In SICP 1.3.2, there is this function
(define (f x y)
((lambda (a b)
(+ (* x (square a))
(* y b)
(* a b)))
(+ 1 (* x y))
(- 1 y)))
Now after 30 minutes chasing after errors, I found this page and it provided this function
def f_lambda(x: Int, y: Int) =
(((a: Int, b: Int) => ((x * square(a)) + (y * b) + (a * b)))
(1 + (x * y), 1 - y))
I do not understand why it is surrounded (like a fort) with parentheses.
Edit: Sorry, my real issue is I don't understand why this function is constructed the way it is. In other words, why are all the parentheses needed in the first place. This looks completely "foreign" compared to the Scala code I've seen so far.
First, it is certainly possible to eliminate a few of the pairs of parentheses in the particular example you give above, although in the case of the outermost pair, this does require putting part or all of the last line on the end of the preceding line:
def f_lambda2(x: Int, y: Int) =
((a: Int, b: Int) => (x * square(a) + y * b + a * b))(1 + x * y, 1 - y)
That said, you may - as with any code - choose to put in extra parentheses to clarify things (eg. around multiplications to make precedence clearer).
Second, there are alternate ways to write such a function which can make what is going on clearer for any reader. This does mean less brevity of code, but I think the clarity gained can definitely be worth it:
def f_lambda3(x: Int, y: Int) = {
def inner(a: Int, b: Int) = (x * square(a)) + (y * b) + (a * b)
inner(1 + x * y, 1 - y)
}
Overall, just because the most efficient, compact representation of a coding concept might arguably involve bracket madness (yay, Lisp!), this doesn't mean this has to be carried over to Scala, which has many much more accessible constructs for writing exressive code.
I don't understand either. You can strip a few off:
def f_lambda(x: Int, y: Int) =
((a: Int, b: Int) => (x * square(a)) + (y * b) + (a * b)) (1 + (x * y), 1 - y)
Or if you want to rely on the fact that multiplication takes precedence over addition:
def f_lambda(x: Int, y: Int) =
((a: Int, b: Int) => x * square(a) + y * b + a * b) (1 + x * y, 1 - y)
Personally I think the first is a bit more readable.
Edit:
Breaking this down a bit, this is declaring an anonymous function which takes two Ints as arguments, styled "a" and "b":
(a: Int, b: Int) => x*square(a) + y*b + a*b
Note this is still using x and y (being the arguments to the outer method). It is then applying this inner function using a = 1 + xy and b = 1 - y.
So substituting, I believe you end up with:
x*square(1 + x*y) + y*(1 - y) + (1 + x*y)*(1 -y)
Why not write it this way in the first place (or using an inner function as Shadowland has)? Well, it's a matter of style and context I guess, so I can't really guess the author's initial intention. The point is that Scala is flexible enough to allow lots of different styles by which you can express the same thing.
Related
In Scala, how can I convert a Seq[Int] to a single number consisting of the numbers in the Seq.
e.g.
Seq(2,3,45,10) to 234510 as a number
A straightforward method is
Seq(2,3,45,10).mkString.toLong
Is there a better and perhaps more performant/functional way?
Seq(2,3,45,10).reduce((x,y) => x * math.pow(10,math.floor(math.log10(y)) + 1).toInt + y)
or
Seq(2,3,45,10).map(BigDecimal(_)).reduce((x,y) => x * BigDecimal(10).pow(y.precision) + y)
But actually i think _.mkString.toLong is the most performant, only problem it will work only for decimal representaion. For arbitrary radix you could do
BigInt(Seq(0x2,0x3,0x45,0x10).map(BigInt(_).toString(16)).mkString, 16)
def toNumber(seq:Seq[Int]):Int = {
def append(scale:Int)(n:Int, m:Int):Int = if(m>=scale) append(scale*10)(n, m) else n*scale + m
seq.foldLeft(0)(append(1))
}
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 ;)
Let me clarify my question by example. This is a standard exponentiation algorithm written with tail recursion in Scala:
def power(x: Double, y: Int): Double = {
def sqr(z: Double): Double = z * z
def loop(xx: Double, yy: Int): Double =
if (yy == 0) xx
else if (yy % 2 == 0) sqr(loop(xx, yy / 2))
else loop(xx * x, yy - 1)
loop(1.0, y)
}
Here sqr method is used to produce the square of loop's result. It doesn't look like a good idea - to define a special function for such a simple operation. But, we can't write just loop(..) * loop(..) instead, since it doubles the calculations.
We also can write it with val and without sqr function:
def power(x: Double, y: Int): Double = {
def loop(xx: Double, yy: Int): Double =
if (yy == 0) xx
else if (yy % 2 == 0) { val s = loop(xx, yy / 2); s * s }
else loop(xx * x, yy - 1)
loop(1.0, y)
}
I can't say that it looks better then variant with sqr, since it uses state variable. The first case is more functional the second way is more Scala-friendly.
Anyway, my question is how to deal with cases when you need to postprocess function's result? Maybe Scala has some other ways to achieve that?
You are using the law that
x^(2n) = x^n * x^n
But this is the same as
x^n * x^n = (x*x)^n
Hence, to avoid squaring after recursion, the value in the case where y is even should be like displayed below in the code listing.
This way, tail-calling will be possible. Here is the full code (not knowing Scala, I hope I get the syntax right by analogy):
def power(x: Double, y: Int): Double = {
def loop(xx: Double, acc: Double, yy: Int): Double =
if (yy == 0) acc
else if (yy % 2 == 0) loop(xx*xx, acc, yy / 2)
else loop(xx, acc * xx, yy - 1)
loop(x, 1.0, y)
}
Here it is in a Haskell like language:
power2 x n = loop x 1 n
where
loop x a 0 = a
loop x a n = if odd n then loop x (a*x) (n-1)
else loop (x*x) a (n `quot` 2)
You could use a "forward pipe". I've got this idea from here: Cache an intermediate variable in an one-liner.
So
val s = loop(xx, yy / 2); s * s
could be rewritten to
loop(xx, yy / 2) |> (s => s * s)
using an implicit conversion like this
implicit class PipedObject[A](value: A) {
def |>[B](f: A => B): B = f(value)
}
As Petr has pointed out: Using an implicit value class
object PipedObjectContainer {
implicit class PipedObject[A](val value: A) extends AnyVal {
def |>[B](f: A => B): B = f(value)
}
}
to be used like this
import PipedObjectContainer._
loop(xx, yy / 2) |> (s => s * s)
is better, since it does not need a temporary instance (requires Scala >= 2.10).
In my comment I pointed out that your implementations can't be tail call optimised, because in the case where yy % 2 == 0, there is a recursive call that is not in tail position. So, for a large input, this can overflow the stack.
A general solution to this is to trampoline your function, replacing recursive calls with data which can be mapped over with "post-processing" such as sqr. The result is then computed by an interpreter, which steps through the return values, storing them on the heap rather than the stack.
The Scalaz library provides an implementation of the data types and interpreter.
import scalaz.Free.Trampoline, scalaz.Trampoline._
def sqr(z: Double): Double = z * z
def power(x: Double, y: Int): Double = {
def loop(xx: Double, yy: Int): Trampoline[Double] =
if (yy == 0)
done(xx)
else if (yy % 2 == 0)
suspend(loop(xx, yy / 2)) map sqr
else
suspend(loop(xx * x, yy - 1))
loop(1.0, y).run
}
There is a considerable performance hit for doing this, though. In this particular case, I would use Igno's solution to avoid the need to call sqr at all. But, the technique described above can be useful when you can't make such optimisations to your algorithm.
In this particular case
No need for utility functions
No need for obtuse piping / implicits
Only need a single standalone recursive call at end - to always give tail recursion
def power(x: Double, y: Int): Double =
if (y == 0) x
else {
val evenPower = y % 2 == 0
power(if (evenPower) x * x else x, if (evenPower) y / 2 else y - 1)
}
Here is a function that takes a pair of Integral
values and divides them:
divide_v1 :: Integral a => (a, a) -> a
divide_v1 (m, n) = (m + n) `div` 2
I invoke the function with a pair of Integral
values and it works as expected:
divide_v1 (1, 3)
Great. That's perfect if my numbers are always Integrals.
Here is a function that takes a pair of Fractional
values and divides them:
divide_v2 :: Fractional a => (a, a) -> a
divide_v2 (m, n) = (m + n) / 2
I invoke the function with a pair of Fractional
values and it works as expected:
divide_v2 (1.0, 3.0)
Great. That's perfect if my numbers are always Fractionals.
I would like a function that works regardless of whether the
numbers are Integrals or Fractionals:
divide_v3 :: Num a => (a, a) -> a
divide_v3 (m, n) = (m + n) ___ 2
What operator do I use for _?
To expand on what AndrewC said, div doesn't have the same properties that / does. For example, in maths, if a divided by b = c, then c times b == a. When working with types like Double and Float, the operations / and * satisfy this property (to the extent that the accuracy of the type allows). But when using div with Ints, the property doesn't hold true. 5 div 3 = 1, but 1*3 /= 5! So if you want to use the same "divide operation" for a variety of numeric types, you need to think about how you want it to behave. Also, you almost certainly wouldn't want to use the same operator /, because that would be misleading.
If you want your "divide operation" to return the same type as its operands, here's one way to accomplish that:
class Divideable a where
mydiv :: a -> a -> a
instance Divideable Int where
mydiv = div
instance Divideable Double where
mydiv = (/)
In GHCi, it looks like this:
λ> 5 `mydiv` 3 :: Int
1
λ> 5 `mydiv` 3 :: Double
1.6666666666666667
λ> 5.0 `mydiv` 3.0 :: Double
1.6666666666666667
On the other hand, if you want to do "true" division, you would need to convert the integral types like this:
class Divideable2 a where
mydiv2 :: a -> a -> Double
instance Divideable2 Int where
mydiv2 a b = fromIntegral a / fromIntegral b
instance Divideable2 Double where
mydiv2 = (/)
In GHCi, this gives:
λ> 5 `mydiv2` 3
1.6666666666666667
λ> 5.0 `mydiv2` 3.0
1.6666666666666667
I think you are looking for Associated Types which allows for implicit type coercion and are explained quite nicely here. Below is an example for the addition of doubles and integers.
class Add a b where
type SumTy a b
add :: a -> b -> SumTy a b
instance Add Integer Double where
type SumTy Integer Double = Double
add x y = fromIntegral x + y
instance Add Double Integer where
type SumTy Double Integer = Double
add x y = x + fromIntegral y
instance (Num a) => Add a a where
type SumTy a a = a
add x y = x + y
Here is some imperative code:
var sum = 0
val spacing = 6
var x = spacing
for(i <- 1 to 10) {
sum += x * x
x += spacing
}
Here are two of my attempts to "functionalize" the above code:
// Attempt 1
(1 to 10).foldLeft((0, 6)) {
case((sum, x), _) => (sum + x * x, x + spacing)
}
// Attempt 2
Stream.iterate ((0, 6)) { case (sum, x) => (sum + x * x, x + spacing) }.take(11).last
I think there might be a cleaner and better functional way to do this. What would be that?
PS: Please note that the above is just an example code intended to illustrate the problem; it is not from the real application code.
Replacing 10 by N, you have spacing * spacing * N * (N + 1) * (2 * N + 1) / 6
This is by noting that you're summing (spacing * i)^2 for the range 1..N. This sum factorizes as spacing^2 * (1^2 + 2^2 + ... + N^2), and the latter sum is well-known to be N * (N + 1) * (2 * N + 1) / 6 (see Square Pyramidal Number)
I actually like idea of lazy sequences in this case. You can split your algorithm in 2 logical steps.
At first you want to work on all natural numbers (ok.. not all, but up to max int), so you define them like this:
val naturals = 0 to Int.MaxValue
Then you need to define knowledge about how numbers, that you want to sum, can be calculated:
val myDoubles = (naturals by 6 tail).view map (x => x * x)
And putting this all together:
val naturals = 0 to Int.MaxValue
val myDoubles = (naturals by 6 tail).view map (x => x * x)
val mySum = myDoubles take 10 sum
I think it's the way mathematician will approach this problem. And because all collections are lazily evaluated - you will not get out of memory.
Edit
If you want to develop idea of mathematical notation further, you can actually define this implicit conversion:
implicit def math[T, R](f: T => R) = new {
def ∀(range: Traversable[T]) = range.view map f
}
and then define myDoubles like this:
val myDoubles = ((x: Int) => x * x) ∀ (naturals by 6 tail)
My personal favourite would have to be:
val x = (6 to 60 by 6) map {x => x*x} sum
Or given spacing as an input variable:
val x = (spacing to 10*spacing by spacing) map {x => x*x} sum
or
val x = (1 to 10) map (spacing*) map {x => x*x} sum
There are two different directions to go. If you want to express yourself, assuming that you can't use the built-in range function (because you actually want something more complicated):
Iterator.iterate(spacing)(x => x+spacing).take(10).map(x => x*x).foldLeft(0)(_ + _)
This is a very general pattern: specify what you start with and how to get the next given the previous; then take the number of items you need; then transform them somehow; then combine them into a single answer. There are shortcuts for almost all of these in simple cases (e.g. the last fold is sum) but this is a way to do it generally.
But I also wonder--what is wrong with the mutable imperative approach for maximal speed? It's really quite clear, and Scala lets you mix the two styles on purpose:
var x = spacing
val last = spacing*10
val sum = 0
while (x <= last) {
sum += x*x
x += spacing
}
(Note that the for is slower than while since the Scala compiler transforms for loops to a construct of maximum generality, not maximum speed.)
Here's a straightforward translation of the loop you wrote to a tail-recursive function, in an SML-like syntax.
val spacing = 6
fun loop (sum: int, x: int, i: int): int =
if i > 0 then loop (sum+x*x, x+spacing, i-1)
else sum
val sum = loop (0, spacing, 10)
Is this what you were looking for? (What do you mean by a "cleaner" and "better" way?)
What about this?
def toSquare(i: Int) = i * i
val spacing = 6
val spaceMultiples = (1 to 10) map (spacing *)
val squares = spaceMultiples map toSquare
println(squares.sum)
You have to split your code in small parts. This can improve readability a lot.
Here is a one-liner:
(0 to 10).reduceLeft((u,v)=>u + spacing*spacing*v*v)
Note that you need to start with 0 in order to get the correct result (else the first value 6 would be added only, but not squared).
Another option is to generate the squares first:
(1 to 2*10 by 2).scanLeft(0)(_+_).sum*spacing*spacing