N-Queens Combinatorial Problem - Polyglot FP for Fun and Profit – Haskell and Scala - Part 1

N-Queens Combinatorial Problem
first see the problem solved using the List monad and a Scala for comprehension
then see the Scala program translated into Haskell, both using a do expressions and using a List comprehension
understand how the Scala for comprehension is desugared, and what role the withFilter function plays
also understand how the Haskell do expressions and List comprehension are desugared, and what role the guard function plays
Miran Lipovača
Polyglot FP for Fun and Profit – Haskell and Scala
@philip_schwarz
slides by https://www.slideshare.net/pjschwarz
Bill Venners
Martin Odersky
Part 1
based on, and inspired by, the work of
Lex Spoon

Yes, I know, Lex Spoon looks a bit too young in that photo, and Bill Venner’s hair doesn’t look realistic in that drawing. Martin
Odersky’s drawing is not bad at all.
By the way, don’t be put off by the fact that the book image below is of the fourth edition. Excerpts from the N-queens section of
that edition are great, and by the way, in the fifth edition (new for Scala 3) I don’t see that section (it may have moved to upcoming
book Advanced Programming in Scala). Also, I do use Scala 3 in this slide deck.
Bill Venners
Martin Odersky
Lex Spoon
@philip_schwarz

A particularly suitable application area of for expressions are
combinatorial puzzles.
An example of such a puzzle is the 8-queens problem: Given a
standard chess-board, place eight queens such that no queen is
in check from any other (a queen can check another piece if they
are on the same column, row, or diagonal).
To find a solution to this problem, it’s actually simpler to generalize it to chess-
boards of arbitrary size.
Hence the problem is to place N queens on a chess-board of N x N squares, where
the size N is arbitrary.
We’ll start numbering cells at one, so the upper-left cell of an N x N board has
coordinate (1,1) and the lower-right cell has coordinate (N,N).
To solve the N-queens problem, note that you need to place a queen in each row.
So you could place queens in successive rows, each time checking that a newly
placed queen is not in check from any other queens that have already been placed.
In the course of this search, it might happen that a queen that needs to be placed in
row k would be in check in all fields of that row from queens in row 1 to k - 1. In
that case, you need to abort that part of the search in order to continue with a
different configuration of queens in columns 1 to k - 1.
Here on the right is a sample
solution to the puzzle.
And below is an example in which there is
nowhere to place queen number 6
because every other cell on the board is in
check from queens 1 to 5.

An imperative solution to this problem would place queens one by one, moving
them around on the board.
But it looks difficult to come up with a scheme that really tries all possibilities.
A more functional approach represents a solution directly, as a value. A solution
consists of a list of coordinates, one for each queen placed on the board.
Note, however, that a full solution can not be found in a single step. It needs to be
built up gradually, by occupying successive rows with queens.
This suggests a recursive algorithm.
Assume you have already generated all solutions of placing k queens on a board
of size N x N, where k is less than N.
Each solution can be represented by a list of length k of coordinates (row,
column), where both row and column numbers range from 1 to N.
It’s convenient to treat these partial solution lists as stacks, where the
coordinates of the queen in row k come first in the list, followed by the
coordinates of the queen in row k - 1, and so on.
The bottom of the stack is the coordinate of the queen placed in the first row of
the board.
All solutions together are represented as a list of lists, with one element for each
solution.
List((8, 6), (7, 3),
(6, 7), (5, 2),
(4, 8), (3, 5),
(2, 1), (1, 4))
See below for the list representing the
above solution to the 8-queens puzzle.
(7, 3)
(8, 6)
(6, 7)
(5, 2)
(4, 8)
(3, 5)
(2, 1)
(1, 4)
R C
O O
W L

Now, to place the next queen in row k + 1, generate all possible extensions of each previous solution
by one more queen.
This yields another list of solutions lists, this time of length k + 1.
Continue the process until you have obtained all solutions of the size of the chess-board N.
This algorithmic idea is embodied in function placeQueens below:
The outer function queens in the program above simply calls placeQueens with the size of the board n
as its argument.
The task of the function application placeQueens(k) is to generate all partial solutions of length k in a
list.
Every element of the list is one solution, represented by a list of length k. So placeQueens returns a list
of lists.
def queens(n: Int): List[List[(Int,Int)]] = {
def placeQueens(k: Int): List[List[(Int,Int)]] =
if (k == 0)
List(List())
else
for {
queens <- placeQueens(k - 1)
column <- 1 to n
queen = (k, column)
if isSafe(queen, queens)
} yield queen :: queens
placeQueens(n)
}
List((8, 6), (7, 3),
(6, 7), (5, 2),
(4, 8), (3, 5),
(2, 1), (1, 4))

If the parameter k to placeQueens is 0, this means that it needs to generate all solutions of
placing zero queens on zero rows. There is only one solution: place no queen at all. This
solution is represented by the empty list. So if k is zero, placeQueens returns List(List()), a
list consisting of a single element that is the empty list.
Note that this is quite different from the empty list List(). If placeQueens returns List(), this
means no solutions, instead of a single solution consisting of no placed queens.
In the other case, where k is not zero, all the work of placeQueens is done in the for
expression.
The first generator of that for expression iterates through all solutions of placing k - 1 queens on
the board.
The second generator iterates through all possible columns on which the kth queen might be
placed.
The third part of the for expression defines the newly considered queen position to be the pair
consisting of row k and each produced column.
The fourth part of the for expression is a filter which checks with isSafe whether the new
queen is safe from check by all previous queens (the definition of isSafe will be discussed a bit
later.)
If the new queen is not in check from any other queens, it can form part of a partial solution, so
placeQueens generates with queen::queens a new solution.
If the new queen is not safe from check, the filter returns false, so no solution is generated.
for {
column <- 1 to n
queen = (k, column)
Here is the for expression
again, for reference.

The only remaining bit is the isSafe method, which is used to check whether a given queen is in check from any other element
in a list of queens.
Here is the definition:
The isSafe method expresses that a queen is safe with respect to some other queens if it is not in check from any other queen.
The inCheck method expresses that queens q1 and q2 are mutually in check.
It returns true in one of three cases:
1. If the two queens have the same row coordinate.
2. If the two queens have the same column coordinate.
3. If the two queens are on the same diagonal (i.e., the difference between their rows and the difference between their
columns are the same).
The first case – that the two queens have the same row coordinate – cannot happen in the application because placeQueens
already takes care to place each queen in a different row. So you could remove the test without changing the functionality of the
program.
def isSafe(queen: (Int, Int), queens: List[(Int, Int)]) =
queens forall (q => !inCheck(queen, q))
def inCheck(q1: (Int, Int), q2: (Int, Int)) =
q1._1 == q2._1 || // same row
q1._2 == q2._2 || // same column
(q1._1 - q2._1).abs == (q1._2 - q2._2).abs // on diagonal

(6, 2)
(7, 5)
(5, 6)
(4, 1)
(3, 7)
(2, 4)
(1, 0)
(0, 3) In online course FP in Scala, the board
coordinates are zero-based, and since
it is obvious that the nth queen is in row
n, a solution is a list of column indices
rather than a list of coordinate pairs.
List(5, 2, 6, 1, 7, 4, 0, 3) @philip_schwarz

Because a column index is sufficient to indicate the coordinates of the nth queen, we can see that the type of isSafe’s first parameter is now Int, i.e. a
column index, rather than (Int, Int) i.e. a coordinate pair.
Also, the type of a solution is List[Int] rather than List[(Int, Int)], and solutions are returned in a Set, rather than a List, to distinguish between a
single solution, which is an ordered list, and the collection of all solutions, which is not ordered (though it could be, if we so wanted).

def queens(n: Int): List[List[(Int,Int)]] = {
if (k == 0)
List(List())
else
for {
column <- 1 to n
queen = (k, column)
placeQueens(n)
}
def queens(n: Int): Set[List[Int]] = {
def placeQueens(k: Int): Set[List[Int]] =
if (k == 0)
Set(List())
else
for {
col <- 0 until n
if isSafe(col, queens)
} yield col :: queens
placeQueens(n)
}
def inCheck(q1: (Int, Int), q2: (Int, Int)) =
q1._1 == q2._1 || // same row
q1._2 == q2._2 || // same column
(q1._1 - q2._1).abs == (q1._2 - q2._2).abs // on diagonal
def isSafe(col: Int, queens: List[Int]): Boolean = {
val row = queens.length
val queensWithRow = (row - 1 to 0 by -1) zip queens
queensWithRow forall
{ case (r, c) => col != c && math.abs(col - c) != row - r }
}
Here are the two versions of
the program side by side.

def queens(n: Int): List[List[(Int,Int)]] =
if k == 0
then List(List())
else
for
column <- 1 to n
queen = (k, column)
yield queen :: queens
placeQueens(n)
def inCheck(queen1: (Int, Int), queen2: (Int, Int)) =
queen1(0) == queen2(0) || // same row
queen1(1) == queen2(1) || // same column
(queen1(0) - queen2(0)).abs == (queen1(1) - queen2(1)).abs // on diagonal
def queens(n: Int): Set[List[Int]] =
def placeQueens(k: Int): Set[List[Int]] =
if k == 0
then Set(List())
else
for
col <- 0 until n
if isSafe(col, queens)
yield col :: queens
placeQueens(n)
def isSafe(col: Int, queens: List[Int]): Boolean =
val row = queens.length
val queensWithRow = (row - 1 to 0 by -1) zip queens
queensWithRow forall
{ case (r, c) => col != c && math.abs(col - c) != row - r }
Same as on the previous slide, but with
really minor tweaks thanks to Scala 3

def queens(n: Int): List[List[Int]] =
def placeQueens(k: Int): List[List[Int]] =
if k == 0
then List(List())
else
for
queen <- 1 to n
if safe(queen, queens)
placeQueens(n)
def onDiagonal(row: Int, column: Int, otherRow: Int, otherColumn: Int) =
math.abs(row - otherRow) == math.abs(column - otherColumn)
def zipWithRows(queens: List[Int]): Iterable[(Int,Int)] =
val rowCount = queens.length
val rowNumbers = rowCount - 1 to 0 by -1
rowNumbers zip queens
def safe(queen: Int, queens: List[Int]): Boolean =
val (row, column) = (queens.length, queen)
val safe: ((Int,Int)) => Boolean = (nextRow, nextColumn) =>
column != nextColumn && !onDiagonal(column, row, nextColumn, nextRow)
zipWithRows(queens) forall safe
In the next part of this slide deck, we are going to
be using the program on the right.
The queens and placeQueens functions are pretty
much the same as in the second version on the
previous slide, but board coordinates are one-
based, and the two functions return a List rather
than a Set.
The logic for determining if a queen is safe from
previously placed queens, is organised in a way that
is in part a hybrid between the two programs on
the previous slide.
(7, 3)
(8, 6)
(6, 7)
(5, 2)
(4, 8)
(3, 5)
(2, 1)
(1, 4)
List(6, 3, 7, 2, 8, 5, 1, 4)
assert(queens(4) == List(List(3, 1, 4, 2), List(2, 4, 1, 3)))
assert(queens(6) == List(List(3, 6, 2, 5, 1, 4),
List(4, 1, 5, 2, 6, 3),
List(5, 3, 1, 6, 4, 2),
List(2, 4, 6, 1, 3, 5)))
@philip_schwarz

The first thing we are going to do, is translate our N-queens
program from Scala into Haskell.
Let’s start with the queens and placeQueens functions.
queens n = placeQueens n
where
placeQueens 0 = [[]]
placeQueens k =
do
queens <- placeQueens(k-1)
queen <- [1..n]
guard (safe queen queens)
return (queen:queens)
if k == 0
then List(List())
else
for
queen <- 1 to n
placeQueens(n)
In Scala we use a for comprehension,
whereas in Haskell we use a do expression.
In Scala, the filtering is done by
whereas in Haskell it is done by
If you are asking yourself what the guard function
is, don’t worry: we’ll be looking at it soon.

where
placeQueens k = do
queen <- [1..n]
where
placeQueens k =
[queen:queens |
queens <- placeQueens(k-1),
queen <- [1..n],
safe queen queens]
We can simplify the Haskell version a little bit by eliminating
the calls to guard and return, which can be achieved by
switching from a do expression to a list comprehension.
If you are asking yourself how the list
comprehension eliminates the need for guard and
return, don’t worry: we’ll be looking at that soon.

where
placeQueens k = [queen:queens |
queen <- [1..n],
safe queen queens]
safe queen queens = all safe (zipWithRows queens)
where
safe (r,c) = c /= col && not (onDiagonal col row c r)
row = length queens
col = queen
onDiagonal row column otherRow otherColumn =
abs (row - otherRow) == abs (column - otherColumn)
zipWithRows queens = zip rowNumbers queens
where
rowCount = length queens
rowNumbers = [rowCount-1,rowCount-2..0]
if k == 0
then List(List())
else
for
queen <- 1 to n
placeQueens(n)
And here is the translation
of the whole program

haskell> queens 4
[[3,1,4,2],[2,4,1,3]]
haskell> queens 5
[[4,2,5,3,1],[3,5,2,4,1],[5,3,1,4,2],[4,1,3,5,2],
[5,2,4,1,3],[1,4,2,5,3],[2,5,3,1,4],[1,3,5,2,4],
[3,1,4,2,5],[2,4,1,3,5]]
haskell> queens 6
[[5,3,1,6,4,2],[4,1,5,2,6,3],
[3,6,2,5,1,4],[2,4,6,1,3,5]]
scala> queens(4)
val res0: List[List[Int]] = List(List(3, 1, 4, 2), List(2, 4, 1, 3))
scala> queens(5)
val res1: List[List[Int]] = List(List(4, 2, 5, 3, 1), List(3, 5, 2, 4, 1), List(5, 3, 1, 4, 2), List(4, 1, 3, 5, 2),
List(5, 2, 4, 1, 3), List(1, 4, 2, 5, 3), List(2, 5, 3, 1, 4), List(1, 3, 5, 2, 4),
List(3, 1, 4, 2, 5), List(2, 4, 1, 3, 5))
scala> queens(6)
val res2: List[List[Int]] = List(List(5, 3, 1, 6, 4, 2), List(4, 1, 5, 2, 6, 3),
List(3, 6, 2, 5, 1, 4), List(2, 4, 6, 1, 3, 5))
Let’s try out both the Scala
program and the Haskell one.
@philip_schwarz

N N-queens solution count
1 1
2 0
3 0
4 2
5 10
6 4
7 40
8 92
9 352
10 724
11 2,680
12 14,200
13 73,712
14 365,596
15 2,279,184
16 14,772,512
17 95,815,104
18 666,090,624
19 4,968,057,848
20 39,029,188,884
21 314,666,222,712
22 2,691,008,701,644
23 24,233,937,684,440
24 227,514,171,973,736
25 2,207,893,435,808,352
26 22,317,699,616,364,044
27 234,907,967,154,122,528
data source: https://en.wikipedia.org/wiki/Eight_queens_puzzle
scala> queens(8).size
val res0: Int = 92
val res1: Int = 352
val res2: Int = 724
val res3: Int = 2680
Haskell> length (queens 8)
92
352
724
2680
14200
73712
365596
For N > 6, the number of solutions is too large for the solutions to
be displayed simply by printing them to the screen.
Let’s at least check that we get the expected number of solutions.

In the next five slides, we are going to look at how the Scala
for comprehension is implemented under the hood.
@philip_schwarz

for
queen <- 1 to n
placeQueens(k - 1) flatMap { queens =>
(1 to n) map { queen =>
queen :: queens
}
}
for
nums <- List(List(1,2),List(3,4))
num <- List(10,20)
yield num :: nums
List(List(10, 1, 2), List(20, 1, 2), List(10, 3, 4), List(20, 3, 4))
List(List(1,2),List(3,4)) flatMap { nums =>
List(10,20) map { num =>
num :: nums
}
}
Here is how the N-queens for comprehension, without
the filter, is implemented, i.e. what it looks like when it
is desugared (when its syntactic sugar is removed).
To aid our understanding, here is a similar for
comprehension that is simpler to understand, but can
be run at the REPL, together with its desugared version.
And here is the value produced by both the for
comprehension and its desugared equivalent.

for
queen <- 1 to n
(1 to n) withFilter { queen =>
safe(queen, queens)
} map { queen =>
queen :: queens
}
}
Same as on the previous slide, except that now we are adding a filter to the for comprehension.
When the filter is desugared, it results in a call to List’s withFilter function.
for
num <- List(10,20)
if (num > 15)
yield num :: nums
List(10,20) withFilter { num =>
num > 15
} map { num =>
num :: nums
}
}
To aid our understanding, here is a similar for
comprehension that is simpler to understand, and that can
be run at the REPL, together with its desugared version.
List(List(20, 1, 2), List(20, 3, 4))
And here is the value produced by both the for comprehension and its desugared equivalent.

Let’s see what the withFilter function does:
• it is applied to a monad, in this case a List
• it takes a predicate, i.e. a function that takes a parameter and returns either true or false.
• it applies the predicate to each value yielded by the monad (in this case, each value contained in a List)
• it filters out the values for which the predicate returns false
• It lets through the values for which the predicate returns true
On the next slide, we look at an example of how the behaviour of a for comprehension (or its desugared equivalent) is
affected by the behaviour of the withFilter function.
@philip_schwarz

for
num <- List(10,20)
if (num > 15)
yield num :: nums
num > 15
} map { num =>
num :: nums
}
}
List(List(20, 1, 2), List(20, 3, 4))
for
num <- List(10,20)
if (true)
yield num :: nums
true
} map { num =>
num :: nums
}
}
for
num <- List(10,20)
if (false)
yield num :: nums
false
} map { num =>
num :: nums
}
}
List()
Notice that the predicate function passed to withFilter may or may not
use its parameter. Like the map and flatMap functions, withFilter is
used to bind the value(s) yielded by a monad (if any) to a variable name
which is then available for use in subsequent computations.
withFilter
lets through
some elements
withFilter
lets through
all elements
withFilter
lets through
no elements
The difference is that while the functions passed to map
and flatMap are almost certain to use the variable, in the
case of withFilter, it may choose to influence the
overall computation without actually using the variable.

scala> List(1,2,3).withFilter(n => true).map(identity)
val res0: List[Int] = List(1, 2, 3)
scala> List(1,2,3).withFilter(n => n > 1).map(identity)
val res1: List[Int] = List(2, 3)
scala> List(1,2,3).withFilter(n => false).map(identity)
val res2: List[Int] = List()
Before we move on, here is a final, illustration
of what the withFilter function does.

Earlier we saw that the do expression that is the Haskell translation of the Scala for
comprehension, involves the use of the guard function.
Also, later we intend to look at how the Haskell do expression is implemented under the hood,
which involves both the guard function and the (>>) function.
So, in the next three slides, we are going to look at how Miran Lipovača explains guard and (>>).

Here is the default implementation of the (>>) function:
(>>) :: (Monad m) => ma -> mb -> mb
m >> n = m >>= _ -> n
Normally, passing some value to a function that ignores its parameter and always
returns some predetermined value always results in that predetermined value. With
monads however, their context and meaning must be considered as well.
Here is how >> acts with Maybe:
ghci> Nothing >> Just 3
Nothing
ghci> Just 3 >> Just 4
Just 4
ghci> Just 3 >> Nothing
Nothing
Miran Lipovača
ghci> [] >> [3]
[]
ghci> [3] >> [4]
[4]
ghci> [3] >> []
[]
Here on the
right is how >>
acts with lists.
>>= is Haskell’s equivalent of Scala’s
flatMap and _ -> n is Haskell’s
equivalent of Scala’s anonymous
function _ => n.
scala> List() >> List(3)
val res0: List[Int] = List()
scala> List(3) >> List(4)
val res1: List[Int] = List(4)
scala> List(3) >> List()
val res2: List[Nothing] = List()
scala> List() >> List()
val res3: List[Nothing] = List()
scala> None >> 3.some
val res0: Option[Int] = None
scala> 3.some >> 4.some
val res1: Option[Int] = Some(4)
scala> 3.some >> None
val res2: Option[Nothing] = None
scala> None >> None
val res3: Option[Nothing] = None
Same as above, but using
the Scala Cats library:
import cats._
import cats.implicits._

MonadPlus and the guard Function
List comprehensions allow us to filter our output. For instance, we can filter a list of numbers to search only for numbers
whose digit contains a 7:
ghci> [x | x <- [1..50], '7' `elem` show x]
[7,17,27,37,47]
We apply show to x to turn our number into a string, and then we check if the character ‘7’ is part of that string. To see how
filtering in list comprehensions translates to the list monad, we need to check out the guard function and the MonadPlus type
class. The MonadPlus type class is for monads that can also act as monoids. Here is its definition:
class Monad m => MonadPlus m where
mzero :: ma
mplus :: ma -> ma -> ma
mzero is synonymous with mempty from the Monoid type class, and mplus corresponds to mappend. Because lists are monoids
as well as monads, they can be made an instance of this type class:
instance MonadPlus [] where
mzero = []
mplus = (++)
For lists, mzero represents a nondeterministic computation that has no results at all – a failed computation. mplus joins two
nondeterministic values into one. The guard function is defined like this:
guard :: Bool -> m ()
guard True = return ()
guard False = mzero
guard takes a Boolean value. If that value is True, guard takes a () and puts it in a minimal default context that succeeds. If
the Boolean value is False, guard makes a failed monadic value.
Miran Lipovača

Here it is in action:
ghci> guard (5 > 2) :: Maybe ()
Just ()
ghci> guard (1 > 2) :: Maybe ()
Nothing
ghci> guard (5 > 2) :: [()]
[()]
ghci> guard (1 > 2) :: [()]
[]
This looks interesting, but how is it useful? In the list monad, we use it to filter out nondeterministic computations:
ghci> [1..50] >>= (x -> guard ('7' `elem` show x) >> return x)
[7,17,27,37,47]
The result here is the same as the result of our previous list comprehension. How does guard achieve this? Let’s first see
how guard functions in conjunction with >>:
ghci> guard (5 > 2) >> return "cool" :: [String]
["cool"]
ghci> guard (1 > 2) >> return "cool" :: [String]
[]
If guard succeeds, the result contained within it is the empty tuple. So then we use >> to ignore the empty tuple and present
something else as the result. However, if guard fails, then so will the return later on, because feeding an empty list to a
function with >>= always results in an empty list. guard basically says, “If this Boolean is False, then produce a failure right
here. Otherwise, make a successful value that has a dummy result of () inside it.” All this does is to allow the computation to
continue.
Miran Lipovača
ghci> [x | x <- [1..50], '7' `elem` show x]
[7,17,27,37,47]
same result

for
queen <- 1 to n
(1 to n) withFilter { queen =>
safe(queen, queens)
} map { queen =>
queen :: queens
}
}
do
queen <- [1..n]
placeQueens(k-1) >>= queens ->
[1..n] >>= queen ->
guard (safe queen queens) >>
Here we see the desugaring of the Scala for comprehension
again, but we also desugar the equivalent Haskell do expression.
Note how, while the desugared Scala code introduces the use of flatMap, withFilter and map, the desugared Haskell
code introduces the use of >>= (Haskell’s equivalent of flatMap) and >>.
Note also how, in Haskell, the guard function is used in both the do expression and in its desugared equivalent.

for
num <- List(10,20)
if (num > 15)
yield num :: nums
num > 15
} map { num =>
num :: nums
}
}
do
nums <- [[1,2],[3,4]]
num <- [10,20]
guard (num > 15)
return (num:nums)
[[1,2],[3,4]] >>= nums ->
[10,20] >>= num ->
guard (num > 15) >>
return (num:nums)
[[20,1,2],[20,3,4]]
List(List(20, 1, 2), List(20, 3, 4))
Same as on the previous slide, except that here the logic
is simpler to understand and can be run at the REPL.
@philip_schwarz

do
nums <- [[1,2],[3,4]]
num <- [10,20]
guard (num > 15)
return (num:nums)
[[1,2],[3,4]] >>= nums ->
[10,20] >>= num ->
guard (num > 15) >>
return (num:nums)
List(List(20, 1, 2), List(20, 3, 4))
do
nums <- [[1,2],[3,4]]
num <- [10,20]
guard (True)
return (num:nums)
[[1,2],[3,4]] >>= nums ->
[10,20] >>= num ->
guard (True) >>
return (num:nums)
do
nums <- [[1,2],[3,4]]
num <- [10,20]
guard (False)
return (num:nums)
[[1,2],[3,4]] >>= nums ->
[10,20] >>= num ->
guard (False) >>
return (num:nums)
List()
guard lets
through some
elements
guard lets
through all
elements
guard lets
through no
elements

ghci> [1,2] >>= n -> ['a',’b’] >>= ch -> return (n,ch)
[(1,'a'),(1,'b'),(2,'a'),(2,'b’)]
…
…
Here is the previous expression rewritten in do notation:
listOfTuples :: [(Int,Char)]
listOfTuples do
n <- [1,2]
ch <- [’a’,’b’]
return (n,ch)
Do Notation and List Comprehensions
Using lists with do notation might remind you of something you’ve seen before. For instance, check out the following piece
of code:
ghci> [(n, ch) | n <- [1,2], ch <- ['a','b’]]
[(1,'a'),(1,'b'),(2,'a'),(2,'b’)]
Yes, list comprehensions! In our do notation example, n became every result from [1,2]. For every such result, ch was
assigned a result from ['a','b’], and then the final line put (n,ch) into a default context (a singleton list) to present it as the
result without introducing any additional nondeterminism.
In this list comprehension, the same thing happened, but we didn’t need to write return at the end to present (n,ch) as the
result, because the output part of a list comprehension did that for us.
In fact, list comprehensions are just syntactic sugar for using lists as monads. In the end, list comprehensions and lists in
do notation translate to using >>= to do computations that feature nondeterminism. Miran Lipovača

do queens <- placeQueens(k-1)
queen <- [1..n]
[queen:queens |
queen <- [1..n],
safe queen queens]
placeQueens(k-1) >>= queens ->
[1..n] >>= queen ->
guard (safe queen queens) >>
We saw earlier what the Haskell do expression looks like when it is desugared.
Desugaring the list comprehension produces the same result.

where
placeQueens k = [queen:queens |
queen <- [1..n],
safe queen queens]
safe queen queens = all safe (zipWithRows queens)
where
safe (r,c) = c /= col && not (onDiagonal col row c r)
row = length queens
col = queen
onDiagonal row column otherRow otherColumn =
abs (row - otherRow) == abs (column - otherColumn)
zipWithRows queens = zip rowNumbers queens
where
rowCount = length queens
rowNumbers = [rowCount-1,rowCount-2..0]
if k == 0
then List(List())
else
for
queen <- 1 to n
placeQueens(n)
As a recap, let’s see again the translation of
the whole program from Scala to Haskell.

There is lots to do in part two. We’ll kick off by taking
the Scala version of the program, and extending it so
that it can display a single solution board as follows:
To conclude the first part of this slide deck,
see below for a simple way of displaying
the solutions to the N-queens problem.
We’ll then get the program to display,
all together, the results of queens(N)
for N = 4, 5, 6, 7, 8. Plus much more.
See you then.
@philip_schwarz

N-Queens Combinatorial Problem - Polyglot FP for Fun and Profit – Haskell and Scala - Part 1

More Related Content

What's hot (20)

Similar to N-Queens Combinatorial Problem - Polyglot FP for Fun and Profit – Haskell and Scala - Part 1 (20)

More from Philip Schwarz (20)

Recently uploaded (20)

N-Queens Combinatorial Problem - Polyglot FP for Fun and Profit – Haskell and Scala - Part 1