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Week 6: if you have read this and you have no questions, please close this issue.

Clear concise report (++) with beauties of Haskell functions like

-- Exercise 3: Composites sieve.
composites :: [Integer]
composites = map fst (filter (not . snd) (sieve' [2..]))

-- characterize numbers by prime property
sieve' :: [Integer] -> [(Integer,Bool)]
sieve' ns = let ps = (sieve [2..])
            in map (\ n -> (n, (inlist n ps))) ns
            where inlist n (p:ps) | p < n = inlist n ps
                                  | p == n = True
                                  | otherwise = False

The test results are very readable. One remark:
There are more carmichael numbers than {(6k+1)(12k+1)(18k+1)}
(see solutions Lab6)

Week 4: if you have read this and you have no questions, please close this issue.

Opg 3.

--intersection:
intersectSet :: (Ord a) => Set a -> Set a -> Set a
intersectSet (Set []) _ = Set []
intersectSet (Set (x:xs)) s | inSet x s = insertSet x (intersectSet (Set xs) s)
                            | otherwise = intersectSet (Set xs) s

--in a commercial environment we would do this:
intersectSet' (Set xs) (Set ys) = list2set (intersect xs ys)

Good remark.

Opg 4.

rClos :: Ord a => Rel a -> Rel a
trClos r | isTransitive [] r = sort r
         | otherwise = trClos (union r (r @@ r))

-- property test functions
isTransitive, isSubset, isMinimal, hasSameDomain, hasSameRange :: Ord a => Rel a -> Rel a -> Bool

-- checks if transitivity holds for the given relation
isTransitive _ r = and [elem (x,z) r | (x,y) <- r, (y',z) <- r, y == y']

Why has isTransitive 2 arguments? The first argument will never be used.
trClos can be written without the (expensive) call of isTransitive.

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