We work through the code giving more sensible names to variables and commenting where necessary.
183 lines
6.0 KiB
OCaml
183 lines
6.0 KiB
OCaml
(*let debugf = Format.ifprintf Format.std_formatter*)
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type pos = int * int
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(** A position on the grid, pair of x, y co-ordinates *)
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(** Pretty print a position to Format.formatter [out] *)
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let pp_pos out ((x, y) : pos) = Format.fprintf out "(%d,@ %d)" x y
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(** Get the x co-ordinate of a position *)
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let pos_x = fst
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(** Get the y co-ordinate of a position *)
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let pos_y = snd
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type square = { pos : pos; length : int }
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(** A type representing a square, consisitng of the bottom-left corner of the
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square and the length of each side. *)
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(** Pretty print a square to Format.formatter [out] *)
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let pp_square out (sq : square) =
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Format.fprintf out "{%a@ len:%d}" pp_pos sq.pos sq.length
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(** Returns true if the squares [sq1] and [sq2] intersect, and false otherwise.
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*)
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let intersects sq1 sq2 =
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let sq1l = pos_x sq1.pos in
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let sq1r = sq1.length + sq1l in
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let sq1b = pos_y sq1.pos in
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let sq1t = sq1.length + sq1b in
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let sq2l = pos_x sq2.pos in
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let sq2r = sq2.length + sq2l in
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let sq2b = pos_y sq2.pos in
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let sq2t = sq2.length + sq2b in
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sq1l < sq2r && sq1r > sq2l && sq1t > sq2b && sq1b < sq2t
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(** Returns true if we can place the square [sq] without overlapping any already
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placed squares in [sqs] and without exceeding the bounds of the grid which
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is [length] along each side. *)
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let square_fits sq length sqs =
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let rec impl sq sqs =
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match sqs with
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| [] -> true
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| h :: t -> if intersects h sq then false else impl sq t
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in
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if pos_x sq.pos + sq.length > length then false
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else if pos_y sq.pos + sq.length > length then false
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else impl sq sqs
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(** Returns true if the position [(x, y)] is in one of the squares [sqs]. *)
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let rec in_squares ((x, y) : pos) sqs =
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match sqs with
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| [] -> false
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| h :: t ->
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let sqx = pos_x h.pos in
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let sqy = pos_y h.pos in
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let len = h.length in
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if x >= sqx && x < sqx + len && y >= sqy && y < sqy + len then true
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else in_squares (x, y) t
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(** Returns the next position to consider when working through the grid we are
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placing squares on. [(x, y)] is the current position, and [sqs] is a list of
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already placed squares.
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Returns (0, length) when we have filled the grid. *)
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let next_pos length ((x, y) : pos) sqs =
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(* We basically walk along each row looking for an empty space. *)
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let rec impl x y =
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if x >= length then impl 0 (y + 1)
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else if in_squares (x, y) sqs then impl (x + 1) y
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else (x, y)
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in
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impl (x + 1) y
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let triangle_num n = n * (n + 1) / 2
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(** Find a solution to the [n]th Partridge problem. Returns a list of squares
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giving the position on the grid. *)
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let find_solution n =
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(* recursive implementation:
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[avail_sqs] is an array where avail_sqs.(x) returns how many sqs of size
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x are available to be placed. [length] is the side length of the grid
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we are placing the squares into.
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[impl idx pos sqs] implements the recursive implementation. [idx] is the
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current index in [avail_sqs] that we are looking at.
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If there is a square available of size [idx] (i.e. avail_sqs.(idx) > 0)
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then we try to place a square of size [idx] at [pos]. If this is
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successful it adds that to the list [sqs] and tries to find a square that
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fits in the next position.
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If [impl] is not successful it tries again at the current position with
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a square of size [idx - 1].
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If we reach an [idx] of 0 we have failed and return an empty list.
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If we reach the position [(0, length)] we have succeeded and return [sqs].
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*)
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let avail_sqs = Array.init (n + 1) Fun.id in
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let length = triangle_num n in
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let rec impl idx pos sqs =
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let sq = { pos; length = idx } in
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if pos = (0, length) then sqs
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else if idx = 0 then []
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else if avail_sqs.(idx) = 0 then impl (idx - 1) pos sqs
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else if square_fits sq length sqs then begin
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Array.set avail_sqs idx (avail_sqs.(idx) - 1);
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let new_sqs = sq :: sqs in
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let new_pos = next_pos length pos new_sqs in
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let result = impl n new_pos new_sqs in
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Array.set avail_sqs idx (avail_sqs.(idx) + 1);
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if List.is_empty result then impl (idx - 1) pos sqs else result
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end
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else impl (idx - 1) pos sqs
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in
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impl n (0, 0) []
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(** Exception raised if we find we have overlapping squares when printing the
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solution. *)
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exception Overlapping_squares of pos
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(** Print the layout given in [sqs] for a grid with side-length [size]. *)
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let print_solution length sqs =
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let array = Array.make (length * length) '.' in
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let set_pos x y c =
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if array.(x + (y * length)) <> '.' then raise (Overlapping_squares (x, y))
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else Array.set array (x + (y * length)) c
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in
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let rec write_length x y n =
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if n = 0 then ()
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else begin
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Array.set array (x + (y * length)) (Char.chr (48 + (n mod 10)));
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write_length (x - 1) y (n / 10)
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end
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in
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let rec impl sqs =
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match sqs with
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| [] -> ()
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| { pos = x, y; length = n } :: t -> begin
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if n = 1 then set_pos x y '*'
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else if n = 2 then begin
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set_pos x y '+';
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set_pos (x + 1) y '+';
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set_pos x (y + 1) '+';
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set_pos (x + 1) (y + 1) '+'
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end
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else begin
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set_pos x y '+';
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set_pos (x + n - 1) y '+';
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set_pos x (y + n - 1) '+';
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set_pos (x + n - 1) (y + n - 1) '+';
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for a = 1 to n - 2 do
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set_pos (x + a) y '-';
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set_pos (x + a) (y + n - 1) '-';
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for b = 1 to n - 2 do
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set_pos (x + a) (y + b) ' '
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done;
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set_pos x (y + a) '|';
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set_pos (x + n - 1) (y + a) '|'
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done;
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write_length (x + n - 2) (y + 1) n
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end;
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impl t
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end
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in
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impl sqs;
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for y = 0 to length - 1 do
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for x = 0 to length - 1 do
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Format.printf "%c" array.(x + (y * length))
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done;
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Format.printf "\n"
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done
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let n = 8
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let tri_n = triangle_num n
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let soln = find_solution n
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let () = Format.printf "@[<hov>Base number: %d,@;side length: %d@;" n tri_n
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let () = Format.printf "Solution: %a@]@\n" (Format.pp_print_list pp_square) soln
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let () = print_solution tri_n soln
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