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(*
* Written by sweeks@sweeks.com on 1999-08-31.
*
* A solution to mpuz. (Try M-x mpuz in emacs.)
* This solution is very loosely based on an OCAML solution posted to
* comp.lang.ml by Laurent Vaucher <blo.b@infonie.fr>.
*)
(* override print so the benchmark is silent *)
fun print _ = ()
structure List =
struct
open List
fun exists(l, p) = List.exists p l
fun map(l, f) = List.map f l
fun fold(l, b, f) =
let
fun loop(l, b) =
case l of
[] => b
| x :: l => loop(l, f(x, b))
in loop(l, b)
end
fun foreach(l, f) = fold(l, (), fn (x, ()) => f x)
end
structure String =
struct
open String
fun fold(s, b, f) =
let
val n = size s
fun loop(i, b) =
if i = n
then b
else loop(i + 1, f(String.sub(s, i), b))
in loop(0, b)
end
end
structure Mpuz =
struct
fun solve(a, b, c, d, e) =
let
fun printNewline() = print "\n"
val sub = Array.sub
val update = Array.update
val letters =
List.fold
([a, b, c, d, e], [], fn (s, letters) =>
String.fold
(s, letters, fn (c, letters) =>
if List.exists(letters, fn c' => c = c')
then letters
else c :: letters))
val letterValues =
Array.array(Char.ord Char.maxChar + 1, 0)
fun letterValue(c) =
Array.sub(letterValues, ord c)
fun setLetterValue(c, v) =
Array.update(letterValues, ord c, v)
fun stringValue(s) =
String.fold(s, 0, fn (c, v) => v * 10 + letterValue c)
fun printResult() =
(List.foreach
(letters, fn c =>
print(concat[String.str(c), " = ",
Int.toString(letterValue(c)), " "]))
; print "\n")
fun testOk() =
let
val b0 = letterValue(String.sub(b, 1))
val b1 = letterValue(String.sub(b, 0))
val a = stringValue a
val b = stringValue b
val c = stringValue c
val d = stringValue d
val e = stringValue e
in if a * b0 = c
andalso a * b1 = d
andalso a * b = e
andalso c + d * 10 = e
then printResult()
else ()
end
val values = List.map([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], fn v =>
(v, ref false))
(* Try all assignments of values to letters. *)
fun loop(letters) =
case letters of
[] => testOk()
| c :: letters =>
List.foreach
(values, fn (v, r) =>
if !r
then ()
else (r := true
; setLetterValue(c, v)
; loop(letters)
; r := false))
in loop(letters)
end
end
structure Main =
struct
fun doit() =
Mpuz.solve("AGH", "FB", "CBEE", "GHFD", "FGIJE")
(*
* Solution:
* J = 0 I = 1 D = 8 E = 2 C = 5 B = 6 F = 4 H = 7 G = 3 A = 9
*)
val doit =
fn size =>
let
fun loop n =
if n = 0
then ()
else (doit();
loop(n-1))
in
loop size
end
end
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