JaKu - 2 months ago 4x

Scala Question

I am trying to parallelize recursive calls of sudoku solver from 25 lines Sudoku solver in Scala. I've changed their

`Fold`

`reduce`

`def reduce(f: (Int, Int) => Int, accu: Int, l: Int, u: Int): Int = {`

accu + (l until u).toArray.reduce(f(accu, _) + f(accu, _))

}

which if run sequentially works fine, but when I change it into

`accu + (l until u).toArray.par.reduce(f(accu, _) + f(accu, _))`

the recursion reaches the bottom much more often and generates false solutions. I thought, that it will execute the bottom level recursion and work it's way up, but doesn't seem to do so.

I've also tried futures

`def parForFut2(f: (Int, Int) => Int, accu: Int, l: Int, u: Int): Int = {`

var sum: Int = accu

val vals = l until u

vals.foreach(t => scala.actors.Futures.future(sum + f(accu, t)))

sum

}

which appears to have the same problem as the

`par.reduce`

`object SudokuSolver extends App {`

// The board is represented by an array of string

val source = scala.io.Source.fromFile("./puzzle")

val lines = (source.getLines).toArray

var m: Array[Array[Char]] = for (

str <- lines;

line: Array[Char] = str.toArray

) yield line

source.close()

// For printing m

def print = {

Console.println("");

refArrayOps(m) map (carr => Console.println(new String(carr)))

}

// The test for validity of n on position x,y

def invalid(i: Int, x: Int, y: Int, n: Char): Boolean =

i < 9 && (m(y)(i) == n || m(i)(x) == n ||

m(y / 3 * 3 + i / 3)(x / 3 * 3 + i % 3) == n || invalid(i + 1, x, y, n))

// Looping over a half-closed range of consecutive Integers [l..u)

// is factored out Into a higher-order function

def parReduce(f: (Int, Int) => Int, accu: Int, l: Int, u: Int): Int = {

accu + (l until u).toArray.par.reduce(f(accu, _) + f(accu, _))

}

// The search function examines each position on the board in turn,

// trying the numbers 1..9 in each unfilled position

// The function is itself a higher-order fold, accumulating the value

// accu by applying the given function f to it whenever a solution m

// is found

def search(x: Int, y: Int, f: (Int) => Int, accu: Int): Int = Pair(x, y) match {

case Pair(9, y) => search(0, y + 1, f, accu) // next row

case Pair(0, 9) => f(accu) // found a solution - print it and continue

case Pair(x, y) => if (m(y)(x) != '0') search(x + 1, y, f, accu) else

parForFut1((accu: Int, n: Int) =>

if (invalid(0, x, y, (n + 48).asInstanceOf[Char])) accu else {

m(y)(x) = (n + 48).asInstanceOf[Char];

val newaccu = search(x + 1, y, f, accu);

m(y)(x) = '0';

newaccu

}, accu, 1, 10)

}

// The main part of the program uses the search function to accumulate

// the total number of solutions

Console.println("\n" + search(0, 0, i => { print; i + 1 }, 0) + " solution(s)")

}

Answer

After Andreas comment I changed the `m: Array[Array[Char]]`

into `m: List[List[Char]]`

which prevents any unnecessary and unwanted changes to it. The final looping method is

```
def reduc(f: (Int, Int) => Int,
accu: Int, l: Int, u: Int, m1: List[List[Char]]):Int =
accu + (l until u).toArray.par.reduce(f(accu, _) + f(accu, _))
```

and I had to pass m as an argument to each used function, so every one of them had its own instance of it. The whole code:

```
object SudokuSolver extends App{
// The board is represented by an Array of strings (Arrays of Chars),
val source = scala.io.Source.fromFile("./puzzle")
val lines = source.getLines.toList
val m: List[List[Char]] = for (
str <- lines;
line: List[Char] = str.toList
) yield line
source.close()
// For prInting m
def printSud(m: List[List[Char]]) = {
Console.println("")
m map (println)
}
Console.println("\nINPUT:")
printSud(m)
def invalid(i:Int, x:Int, y:Int, n:Char,m1: List[List[Char]]): Boolean =
i < 9 && (m1(y)(i) == n || m1(i)(x) == n ||
m1(y / 3 * 3 + i / 3)(x / 3 * 3 + i % 3) == n ||
invalid(i + 1, x, y, n, m1))
def reduc(f: (Int, Int) => Int, accu: Int, l: Int, u: Int,
m1: List[List[Char]]): Int =
accu + (l until u).toArray.par.reduce(f(accu, _) + f(accu, _))
def search(x: Int, y: Int, accu: Int, m1: List[List[Char]]): Int =
Pair(x, y) match {
case Pair(9, y) => search(0, y + 1, accu, m1) // next row
case Pair(0, 9) => { printSud(m1); accu + 1 } // found a solution
case Pair(x, y) =>
if (m1(y)(x) != '0')
search(x + 1, y, accu, m1) // place is filled, we skip it.
else // box is not filled, we try all n in {1,...,9}
reduc((accu: Int, n: Int) => {
if (invalid(0, x, y, (n + 48).asInstanceOf[Char], m1))
accu
else { // n fits here
val line = List(m1(y).patch(x, Seq((n + 48).asInstanceOf[Char]), 1))
val m2 = m1.patch(y, line, 1)
val newaccu = search(x + 1, y, accu, m2);
val m3 = m1.patch(y, m1(y).patch(x, Seq(0), 1), 1)
newaccu
}
}, accu, 1, 10, m1)
}
Console.println("\n" + search(0, 0, 0, m) + " solution(s)")
}
```

Source (Stackoverflow)

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