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I'm trying to get this Recursive Function to work:
#tailrec
def rec(n: BigInt): BigInt = {
if (n == 0) 0
else if (n == 1) 1
else (rec(n - 1) + rec(n - 2))
}
Error:(13, 24) could not optimize #tailrec annotated method rec: it contains a recursive call not in tail position
else (rec(n - 1) + rec(n - 2))
How can this be optimized to work with tailrec?
You're going to have to write a tail recursive helper function:
def rec(n: BigInt): BigInt = {
#tailrec
def helper(n: BigInt, previous: BigInt, next: BigInt): BigInt = {
if (n == 0) previous
else if (n == 1) next
else helper(n - 1, next, (next + previous))
}
helper(n,0,1)
}
This is, so you can pass the previous and next values of your sequence to the function, thereby allowing you to get only one call to your function.
This is a common pattern, as many recursive algorithms can only be made tail recursive through additional control flow mechanisms (for example: continuation passing).
Factorial is another great example, where writing a helper function with an extra parameter is needed to make it tail recursive.
New to Scala and trying to figure out recursion.
Having the fallowing definitions in my session:
def inc(n: Int) = n + 1
def dec(n: Int) = n – 1
How could I redefine function below to use recursion inc and dec?
add(n: Int, m: Int) = n + m
I'm interested in learning both regular recursion and tail recursion.
Thanks
How about this:
scala> def inc(n: Int) = n + 1
inc: (n: Int)Int
scala> def dec(n: Int) = n - 1
dec: (n: Int)Int
scala> def add(n: Int, m: Int): Int = m match {
| case 0 => n
| case _ if m > 0 => add(inc(n), dec(m))
| case _ => add(dec(n), inc(m))
| }
add: (n: Int, m: Int)Int
scala> add(100, 99)
res0: Int = 199
scala> add(100, -99)
res1: Int = 1
Or there is another solution, which is an implementation of the Peano axioms.
scala> def add2(n: Int, m: Int): Int = m match {
| case 0 => n
| case _ if m > 0 => inc(add2(n, dec(m)))
| case _ => dec(add2(n, inc(m)))
| }
add2: (n: Int, m: Int)Int
Tail Recursion has 3 parts as far as I'm concerning:
Condition to end recursion
return value if the condition is met, the returned value is one (or derived from) the parameters of the tail recursive function
and the call to itself if the condition is unmet.
sample:
def inc(n: Int) = n + 1
def dec(n: Int) = n - 1
def add(n:Int, m:Int, sum: Int):Int = {
//condition to break/end the recursion
if (m <= 0) {
// returned value once condition is met. This is the final output of the recursion
sum
} else {
//call to itself once condition is unmet
add(inc(n), dec(m), n + m)
}
}
as you can see, it feels like you are doing while loop but more functional way.
on recursion, calls are stack which result to having it's call stack size as depth of the recursive calls (which can result to stackoverflowexception) on tail recursion it is like how while loop is interpreted.
sample of recursion:
def addAllNumberFromNToZero(n:Int):Int = {
if (m <= 0) {
sum
} else {
n + add(n - 1)
}
}
Using regular recursion, you could try something like:
def inc(n: Int) = n + 1
def dec(n: Int) = n - 1
def add(n: Int, m: Int): Int = {
if (m == 0) n
else add(inc(n), dec(m))
}
The add function recursively calls itself add, each time incrementing n and reducing m. The recursion stops when m reaches zero, at which point m is returned.
I am editing following code in eclipse but it complains that "recursive method loop needs result type", so what's the error? Thanks.
package week2
object exercise {
def factorial(n: Int): Int = {
def loop(acc: Int, n: Int) =
if (n == 0) acc
else loop(acc * n, n - 1)
loop(1, n)
}
factorial(4)
}
Thanks for the answer. Yes, I missed something here. The correct is:
package week2
object exercise {
def factorial(n: Int): Int = {
def loop(acc: Int, n: Int): Int =
if (n == 0) acc
else loop(acc * n, n - 1)
loop(1, n)
} //> factorial: (n: Int)Int
factorial(4) //> res0: Int = 24
}
As sepp2k indicated, it's restricted by compiler. Just wanted to add more info for clarification.
According to Scala's local type inference mechanism
return types of methods can often be omitted since they correspond to the type of the body, which gets inferred by the compiler
Consider non-recursive function
def f(n: Int) = n % 2 == 0
where compiler can infer a result type, since it knows that expression x % 2 == 0 will always produce type Boolean.
But when you deal with a recursive function, like
def f(n: Int) = if (n == 0) 1 else n * f(n-1)
Scala compiler won't infer the result type of if (x == 0) 1 else x * f(x-1) since that leads to an infinite loop. It can't even know that f(x-1) is an eligible operand for * operation, thus will give you an error Cannot resolve reference * with such signature.
The rules of the Scala language that any recursive method needs to be declared with a return type. loop is a recursive method and you didn't declare a return type for it (you only declared one for factorial, which doesn't necessarily need one). Therefore scalac (and by extension Eclipse) is telling you that the recursive method loop needs a return type.
I was wondering if there is some general method to convert a "normal" recursion with foo(...) + foo(...) as the last call to a tail-recursion.
For example (scala):
def pascal(c: Int, r: Int): Int = {
if (c == 0 || c == r) 1
else pascal(c - 1, r - 1) + pascal(c, r - 1)
}
A general solution for functional languages to convert recursive function to a tail-call equivalent:
A simple way is to wrap the non tail-recursive function in the Trampoline monad.
def pascalM(c: Int, r: Int): Trampoline[Int] = {
if (c == 0 || c == r) Trampoline.done(1)
else for {
a <- Trampoline.suspend(pascal(c - 1, r - 1))
b <- Trampoline.suspend(pascal(c, r - 1))
} yield a + b
}
val pascal = pascalM(10, 5).run
So the pascal function is not a recursive function anymore. However, the Trampoline monad is a nested structure of the computation that need to be done. Finally, run is a tail-recursive function that walks through the tree-like structure, interpreting it, and finally at the base case returns the value.
A paper from Rúnar Bjanarson on the subject of Trampolines: Stackless Scala With Free Monads
In cases where there is a simple modification to the value of a recursive call, that operation can be moved to the front of the recursive function. The classic example of this is Tail recursion modulo cons, where a simple recursive function in this form:
def recur[A](...):List[A] = {
...
x :: recur(...)
}
which is not tail recursive, is transformed into
def recur[A]{...): List[A] = {
def consRecur(..., consA: A): List[A] = {
consA :: ...
...
consrecur(..., ...)
}
...
consrecur(...,...)
}
Alexlv's example is a variant of this.
This is such a well known situation that some compilers (I know of Prolog and Scheme examples but Scalac does not do this) can detect simple cases and perform this optimisation automatically.
Problems combining multiple calls to recursive functions have no such simple solution. TMRC optimisatin is useless, as you are simply moving the first recursive call to another non-tail position. The only way to reach a tail-recursive solution is remove all but one of the recursive calls; how to do this is entirely context dependent but requires finding an entirely different approach to solving the problem.
As it happens, in some ways your example is similar to the classic Fibonnaci sequence problem; in that case the naive but elegant doubly-recursive solution can be replaced by one which loops forward from the 0th number.
def fib (n: Long): Long = n match {
case 0 | 1 => n
case _ => fib( n - 2) + fib( n - 1 )
}
def fib (n: Long): Long = {
def loop(current: Long, next: => Long, iteration: Long): Long = {
if (n == iteration)
current
else
loop(next, current + next, iteration + 1)
}
loop(0, 1, 0)
}
For the Fibonnaci sequence, this is the most efficient approach (a streams based solution is just a different expression of this solution that can cache results for subsequent calls). Now,
you can also solve your problem by looping forward from c0/r0 (well, c0/r2) and calculating each row in sequence - the difference being that you need to cache the entire previous row. So while this has a similarity to fib, it differs dramatically in the specifics and is also significantly less efficient than your original, doubly-recursive solution.
Here's an approach for your pascal triangle example which can calculate pascal(30,60) efficiently:
def pascal(column: Long, row: Long):Long = {
type Point = (Long, Long)
type Points = List[Point]
type Triangle = Map[Point,Long]
def above(p: Point) = (p._1, p._2 - 1)
def aboveLeft(p: Point) = (p._1 - 1, p._2 - 1)
def find(ps: Points, t: Triangle): Long = ps match {
// Found the ultimate goal
case (p :: Nil) if t contains p => t(p)
// Found an intermediate point: pop the stack and carry on
case (p :: rest) if t contains p => find(rest, t)
// Hit a triangle edge, add it to the triangle
case ((c, r) :: _) if (c == 0) || (c == r) => find(ps, t + ((c,r) -> 1))
// Triangle contains (c - 1, r - 1)...
case (p :: _) if t contains aboveLeft(p) => if (t contains above(p))
// And it contains (c, r - 1)! Add to the triangle
find(ps, t + (p -> (t(aboveLeft(p)) + t(above(p)))))
else
// Does not contain(c, r -1). So find that
find(above(p) :: ps, t)
// If we get here, we don't have (c - 1, r - 1). Find that.
case (p :: _) => find(aboveLeft(p) :: ps, t)
}
require(column >= 0 && row >= 0 && column <= row)
(column, row) match {
case (c, r) if (c == 0) || (c == r) => 1
case p => find(List(p), Map())
}
}
It's efficient, but I think it shows how ugly complex recursive solutions can become as you deform them to become tail recursive. At this point, it may be worth moving to a different model entirely. Continuations or monadic gymnastics might be better.
You want a generic way to transform your function. There isn't one. There are helpful approaches, that's all.
I don't know how theoretical this question is, but a recursive implementation won't be efficient even with tail-recursion. Try computing pascal(30, 60), for example. I don't think you'll get a stack overflow, but be prepared to take a long coffee break.
Instead, consider using a Stream or memoization:
val pascal: Stream[Stream[Long]] =
(Stream(1L)
#:: (Stream from 1 map { i =>
// compute row i
(1L
#:: (pascal(i-1) // take the previous row
sliding 2 // and add adjacent values pairwise
collect { case Stream(a,b) => a + b }).toStream
++ Stream(1L))
}))
The accumulator approach
def pascal(c: Int, r: Int): Int = {
def pascalAcc(acc:Int, leftover: List[(Int, Int)]):Int = {
if (leftover.isEmpty) acc
else {
val (c1, r1) = leftover.head
// Edge.
if (c1 == 0 || c1 == r1) pascalAcc(acc + 1, leftover.tail)
// Safe checks.
else if (c1 < 0 || r1 < 0 || c1 > r1) pascalAcc(acc, leftover.tail)
// Add 2 other points to accumulator.
else pascalAcc(acc, (c1 , r1 - 1) :: ((c1 - 1, r1 - 1) :: leftover.tail ))
}
}
pascalAcc(0, List ((c,r) ))
}
It does not overflow the stack but as on big row and column but Aaron mentioned it's not fast.
Yes it's possible. Usually it's done with accumulator pattern through some internally defined function, which has one additional argument with so called accumulator logic, example with counting length of a list.
For example normal recursive version would look like this:
def length[A](xs: List[A]): Int = if (xs.isEmpty) 0 else 1 + length(xs.tail)
that's not a tail recursive version, in order to eliminate last addition operation we have to accumulate values while somehow, for example with accumulator pattern:
def length[A](xs: List[A]) = {
def inner(ys: List[A], acc: Int): Int = {
if (ys.isEmpty) acc else inner(ys.tail, acc + 1)
}
inner(xs, 0)
}
a bit longer to code, but i think the idea i clear. Of cause you can do it without inner function, but in such case you should provide acc initial value manually.
I'm pretty sure it's not possible in the simple way you're looking for the general case, but it would depend on how elaborate you permit the changes to be.
A tail-recursive function must be re-writable as a while-loop, but try implementing for example a Fractal Tree using while-loops. It's possble, but you need to use an array or collection to store the state for each point, which susbstitutes for the data otherwise stored in the call-stack.
It's also possible to use trampolining.
It is indeed possible. The way I'd do this is to
begin with List(1) and keep recursing till you get to the
row you want.
Worth noticing that you can optimize it: if c==0 or c==r the value is one, and to calculate let's say column 3 of the 100th row you still only need to calculate the first three elements of the previous rows.
A working tail recursive solution would be this:
def pascal(c: Int, r: Int): Int = {
#tailrec
def pascalAcc(c: Int, r: Int, acc: List[Int]): List[Int] = {
if (r == 0) acc
else pascalAcc(c, r - 1,
// from let's say 1 3 3 1 builds 0 1 3 3 1 0 , takes only the
// subset that matters (if asking for col c, no cols after c are
// used) and uses sliding to build (0 1) (1 3) (3 3) etc.
(0 +: acc :+ 0).take(c + 2)
.sliding(2, 1).map { x => x.reduce(_ + _) }.toList)
}
if (c == 0 || c == r) 1
else pascalAcc(c, r, List(1))(c)
}
The annotation #tailrec actually makes the compiler check the function
is actually tail recursive.
It could be probably be further optimized since given that the rows are symmetric, if c > r/2, pascal(c,r) == pascal ( r-c,r).. but left to the reader ;)
I've looked over a few implementations of Fibonacci function in Scala starting from a very simple one, to the more complicated ones.
I'm not entirely sure which one is the fastest. I'm leaning towards the impression that the ones that uses memoization is faster, however I wonder why Scala doesn't have a native memoization.
Can anyone enlighten me toward the best and fastest (and cleanest) way to write a fibonacci function?
The fastest versions are the ones that deviate from the usual addition scheme in some way. Very fast is the calculation somehow similar to a fast binary exponentiation based on these formulas:
F(2n-1) = F(n)² + F(n-1)²
F(2n) = (2F(n-1) + F(n))*F(n)
Here is some code using it:
def fib(n:Int):BigInt = {
def fibs(n:Int):(BigInt,BigInt) = if (n == 1) (1,0) else {
val (a,b) = fibs(n/2)
val p = (2*b+a)*a
val q = a*a + b*b
if(n % 2 == 0) (p,q) else (p+q,p)
}
fibs(n)._1
}
Even though this is not very optimized (e.g. the inner loop is not tail recursive), it will beat the usual additive implementations.
for me the simplest defines a recursive inner tail function:
def fib: Stream[Long] = {
def tail(h: Long, n: Long): Stream[Long] = h #:: tail(n, h + n)
tail(0, 1)
}
This doesn't need to build any Tuple objects for the zip and is easy to understand syntactically.
Scala does have memoization in the form of Streams.
val fib: Stream[BigInt] = 0 #:: 1 #:: fib.zip(fib.tail).map(p => p._1 + p._2)
scala> fib take 100 mkString " "
res22: String = 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 ...
Stream is a LinearSeq so you might like to convert it to an IndexedSeq if you're doing a lot of fib(42) type calls.
However I would question what your use-case is for a fibbonaci function. It will overflow Long in less than 100 terms so larger terms aren't much use for anything. The smaller terms you can just stick in a table and look them up if speed is paramount. So the details of the computation probably don't matter much since for the smaller terms they're all quick.
If you really want to know the results for very big terms, then it depends on whether you just want one-off values (use Landei's solution) or, if you're making a sufficient number of calls, you may want to pre-compute the whole lot. The problem here is that, for example, the 100,000th element is over 20,000 digits long. So we're talking gigabytes of BigInt values which will crash your JVM if you try to hold them in memory. You could sacrifice accuracy and make things more manageable. You could have a partial-memoization strategy (say, memoize every 100th term) which makes a suitable memory / speed trade-off. There is no clear anwser for what is the fastest: it depends on your usage and resources.
This could work. it takes O(1) space O(n) time to calculate a number, but has no caching.
object Fibonacci {
def fibonacci(i : Int) : Int = {
def h(last : Int, cur: Int, num : Int) : Int = {
if ( num == 0) cur
else h(cur, last + cur, num - 1)
}
if (i < 0) - 1
else if (i == 0 || i == 1) 1
else h(1,2,i - 2)
}
def main(args: Array[String]){
(0 to 10).foreach( (x : Int) => print(fibonacci(x) + " "))
}
}
The answers using Stream (including the accepted answer) are very short and idiomatic, but they aren't the fastest. Streams memoize their values (which isn't necessary in iterative solutions), and even if you don't keep the reference to the stream, a lot of memory may be allocated and then immediately garbage-collected. A good alternative is to use an Iterator: it doesn't cause memory allocations, is functional in style, short and readable.
def fib(n: Int) = Iterator.iterate(BigInt(0), BigInt(1)) { case (a, b) => (b, a+b) }.
map(_._1).drop(n).next
A little simpler tail Recursive solution that can calculate Fibonacci for large values of n. The Int version is faster but is limited, when n > 46 integer overflow occurs
def tailRecursiveBig(n :Int) : BigInt = {
#tailrec
def aux(n : Int, next :BigInt, acc :BigInt) :BigInt ={
if(n == 0) acc
else aux(n-1, acc + next,next)
}
aux(n,1,0)
}
This has already been answered, but hopefully you will find my experience helpful. I had a lot of trouble getting my mind around scala infinite streams. Then, I watched Paul Agron's presentation where he gave very good suggestions: (1) implement your solution with basic Lists first, then if you are going to generify your solution with parameterized types, create a solution with simple types like Int's first.
using that approach I came up with a real simple (and for me, easy to understand solution):
def fib(h: Int, n: Int) : Stream[Int] = { h #:: fib(n, h + n) }
var x = fib(0,1)
println (s"results: ${(x take 10).toList}")
To get to the above solution I first created, as per Paul's advice, the "for-dummy's" version, based on simple lists:
def fib(h: Int, n: Int) : List[Int] = {
if (h > 100) {
Nil
} else {
h :: fib(n, h + n)
}
}
Notice that I short circuited the list version, because if i didn't it would run forever.. But.. who cares? ;^) since it is just an exploratory bit of code.
The code below is both fast and able to compute with high input indices. On my computer it returns the 10^6:th Fibonacci number in less than two seconds. The algorithm is in a functional style but does not use lists or streams. Rather, it is based on the equality \phi^n = F_{n-1} + F_n*\phi, for \phi the golden ratio. (This is a version of "Binet's formula".) The problem with using this equality is that \phi is irrational (involving the square root of five) so it will diverge due to finite-precision arithmetics if interpreted naively using Float-numbers. However, since \phi^2 = 1 + \phi it is easy to implement exact computations with numbers of the form a + b\phi for a and b integers, and this is what the algorithm below does. (The "power" function has a bit of optimization in it but is really just iteration of the "mult"-multiplication on such numbers.)
type Zphi = (BigInt, BigInt)
val phi = (0, 1): Zphi
val mult: (Zphi, Zphi) => Zphi = {
(z, w) => (z._1*w._1 + z._2*w._2, z._1*w._2 + z._2*w._1 + z._2*w._2)
}
val power: (Zphi, Int) => Zphi = {
case (base, ex) if (ex >= 0) => _power((1, 0), base, ex)
case _ => sys.error("no negative power plz")
}
val _power: (Zphi, Zphi, Int) => Zphi = {
case (t, b, e) if (e == 0) => t
case (t, b, e) if ((e & 1) == 1) => _power(mult(t, b), mult(b, b), e >> 1)
case (t, b, e) => _power(t, mult(b, b), e >> 1)
}
val fib: Int => BigInt = {
case n if (n < 0) => 0
case n => power(phi, n)._2
}
EDIT: An implementation which is more efficient and in a sense also more idiomatic is based on Typelevel's Spire library for numeric computations and abstract algebra. One can then paraphrase the above code in a way much closer to the mathematical argument (We do not need the whole ring-structure but I think it's "morally correct" to include it). Try running the following code:
import spire.implicits._
import spire.algebra._
case class S(fst: BigInt, snd: BigInt) {
override def toString = s"$fst + $snd"++"φ"
}
object S {
implicit object SRing extends Ring[S] {
def zero = S(0, 0): S
def one = S(1, 0): S
def plus(z: S, w: S) = S(z.fst + w.fst, z.snd + w.snd): S
def negate(z: S) = S(-z.fst, -z.snd): S
def times(z: S, w: S) = S(z.fst * w.fst + z.snd * w.snd
, z.fst * w.snd + z.snd * w.fst + z.snd * w.snd)
}
}
object Fibo {
val phi = S(0, 1)
val fib: Int => BigInt = n => (phi pow n).snd
def main(arg: Array[String]) {
println( fib(1000000) )
}
}