Solving Boggle
1 Setup
2 Racket checker
3 Testing
4 Do it again!
5 And a third time
6 Outro

Solving Boggle

D. Ben Knoble

Having recently read Jay McCarthy’s A Boggle Solver, and remembering my brother’s similar C assignment for an NC State course, I decided to give it a go.

The part I really remember from working with my brother was the flood-fill "is the input a word in the grid" algorithm. This relates to McCarthy’s solver, which generates words with a similar flood-fill and tests for their presence in a trie-like dictionary.

1 Setup

We’ll use McCarthy’s board representation, which is a hash mapping (x, y) coordinates to letters.

Aside: I would like to find a way to express this without for*/fold, but I have a feeling it would still involve computing the cartesian product.

(define board-n 4)
(define board
  (for*/fold ([cell->char (hash)])
    ([row (in-range board-n)]
     [col (in-range board-n)])
    (hash-set cell->char (cons row col) (random-letter))))

We’ll also keep the printer,

(for ([row (in-range board-n)])
  (for ([col (in-range board-n)])
    (display (hash-ref board (cons row col))))

which can be re-written with for-each.

(for-each (λ (row)
            (for-each (λ (col)
                        (display (hash-ref board (cons row col))))
                      (range board-n))
          (range board-n))

2 Racket checker

Now we need to implement our flood-fill for checking an input. We have to check every possible starting place, and we cannot re-use blocks.

(define (check word board)
  (ormap (curry check-start word board)
         (hash-keys board)))

Given a word, a board, and a starting location, how do we know if the word is there? First, we cannot re-use a block. We handle that with a slight trick: hash-ref takes a default argument, which we can use to build a test that will always fail when the coordinate is not in the board. If we (functionally) remove coordinates from the board before recurring, this will do it.

Then, we have to have the same letter at the starting location and the start of the word. The default argument to hash-ref also takes care of the out-of-bounds cases.

(eq? (hash-ref board c #false) (string-ref word 0))

Then, and only then, can we recur on the other possible directions with a shortened word. Note that we do re-visit the same location, but that should immediately bail out as false because it will already have been seen (<step-starting-letter> takes care of it).

(let ([new-board (hash-remove board c)])
  (for*/or ([dx (in-list '(-1 0 1))]
            [dy (in-list '(-1 0 1))])
    (check-start (substring word 1)
                 (cons (+ (x c) dx)
                       (+ (y c) dy)))))

But wait! If we’re shortening the word, we need to check for an empty case to bail out: that’s how we know we found the word along a path.

(zero? (string-length word))

Putting it all together gives us the following, which embeds the rules that

(define (check-start word board c)
  (or <step-found-it>
      (and <step-starting-letter>

Altogether, aside from the gnarly C details, this is not unlike the solution my brother and I came up with. Of course, the for*/or macro hides the insane code-duplication that our C program exhibited, and for that I am grateful.

3 Testing

Right, now we’d like some tests. We want to make sure our checker works on every word in the board (up to some maximum length), and on no words in the board. With the randomization we’re doing, and the lack of a dictionary, these "words" won’t necessarily be real words. But they will be (in)valid paths in the Boggle board.

The general structure will be something like this.

(for ([word (find-words board maximum-word-length)])
  (check-in-board word board))

And check-in-board is straightforward.

(define (check-in-board word board)
  (check-true (check word board)))

Similarly, we’ll have

(define (check-not-in-board word board)
  (check-false (check word board)))

Now, how can we program find-words? We’re going to use the same idea, though this time by generating a sequence, since we don’t need to materialize the entire list at once. find-words will find all the words on a board up to length maximum-word-length and below length minimum-word-length, since Boggle words are at least 3 letters. It does so by chaining the sequences of words found from each starting position.

(define (find-words board maximum-word-length)
  ; chain the list of sequences
  (apply in-sequences
         (map (curry find-words-at board maximum-word-length)
              (hash-keys board))))

To find the words from each starting position, we’ll lean on McCarthy’s code some more, though instead of printing words from a dictionary we’ll yield all of them. We’ll use in-generator to produce a sequence directly, which we then chain with in-sequences above.

(define (find-words-at board maximum-word-length c)
    (let loop ([board board]
               [path empty]
               [c c])
      (define char (hash-ref board c #f))
      (when char

A quick test on my machine says it takes about 1.5s to find all the words in a 4x4 grid of length between 3 and 8. Not slow, but not as fast as I would like. (Without printing, it’s about half a second.) Following McCarthy’s approach, we first get the new path.

(define new-path (cons char path))

Then, we yield the word if we have one.

(when (<= minimum-word-length (length new-path) maximum-word-length)
  (yield (list->string (reverse new-path))))

Lastly, unless the board is empty or we have reached our maximum path length, we loop over all the new possibilities.

(unless (or (zero? (hash-count board))
            (>= (length new-path) maximum-word-length))
  (define new-board (hash-remove board c))
  (for* ([dx (in-list '(-1 0 1))]
         [dy (in-list '(-1 0 1))])
    (loop new-board new-path
          (cons (+ (x c) dx)
                (+ (y c) dy)))))

Testing all those words does take some time! Roughly 6s, for me. But there is a lot of duplicated work, since some words are the same as others in terms of which paths they take. Also, the work is embarrassingly parallel (i.e., sub-tasks are independent), so parallelization might make a difference here.

At this point, it would be interesting to try to generate words not in the board to make sure the implementation of check isn’t just

(define check (const #t))

But I think this is enough for now.

4 Do it again!

Now that I have that working in Racket, I want to see if I can make a Prolog version work. Why? Because the same relation can determine if a word is on the board or produce a set of possible words: bidirectional Prolog relations capture the backtracking in our solution (from recursion) as well as handling the duality of the checker and the generator. Let’s see it in SWI Prolog first, then we’ll try building it in Racket.

Some example queries, showing the timing as well as the bidirectional use:


% load the program

?- ["boggle"].



% make a 3x3 board

?- boggle:make_board([[a,b,c], [d,e,f], [g,h,i]], B).

% I won't show the large board structure defined by the association lists.

% The key is that the mapping is correct.

B = … .


% make a 3x3 board and check the element at (1,2), which is f

?- boggle:make_board([[a,b,c], [d,e,f], [g,h,i]], B), get_assoc(1-2, B, V).

B = … , V = f.


% check a particular word "adhf"

?- boggle:make_board([[a,b,c], [d,e,f], [g,h,i]], B), get_assoc(1-2, B, V),

    boggle:board_has_word(B, [a,d,h,f]).


% find all the words in a 3x3

?- boggle:make_board([[a,b,c], [d,e,f], [g,h,i]], B),

    setof(W, N^(boggle:board_has_word(B, W),length(W, N), N #>= 3), S).

B = … ,

S = [[a, b, c], [a, b, c, e], [a, b, c, e, d], [a, b, c, e, d|...], [a, b, c, e|...], [a, b, c| ...], [a, b|...], [a|...], [...|...]|...].


% timing the solver

?- boggle:make_board([[a,b,c], [d,e,f], [g,h,i]], B),

    time(setof(W, N^(boggle:board_has_word(B, W),length(W, N), N #>= 3), S)).

% 2,502,809 inferences, 0.864 CPU in 0.880 seconds (98% CPU, 2896937 Lips)

B = … ,

S = [[a, b, c], [a, b, c, e], [a, b, c, e, d], [a, b, c, e, d|...], [a, b, c, e|...], [a, b, c| ...], [a, b|...], [a|...], [...|...]|...].


Running the entire solver on a 4x4 with setof runs out of memory after about 40 seconds, even with a 2-gigabyte limit, but can find solutions one-at-a-time via backtracking more quickly than can be measured, starting with a depth of only about 30 inferences after the first 500. If we constrain the maximum length of the word, too, we obtain results with a 2-gigabyte limit:

?- boggle:make_board([[a,b,c,d], [e,f,g,h], [h,i,j,k], [l,m,n,o]], B),

    time(setof(W, N^(boggle:board_has_word(B, W),length(W, N), N #>= 3, N #=< 8), S)).

% 2,982,360,271 inferences, 698.309 CPU in 702.722 seconds (99% CPU, 4270830 Lips)

B = … ,

S = [[a, b, c], [a, b, c, d], [a, b, c, d, g], [a, b, c, d, g|...], [a, b, c, d|...], [a, b, c| ...], [a, b|...], [a|...], [...|...]|...].

And yes, that’s 11.7 minutes. Impractical. But it made almost three billion inferences at a rate nearly double that of the previous queries, so I forgive it.

Interestingly, re-ordering the solver clauses on the 3x3 to do the length check first slows the query down significantly: I suspect we spend a lot of time trying to match specific lengths with real words, so we duplicate certain work again and again, rather than gathering up the possible words upfront and filtering them down. The number of inferences appears to confirm this.

I did abort this query after 43.45 minutes; in that time, it worked through eleven billion inferences and wasn’t anywhere near done as best I can tell. Compare that to previous versions that only needed 2.5 million inferences to run in less than a second.

?- boggle:make_board([[a,b,c], [d,e,f], [g,h,i]], B),

    time(setof(W, N^(length(W, N), N #>= 3, boggle:board_has_word(B, W)), S)).

^CAction (h for help) ? abort

% 11,123,831,833 inferences, 2577.157 CPU in 2607.577 seconds (99% CPU, 4316318 Lips)

% Execution Aborted

Here’s the full listing. The important pieces are make_board, and board_has_word.


% vim: ft=prolog


:- module(boggle, [make_board/2, board_has_word/2]).


:- use_module(library(assoc)). % assoc-list (AVL trees) predicates

:- use_module(library(clpfd)). % #= and related constraints

:- use_module(library(lists)). % member

:- use_module(library(apply)). % maplist, foldl

:- use_module(library(yall)). % [x]>>f(x) lambda syntax


product(As, Bs, Cs) :-

    findall(A-B, (member(A, As), member(B, Bs)), Cs).


range(N, L) :-

    NN #= N-1,

    bagof(X, between(0, NN, X), L).


coords(N, Coords) :-

    range(N, ToN),

    product(ToN, ToN, Coords).


square(Xss, N) :-

    length(Xss, N),

    maplist([Xs]>>length(Xs, N), Xss).


make_board_fold_helper(Chars, X-Y, Acc, New) :-

    nth0(X, Chars, Row),

    nth0(Y, Row, Char),

    put_assoc(X-Y, Acc, Char, New).


make_board(Chars, B) :-

    square(Chars, N),

    coords(N, Coords),


    foldl(make_board_fold_helper(Chars), Coords, B0, B).


% +B, ?W

board_has_word(B, W) :-

    assoc_to_keys(B, Cs),

    member(C, Cs),

    board_has_word(B, W, C).


board_has_word(_, [], _).


board_has_word(B, [First|W], X-Y) :-

    get_assoc(X-Y, B, First),

    member(DX, [-1, 0, 1]),

    member(DY, [-1, 0, 1]),

    NewX #= X + DX,

    NewY #= Y + DY,

    del_assoc(X-Y, B, _, NewB),

    board_has_word(NewB, W, NewX-NewY).

Let’s work through board_has_word first, since it’s the interesting core of the solver.

The binary relation board_has_word/2 relates a word W to a board B if-and-only-if there is some C (a coordinate) in the board such that the trinary relation board_has_word/3 holds. That trinary relation is like our helper check-start: it holds if the word is in the board starting at that coordinate.

Just like check-start, board_has_word/3 has two cases:

In the latter case, we need that the current position matches (à la <step-starting-letter>), and that the rest of the word starts from one of the adjacent positions (à la <step-recur>).

It’s a remarkably concise spec, considering it’s able to encompass both the checker and the generator. It’s not quite as fast as the Racket version, but could possibly be improved. Combine a solver query from above with a clause asserting the found word is in some dictionary, and you have something like McCarthy’s solver!

Now, make_board relates a (necessarily ground) list of characters (which can be anything; here I use simple atoms), as long as it is a square list-of-lists, to a board structure. It does so with a list of coordinates computed from a cartesian product, which it folds over to repeatedly add characters to the board. This is essentially what <board> does, but without randomness.

The rest is the helper predicates and the boiler-plate of declaring a Prolog module and importing libraries (some of which is SWI Prolog-specific).

5 And a third time

Prolog down, if inefficiently. Can we make this available to Racket? racklog looks promising, but has a very small set of primitives. If we can implement enough of the required predicates (with the right modes) using Racket functions, we can build this thing in racklog.

Here are some queries similar to those from the Prolog version. This first version uses association lists with O(n) access, so the programs are much slower than the Prolog equivalents (which use AVL trees). So I’m showing 2x2 and 3x3s. Times reported are in milli-seconds.

> (time (void (%which (S)

                (%let (B W N)

                  (%set-of W (%and (%make-board '((a b) (d e)) B)

                                   (%board-has-word B W)

                                   (%is N (length W))

                                   (%>= N 3))


cpu time: 166 real time: 166 gc time: 14

> (time (void (%which (S)

                (%let (B W N)

                  (%set-of W (%and (%make-board '((a b c) (d e f) (g h i)) B)

                                   (%board-has-word B W)

                                   (%is N (length W))

                                   (%>= N 3))


cpu time: 68845 real time: 68909 gc time: 5595

> (time (void (%which (S)

                (%let (B W N)

                  (%set-of W (%and (%make-board '((a b c) (d e f) (g h i)) B)

                                   (%board-has-word B W)

                                   (%is N (length W))

                                   (%>= N 3)

                                   (%<= N 8))

cpu time: 69356 real time: 69437 gc time: 5868

After switching to hashes, I get the following times. They aren’t really any faster, though I’ll still present the hash-based version.

> (time (void (%which (S)

                (%let (B W N)

                  (%set-of W (%and (%make-board '((a b) (d e)) B)

                                   (%board-has-word B W)

                                   (%is N (length W))

                                   (%>= N 3))


cpu time: 191 real time: 191 gc time: 34

> (time (void (%which (S)

                (%let (B W N)

                  (%set-of W (%and (%make-board '((a b c) (d e f) (g h i)) B)

                                   (%board-has-word B W)

                                   (%is N (length W))

                                   (%>= N 3))


cpu time: 68330 real time: 69054 gc time: 4990

Mutable hashes would probably be faster, but I would have to be more careful to not mutate logical values that are referred to in other places (or else to undo the mutation). That’s too tricky for this version.

Let’s take a look. The core functionality is strikingly similar to the Prolog code.

(define %make-board-fold-helper
  (%rel (Chars X Y Acc New Row Char)
    [(Chars (cons X Y) Acc New) ; -
     (%is Row (list-ref Chars X))
     (%is Char (list-ref Row Y))
     (%put-assoc (cons X Y) Acc Char New)]))
(define %make-board
  (%rel (Chars B N Coords B0)
    [(Chars B) ; -
     (%square Chars N)
     (%coords N Coords)
     (%empty-assoc B0)
     (%foldl (curry %make-board-fold-helper Chars)
             Coords B0 B)]))
(define %board-has-word
  (%rel (B W Cs C First X Y DX DY NewX NewY NewB)
    [(B W) ; -
     (%assoc-to-keys B Cs)
     (%member C Cs)
     (%board-has-word B W C)]
    [((_) empty (_))]
    [(B (cons First W) (cons X Y)) ; -
     (%get-assoc (cons X Y) B First)
     (%member DX '(-1 0 1))
     (%member DY '(-1 0 1))
     (%is NewX (+ X DX))
     (%is NewY (+ Y DY))
     (%del-assoc (cons X Y) B (_) NewB)
     (%board-has-word NewB W (cons NewX NewY))]))

I’ve gotten into the habit of putting ;- after my clause heads to remind me of Prolog. Scribble doesn’t render it in the code as well as inline here, where it looks a lot like Prolog’s :-.

But we still have to define quite a bit of machinery to make this work. This is mostly self-explanatory: building (moded) relations out of Racket functions.

(define %product
  (%rel (As Bs Cs A B)
    [(As Bs Cs) ; -
     (%set-of (cons A B)
              (%and (%member A As)
                    (%member B Bs))
; spoofed from
(define %between
  (%rel (Low High Value NewLow)
    [(Low High Low)]
    [(Low High Value) ; -
     (%is NewLow (add1 Low))
     (%<= NewLow High)
     (%between NewLow High Value)]))
(define %range
  (%rel (N L NN X)
    [(N L) ; -
     (%is NN (sub1 N))
     (%bag-of X (%between 0 NN X) L)]))
(define %coords
  (%rel (N Coords ToN)
    [(N Coords) ; -
     (%range N ToN)
     (%product ToN ToN Coords)]))
(define %square
  (%rel (Xss N)
    [(Xss N) ; -
     (%is N (length Xss))
     (%andmap (λ (Xs) (%is N (length Xs))) Xss)]))
(define %empty-assoc
  (%rel ()
(define %put-assoc
  (%rel (Key Assoc Val New)
    [(Key Assoc Val New) ; -
     (%is New (hash-set Assoc Key Val))]))
; more general than swipl's get_assoc, since we have
; -Key +Assoc -Val
; +Key +Assoc ?Val
; though not
; -Key +Assoc +Val
(define %get-assoc
  (%rel (Key Assoc Val Keys)
    [(Key Assoc Val) ; -
     (%assoc-to-keys Assoc Keys)
     (%member Key Keys)
     (%is Val (hash-ref Assoc Key #f))]))
(define %del-assoc
  (%rel (Key Assoc Val New)
    [(Key Assoc Val New) ; -
     (%get-assoc Key Assoc Val)
     (%is New (hash-remove Assoc Key))]))
(define %assoc-to-keys
  (%rel (Assoc Keys)
    [(Assoc Keys) ; -
     (%is Keys (hash-keys Assoc))]))
(define %foldl
  (%rel (Goal H T V0 V1 V)
    [(Goal empty V V)]
    [(Goal (cons H T) V0 V) ; -
     (Goal H V0 V1)
     (%foldl Goal T V1 V)]))

We only need certain functionality from some relations, so I avoid generalizing them here (e.g., %foldl only has the arity-4 definition).

6 Outro

We’ve built a Boggle checker and generator in increasingly sophisticated ways, but with increasingly worse performance! Comments on improvements welcome.

Put the code in this order.

<*> ::=
(require racket/list
(define letters
  (string->list "abcdefghijklmnopqrstuvwxyz"))
(define (random-list-ref l)
  (list-ref l (random (length l))))
(define (random-letter)
  (random-list-ref letters))
(define x car)
(define y cdr)
(define minimum-word-length 3)
(define maximum-word-length 8)