```-----------------------------------------------------------------------------
-- |
-- Module      :  Data.SBV.Examples.Puzzles.Counts
-- Copyright   :  (c) Levent Erkok
-- License     :  BSD3
-- Maintainer  :  erkokl@gmail.com
-- Stability   :  experimental
-- Portability :  portable
--
-- Consider the sentence:
--
-- @
--    In this sentence, the number of occurrences of 0 is _, of 1 is _, of 2 is _,
--    of 3 is _, of 4 is _, of 5 is _, of 6 is _, of 7 is _, of 8 is _, and of 9 is _.
-- @
--
-- The puzzle is to fill the blanks with numbers, such that the sentence
-- will be correct. There are precisely two solutions to this puzzle, both of
-- which are found by SBV successfully.
--
--  References:
--
--    * Douglas Hofstadter, Metamagical Themes, pg. 27.
--
--    * <http://www.lboro.ac.uk/departments/ma/gallery/selfref/index.html>
--
-----------------------------------------------------------------------------

module Data.SBV.Examples.Puzzles.Counts where

import Data.SBV

-- | We will assume each number can be represented by an 8-bit word, i.e., can be at most 128.
type Count  = SWord8
type Counts = [Count]

-- | Given a number, increment the count array depending on the digits of the number
count :: Count -> Counts -> Counts
count n cnts = ite (n .< 10)
(upd n cnts)                           -- only one digit
(ite (n .< 100)
(upd d1 (upd d2 cnts))            -- two digits
(upd d1 (upd d2 (upd d3 cnts))))  -- three digits
where (r1, d1)   = n  `bvQuotRem` 10
(d3, d2)   = r1 `bvQuotRem` 10
upd d vals = zipWith inc [0..] vals
where inc i c = ite (i .== d) (c+1) c

-- | Encoding of the puzzle. The solution is a sequence of 10 numbers
-- for the occurrences of the digits such that if we count each digit,
-- we find these numbers.
puzzle :: Counts -> SBool
puzzle cnt = cnt .== last css
where ones = replicate 10 1  -- all digits occur once to start with
css  = ones : zipWith count cnt css

-- | Finds all two known solutions to this puzzle. We have:
--
-- >>> solve
-- Solution #1
-- In this sentence, the number of occurrences of 0 is 1, of 1 is 11, of 2 is 2, of 3 is 1, of 4 is 1, of 5 is 1, of 6 is 1, of 7 is 1, of 8 is 1, of 9 is 1.
-- Solution #2
-- In this sentence, the number of occurrences of 0 is 1, of 1 is 7, of 2 is 3, of 3 is 2, of 4 is 1, of 5 is 1, of 6 is 1, of 7 is 2, of 8 is 1, of 9 is 1.
-- Found: 2 solution(s).
solve :: IO ()
solve = do res <- allSat \$ puzzle `fmap` mkExistVars 10
cnt <- displayModels disp res
putStrLn \$ "Found: " ++ show cnt ++ " solution(s)."
where disp n (_, s) = do putStrLn \$ "Solution #" ++ show n
dispSolution s
dispSolution :: [Word8] -> IO ()
dispSolution ns = putStrLn soln
where soln =  "In this sentence, the number of occurrences"
++  " of 0 is " ++ show (ns !! 0)
++ ", of 1 is " ++ show (ns !! 1)
++ ", of 2 is " ++ show (ns !! 2)
++ ", of 3 is " ++ show (ns !! 3)
++ ", of 4 is " ++ show (ns !! 4)
++ ", of 5 is " ++ show (ns !! 5)
++ ", of 6 is " ++ show (ns !! 6)
++ ", of 7 is " ++ show (ns !! 7)
++ ", of 8 is " ++ show (ns !! 8)
++ ", of 9 is " ++ show (ns !! 9)
++ "."
```