Posted on September 12, 2020

Last week, Eric came to me with a fun programming challenge: the knight’s tour around the keypad. This by now a well-known problem. I had actually been given it once upon a time in undergrad when interviewing for an internship. Alas back then, I couldn’t solve it, and I didn’t get the internship. In this blog post, I want to set the record straight, and show off how to solve it using Haskell.

## The knight’s tour

The problem is this: if we have a phone keypad such as

1 2 3
4 5 6
7 8 9
0

and a knight chess piece positioned on a given number, then how many phone numbers can the knight dial in N moves?

The knight can only move in a chess-legal way. For example, if the knight is on 1, then it can move only to 8 and 6.

### A first approach

To get a feel for the solution, let’s imagine that we have 4 units of fuel to move with and that the knight is on the 1 key. If we know how many numbers could be dialed with 3 units of fuel starting on 1’s neighbours, 6 and 8, then we could just add those together to obtain the count of numbers that can be dialed in 4 units starting on 1. And what happens when we have no fuel left? We can dial exactly one number – the empty number.

The above suggests a straightforward recursive solution. The only difficulty is in determining the neighbours for a given key. Luckily we can write a very simple function to give us just that information.

module KnightTour where

neighbours :: Int -> [Int]
neighbours n = case n of
0 -> [4, 6]
1 -> [6, 8]
2 -> [7, 9]
3 -> [4, 8]
4 -> [3, 9, 0]
5 -> []
6 -> [1, 7, 0]
7 -> [2, 6]
8 -> [1, 3]
9 -> [4, 2]

Now we can write the recursive solution.

countPaths1 :: Int -> Int -> Int
countPaths1 0 _ = 1
countPaths1 fuel key = sum $map (countPaths1$ fuel - 1) (neighbours key)

The problem with this solution is that it runs in exponential time: except in the base case, we make at least 2 recursive calls, so if we draw a tree of all the recursive calls, it will have at least 2^n nodes, where n is the amount of fuel.

### Dynamic programming

We can significantly improve this algorithm by observing that there is a lot of redundant calculation made in it. For example, let’s say we’re on the 8 key with n units of fuel. We need to make the following calculations.

countPaths1 n 8 = countPaths1 (n-1) 1 + countPaths1 (n-1) 3
| countPaths1 (n-1) 1 = countPaths (n-2) 6 + countPaths1 (n-2) 8
| | countPaths1 (n-2) 6 = countPaths1 (n-3) 0 + countPaths1 (n-3) 1 + countPaths (n-3) 7
| | countPaths1 (n-2) 8 = ...
| countPaths1 (n-1) 3 = countPaths (n-2) 4 + countPaths1 (n-2) 8
| | countPaths1 (n-2) 4 = countPaths1 (n-3) 0 + countPaths1 (n-3) 3 + countPaths (n-3) 9
| | countPaths1 (n-2) 8 = ...

We can see that the calculations for countPaths1 (n-2) 8 and countPaths1 (n-3) 0 both occur twice.

To avoid recalculating anything, we can turn our implementation around. Instead of breaking apart the problem into subproblems to be solved recursively, we can build up the solution starting from the base case. This technique is called dynamic programming.

To use dynamic programming, we tend to need to generalize the problem: rather than just consider the count of numbers that can be dialed starting on one key, we compute the count of numbers that can be dialed starting on each key. Let’s call this collection of counts a stage of the algorithm.

To see that this speeds up the algorithm and eliminates the redundancy from the previous solution, let’s suppose that we know the Nth stage. That is, we have some list L such that indexing it as L[K] gives the answer to the question “how many numbers can be dialed starting on key K with N units of fuel?” (Note N is fixed here.) Then, we can easily compute the N+1th stage L' such that for each K', L'[K'] is precalculated as the sum of L[K] for each neighbour K of K'. The initial stage L0 sends every key to 1.

Now let’s encode this in Haskell.

type Stage = [Int]

stage0 :: Stage
stage0 = map (const 1) [0..9]

next :: Stage -> Stage
next l = map (sum . map (l !!) . neighbours) [0..9]

Now if we want to know how many numbers can be dialed starting on K with N units of fuel, then we let next^N be next composed with itself N times and calculate next^N stage0 !! K.

We can generalize a touch further using the following function from the standard library.

iterate :: (a -> a) -> a -> [a]
iterate f x = x : iterate f (f x)

This function constructs an infinite list such that if L = iterate f x, then L[N] = f^N x where f^N is f composed with itself N times. In other words, the list looks like [x, f x, (f . f) x, (f . f . f) x, ...].

This allows us to define the following.

allStages :: [Stage]
allStages = iterate next stage0

And we can answer the original problem – how many numbers can be dialed starting on K with N units of fuel? – by indexing twice.

countPaths2 :: Int -> Int -> Int
countPaths2 fuel key = allStages !! fuel !! key

Now this algorithm runs in linear time in the amount of fuel N.

To recap, we converted an ordinary recursive algorithm to use dynamic programming by turning it around. Instead of breaking down the problem into subproblems, We generalized the problem to consider stages, and built up to the stage we needed. To do this, we identified an initial stage and defined a function that given one stage gives us the next stage.

## Eight Queens

Now that we’ve adequately warmed up, we can tackle another famous dynamic programming problem: the eight queens problem. In this challenge, we want to place 8 queens on a chessboard so that no two queens could attack each other.

This time around, we won’t bother with writing an ordinary recursive function first and converting it. In fact, it’s not always as illuminating as in the previous problem to do so. (The reason it’s less helpful this time is that the ordinary recursive solution is a backtracking search, whereas the dynamic programming solution is more direct.)

Instead, we’ll figure out what a stage is in this algorithm, what the initial stage looks like, and how to calculate the next stage from the current stage.

Since we will need to decide whether tiles are in danger of attack by a queen, we will need to reason about and manipulate boards, so let’s encode them.

module EightQueens where

import Data.Tuple ( swap )
import Data.List ( sort, uniq )

type Tile = Queen | Safe | Danger
deriving (Eq, Ord, Show)

type Board = [Tile]

boardWidth, boardLength :: Int
boardWidth = 8
boardLength = boardWidth^2

indexToVec :: Int -> (Int, Int)
indexToVec = swap . (divMod boardWidth)

A board is just a list of tiles, and a tile either contains a queen, is safe for a queen, or is dangerous for a queen. And what is a stage in this algorithm? Let’s imagine we want to put down the eighth queen. If only we knew all the boards with seven queens on them, then we could find all the safe positions in those boards and put a queen on each. So a stage N in this algorithm is the list of all boards with N queens.

The initial stage (stage 0) is simply a list containing only the empty board.

type Stage = [Board]

stage0 :: Stage
stage0 = [emptyBoard]

emptyBoard :: Board
emptyBoard = map (const Safe) [1..boardLength]

Next, we need to think about how to get the next stage from the current stage: for each board B in the current stage, generate a new board B' for each safe position P in B, such that B'[P] = Queen and B'[P'] = Danger for all positions P' in danger of position P.

This suggests a small piece of functionality that we need – we need to be able to place a queen on a board, putting Danger on all the tiles that queen can reach. And to implement this small piece of functionality, we need the ability to traverse a board and construct a new one. This traversal needs to be aware of the coordinates of the tiles being processed, because whether a tile is in danger depends on the relationship of its coordinates to those of the queen being placed.

imapBoard :: ((Int, Int) -> Tile -> a) -> Board -> [a]
imapBoard f = zipWith f indices where
indices = map indexToVec [0..]

placeQueen :: (Int, Int) -> Board -> Board
placeQueen pos = imapBoard f where
f pos' tile
| pos' == pos = Queen
| pos' inDangerOf pos = Danger
| otherwise = tile

All that’s missing from the above snippet is inDangerOf. Two positions are in danger of each other if they have the same x coordinate (vertical attack), the same y coordinate (horizontal attack), or their absolute x coordinate difference equals their absolute y coordinate difference (diagonal attack).

inDangerOf :: (Int, Int) -> (Int, Int) -> Bool
(x, y) inDangerOf (qx, qy) = x == qx || y == qy || (qx - x)^2 == (qy - y)^2

Now we can loop back to the game plan: place a queen on every Safe tile in a board to generate new boards.

placeQueens :: Board -> [Board]
placeQueens b = map (\pos -> placeQueen pos b) $empties b empties :: Board -> [(Int, Int)] empties = map fst . filter (\(_, tile) -> tile == Safe) . imapBoard (,) Now we apply placeQueens to every board in the stage, concatenating the results to form the next stage. next :: Stage -> Stage next = concatMap placeQueens There’s one small problem with this implementation: the same board can be generated in different ways, so we end up with duplicate boards in our next stage. To see this, imagine that in stage 2 we have the boards B1 and B2 and the positions A, B and C such that: • B1[A] = Queen and B2[A] = Empty • B1[B] = Empty and B2[B] = Queen • B1[C] = Queen and B2[C] = Queen B1 will give rise to a new board B1' in the next stage with B1'[B] = Queen and B2 will give rise to a new board B2' in the next stage with B2'[A] = Queen. The upshot is that after calling next stage2 to obtain stage 3, we have two identical boards in that new stage! The right thing to do is to avoid generating the duplicate boards in the first place, which is left as an exercise to the reader. Instead, we’ll just remove the duplicate boards after generating the next stage. next :: Stage -> Stage next = uniq . sort . concatMap placeQueens As before, we can get the answer by iterating next and indexing. answer :: Board answer = head$ iterate next stage0 !! 8

And maybe we want to see what this board actually looks like.

main :: IO ()

drawBoard :: Board -> String
drawBoard = concatMap drawRow . groupBySize boardWidth where
drawRow = (++ "\n") . concatMap drawTile where
drawTile t = case t of
Queen -> "Q "
Danger -> "_ "
Safe -> "_ "

groupBySize :: Int -> [a] -> [[a]]
groupBySize _ [] = []
groupBySize n l = xs : groupBySize n xss where
(xs, xss) = splitAt n l

To recap, we thought about what a stage is and figured out what the initial stage is. Then, we implemented a function to get the next stage given the current stage. Both examples discussed in this article were the same in that sense: we only needed the immediately previous stage. In general, we might need access to several previous stages. Or, as we saw in the Eight Queens problem, we might need to eliminate redundancy within the calculation of the next stage. Regardless, dynamic programming is a powerful technique that lends itself surprisingly well to implementation in functional languages, despite commonly being taught using hashtables and memoization, both of which are more suited to imperative implementations.