Related
I've been working on the scala recursion problem. I used to develop the program using loops and then use the concept of recursion to convert the existing loop problem in a recursive solution.
So I have written the following code to find the perfect number using loops.
def isPerfect(n: Int): Boolean = {
var sum = 1
// Find all divisors and add them
var i = 2
while ( {
i * i <= n
}) {
if (n % i == 0) if (i * i != n) sum = sum + i + n / i
else sum = sum + i
i += 1
}
// If sum of divisors is equal to
// n, then n is a perfect number
if (sum == n && n != 1) return true
false
}
Here is my attempt to convert it into a recursive solution. But I'm getting the incorrect result.
def isPerfect(n: Int): Boolean = {
var sum = 1
// Find all divisors and add them
var i = 2
def loop(i:Int, n:Int): Any ={
if(n%i == 0) if (i * i != n) return sum + i + n / i
else
return loop(i+1, sum+i)
}
val sum_ = loop(2, n)
// If sum of divisors is equal to
// n, then n is a perfect number
if (sum_ == n && n != 1) return true
false
}
Thank you in advance.
Here is a tail-recursive solution
def isPerfectNumber(n: Int): Boolean = {
#tailrec def loop(d: Int, acc: List[Int]): List[Int] = {
if (d == 1) 1 :: acc
else if (n % d == 0) loop(d - 1, d :: acc)
else loop(d - 1, acc)
}
loop(n-1, Nil).sum == n
}
As a side-note, functions that have side-effects such as state mutation scoped locally are still considered pure functions as long as the mutation is not visible externally, hence having while loops in such functions might be acceptable.
I want to generate a list of Tuple2 objects. Each tuple (a,b) in the list should meeting the conditions:a and b both are perfect squares,(b/30)<a<b
and a>N and b>N ( N can even be a BigInt)
I am trying to write a scala function to generate the List of Tuples meeting the above requirements?
This is my attempt..it works fine for Ints and Longs..But for BigInt there is sqrt problem I am facing..Here is my approach in coding as below:
scala> def genTups(N:Long) ={
| val x = for(s<- 1L to Math.sqrt(N).toLong) yield s*s;
| val y = x.combinations(2).map{ case Vector(a,b) => (a,b)}.toList
| y.filter(t=> (t._1*30/t._2)>=1)
| }
genTups: (N: Long)List[(Long, Long)]
scala> genTups(30)
res32: List[(Long, Long)] = List((1,4), (1,9), (1,16), (1,25), (4,9), (4,16), (4,25), (9,16), (9,25), (16,25))
Improved this using BigInt square-root algorithm as below:
def genTups(N1:BigInt,N2:BigInt) ={
def sqt(n:BigInt):BigInt = {
var a = BigInt(1)
var b = (n>>5)+BigInt(8)
while((b-a) >= 0) {
var mid:BigInt = (a+b)>>1
if(mid*mid-n> 0) b = mid-1
else a = mid+1
}; a-1 }
val x = for(s<- sqt(N1) to sqt(N2)) yield s*s;
val y = x.combinations(2).map{ case Vector(a,b) => (a,b)}.toList
y.filter(t=> (t._1*30/t._2)>=1)
}
I appreciate any help to improve in my algorithm .
You can avoid sqrt in you algorithm by changing the way you calculate x to this:
val x = (BigInt(1) to N).map(x => x*x).takeWhile(_ <= N)
The final function is then:
def genTups(N: BigInt) = {
val x = (BigInt(1) to N).map(x => x*x).takeWhile(_ <= N)
val y = x.combinations(2).map { case Vector(a, b) if (a < b) => (a, b) }.toList
y.filter(t => (t._1 * 30 / t._2) >= 1)
}
You can also re-write this as a single chain of operations like this:
def genTups(N: BigInt) =
(BigInt(1) to N)
.map(x => x * x)
.takeWhile(_ <= N)
.combinations(2)
.map { case Vector(a, b) if a < b => (a, b) }
.filter(t => (t._1 * 30 / t._2) >= 1)
.toList
In a quest for performance, I came up with this recursive version that appears to be significantly faster
def genTups(N1: BigInt, N2: BigInt) = {
def sqt(n: BigInt): BigInt = {
var a = BigInt(1)
var b = (n >> 5) + BigInt(8)
while ((b - a) >= 0) {
var mid: BigInt = (a + b) >> 1
if (mid * mid - n > 0) {
b = mid - 1
} else {
a = mid + 1
}
}
a - 1
}
#tailrec
def loop(a: BigInt, rem: List[BigInt], res: List[(BigInt, BigInt)]): List[(BigInt, BigInt)] =
rem match {
case Nil => res
case head :: tail =>
val a30 = a * 30
val thisRes = rem.takeWhile(_ <= a30).map(b => (a, b))
loop(head, tail, thisRes.reverse ::: res)
}
val squares = (sqt(N1) to sqt(N2)).map(s => s * s).toList
loop(squares.head, squares.tail, Nil).reverse
}
Each recursion of the loop adds all the matching pairs for a given value of a. The result is built in reverse because adding to the front of a long list is much faster than adding to the tail.
Firstly create a function to check if number if perfect square or not.
def squareRootOfPerfectSquare(a: Int): Option[Int] = {
val sqrt = math.sqrt(a)
if (sqrt % 1 == 0)
Some(sqrt.toInt)
else
None
}
Then, create another func that will calculate this list of tuples according to the conditions mentioned above.
def generateTuples(n1:Int,n2:Int)={
for{
b <- 1 to n2;
a <- 1 to n1 if(b>a && squareRootOfPerfectSquare(b).isDefined && squareRootOfPerfectSquare(a).isDefined)
} yield ( (a,b) )
}
Then on calling the function with parameters generateTuples(5,10)
you will get an output as
res0: scala.collection.immutable.IndexedSeq[(Int, Int)] = Vector((1,4), (1,9), (4,9))
Hope that helps !!!
What is a nice and efficient functional way of solving the following problem? In imperative style, this can be done in linear time.
Given two sorted sequences p and q, f returns a sequence r (or any collection) of triples where for every triple (a,b,c) in r, the following hold:
(a < b < c)
One of the following two holds:
a,c are two consecutive elements p, and b is in q
a,c are two consecutive elements q, and b is in p
Example: Consider the following two sequences.
val p = Seq(1,4,5,7,8,9)
val q = Seq(2,3,6,7,8,10)
Then f(p,s) computes the following sequence:
Seq((1,2,4), (1,3,4), (5,6,7), (3,4,6), (3,5,6), (8,9,10))
Current solution: I do not find this one very elegant. I am looking for a better one.
def consecutiveTriplesOneWay(s1: Seq[Int], s2:Seq[Int]) = {
for {
i <- 0 until s1.size - 1 if s1(i) < s1(i+1)
j <- 0 until s2.size if s1(i) < s2(j) && s2(j) < s1(i+1)
} yield (s1(i), s2(j), s1(i+1))
}
def consecutiveTriples(s1: Seq[Int], s2:Seq[Int]) =
consecutiveTriplesOneWay(s1, s2) ++ consecutiveTriplesOneWay(s2, s1)
def main(args: Array[String]) {
val p = Seq(1,4,5,7,8,9)
val q = Seq(2,3,6,7,8,10)
consecutiveTriples(p, q).foreach(println(_))
}
Edit: My imperative solution
def consecutiveTriplesOneWayImperative(s1: Seq[Int], s2:Seq[Int]) = {
var i = 0
var j = 0
val triples = mutable.MutableList.empty[(Int,Int,Int)]
while (i < s1.size - 1 && j < s2.size) {
if (s1(i) < s2(j) && s2(j) < s1(i + 1)) {
triples += ((s1(i), s2(j), s1(i + 1)))
j += 1
} else if (s1(i) >= s2(j))
j += 1
else
i += 1
}
triples.toSeq
}
def consecutiveTriples(s1: Seq[Int], s2:Seq[Int]) =
consecutiveTriplesOneWayImperative(s1,s2) ++
consecutiveTriplesOneWayImperative(s2,s1)
Imperative solution translated to tailrec. Bit verbose but works
def consecutiveTriplesRec(s1: Seq[Int], s2: Seq[Int]) = {
#tailrec
def consTriplesOneWay(left: Seq[Int], right: Seq[Int],
triples: Seq[(Int, Int, Int)]): Seq[(Int, Int, Int)] = {
(left, right) match {
case (l1 :: l2 :: ls, r :: rs) =>
if (l1 < r && r < l2) consTriplesOneWay(left, rs, (l1, r, l2) +: triples)
else if (l1 >= r) consTriplesOneWay(left, rs, triples)
else consTriplesOneWay(l2 :: ls, right, triples)
case _ => triples
}
}
consTriplesOneWay(s1, s2, Nil) ++ consTriplesOneWay(s2, s1, Nil)
}
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$26#503d0400
// unwrap the results
scala> x.run
res11: (List[String], List[Option[Int]]) = (List(6 was even),List(Some(5), None))
// we may even drop the None-s from the output
scala> val (logs, results) = x.map(_.flatten).run
logs: List[String] = List(6 was even)
results: List[Int] = List(5)
I don't think this can be done easily with a for comprehension. But you could use partition.
def getOffs(N:Int) = {
val (evens, odds) = 0 until N partition { x => x % 2 == 0 }
evens foreach { x => println("skipping " + x) }
odds
}
EDIT: To avoid printing the log messages after the partitioning is done, you can change the first line of the method like this:
val (evens, odds) = (0 until N).view.partition { x => x % 2 == 0 }
Given n ( say 3 people ) and s ( say 100$ ), we'd like to partition s among n people.
So we need all possible n-tuples that sum to s
My Scala code below:
def weights(n:Int,s:Int):List[List[Int]] = {
List.concat( (0 to s).toList.map(List.fill(n)(_)).flatten, (0 to s).toList).
combinations(n).filter(_.sum==s).map(_.permutations.toList).toList.flatten
}
println(weights(3,100))
This works for small values of n. ( n=1, 2, 3 or 4).
Beyond n=4, it takes a very long time, practically unusable.
I'm looking for ways to rework my code using lazy evaluation/ Stream.
My requirements : Must work for n upto 10.
Warning : The problem gets really big really fast. My results from Matlab -
---For s =100, n = 1 thru 5 results are ---
n=1 :1 combinations
n=2 :101 combinations
n=3 :5151 combinations
n=4 :176851 combinations
n=5: 4598126 combinations
---
You need dynamic programming, or memoization. Same concept, anyway.
Let's say you have to divide s among n. Recursively, that's defined like this:
def permutations(s: Int, n: Int): List[List[Int]] = n match {
case 0 => Nil
case 1 => List(List(s))
case _ => (0 to s).toList flatMap (x => permutations(s - x, n - 1) map (x :: _))
}
Now, this will STILL be slow as hell, but there's a catch here... you don't need to recompute permutations(s, n) for numbers you have already computed. So you can do this instead:
val memoP = collection.mutable.Map.empty[(Int, Int), List[List[Int]]]
def permutations(s: Int, n: Int): List[List[Int]] = {
def permutationsWithHead(x: Int) = permutations(s - x, n - 1) map (x :: _)
n match {
case 0 => Nil
case 1 => List(List(s))
case _ =>
memoP getOrElseUpdate ((s, n),
(0 to s).toList flatMap permutationsWithHead)
}
}
And this can be even further improved, because it will compute every permutation. You only need to compute every combination, and then permute that without recomputing.
To compute every combination, we can change the code like this:
val memoC = collection.mutable.Map.empty[(Int, Int, Int), List[List[Int]]]
def combinations(s: Int, n: Int, min: Int = 0): List[List[Int]] = {
def combinationsWithHead(x: Int) = combinations(s - x, n - 1, x) map (x :: _)
n match {
case 0 => Nil
case 1 => List(List(s))
case _ =>
memoC getOrElseUpdate ((s, n, min),
(min to s / 2).toList flatMap combinationsWithHead)
}
}
Running combinations(100, 10) is still slow, given the sheer numbers of combinations alone. The permutations for each combination can be obtained simply calling .permutation on the combination.
Here's a quick and dirty Stream solution:
def weights(n: Int, s: Int) = (1 until s).foldLeft(Stream(Nil: List[Int])) {
(a, _) => a.flatMap(c => Stream.range(0, n - c.sum + 1).map(_ :: c))
}.map(c => (n - c.sum) :: c)
It works for n = 6 in about 15 seconds on my machine:
scala> var x = 0
scala> weights(100, 6).foreach(_ => x += 1)
scala> x
res81: Int = 96560646
As a side note: by the time you get to n = 10, there are 4,263,421,511,271 of these things. That's going to take days just to stream through.
My solution of this problem, it can computer n till 6:
object Partition {
implicit def i2p(n: Int): Partition = new Partition(n)
def main(args : Array[String]) : Unit = {
for(n <- 1 to 6) println(100.partitions(n).size)
}
}
class Partition(n: Int){
def partitions(m: Int):Iterator[List[Int]] = new Iterator[List[Int]] {
val nums = Array.ofDim[Int](m)
nums(0) = n
var hasNext = m > 0 && n > 0
override def next: List[Int] = {
if(hasNext){
val result = nums.toList
var idx = 0
while(idx < m-1 && nums(idx) == 0) idx = idx + 1
if(idx == m-1) hasNext = false
else {
nums(idx+1) = nums(idx+1) + 1
nums(0) = nums(idx) - 1
if(idx != 0) nums(idx) = 0
}
result
}
else Iterator.empty.next
}
}
}
1
101
5151
176851
4598126
96560646
However , we can just show the number of the possible n-tuples:
val pt: (Int,Int) => BigInt = {
val buf = collection.mutable.Map[(Int,Int),BigInt]()
(s,n) => buf.getOrElseUpdate((s,n),
if(n == 0 && s > 0) BigInt(0)
else if(s == 0) BigInt(1)
else (0 to s).map{k => pt(s-k,n-1)}.sum
)
}
for(n <- 1 to 20) printf("%2d :%s%n",n,pt(100,n).toString)
1 :1
2 :101
3 :5151
4 :176851
5 :4598126
6 :96560646
7 :1705904746
8 :26075972546
9 :352025629371
10 :4263421511271
11 :46897636623981
12 :473239787751081
13 :4416904685676756
14 :38393094575497956
15 :312629484400483356
16 :2396826047070372396
17 :17376988841260199871
18 :119594570260437846171
19 :784008849485092547121
20 :4910371215196105953021