Type inference for The Simply Typed Lambda Calculus

%% Type inference for The Simply Typed Lambda Calculus

:- op(150, xfx, ⊢).
:- op(140, xfx, :).
:- op(100, xfy, ->).
:- op(100, yfx, $).

Γ ⊢    Term : Type          :- atom(Term), member(Term : Type, Γ).
Γ ⊢ λ(A, B) : Alpha -> Beta :- [A : Alpha|Γ] ⊢ B : Beta.
Γ ⊢   A $ B : Beta          :- Γ ⊢ A : Alpha -> Beta, Γ ⊢ B : Alpha.

/** Examples:

% Typing fix
?- Γ ⊢ y $ f : Y, Γ ⊢ f $ (y $ f) : Y.
Γ = [y: (Y->Y)->Y, f:Y->Y|_G330]

% Typing the Y combinator
?- Γ ⊢ λ(f, (λ(f,f $ f)) $ (λ(g,f $ (g $ g)))) : Y.
Y = (_G309 -> _G309) -> _G309 

% Typing flip id
?- Id = λ(i, i), Flip = λ(f, λ(y, λ(x, f $ x $ y))),
   Γ ⊢ Flip $ Id : T.
T = _G395 -> (_G395 -> _G411) -> _G411

**/

Gödel's beta function

I read a most incredible proof in the book, Lectures On The Curry Howard Isomorphism that I decided to created a program from it.

(defun factorial (n)
  (if (= n 0) 1
      (* n (factorial (- n 1)))))

(defun extended-euclidean-algorithm (a b)
  ;; function extended_gcd(a, b)
  ;;  if (a mod b = 0)
  ;;     return {0, 1}
  ;;  else {x, y} := extended_gcd(b, a mod b)
  ;;       return {y, x-y*(a div b)}
  (if (= (mod a b) 0) (values 0 1)
      (multiple-value-bind (x y) 
          (extended-euclidean-algorithm b (mod a b))
        (values y (- x (* y (floor a b)))))))

(defun extended-euclidean-algorithm-prime (a b)
  ;; extended-euclidean-algorithm solves:
  ;;  u*a + v*b = gcd
  ;; This one solves:
  ;;  u*a - v*b = gcd
  (multiple-value-bind (u v)
      (extended-euclidean-algorithm a b)
    (if (= b 1)
        (values 1 (- a 1))
        (values u (- v)))))

;; For every finite sequence k_0, k_1, ..., k_r there exist two numbers m, n, such that (β m n j) = k_j, for j = 0..r.
(defun β (m n j)
  (mod m (+ (* n (+ j 1)) 1)))

;; Produce n m, such that (β m n j) = k_j
(defun β-solve (k)
  (let* ((n (factorial (apply #'max (- (length k) 1) k)))
         (a (loop for j below (length k) collect (+ (* n (+ j 1)) 1))))
    (labels ((next-m (l m)
               (if (= l (- (length k) 1)) m
                   (let ((a* (apply #'* (subseq a 0 (min (+ l 1) (length k))))))
                     (multiple-value-bind (u v)
                         (extended-euclidean-algorithm-prime (elt a (+ l 1)) a*)
                       (declare (ignore u))
                       (next-m (+ l 1)
                               (+ m (* (* (- m (elt k (+ l 1))) v) a*))))))))
      (values (next-m 0 (elt k 0)) n))))


;; CL-USER> (multiple-value-bind (n m)
;;              (β-solve (map 'list #'char-code "You win!"))
;;            (map 'string #'code-char (loop for j below 8 collect (β n m j))))
;; "You win!"

Counting Infinity

import Data.List
import Data.Ratio

-- A few utilities and toys for later:

data Void                    {- This needs -XEmptyDataDecls -}
instance Show Void

data Tree = N | Tree :@: Tree deriving Show

-- interleaving append
a      +~~+ []     = a
[]     +~~+ b      = b
(a:as) +~~+ (b:bs) = a : b : as +~~+ bs

diagonalZipWith f as bs = [ f a b
                          | (x,y) <- pairs
                          , a <- as !! x
                          , b <- bs !! y ]
  where []     !! _ = []
        (x:_)  !! 0 = [x]
        (_:xs) !! n = xs !! (n-1)

(*) `on` f = \x y -> f x * f y




-- We shall get started with some basic infinite sets of numbers

nats = [0..]
-- [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,..]

ints = 0 : [1..] +~~+ map negate [1..]
-- [0,1,-1,2,-2,3,-3,4,-4,5,-5,6,-6,7,-7,8,-8,9,-9,10,-10,11,-11,..]

primes = nubBy(((>1).).gcd)[2..]
-- [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,..]
-- (My thanks to the author of this whoever you are (: )

fibs = map fst $ iterate (\(x,y)->(y,x+y)) $ (0,1)
-- [0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,..]




-- Looking at the diagonals of every pair of nats:
--
--   (0,0) (1,0) (2,0) (3,0)
--   (0,1) (1,1) (2,1) (3,1)
--   (0,2) (1,2) (2,2) (3,2)
--   (0,3) (1,3) (2,3) (3,3)
--
-- Each element of the first diagonal, having one element, sums to zero,
-- Each element of the second, two of them, sums to one, ...

pairs = concatMap pairsSummingTo [0..]
  where pairsSummingTo n = map (\i -> (i, n-i)) [0..n]
-- [(0,0),(0,1),(1,0),(0,2),(1,1),(2,0),(0,3),(1,2),(2,1),(3,0),(0,4),
--  (1,3),(2,2),(3,1),(4,0),(0,5),(1,4),(2,3),(3,2),(4,1),(5,0),(0,6),
--  (1,5),(2,4),(3,3),(4,2),(5,1),(6,0),(0,7),(1,6),...]




-- Let us consider every unique rational, They all (a%b) have GCD(a,b) = 1
-- Let's then, start with the GCD 1 and execute Euclid's algorithm backwards.

{-
 - i: 1%1 in Q
 - f: a%b in Q -> (a+b)%b in Q
 - g: a%b in Q -> a%(a+b) in Q
 -}

{- [{I}, {FI}, {GI}, {FFI}, {FGI}, {GFI}, {GGI}, {FFFI}, ...] -}
{- concat [[{I}], [{FI}, {GI}], [{FFI}, {FGI}, {GFI}, {GGI}],
           [{FFFI}, ...], ...] -}

rationals = let rats   0   = [i]
                rats (n+1) = map f (rats n) ++ map g (rats n)
             in concatMap rats nats
  where i   = 1%1
        f r = (numerator r + denominator r) % denominator r
        g r = numerator r % (numerator r + denominator r)
-- [1%1,2%1,1%2,3%1,3%2,2%3,1%3,4%1,5%2,5%3,4%3,3%4,3%5,2%5,1%4,5%1,7%2,
--  8%3,7%3,7%4,8%5,7%5,5%4,4%5,5%7,5%8,4%7,3%7,3%8,2%7,...]





-- Every string that can be made from a list of symbols
strings symbols = concat [ enumerate n symbols | n <- [1..] ]
  where enumerate 0 _ = [""]
        enumerate n d = concatMap (\y->map (:y) d) (enumerate (n-1) d)


-- Every binary sequence
binary = strings ['1', '0']
-- ["1","0","11","01","10","00","111","011","101","001","110",
--  "010","100","000","1111","0111","1011","0011","1101","0101",
--  "1001","0001","1110","0110","1010","0010","1100","0100","1000",...]


-- This is a very slow method to generate all well parenthesized expressions
-- It is a perfectly valid method for proving that a set is enumerable though
-- In general, showing that a set is the subset of all strings composed from
-- some symbols
parenthesized = filter balanced $ strings ['(', ')']
  where split s = map (flip splitAt s) [1..length s-1]
        balanced []     = True
        balanced (_:[]) = False
        balanced s      =     head s == '(' && last s == ')'
                           && balanced (tail $ init $ s)
                        || any (uncurry ((&&) `on` balanced)) (split s)
-- ["()","()()","(())","()()()","(())()","()(())","(()())","((()))",
--  "()()()()","(())()()","()(())()","(()())()","((()))()","()()(())",
--  "(())(())","()(()())","(()()())","((())())","()((()))","(()(()))",
--  "((()()))","(((())))",...]



-- This is fabulous, So I stole it from:
-- http://web.comlab.ox.ac.uk/oucl/work/jeremy.gibbons/publications/spigot.pdf
pi = g(1,180,60,2) where 
  g(q,r,t,i) = let (u,y)=(3*(3*i+1)*(3*i+2),div(q*(27*i-12)+5*r)(5*t)) 
               in y : g(10*q*i*(2*i-1),10*u*(q*(5*i-2)+r-y*t),t*u,i+1)
-- [3,1,4,1,5,9,2,6,5,3,5,8,9,7,9,3,2,3,8,4,6,2,6,4,3,3,8,3,2,7,9,5,0,...]




-- We can use typeclasses to recurse over the structure of a type
-- If a function listing all inhabitants of a type is written,
-- One can be sure that infinity will crop up.

class Inhabitants a where
  inhabitants :: [a]

instance Inhabitants Void where
  inhabitants = []

instance Inhabitants () where
  inhabitants = [()]

instance Inhabitants Bool where
  inhabitants = [True, False]

instance Inhabitants Integer where
  inhabitants = ints

instance Inhabitants a => Inhabitants (Maybe a) where
  inhabitants = [Nothing] ++ map Just inhabitants

instance (Inhabitants a, Inhabitants b) => Inhabitants (Either a b) where
  inhabitants = map Left inhabitants +~~+ map Right inhabitants

instance Inhabitants a => Inhabitants [a] where
  inhabitants = [[]] ++ diagonalZipWith (:) inhabitants inhabitants

instance Inhabitants Tree where
  inhabitants = [N] ++ diagonalZipWith (:@:) inhabitants inhabitants

-- take 600 $ inhabitants :: [ Either [Maybe Integer] Tree ]
--                             ^^ You can try various types here
--                                as long as they are in the
--                                Inhabitants class.

XOR Encryption

#include <stdio.h>
#include <stdlib.h>

FILE * input,
     * key,
     * output;

int main(void) {
  input  = fopen("in.txt",  "r");
  key    = fopen("key.txt", "r");
  output = fopen("out.txt", "w");
  
  if(!input)  { puts("input failure");  return EXIT_FAILURE; }
  if(!key)    { puts("key failure");    return EXIT_FAILURE; }
  if(!output) { puts("output failure"); return EXIT_FAILURE; }
  
  while(!feof(input))
    { if(feof(key)) rewind(key); fputc(fgetc(input) ^ fgetc(key), output); }
  
  return EXIT_SUCCESS;
}

Generate, Test and Intertwine

%% Generate, Test and Intertwine
%% -----------------------------

%% The aim here is to generate 4x4 magic squares. First of all, We start
%% with the 3x3 case, since that's easier.
%% This way any improvements can be reused, assuming they are general enough

%% I'll define a magic square as, an nxn table of numbers such that all the
%% rows, columns and diagonals sum to the same value, each number in the range
%% 1 .. n^2 appears once in the grid.

%% In Prolog, We'll represent squares as lists:

%: [4,8,2,
%:  3,5,7,
%:  8,1,6]

%% So to generate this, We'll select numbers from a list of digits and insert
%% them into the table and check that everything adds up.

saturn_digits([1,2,3,4,5,6,7,8,9]).

saturn([A,B,C,
        D,E,F,
        G,H,I]) :-
        saturn_digits(D1),
        select(A, D1, D2),
        select(B, D2, D3),
        select(C, D3, D4),
        select(D, D4, D5),
        select(E, D5, D6),
        select(F, D6, D7),
        select(G, D7, D8),
        select(H, D8, D9),
        select(I, D9, []),
        S is A + B + C,
        S is D + E + F,
        S is G + H + I,
        S is A + D + G,
        S is B + E + H,
        S is C + F + I,
        S is A + E + I,
        S is C + E + G.

%% That first attempt works fine:

%? ?- saturn(Square).
%? Square = [2, 7, 6, 9, 5, 1, 4, 3, 8] ;
%? Square = [2, 9, 4, 7, 5, 3, 6, 1, 8] ;
%? Square = [4, 3, 8, 9, 5, 1, 2, 7, 6] ;
%? Square = [4, 9, 2, 3, 5, 7, 8, 1, 6] ;
%? Square = [6, 1, 8, 7, 5, 3, 2, 9, 4] ;
%? Square = [6, 7, 2, 1, 5, 9, 8, 3, 4] ;
%? Square = [8, 1, 6, 3, 5, 7, 4, 9, 2] ;
%? Square = [8, 3, 4, 1, 5, 9, 6, 7, 2] ;

%% It's quite clear the program is correct,
%% The two important properties are
%%  * Soundness -- Generates only magic squares
%%  * Completeness -- Generates every magic square
%%
%% Since digits are selected from the list, It's possible for the square to
%% be filled in every possible way, and due to:
%:   select(I, D9, []),
%% It's ensured that every digit is used. Every row, column and diagonal is
%% summed, since S is used for all summations, this ensures that all the sums
%% are equal.


%% There is this pattern in the code though, which I don't very much like:

%: select(A, D1, D2),
%: select(B, D2, D3),
%: select(C, D3, D4),
%: select(D, D4, D5),
%: select(E, D5, D6),
%: select(F, D6, D7),
%: select(G, D7, D8),
%: select(H, D8, D9),
%: select(I, D9, []),

%% Prolog has a macro, DCG style, that lets us erase this pattern.
%% If we use --> instead of :- , then this is implicit, everything is sewn
%% together like we did manually, except automatically.
%% {} is used to escape it.
%% So now, It can be rewritten to the equivalent program:

saturn_2([A,B,C,
          D,E,F,
          G,H,I]) -->
        select(A),
        select(B),
        select(C),
        select(D),
        select(E),
        select(F),
        select(G),
        select(H),
        select(I),
        { S is A + B + C },
        { S is D + E + F },
        { S is G + H + I },
        { S is A + D + G },
        { S is B + E + H },
        { S is C + F + I },
        { S is A + E + I },
        { S is C + E + G }.

%? ?- saturn_digits(D), saturn_2(Square, D, []).
%? D = [1, 2, 3, 4, 5, 6, 7, 8, 9],
%? Square = [2, 7, 6, 9, 5, 1, 4, 3, 8] ;
%? ..

%% It produces all the same answers in the same order. One of the fantastic
%% thing about Prolog is you could actually implement DCG in it, if it wasn't
%% already there, Or even your own macros.

%% Anyway, there is an optmisation which is quite blatant: select always
%% succeeds, It's the tests in {} which cause backtracking, So if we were to
%% intertwine the generating (done by select) with the testing, (as done by
%% summation) then backtracking which does occur, will happen sooner.

saturn_3([A,B,C,
          D,E,F,
          G,H,I]) -->
        select(A),
        select(B),
        select(C),
        { S is A + B + C },
        select(D),
        select(G),
        { S is A + D + G },
        select(E),
        { S is C + E + G },
        select(F),
        { S is D + E + F },
        select(H),
        { S is B + E + H },
        select(I),
        { S is G + H + I },
        { S is C + F + I },
        { S is A + E + I }.


%% So now, It's straightforward to use the same process for the 4x4 case.
%% We use DCG first in this case though.

jupiter_digits([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]).

jupiter([A,B,C,D,
         E,F,G,H,
         I,J,K,L,
         M,N,O,P]) -->
        select(A),
        select(B),
        select(C),
        select(D),
        select(E),
        select(F),
        select(G),
        select(H),
        select(I),
        select(J),
        select(K),
        select(L),
        select(M),
        select(N),
        select(O),
        select(P),
        
        { S is A + B + C + D },
        { S is E + F + G + H },
        { S is I + J + K + L },
        { S is M + N + O + P },
        
        { S is A + E + I + M },
        { S is B + F + J + N },
        { S is C + G + K + O },
        { S is D + H + L + P },
        
        { S is A + F + K + P },
        { S is D + G + J + M }.

%% And then rearrange

jupiter_2([A,B,C,D,
           E,F,G,H,
           I,J,K,L,
           M,N,O,P]) -->
        select(A),
        select(B),
        select(C),
        select(D),
        %% * * * *
        %% o o o o
        %% o o o o
        %% o o o o
        { S is A + B + C + D },
        
        select(E),
        select(I),
        select(M),
        %% * * * *
        %% * o o o
        %% * o o o
        %% * o o o
        { S is A + E + I + M },
        
        select(F),
        select(K),
        select(P),
        %% * * * *
        %% * * o o
        %% * o * o
        %% * o o *
        { S is A + F + K + P },
        
        select(G),
        select(H),
        %% * * * *
        %% * * * *
        %% * o * o
        %% * o o *
        { S is E + F + G + H },
        
        select(L),
        %% * * * *
        %% * * * *
        %% * o * *
        %% * o o *
        { S is D + H + L + P },
        
        select(J),
        %% * * * *
        %% * * * *
        %% * * * *
        %% * o o *
        { S is I + J + K + L },
        { S is D + G + J + M },
        
        select(N),
        %% * * * *
        %% * * * *
        %% * * * *
        %% * * o *
        { S is B + F + J + N },
        
        select(O),
        %% * * * *
        %% * * * *
        %% * * * *
        %% * * * *
        { S is M + N + O + P },
        { S is C + G + K + O }.

%? ?- jupiter_digits(D), jupiter_2(Square, D, []).
%? 
%? D = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16],
%? Square = [1, 2, 15, 16, 12, 14, 3, 5, 13, 7, 10, 4, 8, 11, 6, 9] ;
%? ...

%% There is a huge speed difference in the two versions.
%% Here's the final program without comments:

jupiter_3([A,B,C,D,
           E,F,G,H,
           I,J,K,L,
           M,N,O,P]) -->
        select(A),
        select(B),
        select(C),
        select(D),
        { S is A + B + C + D },
        select(E),
        select(I),
        select(M),
        { S is A + E + I + M },
        select(F),
        select(K),
        select(P),
        { S is A + F + K + P },
        select(G),
        select(H),
        { S is E + F + G + H },
        select(L),
        { S is D + H + L + P },
        select(J),
        { S is I + J + K + L },
        { S is D + G + J + M },
        select(N),
        { S is B + F + J + N },
        select(O),
        { S is M + N + O + P },
        { S is C + G + K + O }.