I have some difficulty in understanding the Scala solution to the n Queens problem, below is the implementation assuming isSafe is defined correctly
def queens(n: Int): Set[List[Int]] = {
def placeQueens(k: Int): Set[List[Int]] = k match {
case 0 => Set(List())
case _ =>
for {
queens <- placeQueens(k - 1)
col <- 0 until n
if isSafe(col, queens )
}yield k :: queens
}
placeQueens(n)
}
The for comprehension, as I have seen, theoretically should return a buffered collection, and I see here it buffers a list of queens with k :: queens, but it indeed returns a Set[List] as defined. Can someone throw some light on how this for comprehension works?
Is my assumption correct that for every time will return a collection of collections and since in this case I deal with a Seq and a Set in the nested for for expression it is returning a Set[List].
The question is more related to the for comprehension in implementing nQueen not nQueen in general.
Recall that a for comprehension is just syntactic sugar for map, flatmap and filter, all three of which you are using in your example. Whatever you yield in the yield block will be added to the mapped collection. You might find this article on how yield works to be interesting.
When you yield k :: queens you are creating a list with k added to the queens list which was generated from a recursive invocation using k-1.
Your assumption is correct. The for comprehension will return the types of lists involved. Since placeQueens returns a Set[List[Int]] so will the for comprehension. The translation of your for comprehension is like this:
placeQueens(k-1).flatMap { queens =>
(0 until n).withFilter { col =>
isSafe(col, queens)
}.map { col =>
k::queens
}
}
Remember, that for comprehension, in essence, when applied to a sequence, describes mapping of elements of a this sequence using a function which is specified inside the body of for-comprehension. Therefore, the end result will be the collection of the original outer type (i.e. Set, List or any other Seq-descendant) over the type returned by the body of for-comprehension.
In your case, the outer type is Set and the inner type is the type of k :: queens (which is List[Int], because queens is an element from the sequence returned by placeQueens, which is Set[List[Int]], and k::queens returns the sequence of the same type as queens.
Related
I am using a for comprehension on a stream and I would like to know how many iterations took to get o the final results.
In code:
var count = 0
for {
xs <- xs_generator
x <- xs
count = count + 1 //doesn't work!!
if (x prop)
yield x
}
Is there a way to achieve this?
Edit: If you don't want to return only the first item, but the entire stream of solutions, take a look at the second part.
Edit-2: Shorter version with zipWithIndex appended.
It's not entirely clear what you are attempting to do. To me it seems as if you are trying to find something in a stream of lists, and additionaly save the number of checked elements.
If this is what you want, consider doing something like this:
/** Returns `x` that satisfies predicate `prop`
* as well the the total number of tested `x`s
*/
def findTheX(): (Int, Int) = {
val xs_generator = Stream.from(1).map(a => (1 to a).toList).take(1000)
var count = 0
def prop(x: Int): Boolean = x % 317 == 0
for (xs <- xs_generator; x <- xs) {
count += 1
if (prop(x)) {
return (x, count)
}
}
throw new Exception("No solution exists")
}
println(findTheX())
// prints:
// (317,50403)
Several important points:
Scala's for-comprehension have nothing to do with Python's "yield". Just in case you thought they did: re-read the documentation on for-comprehensions.
There is no built-in syntax for breaking out of for-comprehensions. It's better to wrap it into a function, and then call return. There is also breakable though, but it works with Exceptions.
The function returns the found item and the total count of checked items, therefore the return type is (Int, Int).
The error in the end after the for-comprehension is to ensure that the return type is Nothing <: (Int, Int) instead of Unit, which is not a subtype of (Int, Int).
Think twice when you want to use Stream for such purposes in this way: after generating the first few elements, the Stream holds them in memory. This might lead to "GC-overhead limit exceeded"-errors if the Stream isn't used properly.
Just to emphasize it again: the yield in Scala for-comprehensions is unrelated to Python's yield. Scala has no built-in support for coroutines and generators. You don't need them as often as you might think, but it requires some readjustment.
EDIT
I've re-read your question again. In case that you want an entire stream of solutions together with a counter of how many different xs have been checked, you might use something like that instead:
val xs_generator = Stream.from(1).map(a => (1 to a).toList)
var count = 0
def prop(x: Int): Boolean = x % 317 == 0
val xsWithCounter = for {
xs <- xs_generator;
x <- xs
_ = { count = count + 1 }
if (prop(x))
} yield (x, count)
println(xsWithCounter.take(10).toList)
// prints:
// List(
// (317,50403), (317,50721), (317,51040), (317,51360), (317,51681),
// (317,52003), (317,52326), (317,52650), (317,52975), (317,53301)
// )
Note the _ = { ... } part. There is a limited number of things that can occur in a for-comprehension:
generators (the x <- things)
filters/guards (if-s)
value definitions
Here, we sort-of abuse the value-definition syntax to update the counter. We use the block { counter += 1 } as the right hand side of the assignment. It returns Unit. Since we don't need the result of the block, we use _ as the left hand side of the assignment. In this way, this block is executed once for every x.
EDIT-2
If mutating the counter is not your main goal, you can of course use the zipWithIndex directly:
val xsWithCounter =
xs_generator.flatten.zipWithIndex.filter{x => prop(x._1)}
It gives almost the same result as the previous version, but the indices are shifted by -1 (it's the indices, not the number of tried x-s).
I'm trying to print out all the factors of every number in a list.
Here is my code:
def main(args: Array[String])
{
val list_of_numbers = List(1,4,6)
def get_factors(list_of_numbers:List[Int]) : Int =
{
return list_of_numbers.foreach{(1 to _).filter {divisor => _ % divisor == 0}}
}
println(get_factors(list_of_numbers));
}
I want the end result to contain a single list that will hold all the numbers which are factors of any of the numbers in the list. So the final result should be (1,2,3,4,6). Right now, I get the following error:
error: missing parameter type for expanded function ((x$1) => 1.to(x$1))
return list_of_numbers.foreach{(1 to _).filter {divisor => _ % divisor == 0}}
How can I fix this?
You can only use _ shorthand once in a function (except for some special cases), and even then not always.
Try spelling it out instead:
list_of_numbers.foreach { n =>
(1 to n).filter { divisor => n % divisor == 0 }
}
This will compile.
There are other problems with your code though.
foreach returns a Unit, but you are requiring an Int for example.
Perhaps, you wanted a .map rather than .foreach, but that would still be a List, not an Int.
A few things are wrong here.
First, foreach takes a function A => Unit as an argument, meaning that it's really just for causing side effects.
Second your use of _, you can use _ when the function uses each argument once.
Lastly your expected output seems to be getting rid of duplicates (1 is a factor for all 3 inputs, but it only appears once).
list_of_numbers flatMap { i => (1 to i) filter {i % _ == 0 }} distinct
will do what you are looking for.
flatMap takes a function from A => List[B] and produces a simple List[B] as output, list.distinct gets rid of the duplicates.
Actually, there are several problems with your code.
First, foreach is a method which yields Unit (like void in Java). You want to yield something so you should use a for comprehension.
Second, in your divisor-test function, you've specified both the unnamed parameter ("_") and the named parameter (divisor).
The third problem is that you expect the result to be Int (in the code) but List[Int] in your description.
The following code will do what you want (although it will repeat factors, so you might want to pass it through distinct before using the result):
def main(args: Array[String]) {
val list_of_numbers = List(1, 4, 6)
def get_factors(list_of_numbers: List[Int]) = for (n <- list_of_numbers; r = 1 to n; f <- r.filter(n%_ == 0)) yield f
println(get_factors(list_of_numbers))
}
Note that you need two generators ("<-") in the for comprehension in order that you end up with simply a List. If you instead implemented the filter part in the yield expression, you would get a List[List[Int]].
To find prime factors of a number I was using this piece of code :
def primeFactors(num: Long): List[Long] = {
val exists = (2L to math.sqrt(num).toLong).find(num % _ == 0)
exists match {
case Some(d) => d :: primeFactors(num/d)
case None => List(num)
}
}
but this I found a cool and more functional approach to solve this using this code:
def factors(n: Long): List[Long] = (2 to math.sqrt(n).toInt)
.find(n % _ == 0).fold(List(n)) ( i => i.toLong :: factors(n / i))
Earlier I was using foldLeft or fold simply to get sum of a list or other simple calculations, but here I can't seem to understand how fold is working and how this is breaking out of the recursive function.Can somebody plz explain how fold functionality is working here.
Option's fold
If you look at the signature of Option's fold function, it takes two parameters:
def fold[B](ifEmpty: => B)(f: A => B): B
What it does is, it applies f on the value of Option if it is not empty. If Option is empty, it simply returns output of ifEmpty (this is termination condition for recursion).
So in your case, i => i.toLong :: factors(n / i) represents f which will be evaluated if Option is not empty. While List(n) is termination condition.
fold used for collection / iterators
The other fold that you are taking about for getting sum of collection, comes from TraversableOnce and it has signature like:
def foldLeft[B](z: B)(op: (B, A) => B): B
Here, z is starting value (suppose incase of sum it's 0) and op is associative binary operator which is applied on z and each value of collection from left to right.
So both folds differ in their implementation.
I've encountered this scala code and I'm trying to work out what its doing except the fact it returns an int. I'm unsure of these three lines :
l match {
case h :: t =>
case _ => 0
Function :
def iterate(l: List[Int]): Int =
l match {
case h :: t =>
if (h > n) 0
case _ => 0
}
First, you define a function called iterate and you specified the return type as Int. It has arity 1, parameter l of type List[Int].
The List type is prominent throughout functional programming, and it's main characteristics being that it has efficient prepend and that it is easy to decompose any List into a head and tail. The head would be the first element of the list (if non-empty) and the tail would be the rest of the List(which itself is a List) - this becomes useful for recursive functions that operate on List.
The match is called pattern matching.. it's essentially a switch statement in the C-ish languages, but much more powerful - the switch restricts you to constants (at least in C it does), but there is no such restriction with match.
Now, your first case you have h :: t - the :: is called a "cons", another term from functional programming. When you create a new List from another List via a prepend, you can use the :: operator to do it.
Example:
val oldList = List(1, 2, 3)
val newList = 0 :: oldList // newList == List(0, 1, 2, 3)
In Scala, operators that end with a : are really a method of the right hand side, so 0 :: oldList is the equivalent of oldList.::(0) - the 0 :: oldList is syntactic sugar that makes it easier to read.
We could've defined oldList like
val oldList = 1 :: 2 :: 3 :: Nil
where Nil represents an empty List. Breaking this down into steps:
3 :: Nil is evaluated first, creating the equivalent of a List(3) which has head 3 and empty tail.
2 is prepended to the above list, creative a new list with head 2 and tail List(3).
1 is prepended, creating a new list with head 1 and tail List(2, 3).
The resulting List of List(1, 2, 3) is assigned to the val oldList.
Now when you use :: to pattern match you essentially decompose a List into a head and tail, like the reverse of how we created the List above. Here when you do
l match {
case h :: t => ...
}
you are saying decompose l into a head and tail if possible. If you decompose successfully, you can then use these h and t variables to do whatever you want.. typically you would do something like act on h and call the recursive function on t.
One thing to note here is that your code will not compile.. you do an if (h > n) 0 but there is no explicit else so what happens is your code looks like this to the compiler:
if (h > n) 0
else { }
which has type AnyVal (the common supertype of 0 and "nothing"), a violation of your Int guarentee - you're going to have to add an else branch with some failure value or something.
The second case _ => is like a default in the switch, it catches anything that failed the head/tail decomposition in your first case.
Your code essentially does this:
Take the l List parameter and see if it can be decomposed into a head and tail.
If it can be, compare the head against (what I assume to be) a variable in the outer scope called n. If it is greater than n, the function returns 0. (You need to add what happens if it's not greater)
If it cannot be decomposed, the function returns 0.
This is called pattern matching. It's like a switch statement, but more powerful.
Some useful resources:
http://www.scala-lang.org/node/120
http://www.codecommit.com/blog/scala/scala-for-java-refugees-part-4
I have learned the basic difference between foldLeft and reduceLeft
foldLeft:
initial value has to be passed
reduceLeft:
takes first element of the collection as initial value
throws exception if collection is empty
Is there any other difference ?
Any specific reason to have two methods with similar functionality?
Few things to mention here, before giving the actual answer:
Your question doesn't have anything to do with left, it's rather about the difference between reducing and folding
The difference is not the implementation at all, just look at the signatures.
The question doesn't have anything to do with Scala in particular, it's rather about the two concepts of functional programming.
Back to your question:
Here is the signature of foldLeft (could also have been foldRight for the point I'm going to make):
def foldLeft [B] (z: B)(f: (B, A) => B): B
And here is the signature of reduceLeft (again the direction doesn't matter here)
def reduceLeft [B >: A] (f: (B, A) => B): B
These two look very similar and thus caused the confusion. reduceLeft is a special case of foldLeft (which by the way means that you sometimes can express the same thing by using either of them).
When you call reduceLeft say on a List[Int] it will literally reduce the whole list of integers into a single value, which is going to be of type Int (or a supertype of Int, hence [B >: A]).
When you call foldLeft say on a List[Int] it will fold the whole list (imagine rolling a piece of paper) into a single value, but this value doesn't have to be even related to Int (hence [B]).
Here is an example:
def listWithSum(numbers: List[Int]) = numbers.foldLeft((List.empty[Int], 0)) {
(resultingTuple, currentInteger) =>
(currentInteger :: resultingTuple._1, currentInteger + resultingTuple._2)
}
This method takes a List[Int] and returns a Tuple2[List[Int], Int] or (List[Int], Int). It calculates the sum and returns a tuple with a list of integers and it's sum. By the way the list is returned backwards, because we used foldLeft instead of foldRight.
Watch One Fold to rule them all for a more in depth explanation.
reduceLeft is just a convenience method. It is equivalent to
list.tail.foldLeft(list.head)(_)
foldLeft is more generic, you can use it to produce something completely different than what you originally put in. Whereas reduceLeft can only produce an end result of the same type or super type of the collection type. For example:
List(1,3,5).foldLeft(0) { _ + _ }
List(1,3,5).foldLeft(List[String]()) { (a, b) => b.toString :: a }
The foldLeft will apply the closure with the last folded result (first time using initial value) and the next value.
reduceLeft on the other hand will first combine two values from the list and apply those to the closure. Next it will combine the rest of the values with the cumulative result. See:
List(1,3,5).reduceLeft { (a, b) => println("a " + a + ", b " + b); a + b }
If the list is empty foldLeft can present the initial value as a legal result. reduceLeft on the other hand does not have a legal value if it can't find at least one value in the list.
For reference, reduceLeft will error if applied to an empty container with the following error.
java.lang.UnsupportedOperationException: empty.reduceLeft
Reworking the code to use
myList foldLeft(List[String]()) {(a,b) => a+b}
is one potential option. Another is to use the reduceLeftOption variant which returns an Option wrapped result.
myList reduceLeftOption {(a,b) => a+b} match {
case None => // handle no result as necessary
case Some(v) => println(v)
}
The basic reason they are both in Scala standard library is probably because they are both in Haskell standard library (called foldl and foldl1). If reduceLeft wasn't, it would quite often be defined as a convenience method in different projects.
From Functional Programming Principles in Scala (Martin Odersky):
The function reduceLeft is defined in terms of a more general function, foldLeft.
foldLeft is like reduceLeft but takes an accumulator z, as an additional parameter, which is returned when foldLeft is called on an empty list:
(List (x1, ..., xn) foldLeft z)(op) = (...(z op x1) op ...) op x
[as opposed to reduceLeft, which throws an exception when called on an empty list.]
The course (see lecture 5.5) provides abstract definitions of these functions, which illustrates their differences, although they are very similar in their use of pattern matching and recursion.
abstract class List[T] { ...
def reduceLeft(op: (T,T)=>T) : T = this match{
case Nil => throw new Error("Nil.reduceLeft")
case x :: xs => (xs foldLeft x)(op)
}
def foldLeft[U](z: U)(op: (U,T)=>U): U = this match{
case Nil => z
case x :: xs => (xs foldLeft op(z, x))(op)
}
}
Note that foldLeft returns a value of type U, which is not necessarily the same type as List[T], but reduceLeft returns a value of the same type as the list).
To really understand what are you doing with fold/reduce,
check this: http://wiki.tcl.tk/17983
very good explanation. once you get the concept of fold,
reduce will come together with the answer above:
list.tail.foldLeft(list.head)(_)
Scala 2.13.3, Demo:
val names = List("Foo", "Bar")
println("ReduceLeft: "+ names.reduceLeft(_+_))
println("ReduceRight: "+ names.reduceRight(_+_))
println("Fold: "+ names.fold("Other")(_+_))
println("FoldLeft: "+ names.foldLeft("Other")(_+_))
println("FoldRight: "+ names.foldRight("Other")(_+_))
outputs:
ReduceLeft: FooBar
ReduceRight: FooBar
Fold: OtherFooBar
FoldLeft: OtherFooBar
FoldRight: FooBarOther