-- Linear Diophantine Equation solver -- -- Copyright (c) 2009 The MITRE Corporation -- -- This program is free software: you can redistribute it and/or modify -- it under the terms of the GNU General Public License as published by -- the Free Software Foundation, either version 3 of the License, or -- (at your option) any later version. -- This program is distributed in the hope that it will be useful, -- but WITHOUT ANY WARRANTY; without even the implied warranty of -- MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the -- GNU General Public License for more details. -- You should have received a copy of the GNU General Public License -- along with this program. If not, see <http://www.gnu.org/licenses/>. -- | -- Module : Algebra.CommutativeMonoid.LinDiophEq -- Copyright : (C) 2009 John D. Ramsdell -- License : GPL -- -- Linear Diophantine Equation solver. -- -- The solver uses the algorithm of Contejean and Devie as specified -- in \"An Efficient Incremental Algorithm for Solving Systems of -- Linear Diophantine Equations\", Information and Computation -- Vol. 113, pp. 143-174, 1994. -- -- The algorithm for systems of homogeneous linear Diophantine -- equations follows. Let e[k] be the kth basis vector for 1 <= k <= -- n. To find the minimal, non-negative solutions M to the system of -- equations sum(i=1,n,a[i]*v[i]) = 0, the algorithm of Contejean and -- Devie is: -- -- 1. [init] A := {e[k] | 1 <= k <= n}; M := {} -- -- 2. [new minimal results] M := M + {a in A | a is a solution} -- -- 3. [unnecessary branches] A := {a in A | all m in M : some -- 1 <= k <= n : m[k] < a[k]} -- -- 4. [breadth-first search] A := {a + e[k] | a in A, 1 <= k <= n, -- \<sum(i=1,n,a[i]*v[i]),v[k]> \< 0} -- -- 5. [test] If A = {}, stop, else go to 2. -- -- This module provides a solver for a single linear Diophantine -- equation a*v = b, where a and v are vectors, not matrices. -- -- When solving an inhomogeneous equation, it uses the homogeneous -- solver after adding -b as the first element of v and by bounding -- the first element of a to be one at each step in the computation. -- The first element of a solution is zero if it is a solution to the -- associated homogeneous equation, and one if it is a solution to the -- inhomogeneous equation. -- -- The algorithm is likely to be Fortenbacher's algorithm, the one -- generalized to systems of equations by Contejean and Devie, but I -- have not been able to verified this fact. module Algebra.CommutativeMonoid.LinDiophEq (linDiophEq) where import Data.Array import Data.Set (Set) import qualified Data.Set as S {-- Debugging hack import System.IO.Unsafe z :: Show a => a -> b -> b z x y = seq (unsafePerformIO (print x)) y zz :: Show a => a -> a zz x = z x x pr :: Set (Vector Int) -> [[Int]] pr s = map elems $ S.toList s zzz :: Set (Vector Int) -> Set (Vector Int) zzz s = z (pr s) s --} type Vector a = Array Int a vector :: Int -> [a] -> Vector a vector n elems = listArray (0, n - 1) elems -- | The 'linDiophEq' function takes a list of integers that specifies -- the coefficients of linear Diophantine equation and a constant, -- and returns the equation's minimal, non-negative solutions. -- -- When solving an inhomogeneous equation, the first element of a -- solution is zero if it solves the associated homogeneous equation, -- and one otherwise. -- -- Thus to solve 2x + y - z = 2, use -- -- @ -- linDiophEq [2,1,-1] 2 = [[0,0,1,1],[1,1,0,0],[0,1,0,2],[1,0,2,0]] -- @ -- -- The two minimal solutions to the homogeneous equation are [0,1,1] -- and [1,0,2], so any linear combinations of these solutions -- contributes to a solution. The solution that corresponds to -- [1,0,0] is x = w + 1, y = v, and z = v + 2w. The solution that -- corresponds to [0,2,0] is x = w, y = v + 2, and z = v + 2w. linDiophEq :: [Int] -> Int -> [[Int]] linDiophEq [] _ = [] linDiophEq v 0 = newMinimalResults True (vector n v) (basis n) S.empty where n = length v linDiophEq v c = newMinimalResults False (vector n (negate c:v)) (basis n) S.empty where n = 1 + length v -- Construct the basis vectors for an n-dimensional space basis :: Int -> Set (Vector Int) basis n = S.fromList [ z // [(k, 1)] | k <- indices z ] where z = vector n $ replicate n 0 -- This is the main loop. -- Add elements of a that solve the equation to m and the output -- Variable hom is true when solving a homogeneous equation newMinimalResults :: Bool -> Vector Int -> Set (Vector Int) -> Set (Vector Int) -> [[Int]] newMinimalResults _ _ a _ | S.null a = [] newMinimalResults hom v a m = loop m (S.toList a) -- Test each element in a where loop m [] = -- When done, prepare for next iteration let a' = unnecessaryBranches a m a'' = breadthFirstSearch hom v a' in newMinimalResults hom v a'' m loop m (x:xs) | prod v x == 0 && S.notMember x m = elems x:loop (S.insert x m) xs -- Answer found | otherwise = loop m xs -- Breadth-first search using the algorithm of Contejean and Devie -- Variable hom is true when solving a homogeneous equation breadthFirstSearch :: Bool -> Vector Int -> Set (Vector Int) -> Set (Vector Int) breadthFirstSearch hom v a = S.fold f S.empty a where f x acc = foldl (flip S.insert) acc [ x // [(k, x!k + 1)] | k <- indices x, -- When not hom, bound first element hom || k > 0 || x!k == 0, -- of x to be no more than one prod v x * v!k < 0 ] -- Fortenbacher contribution -- Inner product prod :: Vector Int -> Vector Int -> Int prod x y = sum [ x!i * y!i | i <- indices x ] -- Remove unnecessary branches. A test vector is not necessary if all -- of its elements are greater than or equal to the elements of some -- minimal solution. unnecessaryBranches :: Set (Vector Int) -> Set (Vector Int) -> Set (Vector Int) unnecessaryBranches a m = S.filter f a where f x = all (g x) (S.toList m) g x y = not (lessEq y x) -- Compare vectors element-wise. lessEq :: Vector Int -> Vector Int -> Bool lessEq x y = all (\i-> x!i <= y!i) (indices x)