So I have this recursive function that multiplies two numbers, simple enough.
def mul(n: Int, m: Int):Int =
if(m > 1) n + mul(n, dec(m))
else n
Now I'm trying to turn it into a tail recursive function, and I tried this:
def mulWithTail(n: Int, m: Int):Int = {
#tailrec
def iter(result: Int, x: Int):Int =
if(x == 0) result
else result + iter(result, dec(x))
iter(n, m)
}
However I get the following error:
error: could not optimize #tailrec annotated method iter: it contains
a recursive call not in tail position
else result + iter(result, dec(x))
Question: Can you explain to me why this error is occurring? How should I refactor my code?
You can make your function tail recursive simply adding an extra parameter acting like an accumulator. Like this.
def mul(n: Int, m: Int, acc: Int): Int =
if (m > 1) mul(n, m - 1, n + acc)
else acc
To make a function tail recursive you must not perform any other operation in the recursion step but calling the function recursively. In your code samples you're performing an addition in the recursion step.
n + mul(n, dec(m))
result + iter(result, dec(x))
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.
Suppose I have this code in scala :
def factorial(accumulator: Int, x: Int) : Int = {
if(x == 1)
return accumulator
factorial(x * accumulator, x - 1)
}
println(factorial(1,0))
And the Output :
0
Now I have two questions :
1) Isn't the definition of this function fundamentally wrong? ( will not give the right answer for zero) I could always wrap this function inside another function and treat zero as special case returning 1 but that does not feel right and in tune with the formal definition.
2) Also why I am returned 0 as the answer in the first place? Why doesn't the code get stuck in an infinite loop?
def factorial(x: Int): Int = {
#annotation.tailrec
def factorial(accumulator: Int, x: Int): Int = {
if (x <= 0)
accumulator
else
factorial(x * accumulator, x - 1)
}
assert(x >= 0,"""argument should be "non-negative integer" """)
factorial(1, x)
}
You should not give user possibility to call factorial in wrong way. So your function should be internal
factorial(0) = 1
Yes, you should hide the accumulator and make it an argument of an internal, tailrec function. The special case for zero should also be handled explicitly, there is nothing 'against formal factorial definition' with it.
It works because integer exceeds the maximum negative value.
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 hope this question is not a dupe. It seems most questions refer to multiple recursive calls in one statement, ie.:return func(n - 1) * func(n - 2). My question involves multiple recursive calls inside an if/else statement. THis is what I have (from doing a Project Euler problem):
def multOfThreeAndFive(n: Double): Double = {
def loop(next: Double, acc: Double): Double = {
if (next < 0) acc
else if (next % 3 == 0 || next % 5 == 0) loop(next - 1, acc + next)
else loop(next - 1, acc)
}
loop(n - 1, 0)
}
My question is, since I'm making two separate recursive calls, one inside else if and the other inside the last else, is this still considered tail recursive?
It's easy to check by yourself with special annotation. Compilation will fail if compiler isn't able to optimize code down to tail-recursion emulation:
def multOfThreeAndFive(n: Double): Double = {
#annotation.tailrec
def loop(next: Double, acc: Double): Double = {
if (next < 0) acc
else if (next % 3 == 0 || next % 5 == 0) loop(next - 1, acc + next)
else loop(next - 1, acc)
}
loop(n - 1, 0)
}
And yes, it is considered as a tail recursive function.