Or, we can print it by using println instead:
gCps(5)(println)
// prints 12
Applying this to your code, we obtain:
def solCps[Result](chars : List[Char], n : Int)(k : Cont[String, Result]) : Result = {
#scala.annotation.tailrec
def recSol[Result](n : Int)(k : Cont[List[List[Char]], Result]) : Result = (chars, n) match {
case (_ , 0) => k(List(Nil))
case (Nil, _) => k(Nil)
case (_ , _) =>
recSol(n - 1)(tail =>
k(chars.map(ch => tail.map(ch :: _)).fold(Nil)(_ ::: _)))
}
recSol(n)(result =>
k(result.map(_.mkString).mkString(",")))
}
As you can see, although recSol is now tail-recursive, it comes with the cost of building a more complex continuation at each iteration. So all we've really done is trade space on the JVM's control stack for space on the heap -- the CPS transform does not magically reduce the space complexity of an algorithm.
Also, recSol is only tail-recursive because the recursive call to recSol happens to be the first (non-trivial) expression recSol performs. In general, though, recursive calls would be take place inside a continuation. In the case where there is one recursive call, we can work around that by transforming only calls to the recursive function to CPS. Even so, in general, we would still just be trading stack space for heap space.
I am curious if can implement a Scala function similar to this Javascript function. Obviously, I can do it easily with an inner function.
Knowing that Scala has to declare the parameter type and arity upfront, I just wonder if there is anything I could use to implement this JS function. Thanks.
function factorial(x) {
if (x < 0) throw Error("Cannot calculate factorial of a negative number");
return (function(f) {
return f(f, x, 1);
})(function(f, i, fact) {
return i === 0 ? fact : f(f, i-1, i*fact);
});
}
If I have understood your question correctly, indeed you can do this and the best known approach is to use what's called a Y-combinator. In short a Y-combinator takes a function as a parameter and keeps applying it. The Y-combinator has no knowledge of the parameter types involved
Copying the example Y-combinator right from rosetta code:
def Y[A,B](f: (A=>B)=>(A=>B)) = {
case class W(wf: W=>A=>B) {
def apply(w: W) = wf(w)
}
val g: W=>A=>B = w => f(w(w))(_)
g(W(g))
}
defines your combinator. You can then pass it your recursive function
val fac = Y[Int, Int](f => i => if (i <= 0) 1 else f(i - 1) * i)
fac: Int => Int = <function1>
And then give that something to evaluate
scala> fac(6)
res0: Int = 720
Note: this is a theoretical question, I am not trying to fix anything, nor am I trying to achieve any effect for a practical purpose
When creating a lambda in Scala using the (arguments)=>expression syntax, can the return type be explicitly provided?
Lambdas are no different than methods on that they both are specified as expressions, but as far as I understand it, the return type of methods is defined easily with the def name(arguments): return type = expression syntax.
Consider this (illustrative) example:
def sequence(start: Int, next: Int=>Int): ()=>Int = {
var x: Int = start
//How can I denote that this function should return an integer?
() => {
var result: Int = x
x = next(x)
result
}
}
You can always declare the type of an expression by appending : and the type. So, for instance:
((x: Int) => x.toString): (Int => String)
This is useful if you, for instance, have a big complicated expression and you don't want to rely upon type inference to get the types straight.
{
if (foo(y)) x => Some(bar(x))
else x => None
}: (Int => Option[Bar])
// Without type ascription, need (x: Int)
But it's probably even clearer if you assign the result to a temporary variable with a specified type:
val fn: Int => Option[Bar] = {
if (foo(y)) x => Some(bar(x))
else _ => None
}
Let say you have this function:
def mulF(a: Int, b: Int): Long = {
a.toLong * b
}
The same function can be written as lambda with defined input and output types:
val mulLambda: (Int, Int) => Long = (x: Int, y: Int) => { x.toLong * y }
x => x:SomeType
Did not know the answer myself as I never had the need for it, but my gut feeling was that this will work. And trying it in a worksheet confirmed it.
Edit: I provided this answer before there was an example above. It is true that this is not needed in the concrete example. But in rare cases where you'd need it, the syntax I showed will work.
I have some expensive computation in a loop, and I need to find max value produced by the calculations, though if, say, it will equal to LIMIT I'd like to stop the calculation and return my accumulator.
It may easily be done by recursion:
val list: List[Int] = ???
val UpperBound = ???
def findMax(ls: List[Int], max: Int): Int = ls match {
case h :: rest =>
val v = expensiveComputation(h)
if (v == UpperBound) v
else findMax(rest, math.max(max, v))
case _ => max
}
findMax(list, 0)
My question: whether this behaviour template has a name and reflected in scala collection library?
Update: Do something up to N times or until condition is met in Scala - There is an interesting idea (using laziness and find or exists at the end) but it is not directly applicable to my particular case or requires mutable var to track accumulator.
I think your recursive function is quite nice, so honestly I wouldn't change that, but here's a way to use the collections library:
list.foldLeft(0) {
case (max, next) =>
if(max == UpperBound)
max
else
math.max(expensiveComputation(next), max)
}
It will iterate over the whole list, but after it has hit the upper bound it won't perform the expensive computation.
Update
Based on your comment I tried adapting foldLeft a bit, based on LinearSeqOptimized's foldLeft implementation.
def foldLeftWithExit[A, B](list: Seq[A])(z: B)(exit: B => Boolean)(f: (B, A) => B): B = {
var acc = z
var remaining = list
while (!remaining.isEmpty && !exit(acc)) {
acc = f(acc, list.head)
remaining = remaining.tail
}
acc
}
Calling it:
foldLeftWithExit(list)(0)(UpperBound==){
case (max, next) => math.max(expensiveComputation(next), max)
}
You could potentially use implicits to omit the first parameter of list.
Hope this helps.