Infinite streams in Scala
Stream.from(1)
creates a stream starting from 1 and incrementing by 1. It's all in the API docs.
A Solution Using Iterators
You can also use an Iterator
instead of a Stream
. The Stream
keeps references of all computed values. So if you plan to visit each value only once, an iterator is a more efficient approach. The downside of the iterator is its mutability, though.
There are some nice convenience methods for creating Iterator
s defined on its companion object.
Edit
Unfortunately there's no short (library supported) way I know of to achieve something like
Stream.from(1) takeWhile (factorial(_) <= Long.MaxValue) last
The approach I take to advance an Iterator
for a certain number of elements is drop(n: Int)
or dropWhile
:
Iterator.from(1).dropWhile( factorial(_) <= Long.MaxValue).next - 1
The - 1
works for this special purpose but is not a general solution. But it should be no problem to implement a last
method on an Iterator
using pimp my library. The problem is taking the last element of an infinite Iterator could be problematic. So it should be implemented as method like lastWith
integrating the takeWhile
.
An ugly workaround can be done using sliding
, which is implemented for Iterator
:
scala> Iterator.from(1).sliding(2).dropWhile(_.tail.head < 10).next.head
res12: Int = 9
as @ziggystar pointed out, Streams
keeps the list of previously computed values in memory, so using Iterator
is a great improvment.
to further improve the answer, I would argue that "infinite streams", are usually computed (or can be computed) based on pre-computed values. if this is the case (and in your factorial stream it definately is), I would suggest using Iterator.iterate
instead.
would look roughly like this:
scala> val it = Iterator.iterate((1,BigInt(1))){case (i,f) => (i+1,f*(i+1))}
it: Iterator[(Int, scala.math.BigInt)] = non-empty iterator
then, you could do something like:
scala> it.find(_._2 >= Long.MaxValue).map(_._1).get - 1
res0: Int = 22
or use @ziggystar sliding
solution...
another easy example that comes to mind, would be fibonacci numbers:
scala> val it = Iterator.iterate((1,1)){case (a,b) => (b,a+b)}.map(_._1)
it: Iterator[Int] = non-empty iterator
in these cases, your'e not computing your new element from scratch every time, but rather do an O(1) work for every new element, which would improve your running time even more.
The original "factorial" function is not optimal, since factorials are computed from scratch every time. The simplest/immutable implementation using memoization is like this:
val f : Stream[BigInt] = 1 #:: (Stream.from(1) zip f).map { case (x,y) => x * y }
And now, the answer can be computed like this:
println( "count: " + (f takeWhile (_<Long.MaxValue)).length )