I was trying to decipher what this actually means
check { (n: Int) =>
n > 1 ==> n / 2 > 0
}
from http://www.scalatest.org/user_guide/writing_scalacheck_style_properties
I am first trying to decipher whether that body is
((n: Int) => n > 1) ==> n/2 > 0
or
(n: Int) => (n > 1 ==> n/2 > 0)
Which one would it be? I am guessing the latter since check method takes a function though the method '==>' could be returning a function as well?
Next, I look at the check method signature at https://searchcode.com/codesearch/view/12336175/
def check[A1,P](f: A1 => P)
(implicit
p: P => Prop,
a1: Arbitrary[A1], s1: Shrink[A1]
) {
check(Prop.property(f)(p, a1, s1))
}
I believe A1 would be the Int unless ==> returns a function and changes the return type(doubtful though I think). I am not sure how to find the implicit function P => Prop in the scalacheck library.
I do notice there is an ExtendedBoolean that has a function ==> https://github.com/rickynils/scalacheck/blob/master/src/main/scala/org/scalacheck/Prop.scala
Perhaps (n > 1) was converted to an ExtendedBoolean assuming the function was (n: Int) => (n > 1 ==> n/2 > 0) and then we get
ExtendedBoolean( n > 1 ).==>( n / 2 > 0 ) is called then.
Since the implementation of ==> for ExtendedBoolean is Prop(b) ==> p, I end up with
Prop( n > 1 ) ==> (n / 2 > 0)
I get really confused here as in call by name, the values are usually captured except there is no value for n at this point since it was originally a function. ignoring my confusion for a second, the implementation of ==> for Prop is thus
def ==>(p: => Prop): Prop = flatMap { r1 =>
if(r1.proved) p map { r2 => mergeRes(r1,r2,r2.status) }
else if(!r1.success) Prop(r1.copy(status = Undecided))
else p map { r2 => provedToTrue(mergeRes(r1,r2,r2.status)) }
}
so, we are calling this like so I think with another implicit version to Prop here
Prop( n > 1 ) ==> Prop(n / 2 > 0)
ok, and flatmap is
Prop(prms => f(this(prms))(prms))
hmmm, there must have been another conversion to Prop . I start to trace prms and then look at the object Prop apply method for Booleans which is
def apply(b: Boolean): Prop = if(b) proved else falsified
but I can't resolve b to true or false since those functions have not been evaluated yet. How is all this working together? I think I am missing a few concepts here just barely. Can someone explain this a bit better?
Yours is really a ScalaCheck question, but one way to avoid the whole question is to use ScalaTest's PropertyChecks syntax instead of ScalaCheck's (which is supported in ScalaTest by Checkers):
import org.scalatest.prop.Checkers._
check { (n: Int) =>
n > 1 ==> n / 2 > 0
}
becomes:
import org.scalatest.MustMatchers._
import org.scalatest.prop.PropertyChecks._
forAll { (n: Int) =>
whenever (n > 1) {
n / 2 must be > 0
}
}
Related
I would like to use Scala's property-based testing tool ScalaCheck to express a property
if an integer n * n = 0 then n = 0
How can I write this property in ScalaCheck? I know for example
val myprop = forAll {(n: Int) => n + 1 - 1 = n}
But I do not know how to express "A implies B" in ScalaCheck (without reducing it to Not-A or B, which can look clumsy).
Use ==> (implication operator)
val prop = forAll { n: Int =>
(n * n == 0) ==> n == 0
}
(see their User Guide )
the catch is: in this particular example the condition is very hard to satisfy so ScalaCheck will give up after several tries (but at least it does tell you so, otherwise you get a false positive because your necessary condition was never checked). In that case you can provide a custom generator so that it will generate values that satisfy your condition.
Consider 2+7 == 3+6 (mod 5). Can you some how use scala syntactic sugar to achieve the same in scala code?
Keep in mind that 2+7 and 3+6 are regular scala Int so overriding the + or == to be mod 5 doesn't work. I'm actually interested in more complex congruences on algebras A. I could do A.congruent(a,b), and write that with some nice symbols like A.~(a,b), but I am interested in a == b (A) or a ==(A) b or perhaps A(a == b). Something where the congruence appears inbetween the terms a and b.
The bottom line of my struggles is that the congruence is defined for type A, and a,b are some elements passed to A but not actually of type A. E.g. A might be a group of matrices and the congruence is if individual matrices a and b differ by a scalar i.e. a*b^-1=sI_n. In particular, a, b will live inside of many groups and the congruence will change based on that. So I it isn't possible to simply add a reference within a and b back to A.
Some how the right solution seems to be the mathematical one, label the equivalence with A not the variables a and b. Yet scala syntactic sugar may not have such a sweetness in mind. Any suggestions appreciated.
Try this:
implicit class ModEquals(a: Int) {
def %%(n: Int) = new { def ===(b: Int) = (a - b) % n == 0 }
}
Usage:
7 %% 3 === 10
This solution enriches Ints with a %% method that takes the congruence. In this example, it's just modulu, but this can easily be extended to anything. The returned object is a class that has an === method defined to implements the equality check.
import scala.languageFeature.implicitConversions
import scala.languageFeature.reflectiveCalls
case class ModuloArg(list: List[Int]) {
assert(list.size > 1)
def ==%%?(m: Int) = {
val hm = list.head % m
list.tail.filter(i => (i % m) != hm).isEmpty
}
def ::%%(n: Int) = ModuloArg(n :: list)
}
implicit class ModuloArgOps(i: Int) {
def ::%%(n: Int) = ModuloArg(n :: i :: Nil)
}
Now, you can use these to check modulo equality,
// 4 == 10 (mod 3)
scala> val mod3Equal4And10 = 4 ::%% 10 ==%%? 3
// mod3Equal4And10: Boolean = true
// 4 == 11 (mod 3)
scala> val mod3Equal4And11 = 4 ::%% 11 ==%%? 3
//mod3Equal4And11: Boolean = false
// 4 == 10 == 13 == 16 (mod 3)
scala> val mod3Equal4And10And13And16 = 4 ::%% 10 ::%% 13 ::%% 16 ==%%? 3
// mod3Equal4And10And13And16: Boolean = true
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 ;)
In Scala language, I want to write a function that yields odd numbers within a given range. The function prints some log when iterating even numbers. The first version of the function is:
def getOdds(N: Int): Traversable[Int] = {
val list = new mutable.MutableList[Int]
for (n <- 0 until N) {
if (n % 2 == 1) {
list += n
} else {
println("skip even number " + n)
}
}
return list
}
If I omit printing logs, the implementation become very simple:
def getOddsWithoutPrint(N: Int) =
for (n <- 0 until N if (n % 2 == 1)) yield n
However, I don't want to miss the logging part. How do I rewrite the first version more compactly? It would be great if it can be rewritten similar to this:
def IWantToDoSomethingSimilar(N: Int) =
for (n <- 0 until N) if (n % 2 == 1) yield n else println("skip even number " + n)
def IWantToDoSomethingSimilar(N: Int) =
for {
n <- 0 until N
if n % 2 != 0 || { println("skip even number " + n); false }
} yield n
Using filter instead of a for expression would be slightly simpler though.
I you want to keep the sequentiality of your traitement (processing odds and evens in order, not separately), you can use something like that (edited) :
def IWantToDoSomethingSimilar(N: Int) =
(for (n <- (0 until N)) yield {
if (n % 2 == 1) {
Option(n)
} else {
println("skip even number " + n)
None
}
// Flatten transforms the Seq[Option[Int]] into Seq[Int]
}).flatten
EDIT, following the same concept, a shorter solution :
def IWantToDoSomethingSimilar(N: Int) =
(0 until N) map {
case n if n % 2 == 0 => println("skip even number "+ n)
case n => n
} collect {case i:Int => i}
If you will to dig into a functional approach, something like the following is a good point to start.
First some common definitions:
// use scalaz 7
import scalaz._, Scalaz._
// transforms a function returning either E or B into a
// function returning an optional B and optionally writing a log of type E
def logged[A, E, B, F[_]](f: A => E \/ B)(
implicit FM: Monoid[F[E]], FP: Pointed[F]): (A => Writer[F[E], Option[B]]) =
(a: A) => f(a).fold(
e => Writer(FP.point(e), None),
b => Writer(FM.zero, Some(b)))
// helper for fixing the log storage format to List
def listLogged[A, E, B](f: A => E \/ B) = logged[A, E, B, List](f)
// shorthand for a String logger with List storage
type W[+A] = Writer[List[String], A]
Now all you have to do is write your filtering function:
def keepOdd(n: Int): String \/ Int =
if (n % 2 == 1) \/.right(n) else \/.left(n + " was even")
You can try it instantly:
scala> List(5, 6) map(keepOdd)
res0: List[scalaz.\/[String,Int]] = List(\/-(5), -\/(6 was even))
Then you can use the traverse function to apply your function to a list of inputs, and collect both the logs written and the results:
scala> val x = List(5, 6).traverse[W, Option[Int]](listLogged(keepOdd))
x: W[List[Option[Int]]] = scalaz.WriterTFunctions$$anon$26#503d0400
// unwrap the results
scala> x.run
res11: (List[String], List[Option[Int]]) = (List(6 was even),List(Some(5), None))
// we may even drop the None-s from the output
scala> val (logs, results) = x.map(_.flatten).run
logs: List[String] = List(6 was even)
results: List[Int] = List(5)
I don't think this can be done easily with a for comprehension. But you could use partition.
def getOffs(N:Int) = {
val (evens, odds) = 0 until N partition { x => x % 2 == 0 }
evens foreach { x => println("skipping " + x) }
odds
}
EDIT: To avoid printing the log messages after the partitioning is done, you can change the first line of the method like this:
val (evens, odds) = (0 until N).view.partition { x => x % 2 == 0 }
I thought I could implement n+k patterns as an active pattern in scala via unapply, but it seems to fail with unspecified value parameter: k
object NPlusK {
def apply(n : Int, k : Int) = {
n + k
}
def unapply(n : Int, k : Int) = {
if (n > 0 && n > k) Some(n - k) else None
}
}
object Main {
def main(args: Array[String]): Unit = {
}
def fac(n: Int) : BigInt = {
n match {
case 0 => 1
case NPlusK(n, 1) => n * fac(n - 1)
}
}
}
Is it possible to implement n+k patterns in Scala and in that event how?
You should look at this question for a longer discussion, but here's a short adaptation for your specific case.
An unapply method can only take one argument, and must decide from that argument how to split it into two parts. Since there are multiple ways to divide some integer x into n and k such that x = n + k, you can't use an unapply for this.
You can get around it by creating a separate extractors for each k. Thus, instead of NplusK you'd have Nplus1, Nplus2, etc since there is exactly one way to get n from x such that x = n + 1.
case class NplusK(k: Int) {
def unapply(n: Int) = if (n > 0 && n > k) Some(n - k) else None
}
val Nplus1 = NplusK(1)
val Nplus1(n) = 5 // n = 4
So your match becomes:
n match {
case 0 => 1
case Nplus1(n) => n * fac(n - 1)
}
Deconstructor unapply does not work this way at all. It takes only one argument, the matched value, and returns an option on a tuple, with as many elements as there are arguments to the your pattern (NPlusK). That is, when you have
(n: Int) match {
...
case NPlusK(n, 1)
It will look for an unapply method with an Int (or supertype) argument. If there is such a method, and if the return type is a Tuple2 (as NPlusK appears with two arguments in the pattern), then it will try to match. Whatever subpattern there are inside NPlusK (here the variable n, and the constant 1), will not be passed to unapply in anyway (what do you expect if you write case NPlusK(NPlusK(1, x), NPlusK(1, y))?). Instead, if unapply returns some tuple, then each element of the tuple will be matched to the corresponding subpattern, here n which always matches, and 1 which will match if the value is equal to 1.
You could write
def unapply(n: Int) = if (n > 0) Some((n-1, 1)) else None.
That would match when your NPlusK(n, 1). But that would not match NPlusK(n, 2), nor NPlusK(1, n) (except if n is 2). This does not make much sense. A pattern should probably have only one possible match. NPlusK(x, y) can match n in many different ways.
What would work would be something Peano integers like, with Succ(n) matching n+1.